Chapter 1 – Electric Circuit Variables
Exercises
Exercise 1.2-1 Find the charge that has entered an element by time t when i = 8t2 – 4t A, t ≥ 0.
Assume q(t) = 0 for t < 0.
8
Answer: q(t ) = t 3 − 2t 2 C
3
Solution:
i (t ) = 8 t 2 − 4 t A
q(t ) =
t
∫0 i dτ + q(0) =
8 3
8
2
2 t
τ
−
τ
τ
+
=
τ
−
τ
= t3 −2 t 2 C
(8
4
)
d
0
2
∫0
0
3
3
t
Exercise 1.2-2 The total charge that has entered a circuit element is q(t) = 4 sin 3t C when t ≥ 0
and q(t) = 0 when t < 0. Determine the current in this circuit element for t > 0.
d
Answer: i (t ) = 4sin 3t = 12 cos 3t A
dt
Solution:
i (t ) =
dq d
= 4sin 3t = 12 cos 3t
dt dt
A
Exercise 1.3-1 Which of the three currents, i1 = 45 μA, i2 = 0.03 mA, and i3 = 25 × 10–4 A, is
largest?
Answer: i3 is largest.
Solution:
i1 = 45 μA = 45 × 10-6 A < i2 = 0.03 mA = .03 × 10-3 A = 3 × 10-5 A < i3 = 25 × 10-4 A
1-1
Exercise 1.5-1 Figure E 1.5-1 shows four circuit elements identified by the letters A, B, C, and
D.
(a)
Which of the devices supply 12 W?
(b)
Which of the devices absorb 12 W?
(c)
What is the value of the power received by device B?
(d)
What is the value of the power delivered by device B?
(e)
What is the value of the power delivered by device D?
Answers: (a) B and C, (b) A and D, (c)–12 W, (c) 12 W, (e)–12 W
+
4V
–
–
2V
+
+
6V
–
–
3V
3A
6A
2A
4A
(A)
(B)
(C)
(D)
+
Figure E 1.5-1
Solution:
(a) B and C. The element voltage and current do not adhere to the passive convention in B and C
so the product of the element voltage and current is the power supplied by these elements.
(b) A and D. The element voltage and current adhere to the passive convention in A and D so the
product of the element voltage and current is the power delivered to, or absorbed by these
elements.
(c) −12 W. The element voltage and current do not adhere to the passive convention in B, so the
product of the element voltage and current is the power received by this element: (2 V)(6 A) =
−12 W. The power supplied by the element is the negative of the power delivered to the element,
12 W.
(d) 12 W
(e) –12 W. The element voltage and current adhere to the passive convention in D, so the product
of the element voltage and current is the power received by this element: (3 V)(4 A) = 12 W. The
power supplied by the element is the negative of the power received to the element, −12 W.
1-2
Problems
Section 1-2 Electric Circuits and Current Flow
P1.2.1 The total charge that has entered a circuit element is q(t) = 1.25(1–e–5t) when t ≥ 0 and
q(t) = 0 when t < 0. Determine the current in this circuit element for t ≥ 0.
Answer: i(t) = 6.25e–5t A
i (t ) =
Solution:
d
1.25 (1 − e−5t ) = 6.25 e−5t A
dt
P 1.2-2 The current in a circuit element is i(t) = 4(1–e–5t) A when t ≥ 0 and i(t) = 0 when t < 0.
Determine the total charge that has entered a circuit element for t ≥ 0.
t
0
−∞
−∞
Hint: q (0) = ∫ i (τ ) = ∫ 0dτ = 0
Answer: q(t) = 4t + 0.8e–5t– 0.8 C for t ≥ 0
Solution:
t
t
t
t
4
4
q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4 (1 − e−5τ ) dτ + 0 = ∫ 4 dτ − ∫ 4 e−5τ dτ = 4 t + e−5t − C
0
0
0
0
5
5
P 1.2-3 The current in a circuit element is i(t) = 4 sin 3t A when t ≥ 0 and i(t) = 0 when t < 0.
Determine the total charge that has entered a circuit element for t ≥ 0.
t
0
−∞
−∞
Hint: q (0) = ∫ i (τ )dτ = ∫ 0 dτ = 0
Solution:
t
t
0
0
q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4sin 5τ dτ + 0 = −
4
4
4
t
cos 3τ 0 = − cos 3 t + C
5
5
5
1-3
⎧0 t < 2
⎪
⎪2 2 < t < 4
where the units of current are A
P 1.2-4 The current in a circuit element is i (t ) = ⎨
t
1
4
8
−
<
<
⎪
⎪⎩0 8 < t
and the units of time are s. Determine the total charge that has entered a circuit element for t ≥ 0.
Answer:
t<2
2<t <4
where the units of charge are C.
4<t <8
8<t
⎧0
⎪2t − 4
⎪
q (t ) = ⎨
⎪8 − t
⎩⎪0
Solution:
t
t
−∞
t
−∞
q ( t ) = ∫ i (τ ) dτ = ∫ 0 dτ = 0 C for t ≤ 2 so q(2) = 0.
t
t
2
t
2
t
2
4
t
4
t
8
8
q ( t ) = ∫ i (τ ) dτ + q ( 2 ) = ∫ 2 dτ = 2τ
= 2t−4 C for 2 ≤ t ≤ 4. In particular, q(4) = 4 C.
q ( t ) = ∫ i (τ ) dτ + q ( 4 ) = ∫ −1 dτ + 4 = − τ 4 + 4 = 8−t C for 4 ≤ t ≤ 8. In particular, q(8) = 0 C.
t
q ( t ) = ∫ i (τ ) dτ + q ( 8 ) = ∫ 0 dτ + 0 = 0 C for 8 ≤ t .
P 1.2-5
The total charge q(t), in coulombs, that enters the terminal of an element is
⎧0
⎪
q (t ) = ⎨2t
⎪3 + e −2t (t −2)
⎩
t<0
0≤t ≤2
t>2
Find the current i(t) and sketch its waveform for t ≥ 0.
Solution:
dq ( t )
i (t ) =
dt
⎧0
⎪
i (t ) = ⎨2
⎪
−2( t − 2 )
⎩−2e
t <0
0< t < 2
t >2
1-4
i
P 1.2-6 An electroplating bath, as shown in
Figure P 1.2-6, is used to plate silver uniformly
onto objects such as kitchen ware and plates. A
current of 600 A flows for 20 minutes, and
each coulomb transports 1.118 mg of silver.
What is the weight of silver deposited in
grams?
i
Object to
be plated
Silver bar
Bath
Figure P 1.2-6
Solution:
i = 450 A = 450
C
s
C
s
mg
Silver deposited = 450 ×20 min×60
= 6.0372×105 mg=603.72 g
×1.118
s
min
C
1-5
P1.2-7 Find the charge q(t) and sketch its
waveform when the current entering a terminal
of ab element is as shown in Figure P1.2-7.
Assume that q(t) = 0 for t < 0.
Figure P1.2-7
Solution:
⎧1 when 0 < t ≤ 2
i (t ) = ⎨
⎩ t − 1 when 2 ≤ t
t
t
0
0
q (t ) = ∫ i (τ )dτ + q (0) = ∫ i (τ )dτ
and
since q(0) =0.
When 0 < t ≤ 2, we have
t
q = ∫ 1 dτ = t C
0
When t ≥ 2, we have
t
2
0
0
t
q = ∫ i (τ )dτ = ∫ 1 dτ + ∫ (τ − 1) dτ
2
2
t
t
= τ 0 + τ2 − τ 2 =
2
2
t2
2
−t + 2 C
The sketch of q(t) is shown to the right:.
1-6
Section 1-3 Systems of Units
P 1.3-1 A constant current of 3.2 μA flows through an element. What is the charge that has
passed through the element in the first millisecond?
Answer: 3.2 nC
Solution:
Δq = i Δt =
( 3.2 × 10
−6
A )(1 × 10 − 3 s ) = 3.2 × 10 − 9 As = 3.2 × 10 − 9 C = 3.2 × 10 − 9 nC
P 1.3-2 A charge of 45 nC passes through a circuit element during a particular interval of time
that is 5 ms in duration. Determine the average current in this circuit element during that interval
of time.
Answer: i = 9 μA
Solution:
Δ q 45 × 10−9
i=
=
= 9 × 10−6 = 9 μA
−3
Δt
5 × 10
P 1.3-3 Ten billion electrons per second pass through a particular circuit element. What is the
average current in that circuit element?
Answer: i = 1.602 nA
Solution
electron ⎤ ⎡
C ⎤
⎡
i = ⎢10 billion
1.602 ×10−19
=
⎥
⎢
s
electron ⎥⎦
⎣
⎦⎣
C ⎤
⎡
9 electron ⎤ ⎡
1.602 × 10−19
⎢⎣10×10
⎥
⎢
s
electron ⎥⎦
⎦⎣
electron
C
= 1010 × 1.602 ×10−19
electron
s
C
= 1.602 × 10−9 = 1.602 nA
s
1-7
P1.3-4 The charge flowing in a wire is plotted in Figure P1.3-4. Sketch the corresponding
current.
Figure P1.3-4
P1.3-4
⎧ 15 × 10−9
−3
= 7.5 mA when 0 < t < 2μ s
⎪
−6 = 7.5 × 10
2
10
×
⎪
⎪ 15 × 10−9
d
−3
= −5 mA when 4μ s < t < 7 μ s
i ( t ) = q ( t ) = the slope of the q versus t plot = ⎨−
−6 = −5 × 10
dt
⎪ 3 × 10
0 otherwise
⎪
⎪
⎩
1-8
P1.3-5 The current in a circuit element is plotted in Figure P1.3-5. Sketch the corresponding
charge flowing through the element for t > 0.
Figure P1.3-5
Solution:
t
⎧
450 μ A dτ when 0 < t < 80 ms
∫
0
⎪
t
t
⎪
q ( t ) = ∫ i (τ ) dτ = ⎨( 450 × 10−6 )( 80 × 10−3 ) + ∫ ( −600μ A ) dτ when 80 ms < t < 140 ms
0
80 ms
⎪
t
⎪
450 ×10−6 )( 80 ×10−3 ) + ( −600 × 10−6 )( 60 × 10−3 ) + ∫
0 dτ
(
140 ms
⎩
⎧
450 ×10−6 ) t when 0 < t < 80 ms
(
⎪
⎪
= ⎨( 36 × 10−6 ) + ( −600 ×10−6 ) t when 80 ms < t < 140 ms
⎪
0 C when 140 ms < t
⎪⎩
While 0 < t < 80 ms q(t) increases linearly from 0 to 36 μC and while 80 < t < 140 ms q(t)
decreases linearly from 36 to 0 μC. Here’s the sketch:
1-9
P1.3-6 The current in a circuit element is plotted in Figure P1.3-6. Determine the total charge
that flows through the circuit element between 300 and 1200 μs.
Figure P1.3-6
Solution:
q (t ) = ∫
1000 μ s
300 μ s
i (τ ) dτ = "area under the curve between 300 μ s and 1000 μ s"
⎛ 360 + 720
⎞
q (t ) = ⎜
×10−9 ⎟ (100 × 10−6 ) + ( 720 ×10−9 )( 600 ×10−6 ) = ( 54 + 432 ) × 10−12 = 486 pC
2
⎝
⎠
1-10
Section 1-5 Power and Energy
P1.5-1 Figure P1.5-1 shows four circuit elements identified by the letters A, B, C, and D.
(a)
Which of the devices supply 30 mW?
(b)
Which of the devices absorb 0.03 W?
(c)
What is the value of the power received by device B?
(d)
What is the value of the power delivered by device B?
(e)
What is the value of the power delivered by device C?
Figure P1.5-1
Solution:
(a) A and D. The element voltage and current do not adhere to the passive convention in Figures
P1.5- A and D so the product of the element voltage and current is the power supplied by these
elements.
(b) B and C. The element voltage and current adhere to the passive convention in Figures P1.5-1
B and C so the product of the element voltage and current is the power delivered to, or absorbed
by these elements.
(c) 30 mW. The element voltage and current adhere to the passive convention in Figure P1.5-1B,
so the product of the element voltage and current is the power received by this element: (5 V)(6
mA) = 30 mW. The power supplied by the element is the negative of the power received to the
element, −30 W.
(d) −30 mW
(e) –30 mW. The element voltage and current adhere to the passive convention in Figure P1.51C, so the product of the element voltage and current is the power received by this element: (5
V)(6 mA) = 30 mW. The power supplied by the element is the negative of the power received to
the element, −30 W.
1-11
P 1.5-2 An electric range has a constant current of 10 A entering the positive voltage terminal
with a voltage of 110 V. The range is operated for two hours. (a) Find the charge in coulombs
that passes through the range. (b) Find the power absorbed by the range. (c) If electric energy
costs 12 cents per kilowatt-hour, determine the cost of operating the range for two hours.
Solution:
a.) q =
∫ i dt = iΔt = (10 A )( 2 hrs )( 3600s/hr ) = 7.2×10
4
C
b.) P = v i = (110 V )(10 A ) = 1100 W
c.) Cost =
0.12$
× 1.1kW × 2 hrs = 0.264 $
kWhr
P 1.5-3 A walker’s cassette tape player uses four AA batteries in series to provide 6 V to the
player circuit. The four alkaline battery cells store a total of 200 watt-seconds of energy. If the
cassette player is drawing a constant 10 mA from the battery pack, how long will the cassette
operate at normal power?
Solution:
P = ( 6 V )(10 mA ) = 0.06 W
Δt =
200 W⋅s
Δw
=
= 3.33×103 s
P
0.06 W
1-12
P 1.5-4 The current through and voltage across an element vary with time as shown in Figure P
1.5-4. Sketch the power delivered to the element for t > 0. What is the total energy delivered to
the element between t = 0 and t = 25 s? The element voltage and current adhere to the passive
convention.
v (volts)
i (amp)
30
30
5
5
0
10
15
25
0
t (s)
10
15
(a)
25
t (s)
(b)
Figure P 1.5-4
Solution:
for 0 ≤ t ≤ 10 s:
v = 30 V and i =
30
t = 2t A ∴ P = 30(2t ) = 60t W
15
25
t + b ⇒ v (10 ) = 30 V ⇒ b = 80 V
5
v(t ) = −5t + 80 and i (t ) = 2t A ⇒ P = ( 2t )( −5t +80 ) = −10t 2 +160t W
for 10 ≤ t ≤ 15 s: v ( t ) = −
30
t +b A
10
⇒ b = 75 ⇒ i (t ) = −3t + 75 A
for 15 ≤ t ≤ 25 s: v = 5 V and i (t ) = −
i (25) = 0
∴ P = ( 5 )( −3t + 75 ) = −15t + 375 W
Energy = ∫ P dt =
= 30t 2
10
0
∫
10
0
60t dt + ∫10 (160t −10t 2 ) dt + ∫15 ( 375−15t ) dt
15
15
25
25
+ 80t 2 − 10 t 3 + 375t − 15 t 2 = 5833.3 J
3 10
2 15
1-13
P 1.5-5 An automobile battery is charged with a constant current of 2 A for five hours. The
terminal voltage of the battery is v = 11 + 0.5t V for t > 0, where t is in hours. (a) Find the energy
delivered to the battery during the five hours. (b) If electric energy costs 15 cents/kWh, find the
cost of charging the battery for five hours.
Answer: (b) 1.84 cents
Solution:
a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully
charged).
5( 3600 )
t
5 ( 3600 ) ⎛
0.5 τ ⎞
0.5 2
2 ⎜11 +
w = ∫ Pdt = ∫0 vi dτ = ∫0
⎟ dτ = 22 t + 3600 τ
3600 ⎠
⎝
0
= 441× 103 J = 441 kJ
b.) Cost = 441kJ ×
1 hr
15¢
×
= 1.84¢
3600s kWhr
1-14
P 1.5-6 Find the power, p(t), supplied by the element shown in Figure P 1.5-6 when
v(t) = 4 sin 3t V and i(t) = (1/12) sin 3t A. Evaluate p(t) at t=0.5 s and t = 1 s. Observe that the
power supplied by this element has a positive value at some times and a negative value at other
times.
1
Hint: (sin at )(sin bt ) = (cos(a − b)t − cos(a + b)t )
2
Answer: p(t) = (1/6)cos(6t) W, p(0.5) = 0.0235 W, p(1) = −0.02466 W
Solution:
1
⎛1
⎞ 1
p ( t ) = v ( t ) i ( t ) = ( 4 cos 3 t ) ⎜ sin 3 t ⎟ = ( sin 0 + sin 6 t ) = sin 6 t
6
⎝ 12
⎠ 6
1
p ( 0.5 ) = sin 3 = 0.0235 W
6
1
p (1) = sin 6 = −0.0466 W
6
W
Here is a MATLAB program to plot p(t):
clear
t0=0;
tf=2;
dt=0.02;
t=t0:dt:tf;
%
%
%
%
v=4*cos(3*t);
i=(1/12)*sin(3*t);
% device voltage
% device current
for k=1:length(t)
p(k)=v(k)*i(k);
end
% power
initial time
final time
time increment
time
plot(t,p)
xlabel('time, s');
ylabel('power, W')
1-15
i
P 1.5-7 Find the power, p(t), supplied by the element shown in Figure P
1.5-6 when v(t) = 8 sin 3t V and i(t) = 2 sin 3t A.
1
Hint: (sin at )(sin bt ) = (cos(a − b)t − cos(a + b)t )
2
Answer: p(t) = 8 – 8cos 6t W
Solution:
+
v
–
Figure P 1.5-7
p ( t ) = v ( t ) i ( t ) = ( 8sin 3 t )( 2sin 3 t ) = 8 ( cos 0 − cos 6 t ) = 8 − 8cos 6 t
W
Here is a MATLAB program to plot p(t):
clear
t0=0;
tf=2;
dt=0.02;
t=t0:dt:tf;
%
%
%
%
initial time
final time
time increment
time
v=8*sin(3*t);
i=2*sin(3*t);
% device voltage
% device current
for k=1:length(t)
p(k)=v(k)*i(k);
end
% power
plot(t,p)
xlabel('time, s');
ylabel('power, W')
1-16
i
P 1.5-8 Find the power, p(t), supplied by the element shown in Figure P
1.5-6. The element voltage is represented as v(t) = 4(1–e–2t)V when t ≥ 0
and v(t) = 0 when t < 0. The element current is represented as i(t) = 2e–2t
A when t ≥ 0 and i(t) = 0 when t < 0.
–2t
–2t
Answer: p(t) = 8(1 – e )e
Solution:
+
v
–
W
Figure P 1.5-7
(
)
(
)
p ( t ) = v ( t ) i ( t ) = 4 1 − e −2 t × 2 e −2 t = 8 1 − e−2 t e−2 t
W
Here is a MATLAB program to plot p(t):
clear
t0=0;
tf=2;
dt=0.02;
t=t0:dt:tf;
%
%
%
%
initial time
final time
time increment
time
v=4*(1-exp(-2*t));
i=2*exp(-2*t);
% device voltage
% device current
for k=1:length(t)
p(k)=v(k)*i(k);
end
% power
plot(t,p)
xlabel('time, s');
ylabel('power, W')
1-17
P 1.5-9 The battery of a flashlight develops 3 V, and the current through the bulb is 200 mA.
What power is absorbed by the bulb? Find the energy absorbed by the bulb in a five-minute
period.
Solution:
The power is P = V I =3 × 0.015=0.045 W . Next, the energy is w = PΔ t = 0.045 × 5 × 60=13.5 J .
1-18
P1.5-10 Medical researchers studying
hypertension often use a technique called “2D
gel electrophoresis” to analyze the protein
content of a tissue sample. An image of a
typical “gel” is shown in Figure 1.5-10a.
The procedure for preparing the gel
uses the electric circuit illustrated in Figure
1.5-10b. The sample consists of a gel and a
filter paper containing ionized proteins. A
voltage source causes a large, constant voltage,
500 V, across the sample. The large, constant
voltage moves the ionized proteins from the
filter paper to the gel. The current in the
sample is given by
Devon Svoboda, Queen’s University
(a)
sample
i ( t ) = 2 + 30 e − at mA
+
500 V
–
i (t)
where t is the time elapsed since the beginning
of the procedure and the value of the constant a
is
1
a = 0.85
hr
Determine the energy supplied by the voltage
source when the gel preparation procedure lasts
3 hours.
500 V
(b)
Figure 1.5-10 (a) An image of a gel and (b)
the electric circuit used to preparation a gel.
Solution:
T
T
0
0
energy = w ( t ) = ∫ p ( t ) d t = ∫ v ( t ) i ( t ) d t
=
500 3
2 + 30 e − 0.85t d t
∫
0
1000
(
)
3
3
0
0
(
)
= ∫ d t + 15∫ e − 0.85t d t
15
e − 2.55 − 1
−0.85
= 3 + 16.3 = 19.3 J
= (3 − 0) +
(
)
1-19
Section 1.7 How Can We Check…?
– 5V +
P 1.7-1 Conservation of energy
requires that the sum of the power
absorbed by all of the elements in a
circuit be zero. Figure P 1.7-1 shows a
circuit. All of the element voltages and
currents are specified. Are these
voltage and currents correct? Justify
your answer.
+
2A
–2 V
–
–5 A
+
+ 4V –
3V
2A
–
5A
3A
+ 1V –
Figure P 1.7-1
Hint: Calculate the power absorbed by each element. Add up all of these powers. If the sum is
zero, conservation of energy is satisfied and the voltages and currents are probably correct. If the
sum is not zero, the element voltages and currents cannot be correct.
Solution:
Notice that the element voltage and current of each branch adhere to the passive convention. The
sum of the powers absorbed by each branch are:
(-2 V)(2 A)+(5 V)(2 A)+(3 V)(3 A)+(4 V)(-5 A)+(1 V)(5 A) = -4 W + 10 W + 9 W -20 W + 5 W
=0W
The element voltages and currents satisfy conservation of energy and may be correct.
1-20
P 1.7-2 Conservation of energy requires that the sum of the power absorbed by all of the
elements in a circuit be zero. Figure P 1.7-2 shows a circuit. All of the element voltages and
currents are specified. Are these voltage and currents correct? Justify your answer.
Hint: Calculate the power absorbed by each element. Add up all of these powers. If the sum is
zero, conservation of energy is satisfied and the voltages and currents are probably correct. If the
sum is not zero, the element voltages and currents cannot be correct.
+ 4V –
3A
+
+
–
3V
3V
3V
–
3A
–
2A +
2A
–
–3 A
–3 V
+ 4V –
+
–3 A
Figure P 1.7-2
Solution:
Notice that the element voltage and current of some branches do not adhere to the passive
convention. The sum of the powers absorbed by each branch are:
-(3 V)(3 A)+(3 V)(2 A)+ (3 V)(2 A)+(4 V)(3 A)+(-3 V)(-3 A)+(4 V)(-3 A)
= -9 W + 6 W + 6 W + 12 W + 9 W -12 W
≠0W
The element voltages and currents do not satisfy conservation of energy and cannot be correct.
1-21
– 3V +
P 1.7-3 The element currents and
voltages shown in Figure P 1.7-3 are
correct with one exception: the
a
reference direction of exactly one of the
element currents is reversed. Determine
+
which reference direction has been
5V
reversed.
–
–3A
– 1V +
4A
7A
– 2V +
b
c
−2A
–
−6V
+
2A
–
−8V
+
–5A
d
Figure P 1.7-3
Solution:
Let’s tabulate the power received by each element. We’ll identify each element by its nodes.
nodes
a c
So
Power received, W
( 3)( −3) = −9
a
b
− (1)( 4 ) = −4
b
c
a
d
b
d
c
d
( 2 )( −2 ) = −4
− ( 5 )( 7 ) = −35
( −6 )( 2 ) = −12
( −8)( −5) = 40
Total power received = − ( 9 + 4 + 4 + 35 + 12 ) + 40 = −24 ≠ 0
Changing the current reference direction for a particular element will change the total power by
twice the power of the particular element. Since the element connected between nodes b and d
receives -12 W, changing the reference direction of its current will increase the total power
received by 24 W, as required. After making that change
Total power received = − ( 9 + 4 + 4 + 35 ) + (12 + 40 ) = 0
We conclude that it is the reference direction of the element connected between nodes and b that
has been reversed.
1-22
Design Problems
DP 1-1 A particular circuit element is available in three grades. Grade A guarantees that the
element can safely absorb 1/2 W continuously. Similarly, Grade B guarantees that 1/4 W can be
absorbed safely and Grade C guarantees that 1/8 W can be absorbed safely. As a rule, elements
that can safely absorb more power are also more expensive and bulkier.
The voltage across an element is expected to be about 20 V and the current in the element
is expected to be about 8 mA. Both estimates are accurate to within 25 percent. The voltage and
current reference adhere to the passive convention.
Specify the grade of this element. Safety is the most important consideration, but don’t
specify an element that is more expensive than necessary.
Solution:
The voltage may be as large as 20(1.25) = 25 V and the current may be as large as (0.008)(1.25)
= 0.01 A. The element needs to be able to absorb (25 V)(0.01 A) = 0.25 W continuously. A
Grade B element is adequate, but without margin for error. Specify a Grade B device if you trust
the estimates of the maximum voltage and current and a Grade A device otherwise.
1-23
DP 1-2 The voltage across a circuit element is v(t) = 20 (1–e–8t)V when t ≥ 0 and v(t) = 0 when
t < 0. The current in this element is i(t) = 30e–8t mA when t ≥ 0 and i(t) = 0 when t < 0. The
element current and voltage adhere to the passive convention. Specify the power that this device
must be able to absorb safely.
Hint: Use MATLAB, or a similar program, to plot the power.
Solution:
(
)
(
)
p ( t ) = 20 1 − e −8t × 0.03 e −8t = 0.6 1 − e −8 t e −8 t
Here is a MATLAB program to plot p(t):
W
Here is the plot:
clear
t0=0;
tf=1;
dt=0.02;
t=t0:dt:tf;
%
%
%
%
initial time
final time
time increment
time
v=20*(1-exp(-8*t));
i=.030*exp(-8*t);
% device voltage
% device current
for k=1:length(t)
p(k)=v(k)*i(k);
end
% power
plot(t,p)
xlabel('time, s');
ylabel('power, W')
The circuit element must be able to absorb 0.15 W.
1-24
Chapter 2 Circuit Elements
Exercises
Exercise 2.4-1 Find the power absorbed by a 100-ohm resistor when it is connected directly
across a constant 10-V source.
Answer: 1-W
v 2 (10 )
P= =
=1 W
R 100
2
Solution:
Exercise 2.4-2 A voltage source v = 10 cos t V is connected across a resistor of 10 ohms. Find
the power delivered to the resistor.
Answer: 10 cos2 t W
Solution:
v 2 (10 cos t ) 2
P= =
= 10 cos 2 t W
R
10
Exercise 2.7-1 Find the power absorbed by the CCCS in Figure E 2.7-1.
Hint: The controlling element of this dependent source is a short circuit. The voltage across a
short circuit is zero. Hence, the power absorbed by the controlling element is zero. How much
power is absorbed by the controlled element?
Answer:–115.2 watts are absorbed by the CCCS. (The CCCS delivers +115.2 watts to the rest of
the circuit.)
– 1. 2 0
+ 2 4. 0
Ammeter
Voltmeter
2Ω
2Ω
ic
12 V
+
–
+
4Ω
id = 4ic
vd
–
Figure E 2.7-1
Solution:
ic = − 1.2
A, vd = 24
id = 4 ( − 1.2) = − 4.8
V
A
id and vd adhere to the passive convention so
P = vd id = (24) (−4.8) = −115.2
is the power received by the dependent source
W
Exercise 2.8-1 For the potentiometer circuit of Figure 2.8-2, calculate the meter voltage, vm,
when θ = 45°, Rp = 20 kΩ, and I = 2 mA.
Answer: vm = 5 V
(1 – a)Rp
Voltmeter
+
I
vm
–
+
vm
aRp
I
Rp
–
(a)
(b)
Figure 2.8-2
Solution:
θ = 45° , I = 2 mA, R p = 20 kΩ
a=
θ
45
⇒ aR =
(20 kΩ) = 2.5 kΩ
p
360
360
vm = (2 ×10−3 )(2.5 ×103 ) = 5 V
Exercise 2.8-2 The voltage and current of an
AD590 temperature sensor of Figure 2.8-3 are 10 V
and 280 μA, respectively. Determine the measured
temperature.
i(t)
+
+
v(t)
AD590
–
v(t)
i(t) = kT
–
Answer: T = 280°K, or approximately 6.85°C
(a)
(b)
Figure 2.8-3
Solution:
v = 10 V, i = 280 μA, k = 1
μA
°K
for AD590
⎛ °K ⎞
i
⎟ = 280° K
i = kT ⇒ T = = (280μA)⎜1
⎜ μA ⎟
k
⎝
⎠
t=5s
t=3s
Exercise 2.9-1 What is the value of the
current i in Figure E 2.9-1 at time t = 4 s?
Answer: i = 0 amperes at t = 4 s (both switches
are open).
12 V
+
–
3 kΩ
6V
+
–
i
Figure E 2.9-1
Solution:
At t = 4 s both switches are open, so i = 0 A.
t=5s
Exercise 2.9-2 What is the value of the voltage v in
Figure E 2.9-2 at time t = 4 s? At t = 6 s?
Answer: v = 6 volts at t = 4 s, and v = 0 volts at t = 6 s.
+
3 kΩ
2 mA
i
Figure E 2.9-2
Solution:
At t = 4 s the switch is in the up position, so v = i R = (2 mA)(3 kΩ) = 6V .
At t = 6 s the switch is in the down position, so v = 0 V.
v
–
Section 2-2 Engineering and Linear Models
P 2.2-1 An element has voltage v and current i as
shown in Figure P 2.2-1a. Values of the current i and
corresponding voltage v have been tabulated as shown
in Figure P 2.2-1b. Determine if the element is linear.
i
+
v
–
(a)
v, V
i, A
–3
–4
0
12
32
60
–3
–2
0
2
4
6
(b)
Figure P 2.2-1
Solution:
The element is not linear. For example, doubling the current from 2 A to 4 A does not double the
voltage. Hence, the property of homogeneity is not satisfied.
P 2.2-2
A linear element has voltage v and current i as
shown in Figure P 2.2-2a. Values of the current i and
i
corresponding voltage v have been tabulated as shown in
+
Figure P 2.2-2b. Represent the element by an equation that
v
expresses v as a function of i. This equation is a model of
–
the element. (a) Verify that the model is linear. (b) Use the
model to predict the value of v corresponding to a current
of i = 40 mA. (c) Use the model to predict the value of i
(a)
v, V
i, A
–3.6
2.4
6.0
–30
20
50
(b)
Figure P 2.2-2a
corresponding to a voltage of v = 4 V.
Hint: Plot the data. We expect the data points to lie on a straight line. Obtain a linear model of
the element by representing that straight line by an equation.
Solution:
(a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and
the line passes through the origin so the equation of the line is v = 0.12 i . The element is indeed
linear.
(b) When i = 40 mA, v = (0.12 V/A)×(40 mA) = (0.12 V/A)×(0.04 A) = 4.8 mV
3
(c) When v = 3 V, i =
= 25 A
0.12
P 2.2-3
A linear element has voltage v and current i as
shown in Figure P 2.2-3a. Values of the current i and
i
corresponding voltage v have been tabulated as shown in
+
Figure P 2.2-3b. Represent the element by an equation that
v
v, V
i, mA
expresses v as a function of i. This equation is a model of
–
3.078
5.13
12.825
12
20
50
the element. (a) Verify that the model is linear. (b) Use the
model to predict the value of v corresponding to a current
of i = 4 mA. (c) Use the model to predict the value of i
(a)
(b)
Figure P 2.2-3
corresponding to a voltage of v = 12 V.
Hint: Plot the data. We expect the data points to lie on a straight line. Obtain a linear model of
the element by representing that straight line by an equation.
Solution:
(a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and
the line passes through the origin so the equation of the line is v = 256.5 i . The element is indeed
linear.
(b) When i = 6 mA, v = (256.5 V/A)×(6 mA) = (256.5 V/A)×(0.006 A) = 1.054 V
12
(c) When v = 12 V, i =
= 0.04678 A = 46.78 mA.
256.5
P 2.2-4
An element is represented by the relation between current and voltage as
v = 3i + 5
Determine whether the element is linear.
Solution:
Let i = 1 A , then v = 3i + 5 = 8 V. Next 2i = 2A but 16 = 2v ≠ 3(2i) + 5 = 11.. Hence,
the property of homogeneity is not satisfied. The element is not linear.
P 2.2-5 The circuit shown in Figure P 2.3-5
consists of a current source, a resistor, and
element A.
+
0.4 A
Consider three cases.
10 Ω
v
A
i
−
Figure P 2.3-5
(a)
When element A is a 40-Ω resistor, described by i = v / 40, then the circuit is represented
v v
by
0.4 = +
10 40
Determine the values of v and i. Notice that the above equation has a unique solution.
(b)
When element A is a nonlinear resistor described by i = v2 / 2, then the circuit is
v v2
0.4 = +
represented by
10 2
Determine the values of v and i. In this case there are two solutions of the above
equation. Nonlinear circuits exhibit more complicated behavior than linear
circuits.
v2
When element A is a nonlinear resistor described by i = 0.8 + , then the circuit is
2
2
v
v
described by
0.4 = + 0.8 +
10
2
Show that this equation has no solution. This result usually indicates a modeling
problem. At least one of the three elements in the circuit has not been modeled
accurately.
(c)
Solution:
(a)
(b)
Using the quadratic formula
v
v v
+
=
⇒
v = 3.2 V
10 40 8
v
i=
= 0.08 A
40
v v2
v
0.4 = +
⇒ v 2 + − 0.8 = 0
10 2
5
0.4 =
v=
−0.2 ± 1.8
= 0.8, − 1.0 V
2
( −1) = 0.5 A .
0.82
= 0.32 A . When v = -1.0 V then i =
When v = 0.8 V then i =
2
2
2
v
v
v
0.4 = + 0.8 +
⇒ v 2 + + 0.8 = 0
(c)
10
2
5
2
−0.2 ± 0.04 − 3.2
2
So there is no real solution to the equation.
Using the quadratic formula
v=
Section 2-4 Resistors
P 2.4-1 A current source and a resistor are connected in
series in the circuit shown in Figure P 2.4-1. Elements
connected in series have the same current, so i = is in this
circuit. Suppose that is = 3 A and R = 7 Ω. Calculate the
voltage v across the resistor and the power absorbed by the
resistor.
is
i
+
R
v
–
Figure P 2.4-1
Answer: v = 21 V and the resistor absorbs 63 W.
Solution:
i = is = 3 A and v = Ri = 7 × 3 = 21 V
v and i adhere to the passive convention
∴ P = v i = 21 × 3 = 63 W
is the power absorbed by the resistor.
P 2.4-2 A current source and a resistor are connected in
series in the circuit shown in Figure P 2.4-1. Elements
connected in series have the same current, so i = is in this
circuit. Suppose that i = 3 mA and v = 24 V. Calculate the
resistance R and the power absorbed by the resistor.
Answer: R = 8 kΩ and the resistor absorbs 72 mW.
is
i
+
R
v
–
Figure P 2.4-1
Solution:
i = i = 3 mA and v = 48 V
s
48
v
=
= 16000 = 16 K Ω
R =
i
0.003
P = (3×10− 3 )×48 = 144×10− 3 = 144 mW
P 2.4-3 A voltage source and a resistor are connected in
parallel in the circuit shown in Figure P 2.4-3. Elements
connected in parallel have the same voltage, so v = vs in this
circuit. Suppose that vs = 10 V and R = 5 Ω. Calculate the
current i in the resistor and the power absorbed by the
resistor.
i
vs
R
+
v
–
Figure P 2.4-3
Answer: i = 2 A and the resistor absorbs 20 W.
Solution:
+
–
v = vs =10 V and R = 5 Ω
v
10
=
=2 A
R
5
v and i adhere to the passive convention
∴ p = v i = 2⋅10 = 20 W
is the power absorbed by the resistor
i =
P 2.4-4 A voltage source and a resistor are connected in
parallel in the circuit shown in Figure P 2.4-3. Elements
connected in parallel have the same voltage, so v = vs in this
circuit. Suppose that vs = 24 V and i = 2 A. Calculate the
resistance R and the power absorbed by the resistor.
i
vs
Answer: R = 12 Ω and the resistor absorbs 48 W.
Solution:
v = vs =24 V and i = 3 A
v
24 V
=
= 8Ω
i
3A
p = v i = 24( 3) = 72 W
R =
+
–
R
+
v
–
Figure P 2.4-3
P 2.4-5 A voltage source and two resistors are connected
in parallel in the circuit shown in Figure P 2.4-5. Elements
connected in parallel have the same voltage, so v1 = vs and
v2 = vs in this circuit. Suppose that vs = 150 V, R1 = 50 Ω,
and R2 = 25 Ω. Calculate the current in each resistor and the
power absorbed by each resistor.
i1
+
–
vs
i2
+
v1
R1
R2
–
+
v2
–
Figure P 2.4-5
Hint: Notice the reference directions of the resistor currents.
Answer: i1 = 3 A and i2 = –6 A. R1 absorbs 450 W and R2 absorbs 900 W.
Solution:
v1 = v 2 = vs = 150 V;
R1 = 50 Ω; R2 = 25 Ω
v 1 and i1 adhere to the passive convention so
v 1 150
=
=3 A
R 1 50
150
v
v2 and i 2 do not adhere to the passive convention so i 2 = − 2 = −
= −6 A
25
R2
i1 =
The power absorbed by R1 is P1 = v1 i1 = 150 ⋅ 3 = 450 W
The power absorbed by R 2 is P 2 = − v 2i 2 = −150(−6) = 900 W
P 2.4-6 A current source and two resistors are connected in
series in the circuit shown in Figure P 2.4-6. Elements
connected in series have the same current, so i1 = is and i2 = is
in this circuit. Suppose that is = 2 A, R1 = 4 Ω, and R2 = 8 Ω.
Calculate the voltage across each resistor and the power
absorbed by each resistor.
– v1 +
i1
is
R1
i2
+
R2
v2
–
Hint: Notice the reference directions of the resistor voltages.
Figure P 2.4-6
Answer: v1 = –8 V and v2 = 16 V. R1 absorbs 16 W and R2 absorbs 32 W.
Solution:
i1 = i2 = is =25 mA and R1 = 4 Ω and R2 = 8 Ω
v1 and i1 do not adhere to the passive convention so
v1 = − R1 i1 = −4 ( 0.025 ) = −0.1 V.
The power absorbed by R1 is
P1 =−v1 i1 =−(−0.1)(0.025) = 2.5 mW.
v 2 and i 2 do adhere to the passive convention so v 2 = R 2 i 2 = 8 ( 0.025 ) = 0.2 V.
The power absorbed by R 2 is P2 = v 2 i 2 =(0.2)(0.025) = 5 mW.
P 2.4-7 An electric heater is connected to a constant 250-V source and absorbs 1000 W.
Subsequently, this heater is connected to a constant 210-V source. What power does it absorb
from the 210-V source? What is the resistance of the heater?
Hint: Model the electric heater as a resistor.
Solution:
v2
v2
(250)2
⇒ R =
=
= 62.5 Ω
R
P
1000
v 2 (220)2
=
= 774.4 W
P=
R
62.5
Model the heater as a resistor, then from P =
with a 220 V source
P 2.4-8 The portable lighting equipment for a mine is located 100 meters from its dc supply
source. The mine lights use a total of 5 kW and operate at 120 V dc. Determine the required
cross-sectional area of the copper wires used to connect the source to the mine lights if we
require that the power lost in the copper wires be less than or equal to 5 percent of the power
required by the mine lights.
Hint: Model both the lighting equipment and the wire as resistors.
Solution:
The current required by the mine lights is: i =
P 5000 125
A
=
=
3
v 120
Power loss in the wire is : i 2 R
Thus the maximum resistance of the copper wire allowed is
0.05P 0.05×5000
=
= 0.144 Ω
(125/3) 2
i2
now since the length of the wire is L = 2×100 = 200 m = 20,000 cm
R=
thus R = ρ L / A
with ρ = 1.7×10−6 Ω⋅ cm from Table 2.5−1
A=
ρL
R
=
1.7×10−6 ×20,000
= 0.236 cm 2
0.144
*P 2.4-9 The resistance of a practical resistor depends on the
nominal resistance and the resistance tolerance as follows:
t ⎞
t ⎞
⎛
⎛
Rnom ⎜1 −
⎟ ≤ R ≤ Rnom ⎜ 1 +
⎟
⎝ 100 ⎠
⎝ 100 ⎠
i
R1
+
vs
where Rnom is the nominal resistance and t is the resistance
tolerance expressed as a percentage. For example, a 100-Ω,
2 percent resistor will have a resistance given by
+
–
R2
vo
–
Figure P 2.4-9
98 Ω ≤ R ≤ 102 Ω
The circuit shown in Figure P 2.4-9 has one input, vs, and one output, vo. The gain of this circuit
is given by
v
R2
gain = 0 =
vs R1 + R2
Determine the range of possible values of the gain when R1 is the resistance of a 100-Ω, 2
percent resistor and R2 is the resistance of a 400-Ω, 5 percent resistor. Express the gain in terms
of a nominal gain and a gain tolerance.
Solution:
0.7884 =
380
420
≤ gain ≤
= 0.8108
102 + 380
98 + 420
nominal gain =
gain tolerance =
So
0.7884 + 0.8108
= 0.7996
2
0.7996 − 0.7884
0.8108 − 0.7996
× 100 =
× 100 = 1.40%
0.7996
0.7996
gain = 0.7996 ± 1.40%
+ va –
P 2.4-10 The voltage source shown in
va, V
vb, V
Figure P 2.4-10 is an adjustable dc voltage
11.75
7.05
source. In other words, the voltage vs is a
40 Ω
+
7.5
4.5
constant voltage, but the value of that constant
vs +–
5.625 3.375
R
vb
can be adjusted. The tabulated data were
10
6
–
4.375 2.625
collected as follows. The voltage, vs, was set to
some value, and the voltages across the
Figure P 2.4-10
resistor, va and vb, were measured and
recorded. Next, the value of vs was changed,
and the voltages across the resistors were measured again and recorded. This procedure was
repeated several times. (The values of vs were not recorded.) Determine the value of the
resistance, R.
Solution:
Label the current i as shown. That current is
the element current in both resistors. First
va
i=
40
va
vb
⇒ R = 40
Next v b = R i = R
40
va
For example,
R = 40
7.05
= 24 Ω
11.75
P2.4-11 Consider the circuit shown in Figure P2.4-11.
(a) Suppose the current source supplies 3.125 W of
power. Determine the value of the resistance R.
Figure P2.4-11
(b) Suppose instead the resistance is R = 12 Ω. Determine
the value of the power supplied by the current source.
Solution:
(a) Suppose the current source supplies 3.125 W of power.
1.25 v 3 = 3.125 ⇒ v 3 = 2.5 V
Then, using KVL
R=
20 + 2.5
= 18 Ω
1.25
(b) Suppose instead the resistance is R = 12 Ω. From KVL
1.25 (12 ) − v 3 − 20 = 0 ⇒ v 3 = −5 V
The value of the power supplied by the current source is
1.25 v 3 = 1.25 ( −5 ) = −6.25 W
P2.4-12.We will encounter “ac circuits” in Chapter 10. Frequently we analyze ac circuits using
“phasors” and “impedances”. Phasors are complex numbers that represents currents and voltages
in an ac circuit. Impedances are complex numbers that describe ac circuit elements. (See
Appendix B for a discussion of complex numbers.) Figure P2.4-11 shows a circuit element in an
ac circuit. I and V are complex numbers representing the element current and voltage. Z is a
complex number describing the element itself. “Ohm’s law for ac circuits” indicates that
V=ZI
(a) Suppose V = 12∠45° V , I = B∠θ A and Z = 18 + j 8 Ω . Determine the values of B and θ.
(b) Suppose V = 48∠135° V , I = 3∠15° A and Z = R + j X Ω . Determine the values of R and
X.
Figure P2.4-12
Solution:
(a)
12∠45° = (18 + j 8 )( B∠θ )
B∠θ =
so
12∠45° 12∠45°
=
= 0.609∠21°
18 + j 8 19.7∠24°
B=0.609 A and θ = 21°
48∠135° = ( R + j X )( 3∠15° )
(b)
R+ j X =
so
48∠75°
= 18∠60° = 9 + j 15.6 Ω
3∠15°
R = 9 Ω and X = 15.6 Ω
Section 2-5 Independent Sources
P 2.5-1 A current source and a voltage source are connected
in parallel with a resistor as shown in Figure P 2.5-1. All of the
elements connected in parallel have the same voltage, vs in this
circuit. Suppose that vs = 15 V, is = 3 A, and R = 5 Ω. (a)
Calculate the current i in the resistor and the power absorbed by
the resistor. (b) Change the current source current to is = 5 A
and recalculate the current, i, in the resistor and the power
absorbed by the resistor.
i
vs +
is
R
–
Figure P 2.5-1
Answer: i = 3 A and the resistor absorbs 45 W both when is = 3
A and when is = 5 A.
Solution:
v s 15
2
=
= 3 A and P = R i 2 = 5 ( 3 ) = 45 W
R
5
(b) i and P do not depend on is .
(a) i =
P 2.5-2 A current source and a voltage source are connected
in series with a resistor as shown in Figure P 2.5-2. All of the
elements connected in series have the same current, is, in this
circuit. Suppose that vs = 10 V, is = 2 A, and R = 5 Ω. (a)
Calculate the voltage v across the resistor and the power
absorbed by the resistor. (b) Change the voltage source voltage
to vs = 5 V and recalculate the voltage, v, across the resistor and
the power absorbed by the resistor.
+
The values of i and P are 3 A and 45 W, both when i s = 3 A and when i s = 5 A.
–
vs
is
R
+
v
–
Figure P 2.5-2
Answer: v = 10 V and the resistor absorbs 20 W both when vs =
10 V and when vs = 5 V.
Solution:
(a) From Ohm’s law v = R i s = 5 ( 3) = 15 V . (The resistor voltage does not depend on
v 2 152
=
= 45 W .
5
R
(b) Since v and P do not depend on v s the values of v and P are 15 V and 45 W
both when vs =10V and when vs =5V .
the voltage source voltage.) Next P =
P 2.5-3 The current source and voltage source in the circuit
shown in Figure P 2.5-3 are connected in parallel so that they
both have the same voltage, vs. The current source and voltage
source are also connected in series so that they both have the
same current, is. Suppose that vs = 12 V and is = 3 A. Calculate
the power supplied by each source.
Answer: The voltage source supplies –36 W, and the current
source supplies 36 W.
Solution:
is
is
+
vs
–
vs +
–
Figure P 2.5-3
Consider the current source:
i s and v s do not adhere to the passive convention,
so Pcs =i s v s =3⋅12 = 36 W
is the power supplied by the current source.
Consider the voltage source:
i s and v s do adhere to the passive convention,
so Pvs = i s vs =3 ⋅12 = 36 W
is the power absorbed by the voltage source.
∴ The voltage source supplies −36 W.
P 2.5-4 The current source and voltage source in the circuit
shown in Figure P 2.5-4 are connected in parallel so that they
both have the same voltage, vs. The current source and voltage
source are also connected in series so that they both have the
same current, is. Suppose that vs = 12 V and is = 3 A. Calculate
the power supplied by each source.
Answer: The voltage source supplies 36 W, and the current
source supplies –36 W.
Solution:
Consider the current source. iS and vS adhere to the
passive convention so PCS = iS vS = 2 (12 ) = 24 W
is the received by the current source. The current source
supplies −24 W.
Consider the voltages source. iS and vS do not adhere to
the passive convention so PCS = iS vS = 2 (12 ) = 24 W
is the supplied by the voltage source.
is
+
vs
–
vs +
–
is
Figure P 2.5-4
P 2.5-5
(a)
Find the power supplied by the voltage source shown in
Figure P 2.5-5 when for t ≥ 0 we have
i
v
v = 2 cos t V
and
(b)
+
–
i =10 cos t mA
Determine the energy supplied by this voltage source for
the period 0 ≤ t ≤ 1 s.
Figure P 2.5-5
Solution:
(a) P = v i = (2 cos t ) (10 cos t ) = 20 cos 2 t mW
⎛1 1
⎞
(b) w = ∫ P dt = ∫ 20 cos t dt = 20⎜ t + sin 2t ⎟
0
0
⎝2 4
⎠
1
1
1
= 10 + 5 sin 2 mJ
2
0
P 2.5-6 Figure P 2.5-6 shows a battery connected to a load. The
load in Figure P 2.5-6 might represent automobile headlights, a
digital camera, or a cell phone. The energy supplied by the battery to
load is given by
i
+
–
v
R
t2
w = ∫ vi dt
t1
When the battery voltage is constant and the load resistance is fixed,
then the battery current will be constant and
battery
load
Figure P 2.5-6
w = vi (t2 − t1 )
The capacity of a battery is the product of the battery current and time required to discharge the
battery. Consequently, the energy stored in a battery is equal to the product of the battery voltage
and the battery capacity. The capacity is usually given with the units of Ampere-hours (Ah). A
new 12-V battery having a capacity of 800 mAh is connected to a load that draws a current of 50
mA. (a) How long will it take for the load to discharge the battery? (b) How much energy will be
supplied to the load during the time required to discharge the battery?
Solution:
(a) time to discharge =
capacity 800 mAh
=
= 32 hours
current
25 mA
(b) energy = (12 V) (0.025 A) (32 * 60*60 seconds) =34.56 kJ
Section 2-6 Voltmeters and Ammeters
P 2.6-1
(a)
For the circuit of Figure P 2.6-1:
+ 5 . 0
– . 5 0
Voltmeter
Ammeter
What is the value of the resistance
R?
(b)
How much power is delivered by
the voltage source?
R
12 V
+
–
1
2
A
Figure P 2.6-1
Solution:
v
5
(a) R = =
= 10 Ω
i 0.5
(b) The voltage, 12 V, and the current, 0.5 A, of the voltage source adhere to the passive
convention so the power
P = 12 (0.5) = 6 W
is the power received by the source. The voltage source delivers -6 W.
P 2.6-2 The current source in Figure P 2.6-2 supplies 40 W. What values do the meters in
Figure P 2.6-2 read?
Ammeter
Voltmeter
4Ω
+
–
+
v
–
i
12 V
2A
Figure P 2.6-2
Solution:
The voltmeter current is zero so the ammeter current is equal to the current source current except
for the reference direction:
i = -2 A
The voltage v is the voltage of the current source. The power supplied by the current source is
40 W so
40 = 2 v ⇒ v = 20 V
P 2.6-3 An ideal voltmeter is modeled as an open circuit. A more realistic model of a voltmeter
is a large resistance. Figure P 2.6-3a shows a circuit with a voltmeter that measures the voltage
vm. In Figure P 2.6-3b the voltmeter is replaced by the model of an ideal voltmeter, an open
circuit. Ideally, there is no current in the 100-Ω resistor and the voltmeter measures vmi = 12 V,
the ideal value of vm. In Figure P 2.6-3c the voltmeter is modeled by the resistance Rm. Now the
voltage measured by the voltmeter is
⎛ Rm ⎞
vm = ⎜
⎟12
⎝ Rm +100 ⎠
As Rm → ∞, the voltmeter becomes an ideal voltmeter and vm → vmi = l2 V. When Rm < ∞, the
voltmeter is not ideal and vm < vmi. The difference between vm and vmi is a measurement error
caused by the fact that the voltmeter is not ideal.
(a)
Express the measurement error that occurs when Rm = 900 Ω as a percent of vmi.
(b)
Determine the minimum value of Rm required to ensure that the measurement error is
smaller than 2 percent of vmi.
100 Ω
12 V
100 Ω
12 V
+
–
+
–
Voltmeter
+
vmi = 12 V
−
+
vm
(b)
100 Ω
−
+
(a)
12 V
+
–
Rm
vm
−
(c)
Figure P 2.6-3
Solution:
⎛ 900 ⎞
vm = ⎜
⎟12 = 10.8 V
⎝ 900 + 100 ⎠
(a)
12 − 10.8
= 0.1 = 10%
12
(b) We require
⎛ Rm ⎞
12 − ⎜
12
⎜ R m + 100 ⎟⎟
⎝
⎠
0.02 ≥
12
⇒
Rm
R m + 100
≥ 0.98
⇒
R m ≥ 4900 Ω
(checked: LNAP 6/16/04)
2.6-4 An ideal ammeter is modeled as a short circuit. A more realistic model of an ammeter is
a small resistance. Figure P 2.6-4a shows a circuit with an ammeter that measures the current im.
In Figure P 2.6-4b the ammeter is replaced by the model of an ideal ammeter, a short circuit.
Ideally, there is no voltage across the 1-kΩ resistor and the ammeter measures imi = 2 A, the ideal
value of im. In Figure P 2.6-4c the ammeter is modeled by the resistance Rm. Now the current
measured by the ammeter is
⎛ 1000 ⎞
im = ⎜
⎟2
⎝ 1000 + Rm ⎠
As Rm → 0, the ammeter becomes an ideal ammeter and im → imi = 2 A. When Rm > 0, the
ammeter is not ideal and im < imi. The difference between im and imi is a measurement error
caused by the fact that the ammeter is not ideal.
(a)
Express the measurement error that occurs when Rm = 10 Ω as a percent of imi.
(b)
Determine the maximum value of Rm required to ensure that the measurement error is
smaller than 5 percent.
imi = 2 A
im
Ammeter
2A
2A
1 kΩ
1 kΩ
(b)
im
(a)
2A
Rm
1 kΩ
(c)
Figure P 2.6-4
Solution:
(a)
⎛ 1000 ⎞
im = ⎜
⎟ 2 = 1.98 A
⎝ 1000 + 10 ⎠
% error =
2 − 1.98
× 100 = 0.99%
2
(b)
⎛ 1000 ⎞
2−⎜
2
⎜ 1000 + R m ⎟⎟
⎝
⎠
0.05 ≥
2
⇒
1000
≥ 0.95
1000 + R m
⇒
R m ≤ 52.63 Ω
(checked: LNAP 6/17/04)
P 2.6-5 The voltmeter in Figure P 2.6-5a
measures the voltage across the current
source. Figure P 2.6-5b shows the circuit
after removing the voltmeter and labeling
the voltage measured by the voltmeter as
vm. Also, the other element voltages and
currents are labeled in Figure P 2.6-5b.
Given that
25 Ω
Voltmeter
12 V
+
–
2A
(a)
12 = vR + vm and − iR = is = 2 A
vR = 25 iR
and
(a)
(b)
25 Ω iR
+ vR –
Determine the value of the voltage
measured by the meter.
Determine the power supplied by
each element.
12 V
+
–
2A
is
vm
–
(b)
Figure P 2.6-5
Solution:
a.)
+
v R = 25 i R = 25 ( −2 ) = −50 V
v m = 12 − v R = 12 − ( −50 ) = 62 V
b.)
Element
voltage source
Power supplied
current source
62 ( 2 ) = 124 W
resistor
−v R × i R = − ( −50 )( −2 ) = −100 W
total
0
( )
−12 i s = −12 ( 2 ) = −24 W
P 2.6-6 The ammeter in Figure P 2.6-6a
measures the current in the voltage source.
Figure P 2.6-6b shows the circuit after
removing the ammeter and labeling the current
measured by the ammeter as im. Also, the other
element voltages and currents are labeled in
Figure P 2.6-6b.
Given that
Ammeter
12 V
+
–
25 Ω
2A
2 + im = iR and vR = vs =12 V
vR = 25iR
and
(a)
(b)
(a)
Determine the value of the current
measured by the meter.
Determine the power supplied by each
element.
+
im
12 V
+
–
25 Ω
iR
vR
–
(b)
Figure P 2.6-6
Solution:
a.)
iR =
vR
25
=
12
= 0.48 A
25
i m = i R − 2 = 0.48 − 2 = −1.52 A
b.)
Element
voltage source
Power supplied
current source
v s ( 2 ) = 12 ( 2 ) =24 W
resistor
−v R × i R = − (12 )( 0.48 ) = −5.76 W
total
0
( )
12 i m = 12 ( −1.52 ) = −18.24 W
+
2A
vs
–
Section 2-7 Dependent Sources
P 2.7-1 The ammeter in the circuit shown in Figure P 2.7-1 indicates that ia = 2 A, and the
voltmeter indicates that vb = 8 V. Determine the value of r, the gain of the CCVS.
Answer: r = 4 V/A
2 . 0 0
Ammeter
8 . 0 0
Voltmeter
R
ia
+
+
–
vs
r ia
+
–
vb
–
Figure P 2.7-1
Solution:
r =
vb 8
= =4 Ω
ia 2
1
P 2.7-2 The ammeter in the circuit shown in Figure P 2.7-2 indicates that ia = 2 A, and the
voltmeter indicates that vb = 8 V. Determine the value of g, the gain of the VCCS.
Answer: g = 0.25 A/V
2 . 0 0
Ammeter
8 . 0 0
Voltmeter
ia
R1
+
vs
g vb
+
–
R2
vb
–
Figure P 2.7-2
Solution:
vb = 8 V ; g v b = i a = 2 A ; g =
P 2.7-3
ia 2
A
= = 0.25
vb 8
V
The ammeters in the circuit shown in Figure P 2.7-3 indicate that ia = 32 A and ib = 8
A. Determine the value of d, the gain of the CCCS.
Answer: d = 4 A/A
3 2 . 0
Ammeter
8 . 0 0
Ammeter
ia
R
d ib
vs
ib
+
–
Figure P 2.7-3
Solution:
i b = 8 A ; d i b = i a = 32A ; d =
i a 32
A
=
=4
ib
8
A
2
The voltmeters in the circuit shown in Figure P 2.7-4 indicate that va = 2 V and vb = 8
P 2.7-4
V. Determine the value of b, the gain of the VCVS.
Answer: b = 4 V/V
2 . 0 0
Ammeter
8 . 0 0
Voltmeter
R
ia
+
+
–
vs
r ia
+
–
vb
–
Figure P 2.7-4
Solution:
va = 2 V ; b va = vb = 8 V ; b =
P 2.7-5
vb 8
V
= =4
va 2
V
The values of the current and voltage of each circuit element are shown in Figure P
2.7-5. Determine the values of the resistance, R, and of the gain of the dependent source, A.
−2V+
ia = −0.5 A
A ia = 2 V
−4V+
+
−2 A
10 V
R
+
–
+
1.5 A
2.5 A
14 V
−
–
3.5 A
+
–
12 V
−4 A
Figure P 2.7-5
Solution:
R=−
4
2
V
= 2 Ω and A =
= −4
−2
−0.5
A
(checked: LNAP 6/6/04)
3
P 2.7-6 Find the power supplied by the VCCS in Figure P 2.7-6.
Answer: 17.6 watts are supplied by the VCCS. (–17.6 watts are absorbed by the VCCS.)
– 2. 0 0
Voltmeter
+ 2. 2 0
+
Voltmeter
–
vc
2Ω
+
0.2 Ω
–15.8 V
+
–
6.9 Ω
id = 4vc
vd
–
Figure P 2.7-6
Solution:
vc = −2 V, id = 4 vc = −8 A and vd = 2.2 V
id and vd adhere to the passive convention so
P = vd id = (2.2) (−8) = −17.6
W
is the power received by the dependent source. The power supplied by the dependent source is
17.6 W.
4
P2.7-7 The circuit shown in Figure
P2.7-7 contains a dependent source.
Determine the value of the gain k of
that dependent source.
Figure P2.7-7
Solution:
v a = 10 ( 0.2 ) = 2 V
k v a = − ( −10 ) = 10 V
k=
k va
va
=
10
V
=5
2
V
P2.7-8 The circuit shown in Figure
P2.7-8 contains a dependent source.
Determine the value of the gain k of
that dependent source.
Figure P2.7-8
Solution:
ia = −
10
= −0.05 A = −50 mA
200
k i a = −450 mA
k=
k ia
ia
=
−450
A
=9
−50
A
5
P2.7-9 The circuit shown in Figure
P2.7-9 contains a dependent source.
The gain of that dependent source is
k = 25
V
A
Determine the value of the voltage vb.
Figure P2.7-9
Solution:
−1
= 0.2 A
5
V
k = 25
A
ia = −
v b = 25 ( 0.2 ) = 5 V
P2.7-10 The circuit shown in Figure
P2.7-10 contains a dependent source.
The gain of that dependent source is
k = 90
mA
A
= 0.09
V
V
Determine the value of the current ib.
Figure P2.7-10
Solution:
v a = 100 ( 0.05 ) = 5 V
k = 90
mA
A
= 0.09
V
V
i b = − ( 0.09 )( 5 ) = −0.45 A = −450 mA
6
Section 2-8 Transducers
P 2.8-1 For the potentiometer circuit of Figure 2.8-2, the current source current and
potentiometer resistance are 1.1 mA and 100 kΩ, respectively. Calculate the required angle, θ, so
that the measured voltage is 23 V.
(1 – a)Rp
Voltmeter
+
I
–
vm
+
vm
aRp
I
Rp
–
(a)
(b)
Figure 2.8-2
Solution:
a=
θ
360
,
θ =
360 v
(360)(23 V)
m=
= 75.27°
(100 kΩ)(1.1 mA)
R i
p
P 2.8-2
An AD590 sensor has an associated constant
k =1
. The sensor has a voltage v = 20 V; and the
μA
°K
measured current, i(t), as shown in Figure 2.8-3, is
i(t)
+
+
v(t)
AD590
–
v(t)
i(t) = kT
–
4 μA < i < 13 μA
in a laboratory setting. Find the range of measured
temperature.
(a)
(b)
Figure 2.8-3
Solution:
AD590 : k =1
μA
,
K
v =20 V (voltage condition satisfied)
°
4 μ A < i < 13 μ A ⎫
⎪
⎬
i
T =
⎪⎭
k
⇒
4 ° K< T <13° K
Section 2-9 Switches
t=2s
P 2.9-1 Determine the current, i, at t = 1 s
and at t = 4 s for the circuit of Figure P 2.9-1.
t=3s
15 V
+
–
5 kΩ
+
–
10 V
i
Figure P 2.9-1
Solution:
At t = 1 s the left switch is open and the
right switch is closed so the voltage
across the resistor is 10 V.
i=
v
10
=
= 2 mA
R
5×103
At t = 4 s the left switch is closed and the right switch is open so the voltage across the resistor is
15 V.
i=
v
15
=
= 3 mA
R
5×103
t=3s
P 2.9-2 Determine the voltage, v, at
t = 1 s and at t = 4 s for the circuit shown
in Figure P 2.9-2.
+
1 mA
5 kΩ
v
–
Figure P 2.9-2
Solution:
At t = 1 s the current in the resistor
is 3 mA so v = 15 V.
At t = 4 s the current in the resistor
is 0 A so v = 0 V.
t=2s
2 mA
Rs
12 V
+
+
−
100 Ω
v
12 V
−
(a)
+
−
+
100 Ω
v
−
(b)
Figure P 2.9-3
P 2.9-3 Ideally an open switch is modeled as an open circuit and a closed switch is modeled as
a closed circuit. More realistically, an open switch is modeled as a large resistance and a closed
switch is modeled as a small resistance.
Figure P 2.9-3a shows a circuit with a switch. In figure P 2.9-3b the switch has been
replaced with a resistance. In figure P 2.9-3b the voltage v is given by
⎛ 100 ⎞
v =⎜
⎟12
R
+
100
s
⎝
⎠
Determine the value of v for each of the following cases.
(a)
The switch is closed and Rs = 0 (a short circuit).
(b)
The switch is closed and Rs = 5 Ω.
(c)
The switch is open and Rs = ∞ (an open circuit).
(d)
The switch is open and Rs = 10 kΩ.
Solution:
(a) v = 12 V
⎛ 100 ⎞
(b) v = ⎜
⎟12 = 11.43 V
⎝ 105 ⎠
(c) v = 0 V
⎛ 100 ⎞
(d) v = ⎜
⎟12 = 0.1188 0.12 V
⎝ 10100 ⎠
Section 2-10 How Can We Check…?
P 2.10-1
The circuit shown in
Figure P 2.10-1 is used to test the
– 2 . 0
4 0 . 0
Ammeter
Voltmeter
CCVS. Your lab partner claims that
this measurement shows that the
gain of the CCVS is –20 V/A
R
+
–
vs
instead of +20 V/A. Do you agree?
Justify your answer.
is
vo
is
CCVS
= 20
V
A
+
vo
–
Figure P 2.10-1
Solution:
vo =40 V and i s = − (−2) = 2 A. (Notice that the ammeter measures − i s rather than i s .)
So
vo
40
V
=
= 20
is
2
A
Your lab partner is wrong.
P 2.10-2 The circuit of Figure P 2.10-2 is used
0 . 0 0
to measure the current in the resistor. Once this
Ammeter
current is known, the resistance can be calculated
i
v
R
as R = s . The circuit is constructed using a
i
voltage source with vs = 12 V and a 25-Ω, 1/2-W
+ v
s
–
resistor. After a puff of smoke and an unpleasant
smell, the ammeter indicates that i = 0 A. The
resistor must be bad. You have more 25-Ω, 1/2-W
resistors. Should you try another resistor? Justify
Figure P 2.10-2
your answer.
Hint: 1/2-W resistors are able to safely dissipate one 1/2 W of power. These resistors may fail if
required to dissipate more than 1/2 watt of power.
Solution:
vs 12
=
= 0.48 A. The power absorbed by
R 25
this resistor will be P = i vs = (0.48) (12) = 5.76 W.
We expect the resistor current to be i =
A half watt resistor can't absorb this much power. You should not try another resistor.
Design Problems
DP 2-1 Specify the resistance R in Figure DP 2-1 so that
both of the following conditions are satisfied:
1.
i > 40 mA.
2.
The power absorbed by the resistor is less than 0.5 W.
i
+
R
10 V –
Figure DP 2-1
Solution:
1.)
10
10
> 0.04 ⇒ R <
= 250 Ω
R
0.04
2.)
102 1
<
⇒ R > 200 Ω
R
2
Therefore 200 < R < 250 Ω. For example, R = 225 Ω.
DP 2-2 Specify the resistance R in Figure DP 2-2 so that
both of the following conditions are satisfied:
1.
v > 40 V.
2.
The power absorbed by the resistor is less than 15 W.
Hint: There is no guarantee that specifications can always be
satisfied.
2A
R
Figure DP 2-2
Solution:
1.) 2 R > 40 ⇒ R > 20 Ω
15
2.) 2 2 R < 15 ⇒ R < = 3.75 Ω
4
Therefore 20 < R < 3.75 Ω. These conditions cannot satisfied simultaneously.
+
v
–
R1
R2
R3
DP 2-3 Resistors are given a power rating.
For example, resistors are available with
ratings of 1/8 W, 1/4 W, 1/2 W, and 1 W. A
ir = is
1/2-W resistor is able to safely dissipate 1/2 W
of power, indefinitely. Resistors with larger
power ratings are more expensive and bulkier
than resistors with lower power ratings. Good
is
engineering practice requires that resistor
Figure DP 2-3
power ratings be specified to be as large as, but
not larger than, necessary.
Consider the circuit shown in Figure DP 2-3. The values of the resistances are
R1 = 1000 Ω, R2 = 2000 Ω, and R3 = 4000 Ω
The value of the current source current is
is = 30 mA
Specify the power rating for each resistor.
Solution::
P1 = ( 30 mA ) ⋅ (1000 Ω ) = (.03) (1000 ) = 0.9 W < 1 W
2
2
P2 = ( 30 mA ) ⋅ ( 2000 Ω ) = (.03) ( 2000 ) = 1.8 W < 2 W
2
2
P3 = ( 30 mA ) ⋅ ( 4000 Ω ) = (.03) ( 4000 ) = 3.6 W < 4 W
2
2
Chapter 3 Resistive Circuits
Exercises
Figure E 3.2-1
Exercise 3.2-1
Determine the values of i3, i4, i6, v2, v4, and v6 in Figure E 3.2-1.
Answer: i3 = –3 A, i4 = 3 A, i6 = 4 A, v2 = –3 V, v4 = –6 V, v6 = 6 V
Solution:
Apply KCL at node a to get
2 + 1 + i3 = 0 ⇒ i3 = -3 A
Apply KCL at node c to get
2 + 1 = i4 ⇒ i4 = 3 A
Apply KCL at node b to get
i3 + i6 = 1 ⇒ -3 + i6 = 1 ⇒ i6 = 4 A
Apply KVL to the loop consisting of elements A and B to get
-v2 – 3 = 0 ⇒ v2 = -3 V
Apply KVL to the loop consisting of elements C, E, D, and A to get
3 + 6 + v4 – 3 = 0 ⇒ v4 = -6 V
Apply KVL to the loop consisting of elements E and F to get
v6 – 6 = 0 ⇒ v6 = 6 V
Check: The sum of the power supplied by all branches is
-(3)(2) + (-3)(1) – (3)(-3) + (-6)(3) – (6)(1) + (6)(4) = -6 - 3 + 9 - 18 - 6 + 24 = 0
Exercise 3.3-1 Determine the voltage measured by the voltmeter in the circuit shown in Figure E 3.3-1a.
Hint: Figure E3.3-1b shows the circuit after the ideal voltmeter has been replaced by the equivalent
open circuit and a label has been added to indicate the voltage measured by the voltmeter, vm.
Answer : vm = 2 V
Figure E 3.3-1
Solution:
25
From voltage division ⇒ v =
(8) = 2 V
m 25+75
Exercise 3.3-2 Determine the voltage measured by the voltmeter in the circuit shown in Figure E 3.32a.
Hint: Figure E 3.3-2b shows the circuit after the ideal voltmeter has been replaced by the equivalent
open circuit and a label has been added to indicate the voltage measured by the voltmeter, vm.
Answer: vm = –2 V
Figure E 3.3-2
Solution:
25
From voltage division ⇒ v =
( −8 ) = −2 V
m 25+75
Exercise 3.4-1 A resistor network consisting of parallel resistors is shown in a package used for
printed circuit board electronics in Figure E 3.4-1a. This package is only 2 cm × 0.7 cm, and each
resistor is 1 kΩ. The circuit is connected to use four resistors as shown in Figure E 3.4-1b. Find the
equivalent circuit for this network. Determine the current in each resistor when is = 1 mA.
Answer: Rp = 250 Ω
Figure E 3.4-1
Solution:
1
R
=
eq
1
1
1
1
4
+ 3+ 3+ 3= 3
3
10 10 10 10 10
⇒ R
eq
=
By current division, the current in each resistor =
103
1
=
kΩ
4
4
1 -3
1
(10 ) =
mA
4
4
Exercise 3.4-2 Determine the current
measured by the ammeter in the circuit
shown in Figure E 3.4-2a.
Hint: Figure E 3.4-2b shows the circuit
after the ideal ammeter has been replaced
by the equivalent short circuit and a label
has been added to indicate the current
measured by the ammeter, im.
Answer: im = –1 A
Solution:
From current division
10
i =
( −5 ) = −1 A
m 10+40
Figure E 3.4-2
Exercise 3.6-1 Determine the resistance measured by the ohmmeter in Figure E 3.6-1.
Figure E 3.6-1
Answer:
(30 + 30) ⋅ 30
+ 30 = 50 Ω
(30 + 30) + 30
Section 3-2 Kirchhoff’s Laws
P 3.2-1
Consider the circuit shown in Figure P 3.2-1. Determine the values of the power supplied by
branch B and the power supplied by branch F.
Figure P 3.2-1
Solution:
Apply KCL at node a to get
2 + 1 = i + 4 ⇒ i = -1 A
The current and voltage of element B adhere to the passive convention so (12)(-1) = -12 W is power
received by element B. The power supplied by element B is 12 W.
Apply KVL to the loop consisting of elements D, F, E, and C to get
4 + v + (-5) – 12 = 0 ⇒ v = 13 V
The current and voltage of element F do not adhere to the passive convention so (13)(1) = 13 W is the
power supplied by element F.
Check: The sum of the power supplied by all branches is
-(2)(-12) + 12 – (4)(12) + (1)(4) + 13 – (-1)(-5) = 24 +12 – 48 + 4 +13 –5 = 0
Determine the values of i2, i4, v2, v3, and v6 in Figure P 3.2-2.
P 3.2-2
Solution:
Apply KCL at node a to get
2 = i2 + 6 = 0 ⇒ i2 = −4 A
Apply KCL at node b to get
3 = i4 + 6 ⇒ i4 = -3 A
Apply KVL to the loop consisting of elements A and B to get
-v2 – 6 = 0 ⇒ v2 = -6 V
Apply KVL to the loop consisting of elements C, D, and A to get
-v3 – (-2) – 6 = 0 ⇒ v4 = -4 V
Apply KVL to the loop consisting of elements E, F and D to get
4 – v6 + (-2) = 0 ⇒ v6 = 2 V
Check: The sum of the power supplied by all branches is
−(6)(2) – (-6)(-4) – (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = -12 - 24 + 24 + 6 + 12 – 6 = 0
P 3.2-3 Consider the circuit shown in Figure P 3.2-3.
(a)
Suppose that R1 = 6 Ω and R2 = 3 Ω. Find the current i and the voltage v.
(b)
Suppose, instead, that i = 1.5 A and v = 2 V. Determine the resistances R1 and R2.
(c)
Suppose, instead, that the voltage source supplies 24 W of power and that the current source
supplies 9 W of power. Determine the current i, the voltage v, and the resistances R1 and R2.
Figure P 3.2-3
Solution:
KVL : −12 − R 2 (3) + v = 0 (outside loop)
v = 12 + 3R 2 or R 2 =
KCL
i+
12
− 3 = 0 (top node)
R1
i = 3−
(a)
(b)
v = 12 + 3 ( 4 ) = 24 V and i = 3 −
R2 =
v − 12
3
12
12
or R1 =
R1
3−i
12
= 1.5 A
8
42 − 12
12
= 10 Ω ; R1 =
= 16 Ω
3
3 − 2.25
(checked using LNAP 7/27/08)
(c)
24 = − 12 i, because 12 and i adhere to the passive convention.
12
= 2.4 Ω
3+ 2
because 3 and v do not adhere to the passive convention
∴ i = − 2 A and R1 =
9 = 3v,
3 − 12
= −3 Ω
3
The situations described in (b) and (c) cannot occur if R1 and R2 are required to be
nonnegative.
∴ v = 3V
and R 2 =
P 3.2-4
Determine the power absorbed by each of the resistors in the circuit shown in Figure P 3.2-4.
Figure P 3.2-4
Solution:
Power absorbed by the 4 Ω resistor = 4 ⋅ i 2 = 100 W
2
Power absorbed by the 6 Ω resistor = 6 ⋅ i 2 = 24 W
1
Power absorbed by the 8 Ω resistor = 8 ⋅ i 2 = 72 W
4
12
i = =2A
1 6
20
i =
= 5A
2
4
i = 3−i = − 2 A
3
2
i =i +i =3A
4 2 3
(checked using LNAP 8/16/02)
P 3.2-5 Determine the power absorbed by each of the resistors in the circuit shown in Figure P 3.2-5.
Answer: The 4-Ω resistor absorbs 16 W, the 6-Ω resistor absorbs 24 W, and the 8-Ω resistor absorbs 8
W.
Figure P 3.2-5
Solution:
v1 = 8 V
v2 = −8 + 8 + 12 = 12 V
(checked using LNAP 8/16/02)
v3 = 2⋅ 4 = 8 V
v2
4Ω : P = 3 = 16 W
4
v2
6Ω : P = 2 = 24 W
6
2
v
8Ω : P = 1 = 8 W
8
P 3.2-6
Determine the power supplied by each voltage source in the circuit of Figure P 3.2-6.
Answer:
The 2-V voltage source supplies 2 mW and the 3-V voltage source supplies –6 mW.
Figure P 3.2-6
Solution:
P2 V = + ⎡⎣ 2 × (1× 10−3 ) ⎤⎦ = 2 × 10−3 = 2 mW
P3 V = + ⎡⎣3 × ( −2 × 10−3 ) ⎤⎦ = −6 × 10−3 = −6 mW
(checked using LNAP 8/16/02)
P 3.2-7 What is the value of the resistance R in Figure P 3.2-7?
Hint: Assume an ideal ammeter. An ideal ammeter is equivalent to a short circuit.
Answer: R = 4 Ω
Figure P 3.2-7
Solution:
KCL: iR = 2 + 1 ⇒ iR = 3 A
KVL: vR + 0 − 12 = 0 ⇒ vR = 12 V
∴ R=
vR 12
=
=4Ω
iR
3
(checked using LNAP 8/16/02)
P 3.2-8 The voltmeter in Figure P 3.2-8 measures the value of the voltage across the current source to
be 56 V. What is the value of the resistance R?
Hint: Assume an ideal voltmeter. An ideal voltmeter is equivalent to an open circuit.
Answer: R = 10 Ω
Figure P 3.2-8
Solution:
KVL: vR + 56 + 24 = 0 ⇒ vR = −80 V
KCL: iR + 8 = 0 ⇒ iR = −8 A
∴ R=
vR −80
=
= 10 Ω
iR
−8
(checked using LNAP 8/16/02)
P 3.2-9
Determine the values of the resistances R1 and R2 in Figure P 3.2-9.
Figure P 3.2-9
Solution:
KCL at node b:
5.61 3.71 − 5.61 12 − 5.61
−1.9
=
+
⇒ 0.801 =
+ 1.278
7
R1
5
R1
⇒ R1 =
KCL at node a:
1.9
= 3.983 ≈ 4 Ω
1.278 − 0.801
3.71 3.71 − 5.61 3.71 − 12
−8.29
+
+
= 0 ⇒ 1.855 + ( −0.475 ) +
=0
2
4
R2
R2
⇒ R2 =
8.29
= 6.007 ≈ 6 Ω
1.855 − 0.475
(checked using LNAP 8/16/02)
P 3.2-10 The circuit shown in Figure P 3.2-10 consists of five voltage sources and four current
sources. Express the power supplied by each source in terms of the voltage source voltages and the
current source currents.
Figure P 3.2-10
Solution:
The subscripts suggest a numbering of the sources. Apply KVL to get
v1 = v 2 + v 5 + v 9 − v 6
i1 and v1 do not adhere to the passive convention, so
p 1 = i1 v 1 = i1 ( v 2 + v 5 + v 9 − v 6 )
is the power supplied by source 1. Next, apply KCL to get
i 2 = − ( i1 + i 4 )
i 2 and v 2 do not adhere to the passive convention, so
p 2 = i 2 v 2 = − ( i1 + i 4 ) v 2
is the power supplied by source 2. Next, apply KVL to get
v3 = v6 − (v5 + v9 )
i 3 and v 3 adhere to the passive convention, so
(
p 3 = −i 3 v 3 = −i 3 v 6 − ( v 5 + v 9 )
is the power supplied by source 3. Next, apply KVL to get
v 4 = v 2 + v5 + v8
i 4 and v 4 do not adhere to the passive convention, so
p 4 = i4 v 4 = i4 (v 2 + v5 + v8 )
is the power supplied by source 4. Next, apply KCL to get
)
(
)
i 5 = i 3 − i 2 = i 3 − − ( i1 + i 4 ) = i1 + i 3 + i 4
i 5 and v 5 adhere to the passive convention, so
p 5 = −i 5 v 5 = − ( i1 + i 3 + i 4 ) v 5
is the power supplied by source 5. Next, apply KCL to get
i 6 = i 7 − ( i1 + i 3 )
i 6 and v 6 adhere to the passive convention, so
(
)
p 6 = −i 6 v 6 = − i 7 − ( i1 + i 3 ) v 6
is the power supplied by source 6. Next, apply KVL to get
v 7 = −v 6
i 7 and v 7 adhere to the passive convention, so
p 7 = −i 7 v 7 = −i 7 ( −v 6 ) = i 7 v 6
is the power supplied by source 7. Next, apply KCL to get
i 8 = −i 4
i 8 and v 8 do not adhere to the passive convention, so
p 8 = i 8 v 8 = ( −i 4 ) v 8 = −i 4 v 8
is the power supplied by source 8. Finally, apply KCL to get
i 9 = i1 + i 3
i 9 and v 9 adhere to the passive convention, so
p 9 = −i 9 v 9 = − ( i1 + i 3 ) v 9
is the power supplied by source 9.
(Check:
9
∑p
n =1
n
= 0 .)
P 3.2-11
Determine the power received by each of the resistors in the circuit shown in Figure P 3.2-11.
Figure P 3.2-11
Solution
The subscripts suggest a numbering of the circuit elements. Apply KCL to get
i 2 + 0.2 + 0.3 = 0 ⇒ i 2 = −0.5 A
The power received by the 6 Ω resistor is
p 2 = 6 i 2 2 = 6 ( −0.5 ) = 1.5 W
2
Next, apply KCL to get
i 5 = 0.2 + 0.3 + 0.5 = 1.0 A
The power received by the 8 Ω resistor is
p 5 = 8 i 5 2 = 8 (1) = 8 W
2
Next, apply KVL to get
v 7 = 15 V
The power received by the 20 Ω resistor is
p7 =
v 72
20
=
15 2
= 11.25 W
20
is the power supplied by source 7. Finally, apply KCL to get
i 9 = 0.2 + 0.5 = 0.7 A
The power received by the 5 Ω resistor is
p 9 = 5 i 9 2 = 5 ( 0.7 ) = 2.45 W
2
P 3.2-12 Determine the voltage and current of each of the circuit elements in the circuit shown in
Figure P 3.2-12.
Figure P 3.2-12
Solution: We can label the circuit as follows:
The subscripts suggest a numbering of the circuit elements. Apply KCL at node b to get
i 4 + 0.25 + 0.75 = 0 ⇒ i 4 = −1.0 A
Next, apply KCL at node d to get
i 3 = i 4 + 0.25 = −1.0 + 0.25 = −0.75 A
Next, apply KVL to the loop consisting of the voltage source and the 60 Ω resistor to get
v 2 − 15 = 0 ⇒ v 2 = 15 V
Apply Ohm’s law to each of the resistors to get
i2 =
and
v2
60
=
15
= 0.25 A ,
60
v 3 = 10 i 3 = 10 ( −0.75 ) = −7.5 V
v 4 = 20 i 4 = 20 ( −1) = −20 V
Next, apply KCL at node c to get
i1 + i 2 = i 3 ⇒ i1 = i 3 − i 2 = −0.75 − 0.25 = −1.0 A
Next, apply KVL to the loop consisting of the 0.75 A current source and three resistors to get
v 6 − v 4 − v 3 − v 2 = 0 ⇒ v 6 = v 4 + v 3 + v 2 = −20 + (−7.5) + 15 = −12.5 V
Finally, apply KVL to the loop consisting of the 0.25 A current source and the 20 Ω resistor to get
v 5 + v 4 = 0 ⇒ v 5 = −v 4 = − ( −20 ) = 20 V
(Checked: LNAPDC 8/28/04)
P 3.2-13
Determine the value of the current that is measured by the meter in Figure P 3.2-13.
Figure P 3.2-13
Solution:
We can label the circuit as shown.
The subscripts suggest a numbering of the
circuit elements. Apply KVL to node the left
mesh to get
15 i1 + 25 i1 − 20 = 0 ⇒ i1 =
20
= 0.5 A
40
Apply KVL to node the left mesh to get
v 2 − 25 i1 = 0 ⇒ v 2 = 25 i1 = 25 ( 0.5 ) = 12.5 V
Apply KCL to get i m = i 2 . Finally, apply Ohm’s law to the 50 Ω resistor to get
im = i2 =
v2
50
=
12.5
= 0.25 A
50
(Checked: LNAPDC 9/1/04)
P 3.2-14
Determine the value of the voltage that is measured by the meter in Figure P 3.2-14.
Figure P 3.2-14
Solution:
We can label the circuit as shown.
The subscripts suggest a numbering of the
circuit elements. Ohm’s law to the 48 Ω
resistor to get
v1 = 48 i1
Apply KCL at the top node of the CCCS to get
i1 + 5 i1 = i 2
⇒ i 2 = 6 i1
Ohm’s law to the 4 Ω resistor to get
v m = 4 i 2 = 4 ( 6 i1 ) = 24 i1
Apply KVL to the outside loop to get
v1 + v m − 24 = 0 ⇒ 48 i1 + 24 i1 = 24 ⇒ i1 =
Finally,
24 1
= A
72 3
⎛1⎞
v m = 24 i1 = 24 ⎜ ⎟ = 8 V
⎝ 3⎠
(Checked: LNAPDC 9/1/04)
P 3.2-15
Determine the value of the voltage that is measured by the meter in Figure P 3.2-15.
Figure P 3.2-15
Solution:
We can label the circuit as shown.
The subscripts suggest a numbering of the
circuit elements. Apply KCL at the top node of
the current source to get
i1 = i 2 + 0.25
Apply Ohm’s law to the resistors to get
v1 = 20 i1 and v 2 = 60 i 2 = 60 ( i1 − 0.25 ) = 60 i1 − 15
Apply KVL to the outside to get
v 2 + 80 i1 + v1 = 0 ⇒
Finally,
( 60 i
1
− 15 ) + 80 i1 + 20 i1 = 0 ⇒ i1 =
15
= 0.09375 A
160
v m = 80 i1 = 80 ( 0.09375 ) = 7.5 V
(Checked: LNAPDC 9/1/04)
P 3.2-16
The voltage source in Figure P 3.2-16 supplies 4.8 W of power. The current source supplies
3.6 W. Determine the values of the resistances, R1 and R2.
Figure P 3.2-16
Solution:
i=
R1 =
3.6
4.8
= 0.3 A and v =
= 9.6 V
12
0.5
12 − 9.6
9.6
= 8 Ω and R 2 =
= 12 Ω
0.3
0.3 + 0.5
(Checked: LNAPDC 7/27/08)
P 3.2-17
Answer:
Determine the current i in Figure P 3.3-17.
i=4A
Figure P 3.3-17
Solution:
Apply KCL at node a to determine the current
in the horizontal resistor as shown.
Apply KVL to the loop consisting of the
voltages source and the two resistors to get
-4(2-i) + 4(i) - 24 = 0 ⇒ i = 4 A
P 3.2-18
Determine the value of the current im in Figure P 3.2-18a.
Hint: Apply KVL to the closed path a-b-d-c-a in Figure P 3.2-18b to determine va. Then apply KCL at
node b to find im.
Answer:
im = 9 A.
Figure P 3.2-18
Solution:
−18 + 0 − 12 − va = 0 ⇒ va = −30 V and im =
2
va + 3 ⇒ im = 9 A
5
P3.2-19 Determine the value of the voltage v6 for the circuit shown in Figure P3.2-19.
Figure P3.2-23
Solution:
Apply KCL at the left node:
0.25 + i 2 + 0.22 = 0 ⇒ i 2 = −0.47 A
Use the element equation of the dependent
source:
v 5 = 10 i 2 = 10 ( −0.47 ) = −4.7 V
Apply KVL to the right mesh
v 5 − v 6 − 12 = 0 ⇒ v 6 = v 5 − 12 = −4.7 − 12 = −16.7 V
P3.2-20 Determine the value of the voltage v6 for the circuit shown in Figure P3.2-20.
Figure P3.2-20
Solution:
Apply KCL at the left node:
0.015 + 0.025 = i 2
⇒ i 2 = 0.040 A
From Ohm’s law
v 2 = 25 i 2 = 25(0.04) = 1 V
Use the element equation of the dependent
source:
v 5 = 10 v 2 = 10 (1) = 10 V
Apply KVL to the right mesh
v 5 + v 6 + 12 = 0 ⇒ v 6 = −v 5 − 12 = −10 − 12 = −22 V
P3.2-21 Determine the value of the voltage v5 for the circuit shown in Figure P3.2-21.
Figure P3.2-21
Solution:
Apply KVL to the left mesh:
v 2 − 18 + 12 = 0 ⇒ v 2 = 6 V
Use the element equation of the dependent
source:
i 6 = 0.10 v 2 = 0.10 ( 6 ) = 0.6 A
Apply KCL at the right node
v5
25
+ i 6 = 0.25 ⇒ v 5 = 25 ( 0.25 − i 6 ) = 25 ( 0.25 − 0.6 ) = −8.75 V
P3.2-22 Determine the value of the voltage v5 for the circuit shown in Figure P3.2-22.
Figure P3.2-22
Solution:
Apply KVL to the left mesh:
48 i 2 − 18 + 12 = 0 ⇒ i 2 = 0.125 A
Use the element equation of the dependent
source:
i 6 = 1.5 i 2 = 1.5 ( 0.125 ) = 0.1875 A
Apply KCL at the right node
v5
40
+ i 6 = 0.25 ⇒ v 5 = 40 ( 0.25 − i 6 ) = 40 ( 0.25 − 0.1875 ) = 2.5 V
P3.2-23 Determine the value of the voltage v6 for the circuit shown in Figure P3.2-23.
Figure P3.2-23
Solution:
Apply KVL to the left mesh:
50 i 2 + 18 + 12 = 0 ⇒ i 2 =
−30
= −0.6 A
50
Use the element equation of the dependent
source:
v 5 = 10 i 2 = 10 ( −0.6 ) = −6 V
Apply KVL to the right mesh
v 5 + v 6 + 12 = 0 ⇒ v 6 = −v 5 − 12 = − ( −6 ) − 12 = −6 V
P3.2-24 Determine the value of the voltage v5 for the circuit shown in Figure P3.2-24.
Figure P3.2-24
Solution:
Apply KCL at the left node:
0.015 = 0.025 +
v2
250
⇒ v 2 = 250 ( −0.01) = −2.5 V
Use the element equation of the dependent source:
i 6 = 0.5 v 2 = 0.5 ( −2.5 ) = −1.25 A
Apply KCL at the right node
v5
25
+ i 6 = 0.25 ⇒ v 5 = 25 ( 0.25 − i 6 ) = 25 ( 0.25 − ( −1.25 ) ) = 37.5 V
P3.2-25 The voltage source in the circuit shown in Figure P3.2-25 supplies 2 W of power. The value of
the voltage across the 25 Ω is v5 = 4 V. Determine the values of the resistance R1 and of the gain, G, of
the CCVS.
Figure P3.2-25
Solution:
The voltage source current is calculated from the
values of the source voltage and power:
2
is =
= 0.1 A
20
Apply KCL at the bottom node to get
v2
4
is + G v2 =
⇒ G v2 =
− 0.1 = 0.06 A
25
25
Then
G v 2 0.06
G=
=
= 0.015 A/V = 15 mA/V
v2
4
Next, use Ohm’s law to determine the value of the resistance R1:
R1 =
20 − v 2 20 − 4
=
= 100 Ω
4
v2
25
25
P3.2-26 Consider the circuit shown in Figure P3.2-26.
Determine the values of
(a) The current i a in the 20-Ω resistor.
(b) The voltage v b across the 10-Ω resistor.
(c) The current i c in the independent voltage source.
Figure P3.2-26
Solution:
25
= −1.25 A .
20
(b) From KVL 4 i a − v b + 20 i a = 0 ⇒ v b = 24 i a = 24 ( −1.25 ) = −30 V
(a) From Ohm’s law i a = −
(c) From KCL i c = i a +
vb
10
= −1.25 +
−30
= −4.25 A
10
P3.2-27 Consider the circuit shown in Figure 3.2-27.
a.) Determine the values of the resistances.
b.) Determine the values of the power supplied by
each current source.
c.) Determine the values of the power received by each
resistor.
Figure 3.2-27
Solution:
Using KCL and Ohm’s law:
va = Ra ( I 2 − I1 ) , vb = − Rb ( I1 + I 3 )
and
vc = Rc ( I 2 + I 3 )
Using KVL, the power supplied the current sources
are:
I 2 ( va + vc ) , − I1 ( va + vb ) and I 3 ( −vb + vc )
The power received the resistors are:
va ( I 2 − I1 ) , −vb ( I1 + I 3 ) and vc ( I 2 + I 3 )
a.)
and
b.)
and
Ra =
4
−3
= 4 Ω , Rb = −
=6Ω
3− 2
2 + ( −1.5 )
Rc =
12
=8 Ω
3 + ( −1.5 )
3 ( 4 + 12 ) = 48 W , −2 ( 4 + ( −3) ) = −2 W
−1.5 ( − ( −3) + 12 ) = −22.5 W
c.) 4 ( 3 − 2 ) = 4 W , − ( −3) ( 2 + ( −1.5 ) ) = 1.5 W
and
12 ( 3 + ( −1.5 ) ) = 18 W
LNAPDC (8/30/10):
P3.2-28 Consider the circuit shown in
Figure 3.2-28.
a. Determine the value of the power
supplied by each independent source.
b. Determine the value of the power
received by each resistor.
c. Is power conserved?
Figure 3.2-28
Solution:
Apply KCL twice and KVL to get
Apply Ohm’s law 3 times to get
Apply KCL twice and KVL to get
a. Notice that the voltage and current reference directions for each independent source do not adhere to
passive convention. Consequently the power supplied by the 2 mA current source is
15 (2) = 30 mW
The power supplied by the 3 mA current source is
9 (3) = 27 mW
The power supplied by the 12 V voltage source is
12 (11/3) = 44 mW
b. Notice that the voltage and current reference directions for each resistor do adhere to passive
convention. Consequently the power received by the 18 kΩ resistor is
12 (2/3) = 8 mW
The power received by the 3 kΩ resistor is
15 (5) = 75 mW
The power received by the 2 kΩ resistor is
6 (3) = 18 mW
c. Power is conserved; the sum of the powers supplied by the independent sources is equal to the sum of
the powers received by the resistors.
30 + 27 + 44 = 101 mW = 8 +75 +18
P3.2-29
P3.2-29 The voltage across the capacitor in Figure P3.2-29 is v ( t ) = 24 − 10 e −25 t V for t ≥ 0. Determine
the voltage source current i(t) for t > 0.
Solution: Notice that i(t) is the current in the 20 Ω resistor. Apply KVL to the left mesh to get
−20 i ( t ) + ⎡⎣ 24 − 10 e
⎡ 24 − 10 e −25 t ⎤ − 30
⎣
⎦
= −0.3 − 0.5 e −25 t for t > 0.
⎦ − 30 = 0 ⇒ i ( t ) =
20
− 25 t ⎤
P3.2-30
P3.2-30 The current the inductor in Figure P3.2-30 is given by i ( t ) = 8 − 6 e −25 t A for t ≥ 0. Determine
the voltage v(t) across the 80 Ω resistor for t > 0.
Solution: Notice that i(t) is the current in the 20 Ω resistor. Apply KCL at the top node of the 80 Ω
resistor to get
v (t )
= 10 − 8 − 6 e −25 t
80
(
)
(
)
⇒ v ( t ) = 80 ⎡10 − 8 − 6 e −25 t ⎤ = 160 + 480 e −25 t for t ≥ 0
⎣
⎦
Section 3-3 Series Resistors and Voltage Division
P 3.3-1 Use voltage division to determine the voltages v1, v2, v3, and v4 in the circuit shown in Figure P
3.3-1.
Figure P 3.3-1.
Solution:
6
6
v =
12 =
12 = 4 V
1 6+3+5+ 4
18
3
5
10
v =
12 = 2 V ; v =
12 =
V
2 18
3 18
3
4
8
v =
12 = V
4 18
3
(checked using LNAP 8/16/02)
P 3.3-2 Consider the circuits shown in Figure P 3.3-2.
(a)
Determine the value of the resistance R in Figure P 3.3-2b that makes the circuit in Figure P 3.32b equivalent to the circuit in Figure P 3.3-2a.
(b)
Determine the current i in Figure P 3.3-2b. Because the circuits are equivalent, the current i in
Figure P 3.3-2a is equal to the current i in Figure P 3.3-2b.
(c)
Determine the power supplied by the voltage source.
Solution:
(a ) R = 6 + 3 + 2 + 4 = 15 Ω
28 28
(b) i =
=
= 1.867 A
R 15
( c ) p = 28 ⋅ i =28(1.867)=52.27 W
(28 V and i do not adhere
to the passive convention.)
(checked using LNAP 8/16/02)
P 3.3-3 The ideal voltmeter in the circuit shown in Figure P 3.3-3 measures the voltage v.
(a)
Suppose R2 = 100 Ω. Determine the value of R1.
(b)
Suppose, instead, R1 = 100 Ω. Determine the value of R2.
(c)
Suppose, instead, that the voltage source supplies 1.2 W of power. Determine the values of both
R1 and R2.
Figure P 3.3-3
Solution:
i R2 = v = 8 V
12 = i R1 + v = i R1 + 8
⇒ 4 = i R1
8
8
4 4 ⋅ 50
; R1 = =
=
= 25 Ω
R 2 50
i
8
4
4
8 8 ⋅ 50
= 100 Ω
(b) i = = ; R2 = =
4
R1 50
i
4
8
( c ) 1.2 = 12 i ⇒ i = 0.1 A ; R1 = = 40 Ω; R2 = = 80 Ω
i
i
(a)
i=
(Checked using LNAP 8/16/02)
P 3.3-4
Determine the voltage v in the circuit shown in Figure P 3.3-4.
Figure P 3.3-4
Solution:
Voltage division
16
v1 =
12 = 8 V
16 + 8
4
v3 =
12 = 4 V
4+8
KVL: v3 − v − v1 = 0
v = −4 V
(checked using LNAP 8/16/02)
P 3.3-5 The model of a cable and load resistor connected to a
source is shown in Figure P 3.3-5. Determine the appropriate
cable resistance, R, so that the output voltage, vo, remains
between 9 V and 13 V when the source voltage, vs, varies
between 20 V and 28 V. The cable resistance can only assume
integer values in the range 20 < R < 100 Ω.
Figure P 3.3-5
Solution:
⎛v
⎞
⇒ R = 50 ⎜ s − 1⎟
⎜v
⎟
⎝ o ⎠
with v = 20 V and v > 9 V, R < 61.1 Ω ⎫
⎪
s
0
⎬ R = 60 Ω
with v = 28 V and v < 13 V, R > 57.7 Ω ⎪
s
0
⎭
⎛ 100 ⎞
using voltage divider: v = ⎜
v
0 ⎝ 100 + 2 R ⎟⎠ s
P 3.3-6
The input to the circuit shown in Figure P 3.3-6 is the voltage of the voltage source, va. The
output of this circuit is the voltage measured by the voltmeter, vb. This circuit produces an output that is
proportional to the input, that is
vb = k va
where k is the constant of proportionality.
(a)
Determine the value of the output, vb, when R = 240 Ω and va = 18 V.
(b)
Determine the value of the power supplied by the voltage source when R = 240 Ω and va = 18 V.
(c)
Determine the value of the resistance, R, required to cause the output to be vb = 2 V when the
input is va = 18 V.
(d)
Determine the value of the resistance, R, required to cause vb = 0.2va (that is, the value of the
constant of proportionality is k = 102 ).
Figure P 3.3-6
Solution:
⎛ 180 ⎞
a.) ⎜
⎟ 18 = 10.8 V
⎝ 120 + 180 ⎠
18
⎛
⎞
b.) 18 ⎜
⎟ = 1.08 W
⎝ 120 + 180 ⎠
⎛ R ⎞
c.) ⎜
⎟ 18 = 2 ⇒ 18 R = 2 R + 2 (120 ) ⇒ R = 15 Ω
⎝ R + 120 ⎠
R
⇒ ( 0.2 )(120 ) = 0.8 R ⇒ R = 30 Ω
d.) 0.2 =
R + 120
(Checked using LNAP 8/16/02)
P 3.3-7
Determine the value of voltage v in the circuit shown in Figure P 3.3-7.
Figure P 3.3-7
Solution:
All of the elements are connected in series.
Replace the series voltage sources with a single equivalent voltage having voltage
12 + 12 – 18 = 6 V.
Replace the series 15 Ω, 5 Ω and 20 Ω resistors by a single equivalent resistance of
15 + 5 + 20 = 40 Ω.
By voltage division
6
⎛ 10 ⎞
v=⎜
⎟ 6 = = 1.2 V
5
⎝ 10 + 40 ⎠
(Checked: LNAP 2/6/07)
P 3.3-8
Determine the power supplied by the dependent source in
the circuit shown in Figure P 3.3-8.
Figure P 3.3-8
Solution:
Use voltage division to get
⎛ 10 ⎞
va = ⎜
⎟ (120 ) = 20 V
⎝ 10 + 50 ⎠
Then
i a = 0.2 ( 20 ) = 4 A
The power supplied by the dependent source is given by
p = (120 ) i a = 480 W
(Checked: LNAP 6/21/04)
P 3.3-9 A potentiometer can be used as a transducer to
convert the rotational position of a dial to an electrical
quantity. Figure P 3.3-9 illustrates this situation. Figure P
3.3-9a shows a potentiometer having resistance Rp
connected to a voltage source. The potentiometer has
three terminals, one at each end and one connected to a
sliding contact called a wiper. A voltmeter measures the
voltage between the wiper and one end of the
potentiometer.
Figure P 3.3-9b shows the circuit after the
potentiometer is replaced by a model of the
potentiometer that consists of two resistors. The
parameter a depends on the angle, θ, of the dial. Here
θ
a = 360
° , and θ is given in degrees. Also, in Figure P 3.39b, the voltmeter has been replaced by an open circuit
and the voltage measured by the voltmeter, vm, has been
labeled. The input to the circuit is the angle θ, and the
output is the voltage measured by the meter, vm.
(a)
Show that the output is proportional to the input.
(b)
Let Rp = 1 kΩ and vs = 24 V. Express the output
as a function of the input. What is the value of
the output when θ = 45º? What is the angle when
vm = 10 V?
Figure P 3.3-9
Solution:
(a) Use voltage division to get
vm =
Therefore
aR p
(1 − a ) R p + R p
v s = av s
⎛ vs ⎞
vm = ⎜
⎟θ
⎝ 360 ⎠
So the input is proportional to the input.
⎛1⎞
(b) When vs = 24 V then v m = ⎜ ⎟ θ . When θ = 45o then vm = 3 V. When vm = 10 V then θ = 150o .
⎝ 15 ⎠
(Checked: LNAP 6/12/04)
P 3.3-10
Determine the value of the voltage measured by the meter in Figure P 3.3-10.
Figure P 3.3-10
Solution:
Replace the (ideal) voltmeter with the equivalent open
circuit. Label the voltage measured by the meter. Label
some other element voltages and currents.
Apply KVL the left mesh to get
8 i a + 4 i a − 24 = 0 ⇒ i a = 2 A
Use voltage division to get
vm =
5
5
4 ia =
4 ( 2) = 5 V
5+3
5+3
(Checked using LNAP 9/11/04)
P 3.3-11 For the circuit of Figure P 3.3-11, find the voltage v3 and the current i and show that the
power delivered to the three resistors is equal to that supplied by the source.
Figure P 3.3-11
Solution:
⎛
3 ⎞
⎟
⎝ 3+9 ⎠
From voltage division v3 = 12 ⎜
then i =
v3
=1A
3
The power absorbed by the resistors is: (12 ) ( 6 ) + (12 ) ( 3) + (12 ) ( 3) = 12 W
The power supplied by the source is (12)(1) = 12 W.
= 3V
P 3.3-12 Consider the voltage divider shown in Figure P 3.3-12 when R1 = 6 Ω. It is desired that the
output power absorbed by R1 = 6 Ω be 6 W. Find the voltage vo and the required source vs.
Answer: vs = 14 V, vo = 6 V
Figure P 3.3-12
Solution:
P = 4.5 W and R1 = 8 Ω
i2 =
P
4.5
=
= 0.5625 or i = 0.75 A
R1
8
v0 = i R1 =(0.75) (8)=6 V
from KVL: − v
+ i (2 + 4 + 8 + 2) = 0
s
⇒ v = 16 i = 16 ( 0.75 ) = 12 V
s
Figure P3.3-13
P3.3-13
Consider the voltage divider circuit shown in Figure P3.3-13. The resistor R represents a temperature
sensor. The resistance R, in Ω, is related to the temperature T, in °C, by the equation
1
R = 50 + T
2
a) Determine the meter voltage, v m , corresponding to temperatures 0 °C, 75 °C and 100 °C.
b) Determine the temperature, T, corresponding to the meter voltages 8 V, 10 V and 15 V.
P3.3-13
Using voltage division
⎛ R ⎞
vm = ⎜
⎟ 20
⎝ 75 + R ⎠
R=
Solving for R yields
75 v m
20 − v m
The temperature can be calculated from the resistance using
⎛ 75 v m
⎞ 150 v m
T = 2 ( R − 50 ) = 2 ⎜
− 50 ⎟ =
− 100
⎜ 20 − v m
⎟ 20 − v m
⎝
⎠
⎛ 50 ⎞
a) At 0 °C the resistance is R = 50 Ω so v m = ⎜
⎟ 20 = 8 V . At 75 °C the resistance is
⎝ 75 + 50 ⎠
⎛ 87.5 ⎞
R = 87.5 Ω so v m = ⎜
⎟ 20 = 10.77 V . At 100 °C the resistance is R = 100 Ω so
⎝ 75 + 87.5 ⎠
⎛ 100 ⎞
vm = ⎜
⎟ 20 = 11.43 V .
⎝ 75 + 100 ⎠
150 ( 8 )
− 100 = 0 °C . When v m = 10 V, the temperature is
20 − 8
150 (10 )
150 (15 )
T=
− 100 = 50 °C . When v m = 15 V, the temperature is T =
− 100 = 350 °C .
20 − 10
20 − 15
b) When v m = 8 V, the temperature is T =
P3.3-14 Consider the circuit shown in Figure P3.3-14.
(a) Determine the value of the resistance R required to cause v o = 17.07 V.
(b) Determine the value of the voltage v o when R = 14 Ω .
(c) Determine the power supplied by the voltage source when v o = 14.22 V.
Figure P3.3-14
Solution:
⎛ 8 ⎞
Voltage division indicates that v o = ⎜
⎟ 32 .
⎝8+ R ⎠
(8 ) 32 − 8 = 6.997 7 Ω .
⎛ 8 ⎞
(a) When v o = 17.07 V, then 17.07 = ⎜
⎟ 32 ⇒ R =
17.07
⎝8+ R ⎠
⎛ 8 ⎞
(b) When R = 14 Ω then v o = ⎜
⎟ 32 = 11.6363 V .
⎝ 8 + 14 ⎠
(c) The power supplied by the voltage source is given by 32
vo
= 4 v o since
vo
8
8
the series elements. When v o = 14.22 V the voltage source supplies 56.88 W.
is the current in each of
Figure P3.3-15
P3.3-15. Figure P3.3-15 shows four similar but slightly different circuits. Determine the values of the
voltages v1 , v 2 , v 3 and v 4 .
Solution: Using voltage division:
⎛ 60 ⎞
⎛ 130 ⎞
v1 = ⎜
⎟ 26 = 12 V , v 2 = − ⎜
⎟ 26 = −16.9 V ,
⎝ 60 + 70 ⎠
⎝ 70 + 130 ⎠
⎛ 30 ⎞
⎛ 70 ⎞
v3 = − ⎜
⎟ 26 = −7.8 V and v 4 = − ⎜
⎟ 26 = 18.2 V
⎝ 70 + 30 ⎠
⎝ 30 + 70 ⎠
Figure P3.3-16
P3.3-16. Figure P3.3-16 shows four similar but slightly different circuits. Determine the values of the
voltages v1 , v 2 , v 3 and v 4 .
Solution: Using voltage division:
⎛ 30 ⎞
⎛ 60 ⎞
v1 = − ⎜
⎟ 28 = −16.8 V , v 2 = − ⎜
⎟ 28 = 16.8 V ,
⎝ 20 + 30 ⎠
⎝ 40 + 60 ⎠
⎛ 30 ⎞
⎛ 20 ⎞
v3 = ⎜
⎟ 28 = 12 V and v 4 = − ⎜
⎟ 28 = −7 V
⎝ 40 + 30 ⎠
⎝ 60 + 20 ⎠
Figure P3.3-17
P3.3-17. The input to the circuit shown in Figure P3.3-19 is the voltage source voltage
v s ( t ) = 12 cos ( 377 t ) mV
The output is the voltage vo(t). Determine vo(t).
P3.3-17
Using voltage division: v a ( t ) =
Using voltage division again:
110
12 cos ( 377 t ) = 11cos ( 377 t ) mV
10 + 110
9900
v o (t ) =
1000 v a ( t )
100 + 9900
Therefore:
v o (t ) =
9900
(1000 )11cos ( 377 t ) = 10890 cos ( 377 t ) mV
100 + 9900
= 10.89 cos ( 377 t ) V
Section 3-4 Parallel Resistors and Current Division
Use current division to determine the currents i1, i2, i3, and i4 in the circuit shown in Figure P
P 3.4-1
3.4-1.
Figure P 3.4-1.
Solution:
i =
1
i =
2
i =
3
i =
4
1
1
1
6
4=
4= A
1 + 1 + 1 +1
1+ 2 + 3 + 6
3
6 3 2 1
1
2
3
4 = A;
1 + 1 + 1 +1
3
6 3 2 1
1
2
4 =1 A
1 + 1 + 1 +1
6 3 2 1
1
4=2 A
1 + 1 + 1 +1
6 3 2
Consider the circuits shown in Figure P 3.4-2.
P 3.4-2
Figure P 3.4-2
(a)
(b)
(c)
Determine the value of the resistance R in Figure P 3.4-2b that makes the circuit in Figure P 3.42b equivalent to the circuit in Figure P 3.4-2a.
Determine the voltage v in Figure P 3.4-2b. Because the circuits are equivalent, the voltage v in
Figure P 3.4-2a is equal to the voltage v in Figure P 3.4-2b.
Determine the power supplied by the current source.
Solution:
(a)
(b)
(c)
1 1 1 1 1
= + + = ⇒ R = 2Ω
R 6 12 4 2
v = 6 ⋅ 2 = 12 V
p = 6 ⋅12 = 72 W
The ideal voltmeter in the circuit shown in Figure P 3.4-3 measures the voltage v.
P 3.4-3
Figure P 3.4-3
(a)
(b)
(c)
Suppose R2 = 12 Ω. Determine the value of R1 and of the current i.
Suppose, instead, R1 = 12 Ω. Determine the value of R2 and of the current i.
Instead, choose R1 and R2 to minimize the power absorbed by any one resistor.
Solution:
i=
8
8
or R1 =
R1
i
8 = R 2 (2 − i ) ⇒ i = 2 −
8
8
or R 2 =
2−i
R2
8 2
8
= A ; R1 = = 12 Ω
2
6 3
3
8 4
8
( b ) i = = A ; R 2 = 4 = 12 Ω
6 3
2−
3
1
( c ) R1 = R 2 will cause i= 2 = 1 A. The current in both R1 and R 2 will be 1 A.
2
R1 R 2
1
= 8 ; R1 = R 2 ⇒ 2 ⋅ R1 = 8 ⇒ R1 = 8 ∴ R1 = R 2 = 8 Ω
2 ⋅
R1 + R 2
2
(a)
i = 2−
P 3.4-4
Determine the current i in the circuit shown in Figure P 3.4-4.
Figure P 3.4-4
Solution:
Current Division:
8
( −6 ) = −2 A
16 + 8
8
i2 =
( −6 ) = −3 A
8+8
i1 =
KCL:
i = i 1 − i 2 = −2 − ( −3 ) = 1 A
P 3.4-5 Consider the circuit shown in Figure P 3.4-5 when 4 Ω
≤ R1 ≤ 6 Ω and R2 = 10 Ω. Select the source is so that vo remains
between 9 V and 13 V
Figure P 3.4-5
Solution:
⎛
⎞
R
1
⎟ i and
current division: i = ⎜
2 ⎜R + R ⎟ s
⎝ 1
2⎠
Ohm's Law: v = i R yields
o
2 2
⎛ v ⎞⎛ R + R ⎞
2⎟
i = ⎜ o ⎟⎜ 1
s
⎜ R ⎟⎜ R
⎟
⎝ 2 ⎠⎝
1 ⎠
plugging in R = 4Ω, v > 9 V gives i > 3.15 A
o
s
1
and R = 6Ω, v < 13 V gives i < 3.47 A
o
s
1
So any 3.15 A < i < 3.47 A keeps 9 V < v < 13 V.
s
o
P 3.4-6 Figure P 3.4-6 shows a transistor amplifier. The values of R1 and R2 are to be selected.
Resistances R1 and R2 are used to bias the transistor, that is, to create useful operating conditions. In this
problem, we want to select R1 and R2 so that vb = 5 V. We expect the value of ib to be approximately 10
μA. When i1 ≤ 10ib, it is customary to treat ib as negligible, that is, to assume ib = 0. In that case R1 and
R2 comprise a voltage divider.
(a)
(b)
Figure P 3.4-6
Select values for R1 and R2 so that vb = 5 V and the total power absorbed by R1 and R2 is no more
than 5 mW.
An inferior transistor could cause ib to be larger than expected. Using the values of R1 and R2
from part (a), determine the value of vb that would result from ib = 15 μA.
Solution:
(a) To insure that ib is negligible we require
15
i1 =
≥ 10 (10 × 10−6 ) = 10−3
R1 + R 2
R1 + R 2 ≤ 150 kΩ
so
To insure that the total power absorbed by R1 and R2 is no more than 5 mW we require
152
R1 + R 2 ≥ 45 kΩ
≤ 5 × 10−3
⇒
R1 + R 2
Next to cause vb = 5 V we require
5 = vb =
R2
R1 + R 2
15
⇒
R1 = 2 R 2
For example, R1 = 40 kΩ, R 2 = 80 kΩ, satisfy all three requirements.
(b)
KVL gives
KCL gives
Therefore
Finally
(80 ×10 ) i
3
i1 =
1
+ v b − 15 = 0
vb
+ 15 × 10−6
40 ×10
⎛ v
⎞
(80 ×103 ) ⎜ 40 ×b103 + 15 ×10−6 ⎟ + v b = 15
⎝
⎠
13.8
3v b + 1.2 = 15
⇒ vb =
= 4.6 V
3
3
P 3.4-7 Determine the value of the current i in the circuit shown in Figure P 3.4-7.
Figure P 3.4-7
Solution:
All of the elements of this circuit are connected in parallel. Replace the parallel current sources by a
single equivalent 2 – 0.5 + 1.5 = 3 A current source. Replace the parallel 12 Ω and 6 Ω resistors by a
12 × 6
single
= 4 Ω resistor.
12 + 6
By current division
12
⎛ 4 ⎞
i = −⎜
⎟ 3 = − = −1.714 A
7
⎝ 3+ 4 ⎠
(checked: LNAP 62/6/07)
P 3.4-8 Determine the value of the voltage v in
Figure P 3.4-8.
Figure P 3.4-8
Solution:
Each of the resistors is connected between
nodes a and b. The resistors are connected in
parallel and the circuit can be redrawn like this:
40 & 20 & 40 = 10 Ω
Then
v = 10 ( 0.003) = 0.03 = 30 mV
so
(checked: LNAP 6/21/04)
P 3.4-9 Determine the power supplied by the dependent
source in Figure P 3.4-9.
Figure P 3.4-9
Solution:
Use current division to get
ia = −
so
75
30 × 10−3 ) = −22.5 mA
(
25 + 75
v b = 50 ( −22.5 × 10−3 ) = −1.125 V
The power supplied by the dependent source is
given by
p = − ( 30 × 10−3 ) ( −1.125 ) = 33.75 mW
(checked: LNAP 6/12/04)
.
P 3.4-10 Determine the values of the resistances R1 and
R2 for the circuit shown in Figure P 3.4-10.
Figure P 3.4-10
Solution:
Using voltage division
R1
× 24
8=
40 R 2
R1 +
R 2 + 40
⇒
Using KVL
R1 ( R 2 + 40 )
1
=
3 R1 R 2 + 40 ( R1 + R 2 )
⇒
R1 R 2 + 40 ( R1 + R 2 ) = 3R1 R 2 + 120 R1
24 = 8 + R 2 (1.6 )
Then
R1 =
⇒
⇒
R1 =
40 R 2
2R 2 + 80
R 2 = 10 Ω
40 (10 )
=4Ω
2 (10 ) + 80
P 3.4-11 Determine the values of the resistances R1 and
R2 for the circuit shown in Figure P 3.4-11.
Figure P 3.4-11
Solution:
Using KCL
.024 = 0.0192 +
0.384
R2
⇒
R2 =
0.384
= 80 Ω
0.0048
Using current division
R1
0.384
=
× 0.024
R2
R1 + ( R 2 + 80 )
⇒
16 =
R1 R 2
R1 + R 2 + 80
=
80 R1
R1 + 160
⇒
R1 = 40 Ω
Determine the value of the current measured by the meter in Figure P 3.4-12.
P 3.4-12
Figure P 3.4-12
Solution:
Replace the (ideal) ammeter with the equivalent short
circuit. Label the current measured by the meter.
Apply KCL at the left node of the VCCS to get
1.2 =
va
10
+ 0.2 v a = 0.3 v a
⇒ va =
1.2
=4V
0.3
Use current division to get
im =
30
30
0.2 v a =
0.2 ( 4 ) = 0.6 A
30 + 10
30 + 10
(checked using LNAP 9/11/04)
P3.4-13 Consider the combination of resistors shown in Figure P3.4-13.
Let R p denote the equivalent resistance.
(a) Suppose 20 Ω ≤ R ≤ 320 Ω. Determine the corresponding range of
values of R p .
(b) Suppose instead R = 0 (a short circuit). Determine the value of R p .
(c) Suppose instead R = ∞ (an open circuit). Determine the value of R p .
(d) Suppose instead the equivalent resistance is R p = 40 Ω. Determine the
Figure P3.4-13
value of R.
Solution:
(a) First, when R = 20 Ω then R p = 80 || 20 =
R p = 80 || 320 =
80 ( 20 )
= 16 Ω . Next, when R = 320 Ω then
80 + 20
80 ( 320 )
= 64 Ω . Consequently
80 + 320
16 Ω ≤ R p ≤ 64 Ω.
(b) When R = 0 then R p = 80 || 0 =
(c) When R = ∞ then R p =
80 ( 0 )
= 0 Ω.
80 + 0
80 ( ∞ )
80
=
= 80 Ω .
80 + ∞ 80 + 1
∞
(d) When R p = 40 Ω then 40 = 80 || R =
80 R
80 + R
⇒ 80 + R = 2 R ⇒ R = 80 Ω
P3.4-14 Consider the combination of resistors shown in Figure P3.4-14.
Let R p denote the equivalent resistance.
(a) Suppose 40 Ω ≤ R ≤ 400 Ω. Determine the corresponding range of
values of R p .
(b) Suppose instead R = 0 (a short circuit). Determine the value of R p .
(c) Suppose instead R = ∞ (an open circuit). Determine the value of R p .
(d) Suppose instead the equivalent resistance is R p = 80 Ω. Determine the
value of R.
Figure P3.4-14
Solution:
160 ( 80 ) 160
=
= 53.33 Ω . Next, when R
160 + 80
3
160 ( 440 ) 16 ( 44 )
=
= 117.33 Ω .
= 400 Ω then R p = 160 || ( 40 + 400 ) = 160 || 440 =
160 + 440
6
(a) First, when R = 40 Ω then R p = 160 || ( 40 + 40 ) = 160 || 80 =
Consequently
53.33 Ω ≤ R p ≤ 117.33 Ω.
(b) When R = 0 then R p = 160 || ( 0 + 40 ) = 160 || 40 =
(c) When R = ∞ then R p = 160 || ( ∞ + 40 ) = 160 || ∞ =
(d) When R p = 80 Ω then
80 = 160 || ( R + 40 ) =
160 ( 40 ) 16 ( 4 )
=
= 32 Ω .
160 + 40
2
1
= 160 Ω .
1
1
+
160 ∞
160 ( R + 40 )
160
⇒ R + 200 =
( R + 40 ) = 2 R + 80 ⇒ R = 120 Ω .
160 + R + 40
80
P3.4-15 Consider the combination of resistors shown in Figure P3.4-15.
Let R p denote the equivalent resistance.
(a) Suppose 50 Ω ≤ R ≤ 800 Ω. Determine the corresponding range of
values of R p .
(b) Suppose instead R = 0 (a short circuit). Determine the value of R p .
(c) Suppose instead R = ∞ (an open circuit). Determine the value of R p .
(d) Suppose instead the equivalent resistance is R p = 150 Ω. Determine
the value of R.
Figure P3.4-15
Solution:
200 ( 50 )
200
= 50 +
= 90 Ω . Next, when R =
200 + 50
5
200 ( 800 )
800
800 Ω then R p = 50 + ( 200 || 800 ) = 50 +
= 50 +
= 210 Ω .
200 + 800
5
(a) First, when R = 50 Ω then R p = 50 + ( 200 || 50 ) = 50 +
Consequently
90 Ω ≤ R p ≤ 210 Ω.
(b) When R = 0 then R p = 50 + ( 200 || 0 ) = 50 +
200 ( 0 )
= 50 Ω .
200 + 0
(c) When R = ∞ then R p = 50 + ( 200 || ∞ ) = 50 +
200 ( ∞ )
200
= 50 +
= 250 Ω .
200
200 + ∞
+1
∞
(d) When R p = 150 Ω then
150 = 50 + ( 200 || R ) = 50 +
200 R
200 + R
⇒ 100 ( 200 + R ) = 200 R ⇒ R = 200 Ω .
P3.4-16 The input to the circuit shown in Figure P3.4-16 is the source current, i s . The output is the
current measured by the meter, i o . A current divider connects the source to the meter. Given the
following observations:
A. The input i s = 5 A causes the output to be i o = 2 A .
B. When i s = 2 A the source supplies 48 W.
Determine the values of the resistances R1 and R 2 .
Figure P3.4-16
Solution:
⎛ R1 ⎞
R1
2
From current division, i o = ⎜
so
i s . When i s = 5 A and i o = 2 A then =
⎟
⎜ R1 + R 2 ⎟
5 R1 + R 2
⎝
⎠
2 ( R1 + R 2 ) = 5 R1 or 2 R 2 = 3 R1 .
⎡⎛ R1 R 2 ⎞ ⎤
The power supplied by the source is given by i s ⎢⎜
⎟ i s ⎥ . When i s = 2 A the source supplies 48
⎢⎣⎜⎝ R1 + R 2 ⎟⎠ ⎥⎦
⎡⎛ R1 R 2 ⎞ ⎤
R1 R 2
.
W, so 48 = 2 ⎢⎜
2 ⎥ ⇒ 12 =
⎟
R1 + R 2
⎢⎣⎜⎝ R1 + R 2 ⎟⎠ ⎥⎦
⎛3 ⎞
3
R1 ⎜ R1 ⎟
R1
3
5
2 ⎠
⎝
2
Combining these results gives 12 =
=
= R1 ⇒ R1 = (12 ) = 20 Ω and
5
5
3
⎛3 ⎞
R1 + ⎜ R1 ⎟
2
⎝2 ⎠
3 R1
= 30 Ω .
2
Figure P3.4-17
P3.4-17. Figure P3.4-17 shows four similar but slightly different circuits. Determine the values of the
currents i1 , i 2 , i 3 and i 4 .
Solution:
Using current division:
⎛ 75 ⎞
⎛ 45 ⎞
i1 = − ⎜
⎟ 320 = −240 mA , i 2 = ⎜
⎟ 300 = 100 mA ,
⎝ 75 + 25 ⎠
⎝ 45 + 90 ⎠
⎛ 15 ⎞
⎛ 120 ⎞
i3 = − ⎜
⎟ 500 = −100 mA and i 4 = ⎜
⎟ 250 = 200 mA
⎝ 15 + 60 ⎠
⎝ 120 + 30 ⎠
Figure P3.4-18
P3.4-18. Figure P3.4-18 shows four similar but slightly different circuits. Determine the values of the
currents i1 , i 2 , i 3 and i 4 .
Solution:
Using current division:
⎛ 30 ⎞
⎛ 60 ⎞
i1 = ⎜
⎟ 240 = 80 mA , i 2 = − ⎜
⎟ 240 = −144 mA ,
⎝ 30 + 60 ⎠
⎝ 60 + 40 ⎠
⎛ 20 ⎞
⎛ 60 ⎞
i3 = ⎜
⎟ 240 = 60 mA and i 4 = − ⎜
⎟ 240 = −192 mA
⎝ 60 + 20 ⎠
⎝ 60 + 15 ⎠
Figure P3.4-19
P3.4-19. The input to the circuit shown in Figure P3.4-19 is the current source current Is. The output is
the current io. The output of this circuit is proportion to the input, that is
io = k I s
Determine the value of the constant of proportionality, k.
Solution:
Replace six resistors at the right of the circuit by an equivalent resistance to get
Using current division:
⎛ 5
⎞
⎜ 2R ⎟
2
i1 = ⎜
Is = Is
⎟
5
7
⎜ R+ R⎟
2 ⎠
⎝
Return to the original circuit
Using current division:
io =
Therefore
R
1
1⎛2 ⎞ 1
i1 = i 1 = ⎜ I s ⎟ = I s
R+R
2
2⎝7 ⎠ 7
1
k=
7
Figure P3.4-20
P3.4-20. The input to the circuit shown in Figure P3.4-20 is the voltage source voltage Vs. The output is
the voltage vo. The output of this circuit is proportion to the input, that is
v o = k Vs
Determine the value of the constant of proportionality, k.
Solution:
Replace six resistors at the right of the circuit by an equivalent resistance to get
Using voltage division:
5
⎞
⎛
⎞ ⎛ 5
⎜ R || 2 R ⎟ ⎜ 7 R ⎟
5
v1 = ⎜
Vs ⎜
Vs = Vs
⎟
⎟
5
5
19
⎜ 2 R + R || R ⎟ ⎜ 2 R + R ⎟
2 ⎠ ⎝
7 ⎠
⎝
Return to the original circuit
Using current division:
vo =
Therefore
R
2
1
1⎛ 5 ⎞ 1
v1 = v1 = ⎜ I s ⎟ = V s
R
5
5 ⎝ 19 ⎠ 19
2R+
2
1
k=
19
Section 3-5 Series Voltage Sources and Parallel Current Sources
P 3.5-1
Determine the power supplied by each source in the circuit shown in Figure P 3.5-1.
.
Figure P 3.5-1
Solution:
The voltage sources are connected in series and
can be replaced by a single equivalent voltage
source. Similarly, the parallel current sources can
be replaced by an equivalent current source.
After doing so, and labeling the resistor currents,
we have the circuit shown.
Apply KCL at the top node of the current source to get
i1 + 1.75 = i 2
Apply KVL to the outside loop to get
5 + 2 i 2 + 2 i1 = 0
so
and
5 + 2 ( i1 + 1.75 ) + 2 i1 = 0 ⇒ i1 = −
8.5
= −2.125 A
4
i 2 = −2.125 + 1.75 = −0.375 A
The power supplied by each sources is:
Source
8-V voltage source
Power delivered
−8 i1 = 17 W
3-V voltage source
3 i1 = −6.375 W
3-A current source
3 × 2 i 2 = −2.25 W
1.25-A current source
−1.25 × 2 i 2 = 0.9375 W
(Checked using LNAP, 9/14/04)
1
P 3.5-2
Determine the power supplied by each source in the circuit shown in Figure P 3.5-2.
Figure P 3.5-2
Solution:
20 × 5
= 4 Ω.
20 + 5
The 7-Ω resistor is connected in parallel with a short circuit, a 0-Ω resistor. The equivalent
0× 7
resistance is
= 0 Ω , a short circuit.
0+7
The 20-Ω and 5-Ω resistors are connected in parallel. The equivalent resistance is
The voltage sources are connected in series and can be
replaced by a single equivalent voltage source.
After doing so, and labeling the resistor currents, we have
the circuit shown.
The parallel current sources can be replaced by an
equivalent current source.
Apply KVL to get
−5 + v1 − 4 ( 3.5 ) = 0 ⇒ v1 = 19 V
The power supplied by each sources is:
Source
8-V voltage source
Power delivered
−2 ( 3.5 ) = −7 W
3-V voltage source
−3 ( 3.5 ) = −10.5 W
3-A current source
0.5-A current source
3 ×19 = 57 W
0.5 ×19 = 9.5 W
(Checked using LNAP, 9/15/04)
2
P 3.5-3
Determine the power received by each resistor in the circuit shown in Figure P 3.5-3.
Figure P 3.5-3
Solution:
The voltage sources are connected in series and
can be replaced by a single equivalent voltage
source. Similarly, the parallel current sources can
be replaced by an equivalent current source.
After doing so, and labeling the resistor currents,
we have the circuit shown.
Apply KCL at the top node of the current source to get
i1 + 1.75 = i 2
Apply KVL to the outside loop to get
5 + 2 i 2 + i1 = 0
so
5 + 2 ( i1 + 1.75 ) + 2 i1 = 0 ⇒ i1 = −
8.5
= −2.125 A
4
and
i 2 = −2.125 + 1.75 = −0.375 A
The power supplied by each sources is:
Source
8-V voltage source
Power delivered
−8 i1 = 17 W
3-V voltage source
3 i1 = −6.375 W
3-A current source
3 × 2 i 2 = −2.25 W
1.25-A current source
−1.25 × 2 i 2 = 0.9375 W
(Checked using LNAP, 9/14/04)
3
Section 3-6 Circuit Analysis
P 3.6-1 The circuit shown in Figure P 3.6-1a
has been divided into two parts. In Figure P 3.61b, the right-hand part has been replaced with an
equivalent circuit. The left-hand part of the circuit
has not been changed.
(a)
Determine the value of the resistance R in
Figure P 3.6-1b that makes the circuit in
Figure P 3.6-1b equivalent to the circuit in
Figure P 3.6-1a.
(b)
Find the current i and the voltage v shown
in Figure P 3.6-1b. Because of the
equivalence, the current i and the voltage
v shown in Figure P 3.6-1a are equal to
the current i and the voltage v shown in
Figure P 3.6-1b.
(c)
Find the current i2 shown in Figure P 3.61a using current division.
Figure P 3.6-1
Solution:
(a)
(b)
(c)
48 ⋅ 24
= 32 Ω
48 + 24
32 ⋅ 32
v = 32 + 32 24 = 16 V ;
32 ⋅ 32
8+
32 + 32
16 1
i=
= A
32 2
48
1
1
i2 =
⋅ = A
48 + 24 2
3
R = 16 +
P 3.6-2 The circuit shown in Figure P 3.6-2a has been divided into three parts. In Figure P 3.6-2b, the
rightmost part has been replaced with an equivalent circuit. The rest of the circuit has not been changed.
The circuit is simplified further in Figure 3.6-2c. Now the middle and rightmost parts have been
replaced by a single equivalent resistance. The leftmost part of the circuit is still unchanged.
Figure P 3.6-2
(a)
(b)
(c)
(d)
(e)
Determine the value of the resistance R1 in Figure P 3.6-2b that makes the circuit in Figure P 3.62b equivalent to the circuit in Figure P 3.6-2a.
Determine the value of the resistance R2 in Figure P 3.6-2c that makes the circuit in Figure P 3.62c equivalent to the circuit in Figure P 3.6-2b.
Find the current i1 and the voltage v1 shown in Figure P 3.6-2c. Because of the equivalence, the
current i1 and the voltage v1 shown in Figure P 3.6-2b are equal to the current i1 and the voltage
v1 shown in Figure P 3.6-2c.
Hint: 24 = 6(i1–2) + i1R2
Find the current i2 and the voltage v2 shown in Figure P 3.6-2b. Because of the equivalence, the
current i2 and the voltage v2 shown in Figure P 3.6-2a are equal to the current i2 and the voltage
v2 shown in Figure P 3.6-2b.
Hint: Use current division to calculate i2 from i1.
Determine the power absorbed by the 3-Ω resistance shown at the right of Figure P 3.6-2a.
Solution:
3⋅ 6
=6Ω
3+ 6
1 1 1
=
+ + ⇒ R p = 2.4 Ω then
12 6 6
(a ) R1 = 4 +
(b)
1
Rp
R 2 = 8 + R p = 10.4 Ω
(c) KCL: i2 + 2 = i1 and − 24 + 6 i2 + R 2i1 = 0
⇒ − 24 + 6 (i1 − 2) + 10.4i1 = 0
⇒
i1 =
36
=2.195 A ⇒ v1 =i1 R 2 =2.2 (10.4)=22.83 V
16.4
1
6
( d ) i2 =
( 2.195 ) = 0.878 A,
1 1 1
+ +
6 6 12
v2 = ( 0.878 ) (6) = 5.3 V
(e) i3 =
6
i2 = 0.585 A ⇒
3+ 6
P = 3 i32 = 1.03 W
P 3.6-3 Find i using appropriate circuit reductions and the current divider principle for the circuit of
Figure P 3.6-3.
Figure P 3.6-3
Solution:
Reduce the circuit from the right side by repeatedly replacing series 1 Ω resistors in parallel with a 2 Ω
resistor by the equivalent 1 Ω resistor
This circuit has become small enough to be easily analyzed. The vertical 1 Ω resistor is equivalent to a 2
Ω resistor connected in parallel with series 1 Ω resistors:
i1 =
1+1
(1.5 ) = 0.75 A
2 + (1 + 1)
P 3.6-4
(a)
Determine values of R1 and R2 in Figure P 3.6-4b that make the circuit in Figure P 3.6-4b
equivalent to the circuit in Figure P 3.6-4a.
(b)
Analyze the circuit in Figure P 3.6-4b to determine the values of the currents ia and ib
(c)
Because the circuits are equivalent, the currents ia and ib shown in Figure P 3.6-4b are equal to
the currents ia and ib shown in Figure P 3.6-4a. Use this fact to determine values of the voltage v1
and current i2 shown in Figure P 3.6-4a.
Figure P 3.6-4
Solution:
(a)
1
1
1 1
=
+ +
⇒ R2 = 4 Ω
R2 24 12 8
and
R1 =
(10 + 8) ⋅ 9
= 6Ω
10 + 8 + 9
b
g
(b)
First, apply KVL to the left mesh to get −27 + 6 ia + 3 ia = 0 ⇒ ia = 3 A . Next,
apply KVL to the left mesh to get 4 ib − 3 ia = 0 ⇒ ib = 2.25 A .
(c)
1
8
i2 =
2.25 = 1125
.
A
1 1 1
+ +
24 8 12
b gLM b10 +98g + 9 3OP = −10 V
Q
N
and v1 = − 10
P 3.6-5 The voltmeter in the circuit shown in Figure P 3.6-5 shows that the voltage across the 30-Ω
resistor is 6 volts. Determine the value of the resistance R1.
Hint: Use the voltage division twice.
Answer: R1 = 40 Ω
Figure P 3.6-5
P3.6-5
30
v1 = 6 ⇒ v1 = 8 V
10 + 30
R2
12 = 8 ⇒ R2 = 20 Ω
R2 + 10
20 =
b
g
R1 10 + 30
R1 + 10 + 30
b
g
⇒ R1 = 40 Ω
Alternate values that can be used to change the numbers in this problem:
meter reading, V
6
4
4
4.8
Right-most resistor, Ω
30
30
20
20
R1, Ω
40
10
15
30
P 3.6-6 Determine the voltages va and vc and the currents ib and id for the circuit shown in
Figure P 3.6-6.
Answer: va = –2 V, vc = 6 V, ib = –16 mA, and id = 2 mA
Figure P 3.6-6
Solution:
P 3.6-7 Determine the value of the resistance R in Figure P 3.6-7.
Answer: R = 28 kΩ
Figure P 3.6-7
Solution:
1× 10−3 =
24
12 ×103 + R p
⇒ R p = 12 × 103 = 12 kΩ
( 21×10 ) R
=
( 21×10 ) + R
3
12 × 10 = R p
3
3
⇒ R = 28 kΩ
P 3.6-8 Most of us are familiar with the
effects of a mild electric shock. The effects of a
severe shock can be devastating and often fatal.
Shock results when current is passed through
the body. A person can be modeled as a network
of resistances. Consider the model circuit shown
in Figure P 3.6-8. Determine the voltage
developed across the heart and the current
Figure P 3.6-8
flowing through the heart of the person when he
or she firmly grasps one end of a voltage source
whose other end is connected to the floor.
The heart is represented by Rh. The floor has resistance to current flow equal to Rf, and the person is
standing barefoot on the floor. This type of accident might occur at a swimming pool or boat dock. The
upper-body resistance Ru and lower-body resistance RL vary from person to person.
Solution:
⎛
⎞
130 500
Voltage division ⇒ v = 50 ⎜
15.963 V
⎜ 130 500 + 200 + 20 ⎟⎟ =
⎝
⎠
⎛ 100 ⎞
⎛ 10 ⎞
∴v = v ⎜
⎟ = (15.963) ⎜ ⎟ = 12.279 V
h
⎝ 100 + 30 ⎠
⎝ 13 ⎠
v
∴ i = h = .12279 A
h 100
P 3.6-9 Determine the value of the
current i in Figure 3.6-9.
Answer: i = 0.5 mA
Figure 3.6-9
Solution:
P 3.6-10
Determine the values of ia, ib, and vc in Figure P 3.6-10.
Figure P 3.6-10
Solution:
Req =
15 ( 20 + 10 )
= 10 Ω
15 + ( 20 + 10 )
ia = −
60
⎛ 30 ⎞ ⎛ 60 ⎞
⎛ 20 ⎞
= −6 A, ib = ⎜
⎟⎟ = 4 A, vc = ⎜
⎟ ⎜⎜
⎟ ( −60 ) = −40 V
Req
⎝ 30 + 15 ⎠ ⎝ Req ⎠
⎝ 20 + 10 ⎠
P 3.6-11 Find i and Req a–b if vab = 40 V in the circuit of Figure P 3.6-11.
Answer: Req a–b = 8 Ω, i = 5/6 A
Figure P 3.6-11
Solution:
a)
Req = 24 12 =
b)
(24)(12)
=8Ω
24 + 12
from voltage division:
100
5
⎛ 20 ⎞ 100
v = 40 ⎜
=
V∴ i = 3 = A
⎟
x
x 20 3
3
⎝ 20 + 4 ⎠
from current division: i = i
5
⎛ 8 ⎞
A
=
x ⎜⎝ 8 + 8 ⎟⎠
6
P 3.6-12 The ohmmeter in Figure P 3.6-12 measures the equivalent resistance, Req, of the resistor
circuit. The value of the equivalent resistance, Req, depends on the value of the resistance R.
Figure P 3.6-12
(a) Determine the value of the equivalent resistance, Req, when R = 18 Ω.
(b) Determine the value of the resistance R required to cause the equivalent resistance to be Req = 18 Ω.
Solution:
9 + 10 + 17 = 36 Ω
a.)
b.)
36 R
= 12 ⇒ 24 R = (12 )( 36 ) ⇒ R = 18 Ω
36+R
36 ( 9 )
= 7.2 Ω
36+9
P 3.6-13 Find the Req at terminals a–b in Figure P 3.6-13. Also determine i, i1, and i2.
Answer: Req = 8 Ω, i = 5 A, i1 = 5/3 A, i2 = 5/2 A
Figure P 3.6-13
Solution:
Req = 2 + 1 +
Using current division
⎛ 6 ⎞
i1 = i ⎜
⎟ =
⎝ 6 + 12 ⎠
( 6 12 ) + ( 2 2 ) = 3 + 4 + 1 = 8 Ω
( 5)
( 13 )
=5
3
A
so
i=
40 40
=
=5 A
8
Req
( )
⎛ 2 ⎞
5
1
and i2 = i ⎜
⎟ = ( 5) 2 = 2 A
⎝ 2+2⎠
P 3.6-14 All of the resistances in the circuit shown in Figure P 3.6-14 are multiples of R. Determine
the value of R.
Figure P 3.6-14
Solution:
4
6
R + R = 2R
5
5
= R + ( 2R & 2R ) = 2R
( R & 4 R ) + ( 2 R & 3R ) =
(
R + 2R & ( R + ( 2R & 2R ))
)
So the circuit is equivalent to
Then
12 = 0.1( R + ( 2 R & 2 R ) ) = 0.1( 2 R )
⇒
R = 60 Ω
(checked: ELAB 5/31/04)
P 3.6-15 The circuit shown in Figure P 3.6-15 contains seven resistors, each having resistance R. The
input to this circuit is the voltage source voltage, vs. The circuit has two outputs, va and vb. Express each
output as a function of the input.
Figure P 3.6-15
Solution:
The circuit can be redrawn as
va =
R & ( R + ( R & 2R ))
2R + R & ( R + ( R & 2R ))
vc =
vs =
5
vs
21
R & 2R
2
2
vs = va = vs
R + ( R & 2R )
5
21
vb =
R
1
1
vc = vc = vs
R+R
2
21
(Checked using LNAP 5/23/04)
P 3.6-16 The circuit shown in Figure P 3.6-16 contains
three 10-Ω, 1/4-W resistors. (Quarter-watt resistors can
dissipate 1/4 W safely.) Determine the range of voltage
source voltages, vs, such that none of the resistors absorbs
more than 1/4 W of power.
Figure P 3.6-16
Solution:
vo =
(10 & 10 ) v = 5 v = v s
s
s
10 + (10 & 10 )
15
3
vR + vo − vs = 0
iR =
vR
10
⇒
=
vR =
2
vs
3
2
vs
30
2
4
1
⎛ 2 ⎞
P = ⎜ v s ⎟ (10 ) = v s 2 ≤
90
4
⎝ 30 ⎠
⇒
vs ≤
90 3 10
=
= 2.37 V
16
4
(checked: LNAP 5/31/04)
P 3.6-17 The four resistors shown in Figure P 3.6-17
represent strain gauges. Strain gauges are transducers that
measure the strain that results when a resistor is stretched or
compressed. Strain gauges are used to measure force,
displacement, or pressure. The four strain gauges in Figure P
3.6-17 each have a nominal (unstrained) resistance of 120 Ω
and can each absorb 0.2 mW safely. Determine the range of
voltage source voltages, vs, such that no strain gauge absorbs
more than 0.2 mW of power.
Figure P 3.6-17
Solution:
vs
so the current in each strain gauge is
vs
. The power
2
400
v s2
v s ⎛ v s ⎞ v s2
−3
dissipated by each resistor is given by
so
we
require
0.5
×
10
≤
or
=
⎜
⎟
800
2 ⎝ 400 ⎠ 800
v s ≤ 0.4 = 0.6325 V .
The voltage across each strain gauge is
(checked: LNAP 6/9/04)
P 3.6-18 The circuit shown in Figure P 3.6-18b has been
obtained from the circuit shown in Figure P 3.6-18a by
replacing series and parallel combinations of resistances by
equivalent resistances.
(a)
(b)
(c)
Determine the values of the resistances R1, R2, and R3
in Figure P 3.6-18b so that the circuit shown in Figure
P 3.6-18b is equivalent to the circuit shown in Figure
P 3.6-18a.
Determine the values of v1, v2, and i in Figure P 3.618b.
Because the circuits are equivalent, the values of v1, v2,
and i in Figure P 3.6-18a are equal to the values of v1,
v2, and i in Figure P 3.6-18b. Determine the values of
v4, i5, i6, and v7 in Figure P 3.6-18a.
Solution:
(a)
R1 = 10 & ( 30 + 10 ) = 8 Ω ,
R 2 = 4 + (18 & 9 ) = 10 Ω and
R3 = 6 & ( 6 + 6 ) = 4 Ω
(b)
i = 1 A , v1 = 8 V and v 2 = 4 V
(c)
10
9
1
8 = −2 V , i 5 = −
1= − A,
10 + 30
9 + 18
3
4 1
⎛ 1⎞
v 7 = −18 ⎜ − ⎟ = +6 V and i 6 = = A
12 3
⎝ 3⎠
v4 = −
(checked: LNAP 6/6/04)
Figure P 3.6-18
P 3.6-19
Determine the values of v1, v2, i3, v4, v5, and i6 in Figure P 3.6-19.
Figure P 3.6-19
Solution:
Replace series and parallel combinations of resistances by
equivalent resistances. Then KVL gives
( 20 + 4 + 8 + 16 ) i = 48 ⇒ i = 0.5 A
v a = 20 i = 10 V , v b = 16 i = 8 V and v c = 8 i = 4 V
Compare the original circuit to the equivalent circuit to get
⎛ 10 || (10 + 30) ⎞
⎛ 8 ⎞
v1 = − ⎜
⎟va = −⎜
⎟10 = −4 V
⎝ 12 + 8 ⎠
⎝ 12 + 10 || (10 + 30) ⎠
v 2 = −v c = −4 V
⎛ 20 ⎞
⎛1⎞
i3 = − ⎜
⎟ i = − ⎜ ⎟ ( 0.5 ) = −0.1 A
⎝ 20 + 80 ⎠
⎝5⎠
⎛ 30 ⎞
⎛1⎞
v4 = −⎜
⎟ v1 = − ⎜ ⎟ ( −4 ) = 1 V
⎝ 10 + 30 ⎠
⎝ 4⎠
4
⎛
⎞
⎛1⎞
v5 = ⎜
⎟ v c = ⎜ ⎟ ( 4) = 1 V
⎝ 5+6+6⎠
⎝ 4⎠
⎛
⎞
16
⎛1⎞
i 6 = − ⎜⎜
⎟⎟ i = − ⎜ ⎟ ( 0.5 ) = −0.25 A
⎝ 2⎠
⎝ 16 + ( 4 + 6 + 6 ) ⎠
(checked: LNAP 6/10/04)
P 3.6-20 Determine the values of i, v, and Req by the circuit model shown in Figure P 3.6-20, given
that vab = 18 V.
Figure P 3.6-20
Solution:
Replace parallel resistors by equivalent
resistors:
6 || 30 = 5 Ω and 72 || 9 = 8 Ω
A short circuit in parallel with a resistor is
equivalent to a short circuit.
R eq = 36 || ( 8 + 10 ) = 12 Ω
Using voltage division when vab = 18 V:
v=
8
4
v ab = (18 ) = 8 V
8 + 10
9
i=
v
=1 A
8
(checked: LNAP 6/21/04)
P 3.6-22 Determine the value of the resistance R in the circuit shown in Figure P 3.6-22, given that
Req = 9 Ω.
Answer: R = 15 Ω
Figure P 3.6-22
Solution:
Replace parallel resistors by an equivalent
resistor:
8 || 24 = 6 Ω
A short circuit in parallel with a resistor is
equivalent to a short circuit.
Replace series resistors by an equivalent
resistor:
4+6 = 10 Ω
Now
9 = R eq = 5 + (12 || R ||10 )
so
60
R×
11 ⇒ R = 15 Ω
4=
60
R+
11
(checked: LNAP 6/21/04)
P 3.6-22 Determine the value of the resistance R in the circuit shown in Figure P 3.6-22, given that
Req = 50 Ω.
Figure P 3.6-22
Solution:
R eq = ( R || ( R + R ) || R ) || ( R || ( R + R) || R )
R || ( R + R) || R = 2 R ||
R eq =
R
2
2
R || R =
5
5
5
R 2
= R
2 5
⇒ R = 5 R eq = 200 Ω
(checked: LNAP 6/21/04)
P 3.6-23 Determine the values of r, the gain of the CCVS, and g, the gain of the VCCS, for the circuit
shown in Figure P 3.6-23.
Figure P 3.6-23
Solution:
ia =
9.74
= 1.2175 A
8
⎛ 9.74 ⎞
9.74 − 6.09 = r i a = r ⎜
⎟
⎝ 8 ⎠
⇒
V
⎛ 9.74-6.09 ⎞
r =⎜
⎟8 = 3
A
⎝ 9.74 ⎠
v b = 12 − 9.74 = 2.26 V
gv b +
6.09 9.74 2.26
+
−
=0
8
8
8
g=
gv b
vb
=
⇒
gv b = −1.696 A
−6.696
= −0.75
2.26
(checked: LNAP 6/21/04)
P 3.6-24 The input to the circuit in Figure P 3.6-24 is the voltage of the voltage source, vs. The output
is the voltage measured by the meter, vo. Show that the output of this circuit is proportional to the input.
Determine the value of the constant of proportionality.
Figure P 3.6-24
Solution:
va =
20 & 20
1
vs = vs
20 + ( 20 & 20 )
3
3
1
⎛ 12 ⎞
vo = ⎜
⎟ (10v a ) = × 10 × v s = 2v s
5
3
⎝ 12 + 8 ⎠
V
.
So vo is proportional to vs and the constant of proportionality is 2
V
P 3.6-25 The input to the circuit in Figure P 3.6-25 is the voltage of the voltage source, vs. The output
is the current measured by the meter, io. Show that the output of this circuit is proportional to the input.
Determine the value of the constant of proportionality.
Figure P 3.6-25
Solution:
vs
⎛ 40 ⎞
⎛ 4 ⎞ ⎛ vs ⎞ 4
ia = ⎜
=
⎟
⎜ ⎟ ⎜ ⎟ = vs
⎝ 40 + 10 ⎠ 2 + ( 40 & 10 ) ⎝ 5 ⎠ ⎝ 10 ⎠ 50
100 ⎛ 4 ⎞
8
⎛ 40 ⎞
io = − ⎜
⎟ ( 50i a ) = −
⎜ ⎟ vs = − vs
3 ⎝ 50 ⎠
3
⎝ 20 + 40 ⎠
The output is proportional to the input and the constant of proportionality is −
8 A
.
3 V
P 3.6-26
Determine the voltage measured by the voltmeter in the circuit shown in Figure P 3.6-26.
Figure P 3.6-26
Solution:
Replace the voltmeter by the equivalent open circuit and
label the voltage measured by the meter as vm.
The 10-Ω resistor at the right of the circuit is in series
with the open circuit that replaced the voltmeter so it’s
current is zero as shown. Ohm’s law indicates that the
voltage across that 10-Ω resistor is also zero. Applying
KVL to the mesh consisting of the dependent voltage
source, 10-Ω resistor and open circuit shows that
vm = 8 ia
The 10-Ω resistor and 40-Ω resistor are connected in
parallel. The parallel combination of these resistors is
equivalent to a single resistor with a resistance equal to
40 ×10
=8 Ω
40 + 10
Figure a shows part of the circuit. In Figure b, an equivalent resistor has replaced the parallel resistors.
Now the 4-Ω resistor and 8-Ω resistor are connected in series. The series combination of these resistors
is equivalent to a single resistor with a resistance equal to 4 + 8 = 12 Ω . In Figure c, an equivalent
resistor has replaced the series resistors.
Here the same three circuits with the order reversed. The earlier sequence of figures illustrates the
process of simplifying the circuit by repeatedly replacing series or parallel resistors by an equivalent
resistor. This sequence of figures illustrates an analysis that starts with the simplified circuit and works
toward the original circuit.
Consider Figure a. Using Ohm’s law, we see that the current in the 12-Ω resistor is 2 A. The current in
the voltage source is also 2 A. Replacing series resistors by an equivalent resistor does not change the
current or voltage of any other element of the circuit, so the current in the voltage source must also be 2
A in Figure b. The currents in resistors in Figure b are equal to the current in the voltage source. Next,
Ohm’s law is used to calculate the resistor voltages as shown in Figure b.
Replacing parallel resistors by an equivalent resistor does not change the current or voltage of any other
element of the circuit, so the current in the 4-Ω resistor in Figure c must be equal to the current in the 4Ω resistor in Figure b. Using current division in Figure c are yields
⎛ 40 ⎞
ia = ⎜
⎟ 2 = 1.6 A
⎝ 40 + 10 ⎠
Finally,
v m = 8 i a = 8 ×1.6 = 12.8 V
P 3.6-27
Determine the current measured by the ammeter in the circuit shown in Figure P 3.6-27.
Figure P 3.6-27
Solution:
Replace the ammeter by the equivalent short circuit and
label the current measured by the meter as im.
The 10-Ω resistor at the right of the circuit is in parallel
with the short circuit that replaced the ammeter so it’s
voltage is zero as shown. Ohm’s law indicates that the
current in that 10-Ω resistor is also zero. Applying KCL
at the top node of that 10-Ω resistor shows that
i m = 0.8 v a
Figure a shows part of the circuit. The 2-Ω resistor and 4-Ω resistor are connected in series. The series
combination of these resistors is equivalent to a single 6-Ω resistor.
In Figure b, an equivalent resistor has replaced the series resistors. Now the 3-Ω resistor and 6-Ω resistor
are connected in parallel. The parallel combination of these resistors is equivalent to a single resistor
with a resistance equal to
3× 6
=2Ω
3+ 6
In Figure c, an equivalent resistor has replaced the parallel resistors.
Here the same three circuits with the order reversed. The earlier sequence of figures illustrates the
process of simplifying the circuit by repeatedly replacing series or parallel resistors by an equivalent
resistor. This sequence of figures illustrates an analysis that starts with the simplified circuit and works
toward the original circuit.
Consider Figure a. Using Ohm’s law, we see that the voltage across the 2-Ω resistor is 6 V. The voltage
across the current source is also 6 V. Replacing parallel resistors by an equivalent resistor does not
change the current or voltage of any other element of the circuit, so the voltage across the current source
must also be 6 V in Figure b. The voltage across each resistor in Figure b is equal to the voltage across
the current source.
Replacing series resistors by an equivalent resistor does not change the current or voltage of any other
element of the circuit, so the voltage across the 3-Ω resistor in Figure c must be equal to the voltage
across the 3-Ω resistor in Figure b. Using voltage division in Figure c yields
⎛ 4 ⎞
va = ⎜
⎟6=4 V
⎝ 2+4⎠
Finally,
i m = 0.8 v a = 0.8 × 4 = 3.2 V
P 3.6-28 Determine the value of the resistance R
that causes the voltage measured by the voltmeter
in the circuit shown in Figure P 3.6-28 to be 6 V.
Figure P 3.6-28
Solution:
Use current division in the top part of the circuit to get
⎛ 40 ⎞
ia = ⎜
⎟ ( −3) = −2.4 A
⎝ 40 + 10 ⎠
Next, denote the voltage measured by the voltmeter as vm and use voltage division in the bottom part of
the circuit to get
⎛ R ⎞
⎛ −5 R ⎞
vm = ⎜
⎟ ( −5 i a ) = ⎜
⎟ ia
⎝ 18 + R ⎠
⎝ 18 + R ⎠
Combining these equations gives:
12 R
⎛ −5 R ⎞
vm = ⎜
⎟ ( −2.4 ) =
18 + R
⎝ 18 + R ⎠
When vm = 6 V,
6=
12 R
18 + R
⇒ R=
6 × 18
= 18 Ω
12 − 6
P 3.6-29 The input to the circuit shown in Figure P 3.6-29 is the voltage of the voltage source, vs. The
output is the current measured by the meter, im.
(a)
(b)
(c)
Figure P 3.6-29
Suppose vs = 15 V. Determine the value of the resistance R that causes the value of the current
measured by the meter to be im = 5 A.
Suppose vs = 15 V and R = 24 Ω. Determine the current measured by the ammeter.
Suppose R = 24 Ω. Determine the value of the input voltage, vs, that causes the value of the
current measured by the meter to be im = 3 A.
Soluton:
Use voltage division in the top part of the circuit to get
2
⎛ 12 ⎞
va = ⎜
⎟ ( −v s ) = − v s
5
⎝ 12 + 18 ⎠
Next, use current division in the bottom part of the circuit to get
80 ⎞
⎛ 16 ⎞
⎛
im = − ⎜
⎟ (5 v a ) = ⎜ −
⎟ va
⎝ 16 + R ⎠
⎝ 16 + R ⎠
Combining these equations gives:
80 ⎞ ⎛ 2 ⎞ ⎛ 32 ⎞
⎛
im = ⎜ −
⎟ ⎜ − vs ⎟ = ⎜
⎟ vs
⎝ 16 + R ⎠ ⎝ 5 ⎠ ⎝ 16 + R ⎠
a. When vs = 15 V and im = 12 A
288
⎛ 32 ⎞
12 = ⎜
= 24 Ω
⎟ 15 ⇒ 192 + 12 R = 480 ⇒ R =
12
⎝ 16 + R ⎠
b. When vs = 15 V and R = 80 Ω
⎛ 32 ⎞
im = ⎜
⎟ 15 = 5 A
⎝ 16 + 80 ⎠
c. When im = 3 A and R = 24 Ω
4
⎛ 32 ⎞
3=⎜
⎟ vs = vs
5
⎝ 16 + 24 ⎠
⇒ vs =
15
= 3.75 V
4
P 3.6-30 The ohmmeter in Figure P 3.6-30 measures the equivalent resistance of the resistor circuit
connected to the meter probes.
Figure P 3.6-31
(a) Determine the value of the resistance R required to cause the equivalent resistance to be Req = 12 Ω.
(b) Determine the value of the equivalent resistance when R = 14 Ω.
Solution:
R eq = ( ( R + 4 ) || 20 ) + 2 =
(a)
(b)
12 =
( R + 4 ) × 20 + 2 = 20 R + 80 + 2
R + 24
( R + 4 ) + 20
20 R + 80
20 R + 80
+ 2 ⇒ 10 =
⇒ R + 24 = 2 R + 8 ⇒ R = 16 Ω
R + 24
R + 24
R eq =
20 (14 ) + 80
+ 2 = 11.5 Ω
14 + 24
(Checked: LNAPDC 9/28/04)
P 3.6-31
The voltmeter in Figure P 3.6-31 measures the voltage across the current source.
Figure P 3.6-31
(a)
(b)
Determine the value of the voltage measured by the meter.
Determine the power supplied by each circuit element.
Solution:
Replace the ideal voltmeter with the equivalent open circuit and label the voltage measured by the
meter. Label the element voltages and currents as shown in (b).
Using units of V, A, Ω and W:
Using units of V, mA, kΩ and mW:
a.) Determine the value of the voltage measured by
the meter.
a.) Determine the value of the voltage measured
by the meter.
Kirchhoff’s laws give
Kirchhoff’s laws give
12 + v R = v m and −i R = −i s = 2 mA
12 + v R = v m and −i R = −i s = 2 ×10−3 A
Ohm’s law gives
Ohm’s law gives
(
)
v R = −25 i R
v R = − 25 ×103 i R
Then
Then
(
)
(
)(
)
v R = − 25 ×103 i R = − 25 ×103 −2 ×10−3 = 50 V
v m = 12 + v R = 12 + 50 = 62 V
v R = −25 i R = −25 ( −2 ) = 50 V
v m = 12 + v R = 12 + 50 = 62 V
b.) Determine the power supplied by each element.
voltage source
current source
resistor
( )
(
12 i s = −12 −2 × 10−3
(
)
= 50 ( −2 × 10 )
62 2 ×10−3 = 124 ×10−3 W
vR iR
−3
P 3.6-32
0
voltage source
( )
12 i s = −12 ( −2 )
= −24 mW
= −24 ×10−3 W
= −100 × 10 −3 W
total
)
b.) Determine the power supplied by each
element.
current source
62 ( 2 ) = 124 mW
resistor
v R i R = 50 ( − 2 )
= −100 mW
total
0
Determine the resistance measured by the ohmmeter in Figure P 3.6-32.
Figure P 3.6-32
Solution:
12 +
P 3.6-33
40 × 10
+ 4 = 12 Ω
40 + 10
Determine the resistance measured by the ohmmeter in Figure P 3.6-33.
Figure P 3.6-33
Solution:
( 60 + 60 + 60 ) × 60 = 45 Ω
( 60 + 60 + 60 ) + 60
P3.6-34
Consider the circuit shown in Figure P3.6-34. Given the values of the following currents and voltages:
i1 = 0.625 A , v 2 = −25 V , i 3 = −1.25 A and v 4 = −18.75 V
Determine the values of R1 , R 2 , R 3 and R 4 .
Figure P3.6-34
Solution:
50 + v 2 − v1 = 0 ⇒ v1 = 50 + ( −25 ) = 25 V
From KVL
R1 =
From Ohm’s law
v1
i1
=
25
= 40 Ω
0.625
From KCL
v ⎞
25 ⎞
⎛
⎛
i1 + i 5 + i 2 = 0 ⇒ i 2 = − i1 + i 5 = − ⎜ 0.625 + 1 ⎟ = − ⎜ 0.625 + ⎟ = −1.25 A
40 ⎠
40 ⎠
⎝
⎝
v2
−25
R2 =
=
= 20 Ω
From Ohm’s law
i 2 −1.25
(
)
From KCL
v2
v
−25
= i 6 + i 3 ⇒ i 6 = −i 3 + 2 = − ( −1.25 ) +
= −5 A
4
4
4
From KVL
v 3 + v 4 − 5 i 6 = 0 ⇒ v 3 = −v 4 + 5 i 6 = − ( −18.75 ) + 5 ( −5 ) = −6.25 V
From Ohm’s law
R3 =
v3
i3
=
v 4 −18.75
−6.25
=
= 15 Ω
= 5 Ω and R 4 =
−1.25
−1.25
i3
P3.6-35
Consider the circuits shown in Figure P3.6-35. The equivalent circuit on the right is obtained from the
original circuit on the left by replacing series and parallel combinations of resistors by equivalent
resistors. The value of the current in the equivalent circuit is is = 0.8 A. Determine the values of R1, R2,
R5, v2 and i3.
Figure P3.6-35
Solution:
R1 ||18 = 6 ⇒
18 R1
18 + R1
= 6 ⇒ 3 R1 = 18 + R1 ⇒ R1 = 9 Ω
R 2 + 10 = 28 ⇒ R 2 = 18 Ω
40 = ( 6 + 28 + R 5 ) i s
⇒
40
= 34 + R 5
⇒ R 5 = 50 − 34 = 16 Ω
v 2 = −10 i s = −10 ( 0.8 ) = −8 V and i 3 = −
32
0.8
is = −
= −0.4 A
32 + 32
2
0.8
P3.6-36 Consider the circuit shown in Figure P3.6-36.
Given
2
1
3
v 2 = v s , i 3 = i1 and v 4 = v 2 .
3
5
8
Determine the values of R1 , R 2 and R 4 .
Figure P3.6-36
2
1
3
Hint: Interpret v 2 = v s , i 3 = i1 and v 4 = v 2 as current and voltage division.
3
5
8
Solution:
From voltage division v 4 =
so
R4
50 + R 4
50 + R 4
=
From current division i 3 =
so
R4
v2
3
150
⇒ 8 R 4 = 3 ( 50 + R 4 ) ⇒ R 4 =
= 30 Ω .
8
8−3
R2
R 2 + ( 50 + R 4 )
R2
R 2 + 80
=
i1 =
R2
R 2 + 80
i1
1
⇒ 5 R 2 = R 2 + 80 ⇒ R 2 = 20 Ω .
5
Notice that R 2 || ( 50 + R 4 ) = 20 || ( 50 + 30 ) = 20 || 80 = 16 Ω . From voltage division
v1 =
so
R 2 || ( 50 + R 4 )
(
R1 + R 2 || ( 50 + R 4 )
)
vs =
16
vs
R1 + 16
16
2
48 − 32
=
⇒ 48 = 2 ( R1 + 16 ) ⇒ R1 =
=8 Ω.
R1 + 16 3
2
P3.6-37 Consider the circuit shown in Figure P3.6-37. Given
2
2
4
i 2 = i s , v 3 = v1 and i 4 = i 2 .
5
3
5
Determine the values of R1 , R 2 and R 4 .
2
2
4
Hint: Interpret i 2 = i s , v 3 = v1 and i 4 = i 2 as current and voltage division.
5
3
5
Figure P3.6-37
Solution:
From current division i 4 =
so
80
4
400 − 320
=
⇒ 400 = 4 ( 80 + R 4 ) ⇒ R 4 =
= 20 Ω .
80 + R 4 5
4
From voltage division v 3 =
so
80
i2
80 + R 4
80 || R 4
R 2 + ( 80 || R 4 )
v1 =
80 || 20
16
v1 =
v1
R 2 + ( 80 || 20 )
R 2 + 16
16
2
48 − 32
=
⇒ 48 = 2 ( R 2 + 16 ) ⇒ R 2 =
=8 Ω.
R 2 + 16 3
2
Notice that R 2 + ( 80 || R 4 ) = 8 + ( 80 || 20 ) = 8 + 16 = 24 Ω . From current division
i1 =
so
R1
R1 + 24
=
(
R1
R1 + R 2 + ( 80 || R 4 )
)
is =
R1
R1 + 24
is
2
48
⇒ 5 R1 = 2 ( R1 + 24 ) ⇒ R1 =
= 16 Ω
5
3
P3.6-38 Consider the circuit shown in Figure P3.6-38.
Figure P3.6-38
1
(a) Suppose i 3 = i1 . What is the value of the resistance R?
3
(b) Suppose instead v 2 = 4.8 V . What is the value of the equivalent resistance of the parallel
resistors?
(c) Suppose instead R = 20 Ω. What is the value of the current in the 40 Ω resistor?
1
Hint: Interpret i 3 = i1 as current division.
3
Solution:
(a) From current division i 3 =
20
20
1
i1 so
=
⇒ 60 = 20 + R ⇒ R = 40 Ω .
20 + R
20 + R 3
(b) From voltage division v 2 =
so
(c)
4.8 =
Rp
40 + R p
Rp
40 + R p
24 ⇒
i1 =
24
4.8
⎡ 40 + R p ⎤⎦ = R p
24 ⎣
⇒ Rp =
24
24
24
=
=
= 0.48 A
40 + ( 20 || 20 ) 40 + 10 50
( 0.2 ) 40 = 10 Ω .
1 − 0.2
P3.6-39 Consider the circuit shown in Figure P3.6-39.
Figure P3.6-39
1
(a) Suppose v 3 = v1 . What is the value of the resistance R?
4
(b) Suppose i 2 = 1.2 A . What is the value of the resistance R?
(c) Suppose R = 70 Ω . What is the voltage across the 20 Ω resistor?
(d) Suppose R = 30 Ω . What is the value of the current in this 30 Ω resistor?
1
Hint: Interpret v 3 = v1 as voltage division.
4
Solution:
(a) From voltage division v 3 =
(b)
1.2 =
10
10
1
v1 so
=
⇒ 40 = 10 + R ⇒ R = 30 Ω .
10 + R
10 + R 4
20 ( 2.4 )
20
20
2.4 =
2.4 ⇒ R + 30 =
= 40 ⇒ R = 10 Ω
20 + ( R + 10 )
R + 30
1.2
(c)
20 || ( 70 + 10 ) =
(d)
i2 =
20 ( 80 )
= 16 Ω so v1 = (16 ) 2.4 = 38.4 V
20 + 80
20
20
20
2.4 =
2.4 =
2.4 = 0.8 A
20 + ( R + 10 )
20 + ( 30 + 10 )
60
P3.6-40 Consider the circuit shown in Figure P3.6-40. Given that the voltage of the dependent voltage
source is v a = 8 V , determine the values of R1 and v o .
Figure P3.6-40
Solution:
First,
vo = −
20
8 = −3.2 V
20 + 30
Next,
⎛
⎞
⎜
⎟
⎞
8
40
40 ⎛
10
40 ⎜
10
400
400
⎟=
ic =
= ib =
=
⎜⎜
⎟⎟ =
40 R1 ⎟ 12 40 + R1 + 40 R1 480 + 52 R1
20
40 + R1
40 + R1 ⎝ 12 + 40 || R1 ⎠ 40 + R1 ⎜
⎜⎜ 12 +
⎟
40 + R1 ⎟⎠
⎝
then
400 ( 20 )
8
400
1000 − 480
=
⇒ 480 + 52 R1 =
= 1000 ⇒
= 10 Ω
20 480 + 52 R1
8
52
(
)
P3.6-41 Consider the circuit shown in Figure P3.6-41. Given that the current of the dependent current
source is i a = 2 A , determine the values of R1 and i o .
Figure P3.6-41
Solution:
First,
io = −
15
2 = −0.5 A
15 + 45
Next,
((
2
25
25
vb =
2 10 || 25 + R1
= vc =
0.2
25 + R1
25 + R1
(
(
then
2
500
=
0.2 35 + R1
)
⎛ 10 25 + R1
⎜
1⎜
⎝ 10 + 25 + R1
) ) = 2550+ R
⇒ 35 + R1 = 50 ⇒ R1 = 15 Ω
)
⎞
500
⎟=
⎟ 35 + R1
⎠
P3.6-42 Determine the values of i a , i b , i 2 , and v1 in the circuit shown in Figure P3.6-42.
Figure P3.6-42
Solution:
Use equivalent resistances to reduce the
circuit to
From KCL i b = 4 i a + i a
⇒ i b = −3 i a .
From KVL
12 i a + 10 i b − 6 = 0 ⇒ 12 i a + 10 ( −3 i a ) = 6
So i a = −
1
A , v a = −4 A and i b = 1 A .
3
Returning our attention to the original circuit, notice that i a and i b were not changed when the circuit
was reduced. Now v1 = ( 5 || 20 ) i a = ( 4 )( −0.333) = −1.333 V and i 2 =
12
i = 0.333 A .
12 + 24 b
P3.6-43 The Ohmmeter in Figure P3.6-43 measures R eq , the equivalent resistance of the part of the
circuit to the left of the terminals.
(a) Suppose R eq = 12 Ω. Determine the value of the resistance R.
(b) Suppose instead that R = 14 Ω. Determine the value of the equivalent resistance R eq .
Figure P3.6-43
Solution:
R eq = 2 + ( 20 || ( 4 + R ) ) .
(a) When R eq = 12 Ω then
12 = 2 + ( 20 || ( 4 + R ) ) ⇒ 10 =
20 ( 4 + R )
⇒ 24 + R = 2 ( 4 + R ) ⇒ R = 16 Ω
20 + ( 4 + R )
(b) When R = 14 Ω then
R eq = 2 + ( 20 || ( 4 + 14 ) ) = 2 + ( 20 ||18 ) = 2 +
20 (18 )
= 11.4736 Ω
20 + 18
P3.6-43. Determine the values of the resistance R
and current i a in the circuit shown in Figure
P3.6-43.
Figure P3.6-43
Solution. Replace the series resistors by an equivalent
resistor. Then use KVL to write
80 i a + 8 − 24 = 0 ⇒ i a =
24 − 8
= 0.2 A
80
Use KCL to write
24 − 8 8
8
= +
⇒
R 200
80
8 16
8
8
= −
= 0.16 ⇒ R =
= 50 Ω
R 80 200
0.16
Figure P3.6-44
P3.6-44 The input to the circuit shown in Figure P3.6-44 is the voltage of the voltage source,
32 V. The output is the current in the 10 Ω resistor, io. Determine the values of the resistance, R1, and of
the gain of the dependent source, G, that cause both the value of voltage across the
12 Ω to be v a = 10.38 V and the value of the output current to be i o = 0.4151 A.
Solution:
Using voltage division
10.38 = v a =
12 ( 32 )
12
= 36.9942 ≈ 37 Ω ⇒ R1 = 25 Ω
( 32 ) ⇒ R1 + 12 =
R1 + 12
10.38
Using current division
0.4151 = i o =
40
0.4151
A
G v a = ( 0.8 ) G (10.38 ) ⇒ G =
= 0.05
40 + 10
V
( 0.8)10.38
P3.6-45 The equivalent circuit in Figure 3.6-45 is obtained from the original circuit by replacing series
and parallel combinations of resistors by equivalent resistors. The values of the currents in the
equivalent circuit are ia = 3.5 A and ib = −1.5 A. Determine the values of the voltages v1 and v2 in the
original circuit.
Figure P3.6-45
Solution:
Label the currents in the equivalent circuit that correspond to the give currents in the equivalent circuit:
Use current division:
Using Ohm’s law:
v1 = −35 i a = −35 ( 3.5 ) = −122.5 V and v 2 = −50 ( −0.75 ) = 37.5 V
P3.6-46 Figure 3.6-46 shows three separate, similar
circuits. In each a 12 V source is connected to a subcircuit
consisting of three resistors. Determine the values of the
voltage source currents i1, i2 and i3. Conclude that while
the voltage source voltage is 12 V in each circuit, the
voltage source current depends on the subcircuit
connected to the voltage source.
Figure P3.6-46
Solution:
Replace the resistor subcircuit by an equivalent resistor in each circuit:
Using Ohm’s law:
i1 =
12
12
12
= 0.75 mA , i 2 =
= 3 mA and i 3 =
= 4 mA
16 kΩ
4 kΩ
3 kΩ
Figure P3.6-47
P3.6-47 Determine the values of the voltages, v1 and v2, and of the current, i3, in the circuit shown in
Figure P3.6-47.
Solution:
Replace series and parallel combinations of resistors by equivalent resistors to get
( 4 + ( 80 || 20 ) = 4 + 16 = 20 Ω and 40|| ( 80 + 80 ) = 40 ||160 = 32 Ω. ) Next, apply KVL to the left mesh to
get
50 + 30 i a − 20 i a = 0 ⇒ i a =
ib =
Ohm’s law gives
30 i a
32
=
50
= −5 A and 30 i a = −150 V
20 − 10
−150
= −4.6875 A
32
Label ib on the original circuit
Finally
and
v1 = ( 80 || 20 ) i a = 16 ( −5 ) = −80 V , v 2 =
i3 =
1
( 30 i a ) = −75 V
2
80 + 80
4
i b = ( −4.6875 ) = −3.75 A
40 + ( 80 + 80 )
5
Section 3-7 Analyzing Resistive Circuits using MATLAB
P3.7-1 Determine the power supplied by each of the sources, independent and dependent, in the circuit
shown in Figure P3.7-1.
Hint: Use the guidelines given in Section 3.7 to label the circuit diagram. Use MATLAB to solve the
equations representing the circuit.
Figure 3.7-1
Solution:
We’ll begin by choosing the bottom node to be the reference node. Next we’ll label the other nodes and
some element voltages:
Notice that the 8 Ω resistor, the 10 Ω resistor and the two independent current sources are all connected
in parallel. Consequently, the element voltages of theses elements can be labeled so that they are equal.
Similarly, the 4 Ω resistor and the dependent current source are connected in parallel so their voltages
can be labeled so as to be equal.
Using Ohm’s Law we see that the current directed downward in the 8 Ω resistor is
downward in the 10 Ω resistor is
v1
v1
8
, current directed
, and the current directed from left to right in the 2 Ω resistor is
10
Applying Kirchhoff’s Current Law (KCL) at node a gives
5=
v1
8
+ 2.5 +
v1
10
+
v2
2
⇒ 0.225 v1 + 0.5 v 2 = 2.5
Using Ohm’s Law we see that the current directed downward in the 4 Ω resistor is
v2
2
.
(1)
v3
4
. Applying
Kirchhoff’s Current Law (KCL) at node a gives
1
v2
+ 1.5 v1 =
v3
⇒ 1.5 v1 + 0.5 v 2 − 0.25 v 3 = 0
(2)
2
4
Applying Kirchhoff’s Voltage Law (KVL) to the mesh consisting of the 10 Ω resistor, the 2 Ω resistor
and the dependent source to get
v 2 + v 3 − v1 = 0
(3)
Equations 1, 2 and 3 comprise a set of three simultaneous equations in the three unknown voltages v1 ,
v 2 and v 3 . We can write these equations in matrix form as
0 ⎤ ⎡ v1 ⎤ ⎡ 2.5⎤
⎡ 0.225 0.5
⎢ 1.5 0.5 −0.25⎥ ⎢ v ⎥ = ⎢ 0 ⎥
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎢⎣ −1
1
1 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ 0 ⎥⎦
We can solve this matrix equation using MATLAB:
Hence
v1 = −4.1096 V, v 2 = 6.8493 V and v 3 = −10.9589 V
2
The power supplied by the 5 A current source is 5 v1 = 5 ( −4.1096 ) = −20.548 W . The power supplied
by the 2.5 A current source is −2.5 v1 = −2.5 ( −4.1096 ) = 10.247 W . The power supplied by the
dependent current source is (1.5 v1 ) v 3 = 1.5 ( −4.1096 )( −10.9589 ) = 67.555 W .
Observation: Changing the order of the 8 Ω resistor, the 10 Ω resistor and the two independent current
sources only changes the order of the terms in the KCL equation at node a. We know that addition is
commutative, so change the order of the terms will not affect the values of the voltages v1 , v 2 and v 3 .
For example, if the positions of the 2.5 A current source and 8 Ω resistor are swithched:
The KCL equation at node a is
5 = 2.5 +
v1
8
+
v2
+ 2.5 +
v2
+
v1
10
2
⇒ 0.225 v1 + 0.5 v 2 = 2.5
Similarly, when the circuit is drawn as
The KCL equation at node a is
5=
v1
10
+
v1
8
2
⇒ 0.225 v1 + 0.5 v 2 = 2.5
The changes do not affect the values of the voltages v1 , v 2 and v 3 .
3
P3.7-2 Determine the power supplied by each of the sources, independent and dependent, in the circuit
shown in Figure P3.7-2.
Hint: Use the guidelines given in Section 3.7 to label the circuit diagram. Use MATLAB to solve the
equations representing the circuit.
Figure 3.7-2
Solution:
We’ll begin by choosing the bottom node to be the reference node. Next we’ll label the other nodes and
some element currents:
Notice that two 4 Ω resistors, an 8 Ω resistor and the two independent voltage sources are all connected
in series. Consequently, the element currents of theses elements can be labeled so that they are equal.
Similarly, a 4 Ω resistor and the dependent voltage source are connected in series so their currents can
be labeled so as to be equal.
The current in each resistor has been labeled so we can use Ohm’s Law to calculate resistor voltages
from the resistor currents and the resistances. Apply Kirchhoff’s Voltage Law (KVL) to the left mesh to
get
6 + 8 i1 + 8 i 2 + 4 i1 − 15 + 4 i1 = 0 ⇒ 16 i1 + 8 i 2 = 9
(1)
Apply Kirchhoff’s Voltage Law (KVL) to the right mesh to get
4 i 3 + 5 i1 − 8 i 2 = 0
(2)
Applying Kirchhoff’s Current Law (KCL) at node a to get
i 1 = i 2 + i 3 ⇒ − i1 + i 2 + i 3 = 0
(3)
4
Equations 1, 2 and 3 comprise a set of three simultaneous equations in the three unknown voltages v1 ,
v 2 and v 3 . We can write these equations in matrix form as
⎡16 8 0 ⎤ ⎡ i1 ⎤ ⎡9 ⎤
⎢ 5 −8 4 ⎥ ⎢i ⎥ = ⎢0 ⎥
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎢⎣ −1 1 1 ⎥⎦ ⎢⎣i 3 ⎥⎦ ⎢⎣0 ⎥⎦
We can solve this matrix equation using MATLAB:
Hence
i1 = 0.4091 A, i 2 = 0.3068 A and i 3 = 0.1023 A
The power supplied by the 15 V voltage source is 15 i1 = 15 ( 0.4091) = 6.1365 W . The power supplied
by the 6 V voltage source is −6 i1 = −6 ( 0.4091) = −2.4546 W . The power supplied by the dependent
voltage source is − ( 5 i 2 ) i 3 = −5 ( 0.3068 )( 0.1023) = 0.1569 W .
Observation: Changing the order of the two 4 Ω resistors, an 8 Ω resistor and the two independent
voltage sources in the left mesh changes the order of the terms in the KVL equation for that mesh. We
5
know that addition is commutative, so change the order of the terms will not affect the values of the
currents i1 , i 2 and i 3 . For example, when the circuit is drawn as
The KVL equation for the left mesh is
6 − 15 + 8 i 2 + 4 i1 + 8 i1 + 4 i1 = 0 ⇒ 16 i1 + 8 i 2 = 9
When the circuit is drawn as
The KVL equation for the left mesh is
4 i1 + 4 i1 + 8 i 2 + 6 − 15 + 8 i1 = 0 ⇒ 16 i1 + 8 i 2 = 9
These changes do not affect the values of the currents i1 , i 2 and i 3 .
6
Figure P3.7-3
P3.7-3. Determine the power supplied by each of the independent sources in the circuit shown in Figure
P3.7-3.
P3.7-3
Label the element currents and voltages as suggested in Table 3.7-1 Guidelines for Labeling Circuit
Variables:
Apply KCL at the top left node:
Apply KVL to the left mesh:
In matrix form:
Solving using MATLAB:
v1
4
+
v1
4
+ 2 + i2 = 0
8 i 2 − 12 + 12 i 2 − v1 = 0
⎡1 1
⎤
1 ⎥ ⎡v1 ⎤ ⎡ −2 ⎤
⎢4 + 6
=
⎢
⎥ ⎢⎣ i 2 ⎥⎦ ⎢⎣12 ⎥⎦
⎣ −1 8 + 12 ⎦
⎡ v1 ⎤ ⎡ −5.5714 ⎤
⎢i ⎥ = ⎢
⎥
⎣ 2 ⎦ ⎣ 0.3214 ⎦
The current source supplies
−2 v1 = −2 ( −5.5714 ) = 11.1428 W
The voltage source supplies
12 i 2 = 12 ( 0.3214 ) = 3.8568 W
7
Figure P3.7-4
P3.7-4. Determine the power supplied by each of the sources in the circuit shown in Figure P3.7-4.
P3.7-4
Label the element currents and voltages as suggested in Table 3.7-1 Guidelines for Labeling Circuit
Variables:
Apply KVL to the loop consisting of 30 Ω, 40 Ω and 50 Ω resistors:
Apply KVL to the right mesh:
v c = 40 i d + 12 v c
Apply KCL at the top node of the current source:
va
50
Apply KCL at the top node of a 40 Ω resistor: i b =
In matrix form:
Solving using MATLAB:
The current source supplies
The VCVS supplies
30 i b + v c − v a = 0
+ i b = 2.4
vc
40
+ id
1
0 ⎤ ⎡v a ⎤ ⎡ 0 ⎤
⎡ −1 30
⎢ 0
⎥ ⎢i ⎥ ⎢ 0 ⎥
0
11
40
⎢
⎥⎢ b ⎥ = ⎢ ⎥
⎢ 0.02 1
0
0 ⎥ ⎢ v c ⎥ ⎢ 2.4 ⎥
⎢
⎥⎢ ⎥ ⎢ ⎥
−1 0.025 1 ⎦ ⎢⎣ i d ⎥⎦ ⎣ 0 ⎦
⎣ 0
⎡ v a ⎤ ⎡ 41.0526 ⎤
⎢i ⎥ ⎢
⎥
⎢ b ⎥ = ⎢ 1.5789 ⎥
⎢ v ⎥ ⎢ −6.3158⎥
⎢ c⎥ ⎢
⎥
⎢⎣ i d ⎥⎦ ⎣ 1.7368 ⎦
2.4 v a = 2.4 ( 41.0526 ) = 98.5262 W
− (12 v c ) i d = −12 ( −6.3158 )(1.7368 ) = 131.6314 W
8
Section 3-8 How Can We Check …
P 3.8-1 A computer analysis program, used for the circuit
of Figure P 3.8-1, provides the following branch currents
and voltages:
i1 = –0.833 A, i2 = –0.333 A, i3 = –1.167 A,
and
v = –2.0 V.
Are these answers correct?
Hint: Verify that KCL is satisfied at the center node and
that KVL is satisfied around the outside loop consisting of
the two 6-Ω resistors and the voltage source.
Solution:
Figure P 3.8-1
KCL at node a: i = i + i
3 1 2
− 1.167 = −0.833 + ( −0.333)
− 1.167 = − 1.166 OK
KVL loop consisting of the vertical 6 Ω resistor, the 3 Ω and4Ω resistors,
and the voltage source:
6i + 3i + v + 12 = 0
3
2
yields v = −4.0 V not v = −2.0 V
The answers are not correct.
The circuit of Figure P 3.8-2 was assigned as
P 3.8-2
a homework problem. The answer in the back of the
textbook says the current, i, is 1.25 A. Verify this
answer using current division.
Figure P 3.8-2
Solution:
Apply current division to get:
1
⎛
⎞
⎜
⎟
⎛1⎞
20
i=⎜
5 = ⎜ ⎟ 5 = 1.25 A √
1
1
1 ⎟
⎝4⎠
⎜
⎟
+ +
⎝ 20 20 5 + 5 ⎠
The answer is correct.
P 3.8-3
The circuit of Figure P 3.8-3 was built in the lab and vo was
measured to be 6.25 V. Verify this measurement using the voltage
divider principle.
Figure P 3.8-3
Solution:
Apply voltage division to get:
320
⎛
⎞
⎛ 320 ⎞
v=⎜
⎟ 24 = ⎜
⎟ 24 = 6.4 V ≠ 6.25 V
⎝ 650 + 320 + 230 ⎠
⎝ 1200 ⎠
The measurement is not correct.
P 3.8-4 The circuit of Figure P 3.8-4 represents an auto’s
electrical system. A report states that
iH = 9 A, iB = –9 A, and iA = 19.1 A.
Verify that this result is correct.
Hint: Verify that KCL is satisfied at each node and that KVL is satisfied
around each loop.
Figure P 3.8-4
Solution:
KVL bottom loop:
KVL right loop:
KCL at left node:
− 14 + 0.1i + 1.2i = 0
A
H
− 12 + 0.05i + 1.2i
= 0
B
H
i +i =i
A B H
Solving the three above equations yields:
i = 16.8 A
i = 10.3 A
i = −6.49 A
and
A
H
B
These are not the given values. Consequently, the report is incorrect.
P 3.8-5 Computer analysis of the circuit in Figure P 3.8-5 shows
that
ia = –0.5 mA and ib = –2 mA.
Was the computer analysis done correctly?
Hint: Verify that the KVL equations for all three meshes are
satisfied when ia = –0.5 mA and ib = –2 mA.
Figure P 3.8-5
Solution:
1
⎛
⎞
Top mesh: 0 = 4 i a + 4 i a + 2 ⎜ i a + − i b ⎟ = 10 ( −0.5 ) + 1 − 2 ( −2 )
2
⎝
⎠
Lower left mesh: vs = 10 + 2 ( i a + 0.5 − i b ) = 10 + 2 ( 2 ) = 14 V
Lower right mesh: vs + 4 i a = 12 ⇒ vs = 12 − 4 (−0.5) = 14 V
The KVL equations are satisfied so the analysis is correct.
P 3.8-6 Computer analysis of the circuit in Figure P 3.8-6.
shows that
ia = 0.5 mA and ib = 4.5 mA.
Was the computer analysis done correctly?
Hint: First, verify that the KCL equations for all five nodes are
satisfied when ia = 0.5 mA and ib = 4.5 mA. Next, verify that the
KVL equation for the lower left mesh (a-e-d-a) is satisfied. (The
KVL equations for the other meshes aren’t useful because each
involves an unknown voltage.)
Solution: Apply KCL at nodes b and c to label the circuit as
Apply KCL to get
i c + 4 = 4.5 ⇒ i c = 0.5 A
i a + i c = 1 ⇒ 0.5 + 0.5 = 1 √
6 = 1 + i a + i b = 1 + 0.5 + 4.5 √
(node d)
(node a)
(node e)
Apply KVL to get to mesh (a-e-d-a)
−3 ( 0.5 ) + 2 ( 4.5 ) − 5 ( 0.5 ) = −1.5 + 9 − 2.5 ≠ 0
The given currents do not satisfy these five Kirchhoff’s laws
equations and therefore are not correct.
P 3.8-7 Verify that the element currents
and voltages shown in Figure P 3.8-7
satisfy Kirchhoff’s laws:
(a)
Verify that the given currents
satisfy the KCL equations
corresponding to nodes a, b, and c.
(b)
Verify that the given voltages
satisfy the KVL equations
corresponding to loops a-b-d-c-a
and a-b-c-d-a.
Figure P 3.8-7
Solution:
(a)
7 + ( −3) = 4
4 + ( −2 ) = 2
− 5 = − 2 + ( −3 )
(b)
−1 − ( −6 ) + ( −8 ) + 3 = 0
−1 − 2 − ( −8 ) − 5 = 0
(node a )
(node b)
(node c)
(loop a - b - d - c - a )
(loop a - b - c - d - a )
The given currents and voltages satisfy these five Kirchhoff’s laws equations
P 3.8-8 Figure P 3.8-8 shows a circuit and some corresponding data.
The tabulated data provides values of the current, i, and voltage, v,
corresponding to several values of the resistance R2.
(a)
Use the data in rows 1 and 2 of the table to find the values of
vs and R1.
(b)
Use the results of part (a) to verify that the tabulated data are
consistent.
(c)
Fill in the missing entries in the table.
Figure P 3.8-8
Solution:
i=
(a)
R1 + R 2
2.4 =
from row 1
1.2 =
from row 2
so
vs
vs
R1
vs
R1 + 10
2.4 R1 = v s = 1.2 ( R1 + 10 )
then
(b)
⇒
R1 = 10 Ω
v s = 2.4 (10 ) = 24 V
i=
24 R 2
24
and v =
10 + R 2
10 + R 2
24
480
= 0.8 A and v =
= 16 V .
30
30
720
When R2 = 30 Ω then v =
= 18 V .
40
24
When R2 = 40 Ω the i =
= 0.48 A .
50
When R2 = 20 Ω then i =
24
= 0.6 A .
40
960
When R2 = 40 W then v =
= 19.2 V .
50
(c) When R2 = 30 Ω then i =
(checked: LNAP 6/21/04)
P 3.8-9 Figure P 3.8-9 shows a circuit and some corresponding data. The tabulated data provide values
of the current, i, and voltage, v, corresponding to several values of the resistance R2.
(a)
Use the data in rows 1 and 2 of the table to find the values of is and R1.
(b)
Use the results of part (a) to verify that the tabulated data are consistent.
(c)
Fill in the missing entries in the table.
Figure P 3.8-9
Solution:
i=
(a)
R1
R1 + R 2
is
From row 1
R1
4
=
is
3 R1 + 10
⇒
4R1 + 40 = 3R1i s
From row 2
R1
6
=
is
7 R1 + 20
⇒
6R1 + 120 = 7 R1i s
⇒
28R1 + 280 = 18 R1 + 360
⇒
is = 3 A
So
4 R1 + 40
3R1
= is =
When R2 = 40 Ω then i =
7 R1
4
8
=
is
3 8 + 10
Then
(b)
6 R1 + 120
i=
⇒
R1 = 8 Ω
24 R 2
8
24
and v = R 2 i =
( 3) =
8 + R2
8 + R2
8 + R2
24
960
= 0.5 A and v =
= 20 V . These are the values in the table so tabulated
48
48
data is consistent.
(c) When R2 = 80 Ω then i =
24 ( 80 ) 240
24 3
=
A and v =
=
V.
88 11
88
11
(checked: LNAP 6/21/04)
Design Problems
DP 3-1 The circuit shown in Figure DP 3.1 uses a potentiometer to produce a variable voltage. The
voltage vm varies as a knob connected to the wiper of the potentiometer is turned. Specify the resistances
R1 and R2 so that the following three requirements are satisfied:
1.
The voltage vm varies from 8 V to 12 V as the wiper moves from one end of the potentiometer to
the other end of the potentiometer.
2.
The voltage source supplies less than 0.5 W of power.
3.
Each of R1, R2, and RP dissipates less than 0.25 W.
Figure DP 3.1
Solution:
Using voltage division:
vm =
R 2 + aR p
R1 + (1 − a ) R p + R 2 + aR p
vm = 8 V when a = 0 ⇒
vm = 12 V when a = 1 ⇒
24 =
R 2 + aR p
R1 + R 2 + R p
R2
R1 + R 2 + R p
R2 + R p
R1 + R 2 + R p
=
1
3
=
1
2
24
The specification on the power of the voltage source indicates
242
1
≤
⇒ R1 + R 2 + R p ≥ 1152 Ω
R1 + R 2 + R p 2
Try Rp = 2000 Ω. Substituting into the equations obtained above using voltage division gives
3R 2 = R1 + R 2 + 2000 and 2 ( R 2 + 2000 ) = R1 + R 2 + 2000 . Solving these equations gives R1 = 6000 Ω
and R 2 = 4000 Ω .
With these resistance values, the voltage source supplies 48 mW while R1, R2 and Rp dissipate
24 mW, 16 mW and 8 mW respectively. Therefore the design is complete.
DP 3-2 The resistance RL in Figure DP 3.2 is the equivalent resistance of a pressure transducer. This
resistance is specified to be 200 Ω ± 5 percent. That is, 190 Ω ≤ RL ≤ 210 Ω. The voltage source is a 12
V ± 1 percent source capable of supplying 5 W. Design this circuit, using 5 percent, 1/8-watt resistors
for R1 and R2, so that the voltage across RL is
vo = 4V ± 10%
(A 5 percent, 1/8-watt 100-Ω resistor has a resistance between 95 and 105 Ω and can safely dissipate
1/8-W continuously.)
Figure DP 3.2
Solution:
Try R1 = ∞. That is, R1 is an open circuit. From KVL, 8 V will appear across R2. Using voltage
200
division,
12 = 4 ⇒ R 2 = 400 Ω . The power required to be dissipated by R2
R 2 + 200
82
1
= 0.16 W < W . To reduce the voltage across any one resistor, let’s implement R2 as the series
is
400
8
combination of two 200 Ω resistors. The power required to be dissipated by each of these resistors is
42
1
= 0.08 W < W .
200
8
Now let’s check the voltage:
190
210
11.88
< v < 12.12
0
190 + 420
210 + 380
3.700 < v0 < 4.314
4 − 7.5% < v0 < 4 + 7.85%
Hence, vo = 4 V ± 8% and the design is complete.
DP 3-3 A phonograph pickup, stereo amplifier, and speaker are shown in Figure DP 3.3a and redrawn
as a circuit model as shown in Figure DP 3.3b. Determine the resistance R so that the voltage v across
the speaker is 16 V. Determine the power delivered to the speaker.
Figure DP 3.3
Solution:
Vab ≅ 200 mV
10
10
120 Vab =
(120) (0.2)
10 + R
10 + R
240
⇒ R=5Ω
let v = 16 =
10 + R
162
= 25.6W
∴P=
10
v=
DP 3-4 A Christmas tree light set is required that will operate from a 6-V battery on a tree in a city
park. The heavy-duty battery can provide 9A for the four-hour period of operation each night. Design a
parallel set of lights (select the maximum number of lights) when the resistance of each bulb is 12 Ω.
Solution:
N 1
N
⎛1⎞
i = G v = v where G = ∑
= N⎜ ⎟
T
T
R
⎝ R⎠
n = 1 Rn
∴N=
iR ( 9 )(12 )
=
= 18 bulbs
6
v
DP 3-5 The input to the circuit shown in Figure DP 3.5 is the voltage source
voltage, vs. The output is the voltage vo. The output is related to the input by
vo =
R2
vs = gvs
R1 + R2
The output of the voltage divider is proportional to the input. The constant of
proportionality, g, is called the gain of the voltage divider and is given by
R2
R1 + R2
The power supplied by the voltage source is
⎛ vs ⎞
vs2
vs2
p = vsis = vs ⎜
=
⎟=
⎝ R1 + R2 ⎠ R1 + R2 Rin
where
Rin = R1 + R2
is called the input resistance of the voltage divider.
g=
(a)
(b)
Figure DP 3.5
Design a voltage divider to have a gain, g = 0.65.
Design a voltage divider to have a gain, g = 0.65, and an input resistance, Rin = 2500 Ω.
Solution:
R2
⇒ g R1 = (1 − g ) R 2
R1 + R 2
Thus either resistance can be determined from the other resistance and the gain of the voltage divider.
Also
R2
R2
g=
=
⇒ R 2 = g R in
R1 + R 2 R in
Notice that
Consequently
g=
g R1 = (1 − g ) R 2 = (1 − g ) g R in
⇒ R1 = (1 − g ) R in
(a) The solution of this problem is not unique. Given any value of R1, we can determine a value of R2
that will cause g = 0.65. Let’s pick a convenient value for R1, say
R1 = 100 Ω
Then
(b)
and
g R1 = (1 − g ) R 2
⇒ R2 =
g R1
1− g
=
0.65 ×100
= 186 Ω
1 − 0.65
R 2 = g R in = 0.65 × 2500 = 1625 Ω
R1 = (1 − g ) R in = (1 − 0.65 ) 2500 = 875 Ω
DP 3-6 The input to the circuit shown in Figure DP 3.6 is
the current source current, is. The output is the current io. The
output is related to the input by
R1
io =
is = gis
R1 + R2
The output of the current divider is proportional to the input.
The constant of proportionality, g, is called the gain of the
current divider and is given by
R1
g=
R1 + R2
The power supplied by the current source is
⎡ ⎛ R R ⎞⎤
RR
p = vsis = ⎢is ⎜ 1 2 ⎟ ⎥ is = 1 2 is2 = Rin is2
R1 + R2
⎣ ⎝ R1 + R2 ⎠ ⎦
RR
where
Rin = 1 2
R1 + R2
is called the input resistance of the current divider.
(a)
(b)
Figure DP 3.6
Design a current divider to have a gain, g = 0.65.
Design a current divider to have a gain, g = 0.65, and an input resistance, Rin = 10000 Ω.
Solution:
Notice that
g=
R1
⇒ g R 2 = (1 − g ) R1
R1 + R 2
Thus either resistance can be determined from the other resistance and the gain of the current divider.
Also
R1
R in
R in
=
⇒ R2 =
g=
R1 + R 2 R 2
g
Consequently
(1 − g ) R1 = g R 2 = R in
⇒ R1 =
R in
(1 − g )
Thus specified values of g and Rin uniquely determine the required values of R1 and R2.
(a) The solution of this problem is not unique. Given any value of R1, we can determine a value of R2
that will cause g = 0.65. Let’s pick a convenient value for R1, say
R1 = 100 Ω
Then
g R 2 = (1 − g ) R1 ⇒ R 2 =
R2 =
(b)
and
R1 =
R in
R in
g
(1 − g )
=
=
(1 − g ) R1 (1 − 0.65 ) ×100
=
= 54 Ω
g
10000
= 15385 Ω
0.65
10000
= 28571 Ω
(1 − 0.65 )
0.65
DP 3-7 Design the circuit shown in Figure DP 3-7 to have an output vo = 8.5 V when the input is vs =
12 V. The circuit should require no more than 1 mW from the voltage source.
Figure DP 3.7
Solution:
vo
g=
The required gain is
8.5
= 0.7083
12
=
vs
0.7083 R1 = (1 − 0.7083) R 2
Consequently,
⇒ R 2 = 2.428 R1
It is also required that
122
0.001 ≥
R1 + R 2
⇒ R1 + R 2 ≥ 144000 ⇒ 3.428 R1 ≥ 144000 ⇒ R1 ≥ 42007 Ω
R1 = 45 kΩ and R 2 = 109.26 kΩ
For example,
DP 3-8 Design the circuit shown in Figure DP 3.8 to have an output io = 1.8 mA when the input is is =
5 mA. The circuit should require no more than 1 mW from the current source.
Figure DP 3.8
Solution:
g=
The required gain is
io
is
=
1.8
= 0.36
5
0.36 R 2 = (1 − 0.36 ) R1 ⇒ R 2 = 1.778 R1
Consequently,
It is also required that
0.001 ≥ 0.005 2
R1 R 2
R1 + R 2
⇒
R1 + R 2
R1 R 2
⇒ R1 ≤
For example,
=
2.778 R1
1.778 R12
≥
0.005 2
= 0.025
0.001
2.778
= 62.5 Ω
1.778 ( 0.025 )
R1 = 60 Ω and R 2 = 106.7 Ω
DP 3.9 A thermistor is a temperature dependent resistor. The thermistor
resistance, RT, is related to the temperature by the equation
β 1 T −1 T o )
RT = RT e (
where T has units of °K and R is in Ohms. R0 is resistance at temperature T0
and the parameter β is in °K. For example, suppose that a particular
thermistor has a resistance R0 = 620 Ω at the temperature T0 = 20 °C = 293
°K and β = 3330 °K. At T = 70 °C = 343 °K the resistance of this thermistor
will be
3330 ( 1 342 −1 293 )
R T = 620 e
= 121.68 Ω
In Figure DP 3-9 this particular thermistor in used in a voltage divider
circuit. Specify the value of the resistor R that will cause the voltage vT
across the thermistor to be 4 V when the temperature is 100 °C.
Solution
At T = 373°K
R T = 620 e
3330 ( 1 372 −1 293 )
= 54.17 Ω
Using voltage division
4 = vT =
54.17
54.17
( 40 ) ⇒ R + 54.17 =
( 40 )
4
R + 54.17
R = 541.7 − 54.17 = 487.54 Ω
Figure DP 3-9
DP3-10 The circuit shown in Figure DP 3-10 contains a thermistor that has a resistance R0 = 620 Ω at
the temperature T0 = 20 °C = 293 °K and β = 3330 °K. (See problem DP 3-9.) Design this circuit (that is,
specify the values of R and Vs) so that the thermistor voltage is vT = 4 V when T = 100 °C and vT = 20 V
when T = 0 °C.
Figure DP 3-10
Solution
3330 ( 1 273 −1 293 )
= 1425.6 Ω
3330 ( 1 373 −1 293 )
= 54.17 Ω
At T = 0°C = 273°K
R T = 620 e
At T = 100°C = 373°K
R T = 620 e
Using voltage division
54.17
54.17
V s ) ⇒ R + 54.17 =
(
(Vs )
4
R + 54.17
1425.6
1425.6
20 = v T =
V s ) ⇒ R + 1425.6 =
(
(Vs )
20
R + 1425.6
4 = vT =
In matrix form:
Solving gives
⎡ 54.17
⎢ 4
⎢
⎢1425.6
⎢⎣ 20
⎤
−1⎥
⎡V s ⎤ ⎡ 54.17 ⎤
⎥⎢ ⎥ = ⎢
R ⎦ ⎣1425.6 ⎥⎦
⎣
⎥
−1
⎥⎦
V s = 23.7528 V and R = 267.5029 Ω
DP3-11 The circuit shown in Figure DP 3-11 is designed
help orange grower protect their crops against frost by
sounding an alarm when the temperature falls below
freezing. It contains a thermistor that has a resistance R0
= 620 Ω at the temperature T0 = 20 °C = 293 °K and β =
3330 °K. (See problem DP 3-9.)
The alarm will sound when the voltage at the
− input of the comparator is less than the voltage at the
+ input. Using voltage division twice, we see that the
alarm sounds whenever
R2
RT + R2
<
R4
R3 + R 4
Figure DP 3-11
Determine values of R2, R3 and R4 that cause the alarm to
sound whenever the temperature is below freezing.
Solution:
The solution is not unique. For example, pick R3 = 30 kΩ and R4 = 10 kΩ. The alarm turns on and off
R2
10
=
= 0.25
when
R T + R 2 30 + 10
At freezing the temperature is T = 0°C = 273°K and the thermistor resistance is
R T = 620 e 3330 ( 1 273 −1 293 ) = 1425.6 Ω
At freezing, we have
R2
1425.6 + R 2
= 0.25 ⇒ 0.75 R 2 = 356.4
R 2 = 475.2 Ω
Let’s check. At −2 °C
R T = 620 e 3330 ( 1 271−1 293 ) = 1560 Ω
475.2
= 0.2335 < 0.25
1560 + 475.2
so the alarm is on. At 2 °C
R T = 620 e 3330 ( 1 275 −1 293 ) = 1305 Ω
475.2
= 0.2669 > 0.25
1305 + 475.2
so the alarm is off.
Chapter 4 Exercises
Exercise 4.2-1 Determine the node voltages, va and vb, for the
circuit of Figure E 4.2-1.
Answer: va = 3 V and vb = 11 V
Figure E 4.2-1
Solution:
KCL at a:
v −v
a
a
b
+
+ 3 = 0 ⇒ 5 v − 3 v = −18
a
b
3
2
KCL at b:
v −v
b
a
− 3 −1 = 0 ⇒ v − v = 8
b
a
2
v
Solving these equations gives:
va = 3 V and vb = 11 V
Exercise 4.2-2 Determine the node voltages, va
and vb, for the circuit of Figure E 4.2-2.
Answer: va = –4/3 V and vb = 4 V
Figure E 4.2-2
Solution:
v −v
a
a
b
+
+ 3 = 0 ⇒ 3 v − 2 v = −12
a
b
4
2
v
v −v
b
a
b
−
−4= 0
⇒ − 3 v + 5 v = 24
a
b
3
2
v
KCL at a:
KCL at b:
Solving:
va = −4/3 V and vb = 4 V
Exercise 4.3-1 Find the node voltages for the circuit
of Figure E 4.3-1.
Hint: Write a KCL equation for the supernode
corresponding to the 10-V voltage source.
Answer:
2+
vb + 10 vb
+
= 5 ⇒ vb = 30 V and va = 40 V
20
30
Figure E 4.3-1
Solution:
Apply KCL to the supernode to get
v + 10 v
2+ b
+ b =5
20
30
Solving:
v = 30 V and v = v + 10 = 40 V
b
a
b
Exercise 4.3-2 Find the voltages va and vb for the circuit of Figure E 4.3-2.
Answer:
(vb + 8) − (−12) vb
+
= 3 ⇒ vb = 8 V and va = 16 V
10
40
Figure E 4.3-2
Solution:
( vb + 8) − ( −12) + vb = 3
10
40
⇒ v = 8 V and v = 16 V
b
a
Exercise 4.4-1 Find the node voltage vb for the circuit
shown in Figure E 4.4-2.
Hint: Apply KCL at node a to express ia as a function of
the node voltages. Substitute the result into
vb = 4ia and solve for vb.
6 v v
Answer: − + b − b = 0 ⇒ vb = 4.5 V
8 4 12
Figure E 4.4-2
Solution:
Apply KCL at node a to express ia as a function of the node voltages. Substitute the result into
vb = 4 ia and solve for vb .
6 vb
+
=i
8 12 a
⎛ 9 + vb ⎞
⇒ v = 4i = 4 ⎜
⎟ ⇒ v = 4.5 V
b
a
b
⎜ 12 ⎟
⎝
⎠
Exercise 4.4-2 Find the node voltages for the circuit
shown in Figure E 4.4-2.
Hint: The controlling voltage of the dependent source
is a node voltage, so it is already expressed as a function
of the node voltages. Apply KCL at node a.
Answer:
va − 6 va − 4va
+
= 0 ⇒ va = −2 V
20
15
Figure E 4.4-2
Solution:
The controlling voltage of the dependent source is a node voltage so it is already expressed as a function
of the node voltages. Apply KCL at node a.
v −6 v −4v
a
a = 0 ⇒ v = −2 V
+ a
a
20
15
Exercise 4.5-1 Determine the
value of the voltage measured by
the voltmeter in Figure E 4.5-1.
Answer: –1 V
Figure E 4.5-1
Solution:
Mesh equations:
−12 + 6 i + 3 ⎛⎜ i − i ⎞⎟ − 8 = 0 ⇒ 9 i − 3 i = 20
1
1
2
⎝ 1 2⎠
8 − 3 ⎛⎜ i − i ⎞⎟ + 6 i = 0 ⇒ − 3 i + 9 i = −8
2
1
2
⎝ 1 2⎠
Solving these equations gives:
13
1
i =
A and i = − A
1 6
2
6
The voltage measured by the meter is 6 i2 = −1 V.
Exercise 4.6-1 Determine the value of the voltage measured by the voltmeter in Figure E 4.6-1.
Hint:
Write and solve a single mesh equation to determine the current in the 3-Ω resistor.
Answer: –4 V
Figure E 4.6-1
Solution:
−12
⎛ 3⎞
9 + 3 i + 2 i + 4 ⎜ i + ⎟ = 0 ⇒ ( 3 + 2 + 4 ) i = −9 − 3 ⇒ i =
A
9
⎝ 4⎠
The voltmeter measures
3 i = −4 V
Mesh equation:
Exercise 4.6-2 Determine the value of the current measured by the ammeter in Figure E 4.6-2.
Hint:
Write and solve a single mesh equation.
Answer: –3.67 A
Figure E 4.6-2
Solution:
Mesh equation:
15 + 3 i + 6 ( i + 3) = 0 ⇒
( 3 + 6 ) i = −15 − 6 ( 3)
⇒ i=
−33
2
= −3 A
9
3
Section 4-2 Node Voltage Analysis of Circuits with Current Sources
P 4.2-1
The node voltages in the circuit of Figure P 4.2-1 are
v1 = –4 V and v2 = 2 V.
Determine i, the current of the current source.
Answer: i = 1.5 A
Figure P 4.2-1
Solution:
v −v
−4 − 4 − 2
1
1 2
0=
+
+i =
+
+ i = −1.5 + i ⇒ i = 1.5 A
8
6
8
6
(checked using LNAP 8/13/02)
v
KCL at node 1:
P 4.2-2 Determine the node voltages for the circuit of Figure
P 4.2-2.
Answer: v1 = 2 V, v2 = 30 V, and v3 = 24 V
Figure P 4.2-2
Solution:
KCL at node 1:
KCL at node 2:
KCL at node 3:
v −v
v
1 2
1
+
+ 1 = 0 ⇒ 5 v − v = −20
1 2
20
5
v −v
v −v
1 2
2
3
+2=
⇒ − v + 3 v − 2 v = 40
1
2
3
20
10
v −v
v
2
3
3
+1 =
⇒ − 3 v + 5 v = 30
2
3
10
15
Solving gives v1 = 2 V, v2 = 30 V and v3 = 24 V.
(checked using LNAP 8/13/02)
Figure P4.2-3
P4.2-3 The encircled numbers in the circuit shown Figure P4.2-3 are node numbers. Determine the
values of the corresponding node voltages, v1 and v2.
Solution:
First, express the resistor currents in terms of the node voltages:
Apply KCL at node 1 to get
v1
15
+ 0.025 +
v1 − v 2
40
=0
Multiply both sides by 40 and simplify to get
Apply KCL at node 2 to get
11
v 1 − v 2 = −1
3
v1 − v 2 v 2
0.025 +
=
40
25
Multiply both sides by 40 and simplify to get
−v1 +
Solving, we get
13
v2 = 1
5
v1 = −0.1875 V and v 2 = 0.3125 V
P 4.2-4 Consider the circuit shown in Figure P 4.2-4. Find values of the resistances R1 and R2 that
cause the voltages v1 and v2 to be v1 = 1 V and v2 = 2 V.
Figure P 4.2-4
Solution:
−.003 +
Write the node equations:
−
When v1 = 1 V, v2 = 2 V
v1 v1 − v2
+
=0
R1
500
v1 − v2 v2
+
− .005 = 0
500
R2
1
1
−1
+
= 0 ⇒ R1 =
= 200 Ω
1
R1 500
.003 +
500
2
−1 2
−
+
− .005 = 0 ⇒ R2 =
= 667 Ω
1
500 R2
.005 −
500
−.003 +
(checked using LNAP 8/13/02)
P 4.2-5 Find the voltage v for the circuit shown in Figure P
4.2-5.
Answer: v = 21.7 mV
Figure P 4.2-5
Solution:
Write node equations:
Solving gives:
Finally:
v1
v − v 2 v1 − v3
+ 1
+
=0
500
125
250
v − v3
v − v2
− 1
− .001 + 2
=0
125
250
v − v3 v1 − v3 v3
− 2
−
+
=0
250
250 500
v1 = 0.261 V, v2 = 0.337 V, v3 = 0.239 V
v = v1 − v3 = 0.022 V
(checked using LNAP 8/13/02)
P 4.2-6 Simplify the circuit shown in
Figure P 4.2-6 by replacing series and
parallel resistors with equivalent
resistors; then analyze the simplified
circuit by writing and solving node
equations.
(a) Determine the power supplied by
each current source.
(b) Determine the power received by
the 12-Ω resistor.
Figure P 4.2-6
Solution: Replacing series and parallel resistors with equivalent resistors we get
12 Ω + ( 40 Ω & 10 Ω ) = 20 Ω
60 Ω & 120 Ω = 40 Ω
The node equations are
3 × 10−3 =
2 × 10−3 +
v2 − v3
v1 − v 2
20
v1 − v 2
+
=
20
v1 − v 3
v1 − v 3
20
v 2 − v3
10
v3
⇒
0.06 = 2v1 − ( v 2 − v 3 )
⇒
0.04 = −v1 + 3v 2 − 2v 3
⇒
0 = − ( 2v1 + 4v 2 ) + 7v 3
⎡ 2 −1 −1⎤ ⎡ v1 ⎤ ⎡.06 ⎤
⎢ −1 3 −2 ⎥ ⎢v ⎥ = ⎢.04 ⎥
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎢⎣ −2 −4 +7 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ 0 ⎥⎦
⎡ v1 ⎤ ⎡0.244 ⎤
⎢ ⎥ ⎢
⎥
⎢v 2 ⎥ = ⎢ 0.228⎥
⎢ v 3 ⎥ ⎢⎣ 0.200 ⎥⎦
⎣ ⎦
10
+
20
=
40
Solving, e.g. using MATLAB, gives
⇒
(a) The power supplied by the 3 mA current source is ( 3 ×10−3 ) ( 0.244 ) = 0.732 mW. The power
supplied by the 2 mA source is ( 2 ×10−3 ) ( 0.228 ) = 0.456 mW.
(b) The current in the 12 Ω resistor is equal to the current i =
power received by the 12 Ω resistor is ( 0.8 ×10
v1 − v 2
20
) (12 ) = 7.68 ×10
−3 2
−b
=
0.244 − 0.228
= 0.8 mA so the
20
= 7.68 μ W.
(checked: LNAP and MATLAB 5/31/04)
P 4.2-7 The node voltages in the circuit shown in Figure P 4.2-7 are va = 7 V and vb = 10 V. Determine
values of the current source current, is, and the resistance, R.
Figure P 4.2-7
Solution
Apply KCL at node a to get
2=
va
R
+
va
4
+
va − vb
2
=
7 7 7 − 10 7 1
+ +
= +
⇒ R=4Ω
R 4
R 4
2
Apply KCL at node b to get
is +
va − vb
2
=
vb
8
+
vb
8
= is +
7 − 10 10 10
= +
⇒ is = 4 A
2
8 8
(checked: LNAP 6/21/04)
Figure P4.2-8
P4.2-8 The encircled numbers in the circuit shown Figure P4.2-8 are node numbers. The corresponding
node voltages are v1 and v2.The node equation representing this circuit is
⎡ 0.225 −0.125⎤ ⎡ v1 ⎤ ⎡ −3⎤
⎢ −0.125 0.125 ⎥ ⎢v ⎥ = ⎢ 2 ⎥
⎣
⎦⎣ 2⎦ ⎣ ⎦
a) Determine the values of R and Is in Figure P4.2-8
b) Determine the value of the power supplied by the 3 A current source.
Solution:
a) The node equations representing the circuit are
v1
R
+
v1 − v 2
8
+ 3 = 0 and
v1 − v 2
8
+ Is = 0
In matrix form, we have
1
⎡
⎢ 0.125 + R
⎢
⎣ 0.125
⎤
−0.125⎥ ⎡ v1 ⎤ ⎡ −3 ⎤
=
⎥ ⎢⎣v 2 ⎥⎦ ⎢⎣ − I s ⎥⎦
−0.125⎦
After multiplying the 2nd row by -1, we have
1
⎡
⎢ 0.125 + R
⎢
⎣ −0.125
⎤
−0.125⎥ ⎡ v1 ⎤ ⎡ −3⎤
=
⎥ ⎢⎣v 2 ⎥⎦ ⎢⎣ I s ⎥⎦
0.125 ⎦
Comparing to the given node equation, we see that
1
= 0.1 ⇒ R = 10 Ω and I s = 2 A
R
b) Solving the given node equations, we get
v1 = −10 V and v 2 = 6 V
The power supplied by the 3 A current source is given by
−v1 ( 3) = − ( −10 )( 3) = 30 W
Section 4-3 Node Voltage Analysis of Circuits with Current and Voltage Sources
P 4.3-1 The voltmeter in Figure P 4.3-1 measures vc, the node voltage at node c. Determine the value of
vc.
Figure P 4.3-1
Solution:
Express the voltage of the voltage source in terms of its node voltages:
0 − va = 6 ⇒ va = −6 V
KCL at node b:
va − vb
v −v
+2= b c
6
10
KCL at node c:
Finally:
⇒
−6 − vb
v −v
+2= b c
6
10
vb − vc vc
=
10
8
⇒ −1−
⇒ 4 vb − 4 vc = 5 vc
⎛9 ⎞
30 = 8 ⎜ vc ⎟ − 3 vc
⎝4 ⎠
vb
v −v
+2= b c
6
10
⇒ vb =
⇒ 30 = 8 vb − 3 vc
9
vc
4
⇒ vc = 2 V
(checked using LNAP 8/13/02)
P 4.3-2 The voltages va, vb, vc, and vd in Figure P 4.3-2 are the node voltages corresponding to nodes a,
b, c, and d. The current i is the current in a short circuit connected between nodes b and c. Determine the
values of va, vb, vc, and vd and of i.
Answer: va = –12 V, vb = vc = 4 V, vd = –4 V, i = 2 mA
Figure P 4.3-2
Solution:
Express the branch voltage of each voltage source in terms of its node voltages to get:
va = −12 V, vb = vc = vd + 8
KCL at node b:
vb − va
= 0.002 + i ⇒
4000
vb − ( −12 )
= 0.002 + i ⇒ vb + 12 = 8 + 4000 i
4000
KCL at the supernode corresponding to the 8 V source:
v
0.001 = d + i ⇒ 4 = vd + 4000 i
4000
so
vb + 4 = 4 − vd
⇒
( vd + 8 ) + 4 = 4 − vd
Consequently vb = vc = vd + 8 = 4 V and i =
⇒ vd = −4 V
4 − vd
= 2 mA
4000
(checked using LNAP 8/13/02)
Figure P4.3-3
P4.3-3. Determine the values of the power supplied by each of the sources in the circuit shown in Figure
P4.3-3.
Solution: First, label the node voltages. Next, express the resistor currents in terms of the node
voltages.
Identify the supernode corresponding to the 24 V source
Apply KCL to the supernode to get
12 − ( v a − 24 )
10
+ 0.6 =
v a − 24
40
+
va
40
⇒ 196 = 6 v a
⇒ v a = 32 V
The 12 V source supplies
⎛ 12 − ( v a − 24 ) ⎞
⎛ 12 − ( 32 − 24 ) ⎞
⎟ = 12 ⎜
12 ⎜
⎟ = 4.8 W
⎜
⎟
10
10
⎝
⎠
⎝
⎠
The 24 V source supplies
va ⎞
⎛
32 ⎞
⎛
24 ⎜ −0.6 + ⎟ = 24 ⎜ −0.6 + ⎟ = 4.8 W
40 ⎠
40 ⎠
⎝
⎝
The current source supplies
0.6 v a = 0.6 ( 32 ) = 19.2 W
Figure P4.3-4
P4.3-4. Determine the values of the node voltages, v1, v2 and v3 in the circuit shown in Figure P4.3-13.
Solution:
First, express the resistor currents in terms of the node voltages:
Apply KCL to the supernode to get
−15 − v1
20
+
v 2 − ( v1 + 15 )
Apply KCL at node 2 to get
In matrix form:
Solving using MATLAB:
50
v1 − v 2
25
=
=
v2
40
v1 − v 2
25
+
+
v1 + 5
10
v 2 − ( v1 + 15 )
50
⇒ 0.21 v1 − 0.06 v 2 = −1.55
⇒ − 0.06 v1 + 0.085 v 2 = 0.3
⎡ 0.21 −0.06 ⎤ ⎡ v1 ⎤ ⎡ −1.55⎤
⎢ −0.06 0.085 ⎥ ⎢v ⎥ = ⎢ 0.3 ⎥
⎣
⎦⎣ 2⎦ ⎣
⎦
v1 = −7.9825 V and v 2 = −2.1053 V
P 4.3-5 The voltages va, vb, and vc in Figure P 4.3-5 are the
node voltages corresponding to nodes a, b, and c. The values
of these voltages are:
va = 12 V, vb = 9.882 V, and vc = 5.294 V
Determine the power supplied by the voltage source.
Figure P 4.3-5
Solution:
The power supplied by the voltage source is
⎛v −v v −v ⎞
⎛ 12 − 9.882 12 − 5.294 ⎞
va ( i1 + i 2 ) = va ⎜ a b + a c ⎟ = 12 ⎜
+
⎟
6 ⎠
4
6
⎝
⎠
⎝ 4
= 12(0.5295 + 1.118) = 12(1.648) = 19.76 W
(checked using LNAP 8/13/02)
P 4.3-6 The voltmeter in the circuit of Figure P
4.3-6 measures a node voltage. The value of that
node voltage depends on the value of the resistance
R.
(a)
Determine the value of the resistance R that
will cause the voltage measured by the
voltmeter to be 4 V.
(b)
Determine the voltage measured by the
voltmeter when R = 1.2 kΩ = 1200 Ω.
Answer: (a) 6 kΩ (b) 2 V
Figure P 4.3-6
Solution:
Label the voltage measured by the meter. Notice that this is a node voltage.
Write a node equation at the node at which
the node voltage is measured.
⎛ 12 − v m ⎞ v m
v −8
−⎜
+ 0.002 + m
=0
⎟+
3000
⎝ 6000 ⎠ R
That is
6000 ⎞
6000
⎛
⎜3 +
⎟ v m = 16 ⇒ R = 16
R ⎠
⎝
−3
vm
(a) The voltage measured by the meter will be 4 volts when R = 6 kΩ.
(b) The voltage measured by the meter will be 2 volts when R = 1.2 kΩ.
P 4.3-7 Determine the values of the node voltages, v1 and v2,
in Figure P 4.3-7. Determine the values of the currents ia and ib.
Figure P 4.3-7
Solution:
Apply KCL at nodes 1 and 2 to get
10 − v1
1000
10 − v 2
v1
=
+
v1 − v 2
3000 5000
v1 − v 3
v3
+
=
4000
5000 2000
⇒
23v1 − 3v 2 = 150
⇒
-4v1 + 19v 3 = 50
Solving, e.g. using MATLAB, gives
⎡ 23 −3⎤ ⎡ v1 ⎤ ⎡150 ⎤
⎢ −4 19 ⎥ ⎢ v ⎥ = ⎢ 50 ⎥
⎣
⎦⎣ 2⎦ ⎣ ⎦
ib =
Then
v1 − v 2
5000
=
⇒
v1 = 7.06 V and v1 = 4.12 V
7.06 − 4.12
= 0.588 mA
5000
Apply KCL at the top node to get
ia =
v 1 − 10
1000
+
v 2 − 10
4000
=
7.06 − 10 4.12 − 10
+
= −4.41 mA
1000
4000
(checked: LNAP 5/31/04)
P 4.3-8 The circuit shown in Figure P 4.3-8 has two inputs, v1 and v2, and one output, vo. The output is
related to the input by the equation
vo = av1 + bv2
where a and b are constants that depend on R1, R2 and R3.
(a)
(b)
Determine the values of the coefficients a and b when R1 = 10 Ω, R2 = 40 Ω and R3 = 8 Ω.
Determine the values of the coefficients a and b when R1 = R2 and R3 = R1 || R2.
Figure P 4.3-8
Solution:
vo
R3
+
v o − v1
R1
+
vo − v2
R2
=0
⇒
vo =
v1
v2
+
R
R
R
R
1+ 1 + 1 1 + 2 + 2
R 2 R3
R1 R 3
(a) When R 1 = 10 Ω, R 2 = 40 Ω and R 3 = 8 Ω
vo =
v1
v2
+
= 0.4v1 + 0.1v 2
1 5 1+ 4 + 5
1+ +
4 4
So a = 0.4 and b = 0.1.
(b) When R 1 = R 2 and R 3 = R1 & R 2 = R1 / 2
vo =
v1
+
v2
1+1+ 2 1+1+ 2
= 0.25v1 + 0.25v 2
So a = 0.25 and b = 0.25.
(checked: LNAP 5/31/04)
P 4.3-9
Determine the values of the node voltages of the circuit shown in Figure P 4.3-9.
Figure P 4.3-9
Solution:
Express the voltage source voltages as functions of the node voltages to get
v 2 − v1 = 5 and v 4 = 15
Apply KCL to the supernode corresponding to the 5 V source to get
1.25 =
v1 − v 3
8
+
v 2 − 15
20
=0
⇒
80 = 5v1 + 2v 2 − 5v 3
Apply KCL at node 3 to get
v1 − v 3
8
=
v3
40
+
v 3 − 15
12
⇒
− 15v1 + 28v 3 = 150
Solving, e.g. using MATLAB, gives
⎡ −1 1 0 ⎤ ⎡ v1 ⎤ ⎡ 5 ⎤
⎢ 5 2 −5⎥ ⎢v ⎥ = ⎢ 80 ⎥
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎢⎣ −15 0 28 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣150 ⎥⎦
⇒
⎡ v1 ⎤ ⎡ 22.4 ⎤
⎢ ⎥ ⎢
⎥
⎢v 2 ⎥ = ⎢ 27.4 ⎥
⎢ v 3 ⎥ ⎢⎣17.4 ⎥⎦
⎣ ⎦
So the node voltages are:
v1 = 22.4 V, v 2 = 27.4 V, v 3 = 17.4 V, and v 4 = 15
(checked: LNAP 6/9/04)
P 4.3-10 Figure P 4.3-10 shows a measurement made in the laboratory. Your lab partner forgot to
record the values of R1, R2, and R3. He thinks that the two resistors were 10-kΩ resistors and the other
was a 5-kΩ resistor. Is this possible? Which resistor is the 5-kΩ resistor?
Figure P 4.3-10
Solution:
Write a node equation to get
⎛ 12 − 4.5 ⎞ 4.5 4.5 − 6
7.5 4.5 1.5
−⎜
+
+
=0 ⇒ −
+
−
=0
⎜ R1 ⎟⎟ R 3
R
R
R
R
2
1
3
2
⎝
⎠
Notice that
7.5
4.5
is either 0.75 mA or 1.5 mA depending on whether R1 is 10 kΩ or 5 kΩ. Similarly,
R1
R3
is either 0.45 mA or 0.9 mA and
1.5
is either 0.15 mA or 0.3 mA. Suppose R1 and R2 are 10 kΩ
R2
resistors and R3 is a 5 kΩ resistor. Then
−
7.5 4.5 1.5
+
−
= −0.75 + 0.9 − 0.15 = 0
R1 R 3 R 2
It is possible that two of the resistors are 10 kΩ and the third is 5 kΩ. R3 is the 5 kΩ resistor.
(checked: LNAP 6/9/04)
Figure P4.3-11
P4.3-11. Determine the values of the power supplied by each of the sources in the circuit shown in
Figure P4.3-11.
Solution: First, label the node voltages. Next, express the resistor currents in terms of the node
voltages.
Identify the supernode corresponding to the 10 V source
Apply KCL to the supernode to get
15 − ( v a + 10 ) 15 − v a v a + 10 v a
+
=
+
4
6
8
3
⇒ 60 = 21 v a
⇒ v a = 2.857 V
The 15 V source supplies
⎛ 15 − ( v a + 10 ) 15 − v a
+
15 ⎜
⎜
4
6
⎝
⎞
15 − 12.857 15 − 2.857 ⎞
⎟ = 15 ⎛⎜
+
⎟ = 15 ( 2.56 ) = 38.4 W
⎟
4
6
⎝
⎠
⎠
⎛ 15 − v a v a ⎞
⎛ 15 − 2.857 2.857 ⎞
The 10 V source supplies 10 ⎜
+ ⎟ = 10 ⎜
+
⎟ = 10 (1.071) = 10.71 W
6
3
6
3
⎝
⎠
⎝
⎠
P 4.3-12
Determine the values of the node voltages of the circuit shown in Figure P 4.3-12.
Figure P 4.3-12
Solution:
Express the voltage source voltages in terms of the node voltages:
v 2 − v1 = 8 and v 3 − v1 = 12
Apply KVL to the supernode to get
v2
10
so
+
v1
4
+
v3
5
=0
⇒
2v 2 + 5v1 + 4v 3 = 0
2 ( 8 + v1 ) + 5v1 + 4 (12 + v1 ) = 0
⇒
v1 = −
64
V
11
The node voltages are
v1 = −5.818 V
v 2 = 2.182 V
v 3 = 6.182 V
(checked: LNAP 6/21/04)
Figure P4.3-13
P4.3-13. Determine the values node voltages, v1 and v2, in the circuit shown in Figure P4.3-13.
Solution: Apply KCL at node 1 to get
v1
50
+
v1 − v 2
65
v1 − 60
+
80
1
1 ⎞
60
⎛ 1
⎛ 1 ⎞
= 0 ⇒ ⎜ + + ⎟ v1 − ⎜ ⎟ v 2 =
80
⎝ 50 65 80 ⎠
⎝ 65 ⎠
Apply KCL at node 2 to get
0.1 =
v 2 − v1
65
+
v 2 − 60
75
1 ⎞
⎛ 1 ⎞
⎛ 1
= ⇒ − ⎜ ⎟ v1 + ⎜ + ⎟ v 2 = 0.1
⎝ 65 ⎠
⎝ 65 75 ⎠
In matrix form
1
1
⎡1
⎢ 50 + 65 + 80
⎢
1
⎢
−
⎢⎣
65
1 ⎤
⎡ 60 ⎤
65 ⎥ ⎡ v1 ⎤ ⎢ ⎥
⎥ ⎢ ⎥ = 80
1
1 ⎥ ⎣v 2 ⎦ ⎢ ⎥
+
⎣0.1⎦
65 75 ⎥⎦
Solving, we get
v1 = 13.2356 V and v2 = 22.3456 V
−
Figure P4.3-14
P4.3-14. The voltage source in the circuit shown in Figure P4.3-14 supplies 83.802 W. The current source
supplies 17.572 W. Determine the values of the node voltages v1 and v2.
Solution: From the power supplied by the current source we calculate
17.572 = v 2 ( 0.25 ) ⇒ v 2 =
i3 =
Using Ohm’s law
17.572
= 70.288 V
0.25
70.288 − 80
= −0.4856 A
20
From the power supplied by the voltage source we calculate
83.802 = 80 i 6
⇒ i6 =
83.802
= 1.0475 V
80
i1 = − ( i 3 + i 6 ) = − ( −0.4856 + 1.0475 ) = −0.5619 A
Using Ohm’s law
In summary
−0.5619 =
v1 − 80
50
⇒ v1 = 80 + 50 ( −0.5619 ) = 51.905 V
v1 = 51.905 V and v2 = 70.288 V
Section 4-4 Node Voltage Analysis with Dependent Sources
P 4.4-1 The voltages va, vb, and vc in Figure P 4.4-1 are
the node voltages corresponding to nodes a, b, and c. The
values of these voltages are:
va = 8.667 V, vb = 2 V,
and vc = 10 V
Determine the value of A, the gain of the dependent source.
Figure P 4.4-1
Solution:
Express the resistor currents in terms of the
node voltages:
va − vc
= 8.667 − 10 = −1.333 A and
1
v −v
2 − 10
= −4 A
i 2= b c =
2
2
i 1=
Apply KCL at node c:
i1 + i 2 = A i1 ⇒ − 1.333 + ( −4 ) = A (−1.333)
⇒
A=
−5.333
=4
−1.333
(checked using LNAP 8/13/02)
P 4.4-2
Find ib for the circuit shown in Figure P 4.4-2.
Answer: ib = –12 mA
Figure P 4.4-2
Solution:
Write and solve a node equation:
va − 6
v
v − 4va
+ a + a
= 0 ⇒ va = 12 V
1000 2000 3000
ib =
va − 4va
= −12 mA
3000
(checked using LNAP 8/13/02)
P 4.4-3 Determine the node voltage vb for the circuit of
Figure P 4.4-3.
Answer: vb = 1.5 V
Figure P 4.4-3
Solution:
First express the controlling current in terms of
the node voltages:
2 − vb
i =
a
4000
Write and solve a node equation:
−
2 − vb
v
⎛ 2 − vb ⎞
+ b − 5⎜
⎟ = 0 ⇒ vb = 1.5 V
4000 2000 ⎝ 4000 ⎠
(checked using LNAP 8/14/02)
P 4.4-4 The circled numbers in Figure P 4.4-4
are node numbers. The node voltages of this circuit
are v1 = 10 V, v2 = 14 V, and v3 = 12 V.
(a)
Determine the value of the current ib.
(b)
Determine the value of r, the gain of the
CCVS.
Answers: (a) –2 A (b) 4 V/A
Figure P 4.4-4
Solution:
Apply KCL to the supernode of the CCVS to get
12 − 10 14 − 10 1
+
− + i b = 0 ⇒ i b = −2 A
4
2
2
Next
10 − 12
1⎫
=− ⎪
V
−2
=4
4
2⎬ ⇒ r =
1
A
−
r i a = 12 − 14 ⎪⎭
2
ia =
(checked using LNAP 8/14/02)
P 4.4-5 Determine the value of the current ix in the circuit of
Figure P 4.4-5.
Answer: ix = 2.4 A
Figure P 4.4-5
Solution:
First, express the controlling current of the CCVS in
v2
terms of the node voltages: i x =
2
Next, express the controlled voltage in terms of the
node voltages:
v2
24
12 − v 2 = 3 i x = 3
V
⇒ v2 =
2
5
so ix = 12/5 A = 2.4 A.
(checked using ELab 9/5/02)
Figure P4.4-6
P4.4-6 The encircled numbers in the circuit shown Figure P4.4-6 are node numbers. Determine the value
of the power supplied by the CCVS .
P4.4-6
First, express the contolling current of the CCVS in terms of the node voltages:
ia =
v2
20
⎛ v2 ⎞
v1 = 12 V and v 3 = 40 i a = 40 ⎜ ⎟ = 2 v 2
⎝ 20 ⎠
Next, express the resistor currents in terms of the node voltages:
Notice that
Apply KCL at node 2 to get
Then
The CCVS supplies
12 − v 2
+
v2
+
v2 − 2v2
= 0 ⇒ v 2 = 16 V
5
20
10
v 2 16
ia =
=
= 0.8 A and v 3 = 40 i a = 40 ( 0.8 ) = 32 V
20 20
⎛ v3 − v2 ⎞
⎛ 32 − 16 ⎞
v3 ⎜
⎟ = 32 ⎜
⎟ = 51.2 W
⎝ 10 ⎠
⎝ 10 ⎠
Figure P4.4-7
P4.4-7 The encircled numbers in the circuit shown Figure 4.4-27 are node numbers. The corresponding
node voltages are:
v1 = 9.74 V and v2 = 6.09 V
Determine the values of the gains of the dependent sources, r and g.
Solution:
Using Ohm’s law, i b =
Then
v1
8
=
9.74
= 1.2175 A . Using KVL, the voltage across the CCVS is
8
r i b = v1 − v 2 = 9.74 − 6.09 = 3.65 V
r=
r ib
ib
=
3.65
= 2.9979 V/A
1.2175
Using KVL, v b = 12 − v1 = 12 − 9.74 = 2.26 V . Apply KCL to the supernode corresponding to the CCVS
to get
12 − v1 v1 v 2
12 − 9.74 9.74 6.09
= + + g vb ⇒
=
+
+ g v b ⇒ g v b = −1.6963 A
8
8 8
8
8
8
g v b −1.6963
g=
=
= −0.7506 A/V
Then
vb
2.26
P 4.4-8 Determine the value of the power supplied by the
dependent source in Figure P 4.4-8.
Figure P 4.4-8
Solution:
Label the node voltages.
First, v2 = 10 V due to the independent voltage
source. Next, express the controlling current of the
dependent source in terms of the node voltages:
ia =
v3 − v 2
16
=
v 3 − 10
16
Now the controlled voltage of the dependent source can be expressed as
⎛ v 3 − 10 ⎞
v1 − v 3 = 8 i a = 8 ⎜
⎟
⎝ 16 ⎠
⇒
v1 =
3
v3 − 5
2
Apply KCL to the supernode corresponding to the dependent source to get
v1 − v 2
+
4
v1
12
+
v3 − v 2
16
+
v3
8
=0
Multiplying by 48 and using v2 = 10 V gives
16v1 + 9v 3 = 150
Substituting the earlier expression for v1
⎛3
⎞
16 ⎜ v 3 − 5 ⎟ + 9v 3 = 150
⎝2
⎠
⇒
v 3 = 6.970 V
Then v1 = 5.455 V and ia = -0.1894 A. Applying KCL at node 2 gives
v1
12
So
= ib +
10 − v1
4
⇒
12 i b = −3 + 4 v1 = −30 + 4 ( 5.455 )
i b = −0.6817 A.
Finally, the power supplied by the dependent source is
p = ( 8 i a ) i b = 8 ( −0.1894 ) ( −0.6817 ) = 1.033 W
(checked: LNAP 5/24/04)
P 4.4-9 The node voltages in the circuit shown in Figure
P 4.4-9 are
v1 = 4 V, v2 = 0 V, and v3 = –6 V
Determine the values of the resistance, R, and of the gain,
b, of the CCCS.
Figure P 4.4-9
Solution:
Apply KCL at node 2:
i a + bi a = i b =
but
ia =
v3 − v 2
20
v 2 − v1
40
=
=
−6 − ( 0 )
= −0.3 A
20
0−4
= −0.1
40
so
(1 + b )( −0.1) = ( −0.3)
⇒
b=2
A
A
Next apply KCL to the supernode corresponding to the voltage source.
v1
10
+ 2 ia +
v3
R
=0
⇒
4
−6
+ 2 ( −0.1) +
=0
10
R
⇒
R=
6
= 30 Ω
.2
(checked: LNAP 6/9/04)
P 4.4-10 The value of the node voltage at node b in the
circuit shown in Figure P 4.4-10 is vb = 18 V.
(a)
Determine the value of A, the gain of the dependent
source.
(b)
Determine the power supplied by the dependent
source.
Figure P 4.4-10
Solution:
(a) Express the controlling voltage of the dependent source in terms of the node voltages:
va = 9 − vb
Apply KCL at node b to get
9 − vb
100
= A (9 − v b ) +
vb
200
⇒
A=
18 − 3v b
200 ( 9 − v b )
= 0.02
(b) The power supplied by the dependent source is
− ( Av a ) v b = − ( 0.02 ( 9 − 18 ) ) (18 ) = 3.24 W
(checked: LNAP 6/06/04)
P 4.4-11 Determine the power supplied by the
dependent source in the circuit shown in Figure P 4.4-11.
Figure P 4.4-11
Solution:
This circuit contains two ungrounded voltage sources, both incident to node x. In such a circuit it is
necessary to merge the supernodes corresponding to the two ungrounded voltage sources into a single
supernode. That single supernode separates the two voltage sources and their nodes from the rest of the
circuit. It consists of the two resistors and the current source. Apply KCL to this supernode to get
v x − 20 v x
+ +4=0
⇒ v x = 10 V .
2
10
The power supplied by the dependent source is
( 0.1 v ) ( −30 ) = −30 W .
x
(checked: LNAP 6/6/04)
P 4.4-12 Determine values of the node voltages,
v1, v2, v3, v4, and v5, in the circuit shown in Figure P
4.4-12.
Figure P 4.4-12.
Solution:
Express the voltages of the independent voltage sources in terms of the node voltages
v1 − v 2 = 16 and v 4 − v 5 = 8
Express the controlling current of the dependent source in terms of the node voltages
ix =
v3
6
Express the controlled voltage of the dependent source in terms of the node voltages
⎛ v3 ⎞
v 2 − v 4 = 4i x = 4 ⎜ ⎟
⎝6⎠
⇒
− 6v 2 + 4v 3 + 6v 4 = 0
Apply KCL to the supernode to get
v1 − v 3
2
+
v 4 − v3
3
+
v5
8
=1
⇒
12v1 − 20v 3 + 8v 4 + 3v 5 = 24
Apply KCL at node 3 to get
v 3 − v1
2
+
v3
6
+
v3 − v 4
3
=0
⇒
− 3v1 + 6v 2 − 2v 4 = 0
Solving, e.g. using MATLAB, gives
0 0 ⎤ ⎡ v1 ⎤ ⎡16 ⎤
⎡ 1 −1 0
⎢ ⎥
⎢0 0
0
1 −1⎥⎥ ⎢v 2 ⎥ ⎢⎢ 8 ⎥⎥
⎢
⎢ 0 −6 4
6 0 ⎥ ⎢v 3 ⎥ = ⎢ 0 ⎥
⎢
⎥⎢ ⎥ ⎢ ⎥
12
0
20
8
3
−
⎢
⎥ ⎢v 4 ⎥ ⎢ 24 ⎥
⎢⎣ −3 0
6 −2 0 ⎥⎦ ⎢⎣ v 5 ⎥⎦ ⎢⎣ 0 ⎥⎦
⇒
⎡ v1 ⎤ ⎡ 24 ⎤
⎢v ⎥ ⎢ ⎥
⎢ 2⎥ ⎢ 8 ⎥
⎢ v 3 ⎥ = ⎢12 ⎥
⎢ ⎥ ⎢ ⎥
⎢v 4 ⎥ ⎢ 0 ⎥
⎢ v 5 ⎥ ⎢⎣ −8⎥⎦
⎣ ⎦
(checked: LNAP 6/13/04)
P 4.4-13 Determine values of the node voltages,
v1, v2, v3, v4, and v5, in the circuit shown in Figure
P 4.4-13.
Figure P 4.4-13
Solution:
Express the voltage source voltages in terms of the node voltages:
v1 − v 2 = 8 and v 4 − v 3 = 16
Express the controlling current of the dependent source in terms of the node voltages:
ix =
v 2 − v3
10
+
v1 − v 3
5
= 0.2v1 + 0.1v 2 − 0.3v 3
Express the controlled voltage of the dependent source in terms of the node voltages:
v 5 = 4i x = 0.8v1 = 0.4v 2 − 1.2v 3
⇒
0.8v1 + 0.4v 2 − 1.2v 3 − v 5 = 0
Apply KVL to the supernodes
v1 − v 5
+
2
v2 − v4
v4 − v2
4
4
+
v3
8
+
+
v 2 − v3
10
v3 − v 2
10
+
+
v1 − v 3
5
v 3 − v1
5
=0
=2
⇒
⇒
14v1 + 7v 2 − 6v 3 − 5v 4 − 10v 5 = 0
− 8v1 − 14v 2 + 17v 3 + 10v 4 = 80
Solving, e.g. using MATLAB, gives
0
0
0 ⎤ ⎡ v1 ⎤ ⎡ 8 ⎤
−1
⎡1
⎢ ⎥
⎢0
0
1
0 ⎥⎥ ⎢v 2 ⎥ ⎢⎢16 ⎥⎥
−1
⎢
⎢ 0.8 0.4 −1.2 0 −1 ⎥ ⎢ v 3 ⎥ = ⎢ 0 ⎥
⎢
⎥⎢ ⎥ ⎢ ⎥
7
−6 −5 −10 ⎥ ⎢v 4 ⎥ ⎢ 0 ⎥
⎢ 14
⎢⎣ −8 −14 17 10
0 ⎥⎦ ⎢⎣ v 5 ⎥⎦ ⎢⎣80 ⎥⎦
⇒
⎡ v1 ⎤ ⎡11.32 ⎤
⎢v ⎥ ⎢
⎥
⎢ 2 ⎥ ⎢ 3.32 ⎥
⎢ v 3 ⎥ = ⎢ 2.11 ⎥
⎢ ⎥ ⎢
⎥
⎢v 4 ⎥ ⎢18.11⎥
⎢ v 5 ⎥ ⎢⎣ 7.85 ⎥⎦
⎣ ⎦
(checked: LNAP 6/13/04)
P 4.4-14 The voltages v1, v2, v3, and v4 are
the node voltages corresponding to nodes 1, 2,
3, and 4 in Figure P 4.4-14. Determine the
values of these node voltages.
Figure P 4.4-14
Solution:
Express the controlling voltage and current of
the dependent sources in terms of the node
voltages:
v3 − v4
v a = v 4 and i b =
R2
Express the voltage source voltages in terms of
the node voltages:
v1 = V s and v 2 − v 3 = Av a = Av 4
Apply KCL to the supernode corresponding to the dependent voltage source
v 2 − v1
R1
+
v3 − v4
R2
= Is
⇒ − R 2 v1 + R 2 v 2 + R1 v 3 − R1 v 4 = R1 R 2 I s
Apply KCL at node 4:
B
v3 − v4
R2
+
v3 − v4
R2
=
v4
R3
⇒
⎛
( B + 1) v 3 − ⎜⎜ B + 1 +
⎝
R2 ⎞
⎟ v4 = 0
R 3 ⎟⎠
Organizing these equations into matrix form:
⎡ 1
⎢ 0
⎢
⎢− R2
⎢
⎢
⎢ 0
⎢⎣
With the given values:
0
1
R2
0
⎤
⎥ ⎡ v1 ⎤ ⎡ V s ⎤
⎥ ⎢v ⎥ ⎢
⎥
⎥ ⎢ 2⎥ = ⎢ 0 ⎥
⎥ ⎢v 3 ⎥ ⎢ R R I ⎥
⎛
R 2 ⎞⎥ ⎢ ⎥ ⎢ 1 2 s ⎥
B +1 − ⎜ B +1+
⎟⎟ ⎥ ⎣⎢v 4 ⎦⎥ ⎣ 0 ⎦
⎜
R
3
⎝
⎠ ⎥⎦
0
−1
R1
0
−A
− R1
⎡ v1 ⎤ ⎡ 25 ⎤
0 0
0 ⎤ ⎡ v1 ⎤ ⎡ 25 ⎤
⎡ 1
⎢v ⎥ ⎢
⎢v ⎥ ⎢
⎢ 0
⎥
⎥
−5 ⎥ ⎢ 2 ⎥ ⎢ 0 ⎥
1 −1
44.4 ⎥⎥
2⎥
⎢
⎢
⎢
=
⇒
=
⎢ v 3 ⎥ ⎢ 8.4 ⎥
⎢ −20 20 10
−10 ⎥ ⎢ v 3 ⎥ ⎢ 400 ⎥
⎢ ⎥ ⎢
⎢
⎥⎢ ⎥ ⎢
⎥
⎥
0 4 −4.667 ⎦ ⎢⎣v 4 ⎥⎦ ⎣ 0 ⎦
⎢⎣ v 4 ⎥⎦ ⎣ 7.2 ⎦
⎣ 0
(Checked using LNAP 9/29/04)
P 4.4-15 The voltages v1, v2, v3, and v4 in Figure
P 4.4-15 are the node voltages corresponding to
nodes 1, 2, 3, and 4. The values of these voltages
are
v1 = 10 V, v2 = 75 V, v3 = –15 V, and v4 = 22.5 V
Determine the values of the gains of the dependent
sources, A and B, and of the resistance R1.
Figure P 4.4-15
Solution:
Express the controlling voltage and current of the dependent sources in terms of the node voltages:
v a = v 4 = 22.5 V
ib =
and
v3 − v4
R2
=
−15 − 22.5
= −0.75
50
Express the dependent voltage source voltage in terms of the node voltages:
v 2 − v 3 = Av a = Av 4
A=
so
v 2 − v3
v4
=
75 − ( −15 )
= 4 V/V
22.5
Apply KCL to the supernode corresponding to the dependent voltage source
v 2 − v1
R1
+
v3 − v4
R2
= Is
⇒
75 − 10 −15 − 22.5
+
= 2.5 ⇒ R1 = 20 Ω
R1
50
Apply KCL at node 4:
v3 − v 4
R2
=
v4
R3
+B
v3 − v 4
R2
⇒
−15 − 22.5 22.5
−15 − 22.5
=
+B
⇒ B = 2.5 A/A
50
20
50
(Checked using LNAP 9/29/04)
P 4.4-16 The voltages v1, v2, and v3 in Figure P 4.4-16 are
the node voltages corresponding to nodes 1, 2, and 3. The
values of these voltages are
v1 = 12 V, v2 = 21 V, and v3 = –3 V
(a)
Determine the values of the resistances R1 and R2.
(b)
Determine the power supplied by each source.
Figure P 4.4-16
Solution:
v 2 − v1
R1 =
(b)
The power supplied by the voltage source is 12 ( 0.5 + 1.25 − 2 ) = −3 W . The power supplied by
2 − 0.5
=
v2
21 − 12
−3
= 6 Ω and R 2 =
=
=4Ω
1.5
1.25 − 2 −0.75
(a)
the 1.25-A current source is 1.25 ( −3 − 12 ) = −18.75 W . The power supplied by the 0.5-A current source
is −0.5 ( 21) = −10.5 W . The power supplied by the 2-A current source is 2 ( 21 − ( −3) ) = 48 W .
P 4.4-17 The voltages v1, v2, and v3 in Figure P 4.4-17
are the node voltages corresponding to nodes 1, 2, and 3.
The values of these voltages are
v1 = 12 V, v2 = 9.6 V,
and v3= –1.33 V
(a)
Determine the values of the resistances R1 and R2.
(b)
Determine the power supplied by each source.
Figure P 4.4-18
Solution:
12 − ( −1.33)
= 1.666 A
8
9.6
i2 =
= 2.4 A
4
v 2 − v1 9.6 − 12
=
= 6 Ω and
R1 =
2 − i1
2 − 2.4
i1 =
and
(a)
R2 =
v3
i1 − 2
=
−1.33
= 3.98 4 Ω
1.666 − 2
(b) The power supplied by the voltage source is 12 ( 2.4 + 1.66 − 2 ) = 24.7 W . The power supplied by
the current source is 2 ( 9.6 − ( −1.33) ) = 21.9 W .
(Checked using LNAP 10/2/04)
P4.4-18
The voltages v 2 , v 3 and v 4 for the circuit shown in Figure P4.4-18 are:
v 2 = 16 V, v 3 = 8 V and v 4 = 6 V
Determine the values of the following:
(a) The gain, A, of the VCVS
(b) The resistance R 5
(c) The currents i b and i c
(d) The power received by resistor R 4
Figure P4.4-18
Solution:
Given the node voltages
v 2 = 16 V, v 3 = 8 V and v 4 = 6 V
A=
Av a
va
=
⎛ v3 − v4 ⎞
R5 ⎜
⎟ = v4
⎝ 15 ⎠
ib =
16 − 8
V
=4
8−6
V
⇒ R5 =
15 ( 6 )
= 45 Ω ,
8−6
40 − 24
40 − 16 16
= 2 A and i c =
− = 0.6667 A
12
12
12
p4 =
va2
15
=
22
= 0.2667 W
15
P4.4-19
Determine the values of the node voltages v1 and v 2 for the circuit shown in Figure P4.4-19.
Figure P4.4-19
P4.4-19
The node equations are
28 − v1
4
and
+ 3 v1 =
v1 − v 2
5
v1 − v 2
5
= 3 v1 +
v2
6
⇒ 5 ( 28 − v1 ) + 20 ( 3 v1 ) = 4 ( v1 − v 2 ) ⇒ 140 = −51 v1 − 4 v 2
+ 4 v 3 = 3 v1 +
v2
6
+ 4 ( v1 − v 2 ) ⇒ 0 = 204 v1 − 109 v 2
Using MATLAB to solve these equations:
Consequently
v1 = −2.3937 V and v 2 = −4.4800 V
P4.4-20
The encircled numbers in Figure P4.4-20 are node numbers. Determine the values of v1 , v 2 and v 3 , the
node voltages corresponding to nodes 1, 2 and 3.
Figure P4.4-20
Solution:
Apply KCL to the supernode corresponding to the horizontal voltage source to get
v1
10
=
va
2
=
v3 − v 2
2
=
v 3 − ( v1 + 10 )
2
(
⇒ v1 = 5 v 3 − ( v1 + 10 )
)
⇒ 50 = −6 v1 + 5 v 3
Looking at the dependent source we notice that
(
v 3 = 5 v a = 5 ( v 3 − v 2 ) = 5 v 3 − ( v1 + 10 )
)
Using MATLAB to solve these equations:
Consequently
Then
v1 = −50 V and v 3 = −50 V
v 2 = v1 + 10 = −40 V
⇒ 50 = −5 v1 + 4 v 3
P4.4-21
Determine the values of the node voltages v1 , v 2 and v 3 for the circuit shown in Figure P4.4-21.
Figure P4.4-21
Solution:
Represent the controlling current of the dependent source in terms of the node voltages: i a =
Represent the controlled voltage of the dependent source in terms of the node voltages:
4 i a = v1 − v 2
⎛ v3 ⎞
⇒ = 4 ⎜ ⎟ = v1 − v 2
⎝ 2⎠
⇒ 0 = v1 − v 2 − 2 v 3
Apply KCL to the supernode corresponding to the dependent voltage source:
12 − v1
2
=
v2
2
+
v 2 − v3
2
⇒ 12 − v1 = v 2 + v 2 − v 3 ⇒ 12 = v1 + 2 v 2 − v 3
Apply KCL to top node of the current source:
v 2 − v3
2
+1 =
v3
2
⇒ v 2 − v 3 + 2 = v 3 ⇒ 2 = −v 2 + 2 v 3
Solving these equations using MATLAB gives
v1 = 8.2857 V ,
v 2 = 3.1459 V
and
v 3 = 2.5714 V
v3
2
P4.4-22
Determine the values of the node voltages v1 , v 2 and v 3 for the circuit shown in Figure P4.4-22.
Figure P4.4-22
Solution:
The node equations are:
12 − v1
2
4 ia =
v2
2
⎛ v3 ⎞
= 4 i a = 4 ⎜ ⎟ ⇒ 12 − v1 = 4 v 3 ⇒ 12 = v1 + 4 v 3
⎝ 2⎠
+
v 2 − v3
v 2 − v3
2
⎛ v3 ⎞ v 2 v 2 − v3
⇒ 4⎜ ⎟ =
+
2
⎝ 2⎠ 2
2
+1 =
v3
2
⇒ v 2 − v 3 + 2 = v 3 ⇒ 2 = −v 2 + 2 v 3
Solving these equations using MATLAB gives
v1 = 28 V ,
v 2 = −10 V
and
v 3 = −4 V
⇒ 0 = 2 v 2 − 5 v3
Section 4-5 Mesh Current Analysis with Independent Voltage Sources
P 4.5-1 Determine the mesh currents, i1, i2, and
i3, for the circuit shown in Figure P 4.5-1.
Answers: i1 = 3 A, i2 = 2 A, and i3 = 4 A
Figure P 4.5-1
Solution:
The mesh equations are
2 i1 + 9 (i1 − i 3 ) + 3(i1 − i 2 ) = 0
15 − 3 (i1 − i 2 ) + 6 (i 2 − i 3 ) = 0
−6 (i 2 − i 3 ) − 9 (i1 − i 3 ) − 21 = 0
or
14 i1 − 3 i 2 − 9 i 3 = 0
−3 i1 + 9 i 2 − 6 i 3 = −15
−9 i1 − 6 i 2 + 15 i 3 = 21
so
i1 = 3 A, i2 = 2 A and i3 = 4 A.
(checked using LNAP 8/14/02)
P 4.5-2 The values of the mesh currents in
the circuit shown in Figure P 4.5-2 are
i1 = 2 A, i2 = 3 A, and i3 = 4 A.
Determine the values of the resistance R and of
the voltages v1 and v2 of the voltage sources.
Answers: R = 12 Ω, v1 = –4 V, and v2 = –28 V
Figure P 4.5-2
Solution:
The mesh equations are:
Top mesh:
so
Bottom, right mesh:
so
Bottom left mesh
so
4 (2 − 3) + R(2) + 10 (2 − 4) = 0
R = 12 Ω.
8 (4 − 3) + 10 (4 − 2) + v 2 = 0
v2 = −28 V.
−v1 + 4 (3 − 2) + 8 (3 − 4) = 0
v1 = −4 V.
(checked using LNAP 8/14/02)
P 4.5-3 The currents i1 and i2 in Figure P 4.5-3
are the mesh currents. Determine the value of
the resistance R required to cause va = –6 V.
Answer: R = 4 Ω
Figure P 4.5-3
Solution:
Ohm’s Law: i 2 =
−6
= −0.75 A
8
KVL for loop 1:
R i1 + 4 ( i1 − i 2 ) + 3 + 18 = 0
KVL for loop 2
+ (−6) − 3 − 4 ( i1 − i 2 ) = 0
⇒ − 9 − 4 ( i1 − ( −0.75 ) ) = 0
⇒ i1 = −3 A
R ( −3) + 4 ( −3 − ( −0.75 ) ) + 21 = 0 ⇒ R = 4 Ω
(checked using LNAP 8/14/02)
P 4.5-4 Determine the mesh currents, ia and
ib, in the circuit shown in
Figure P 4.5-4.
Figure P 4.5-4
Solution:
KVL loop 1:
KVL loop 2:
Solving these equations:
25 ia − 2 + 250 ia + 75 ia + 4 + 100 (ia − ib ) = 0
450 ia −100 ib = −2
−100(ia − ib ) − 4 + 100 ib + 100 ib + 8 + 200 ib = 0
−100 ia + 500 ib = − 4
ia = − 6.5 mA , ib = − 9.3 mA
(checked using LNAP 8/14/02)
Find the current i for the circuit of Figure P 4.5-5.
P 4.5-5
Hint:
A short circuit can be treated as a 0-V voltage source.
Figure P 4.5-5
Solution:
Mesh Equations:
mesh 1 : 2i1 + 2 (i1 − i2 ) + 10 = 0
mesh 2 : 2(i2 − i1 ) + 4 (i2 − i3 ) = 0
mesh 3 : − 10 + 4 (i3 − i2 ) + 6 i3 = 0
Solving:
5
i = i2 ⇒ i = − = −0.294 A
17
(checked using LNAP 8/14/02)
P 4.5-6 Simplify the circuit shown in Figure P
4.5-6 by replacing series and parallel resistors by
equivalent resistors. Next, analyze the simplified
circuit by writing and solving mesh equations.
(a) Determine the power supplied by each source.
(b) Determine the power absorbed by the 30-Ω
resistor.
Figure P 4.5-6
Solution: Replace series and parallel resistors with equivalent resistors:
60 Ω & 300 Ω = 50 Ω , 40 Ω + 60 Ω = 100 Ω and 100 Ω + 30 Ω + ( 80 Ω & 560 Ω ) = 200 Ω
so the simplified circuit is
The mesh equations are
200 i1 + 50 ( i1 − i 2 ) − 12 = 0
100 i 2 + 8 − 50 ( i1 − i 2 ) = 0
or
⎡ 250 −50 ⎤ ⎡ i1 ⎤ ⎡12 ⎤
⎢ −50 150 ⎥ ⎢i ⎥ = ⎢ −8⎥
⎣
⎦⎣ 2⎦ ⎣ ⎦
⇒
⎡ i1 ⎤ ⎡ 0.04 ⎤
⎢i ⎥ = ⎢
⎥
⎣ 2 ⎦ ⎣ −0.04 ⎦
The power supplied by the 12 V source is 12 i1 = 12 ( 0.04 ) = 0.48 W . The power supplied by the 8 V
source is −8i 2 = −8 ( −0.04 ) = 0.32 W . The power absorbed by the 30 Ω resistor is
i12 ( 30 ) = ( 0.04 ) ( 30 ) = 0.048 W .
2
(checked: LNAP 5/31/04)
Section 4-6 Mesh Current Analysis with Voltage and Current Sources
P 4.6-1 Find ib for the circuit shown in
Figure P 4.6-1.
Answer: ib = 0.6 A
Figure P 4.6-1
Solution:
1
A
2
mesh 2: 75 i2 + 10 + 25 i2 = 0
mesh 1: i1 =
⇒ i2 = − 0.1 A
ib = i1 − i2 = 0.6 A
(checked using LNAP 8/14/02)
P 4.6-2 Find vc for the circuit shown in
Figure P 4.6-2.
Answer: vc = 15 V
Figure P 4.6-2
Solution:
Mesh currents:
mesh a: ia = − 0.25 A
mesh b: ib = − 0.4 A
Ohm’s Law:
vc = 100(ia − ib ) = 100(0.15) =15 V
(checked using LNAP 8/14/02)
P 4.6-3 Find v2 for the circuit shown in Figure P 4.6-3.
Answer: v2 = 2 V
Figure P 4.6-3
Solution:
Express the current source current as a function of the mesh currents:
i1 − i2 = − 0.5 ⇒ i1 = i2 − 0.5
Apply KVL to the supermesh:
30 i1 + 20 i2 + 10 = 0 ⇒ 30 (i2 − 0.5) + 20i2 = − 10
50 i2 − 15 = − 10 ⇒ i2 =
5
= .1 A
50
i1 =−.4 A and v2 = 20 i2 = 2 V
(checked using LNAP 8/14/02)
Find vc for the circuit shown in
P 4.6-4
Figure P 4.6-4.
Figure P 4.6-4
Solution:
Express the current source current in terms of
the mesh currents:
ib = ia − 0.02
Apply KVL to the supermesh:
250 ia + 100 (ia − 0.02) + 9 = 0
∴ ia = − .02 A = − 20 mA
vc = 100(ia − 0.02) = −4 V
(checked using LNAP 8/14/02)
P 4.6-5
Determine the value of the voltage measured by the voltmeter in Figure P 4.6-5.
Answer: 8 V
Figure P 4.6-5
Solution: Label the mesh currents:
Express the current source current in terms of the mesh currents:
i 3 − i 1 = 2 ⇒ i1 = i 3 − 2
Supermesh:
Lower, left mesh:
6 i1 + 3 i 3 − 5 ( i 2 − i 3 ) − 8 = 0 ⇒ 6 i1 − 5 i 2 + 8 i 3 = 8
−12 + 8 + 5 ( i 2 − i 3 ) = 0 ⇒ 5 i 2 = 4 + 5 i 3
Eliminating i1 and i2 from the supermesh equation:
6 ( i 3 − 2 ) − ( 4 + 5 i 3 ) + 8 i 3 = 8 ⇒ 9 i 3 = 24
⎛ 24 ⎞
The voltage measured by the meter is: 3 i 3 = 3 ⎜ ⎟ = 8 V
⎝ 9 ⎠
(checked using LNAP 8/14/02)
Determine the value of the current measured by the ammeter in Figure P 4.6-6.
P 4.6-6
Hint:
Write and solve a single mesh equation.
Answer: –5/6 A
Figure P 4.6-6
Solution:
Mesh equation for right mesh:
4 ( i − 2 ) + 2 i + 6 ( i + 3) = 0 ⇒ 12 i − 8 + 18 = 0 ⇒ i = −
10
5
A=− A
12
6
(checked using LNAP 8/14/02)
Figure P4.6-7
P4.6-7 The mesh currents are labeled in the circuit shown Figure 4.6-7. The value of these mesh currents
are:
i1 = −1.1014 A, i 2 = 0.8986 A and i 3 = −0.2899 A
a.) Determine the values of the resistances R1 and R 3 .
b.) Detemine the value of the current sourc current.
c.) Determine the value of the power supplied by the 12 V voltage source.
Solution: Label the resistor currents and the current source currrents in terms of the mesh currents:
a.) Apply KVL to the supermesh corresponding to the current source to get
R1 i1 + 12 + 24 ( i 2 − i 3 ) − 24 = 0 ⇒ R1 =
12 − 24 ( i 2 − i 3 )
i1
=
12 − 24 ( 0.8986 − ( −0.2899 ) )
−1.1014
= 15 Ω
Apply KVL to the rightmost mesh to get
R 3 i 3 + 32 − 24 ( i 2 − i 3 ) = 0 ⇒ R 3 =
b.)
−32 + 24 ( i 2 − i 3 )
i3
=
−32 + 24 ( 0.8986 − ( −0.2899 ) )
−0.2899
= 12 Ω
I s = i 2 − i1 = 0.8986 − ( −1.1014 ) = 2 A
c.) Noticing that 12 V and i 2 adhere to the passive convention, the power supplied by the 12 V voltage
source is
−12 i 2 = −12 ( 0.8986 ) = −10.783 W .
P 4.6-8
Determine values of the mesh currents, i1,
i2, and i3, in the circuit shown in Figure P 4.6-8.
Figure P 4.6-8
Solution: Use units of V, mA and kΩ. Express the currents to the supermesh to get
i1 − i 3 = 2
Apply KVL to the supermesh to get
4 ( i 3 − i 3 ) + (1) i 3 − 3 + (1) ( i1 − i 2 ) = 0
⇒
i1 − 5 i 2 + 5 i 3 = 3
Apply KVL to mesh 2 to get
2i 2 + 4 ( i 2 − i 3 ) + (1) ( i 2 − i1 ) = 0
( −1) i1 + 7i 2 − 4i 3 = 0
⇒
Solving, e.g. using MATLAB, gives
⎡ 1 0 −1⎤ ⎡ i1 ⎤ ⎡ 2 ⎤
⎢
⎥⎢ ⎥ ⎢ ⎥
⎢ 1 −5 5 ⎥ ⎢i 2 ⎥ = ⎢ 3 ⎥
⎢⎣ −1 7 −4 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ 0 ⎥⎦
⇒
⎡ i 1 ⎤ ⎡ 3⎤
⎢ ⎥ ⎢ ⎥
⎢i 2 ⎥ = ⎢1⎥
⎢ i 3 ⎥ ⎢⎣1⎥⎦
⎣ ⎦
(checked: LNAP 6/21/04)
Figure P4.6-9
P4.6-9 The mesh currents are labeled in the circuit shown Figure 4.6-9. Determine the value of the mesh
currents i1 and i 2 .
Solution: Determine the value of the mesh currents i1 and i 2 .
i 2 = i1 + 4 A , 8 i1 + 12 i1 + 5 i 2 = 0 ⇒ 8 i1 + 12 i1 + 5 ( 4 + i1 ) = 0 ⇒ i1 = −0.8 A
i 2 = i1 + 4 = −0.8 + 4 = 3.2 A
and
P 4.6-10 The mesh currents in the circuit shown in
Figure P 4.6-10 are
i1 = –2.2213 A, i2 = 0.7787 A, and i3 = 0.0770 A
(a) Determine the values of the resistances R1 and R3.
(b) Determine the value of the power supplied by the
current source.
Figure P 4.6-10
Solution:
(a)
50 ( i 3 − i 2 ) + R 3i 3 + 32 = 0 ⇒ 50 ( 0.0770 − 0.7787 ) + R 3 ( 0.0770 ) + 32 = 0
⇒ R 3 = 40 Ω
i1 R1 + 20i 2 + 50 ( i 2 − i 3 ) − 24 = 0 ⇒
R1 ( −2.2213) + 20 ( 0.7787 ) + 50 ( 0.7787 − 0.0770 ) = 24
⇒
(b)
R1 = 12 Ω
I s = i 2 − i1 = 0.7787 − ( −2.2213) = 3 A
The power supplied by the current source is
p = I s ( 24 − R1 i1 ) = 3 ( 24 − 12 ( −2.2213) ) = 152 W
(checked: LNAP 6/19/04)
P 4.6-11
Determine the value of the voltage measured by the voltmeter in Figure P 4.6-11.
Hint: Apply KVL to a supermesh to determine the current in the 2-Ω resistor.
Answer: 4/3 V
Figure P 4.6-11
Solution:
3
3
= i1 − i 2 ⇒ i1 = + i 2 .
4
4
⎛3
⎞
Apply KVL to the supermesh: −9 + 4i1 + 3 i 2 + 2 i 2 = 0 ⇒ 4 ⎜ + i 2 ⎟ + 5 i 2 = 9 ⇒ 9 i 2 = 6
⎝4
⎠
2
4
so i 2 = A and the voltmeter reading is 2 i 2 = V
3
3
Express the current source current in terms of the mesh currents:
P 4.6-12
Hint:
Determine the value of the current measured by the ammeter in Figure P 4.6-12.
Apply KVL to a supermesh.
Answer: –0.333 A
Figure P 4.6-12
Solution:
Express the current source current in terms of the mesh currents: 3 = i1 − i 2
⇒ i1 = 3 + i 2 .
Apply KVL to the supermesh: −15 + 6 i1 + 3 i 2 = 0 ⇒ 6 ( 3 + i 2 ) + 3 i 2 = 15 ⇒ 9 i 2 = −3
Finally, i 2 = −
1
A is the current measured by the ammeter.
3
Figure P4.6-13
P4.6-13 Determine the values of the mesh currents i1, i2 and i3 and the output voltage vo in the circuit
shown Figure 4.6-13.
Solution: Notice that the current source are each in a single mesh. Consequently, i1 = 2.4 A and
i2 = 1.2 A. Label the resistor currents in terms of the mesh currents:
Apply KVL to mesh 3 to get
12 i 3 + 15 − 16 (1.2 − i 3 ) − 18 ( 2.4 − i 3 ) = 0 ⇒ 46 i 3 = 47.4 ⇒ i 3 = 1.0304 A
Apply KVL to the rightmost mesh to get
v o − 15 − 12 i 3 = 0 ⇒ v o = 15 + 12 (1.0304 ) = 27.3648 V
Figure P4.6-14
P4.6-14 Determine the values of the power supplied by the sources in the circuit shown Figure P4.6-14.
Solution: First, label the mesh currents, taking advantage of the current sources. Next, express the
resistor currents in terms of the mesh currents:
Apply KVL to the middle mesh:
15 i + 25 ( i + 3) + 10 ( i − 5 ) = 0 ⇒ i = −
⎛ 1
⎞
The 5 A current source supplies 5 (10 )( i − 5 ) = 5 (10 ) ⎜ − − 5 ⎟ = 275 W
⎝ 2 ⎠
⎛ 1
⎞
The 3 A current source supplies 3 ( 25 )( i + 3) = 3 ( 25 ) ⎜ − + 3 ⎟ = 187.5 W
⎝ 2
⎠
1
A
2
Figure P4.6-15
P4.6-15 Determine the values of the resistance R and of the power supplied by the 6 A current source in
the circuit shown Figure P4.6-15.
Solution: First, label the mesh currents:
Notice that
i1 = −2.5 A, i 2 = 1 A and 6 = i 3 + 2.5 ⇒ i 3 = 3.5 A
Next, express the resistor currents in terms of the mesh currents:
Apply KVL to the bottom, left mesh:
Apply KVL to the right mesh
The 6 A current source supplies
4 ( 3.5 ) − 10 ( 2.5 ) + R (1) = 0 ⇒ R = 11 Ω
3.5 ( 5 ) + 2.5 (10 ) − v = 0 ⇒ v = 42.5 V
6 v = 6 ( 42.5 ) = 255 W
Section 4-7 Mesh Current Analysis with Dependent Sources
P 4.7-1 Find v2 for the circuit shown in
Figure P 4.7-1.
Answer: v2 = 10 V
Figure P 4.7-1
Solution:
Express the controlling voltage of the dependent source as a function of the mesh current
v2 = 50 i1
Apply KVL to the right mesh:
−100 (0.04(50i1 ) − i1 ) + 50i1 + 10 = 0 ⇒ i1 = 0.2 A
v2 = 50 i1 = 10 V
(checked using LNAP 8/14/02)
P4.7-2 Determine the values of the power supplied
by the voltage source and by the CCCS in the
circuit shown Figure P4.7-2
Figure P4.7-2
Solution: First, label the mesh currents, taking advantage of the current sources. Next, express the
resistor currents in terms of the mesh currents:
Apply KVL to the left mesh:
4000 i a + 2000 ( 6 i a ) − 2 = 0 ⇒ i a =
The 2 A voltage source supplies
The CCCS supplies
2 i a = 2 ( 0.125 × 10−3 ) = 0.25 mW
( 5 i a ) ⎡⎣( 2000 ) ( 6 i a )⎤⎦ = ( 60 ×10 )( 0.125 ×10 )
3
−3 2
1
= 0.125 mA
8
= 0.9375 × 10−3 = 0.9375 mW
P 4.7-3 Find vo for the circuit shown in
Figure P 4.7-3.
Answer: vo = 2.5 V
Figure P 4.7-3
Solution: Express the controlling current of the dependent source as a function of the mesh current:
ib = .06 − ia
Apply KVL to the right mesh:
−100 (0.06 − i a ) + 50 (0.06 − i a ) + 250 i a = 0 ⇒
ia = 10 mA
vo = 50 i b = 50 (0.06 − 0.01) = 2.5 V
Finally:
(checked using LNAP 8/14/02)
P 4.7-4 Determine the mesh current ia for the
circuit shown in Figure P 4.7-4.
Answer: ia = –24 mA
Figure P 4.7-4
Solution: Express the controlling voltage of the dependent source as a function of the mesh current:
vb = 100 (.006 − ia )
Apply KVL to the right mesh:
−100 (.006 − ia ) + 3 [100(.006 − ia )] + 250 ia = 0 ⇒ ia = −24 mA
(checked using LNAP 8/14/02)
P 4.7-5 Although scientists continue to debate exactly why and how it works, the process of utilizing
electricity to aid in the repair and growth of bones—which has been used mainly with fractures—may
soon be extended to an array of other problems, ranging from osteoporosis and osteoarthritis to spinal
fusions and skin ulcers.
An electric current is applied to bone fractures that have not healed in the normal period of time.
The process seeks to imitate natural electrical forces within the body. It takes only a small amount of
electric stimulation to accelerate bone recovery. The direct current method uses an electrode that is
implanted at the bone. This method has a success rate approaching 80 percent.
The implant is shown in Figure P 4.7-5a and the circuit model is shown in Figure P 4.7-5b. Find
the energy delivered to the cathode during a 24-hour period. The cathode is represented by the
dependent voltage source and the 100-kΩ resistor.
Figure P 4.7-5
Solution:
Apply KVL to left mesh : − 3 + 10 × 103 i1 + 20 × 103 ( i1 − i2 ) = 0 ⇒ 30 × 103 i1 − 20 × 103 i2 = 3
Apply KVL to right mesh : 5 ×103 i1 + 100 ×103 i2 + 20 × 103 ( i2 − i1 ) = 0 ⇒ i1 = 8i2
Solving (1) & ( 2 ) simultaneously
⇒ i1 =
6
3
mA, i2 =
mA
55
220
(1)
( 2)
Power delivered to cathode =
( 5 i1 ) ( i2 ) + 100 ( i2 )2
( 55)( 3 220) + 100 ( 3 220) = 0.026 mW
( 2.6 ×10−5 W ) ( 24 hr ) (3600 s hr ) = 2.25 J
2
= 5 6
∴ Energy in 24 hr. =
Figure P4.7-6
P4.7-6 Determine the value of the power supplied by the VCCS in the circuit shown Figure P4.7-6.
Solution: First, label the mesh currents.
Next, express the controlling voltage of the VCCS in terms of the mesh currents:
v a = 20 i 2
i1 = 2 A and i 3 =
va
= 10 i 2
2
Next, express the resistor currents in terms of the mesh currents:
Notice that
Apply KVL to the middle mesh:
Consequently
The VCCS supplies
20 i 2 + 2 ( i 2 − 10 i 2 ) + 8 ( i 2 − 2 ) = 0 ⇒ i 2 = 1.6 A
v a = 20 i 2 = 20 (1.6 ) = 32 V and i 3 =
va
2
=
32
= 16 A
2
⎡ 2 ( i 3 − i 2 ) ⎤ = 32 ( 2 )(16 − 1.6 ) = 460.8 W
⎦ 2
2 ⎣
va
P 4.7-7
The currents i1, i2 and i3 are the mesh
currents of the circuit shown in Figure P 4.7-7.
Determine the values of i1, i2, and i3.
Figure P 4.7-7
Solution:
Express va and ib, the controlling voltage and current of the dependent sources, in terms of the mesh
currents
v a = 5 ( i 2 − i 3 ) and i b = −i 2
Next express 20 ib and 3 va, the controlled voltages of the dependent sources, in terms of the mesh
currents
20 i b = −20 i 2 and 3 v a = 15 ( i 2 − i 3 )
Apply KVL to the meshes
−15 ( i 2 − i 3 ) + ( −20 i 2 ) + 10 i1 = 0
− ( −20 i 2 ) + 5 ( i 2 − i 3 ) + 20 i 2 = 0
10 − 5 ( i 2 − i 3 ) + 15 ( i 2 − i 3 ) = 0
These equations can be written in matrix form
⎡10 −35 15 ⎤ ⎡ i1 ⎤ ⎡ 0 ⎤
⎢ 0 45 −5 ⎥ ⎢i ⎥ = ⎢ 0 ⎥
⎢
⎥⎢ 2⎥ ⎢
⎥
⎢⎣ 0 10 −10 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ −10 ⎥⎦
Solving, e.g. using MATLAB, gives
i1 = −1.25 A, i 2 = +0.125 A, and i 3 = +1.125 A
(checked: MATLAB & LNAP 5/19/04)
P 4.7-8
Determine the value of the power supplied by
the dependent source in Figure P 4.7-8.
Figure P 4.7-8
Solution: Label the mesh currents:
Express ia, the controlling current of the CCCS,
in terms of the mesh currents
i a = i 3 − i1
Express 2 ia, the controlled current of the
CCCS, in terms of the mesh currents:
i 1 − i 2 = 2 i a = 2 ( i 3 − i 1 ) ⇒ 3 i1 − i 2 − 2 i 3 = 0
Apply KVL to the supermesh corresponding to the CCCS:
80 ( i1 − i 3 ) + 40 ( i 2 − i 3 ) + 60 i 2 + 20 i1 = 0
⇒
100i1 + 100i 2 − 120i 3 = 0
Apply KVL to mesh 3
10 + 40 ( i 3 − i 2 ) + 80 ( i 3 − i1 ) = 0
⇒
-80 i1 − 40 i 2 + 120 i 3 = −10
These three equations can be written in matrix form
−1
−2 ⎤ ⎡ i1 ⎤ ⎡ 0 ⎤
⎡ 3
⎢100 100 −120 ⎥ ⎢i ⎥ = ⎢ 0 ⎥
⎢
⎥⎢ 2⎥ ⎢
⎥
⎢⎣ −80 −40 120 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ −10 ⎥⎦
Solving, e.g. using MATLAB, gives
i1 = −0.2 A, i 2 = −0.1 A and i 3 = −0.25 A
Apply KVL to mesh 2 to get
v b + 40 ( i 2 − i 3 ) + 60 i 2 = 0 ⇒ v b = −40 ( −0.1 − ( −0.25 ) ) − 60 ( −0.1) = 0 V
So the power supplied by the dependent source is p = v b ( 2i a ) = 0 W .
(checked: LNAP 6/7/04)
P 4.7-9
Determine the value of the resistance R
in the circuit shown in Figure P 4.7-9.
Figure P 4.7-9
Solution:
Notice that i b and 0.5 mA are the mesh currents. Apply KCL at the top node of the dependent source to
get
1
mA
6
Apply KVL to the supermesh corresponding to the dependent source to get
i b + 0.5 × 10−3 = 4 i b
⇒ ib =
(
)
−5000 i b + (10000 + R ) 0.5 × 10−3 − 25 = 0
(
)
⎛1
⎞
−5000 ⎜ × 10−3 ⎟ + (10000 + R ) 0.5 × 10−3 = 25
⎝6
⎠
125
6
R=
= 41.67 kΩ
0.5 ×10−3
P 4.7-10 The circuit shown in Figure P 4.7-10 is the
small signal model of an amplifier. The input to the
amplifier is the voltage source voltage, vs. The output of
the amplifier is the voltage vo.
(a)
The ratio of the output to the input, vo/vs, is
called the gain of the amplifier. Determine the
gain of the amplifier.
(b)
The ratio of the current of the input source to the
input voltage, ib/vs, is called the input resistance
of the amplifier. Determine the input resistance.
(checked: LNAP 6/21/04)
Figure P 4.7-10
Solution: The controlling and controlled currents of the CCCS, i b and 40 i b, are the mesh currents.
Apply KVL to the left mesh to get
1000 i b + 2000 i b + 300 ( i b + 40i b ) − v s = 0
⇒
15300i b = v s
The output is given by
(a) The gain is
(b) The input resistance is
v o = −3000 ( 40 i b ) = −120000 i b
vo
vs
=−
120000
= −7.84 V/V
15300
vs
ib
= 15300 Ω
(checked: LNAP 5/24/04)
P 4.7-11
Determine the values of the mesh currents of
the circuit shown in Figure P 4.7-11.
Figure P 4.7-11
Solution:
Label the mesh currents.
Express ix in terms of the mesh currents:
i x = i1
Express 4ix in terms of the mesh currents:
4i x = i 3
Express the current source current in terms of the mesh currents to get:
0.5 = i1 − i 2
⇒
i 2 = i x − 0.5
Apply KVL to supermesh corresponding to the current source to get
5i1 + 20 ( i1 − i 3 ) + 10 ( i 2 − i 3 ) + 25i 2 = 0
Substituting gives
5i x + 20 ( −3i x ) + 10 ( i x − 0.5 − 4i x ) + 25 ( i x − 0.5 ) = 0
⇒
ix = −
35
= −0.29167
120
So the mesh currents are
i1 = i x = −0.29167 A
i 2 = i x − 0.5 = −0.79167 A
i 3 = 4i x = −1.1667 A
(checked: LNAP 6/21/04)
P 4.7-12
The currents i1, i2, and i3 are the
mesh currents corresponding to meshes 1, 2, and
3 in Figure P 4.7-12. Determine the values of
these mesh currents.
Figure P 4.7-12
Solution:
Express the controlling voltage and current of
the dependent sources in terms of the mesh
currents:
v a = R 3 ( i1 − i 2 ) and i b = i 3 − i 2
Express the current source currents in terms of
the mesh currents:
i 2 = − I s and i1 − i 3 = B i b = B ( i 3 − i 2 )
Consequently
i1 − ( B + 1) i 3 = B I s
Apply KVL to the supermesh corresponding to the dependent current source
R1 i 3 + A R 3 ( i1 − i 2 ) + R 2 ( i 3 − i 2 ) + R 3 ( i1 − i 2 ) − V s = 0
or
( A + 1) R 3 i1 − ( R 2 + ( A + 1) R3 ) i 2 + ( R1 + R 2 ) i 3 = Vs
Organizing these equations into matrix form:
⎡
0
⎢
1
⎢
⎢
⎢⎣( A + 1) R 3
1
0
− ( R 2 + ( A + 1) R 3 )
0 ⎤ ⎡ i1 ⎤ ⎡ − I s ⎤
⎥⎢ ⎥ ⎢
⎥
− ( B + 1) ⎥ ⎢i 2 ⎥ = ⎢ B I s ⎥
⎥
R1 + R 2 ⎥⎦ ⎢⎣i 3 ⎥⎦ ⎢⎣ V s ⎥⎦
With the given values:
⎡ i1 ⎤ ⎡ −0.8276 ⎤
1
0 ⎤ ⎡ i1 ⎤ ⎡ −2 ⎤
⎡0
⎢
⎥
⎢1
⎥ i = ⎢ 6 ⎥ ⇒ ⎢i ⎥ = ⎢ −2 ⎥ A
−
0
4
⎢ 2⎥ ⎢
⎢
⎥⎢ 2⎥ ⎢ ⎥
⎥
⎢ i 3 ⎥ ⎢⎣ −1.7069 ⎥⎦
⎢⎣ 60 −80 50 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ 25 ⎥⎦
⎣ ⎦
(Checked using LNAP 9/29/04)
P 4.7-13 The currents i1, i2, and i3 are the
mesh currents corresponding to meshes 1, 2,
and 3 in Figure P 4.7-13. The values of these
currents are
i1 = –1.375 A, i2 = –2.5 A, and i3 = –3.25 A
Determine the values of the gains of the
dependent source, A and B.
Figure P 4.7-13
Solution: Express the controlling voltage and current of the dependent sources in terms of the mesh
currents:
v a = 20 ( i1 − i 2 ) = 20 ( −1.375 − ( −2.5 ) ) = 22.5
and
i b = i 3 − i 2 = −3.25 − ( −2.5 ) = −0.75 A
Express the current source currents in terms of the mesh currents:
i 2 = −2.5 A
and
i 3 − i1 = B i b
⇒ − 1.375 − ( −2.5 ) = B ( −0.75 ) ⇒ B = 2.5 A/A
Apply KVL to the supermesh corresponding to the dependent current source
0 = 20 i 3 + Av a + 50 i b + v a − 10 = 20 ( −3.25 ) + A ( 22.5 ) + 50 ( −0.75 ) + 22.5 − 10 ⇒
A = 4 V/V
(Checked using LNAP 9/29/04)
P 4.7-14 Determine the current i in the circuit shown in
Figure P 4.7-14.
Answer: i = 3 A
Figure P 4.7-14
Solution:
Label the node voltages as shown. The controlling currents of the CCCS is expressed as i =
The node equations are
and
12 =
va
28
+
va − vb
4
va − vb
4
+
va
14
=
+
va
28
.
va
14
vb
8
Solving the node equations gives v a = 84 V and v b = 72 V . Then i =
va
28
=
84
=3A .
28
(checked using LNAP 6/16/05)
P4.7-15 Determine the values of the mesh currents i1 and i2 for the circuit shown in Figure P4.7-15.
Figure P4.7-15
Solution: Expressing the dependent source currents in terms of the mesh currents we get:
i1 = 4 i a = 4 ( i 2 + 1) ⇒ 4 = i1 − 4 i 2
Apply KVL to mesh 2 to get
2 i 2 + 2 ( i 2 + 1) − 2 ( i1 − i 2 ) = 0 ⇒ − 2 = −2 i1 + 6 i 2
Solving these equations using MATLAB we
get
i1 = −8 A and i2 = −3 A
P4.7-16 Determine the values of the mesh currents i1 and i2 for the circuit shown in Figure P4.7-16 .
Figure P4.7-16
Apply KVL to mesh 1 to get
2 i1 + 4 i a + 2 ( i1 − i 2 ) − 12 = 0 ⇒ 2 i1 + 4 ( i 2 + 1) + 2 ( i1 − i 2 ) − 12 = 0 ⇒ 8 = 4 i1 + 2 i 2
Apply KVL to mesh 2 to get
2 i 2 + 2 ( i 2 + 1) − 2 ( i1 − i 2 ) = 0 ⇒ − 2 = −2 i1 + 6 i 2
Solving these equations using MATLAB we
get
i1 = 1.8571 A and i2 = 0.2857 A
Section 4.8 The Node Voltage Method and Mesh Current Method Compared
P 4.8-2 The circuit shown in Figure P 4.8-2 has two
inputs, vs and is, and one output vo. The output is related
to the inputs by the equation
vo = ais + bvs
where a and b are constants to be determined. Determine
the values a and b by
(a) writing and solving mesh equations and
(b) writing and solving node equations.
Figure P 4.8-2
Solution:
(a)
Apply KVL to meshes 1 and 2:
32i1 − v s + 96 ( i1 − i s ) = 0
v s + 30i 2 + 120 ( i 2 − i s ) = 0
150i 2 = +120i s − v s
vs
4
i2 = is −
5
150
1
v o = 30i 2 = 24i s − v s
5
So a = 24 and b = -.02.
(b)
Apply KCL to the supernode corresponding to
the voltage source to get
va − (vs + vo )
96
So
is =
+
vs + vo
120
va − vo
32
+
vo
30
=
=
vs + vo
120
vs
120
+
+
vo
30
vo
24
Then
1
v o = 24i s − v s
5
So a = 24 and b = -0.2.
(checked: LNAP 5/24/04)
P 4.8-3 Determine the power supplied by the dependent
source in the circuit shown in Figure P 4.8-3 by writing and
solving
(a) node equations and
(b) mesh equations.
Figure P 4.8-3
Solution:
(a) Label the reference node and node voltages.
v b = 120 V
due to the voltage source.
Apply KCL at the node between the resistors to get
vb − va
50
=
va
10
⇒
v a = 20 V
and the power supplied by the dependent source is
p = v bi a = (120 ) ⎡⎣0.2 ( 20 ) ⎤⎦ = 480 W
(b) Label the mesh currents. Express the controlling voltage
of the dependent source in terms of the mesh current to get
v a = 10 ( i 2 − i1 )
Express the controlled current of the dependent source in
terms of the mesh currents to get
−i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1
⇒
i1 = 2i 2
Apply KVL to the bottom mesh to get
50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0
⇒
⇒
i 2 − i1 = 2
So
i 2 − 2i 2 = 2
i 2 = −2 A
⇒
i 1 = −4 A
Then
v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A
The power supplied by the dependent source is
p = 120 ( i a ) = 120 ( 4 ) 480 W
(checked: LNAP 6/21/04)
Section 4.8 The Node Voltage Method and Mesh Current Method Compared
P 4.8-2 The circuit shown in Figure P 4.8-2 has two
inputs, vs and is, and one output vo. The output is related
to the inputs by the equation
vo = ais + bvs
where a and b are constants to be determined. Determine
the values a and b by
(a) writing and solving mesh equations and
(b) writing and solving node equations.
Figure P 4.8-2
Solution:
(a)
Apply KVL to meshes 1 and 2:
32i1 − v s + 96 ( i1 − i s ) = 0
v s + 30i 2 + 120 ( i 2 − i s ) = 0
150i 2 = +120i s − v s
vs
4
i2 = is −
5
150
1
v o = 30i 2 = 24i s − v s
5
So a = 24 and b = -.02.
(b)
Apply KCL to the supernode corresponding to
the voltage source to get
va − (vs + vo )
96
So
is =
+
vs + vo
120
va − vo
32
+
vo
30
=
=
vs + vo
120
vs
120
+
+
vo
30
vo
24
Then
1
v o = 24i s − v s
5
So a = 24 and b = -0.2.
(checked: LNAP 5/24/04)
P 4.8-3 Determine the power supplied by the dependent
source in the circuit shown in Figure P 4.8-3 by writing and
solving
(a) node equations and
(b) mesh equations.
Figure P 4.8-3
Solution:
(a) Label the reference node and node voltages.
v b = 120 V
due to the voltage source.
Apply KCL at the node between the resistors to get
vb − va
50
=
va
10
⇒
v a = 20 V
and the power supplied by the dependent source is
p = v bi a = (120 ) ⎡⎣0.2 ( 20 ) ⎤⎦ = 480 W
(b) Label the mesh currents. Express the controlling voltage
of the dependent source in terms of the mesh current to get
v a = 10 ( i 2 − i1 )
Express the controlled current of the dependent source in
terms of the mesh currents to get
−i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1
⇒
i1 = 2i 2
Apply KVL to the bottom mesh to get
50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0
⇒
⇒
i 2 − i1 = 2
So
i 2 − 2i 2 = 2
i 2 = −2 A
⇒
i 1 = −4 A
Then
v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A
The power supplied by the dependent source is
p = 120 ( i a ) = 120 ( 4 ) 480 W
(checked: LNAP 6/21/04)
Section 4.8 The Node Voltage Method and Mesh Current Method Compared
P 4.8-2 The circuit shown in Figure P 4.8-2 has two
inputs, vs and is, and one output vo. The output is related
to the inputs by the equation
vo = ais + bvs
where a and b are constants to be determined. Determine
the values a and b by
(a) writing and solving mesh equations and
(b) writing and solving node equations.
Figure P 4.8-2
Solution:
(a)
Apply KVL to meshes 1 and 2:
32i1 − v s + 96 ( i1 − i s ) = 0
v s + 30i 2 + 120 ( i 2 − i s ) = 0
150i 2 = +120i s − v s
vs
4
i2 = is −
5
150
1
v o = 30i 2 = 24i s − v s
5
So a = 24 and b = -.02.
(b)
Apply KCL to the supernode corresponding to
the voltage source to get
va − (vs + vo )
96
So
is =
+
vs + vo
120
va − vo
32
+
vo
30
=
=
vs + vo
120
vs
120
+
+
vo
30
vo
24
Then
1
v o = 24i s − v s
5
So a = 24 and b = -0.2.
(checked: LNAP 5/24/04)
P 4.8-3 Determine the power supplied by the dependent
source in the circuit shown in Figure P 4.8-3 by writing and
solving
(a) node equations and
(b) mesh equations.
Figure P 4.8-3
Solution:
(a) Label the reference node and node voltages.
v b = 120 V
due to the voltage source.
Apply KCL at the node between the resistors to get
vb − va
50
=
va
10
⇒
v a = 20 V
and the power supplied by the dependent source is
p = v bi a = (120 ) ⎡⎣0.2 ( 20 ) ⎤⎦ = 480 W
(b) Label the mesh currents. Express the controlling voltage
of the dependent source in terms of the mesh current to get
v a = 10 ( i 2 − i1 )
Express the controlled current of the dependent source in
terms of the mesh currents to get
−i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1
⇒
i1 = 2i 2
Apply KVL to the bottom mesh to get
50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0
⇒
⇒
i 2 − i1 = 2
So
i 2 − 2i 2 = 2
i 2 = −2 A
⇒
i 1 = −4 A
Then
v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A
The power supplied by the dependent source is
p = 120 ( i a ) = 120 ( 4 ) 480 W
(checked: LNAP 6/21/04)
Section 4.8 The Node Voltage Method and Mesh Current Method Compared
P 4.8-2 The circuit shown in Figure P 4.8-2 has two
inputs, vs and is, and one output vo. The output is related
to the inputs by the equation
vo = ais + bvs
where a and b are constants to be determined. Determine
the values a and b by
(a) writing and solving mesh equations and
(b) writing and solving node equations.
Figure P 4.8-2
Solution:
(a)
Apply KVL to meshes 1 and 2:
32i1 − v s + 96 ( i1 − i s ) = 0
v s + 30i 2 + 120 ( i 2 − i s ) = 0
150i 2 = +120i s − v s
vs
4
i2 = is −
5
150
1
v o = 30i 2 = 24i s − v s
5
So a = 24 and b = -.02.
(b)
Apply KCL to the supernode corresponding to
the voltage source to get
va − (vs + vo )
96
So
is =
+
vs + vo
120
va − vo
32
+
vo
30
=
=
vs + vo
120
vs
120
+
+
vo
30
vo
24
Then
1
v o = 24i s − v s
5
So a = 24 and b = -0.2.
(checked: LNAP 5/24/04)
P 4.8-3 Determine the power supplied by the dependent
source in the circuit shown in Figure P 4.8-3 by writing and
solving
(a) node equations and
(b) mesh equations.
Figure P 4.8-3
Solution:
(a) Label the reference node and node voltages.
v b = 120 V
due to the voltage source.
Apply KCL at the node between the resistors to get
vb − va
50
=
va
10
⇒
v a = 20 V
and the power supplied by the dependent source is
p = v bi a = (120 ) ⎡⎣0.2 ( 20 ) ⎤⎦ = 480 W
(b) Label the mesh currents. Express the controlling voltage
of the dependent source in terms of the mesh current to get
v a = 10 ( i 2 − i1 )
Express the controlled current of the dependent source in
terms of the mesh currents to get
−i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1
⇒
i1 = 2i 2
Apply KVL to the bottom mesh to get
50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0
⇒
⇒
i 2 − i1 = 2
So
i 2 − 2i 2 = 2
i 2 = −2 A
⇒
i 1 = −4 A
Then
v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A
The power supplied by the dependent source is
p = 120 ( i a ) = 120 ( 4 ) 480 W
(checked: LNAP 6/21/04)
Section 4.9 Circuit Analysis Using MATLAB
Figure P4.9-1
P4.9-1. The encircled numbers in the circuit shown Figure P4.9-1 are node numbers. Determine the
values of the corresponding node voltages, v1, v2 and v3.
Solution: First, express the resistor currents in terms of the node voltages:
Apply KCL at node 1 to get 5 =
Apply KCL at node 2 to get
Apply KCL at node 3 to get
In matrix form:
Solving using MATLAB:
v1 − v 2
5
v1 − v 2
5
v2 − v3
4
=
+
+
v2
10
v1 − v 3
2
+
v2 − v3
v1 − v 3
2
4
⇒ 0.7 v1 − 0.2 v 2 − 0.5 v 3 = 5
⇒ − 0.2 v1 + 0.55 v 2 − 0.25 v 3 = 0
= 3 ⇒ − 0.5 v1 − 0.25 v 2 + 0.75 v 3 = −3
⎡ 0.7 −0.2 −0.5 ⎤ ⎡ v1 ⎤ ⎡ 5 ⎤
⎢ −0.2 0.55 −0.25⎥ ⎢v ⎥ = ⎢ 0 ⎥
⎢
⎥ ⎢ 2⎥ ⎢ ⎥
⎢⎣ −0.5 −0.25 0.75 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ −3⎥⎦
v1 = 28.1818 V, v 2 = 20 V and v 3 = 21.4545
Figure P4.9-2
P4.9-2. Determine the values of the node voltages, v1 and v2, in the circuit shown Figure P4.9-2.
Solution: First, express the resistor currents in terms of the node voltages:
Apply KCL at node 1 to get
Apply KCL at node 2 to get
In matrix form:
Solving using MATLAB:
15 − v1
25
v1 − v 2
50
=
=
v1 − v 2
50
v2
10
+
+
v1 + 8
20
v2 + 8
40
⇒ 0.11v1 − 0.02 v 2 = 0.2
⇒ − 0.02 v1 + 0.145 v 2 = −0.2
⎡ 0.11 −0.02 ⎤ ⎡ v1 ⎤ ⎡ 0.2 ⎤
⎢ −0.02 0.145 ⎥ ⎢v ⎥ = ⎢ −0.2 ⎥
⎣
⎦ ⎣ 2⎦ ⎣
⎦
v1 = 1.6077 V and v 2 = −1.1576 V
Figure P4.9-3
P4.9-3. Determine the values of the node voltages, v1, v2 and v3 in the circuit shown in Figure P4.9-3.
Solution: First, express the resistor currents in terms of the node voltages:
Apply KCL at node 1 to get
Apply KCL at node 2 to get
Apply KCL at node 3 to get
In matrix form:
Solving using MATLAB:
−15 − v1
25
v1 − v 2
20
v1 − v 2
50
=
+
=
v1 − v 2
20
v2
25
+
v1 − v 3
50
v2 − v3
v 2 − v3
40
+
40
=
⇒ − 0.05 v1 + 0.115 v 2 − 0.025 v 3 = 0
v 3 − 10
10
⇒ 0.11v1 − 0.05 v 2 − 0.02 v 3 = −0.6
⇒ − 0.02 v1 − 0.025 v 2 + 0.145 v 3 = 1
⎡ 0.11 −0.05 −0.02 ⎤ ⎡ v1 ⎤ ⎡ −0.6 ⎤
⎢ −0.05 0.115 −0.025⎥ ⎢v ⎥ = ⎢ 0 ⎥
⎢
⎥ ⎢ 2⎥ ⎢
⎥
⎢⎣ −0.02 −0.025 0.145 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ 1 ⎥⎦
v1 = 1.6077 V and v 2 = −1.1576 V
P4.9-4 Determine the node voltages, v1 and v2, for the circuit shown in Figure P4.9-4.
Figure P.4.9-4
Solution: Emphasize and label the nodes:
Notice the 24 V source connected between node 3 and the reference node. Consequently
v 3 = 24 V
Apply KCL at node 1 to get
v1
8
v1
+
v1 − v 2
25
+2=0
is the current directed downward in the 8 Ω resistor and
v1 − v 2
is the current
8
25
directed from left to right in the 25 Ω resistor. We will simplify this equation by doing two things:
In this equation
1. Multiplying each side by 8 × 25 = 200 to eliminate fractions.
2. Move the terms that don’t involve node voltages to the right side of the equation.
The result is
33 v1 − 8 v 2 = −400
Next, apply KCL at node 2 to get
v2
9
+
v 2 − 24
14
=
v1 − v 2
25
In this equation
v2
9
is the current directed downward in the 9 Ω resistor,
v1 − v 2
v 2 − 24
14
is the current directed
is the current directed from left to right in the 25 Ω
25
resistor. We will simplify this equation by doing two things:
from left to right in the 14 Ω resistor and
1. Multiplying each side by 8 × 25 × 14 = 2800 to eliminate fractions.
2. Move the terms that involve node voltages to the left side of the equation and move the terms
that don’t involve node voltages to the right side of the equation.
The result is
− ( 9 ×14 ) v1 + ( 9 × 14 + 25 × 14 + 25 × 9 ) v 2 = 24 × 25 × 9 ⇒ − 126 v1 + 701v 2 = 5400
The simultaneous equations can be written in matrix form
33 v1 − 8 v 2 = −400
−8 ⎤ ⎡ v1 ⎤ ⎡ −400 ⎤
⎡ 33
⇒ ⎢
⎥⎢ ⎥ = ⎢
⎥
−126 v1 + 701 v 2 = 5400
⎣ −126 701⎦ ⎣ v 2 ⎦ ⎣ 5400 ⎦
We can use MATLAB to solve the matrix equation:
Then
⎡ v1 ⎤ ⎡ −10.7209 ⎤
⎢v ⎥ = ⎢
⎥
⎣ 2 ⎦ ⎣ 5.7763 ⎦
That is, the node voltages are v1 = −10.7209 V and v 2 = 5.7763 V .
P4.9-5 Determine the mesh currents, i1 and i2, for the circuit shown in Figure P4.9-5.
Figure P4.9.5
Solution: Label the label the mesh currents. Then, label the element currents in terms of the mesh
currents:
Notice that the 2 A source on the outside of the circuit is in mesh 3 and that the currents 2 A and i 3 have
the same direction. Consequently
i3 = 2 A
Apply KVL to mesh 1 to get
25 ( i1 − i 3 ) + 9 ( i1 − i 2 ) + 8 i1 = 0
In this equation 25 ( i1 − i 3 ) is the voltage across the 25 Ω resistor (+ on the left), 9 ( i1 − i 2 ) is the
voltage across the 9 Ω resistor (+ on top) and 8i1 is the voltage across the 8 Ω resistor (+ on bottom).
Substituting i 3 = 2 A and doing a little algebra gives
42 i1 − 9 i 2 = 50
Next, apply KVL to mesh 2 to get
14 ( i 2 − i 3 ) + 24 − 9 ( i1 − i 2 ) = 0
In this equation 14 ( i 2 − i 3 ) is the voltage across the 14 Ω resistor (+ on the left), 24 is the voltage source
voltage and 9 ( i1 − i 2 ) is the voltage across the 9 Ω resistor (+ on top). Substituting i 3 = 2 A and doing
a little algebra gives
−9 i1 + 23 i 2 = −24 + 14 ( 2 ) = 4
The simultaneous equations can be written in matrix form
42 i1 − 9 i 2 = 50
⎡ 42 −9 ⎤ ⎡ i1 ⎤ ⎡50 ⎤
⇒ ⎢
⎥⎢ ⎥ = ⎢ ⎥
−9 i1 + 23 i 2 = 4
⎣ −9 23 ⎦ ⎣i 2 ⎦ ⎣ 4 ⎦
We can use MATLAB to solve the matrix equation:
Then
⎡ i1 ⎤ ⎡1.3401 ⎤
⎢ ⎥=⎢
⎥
⎣i 2 ⎦ ⎣0.6983⎦
That is, the mesh currents are i1 = 1.3401 A and i 2 = 0.6983 A .
P4.9-6 Represent the circuit shown in Figure P4.9-6 by the matrix equation
⎡ a 11
⎢
⎣ a 21
a 12 ⎤ ⎡ v1 ⎤ ⎡ −40 ⎤
=
a 22 ⎦⎥ ⎣⎢ v 2 ⎦⎥ ⎢⎣ −228⎥⎦
Determine the values of the coefficients a11 , a12 , a 21 and a 22 .
Figure P4.9-6
Solution: Emphasize and label the nodes:
Noticing the 10 V source connected between node 3 and the reference node, we determine that node
voltage at node 3 is
v 3 = −10 V
Apply KCL at node 1 to get
In this equation
v1
v1
10
+ 0.4 +
v1 − v 2
10
=0
is the current directed downward in the vertical 10 Ω resistor and
v1 − v 2
is the
10
10
current directed from left to right in the horizontal 10 Ω resistor. We will simplify this equation by doing
two things:
1. Multiplying each side by 10 to eliminate fractions.
2. Move the terms that don’t involve node voltages to the right side of the equation.
2 v1 − v 2 = −4
The result is
Next, apply KCL at node 2 to get
v2
19
In this equation
v2
19
+
v 2 − ( −10 )
22
=
v1 − v 2
10
+ 0.4
is the current directed downward in the 19 Ω resistor,
directed from left to right in the 22 Ω resistor and
v 2 − ( −10 )
v1 − v 2
22
is the current
is the current directed from left to right in the
10
horizontal 10 Ω resistor. We will simplify this equation by doing two things:
1. Multiplying each side by 19 × 22 × 10 = 4180 to eliminate fractions.
2. Move the terms that involve node voltages to the left side of the equation and move the terms
that don’t involve node voltages to the right side of the equation.
The result is
− (19 × 22 ) v1 + (19 × 10 + 22 ×10 + 19 × 22 ) v 2 = −10 × 10 × 19 + 0.4 × 19 × 22 × 10
⇒ − 418 v1 + 828 v 2 = −228
Comparing our equations to the given equations, we see that we need to multiply both sides of our first
equation by 10. Then
20 v1 − 10 v 2 = −40
⎡ 20 −10 ⎤ ⎡ v1 ⎤ ⎡ −40 ⎤
⇒ ⎢
⎥⎢ ⎥ = ⎢
⎥
−418 v1 + 828 v 2 = −228
⎣ −418 828 ⎦ ⎣v 2 ⎦ ⎣ −228⎦
Comparing coefficients gives
a 11 = 20 , a 12 = −10 , a 21 = −418 and a 22 = 828 .
P4.9-7 Represent the circuit shown in Figure P4.9-7 by the matrix equation
⎡ a 11
⎢
⎣ a 21
a12 ⎤ ⎡ i1 ⎤ ⎡ 4 ⎤
=
a 22 ⎦⎥ ⎣⎢i 2 ⎦⎥ ⎢⎣10 ⎥⎦
Determine the values of the coefficients a11 , a12 , a 21 and a 22 .
Figure P4.9-7
Solution: Label the label the mesh currents. Then, label the element currents in terms of the mesh
currents:
Notice that the 0.4 A source on the inside of the circuit is in both mesh 1 and mesh 3. Mesh current i1 is
directed in the same way as current source current but the mesh current i 3 is directed opposite to the
current source current. Consequently
i1 − i 3 = 0.4 A
The current source is in both mesh 1 and mesh 3 so we apply KVL to the supermesh corresponding to
the current source (i.e. the perimeter of meshes 1 and 3). The result is
10 i 3 + 19 ( i1 − i 2 ) + 10 i1 = 0
In this equation 10i 3 is the voltage across the horizontal 10 Ω resistor (+ on the left), 19 ( i1 − i 2 ) is the
voltage across the 19 Ω resistor (+ on top) and 10i1 is the voltage across the vertical 10 Ω resistor (+ on
bottom). Substituting i 3 = i1 − 0.4 and doing a little algebra gives
39 i1 − 19 i 2 = 4
Next, apply KVL to mesh 2 to get
22 i 2 − 10 − 19 ( i1 − i 2 ) = 0
In this equation 22i 2 is the voltage across the 22 Ω resistor (+ on the left), 10 is the voltage source
voltage and 19 ( i1 − i 2 ) is the voltage across the 19 Ω resistor (+ on top). Doing a little algebra gives
−19 i1 + 41i 2 = 10
To summarize, the circuit is represented by the simultaneous equations:
39 i1 − 19 i 2 = 4
⎡ 39 −19 ⎤ ⎡ i1 ⎤ ⎡ 4 ⎤
⇒ ⎢
⎥⎢ ⎥ = ⎢ ⎥
−19 i1 + 41i 2 = 10
⎣ −19 41 ⎦ ⎣i 2 ⎦ ⎣10 ⎦
Comparing these equations to the given equations shows
a 11 = 39 , a 12 = −19 , a 21 = −19 and a 22 = 41 .
P4.9-8 Determine the values of the power supplied by each of the sources for the circuit shown in Figure
P4.9-8.
Figure P.4.9-8
Solution: First, label the mesh currents and then label the element currents:
Notice the 2.4 A source in both mesh 2 and mesh 3. We have
i 3 − i 2 = 2.4 A
Apply KVL to mesh 1 to get
40 i1 − 5 ( i 3 − i1 ) − 5 ( i 2 − i1 ) = 0 ⇒ 50 i1 − 5 i 2 − 5 i 3 = 0
Identify the supermesh corresponding to the 2.4 A current source:
Apply KVL to the supermesh to get
5 ( i 2 − i1 ) + 5 ( i 3 − i1 ) + 24 + 40 i 2 = 0 ⇒ − 10 i1 + 45 i 2 + 5 i 3 = −24
Writing the mesh equations in matrix form gives
⎡ 0 −1 1 ⎤ ⎡ i1 ⎤ ⎡ 2.4 ⎤
⎢ 50 −5 −5⎥ ⎢i ⎥ = ⎢ 0 ⎥
⎢
⎥⎢ 2⎥ ⎢
⎥
⎢⎣ −10 45 5 ⎥⎦ ⎢⎣i 3 ⎥⎦ ⎢⎣ −24 ⎥⎦
Solving using MATLAB:
That is, the mesh currents are i1 = 0.1 A, i 2 = −0.7 A and i 3 = 1.7 A.
The 24 V source supplies
−24 i 3 = ( −24 )(1.7 ) = −40.8 W
The power supplied by the current source depends on vs, the voltage across the current source. Apply
KVL to mesh 3 to get
5 ( i 3 − i1 ) + 24 − v s = 0 ⇒ v s = 5 (1.7 − 0.1) + 24 = 32 V
The current source supplies
2.4 vs = 2.4 ( 32 ) = 76.8 W
Figure P4.9-9
P4.9-9 The mesh currents are labeled in the circuit shown Figure 4.9-9. Determine the value of the mesh
currents i1 and i 2 .
Solution: Determine the value of the mesh currents i1 and i 2 .
Replace series resistors with an equivalent resistor and series voltage sources with and equivalent
voltage source to get
Apply KVL to mesh 1
16 i1 + 8 ( i1 − i 2 ) − 9 = 0 ⇒ 24 i1 − 8 i 2 = 9
Apply KVL to mesh 2
4 i 2 + 5 i1 − 8 ( i1 − i 2 ) = 0 ⇒ − 3 i1 + 12 i 2 = 0
In matrix form
Solving using MATLAB
⎡ 24 −8⎤ ⎡ i1 ⎤ ⎡9 ⎤
⎢ −3 12 ⎥ ⎢i ⎥ = ⎢ 0 ⎥
⎣
⎦⎣ 2⎦ ⎣ ⎦
So the mesh currents are
i1 = 0.4091 A and i 2 = 0.1023 A
Figure P4.9-10
P4.9-10 The encircled numbers in the circuit shown Figure P4.9-10 are node numbers. Determine the
values of the corresponding node voltages, v1 and v2.
Solution: Determine the value of the node voltages, v1 and v2.
Replace parallel resistors with an equivalent resistor and parallel sources with and equivalent current
source to get
Apply KCL at node 1
Apply KCL at node 2
In matrix form
Solving using MATLAB
2.5 =
v1
4.44
v1 − v 2
2
1
⎡ 1
⎢ 4.44 + 2
⎢
⎢ 1 + 1.5
⎣⎢ 2
+
v1 − v 2
2
+ 1.5 v1 =
=0
v2
4
1 ⎤
2 ⎥ ⎡ v1 ⎤ ⎡ 2.5⎤
⎥⎢ ⎥ =
1 1 ⎥ ⎣v 2 ⎦ ⎢⎣ 0 ⎥⎦
− −
2 4 ⎦⎥
−
So the node voltages are
v1 = −4.1111 V and v 2 = −10.9630 V
Section 4.10 How Can We Check … ?
P 4.10-1 Computer analysis of the circuit shown in
Figure P 4.10-1 indicates that the node voltages are
va = 5.2 V, vb = –4.8 V, and vc = 3.0 V.
Is this analysis correct?
Hint: Use the node voltages to calculate all the element
currents. Check to see that KCL is satisfied at each node.
Figure P 4.10-1
Solution:
Apply KCL at node b:
vb − va
v −v
1
−
+ b c = 0
4
2
5
−4.8 − 5.2
1 − 4.8 − 3.0
− +
≠0
4
2
5
The given voltages do not satisfy the KCL equation at node b. They are not correct.
P 4.10-2 An old lab report asserts that the node
voltages of the circuit of Figure P 4.10-2 are
va = 4 V, vb = 20 V, and vc = 12 V.
Are these correct?
Solution:
Apply KCL at node a:
v
⎛v −v ⎞
−⎜ b a ⎟ − 2 + a = 0
2
⎝ 4 ⎠
4
⎛ 20 − 4 ⎞
−⎜
= −4≠ 0
⎟−2+
2
⎝ 4 ⎠
The given voltages do not satisfy the KCL equation at node a. They are not correct.
P 4.10-3
Your lab partner forgot to record the values of R1,
R2, and R3. He thinks that two of the resistors in Figure P
4.10-3 had values of 10 kΩ and that the other had a value of
5 kΩ. Is this possible? Which resistor is the 5-kΩ resistor?
Figure P 4.10-3
Solution:
⎛ 12 − 7.5 ⎞ 7.5 7.5 − 6
−⎜
+
=0
⎟+
R2
⎝ R1 ⎠ R3
Writing a node equation:
4.5 7.5 1.5
+
+
=0
R1 R3 R2
There are only three cases to consider. Suppose R1 = 5 kΩ and R 2 = R 3 = 10 kΩ. Then
−
So
−
4.5 7.5 1.5 −0.9 + 0.75 + 0.15
+
+
=
= 0
R1 R3 R2
1000
This choice of resistance values corresponds to branch currents that satisfy KCL. Therefore, it is indeed
possible that two of the resistances are 10 kΩ and the other resistance is 5 kΩ. The 5 kΩ is R1.
P 4.10-4 Computer analysis of the circuit
shown in Figure P 4.10-4 indicates that the mesh
currents are
i1 = 2 A, i2 = 4 A, and i3 = 3 A.
Verify that this analysis is correct.
Hint: Use the mesh currents to calculate the
element voltages. Verify that KVL is satisfied for
each mesh.
Figure P 4.10-4
Solution: Applying KVL to each mesh:
Top mesh:
Bottom right mesh
10 (2 − 4) + 12(2) + 4 (2 − 3) = 0
8 (3 − 4) + 4 (3 − 2) + 4 = 0
Bottom, left mesh:
28 + 10 (4 − 2) + 8 (4 − 3) ≠ 0
(Perhaps the polarity of the 28 V source was entered incorrectly.)
KVL is not satified for the bottom, left mesh so the computer analysis is not correct.
PSpice Problems
SP 4-1
Use PSpice to determine the node voltages of the circuit shown in Figure SP 4-1
Figure SP 4-1
Solution: The PSpice schematic after running a “Bias Point” simulation:
SP 4-2
Use PSpice to determine the mesh currents of the circuit shown in Figure SP 4-2.
Figure SP 4-2
Solution: The PSpice schematic after running a “Bias Point” simulation:
From the PSpice output file:
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V1
V_V2
-3.000E+00
-2.250E+00
V_V3
-7.500E-01
The voltage source labeled V3 is a short circuit used to measure the mesh current. The mesh
currents are i1 = −3 A (the current in the voltage source labeled V1) and i2 = −0.75 A (the current
in the voltage source labeled V3).
SP 4-3
The voltages va, vb, vc, and vd in Figure SP 4-3 are the node voltages corresponding to
nodes a, b, c and d. The current i is the current in a short circuit connected between nodes b and
c. Use PSpice to determine the values of va, vb, vc, and vd and of i.
Solution: The PSpice schematic after running a “Bias Point” simulation:
The PSpice output file:
**** INCLUDING sp4_2-SCHEMATIC1.net ****
* source SP4_2
V_V4
0 N01588 12Vdc
R_R4
N01588 N01565 4k
V_V5
N01542 N01565 0Vdc
R_R5
0 N01516 4k
V_V6
N01542 N01516 8Vdc
I_I1
0 N01565 DC 2mAdc
I_I2
0 N01542 DC 1mAdc
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V4
V_V5
V_V6
-4.000E-03
2.000E-03
-1.000E-03
From the PSpice schematic: va = −12 V, vb = vc = 4 V, vd = −4 V. From the output file: i = 2 mA.
SP 4-4
Determine the current, i, shown in Figure SP 4-4.
Answer: i = 0.56 A
Figure SP 4-4
Solution: The PSpice schematic after running a “Bias Point” simulation:
The PSpice output file:
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V7
V_V8
-5.613E-01
-6.008E-01
The current of the voltage source labeled V7 is also the current of the 2 Ω resistor at the top of
the circuit. However this current is directed from right to left in the 2 Ω resistor while the current
i is directed from left to right. Consequently, i = +5.613 A.
Design Problems
DP 4-1 An electronic instrument incorporates a 15-V power supply. A digital display is added
that requires a 5-V power supply. Unfortunately, the project is over budget and you are
instructed to use the existing power supply. Using a voltage divider, as shown in Figure DP 4-1,
you are able to obtain 5 V. The specification sheet for the digital display shows that the display
will operate properly over a supply voltage range of 4.8 V to 5.4 V. Furthermore, the display will
draw 300 mA (I) when the display is active and 100 mA when quiescent (no activity).
(a)
Select values of R1 and R2 so that the display will be supplied with 4.8 V to 5.4 V under
all conditions of current I.
(b)
Calculate the maximum power dissipated by each resistor, R1 and R2, and the maximum
current drawn from the 15-V supply.
(c)
Is the use of the voltage divider a good engineering solution? If not, why? What problems
might arise?
Figure DP 4-1
Solution:
Model the circuit as:
(a) We need to keep v2 across R2 in the range 4.8 ≤ v2 ≤ 5.4
display is active
⎧0.3 A
For I = ⎨
⎩ 0.1 A display is not active
KCL at a:
v2 − 15 v2
+
+I =0
R1
R2
Assumed that maximum I results in minimum v2 and visa-versa.
Then
⎧4.8 V
v2 = ⎨
⎩5.4 V
when I = 0.3 A
when I = 0.1 A
Substitute these corresponding values of v2 and I into the KCL equation and solve for the resistances
4.8 − 15 4.8
+
+ 0.3 = 0
R1
R2
5.4 − 15 5.4
+
+ 0.1 = 0
R1
R2
⇒ R1 = 7.89 Ω, R2 = 4.83 Ω
(b)
15 − 4.8
= 1.292 A ⇒ PR
= (1.292)2 (7.89) = 13.17 W
1max
1max
7.89
5.4 )2
(
5.4
IR
=
= 1.118 A
⇒ PR
=
= 6.03 W
2max
2 max
4.83
4.83
maximum supply current = I R
= 1.292 A
IR
=
1max
(c) No; if the supply voltage (15V) were to rise or drop, the voltage at the display would drop
below 4.8V or rise above 5.4V.
The power dissipated in the resistors is excessive. Most of the power from the supply is
dissipated in the resistors, not the display.
DP 4-2 For the circuit shown in Figure DP 4-2, it is desired to set the voltage at node a equal to
0 V in order to control an electric motor. Select voltages v1 and v2 in order to achieve va = 0 V
when v1 and v2 are less than 20 V and greater than zero and R = 2 Ω.
Figure DP 4-2
Solution:
Express the voltage of the 8 V source in terms of its node voltages to get vb − va = 8 . Apply KCL
to the supernode corresponding to the 8 V source:
va − v1
R
+
va vb vb − ( −v 2 )
+ +
= 0 ⇒ 2 va − v1 + 2 vb + v 2 = 0
R R
R
⇒ 2 va − v1 + 2 ( va + 8 ) + v 2 = 0
⇒ 4 va − v1 + v 2 + 16 = 0
⇒ va =
Next set va = 0 to get
0=
v1 − v 2
For example, v1 = 18 V and v2 = 2 V.
4
v1 − v 2
4
− 4 ⇒ v1 − v 2 = 16 V
−4
DP 4-3 A wiring circuit for a special lamp in a
home is shown in Figure DP 4-3. The lamp has a
resistance of 2 Ω, and the designer selects R = 100
Ω. The lamp will light when I ≥ 50 mA but will
Figure DP 4-3
burn out when I > 75 mA.
(a)
Determine the current in the lamp and determine if it will light for R = 100 Ω.
(b)
Select R so that the lamp will light but will not burn out if R changes by ± 10 percent
because of temperature changes in the home.
Solution:
(a)
Apply KCL to left mesh:
−5 + 50 i1 + 300 (i1 − I ) = 0
Apply KCL to right mesh:
( R + 2) I + 300 ( I − i1 ) = 0
150
1570 + 35 R
We desire 50 mA ≤ I ≤ 75 mA so if R = 100 Ω, then I = 29.59 mA ⇒ l amp so the lamp will not
light.
Solving for I:
I=
(b) From the equation for I, we see that decreasing R increases I:
try R = 50 Ω ⇒ I = 45 mA (won't light)
try R = 25Ω ⇒ I = 61 mA ⇒ will light
Now check R±10% to see if the lamp will light and not burn out:
−10% → 22.5Ω → I = 63.63 mA ⎫ lamp will
⎬
+10% → 27.5Ω → I = 59.23 mA ⎭ stay on
DP 4-4 In order to control a device using the circuit shown in Figure DP 4-4, it is necessary
that vab = 10 V. Select the resistors when it is required that all resistors be greater than 1 Ω and R3
+ R4 = 20 Ω.
Figure DP 4-4
Solution:
R = R1 || R 2 || ( R 3 + R 4 )
Equivalent resistance:
R
( 25 )
10 + R
We require vab = 10 V. Apply the voltage division principle in the left circuit to get:
Voltage division in the equivalent circuit: v1 =
10 =
(
)
R1 R 2 ( R3 + R4 )
R4
R4
v1 =
×
× 25
R3 + R4
R3 + R4 10 + R1 R 2 ( R3 + R4 )
(
)
This equation does not have a unique solution. Here’s one solution:
choose R1 = R2 = 25 Ω and R3 + R4 = 20 Ω
then 10 =
(12.5 20 ) × 25 ⇒ R = 18.4Ω
R4
×
4
20 10 + (12.5 20 )
and R3 + R4 = 20 ⇒ R3 = 1.6 Ω
DP 4-5 The current i shown in the circuit of Figure DP 4-5 is used to measure the stress
between two sides of an earth fault line. Voltage v1 is obtained from one side of the fault, and v2
is obtained from the other side of the fault. Select the resistances R1, R2, and R3 so that the
magnitude of the current i will remain in the range between 0.5 mA and 2 mA when v1 and v2
may each vary independently between +1 V and +2 V (1 V ≤ vn ≤ 2 V).
Solution:
Apply KCL to the left mesh:
(R
1
+ R 3 ) i1 − R 3 i2 − v1 = 0
Apply KCL to the left mesh:
(R
− R 3 i1 +
2
+ R 3 ) i2 + v2 = 0
Solving for the mesh currents using Cramer’s rule:
− R3
⎡v1
⎤
⎡( R1 + R 3 )
⎢
⎥
⎢−R
⎢⎣ −v2 ( R 2 + R 3 ) ⎥⎦
3
⎣
and i2 =
i1 =
Δ
Δ
2
where Δ = ( R1 + R 3 ) ( R 2 + R 3 ) − R 3
v1 ⎤
− v2 ⎦⎥
Try R1 = R2 = R3 = 1 kΩ = 1000 Ω. Then Δ = 3 MΩ. The mesh currents will be given by
i
=
[ 2v1 − v2 ] 1000
and i 2 =
[ −2v2 + v1 ] 1000
3 × 10
3 ×10
Now check the extreme values of the source voltages:
1
6
6
if v1 = v2 = 1 V ⇒ i = 2
mA
3
if v1 = v2 = 2 V ⇒ i = 4 mA
3
⇒ i = i1 − i2 =
okay
okay
v1 + v2
3000
Chapter 5 Circuit Theorems
Exercises
R
Exercise 5.2-1 Determine values of R and is
so that the circuits shown in Figures E 5.2-1a,b
are equivalent to each other due to a source
transformation.
+
–
Answer: R = 10 Ω and is = 1.2 A
12 V
10 Ω
is
(a)
(b)
Figures E 5.2-1
Exercise 5.2-2 Determine values of R and is
so that the circuits shown in Figures E 5.2-2a,b
are equivalent to each other due to a source
transformation.
R
–
+
Hint: Notice that the polarity of the voltage
source in Figure E 5.2-2a is not the same as in
Figure E 5.2-1a.
12 V
10 Ω
is
(b)
(a)
Figures E 5.2-2
Answer: R = 10 Ω and is = –1.2 A
8Ω
Exercise 5.2-3 Determine values of R and vs
so that the circuits shown in Figures E 5.2-3a,b
are equivalent to each other due to a source
transformation.
+
–
Answer: R = 8 Ω and vs = 24 V
vs
R
3A
(a)
(b)
Figure E 5.2-3
Exercise 5.2-4 Determine values of R and vs
so that the circuits shown in Figures E 5.2-4a,b
are equivalent to each other due to a source
transformation.
Hint: Notice that the reference direction of
the current source in Figure E 5.2-4b is not the
same as in Figure E 5.2-3b.
Answer: R = 8 Ω and vs = –24 V
8Ω
+
–
vs
R
3A
(b)
(a)
Figure E 5.2-4
3Ω
Exercise 5.4-1 Determine values of Rt and voc
that cause the circuit shown in Figure E 5.4-1b
to be the Thévenin equivalent circuit of the
circuit in Figure E 5.4-1a.
+
–
6Ω
+
–
6Ω
3V
Answer: Rt = 8 Ω and voc = 2 V
b
(b)
Figure E 5.2-1
a
voc
b
(a)
Solution:
Rt
a
6Ω
Exercise 5.4-2 Determine values of Rt
and voc that cause the circuit shown in
Figure E 5.4-2b to be the Thévenin
equivalent circuit of the circuit in
Figure E 5.4-2a.
12 V
3Ω
+
–
+
–
ia
Rt
a
+
–
2ia
voc
b
Answer: Rt = 3 Ω and voc = –6 V
(a)
Solution:
ia =
2 i a − 12
⇒ i a = −3 A
6
voc = 2 i a = −6 V
12 + 6 i a = 2 i a
3 i sc = 2 i a
Rt =
−6
=3Ω
−2
b
(b)
Figure E 5.2-2
⇒ i a = −3 A
⇒ i sc =
2
( − 3 ) = −2 A
3
a
3Ω
Exercise 5.5-1 Determine values of Rt
and isc that cause the circuit shown in
Figure E 5.5-1b to be the Norton
equivalent circuit of the circuit in
Figure E 5.5-1a.
Answer: Rt = 8 Ω and isc = 0.25 A
+
–
6Ω
a
6Ω
3V
isc
Rt
b
(a)
b
(b)
Figure E 5.5-1
Solution:
a
3Ω
Exercise 5.6-1 Find the maximum power that
can be delivered to RL for the circuit of
Figure E 5.6-1 using a Thévenin equivalent
circuit.
Answer: 9 W when RL = 4 Ω
18 V
+
–
2Ω
6Ω
Figure E 5.6-1
Solution:
voc =
6
(18) = 12 V
6+3
Rt = 2 +
( 3)( 6 ) = 4 Ω
3+ 6
For maximum power, we require
R L = Rt = 4 Ω
Then
2
pmax =
voc
122
=
=9 W
4 Rt 4 ( 4 )
RL
Section 5-2: Source Transformations
P 5.2-1 The circuit shown in Figure P
5.2-1a has been divided into two parts.
The circuit shown in Figure P 5.2-1b was
obtained by simplifying the part to the
right of the terminals using source
transformations. The part of the circuit to
the left of the terminals was not changed.
4Ω
i
2V
2Ω
– +
+
9V
+
–
v
ia
4Ω
0.5 A
–
(a)
(a)
Determine the values of Rt and vt
in Figure P 5.2-1b.
(b)
+
Determine the values of the
+
v
vt
current i and the voltage v in
9 V +–
–
Figure P 5.2-1b. The circuit in
ia
–
Figure P 5.2-1b is equivalent to
the circuit in Figure P 5.2-1a.
(b)
Consequently, the current i and
Figure P 5.2-1
the voltage v in Figure P 5.2-1a
have the same values as do the current i and the voltage v in Figure P 5.2-1b.
(c)
4Ω
Determine the value of the current ia in Figure P 5.2-1a.
Solution:
(a)
i
Rt
2Ω
∴ Rt = 2 Ω
(b)
−9 − 4i − 2i + (−0.5) = 0
−9 + (−0.5)
i =
= −1.58 A
4+2
v = 9 + 4 i = 9 + 4(−1.58) = 2.67 V
(c)
ia = i = − 1.58 A
vt = − 0.5 V
(checked using LNAP 8/15/02)
8Ω
P 5.2-2 Consider the circuit of Figure P 5.2-2.
Find ia by simplifying the circuit (using source
transformations) to a single-loop circuit so that
you need to write only one KVL equation to
find ia.
3Ω
10 V
+
–
ia
6Ω
2A
4Ω
Figure P 5.2-2
Solution:
Finally, apply KVL:
−10 + 3 ia + 4 ia −
16
=0
3
∴ ia = 2.19 A
(checked using LNAP 8/15/02)
3A
P 5.2-3 Find vo using source
transformations if i = 5/2 A in the
circuit shown in Figure P 5.2-3.
Hint: Reduce the circuit to a
single mesh that contains the
voltage source labeled vo.
6Ω
8V
3Ω
10 Ω
+–
2A
16 Ω
Answer: vo = 28 V
12 Ω
20 Ω
v0
7Ω
i
+–
Figure P 5.2-3
Solution:
Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors:
Equivalents for series resistors, series voltage source at left; series resistors, then source
transformation at top:
Source transformation at left; series resistors at right:
Parallel resistors, then source transformation at left:
Finally, apply KVL to loop
− 6 + i (9 + 19) − 36 − vo = 0
i = 5 / 2 ⇒ vo = −42 + 28 (5 / 2) = 28 V
(checked using LNAP 8/15/02)
P 5.2-4 Determine the value of
the current ia in the circuit shown
in Figure P 5.2-4.
6 kΩ
10 V
4 kΩ
4 kΩ
+–
ia
+
–
12 V
4 kΩ
3 kΩ
–
+
6V
Figure P 5.2-4
Solution:
− 4 − 2000 ia − 4000 ia + 10 − 2000 ia − 3 = 0
∴ ia = 375 μ A
(checked using LNAP 8/15/02)
4A
P 5.2-5 Use source transformations to find the
current ia in the circuit shown in Figure P 5.2-5.
6Ω
6V
– +
Answer: ia = 1 A
12 V
+
–
1A
ia
3Ω
Figure P 5.2-5.
Solution:
−12 − 6 ia + 24 − 3 ia − 3 = 0 ⇒ ia = 1 A
(checked using LNAP 8/15/02)
P 5.2-6 Use source transformations to find
the value of the voltage va in Figure P 5.2-6.
Answer: va = 7 V
8V
100 Ω
+ –
+
–
10 V
+
va
–
100 Ω
100 Ω
30 mA
Figure P 5.2-6
Solution:
A source transformation on the right side of the circuit, followed by replacing series resistors
with an equivalent resistor:
Source transformations on both the right side and the left side of the circuit:
Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources
with an equivalent current source:
Finally,
va =
50 (100 )
100
( 0.21) = ( 0.21) = 7 V
50 + 100
3
(checked using LNAP 8/15/02)
P5.2-7
Figure P5.2-7
The equivalent circuit in Figure P5.2-7 is obtained from the original circuit using source
transformations and equivalent resistances. (The lower case letters a and b identify the nodes of
the capacitor in both the original and equivalent circuits.) Determine the values of Ra, Va, Rb
and I b in the equivalent circuit.
Solution
Performing a source transformation at each end of the circuit yields
Thenx
where
V a = 2.2 (10 ) − 32 = −10 V , R a = 18 + 10 = 28 Ω , R b = 18 || 9 = 6 Ω and I b = 2.5 +
36
18
= 4.5 A
P 5.2-8 The circuit shown in Figure P 5.2-8 contains an unspecified resistance R.
(a)
Determine the value of the current i when R = 4 Ω.
(b)
Determine the value of the voltage v when R = 8 Ω.
(c)
Determine the value of R that will cause i = 1 A.
(d)
Determine the value of R that will cause v = 16 V.
i
R
+
24 Ω
+
–
12 V
v
–
18 Ω
2A
24 Ω
Figure P 5.2-8
Solution:
Replace series and parallel resistors by an
equivalent resistor.
18 & (12 + 24 ) = 12 Ω
Do a source transformation, then replace
series voltage sources by an equivalent
voltage source.
12 Ω
Do two more source transformations
Now current division gives
24
⎛ 8 ⎞
i=⎜
⎟3 =
8+ R
⎝8+ R ⎠
Then Ohm’s Law gives
24 R
v = Ri =
8+ R
(a )
i=
24
=2A
8+ 4
(b) v =
24 ( 8 )
= 12 V
8+8
(c) 1 =
24
8+ R
(d) 16 =
24 R
8+ R
⇒
⇒
R = 16 Ω
R = 16 Ω
(checked: LNAP 6/9/04)
15 Ω
P 5.2-9 Determine the value of the power
supplied by the current source in the circuit
shown in Figure P 5.2-9.
+
24 V –
2A
25 Ω
24 Ω
–
32 V +
Figure P 5.2-9
Solution:
Use source transformations and equivalent resistances to reduce the circuit as follows
The power supplied by the current source is given by
p = ⎡⎣ 23.1 + 2 (10.3125 ) ⎤⎦ 2 = 87.45 W
12 Ω
Section 5-3 Superposition
P5.3-1
The inputs to the circuit shown in Figure P5.3-1 are the voltage source voltages v1 and v2.. The
output of the circuit is the voltage vo. The output is related to the inputs by
vo = a v1 + b v2
where a and b are constants. Determine the values of a and b.
Figure P5.3-1
Solution:
Let vo1 = a v1 be the output when v2 = 0. In this case,
the right voltage source acts like a short circuit so we
have the circuit show to the right. Then
v o1 =
20 || 5
4
1
1
v1 =
v1 = v1 ⇒ a =
20 + ( 20 || 5 )
20 + 4
6
6
Let vo2 = b v2 be the output when v1 = 0. In this case,
the left voltage source acts like a short circuit so we
have the circuit show to the right. Then
v o2 =
20 || 20
10
2
v2 =
v2 = v2
5 + ( 20 || 20 )
5 + 10
3
⇒ b=
2
3
P5.3-2
A particular linear circuit has two inputs, v1 and v2, and one output, vo. Three measurements are
made. The first measurement shows that the output is vo = 4 V when the inputs are v1 = 2 V and
v2 = 0. The second measurement shows that the output is vo = 10 V when the inputs are v1 = 0
and v2 = −2.5 V. In the third measurement the inputs are v1 = 3 V and v2 = 3 V. What is the
value of the output in the third measurement?
Solution:
The output of a linear circuit is a linear combination of the inputs:
v o = a1v1 + a 2 v 2
From the first two measurements we have:
⎛ a1 ⎞ ⎛ 2 ⎞
0 ⎞ ⎛ a1 ⎞
⎛ 4 ⎞ ⎛2
⎜ ⎟=⎜
⎟⎜ a ⎟ ⇒ ⎜ a ⎟ = ⎜ ⎟
⎝10 ⎠ ⎝ 0 −2.5 ⎠ ⎝ 2 ⎠
⎝ 2 ⎠ ⎝ −4 ⎠
Now the output of the third measurement can be determine to be
v o = a 1 ( 3) + a 2 ( 3) = ( 2 )( 3) + ( −4 )( 3) = −6 V
P5.3-3
The circuit shown in Figure P5.3-3 has two inputs, vs and is, and one output io. The output is
related to the inputs by the equation
io = a is + b vs
Given the following two facts:
The output is io = 0.45 A when the inputs are is = 0.25 A and vs = 15 V.
and
The output is io = 0.30 A when the inputs are is = 0.50 A and vs = 0 V.
Determine the values of the constants a and b and the values of the resistances are R 1 and R 2.
Answers: a = 0.6 A/A, b = 0.02 A/V, R 1 = 30 Ω and R 2 = 20 Ω.
Figure P5.3-3
Solution:
From the 1st fact:
0.45 = a ( 0.25 ) + b (15 )
From the 2nd fact:
0.30 = a ( 0.50 ) + b ( 0 ) ⇒ a =
0.30
= 0.60
0.50
0.45 = ( 0.60 )( 0.25 ) + b (15 ) ⇒ b =
Substituting gives
Next, consider the circuit:
a i s = i o1 = i o
so
0.60 =
R1 + R 2
b v s = i o2 = i o
and
so
R1
0.02 =
1
R1 + R 2
v s =0
⎛ R1 ⎞
i
=⎜
⎜ R1 + R 2 ⎟⎟ s
⎝
⎠
⇒ 2 R1 = 3 R 2
i s =0
=
vs
R1 + R 2
⇒ R1 + R 2 =
1
= 50 Ω
0.02
Solving these equations gives R 1 = 30 Ω and R 2 = 20 Ω.
0.45 − ( 0.60 )( 0.25 )
= 0.02
15
P 5.3-4 Use superposition to find the value of 10 Ω
the voltage v in Figure P 5.3-4.
9A
20 Ω
+
v
6A
15 Ω
–
Figure P 5.3-4
Solution:
Consider 6 A source only (open 9 A source)
Use current division:
v1
⎡ 15 ⎤
= 6 ⎢
⇒ v1 = 40 V
20
⎣15 + 30 ⎥⎦
Consider 9 A source only (open 6 A source)
Use current division:
v2
⎡ 10 ⎤
= 9 ⎢
⇒ v2 = 40 V
20
⎣10 + 35 ⎥⎦
∴ v = v1 + v2 = 40 + 40 = 80 V
(checked using LNAP 8/15/02)
Figure P5.3-5
P5.3-5 Determine v(t), the voltage across the vertical resistor in the circuit in Figure P5.3-5.
Solution;
We’ll use superposition. Let v1(t) the be the part of v(t) due to the voltage source acting alone.
Similarly, let v2(t) the be the part of v(t) due to the voltage source acting alone. We can use these
circuits to calculate v1(t) and v2(t).
Notice that v1(t) is the voltage across parallel resistors. Using equivalent resistance, we calculate
40||10 = 8 Ω. Next, using voltage division we calculate
8
v1 ( t ) =
(12 ) = 2 V
8 + 40
Similarly v2(t) is the voltage across parallel resistors Using equivalent resistance we first
determine 40||40 = 20 Ω and then calculate
20
(12 cos ( 5t ) ) = 8cos ( 5t ) V
10 + 20
v ( t ) = v1 ( t ) + v 2 ( t ) = 2 + 8cos ( 5 t ) V
v 2 (t ) =
Using superposition
15 mA
P 5.3-6 Use superposition to find the
value of the current i in Figure P 5.3-6.
4 kΩ
Answer: i = 3.5 mA
15 V
+–
2 kΩ
30 mA
12 kΩ
6 kΩ
i
Figure P5.3-6
Solution:
Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due to
the 30 mA current source.
⎛ 2 ⎞
⎛ 6 ⎞
ia = 30 ⎜
⎟ = 6 mA ⇒ i1 = ia ⎜
⎟ = 2 mA
⎝ 2+8⎠
⎝ 6 + 12 ⎠
Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i
due to the 15 mA current source.
⎛ 4 ⎞
ib = 15 ⎜
⎟ = 6 mA ⇒
⎝ 4+6⎠
⎛ 6 ⎞
i2 = ib ⎜
⎟ = 2 mA
⎝ 6 + 12 ⎠
Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 V
voltage source.
⎛ 6 || 6 ⎞
⎛ 3 ⎞
i3 = − 2.5 ⎜⎜
⎟⎟ = − 10 ⎜
⎟ = −0.5 mA
⎝ 3 + 12 ⎠
⎝ ( 6 || 6 ) + 12 ⎠
Finally,
i = i1 + i2 + i3 = 2 + 2 − 0.5 = 3.5 mA
(checked using LNAP 8/15/02)
Figure P5.3-7
P5.3-7 Determine v(t), the voltage across the 40 Ω resistor in the circuit in Figure P5.3-7.
Solution:
We’ll use superposition. Let v1(t) the be the part of v(t) due to the voltage source acting alone.
Similarly, let v2(t) the be the part of v(t) due to the voltage source acting alone. We can use these
circuits to calculate v1(t) and v2(t).
Using voltage division we calculate
v1 ( t ) = −
40
(12 + 15cos (8 t ) ) = −9.6 − 12 cos (8t )
10 + 40
Using equivalent resistance we first determine 10||40 = 8 Ω and then calculate
v 2 ( t ) = 8 (1 + sin ( 5t ) ) = 8 + 8sin ( 5t )
Using superposition
v ( t ) = v1 ( t ) + v 2 ( t ) = −1.6 + 8sin ( 5 t ) − 12 cos ( 8 t ) V
ix
P 5.3-8 Use superposition to find the value of
the current ix in Figure P 5.3-8.
Answer: i = 3.5 mA
+
–
6Ω
3Ω
8V
2A
+
–
3ix
Figure P5.3-8
Solution:
Consider 8 V source only (open the 2 A source)
Let i1 be the part of ix due to the 8 V
voltage source.
Apply KVL to the supermesh:
6 ( i1 ) + 3 ( i 1 ) + 3 ( i 1 ) − 8 = 0
8 2
= A
12 3
Let i2 be the part of ix due to the 2 A
current source.
i1 =
Consider 2 A source only (short the 8 V source)
Apply KVL to the supermesh:
6 ( i 2 ) + 3 ( i 2 + 2 ) + 3 i2 = 0
i2 =
Finally,
i x = i1 + i 2 =
2 1 1
− = A
3 2 6
−6
1
=− A
12
2
P 5.3-9 The input to the circuit shown in Figure P 5.3-9 is the voltage source voltage, vs. The
output is the voltage vo. The current source current, ia, is used to adjust the relationship between
the input and output. Design the circuit so that input and output are related by the equation vo =
2vs + 9.
Hint: Determine the required values of A and ia.
A ix
6Ω
+ –
ix
+
–
vs
12 Ω
12 Ω
ia
+
vo
−
Figure P 5.3-9
Soluton:
ix =
vs − va
R1
va − vo = A ix = A
va =
vs − va
R1
R1 v o + A v s
R1 + A
Apply KCL to the supernode corresponding to the CCVS to get
va − vs
R1
R1 + R 2
R1 R 2
+
va
R2
va −
+ ia +
vs
R1
vo
R3
+ ia +
=0
vo
R3
=0
R1 + R 2 ⎛ R 1 v o + A v s ⎞ v s
vo
+ ia +
=0
⎜
⎟−
R1 R 2 ⎜⎝ R1 + A ⎟⎠ R1
R3
(
)
⎛ R +R
⎛ R1 + R 2 A
1 ⎞⎟
1 ⎞
1
2
⎜
+
vo + ⎜
− ⎟ vs + ia = 0
⎜ R 2 R1 + A R 3 ⎟
⎜ R1 R 2 R1 + A R1 ⎟
⎝
⎠
⎝
⎠
(
(
)
) (
R 2 R 3 ( R1 + A )
R 3 R1 + R 2 + R 2 R 1 + A
)v
(
o+
)
A − R2
(
R 2 R1 + A
)
vs + ia = 0
vo =
(
R 2 R 3 ( R1 + A )
)
vs −
ia
R 3 ( R1 + R 2 ) + R 2 ( R 1 + A )
R 3 ( R 1 + R 2 ) + R 2 ( R1 + A )
R3 R 2 − A
When R1 = 6 Ω, R 2 = 12 Ω and R 3 = 12 Ω
vo =
12 ( 6 + A )
12 − A
vs −
ia
24 + A
24 + A
Comparing this equation to v o = 2 v s + 9 , we requires
12 − A
=2 ⇔
24 + A
A = −12
V
A
Then 2 v s + 9 = v o = 2v s + 6i a so we require
9 = 6ia
⇒ i a = 1.5 A
(checked: LNAP 6/22/04)
P 5.3-10 The circuit shown in Figure P 5.3-10 has three inputs: v1, v2, and i3. The output of the
circuit is vo. The output is related to the inputs by
vo = av1 + bv2 + ci3
where a, b, and c are constants. Determine the values of a, b, and c.
v2
+
8Ω
–
+
+
–
v1
40 Ω
10 Ω
vo
i3
–
Figure P 5.3-10
Solution:
vo1 =
vo2 = −
40 ||10
1
1
v1 = v1 ⇒ a =
8 + 40 ||10
2
2
10
3
v1 = − v 2
8 || 40 + 10
5
⇒ b=−
3
5
vo3 = ( 8 ||10 || 40 ) i 3 = 4 i 3 ⇒ c = 4
(checked: LNAP 6/22/04)
P 5.3-11 Determine the voltage vo(t) for the
circuit shown in Figure P 5.3-11.
–
+
12 cos 2t V
4 ix
10 Ω
+
40 Ω
+
–
2 V 10 Ω
ix
5Ω
vo(t)
–
Figure P 5.3-11
Solution: Using superposition:
v x = 10 i x
and
v x − 12 cos 2t
40
+
vx
10
+
vx
10
= 4ix
so
10 i x − 12 cos 2t
40
= 2ix
⇒ ix = −
12
cos 2t
70
Finally,
v o1 = −5 4 i x = 3.429 cos 2t V
( )
v x = 10 i x
and
vx
40
+
vx − 2
10
+
vx
10
= 4ix
so
−0.2 = 1.75 i x
Finally,
⇒ i x = −0.11429 A
( )
v o1 = −5 4 i x = 2.286 V
v o = v o1 + v o2 = 3.429 cos 2t + 2.286 V
(checked: LNAP 6/22/04)
P 5.3-12 Determine the value of the voltage vo in the
circuit shown in Figure P 5.3-12.
96 Ω
32 Ω
20 V
+
0.3 A
–
120 Ω
30 Ω
+
vo
–
Figure P 5.3-12
Solution: Using superposition:
v o1 = 24 ( 0.3) = 7.2 V
v o2 = −
30
20 = −4 V
120 + 30
v o = v o1 + v o 2 = 3.2 V
(checked: LNAP 5/24/04)
P 5.3-13 Determine the value of the
voltage vo in the circuit shown in
Figure P 5.3-13.
–
a
vo
+
b
2Ω
i1
R
4Ω
8Ω
8Ω
i2
Figure P 5.3-13
Solution:
Using superposition
⎛ R || 4 ⎞
⎛
⎞
4
v o = −2 ⎜⎜
⎟⎟ i1 + 2 ⎜⎜
⎟⎟ i 2
6
+
R
||
4
2
+
R
||
4
+
4
(
)
(
)
⎝
⎠
⎝
⎠
Comparing to v o = −0.5 i1 + 4 , we require
⎛ R || 4 ⎞
−2 ⎜⎜
⎟⎟ = −0.5 ⇒ 4 ( R || 4 ) = 6 + ( R || 4 ) ⇒ R || 4 = 2 ⇒ R = 4 Ω
⎝ 6 + ( R || 4 ) ⎠
and
⎛
⎞
⎛
⎞
4
4
2 ⎜⎜
⎟⎟ i 2 = 4 ⇒ 2 ⎜⎜
⎟⎟ i 2 = 4 ⇒ i 2 = 4 A
⎝ 2 + ( R || 4 ) + 4 ⎠
⎝ 2 + ( 4 || 4 ) + 4 ⎠
(checked LNAP 6/12/04)
P 5.3-14 Determine values of the current,
ia, and the resistance, R, for the circuit shown
in Figure P 5.3-14.
+
8V
–
ia
5 kΩ
20 kΩ
7 mA
4 kΩ
Figure P 5.3-14
Solution:
Use units of mA, kΩ and V.
4 + (5||20) = 8 kΩ
(a) Using superposition
8
⎛ 8 ⎞
2=⎜
⇒ 2 ( R + 8 ) = 48 ⇒ R = 16 kΩ
⎟7−
R +8
⎝ R +8⎠
(b) Using superposition again
8 ⎤ 4⎛ 2
1⎞
⎛ 5 ⎞ ⎡⎛ 16 ⎞
ia = ⎜
= ⎜ × 7 + ⎟ = 4 mA
⎟ ⎢⎜
⎟7+
⎥
8 + 16 ⎦ 5 ⎝ 3
3⎠
⎝ 5 + 20 ⎠ ⎣⎝ 8 + 16 ⎠
R
2 mA
20 Ω
P 5.3-15 The circuit shown in Figure P 5.3-15 has
three inputs: v1, i2, and v3. The output of the circuit is the
v1
current io. The output of the circuit is related to the
inputs by
+
–
i2
io
12 Ω
i1 = avo + bv2 + ci3
where a, b, and c are constants. Determine the values of
a, b, and c.
40 Ω
v3 –
+
10 Ω
Figure P 5.3-15
Solution:
⎞
⎞ ⎛
v1
10 ⎞ ⎛
10 ⎞ ⎛
20
⎛
+
−
io = ⎜ −
⎜
⎟ i2
⎜
⎟
⎟
⎜
⎟
⎝ 10 + 40 ⎠ ⎜⎝ 20 + 12 + ( 40 & 10 ) ⎟⎠ ⎝ 10 + 40 ⎠ ⎜⎝ 20 + ⎡⎣12 + ( 40 & 10 ) ⎤⎦ ⎟⎠
⎞
⎛
⎞⎛
v3
20 + 12
+ ⎜⎜ −
⎟
⎟⎟ ⎜⎜
⎟
⎝ 40 + ( 20 + 12 ) ⎠ ⎝ 10 + ⎡⎣ 40 & ( 20 + 12 ) ⎤⎦ ⎠
⎛ 1 ⎞
⎛ 1⎞
⎛ 1 ⎞
io = ⎜ −
⎟ v1 + ⎜ − ⎟ i 2 + ⎜ −
⎟ v3
⎝ 200 ⎠
⎝ 10 ⎠
⎝ 62.5 ⎠
So
a = −0.05, b = −0.1 and c = −0.016
(checked: LNAP 6/19/04)
25 V
P 5.3-16 Using the superposition
principle, find the value of the
current measured by the ammeter in
Figure P 5.3-16a.
5A
Hint: Figure P 5.3-16b shows the
circuit after the ideal ammeter has
been replaced by the equivalent short
circuit and a label has been added to
indicate the current measured by the
ammeter, im.
Answer:
25
3
im =
−
5 = 5−3 = 2 A
3+ 2 2+3
Ammeter
–+
3Ω
2Ω
(a)
25 V
–+
3Ω
5A
2Ω
(b)
Figure P 5.3-16
Solution:
im =
25
3
−
( 5) = 5 − 3 = 2 A
3+ 2 2+3
im
Section 5-4: Thèvenin’s Theorem
P 5.4-1 Determine values of Rt and voc that cause the circuit shown in Figure P 5.4-1b to be the
Thévenin equivalent circuit of the circuit in Figure P 5.4-1a.
Hint: Use source transformations and equivalent resistances to reduce the circuit in Figure P
5.4-1a until it is the circuit in Figure P 5.4-1b.
Answer: Rt = 5 Ω and voc = 2 V
3Ω
+
– 12 V
3Ω
6Ω
Rt
a
+
–
3A
voc
b
(a)
a
b
(b)
Figure P 5.4-1
Solution:
(checked using LNAP 8/15/02)
P 5.4-2 The circuit shown in Figure P 5.4-2b is the Thévenin equivalent circuit of the circuit
shown in Figure P 5.4-2a. Find the value of the open-circuit voltage, voc, and Thévenin
resistance, Rt.
Answer: voc = –12 V and Rt = 16 Ω
10 Ω
15 V
8Ω
40 Ω
+
–
(a)
Rt
+
–
voc
(b)
Figure P 5.4-2
Solution:
The circuit from Figure P5.4-2a can be reduced to its Thevenin equivalent circuit in four steps:
(a)
(b)
(c)
(d)
Comparing (d) to Figure P5.4-2b shows that the Thevenin resistance is Rt = 16 Ω and the open
circuit voltage, voc = −12 V.
P 5.4-3 The circuit shown in Figure P 5.4-3b is the Thévenin equivalent circuit of the circuit
shown in Figure P 5.4-3a. Find the value of the open-circuit voltage, voc, and Thévenin
resistance, Rt.
Answer: voc = 2 V and Rt = 4 Ω
12 V
Rt
–+
1A
6Ω
6Ω
+
–
voc
6Ω
(a)
(b)
Figure P 5.4-3
Solution:
The circuit from Figure P5.4-3a can be reduced to its Thevenin equivalent circuit in five steps:
(a)
(c)
(b)
(d)
(e)
Comparing (e) to Figure P5.4-3b shows that the Thevenin resistance is Rt = 4 Ω and the open
circuit voltage, voc = 2 V.
(checked using LNAP 8/15/02)
12 Ω
P 5.4-4 Find the Thévenin equivalent circuit
for the circuit shown in Figure P 5.4-4.
6Ω
+
–
10 Ω
a
3Ω
18 V
b
Figure P 5.4-4
Find Rt:
Rt =
Write mesh equations to find voc:
12 (10 + 2 )
=6Ω
12 + (10 + 2 )
Mesh equations:
12 i1 + 10 i1 − 6 ( i2 − i1 ) = 0
6 ( i2 − i1 ) + 3 i 2 − 18 = 0
28 i1 = 6 i 2
9 i 2 − 6 i1 = 18
36 i1 = 18 ⇒ i1 =
i2 =
Finally,
1
A
2
14 ⎛ 1 ⎞ 7
⎜ ⎟= A
3 ⎝2⎠ 3
⎛7⎞
⎛1⎞
voc = 3 i 2 + 10 i1 = 3 ⎜ ⎟ + 10 ⎜ ⎟ = 12 V
⎝ 3⎠
⎝ 2⎠
(checked using LNAP 8/15/02)
0.75va
P 5.4-5 Find the Thévenin equivalent circuit for the circuit
shown in Figure P 5.4-5.
8Ω
Answer: voc = –2 V and Rt = –8/3 Ω
+
–
6V
a
–
va
+
4Ω
b
Figure P 5.4-4
Solution:
Find voc:
Notice that voc is the node voltage at node a. Express
the controlling voltage of the dependent source as a
function of the node voltage:
va = −voc
Apply KCL at node a:
⎛ 6 − voc ⎞ voc ⎛ 3 ⎞
−⎜
+ ⎜ − voc ⎟ = 0
⎟+
⎝ 8 ⎠ 4 ⎝ 4 ⎠
−6 + voc + 2 voc − 6 voc = 0 ⇒ voc = −2 V
Find Rt:
We’ll find isc and use it to calculate Rt. Notice that
the short circuit forces
Apply KCL at node a:
va = 0
⎛ 6−0⎞ 0 ⎛ 3 ⎞
−⎜
⎟ + + ⎜ − 0 ⎟ + i sc = 0
⎝ 8 ⎠ 4 ⎝ 4 ⎠
i sc =
Rt =
6 3
= A
8 4
voc −2
8
=
=− Ω
3
i sc 3 4
(checked using LNAP 8/15/02)
3Ω
3Ω
P 5.4-6 Find the Thévenin equivalent circuit
for the circuit shown in Figure P 5.4-6.
2va
+
–
6Ω
+
va
–
a
3A
b
Figure P 5.4-6
Solution:
Find voc:
2 va − va va
= + 3 + 0 ⇒ va = 18 V
3
6
The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V
Apply KCL at the top, middle node:
Find isc:
2 va − va va
v
= + 3 + a ⇒ va = −18 V
3
6
3
v
−18
Apply Ohm’s law to the right-hand 3 Ω resistor :
= −6 V
i sc = a =
3
3
v
18
R t = oc =
= −3 Ω
Finally:
i sc −6
Apply KCL at the top, middle node:
(checked using LNAP 8/15/02)
P5.4-7 The equivalent circuit in Figure P5.4-7 is
obtained by replacing part of the original circuit by
its Thevenin equivalent circuit. The values of the
parameters of the Thevenin equivalent circuit are
v oc = 15 V and R t = 60 Ω
Determine the following:
a.) The values of V s and R a . (Three resistors in the
original circuit have equal resistance, Ra.)
b.) The value of R b required to cause i = 0.2 A.
c.) The value of R b required to cause v = 12 V.
Figure P5.4-7
Solution: a.) From
We see that v oc =
b.) i =
c.) v =
v oc
Rt + Rb
Rb
Rt + Rb
Vs
2
and R t = R a . With the given values of voc and Rt we calculate
5
5
V
2
15 = s ⇒ Vs = 75 V and 60 = R a ⇒ R a = 150 Ω .
5
5
⇒ 0.2 =
v oc
15
60 + R b
⇒ 12 =
⇒ R b = 15 Ω
15 R b
60 + R b
⇒ R b = 240 Ω
P 5.4-8 A resistor, R, was connected to a circuit
box as shown in Figure P 5.4-8. The voltage, v,
was measured. The resistance was changed, and
the voltage was measured again. The results are
shown in the table. Determine the Thévenin
equivalent of the circuit within the box and predict
the voltage, v, when R = 8 kΩ.
i
+
Circuit
v
–
R
R
v
2 kΩ
4 kΩ
6V
2V
Figure P 5.4-8
Solution:
From the given data:
2000
⎫
voc ⎪
R t + 2000 ⎪
⎧ voc = 1.2 V
⎬ ⇒ ⎨
4000
⎩ R t = −1600 Ω
voc ⎪
2=
R t + 4000 ⎪⎭
6=
When R = 8000 Ω,
v=
R
voc
Rt + R
v=
8000
(1.2 ) = 1.5 V
−1600 + 8000
P 5.4-9 A resistor, R, was connected to a circuit
box as shown in Figure P 5.4-9. The current, i, was
measured. The resistance was changed, and the
current was measured again. The results are shown
in the table.
(a)
(b)
Hint:
i
R
i
2 kΩ
4 kΩ
4 mA
3 mA
+
Circuit
Specify the value of R required to cause i =
2 mA.
Given that R > 0, determine the maximum
possible value of the current i.
v
R
–
Figure P 5.4-9
Use the data in the table to represent the circuit by a Thévenin equivalent.
Solution:
From the given data:
voc
⎫
R t + 2000 ⎪⎪
⎧ voc = 24 V
⎬ ⇒ ⎨
voc
⎩ R t = 4000 Ω
⎪
0.003 =
R t + 4000 ⎪⎭
0.004 =
i=
voc
Rt + R
(a) When i = 0.002 A:
24
⇒ R = 8000 Ω
0.002 =
4000 + R
(b) Maximum i occurs when R = 0:
24
= 0.006 = 6 mA ⇒ i ≤ 6 mA
4000
P 5.4-10 For the circuit of Figure P 5.4-10, specify the resistance R that will cause current ib to
be 2 mA. The current ia has units of amps.
Hint: Find the Thévenin equivalent circuit of the circuit connected to R.
2000ia
6 kΩ
+ –
12 V
+
–
1 kΩ
ia
ib
R
Figure P 5.4-10
Solution:
−12 + 6000 ia + 2000 ia + 1000 ia = 0
ia = 4 3000 A
voc = 1000 ia =
4
V
3
ia = 0 due to the short circuit
−12 + 6000 isc = 0 ⇒ isc = 2 mA
4
voc
Rt =
= 3 = 667 Ω
isc .002
4
3
ib =
667 + R
ib = 0.002 A requires
4
3 − 667 = 0
R =
0.002
(checked using LNAP 8/15/02)
4i
a
P 5.4-11 For the circuit of Figure P 5.4-11, specify
the value of the resistance RL that will cause current
iL to be –2 A.
+ –
10 A
2Ω
iL
RL
i
Answer: RL = 12 Ω
b
Figure P 5.4-11
Solution:
10 = i + 0 ⇒ i = 10 A
voc + 4 i − 2 i = 0
⇒ voc = −2 i = −20 V
i + i sc = 10 ⇒ i = 10 − i sc
4 i + 0 − 2 i = 0 ⇒ i = 0 ⇒ i sc = 10 A
Rt =
−2 = iL =
voc −20
=
= −2 Ω
isc
10
−20
⇒ RL = 12 Ω
RL − 2
(checked using LNAP 8/15/02)
P 5.4-12 The circuit shown in Figure P 5.4-12 contains an adjustable resistor. The resistance R
can be set to any value in the range 0 ≤ R ≤ 100 kΩ.
(a)
Determine the maximum value of the current ia that can be obtained by adjusting R.
Determine the corresponding value of R.
(b)
Determine the maximum value of the voltage va that can be obtained by adjusting R.
Determine the corresponding value of R.
(c)
Determine the maximum value of the power supplied to the adjustable resistor that can be
obtained by adjusting R. Determine the corresponding value of R.
ia
+
+
–
12 kΩ
R
12 V
va
−
2 mA
18 kΩ
24 kΩ
Figure P 5.4-12
Solution: Replace the part of the circuit that is connected to the variable resistor by its Thevenin
equivalent circuit:
18 kΩ || (12 kΩ + 24 kΩ ) = 18 kΩ || 36 kΩ = 12 kΩ
ia =
R
36
36
and v a =
R + 12000
R + 12000
2
36
⎛
⎞
p = ia va = ⎜
⎟ R
⎝ R + 12000 ⎠
36
= 3 mA when R = 0 Ω (a short circuit).
0 + 12000
105
36 = 32.14 V when R is as large as possible, i.e. R = 100 kΩ.
(b) v a = 5
10 + 12000
(c) Maximum power is delivered to the adjustable resistor when R = R t = 12 kΩ . Then
(a) i a =
2
36
⎛
⎞
p = ia va = ⎜
⎟ 12000 = 0.027 = 27 mW
⎝ 12000 + 12000 ⎠
(checked: LNAP 6/22/04)
P 5.4-13 The circuit shown in Figure P 5.4-13
consists of two parts, the source (to the left of the
terminals) and the load. The load consists of a single
adjustable resistor having resistance 0 ≤ RL ≤ 20 Ω.
The resistance R is fixed, but unspecified. When
RL = 4 Ω, the load current is measured to be io = 0.375
A. When RL = 8 Ω, the value of the load current is
io = 0.300 A.
(a)
Determine the value of the load current when
RL = 10 Ω.
(b)
Determine the value of R.
48 Ω
io
24 V
+
–
R
source
RL
load
Figure P 5.4-13
Solution:
Replace the source by it’s Thevenin equivalent circuit to get
io =
v oc
R t +R L
Using the given formation
v oc ⎫
⎪
R t + 4⎪
⎬
v oc ⎪
0.300 =
R t + 8 ⎪⎭
0.375 =
So
Rt =
⇒
0.375 ( R t + 4 ) = 0.300 ( R t + 8 )
( 0.300 ) 8 − ( 0.375) 4 = 12 Ω and v
0.075
6
(a) When R L = 10 Ω, i o =
= 0.2727 A.
12 + 10
(b) 12 Ω = R t = 48 11R ⇒ R = 16 Ω .
oc
= 0.3 (12 + 8 ) = 6 V
(checked: LNAP 5/24/04)
P 5.4-14 The circuit shown in
Figure P 5.4-14 contains an unspecified
resistance, R. Determine the value of R in each
of the following two ways.
(a)
(b)
Write and solve mesh equations.
Replace the part of the circuit
connected to the resistor R by a
Thévenin equivalent circuit. Analyze
the resulting circuit.
40 Ω
20 Ω
R
+
–
40 V
0.25 A
20 Ω
10 Ω
Figure P 5.4-14
Solution:
(a)
i 3 − i 2 = 0.25 A
Apply KVL to mesh 1 to get
20 ( i1 − i 2 ) + 20 ( i1 − i 3 ) − 40 = 0
Apply KVL to the supermesh corresponding to the unspecified resistance to get
40i 2 + 10i 3 − 20 ( i1 − i 3 ) − 20 ( i1 − i 2 ) = 0
Solving, for example using MATLAB, gives
−1
1 ⎤ ⎡ i1 ⎤ ⎡0.25⎤
⎡ 0
⎢ 40 −20 −20 ⎥ ⎢i ⎥ = ⎢ 40 ⎥
⎢
⎥⎢ 2⎥ ⎢
⎥
⎢⎣ −40 60 30 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ 0 ⎥⎦
⇒
Apply KVL to mesh 2 to get
40i 2 + R ( i 2 − i 3 ) − 20 ( i1 − i 2 ) = 0
⇒
R=
⎡ i1 ⎤ ⎡1.875 ⎤
⎢ ⎥ ⎢
⎥
⎢i 2 ⎥ = ⎢0.750 ⎥
⎢ i 3 ⎥ ⎢⎣1.000 ⎥⎦
⎣ ⎦
20 ( i1 − i 2 ) − 40i 2
i2 − i3
= 30 Ω
(b)
⎛ 20 ⎞
⎛ 40 ⎞
v oc = ⎜
⎟ 40 − ⎜
⎟ 40 = −12 V
⎝ 20 + 20 ⎠
⎝ 10 + 40 ⎠
R t = 18 Ω
0.25 =
12
18 + R
⇒
R = 30 Ω
(checked: LNAP 5/25/04)
a
96 Ω
io
32 Ω
20 V
+
P 5.4-15 Consider the circuit shown in Figure P 5.4-15.
Replace the part of the circuit to the left of terminals a–b by
its Thévenin equivalent circuit. Determine the value of the
current io.
32 Ω
–
+
120 Ω
30 Ω
vo
–
b
Figure P 5.4-15
Solution:
Find the Thevenin equivalent circuit for the part of the circuit to the left of the terminals a-b.
Using voltage division twice
v oc =
32
30
20 −
20 = 5 − 4 = 1 V
32 + 96
120 + 30
R t = ( 96 || 32 ) + (120 || 30 ) = 24 + 24 = 48 Ω
Replacing the part of the circuit to the left of terminals a-b by its Thevenin equivalent circuit
gives
io =
1
= 0.0125 A = 12.5 mA
48 + 32
(checked: LNAP 5/24/04)
P 5.4-16 An ideal voltmeter is
modeled as an open circuit. A more
realistic model of a voltmeter is a large
resistance. Figure P 5.4-16a shows a
circuit with a voltmeter that measures
the voltage vm. In Figure P 5.4-16b the
voltmeter is replaced by the model of
an ideal voltmeter, an open circuit.
The voltmeter measures vmi, the ideal
value of vm.
Determine the value of vmi.
(b)
Express the measurement error
that occurs when Rm = 1000 Ω
as a percentage of vmi.
(c)
10 Ω
+
–
25 V
50 Ω
vm
–
(a)
200 Ω
10 Ω
+
–
25 V
50 Ω
vmi
–
(b)
200 Ω
10 Ω
+
+
–
25 V
50 Ω
Rm
vm
–
(c)
Figure P 5.4-16
Determine the minimum value
of Rm required to ensure that
the measurement error is
smaller than 2 percent of vmi.
Solution: Replace the circuit by its Thevenin equivalent circuit:
(a)
Voltmeter
+
+
As Rm → ∞, the voltmeter becomes an
ideal voltmeter and vm → vmi. When
Rm < ∞, the voltmeter is not ideal and
vm > vmi. The difference between vm
and vmi is a measurement error caused
by the fact that the voltmeter is not
ideal.
(a)
200 Ω
⎛ Rm ⎞
vm = ⎜
5
⎜ R m + 50 ⎟⎟
⎝
⎠
v mi = lim v m = 5 V
R m →∞
(b) When R m = 1000 Ω, v m = 4.763 V so
% error =
5 − 4.762
× 100 = 4.76%
5
(c)
⎛ Rm ⎞
5−⎜
5
⎜ R m + 50 ⎟⎟
⎝
⎠
0.02 ≥
5
⇒
Rm
R m + 50
≥ 0.98
⇒
R m ≥ 2450 Ω
(checked: LNAP 6/16/04)
P5.4-17
Given that 0 ≤ R ≤ ∞ in the circuit shown in Figure P5.4-17, consider these two observations:
Observation 1:
When R = 2 Ω then v R = 4 V and i R = 2 A.
Observation 1:
When R = 6 Ω then v R = 6 V and i R = 1 A.
Determine the following
a) The maximum value of i R and the value of R that causes i R to be maximal.
b) The maximum value of v R and the value of R that causes v R to be maximal.
c) The maximum value of p R = i R v R and the value of R that causes p R to be maximal.
Figure P5.4-17
Solution:
We can replace the part of the circuit to the left of the terminals by its Thevenin equivalent
circuit:
Using voltage division v R =
iR =
v oc
R + Rt
R
v oc and using Ohm’s law
R + Rt
.
v oc
R
v oc =
will be maximum when
Rt
R + Rt
1+
R
R = ∞. The maximum value of v R will be v oc . Similarly,
By inspection , v R =
iR =
v oc
R + Rt
will be maximum when R = 0. The maximum value
of i R will be
v oc
Rt
= i sc .
The maximum power transfer theorem tells use that p R = i R v R will be maximum when R = R t .
⎛ v oc
Then p R = i R v R = ⎜
⎜ R + Rt
⎝
⎞⎛ R
⎞
⎛ v oc
v oc ⎟ = R ⎜
⎟⎜
⎟⎜ R + R t
⎟
⎜ R + Rt
⎠⎝
⎠
⎝
2
⎞
⎟⎟ .
⎠
Let’s substitute the given data into the equation i R =
When R = 2 Ω we get 2 =
1=
v oc
6 + Rt
v oc
2 + Rt
v oc
R + Rt
.
⇒ 4 + 2 R t = v oc . When R = 6 Ω we get
⇒ 6 + R t = v oc .
So 6 + R t = 4 + 2 R t
⇒ R t = 2 Ω and v oc = 4 + 2 R t = 8 V . Also i sc =
v oc
Rt
=
8
= 4 A.
2
P5.4-18 Consider the circuit shown in Figure P5.4-18. Determine
a) The value of v R that occurs when R = 9 Ω.
b) The value of R that causes v R = 5.4 V.
c) The value of R that causes i R = 300 mA.
Figure P5.4-18
Solution: Reduce this circuit using source transformations and equivalent resistance:
9
⎛ R ⎞
so the questions can be easily answered:
Now v R = ⎜
⎟ 9 and i R =
R + 18
⎝ R + 18 ⎠
a) When R = 9 Ω then v R = 3 V.
b) When R = 27 Ω then v R = 5.4 V.
c) When R = 12 Ω then i R = 300 mA.
P5.4-19 The circuit shown in Figure P5.4-19a can be
reduced to the circuit shown in Figure P5.4-19b using
source transformations and equivalent resistances.
Determine the values of the source voltage v oc and the
(a)
resistance R.
(b)
Figure P5.4-19
Solution
46 = R t = R + ( 42 || 84 ) = R + 28 ⇒ R = 18 Ω
v oc =
84
(18) = 12 V
42 + 84
P5.4-20 The equivalent circuit in Figure P5.4-20
is obtained by replacing part of the original
circuit by its Thevenin equivalent circuit. The
values of the parameters of the Thevenin
equivalent circuit are
v oc = 15 V and R t = 60 Ω
Determine the following:
a.) The values of V s and R a . (Three resistors in
the original circuit have equal resistance, Ra.)
b.) The value of R b required to cause i = 0.2 A.
c.) The value of R b required to cause v = 5 V.
Figure P5.4-20
Solution
a.) From
We see that v oc =
b.) i =
c.) v =
v oc
Rt + Rb
Rb
Rt + Rb
Vs
3
and R t = R a . With the given values of voc and Rt we calculate
2
2
V
3
15 = s ⇒ Vs = 30 V and 60 = R a ⇒ R a = 40 Ω .
2
2
⇒ 0.2 =
v oc
15
60 + R b
⇒ 5=
15 R b
60 + R b
⇒ R b = 15 Ω
⇒ R b = 30 Ω
Section 5-5: Norton’s Theorem
P5.5-1 The part of the circuit shown in Figure P5.3-1a to the left of the terminals can be reduced
to its Norton equivalent circuit using source transformations and equivalent resistance. The
resulting Norton equivalent circuit, shown in Figure P5.3-1b, will be characterized by the
parameters:
i sc = 0.5 A and R t = 20 Ω
a) Determine the values of v s and R1 .
b) Given that 0 ≤ R 2 ≤ ∞ , determine the maximum values of the voltage, v, and of the
power, p = vi.
Answers: v s = 37.5 V, R1 = 25 Ω, max v = 10 V and max p = 1.25 W
(a)
(b)
Figure P5.5-1
Solution: Two source transformations reduce the circuit as follows:
Recognizing the parameters of the Norton equivalent circuit gives:
0.5 = i sc =
12.5 + v s
100
⇒ v s = 37.5 V and 20 = R t = 100 || R1 =
(
)
Next, the voltage across resistor R 2 is given by v = i sc R t || R 2 =
100 R1
100 + R1
R t R 2 i sc
Rt + R2
=
⇒ R1 = 25 Ω
R t i sc
so this
Rt
+1
R2
voltage is maximum when R 2 = ∞ and max v = R t i sc =10 V. The power p = vi will be
maximum when R 2 = R t = 20 Ω . Then v =
R t i sc
2
=
20 ( 0.5 )
2
=5 V, i=
v
5
=
= 0.25 A and
R 2 20
p = v i = 5 ( 0.25 ) = 1.25 W.
P 5.5-2 Two black boxes are
shown in Figure P 5.5-2. Box A
contains the Thévenin equivalent of
some linear circuit, and box B
contains the Norton equivalent of the
same circuit. With access to just the
outsides of the boxes and their
terminals, how can you determine
which is which, using only one
shorting wire?
Box A
1Ω
1V
Box B
a
+
–
a
1A
b
1Ω
b
Figure P 5.5-2
Solution:
When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in
Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of
each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will
be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will
cool off.
(a)
(b)
Figure P5.5-3
P5.5-3 The circuit shown in Figure P5.5-3a can be reduced to the circuit shown in Figure P5.53b using source transformations and equivalent resistances. Determine the values of the source
current i sc and the resistance R.
Solution:
48 = R t = R || ( 80 + 160 ) =
i sc =
240 R
⇒ R = 60 Ω
R + 240
80
( 4.8 ) = 1.6 A
80 + 160
P 5.5-4
Find the Norton equivalent circuit for the circuit shown in Figure P 5.5-4.
3Ω
5Ω
a
4A
8Ω
5A
b
Figure P 5.5-4
Solution:
P 5.5-5 The circuit shown in Figure P 5.5-5b is the Norton equivalent circuit of the circuit
shown in Figure P 5.5-5a. Find the value of the short-circuit current, isc, and Thévenin resistance,
Rt.
Answer: isc = 1.13 A and Rt = 7.57 Ω
3Ω
5Ω
– +
+
–
10 V
2ia
6Ω
isc
Rt
ia
(a)
(b)
Figure P 5.5-5
Solution:
To determine the value of the short circuit current, isc, we connect a short circuit across the
terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a)
shows the circuit from Figure 5.6-4a after adding the short circuit and labeling the short circuit
current. Also, the meshes have been identified and labeled in anticipation of writing mesh
equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently,
i2 = isc . The controlling current of the CCVS is expressed in terms of the mesh currents as
i a = i1 − i 2 = i1 − isc
Apply KVL to mesh 1 to get
3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 7 i1 − 4 i 2 = 10
Apply KVL to mesh 2 to get
5 i 2 − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = 0 ⇒ i1 =
11
i2
6
Substituting into equation 1 gives
⎛ 11 ⎞
7 ⎜ i 2 ⎟ − 4 i 2 = 10 ⇒ i 2 = 1.13 A ⇒ i sc = 1.13 A
⎝6 ⎠
Figure (a) Calculating the short circuit current, isc, using mesh equations.
(1)
To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltage
source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the
terminals of the circuit and then label the voltage across that current source as shown in Figure
(b). The Thevenin resistance will be calculated from the current and voltage of the current source
as
v
Rt = T
iT
In Figure (b), the meshes have been identified and labeled in anticipation of writing mesh
equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (b), mesh current i2 is equal to the negative of the current source current.
Consequently, i2 = i T . The controlling current of the CCVS is expressed in terms of the mesh
currents as
i a = i1 − i 2 = i1 + i T
Apply KVL to mesh 1 to get
3 i 1 − 2 ( i 1 − i 2 ) + 6 ( i1 − i 2 ) = 0 ⇒ 7 i 1 − 4 i 2 = 0 ⇒ i 1 =
4
i2
7
(2)
Apply KVL to mesh 2 to get
5 i 2 + vT − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = −vT
Substituting for i1 using equation 2 gives
⎛4 ⎞
−6 ⎜ i 2 ⎟ + 11 i 2 = −vT
⎝7 ⎠
Finally,
Rt =
⇒ 7.57 i 2 = −vT
vT −vT −vT
=
=
= 7.57 Ω
iT
i2
−iT
Figure (b) Calculating the Thevenin resistance, R t =
vT
, using mesh equations.
iT
To determine the value of the open circuit voltage, voc, we connect an open circuit across
the terminals of the circuit and then calculate the value of the voltage across that open circuit.
Figure (c) shows the circuit from Figure 4.6-4a after adding the open circuit and labeling the
open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing
mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively.
In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently,
i2 = 0 A . The controlling current of the CCVS is expressed in terms of the mesh currents as
i a = i1 − i 2 = i1 − 0 = i1
Apply KVL to mesh 1 to get
3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 3 i1 − 2 ( i1 − 0 ) + 6 ( i1 − 0 ) − 10 = 0
⇒ i1 =
10
= 1.43 A
7
Apply KVL to mesh 2 to get
5 i 2 + voc − 6 ( i1 − i 2 ) = 0 ⇒ voc = 6 ( i1 ) = 6 (1.43) = 8.58 V
Figure (c) Calculating the open circuit voltage, voc, using mesh equations.
As a check, notice that R t isc = ( 7.57 )(1.13) = 8.55 ≈ voc
(checked using LNAP 8/16/02)
3Ω
P 5.5-6 The circuit shown in
Figure P 5.5-6b is the Norton
equivalent circuit of the circuit shown
in Figure P 5.5-6a. Find the value of
the short-circuit current, isc, and
Thévenin resistance, Rt.
6Ω
+
–
+
24 V
va
1.33va
isc
Rt
–
Answer: isc = –24 A and Rt = –3 Ω
(a)
(b)
Figure P 5.5-6
Solution:
To determine the value of the short circuit current, Isc, we connect a short circuit across the
terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a)
shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuit
current. Also, the nodes have been identified and labeled in anticipation of writing node
equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage.
Consequently, v1 = −24 V . The voltage at node 3 is equal to the voltage across a short, v3 = 0 .
The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The
voltage at node 3 is equal to the voltage across a short, i.e. v3 = 0 .
Apply KCL at node 2 to get
v1 − v 2
3
=
v 2 − v3
6
⇒ 2 v1 + v 3 = 3 v 2
⇒ − 48 = 3 v a
⇒ v a = −16 V
Apply KCL at node 3 to get
v 2 − v3
6
+
4
v 2 = isc
3
⇒
9
v a = isc
6
⇒ isc =
9
( −16 ) = −24 A
6
Figure (a) Calculating the short circuit current, Isc, using mesh equations.
To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage
source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across
the terminals of the circuit and then label the voltage across that current source as shown in
Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current
source as
v
R th = T
iT
Also, the nodes have been identified and labeled in anticipation of writing node equations. Let
v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. v1 = 0 . The
controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The
voltage at node 3 is equal to the voltage across the current source, i.e. v3 = vT .
Apply KCL at node 2 to get
v1 − v 2
3
=
v2 − v3
6
⇒ 2 v1 + v 3 = 3 v 2
⇒ vT = 3 v a
Apply KCL at node 3 to get
v2 − v3
6
+
4
v 2 + iT = 0 ⇒ 9 v 2 − v3 + 6 iT = 0
3
⇒ 9 v a − vT + 6 iT = 0
⇒ 3 v T − vT + 6 iT = 0 ⇒ 2 vT = −6 iT
Finally,
Rt =
vT
= −3 Ω
iT
vT
, using mesh equations.
iT
To determine the value of the open circuit voltage, voc, we connect an open circuit across
the terminals of the circuit and then calculate the value of the voltage across that open circuit.
Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling the
open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing
node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively.
Figure (b) Calculating the Thevenin resistance, R th =
In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage.
Consequently, v1 = −24 V . The controlling voltage of the VCCS, va, is equal to the node voltage
at node 2, i.e. va = v2 . The voltage at node 3 is equal to the open circuit voltage, i.e. v3 = voc .
Apply KCL at node 2 to get
v1 − v 2
=
v 2 − v3
3
Apply KCL at node 3 to get
6
v2 − v3
⇒ 2 v1 + v 3 = 3 v 2
⇒ − 48 + v oc = 3 v a
4
v 2 = 0 ⇒ 9 v 2 − v 3 = 0 ⇒ 9 v a = v oc
6
3
Combining these equations gives
+
3 ( −48 + voc ) = 9 v a = voc
⇒ voc = 72 V
Figure (c) Calculating the open circuit voltage, voc, using node equations.
As a check, notice that
R th I sc = ( −3)( −24 ) = 72 = Voc
(checked using LNAP 8/16/02)
P 5.5-7 Determine the value of the
resistance R in the circuit shown in
Figure P 5.5-7 by each of the following
methods:
(a)
(b)
5 kΩ
10 kΩ
ib
25 V +–
Replace the part of the circuit to
the left of terminals a–b by its the
Norton equivalent circuit. Use
current division to determine the
value of R.
4 ib
a
R
0.5 mA
b
Figure P 5.5-7
Analyze the circuit shown Figure P 5.5-6 using mesh equations. Solve the mesh
equations to determine the value of R.
Solution: (a) Replace the part of the circuit that is connected to the left of terminals a-b by its
Norton equivalent circuit:
Apply KCL at the top node of the dependent
source to see that i b = 0 A . Then
( )
v oc = 25 + 5000 i b = 25 V
Apply KVL to the supermesh corresponding to the
dependent source to get
( )
−5000 i b + 10000 3 i b − 25 = 0 ⇒ i b = 1 mA
Apply KCL to get
i sc = 3 i b = 3 mA
Then
Rt =
v oc
i sc
= 8.33 kΩ
Current division gives
0.5 =
8333
3 ⇒ R = 41.67 kΩ
R + 8333
(b)
Notice that i b and 0.5 mA are the mesh currents.
Apply KCL at the top node of the dependent
source to get
1
i b + 0.5 × 10−3 = 4 i b ⇒ i b = mA
6
Apply KVL to the supermesh corresponding to
the dependent source to get
(
)
−5000 i b + (10000 + R ) 0.5 × 10−3 − 25 = 0
(
)
⎛1
⎞
−5000 ⎜ × 10−3 ⎟ + (10000 + R ) 0.5 × 10−3 = 25
⎝6
⎠
125
6
R=
= 41.67 kΩ
0.5 ×10−3
P5.5-8 Find the Norton equivalent circuit of this circuit:
Solution
Simplify the circuit using a source transformation:
Identify the open circuit voltage and short circuit current.
Apply KVL to the mesh to get:
(10 + 2 + 3) i x − 15 = 0
⇒ ix = 1 A
Then
v oc = 3 i x = 3 V
Express the controlling current of the
CCVS in terms of the mesh currents:
i x = i1 − i sc
The mesh equations are
10 i1 + 2 ( i1 − i sc ) + 3 ( i1 − i sc ) − 15 = 0 ⇒ 15 i1 − 5 i sc = 15
and
i sc − 3 ( i1 − i sc ) = 0 ⇒ i1 =
4
i sc
3
so
⎛4 ⎞
15 ⎜ i sc ⎟ − 5 i sc = 15 ⇒ i sc = 1 A
⎝3 ⎠
The Thevenin resistance is
Rt =
3
=3Ω
1
Finally, the Norton equivalent circuit is
(checked: LNAP 6/21/04)
P5.5-9 Find the Norton equivalent circuit of this circuit:
Solution
Identify the open circuit voltage and short circuit current.
⎛1⎞
v1 = ⎜ ⎟ 3 = 1 V
⎝ 3⎠
Then
v oc = v1 − 4 ( 2.5 v 1 ) = −9 V
⎛1
⎞
v1 = 3 ⎜ − i sc ⎟ = 1 − 3 i sc
⎝3
⎠
4 ( 2.5 v1 + i sc ) + 5 i sc − v1 = 0
⇒ 9 v1 + 9 i sc = 0
9 (1 − 3 i sc ) + 9 i sc = 0 ⇒ i sc =
The Thevenin resistance is
Rt =
1
A
2
−9
= −18 Ω
0.5
Finally, the Norton equivalent circuit is
(checked: LNAP 6/21/04)
P 5.5-10 An ideal ammeter is
modeled as a short circuit. A more
realistic model of an ammeter is a
small resistance. Figure P 5.5-10a
shows a circuit with an ammeter
that measures the current im. In
Figure P 5.5-10b the ammeter is
replaced by the model of an ideal
ammeter, a short circuit. The
ammeter measures imi, the ideal
value of im.
As Rm → 0, the ammeter
becomes an ideal ammeter and im
→ imi. When Rm > 0, the ammeter is
not ideal and im < imi. The
difference between im and imi is a
measurement error caused by the
fact that the ammeter is not ideal.
(a)
Determine the value of imi.
(b)
Express the measurement
error that occurs when
Rm = 20 Ω as a percentage
of imi.
(c)
im
4 kΩ
3 mA
Ammeter
4 kΩ
2 kΩ
(a)
imi
4 kΩ
3 mA
4 kΩ
2 kΩ
(b)
im
4 kΩ
3 mA
Determine the maximum
value of Rm required to
ensure that the measurement
error is smaller than 2
percent of imi.
4 kΩ
2 kΩ
(c)
Figure P 5.5-10
Solution:
Replace the circuit by its Norton equivalent circuit:
⎛ 1600 ⎞
im = ⎜
1.5 × 10−3 )
⎜ 1600 + R m ⎟⎟ (
⎝
⎠
(a)
Rm
i mi = lim
R m →0
i m = 1.5 mA
(b) When Rm = 20 Ω then i m = 1.48 mA so
% error =
1.5 − 1.48
× 100 = 1.23%
1.5
(c)
⎛ 1600 ⎞
0.015 − ⎜
0.015 )
⎜ 1600 + R m ⎟⎟ (
⎝
⎠
0.02 ≥
0.015
⇒
1600
≥ 0.98 ⇒ R m ≤ 32.65 Ω
1600 + R m
(checked: LNAP 6/18/04)
3Ω
6Ω
P 5.5-11 Determine values of Rt and isc
that cause the circuit shown in
Figure P 5.5-11b to be the Norton
equivalent circuit of the circuit in
Figure P 5.5-11a.
+
–
+
–
12 V
ia
a
a
isc
2ia
Rt
b
Answer: Rt = 3 Ω and isc = –2 A
b
(a)
(b)
Figure P 5.5-11
Solution:
ia =
2 i a − 12
⇒ i a = −3 A
6
voc = 2 i a = −6 V
12 + 6 i a = 2 i a
3 i sc = 2 i a
Rt =
−6
=3Ω
−2
⇒ i a = −3 A
⇒ i sc =
2
( − 3 ) = −2 A
3
12 Ω
P 5.5-12 Use Norton’s theorem to formulate a general
expression for the current i in terms of the variable
resistance R shown in Figure P 5.5-12.
8 Ω
a
30 V
+
–
i
R
b
Answer: i = 20/(8 + R) A
Figure P 5.5-12
Solution:
12 × 24 12 × 24
=
= 8Ω
12 + 24
36
24
voc =
( 30 ) = 20 V
12 + 24
Rt =
i=
20
8+ R
16 Ω
Section 5-6: Maximum Power Transfer
P 5.6-1 The circuit model for a photovoltaic cell is given
in Figure P 5.6-1 (Edelson, 1992). The current is is
proportional to the solar insolation (kW/m2).
(a)
(b)
Find the load resistance, RL, for maximum power
transfer.
Find the maximum power transferred when is = 1 A.
8Ω
ia
R2
+
+
–
vs
+
–
4 ia
v
R
–
i
Figure P 5.6-1
Solution:
(a) The value of the current in R 2 is 0 A so v oc = 4 i a . Then
KVL gives
8 i a + 4 i a − V s = 0 ⇒ V s = 12 i a = 3 ( 4 i a ) = 3 ( v oc ) = 24 V
Next, KVL gives
8 i a + 4 i a − 24 = 0 ⇒ i a = 2 A
and
4 i a = R 2 i sc
⇒ 4 ( 2) = R2 ( 2) ⇒ R2 = 4 Ω
(b) The power delivered to the resistor to the right of the terminals is maximized by setting R
equal to the Thevenin resistance of the part of the circuit to the left of the terminals:
R = Rt =
Then
p max =
v oc
i sc
v oc 2
4 Rt
=
=
8
=4Ω
2
82
=4W
4 ( 4)
1Ω
P 5.6-2 For the circuit in Figure P 5.6-2, (a) find R such
that maximum power is dissipated in R and (b) calculate
is
the value of maximum power.
100 Ω
Answer: R = 60 Ω and Pmax = 54 mW
Figure P 5.6-2
Solution:
a) For maximum power transfer, set RL equal
to the Thevenin resistance:
R L = R t = 100 + 1 = 101 Ω
b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin
equivalent circuit:
The voltage across RL is
vL =
Then
pmax
101
(100 ) = 50 V
101 + 101
2
vL
502
=
=
= 24.75 W
R L 101
RL
150 Ω
P 5.6-3 For the circuit in Figure P 5.6-3, prove
that for Rs variable and RL fixed, the power
dissipated in RL is maximum when Rs = 0.
6V
+
–
100 Ω
R
+
–
2V
Figure P 5.6-3
Solution:
Reduce the circuit using source transformations:
Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the value
of that maximum power is
P
= i 2 ( R) = (0.03)2 (60) = 54 mW
max R
Rs
P 5.6-4 Find the maximum power to the load RL if the
maximum power transfer condition is met for the circuit of
Figure P 5.6-4.
vs
+
–
RL
Answer: max pL = 0.75 W
source
network
load
Figure P 5.6-4
Solution:
⎡ RL ⎤
v L = vS ⎢
⎥
⎣⎢ R S + R L ⎦⎥
∴ pL =
v L2
RL
=
v S2 R L
( RS + R L )2
By inspection, pL is max when you reduce RS to get the
smallest denominator.
∴ set RS = 0
8Ω
P 5.6-5 Determine the maximum power that can be
absorbed by a resistor, R, connected to terminals a–b
of the circuit shown in Figure P 5.6-5. Specify the
required value of R.
20 Ω
a
10 Ω
20 A
120 Ω
50 Ω
b
Figure P 5.6-5
Solution:
The required value of R is
R = Rt = 8 +
( 20 + 120 ) (10 + 50 ) = 50 Ω
( 20 + 120 ) + (10 + 50 )
30
⎡ 170
voc = ⎢
( 20 )⎤⎥ 10 − ⎡⎢
( 20 )⎤⎥ 50
⎣170 + 30
⎦
⎣170 + 30
⎦
170(20)(10) − 30(20)(50) 4000
=
=
= 20 V
200
200
The maximum power is given by
2
voc
202
pmax =
=
=2W
4 R t 4 ( 50 )
P 5.6-6 Figure P 5.6-6 shows a source
Ro
connected to a load through an
amplifier. The load can safely receive
+
up to 15 W of power. Consider three
+
+
Ava
v
500
mV
100 kΩ
RL
a
cases:
–
–
–
(a)
A = 20 V/V and Ro = 10 Ω.
Determine the value of RL that
source
amplifier
load
maximizes the power delivered
Figure P 5.6-6
to the load and the
corresponding maximum load
power.
(b)
A = 20 V/V and RL = 8 Ω. Determine the value of Ro that maximizes the power delivered
to the load and the corresponding maximum load power.
(c)
Ro = 10 Ω and RL = 8 Ω. Determine the value of A that maximizes the power delivered to
the load and the corresponding maximum load power.
Solution:
iL =
A
vs
Ro +RL
PL = i L R L =
2
A 2v s 2 R L
(R
o
+RL)
2
(a) R t =R o so R L =R o = 10 Ω maximizes the power delivered to the load. The corresponding
load power is
2
⎛1⎞
20 ⎜ ⎟ 10
⎝2⎠
PL =
= 2.5 W .
2
(10 + 10 )
2
(b) Ro = 0 maximizes PL (The numerator of PL does not depend on Ro so PL can be maximized
by making the denominator as small as possible.) The corresponding load power is
2
PL =
A 2v s 2 R L
R L2
=
A 2v s 2
RL
⎛1⎞
20 ⎜ ⎟
⎝ 2 ⎠ = 12.5 W.
=
8
2
(c) PL is proportional to A2 so the load power continues to increase as A increases. The load can
safely receive 15 W. This limit corresponds to
2
⎛1⎞
A ⎜ ⎟ 8
⎝2⎠
15 =
2
(18)
2
⇒
A = 36
15
= 49.3 V.
8
(checked: LNAP 6/9/04)
P 5.6-7 The circuit in Figure P 5.6-7 contains a variable resistance, R, implemented using a
potentiometer. The resistance of the variable resistor varies over the range 0 ≤ R ≤ 1000 Ω. The
variable resistor can safely receive 1/4 W power. Determine the maximum power received by the
variable resistor. Is the circuit safe?
180 Ω
+
–
10 V
R
120 Ω
150 Ω
470 Ω
+
–
20 V
Figure P 5.6-7
Solution:
Replace the part of the circuit connected to the variable resistor by its Thevenin equivalent
circuit. First, replace the left part of the circuit by its Thevenin equivalent:
⎛ 150 ⎞
v oc1 = ⎜
⎟10 = 4.545 V
⎝ 150 + 180 ⎠
R t1 = 180 & 150 = 81.8 Ω
Next, replace the right part of the circuit by its Thevenin equivalent:
⎛ 470 ⎞
v oc2 = ⎜
⎟ 20 = 15.932 V
⎝ 470 + 120 ⎠
R t2 = 120 & 470 = 95.6 Ω
Now, combine the two partial Thevenin equivalents:
v oc = v oc1 − v oc2 = −10.387 V and R t = R t1 + R t2 = 177.4 Ω
So
The power received by the adjustable resistor
will be maximum when R = Rt = 177.4 Ω. The
maximum power received by the adjustable
( −11.387 ) = 0.183 W .
p=
4 (177.4 Ω )
2
resistor will be
(checked LNAPDC 7/24/04)
Rt
P 5.6-8 For the circuit of Figure P 5.6-8, find the
power delivered to the load when RL is fixed and
Rt may be varied between 1 Ω and 5 Ω. Select Rt
so that maximum power is delivered to RL.
10 V
Answer: 13.9 W
+
–
RL = 5 Ω
Figure P 5.6-8
Solution:
⎛ 10 ⎞ ⎡ R L
⎤
100 R L
p=iv=⎜
10 ) ⎥ =
(
⎟
⎢
2
⎜ Rt + R L ⎟ ⎢ Rt + R L
⎝
⎠⎣
⎦⎥ ( R t + R L )
The power increases as Rt decreases so choose Rt = 1 Ω. Then
pmax = i v =
100 ( 5 )
(1 + 5)
2
= 13.9 W
5
P 5.6-9 A resistive circuit was connected
to a variable resistor, and the power
delivered to the resistor was measured as
shown in Figure P 5.6-9. Determine the
Thévenin equivalent circuit.
Power
(W)
0
Answer: Rt = 20 Ω and voc = 20 V
10
20
30
R (ohms)
Figure P 5.6-9
Solution:
From the plot, the maximum power is 5 W when R = 20 Ω. Therefore:
Rt = 20 Ω
and
2
pmax
v
= oc
4 Rt
⇒ voc =
pmax 4 Rt = 5 ( 4 ) 20 = 20 V
40
Figure P5.6-10
P5.6-10 The part circuit shown in Figure P5.6-10a to left of the terminals can be reduced to its
Norton equivalent circuit using source transformations and equivalent resistance. The resulting
Norton equivalent circuit, shown in Figure P5.6-10b, will be characterized by the parameters:
i sc = 1.5 A and
R t = 80 Ω
(a) Determine the values of i s and R1 .
(b) Given that 0 ≤ R 2 ≤ ∞ , determine the maximum value of p = vi, the power delivered to R2.
Solution: Two source transformations reduce the circuit as follows:
(a) Recognizing the parameters of the Norton equivalent circuit gives:
1.5 = i sc = i s + 0.25 ⇒ i s = 1.25 A and 80 = R t = 100 || R1 =
100 R1
100 + R1
⇒ R1 = 400 Ω
(b) The maximum value of the power delivered to R2 occurs when R 2 = R t = 80 Ω . Then
2
1
⎛1 ⎞
i = i sc = 0.75 A and p = ⎜ i sc ⎟ R t = ( 0.6252 ) 80 = 45 W
2
⎝2 ⎠
P5.6-11. Given that 0 ≤ R ≤ ∞ in the circuit shown in
Figure P5.6-12, determine (a) maximum value of i a , (b)
the maximum value of v a , and (c) the maximum value of
pa = iava .
Figure P5.6-11
Solution:
Replace the parallel combination of resistor R and the 8 Ω resistor by an equivalent resistance.
Using voltage division
va =
R eq
4 + R eq
(12 ) =
1
(12 )
4
+1
R eq
Consequently, the maximum value of va corresponds to the is obtained by maximizing Req. The
maximum of Req is obtained by maximizing R. Given that 0 ≤ R ≤ ∞, the maximum value of Req
is 8 Ω and the maximum value of va is
v a max =
Using Ohm’s law
1
4
+1
8
(12 ) = 8 V
ia =
12
4 + R eq
Consequently, the maximum value of ia corresponds to the is obtained by minimizing Req. The
minimum of Req is obtained by maximizing R. Given that 0 ≤ R ≤ ∞, the minimum value of Req
is 0 Ω and the maximum value of ia is
i a max =
12
=3A
4+0
The maximum power theorem indicates that the maximum value of p a = i a v a occurs when Req =
Rt. In this case, Rt = 4 Ω. We require Req = 4 Ω which is accomplished by making R = 8 Ω, an
acceptable value since
0 ≤ 8 ≤ ∞. Then
2
2
⎛ 12 ⎞
⎛ 12 ⎞
⎜ ⎟
⎜ ⎟
2
2
pa = ⎝ ⎠ = ⎝ ⎠ = 9 W
R eq
4
P5.6-12. Given that 0 ≤ R ≤ ∞ in the circuit shown in
Figure P5.6-12, determine value of R that maximizes the
power p a = i a v a and the corresponding maximum value
of p a .
Figure P5.6-12
Solution:
Replace the combination of resistor R and the 20 Ω and 2 Ω resistors by an equivalent resistance.
The maximum power theorem indicates that the maximum value of p a = i a v a occurs when Req =
Rt. In this case, Rt = 8 Ω. We require
8 = R eq =
20 ( R + 2 )
20 R + 40
=
20 + ( R + 2 )
R + 22
8 ( R + 22 ) = 20 R + 40 ⇒ R =
8 ( 22 ) − 40
= 11.333 Ω
20 − 8
This isn’t a standard resistance value but it is an acceptable value for this problem since 0 ≤
11.333 ≤ ∞. Then
2
2
⎛6⎞
⎛6⎞
⎜ ⎟
⎜ ⎟
2
2
p a = ⎝ ⎠ = ⎝ ⎠ = 1.125 W
8
R eq
Section 5.8 Using PSpice to Determine the Thevenin Equivalent Circuit
P5.8-1
a) Here are the results of simulating the circuit in PSpice. The numbers shown in white on a
black background are the node voltages.
b) Add a resistor across the terminals of
Circuit A. Then
v oc = v R
i sc =
vR
R
when
R≈∞
when
R≈0
1
Here are the PSpice simulation results:
v oc = 10 V
i sc =
Rt =
1× 10−6
=1 A
1× 10−6
v oc
i sc
=
10
= 10 Ω
1
c) Here is the result of simulation the circuit after replacing Circuit A by its Thevenin equivalent:
d) The node voltages of Circuit B are the same before and after replacing Circuit A by its
Thevenin equivalent circuit.
2
Section 5-9 How Can We Check…?
P 5.9-1 For the circuit of Figure P 5.9-1, the current i has been measured for three different
values of R and is listed in the table. Are the data consistent?
1 kΩ
ix
R(Ω)
i(mA)
5000ix
5000
500
0
16.5
43.8
97.2
+ –
R
i
4 kΩ
+
–
10 V
4 kΩ
Figure P 5.9-1
Solution:
Use the data in the first two lines of the table to determine voc
and Rt:
voc ⎫
R t + 0 ⎪⎪
⎧voc = 39.9 V
⎬ ⇒ ⎨
voc ⎪
⎩ R t = 410 Ω
0.0438 =
R t + 500 ⎪⎭
0.0972 =
Now check the third line of the table. When R= 5000 Ω:
v
39.9
i = oc =
= 7.37 mA
R t + R 410 + 5000
which disagree with the data in the table.
The data is not consistent.
i
P 5.9-2
R
v –
+
Your lab partner built the circuit
shown in Figure P 5.9-2 and measured the
6 kΩ
current i and voltage v corresponding to
several values of the resistance R. The results
12 V
+
–
2 mA
18 kΩ
are shown in the table in Figure P 5.9-2. Your
lab partner says that RL = 8000 Ω is required
to cause i = 1 mA. Do you agree? Justify your
answer.
24 kΩ
R
i
v
open
10 kΩ
short
0 mA
0.857 mA
3 mA
12 V
8.57 V
0V
Figure P 5.9-2
Solution:
Use the data in the table to determine voc and isc:
voc = 12 V
(line 1 of the table)
isc = 3 mA
so
Rt =
(line 3 of the table)
voc
= 4 kΩ
isc
Next, check line 2 of the table. When R = 10 kΩ:
v
12
i = oc =
= 0.857 mA
3
R t + R 10 (10 ) + 5 (103 )
To cause i = 1 mA requires
which agrees with the data in the table.
v
12
0.001 = i = oc =
⇒ R = 8000 Ω
R t + R 10 (103 ) + R
I agree with my lab partner’s claim that R = 8000 causes i = 1 mA.
12 kΩ
P 5.9-3
In preparation for lab, your lab
3R
partner determined the Thévenin
equivalent of the circuit connected to RL in
2R
Figure P 5.9-3. She says that the Thévenin
resistance is Rt =
voltage is voc =
60
11
6
R
11
and the open-circuit
30 V
20 V
V. In lab, you built the
i
R
+
–
+
–
Load
10 V
+
–
circuit using R = 110 Ω and RL = 40 Ω and
measured that i = 54.5 mA. Is this
Figure P 5.9-3
measurement consistent with the prelab
calculations? Justify your answers.
Solution:
1
1 1
1
11
= +
+
=
R t R 2 R 3R 6 R
⇒ Rt =
6R
11
and
⎛ 23 ⎞
⎛ 34 ⎞
⎛ 65 ⎞
180
voc = ⎜
⎟ 30 + ⎜
⎟ 20 + ⎜
⎟ 10 =
11
⎝ 3+ 2 3⎠
⎝ 2+3 4⎠
⎝ 1+ 6 5 ⎠
so the prelab calculation isn’t correct.
But then
180
180
voc
11
i=
=
= 11 = 163 mA ≠ 54.5 mA
6
Rt + R
(110 ) + 40 60 + 40
11
so the measurement does not agree with the corrected prelab calculation.
RL
P 5.9-4 Your lab partner claims that the current i in Figure P 5.9-4 will be no greater than 12.0
mA, regardless of the value of the resistance R. Do you agree?
i
500 Ω
R
12 V
+
–
3 kΩ
6 kΩ
1500 Ω
Figure P 5.9-4
Solution:
6000 & 3000 & ( 500 + 1500 ) = 2000 & 2000 = 1000 Ω
i=
12
12
≤
= 12 mA
R + 1000 1000
How about that?! Your lab partner is right.
(checked using LNAP 6/21/05)
P 5.9-5 Figure P 5.9-5 shows a circuit and some corresponding data. Two resistances, R1 and
R, and the current source current are unspecified. The tabulated data provide values of the
current, i, and voltage, v, corresponding to several values of the resistance R.
(a)
Consider replacing the part of the circuit connected to the resistor R by a Thévenin
equivalent circuit. Use the data in rows 2 and 3 of the table to find the values of Rt and
voc, the Thévenin resistance and the open-circuit voltage.
(b)
Use the results of part (a) to verify that the tabulated data are consistent.
(c)
Fill in the blanks in the table.
(d)
Determine the values of R1 and is.
12 V
+
–
24 Ω
is
i
+
v
R
–
R, Ω
i, A
v, V
0
10
20
40
80
3
1.333
0.857
0.5
?
0
13.33
17.14
?
21.82
18 Ω
(b)
R1
12 Ω
(a)
Figure P5.9-5
Solution:
(a)
KVL gives
v oc = ( R t + R ) i
from row 2
from row 3
So
(R
t
v oc = ( R t + 10 ) (1.333)
v oc = ( R t + 20 ) ( 0.857 )
+ 10 ) (1.333) = ( R t + 20 ) ( 0.857 )
28 ( R t + 10 ) = 18 ( R t + 20 )
Solving gives
10 R t = 360 − 280 = 80
and
⇒
Rt = 8 Ω
v oc = ( 8 + 10 )(1.333) = 24 V
(b)
i=
v oc
Rt + R
=
24
R
24 R
and v =
v oc =
8+ R
R + Rt
R+8
When R = 0, i = 3 A, and v = 0 V.
1
When R = 40 Ω, i = A .
2
24 ( 80 ) 240
=
= 21.82 .
When R = 80 Ω, v =
88
11
These are the values given in the tabulated data so the data is consistent.
24 ( 40 )
(c) When R = 40 Ω, v =
= 20 V .
48
24
= 0.2727 A .
When R = 80 Ω, i =
88
(d) First
8 = R t = 24 & 18 & ( R1 + 12 )
⇒
R1 = 24 Ω
the, using superposition,
24 = v oc =
(
24
24 + 18 & ( R1 + 12 )
)
(
)
12 + 24 & 18 ( R1 + 12 ) i s = 8 + 8i s
⇒
is = 2 A
(checked using LNAP 6/21/05)
PSpice Problems
SP 5-1 The circuit in Figure SP 5.1 has three inputs: v1, v2, and i3. The circuit has one output,
vo. The equation
vo = av1 + bv2 + ci3
expresses the output as a function of the inputs. The coefficients a, b, and c are real constants.
(a)
Use PSpice, and the principle of superposition, to determine the values of a, b, and c.
(b)
Suppose v1 = 10 V, v2 = 8 V, and we want the output to be vo = 7 V. What is the required
value of i3?
Hint:
The output is given by vo = a when v1 = 1 V, v2 = 0 V, and i3 = 0 A.
Answer: (a) vo = 0.3333v1 + 0.3333v2 + 33.33i3,
100 Ω
(b) i3 = 30 mA
v2
+–
+
–
v1
+
vo
–
100 Ω
100 Ω
i3
Figure SP 5.1
Solution:
a = 0.3333
b = 0.3333
c =33.33 V/A
(a)
(b)
vo = 0.3333 v1 + 0.3333 v2 + 33.33 i 3
7 = 0.3333 (10 ) + 0.3333 ( 8 ) + 33.33 i 3
18
3 = 3 = 30 mA
⇒ i3 =
100
100
3
7−
SP 5-2 The pair of terminals a–b partitions the circuit in Figure SP 5.2 into two parts. Denote
the node voltages at nodes 1 and 2 as v1 and v2. Use PSpice to demonstrate that performing a
source transformation on the part of the circuit to the left of the terminal does not change
anything to the right of the terminals. In particular, show that the current, io, and the node
voltages, v1 and v2, have the same values after the source transformation as before the source
transformation.
100 Ω
a
1
8V
2
+–
10 V
+
–
100 Ω
io
100 Ω
b
Figure SP 5.2
Solution:
Before the source transformation:
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V1
V_V2
-3.000E-02
-4.000E-02
After the source transformation:
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V2
-4.000E-02
30 mA
0.75va
SP 5-3 Use PSpice to find the Thévenin equivalent circuit for
the circuit shown in Figure SP 5.3.
8Ω
Answer: voc = –2 V and Rt = –8/3 Ω
+
–
Solution:
6V
a
–
va
+
4Ω
b
Figure SP 5.3
voc = −2 V
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V3
V_V4
-7.500E-01
7.500E-01
isc = 0.75 A
Rt = −2.66 Ω
SP 5-4 The circuit shown in Figure SP 5-4b is the Norton equivalent circuit of the circuit
shown in Figure SP 5-4a. Find the value of the short-circuit current, isc, and Thévenin resistance,
Rt.
Answer: isc = 1.13 V and Rt = 7.57 Ω
5Ω
3Ω
– +
+
–
10 V
2ia
6Ω
isc
Rt
ia
(a)
(b)
Figure SP 5-4
Solution:
voc = 8.571 V
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V5
-2.075E+00
V_V6
1.132E+00
X_H1.VH_H1 9.434E-01
isc = 1.132 A
Rt = 7.571 Ω
Design Problems
R1
i
+
DP 5-1 The circuit shown in Figure DP 51a has four unspecified circuit parameters:
vs, R1, R2, and R3. To design this circuit, we
must specify the values of these four
parameters. The graph shown in Figure DP
5-1b describes a relationship between the
current i and the voltage v.
Specify values of vs, R1, R2, and R3
that cause the current i and the voltage v in
Figure DP 5-1a to satisfy the relationship
described by the graph in Figure DP 5-1b.
First Hint: The equation representing the
straight line in Figure DP 5-1b is
v = –Rti + voc
That is, the slope of the line is equal to –1
times the Thévenin resistance and the “vintercept” is equal to the open-circuit
voltage.
Second Hint: There is more than one correct
answer to this problem. Try setting R1 = R2.
R3
+
–
R2
vs
v
–
(a)
v, V
6
4
2
–6
–4
–2
2
4
6
8
i, mA
–2
–4
–6
–8
(b)
Figure DP 5-1
Solution:
The equation of representing the straight line in Figure DP 5-1b is v = − R t i + voc . That is, the
slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the
0−5
open circuit voltage. Therefore: R t = −
= 625 Ω and voc = 5 V.
0.008 − 0
Try R1 = R 2 = 1 kΩ . (R1 || R2 must be smaller than Rt = 625 Ω.) Then
5=
R2
1
vs = vs
R1 + R 2
2
⇒ vs = 10 V
and
R1 R2
= R3 + 500 ⇒ R3 = 125 Ω
R1 + R2
Now vs, R1, R2 and R3 have all been specified so the design is complete.
625 = R 3 +
DP 5-2 The circuit shown in Figure DP 5.2a has four unspecified circuit parameters: is, R1, R2,
and R3. To design this circuit, we must specify the values of these four parameters. The graph
shown in Figure DP 5-2b describes a relationship between the current i and the voltage v.
Specify values of is, R1, R2, and R3 that cause the current i and the voltage v in Figure DP
5-2a to satisfy the relationship described by the graph in Figure DP 5-2b.
First Hint: Calculate the open-circuit voltage, voc, and the Thévenin resistance, Rt, of the part of
the circuit to the left of the terminals in Figure DP 5-2a.
Second Hint: The equation representing the straight line in Figure DP 5-2b is
v = –Rti + voc
That is, the slope of the line is equal to –1 times the Thévenin resistance and the “v-intercept” is
equal to the open-circuit voltage.
Third Hint: There is more than one correct answer to this problem. Try setting both R3 and R1 +
R2 equal to twice the slope of the graph in Figure DP 5-2b.
v, V
6
4
R2
i
2
+
is
R1
R3
v
–6
–
–4
–2
2
4
6
8
i, mA
–2
(a)
–4
–6
–8
(b)
Figure DP 5-2
Solution:
The equation of representing the straight line in Figure DP 5-2b is v = − R t i + voc . That is, the
slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the
0 − ( −3 )
open circuit voltage. Therefore: R t = −
= 500 Ω and voc = −3 V.
−0.006 − 0
From the circuit we calculate
R 3 ( R1 + R 2 )
R1 R 3
is
and voc = −
Rt =
R1 + R 2 + R 3
R1 + R 2 + R 3
so
500 Ω =
R 3 ( R1 + R 2 )
R1 + R 2 + R 3
and −3 V = −
R1 R 3
R1 + R 2 + R 3
is
Try R 3 = 1kΩ and R1 + R 2 = 1kΩ . Then R t = 500 Ω and
−3 = −
1000 R1
is = −
R1
i s ⇒ 6 = R1 i s
2000
2
This equation can be satisfied by taking R1 = 600 Ω and is = 10 mA. Finally, R2 = 1 kΩ - 400 Ω =
400 Ω. Now is, R1, R2 and R3 have all been specified so the design is complete.
R1
i
R3
+
DP 5-3 The circuit shown in
Figure DP 5-3a has four unspecified circuit
parameters: vs, R1, R2, and R3. To design this
circuit, we must specify the values of these
four parameters. The graph shown in
Figure DP 5-3b describes a relationship
between the current i and the voltage v.
Is it possible to specify values of vs,
R1, R2, and R3 that cause the current i and the
voltage v in Figure DP 5-1a to satisfy the
relationship described by the graph in
Figure DP 5-3b? Justify your answer.
+
–
R2
vs
v
–
(a)
v, V
6
4
2
–6
–4
–2
2
4
6
8
i, mA
–2
–4
–6
–8
(b)
Figure DP 5-3
Solution:
The slope of the graph is positive so the Thevenin resistance is negative. This would require
R1 R 2
R3 +
< 0 , which is not possible since R1, R2 and R3 will all be non-negative.
R1 + R 2
Is it not possible to specify values of vs, R1, R2 and R3 that cause the current i and the
voltage v in Figure DP 5-3a to satisfy the relationship described by the graph in Figure DP 5-3b.
DP 5-4 The circuit shown in
Figure DP 5-4a has four unspecified circuit
parameters: vs, R1, R2, and d, where d is the
gain of the CCCS. To design this circuit, we
must specify the values of these four
parameters. The graph shown in
Figure DP 5-4b describes a relationship
between the current i and the voltage v.
Specify values of vs, R1, R2, and d that
cause the current i and the voltage v in Figure
DP 5-4a to satisfy the relationship described
by the graph in Figure DP 5-4b.
+
–
ia
i
R1
+
That is, the slope of the line is equal to –1
times the Thévenin resistance and the
“v-intercept” is equal to the open-circuit
voltage.
Second Hint: There is more than one correct
answer to this problem. Try setting R1 = R2.
v
–
(a)
v, V
6
4
First Hint: The equation representing the
straight line in Figure DP 5-4b is
v = –Rti + voc
R2
dia
vs
2
–6
–4
–2
2
4
6
8
i, mA
–2
–4
–6
–8
(b)
Figure DP 5-4
Solution:
The equation of representing the straight line in Figure DP 5-4b is v = − R t i + voc . That is, the
slope of the line is equal to the Thevenin impedance and the "v - intercept" is equal to the open
−5 − 0
= −625 Ω and voc = −5 V.
circuit voltage. Therefore: R t = −
0 − 0.008
The open circuit voltage, voc, the short circuit current, isc, and the Thevenin resistance, Rt,
of this circuit are given by
R 2 ( d + 1)
voc =
vs
R1 + ( d + 1) R 2
,
( d + 1) v
isc =
s
R1
and
Rt =
R1 R 2
R1 + ( d + 1) R 2
Let R1 = R2 = 1 kΩ. Then
−625 Ω = R t =
and
−5 =
( d + 1) vs
d +2
1000
1000
⇒ d=
− 2 = −3.6 A/A
−625
d +2
⇒ vs =
−3.6 + 2
( − 5) = −3.077 V
−3.6 + 1
Now vs, R1, R2 and d have all been specified so the design is complete.
Chapter 6 The Operational Amplifier
Exercises
Exercise 6.6-1 Specify the values of R1 and R2 in Figure E 6.6-1 that are required to cause v3 to
be related to v1 and v2 by the equation v3 = (4)v1 – ( 15 )v2.
Answer: R1 = 10 kΩ and R2 = 2.5 kΩ
Figure E 6.6-1
Solution:
⎞
R2
R1 ⎞
⎛ 10 × 103 ⎞ ⎛
⎛ 10 ×103 ⎞ ⎛
v3 = ⎜ −
v
+
+
+
v
1
1
⎜
⎟
⎜
⎟
⎜
⎟
2
3
3
3
3 ⎟ 1
⎝ 10 × 10 ⎠ ⎜⎝ R 2 + 10 × 10 ⎟⎠
⎝ 10 × 10 ⎠ ⎝ 10 × 10 ⎠
⎛
⎞
R2
R1 ⎞
⎛
= −⎜
v2 + 2 ⎜1 +
v
⎟
3
3 ⎟ 1
⎜
⎟
⎝ 10 × 10 ⎠
⎝ R 2 + 10 × 10 ⎠
⎛1⎞
We require v3 = ( 4 ) v1 − ⎜ ⎟ v2 , so
⎝5⎠
R1 ⎞
⎛
4 = 2 ⎜1 +
⇒ R1 = 10 × 103 = 10 kΩ
3 ⎟
10
×
10
⎝
⎠
and
R2
1
=
5 R 2 + 10 × 103
⇒ R 2 + 10 × 103 = 5 R 2
⇒
R 2 = 2.5 kΩ
Exercise 6.6-2 Specify the values of R1 and R2 in Figure E 6.6-1 that are required to cause v3 to
be related to v1 and v2 by the equation v3 = (6)v1 – ( 54 )v2.
Answer: R1 = 20 kΩ and R2 = 40 kΩ
Figure E 6.6-1
Solution: As in Ex 6.7-1
⎛
⎞
R2
R1 ⎞
⎛
v3 = − ⎜
v
+
2
1
+
v
⎟
⎜
2
3 ⎟ 1
⎜ R 2 + 10 × 103 ⎟
⎝ 10 × 10 ⎠
⎝
⎠
⎛ 4⎞
We require v3 = ( 6 ) v1 − ⎜ ⎟ v2 , so
⎝5⎠
R1 ⎞
⎛
6 = 2 ⎜1 +
⇒ R1 = 20 ×103 = 20 kΩ
3 ⎟
⎝ 10 ×10 ⎠
and
R2
4
=
5 R 2 + 10 × 103
⇒ 4 R 2 + 40 × 103 = 5 R 2
⇒ R 2 = 40 kΩ
Exercise 6.7-1 The input offset voltage of a typical
μA741 operational amplifier is 1 mV and the bias current
is 80 nA. Suppose the operational amplifier in Figure 6.72a is a typical μA741. Show that the output offset voltage
of the inverting amplifier will be at most 10 mV.
Figure 6.7-2a
Solution:
Analysis of the circuit in Section 6.7 showed that output offset voltage = 6 vos + (50 × 103 ) ib1
For a μ A741 op amp, vos ≤ 1 mV and ib1 ≤ 80 nA so
output offset voltage = 6 vos + (50 × 103 )ib1 ≤ 6 (10−3 )+(50.103 )(80×10−9 ) = 10 mV
Exercise 6.7-2 Suppose the 10-kΩ resistor in Figure 6.72a is changed to 2 kΩ and the 50-kΩ resistor is changed to
10 kΩ. (These changes will not change the gain of the
inverting amplifier. It will still be – 5.) Show that the
maximum output offset voltage is reduced to 35 mV. (Use
ib = 500 nA and vos = 5 mV to calculate the maximum
output offset voltage that could be caused by the μA741
amplifier.)
Figure 6.7-2a
Solution:
vo = −
⎛ R ⎞
R2
vin + ⎜1 + 2 ⎟ vos + R2ib1
R1
⎝ R1 ⎠
When R2 = 10 kΩ, R1 = 2 kΩ, vos ≤ 5 mV and
(
) (
output offset voltage ≤ 6 5 × 10−3 + 10 × 103
ib1 ≤ 500 nA then
) ( 500.10 ) ≤ 35×10
−9
−3
= 35 mV
Exercise 6.7-3 Suppose the μA741 operational amplifier
in Figure 6.7-2a is replaced with a typical OPA101AM
operational amplifier. Show that the output offset voltage
of the inverting amplifier will be at most 0.6 mV.
Figure 6.7-2a
Solution:
Analysis of this circuit in Section 6.7 showed that output offset voltage = 6 vos + ( 50 ×103 ) ib1
For a typical OPA1O1AM, vos = 0.1 mV and ib = 0.012 nA so
output offset voltage ≤ 6 ⎡⎣0.1×10−3 ⎤⎦ + ( 50 ×103 ) ⎡⎣0.012 ×10−9 ⎤⎦
≤ 0.6 ×10−3 + 0.6 ×10−6 − 0.6 ×10−3 = 0.6 mV
Rf
Exercise 6.7-4
(a)
(b)
Ra
Determine the voltage ratio vo/vs for the op amp
circuit shown in Figure E 6.7-4
5
Calculate vo/vs for a practical op amp with A = 10 ,
Ro = 100 Ω, and Ri = 500 kΩ. The circuit resistors
are Rs = 10 kΩ, Rf = 50 kΩ, and Ra = 25 kΩ.
–
+
+
–
vs
Rs
Figure E 6.7-4
Solution:
Writing node equations
v− − vs v− − vo
v−
+
+
=0
Ra
Rb
Ri + Rs
⎛
⎞
R
i v ⎟
vo − ⎜ − A
−
⎜
⎟
R +R
i
s ⎠ + vo − v− = 0
⎝
R0
Rb
Av =
R0 ( Ri + Rs ) + ARi R f
vo
=
vs ( R f + R0 ) ( Ri + Rs ) + Ra ( R f + R0 + Ri + Rs ) − ARi Ra
For the given values, Av = −2.00006 V/V.
vo
–
Answer: (b) vo/vs = – 2
After some algebra
+
Section 6-3: The Ideal Operational Amplifier
20 kΩ
P 6.3-1 Determine the value of voltage
measured by the voltmeter in Figure P 6.3-1.
–
Voltmeter
+
Answer: – 4 V
20 kΩ
50 kΩ
–
+
4V
Figure P 6.3-1
Solution:
(checked using LNAP 8/16/02)
3 kΩ
P 6.3-2
4 kΩ
Find vo and io for the circuit of Figure P 6.3-2.
12 V
io
+
–
2 kΩ
–
+
+
1 kΩ
R
vo
–
Figure P 6.3-2
Solution:
Apply KVL to loop 1:
− 12 + 3000 i1 + 0 + 2000 i1 = 0
12
= 2.4 mA
5000
The currents into the inputs of an ideal op
amp are zero so
io = i1 = 2.4 mA
⇒ i1 =
i2 = − i1 = − 2.4 mA
va = i2 (1000 ) + 0 = −2.4 V
Apply Ohm’s law to the 4 kΩ resistor
vo = va − io ( 4000 )
= −2.4 − ( 2.4 ×10−3 ) ( 4000 ) = −12 V
(checked using LNAP 8/16/02)
4 kΩ
8 kΩ
P 6.3-3 Find vo and io for the circuit of Figure P 6.3-3.
Answer: vo = – 30 V and io = 3.5 mA
io
–
12 V
+
–
+
2V
–
+
+
20 kΩ
vo
–
Figure P 6.3-3
Solution:
The voltages at the input nodes of an ideal op
amp are equal so va = −2 V .
Apply KCL at node a:
vo − ( −2 ) 12 − ( −2 )
+
= 0 ⇒ vo = −30 V
8000
4000
Apply Ohm’s law to the 8 kΩ resistor
io =
−2 − vo
= 3.5 mA
8000
(checked using LNAP 8/16/02)
10 kΩ
P 6.3-4
Find v and i for the circuit of Figure P 6.3-4.
–
0.1 mA
+
v
–
+
+
–
5V
20 kΩ
i
Figure P 6.3-4
Solution:
The voltages at the input nodes of an ideal
op amp are equal so v = 5 V .
Apply KCL at the inverting input node of
the op amp:
⎛ v −5 ⎞
−3
−⎜ a
⎟ − 0.1× 10 − 0 = 0 ⇒ va = 4 V
⎝ 10000 ⎠
Apply Ohm’s law to the 20 kΩ resistor
va
1
i =
= mA
20000
5
(checked using LNAP 8/16/02)
3 kΩ
4 kΩ
P 6.3-5 Find vo and io for the circuit of
Figure P 6.3-5.
Answer: vo = – 15 V and io = 7.5 mA
–
io
+
12 V
+
–
2 mA
6 kΩ
+
vo
–
Figure P 6.3-5
Solution:
The voltages at the input nodes of
an ideal op amp are equal, so
va = 0 V . Apply KCL at node a:
⎛ v − 0 ⎞ ⎛ 12 − 0 ⎞
−3
−⎜ o
⎟−⎜
⎟ − 2 ⋅10 = 0
3000
4000
⎝
⎠ ⎝
⎠
⇒ vo = − 15 V
Apply KCL at the output node of
the op amp:
v
v
io + o + o = 0 ⇒ io = 7.5 mA
6000 3000
(checked using LNAP 8/16/02)
6 kΩ
P 6.3-6 Determine the value of voltage
measured by the voltmeter in
Figure P 6.3-6.
Answer: 7.5 V
+
Voltmeter
–
8 kΩ
+
–
2.5 V
6 kΩ
4 kΩ
Figure P 6.3-6
Soluton:
The currents into the inputs of an ideal op amp
are zero and the voltages at the input nodes of
an ideal op amp are equal so va = 2.5 V .
Apply Ohm’s law to the 4 kΩ resistor:
v
2.5
ia = a =
= 0.625 mA
4000 4000
Apply KCL at node a:
ib = ia = 0.625 mA
Apply KVL:
vo = 8000 ib + 4000 ia
= (12 × 103 )( 0.625 × 10−3 ) = 7.5 V
(checked using LNAP 8/16/02)
P 6.3-7
Find vo and io for the circuit of Figure P 6.3-7.
R1
R2
R4
io
–
+
–
R3
–
+
vs
+
+
R5
vo
–
Figure P 6.3-7
Solution:
Label the circuit to account for the properties of the ideal op amps:
Apply KCL at the inverting node of the left op amp to get:
R2
⎛ vs − 0 ⎞ ⎛ va − 0 ⎞
−⎜
vs
⎟⎟ + 0 = 0 ⇒ v a = −
⎟ − ⎜⎜
R1
⎝ R1 ⎠ ⎝ R 2 ⎠
Apply KCL at the output node of the left op amp to get:
io =
0 − va
R2
+
0 − va
R3
=−
R 2 + R3
R 2 R3
⎛ R 2 + R3 ⎞
va = ⎜
v
⎜ R1 R 3 ⎟⎟ s
⎝
⎠
Apply KCL at the inverting node of the right op amp to get:
⎛ vo − 0 ⎞ ⎛ va − 0 ⎞
R4
R2 R4
−⎜
−⎜
+ 0 = 0 ⇒ vo = −
va =
vs
⎟
⎟
⎜ R 4 ⎟ ⎜ R3 ⎟
R3
R1 R 3
⎝
⎠ ⎝
⎠
6 kΩ
+
P 6.3-8 Determine the current io for the circuit
shown in Figure P 6.3-8.
–
8 kΩ
Answer: io = 2.5 mA
io
6 kΩ
+
–
4 kΩ
2V
4 kΩ
6 kΩ
8 kΩ
–
+
+
–
6 kΩ
5.8 V
Figure P 6.3-8
Solution:
The node voltages have been
labeled using:
1. The currents into the inputs of an
ideal op amp are zero and the
voltages at the input nodes of an
ideal op amp are equal.
2. KCL
3. Ohm’s law
then
and
v0 = 11.8 − 1.8 = 10 V
10
io =
= 2.5 mA
4000
(checked using LNAP 8/16/02)
P 6.3-9 Determine the current io for the
circuit shown in Figure P 6.3-9.
Answer: io = 2.5 mA
4 kΩ
–
+
18 V
4 kΩ
–
a
+
+
b
8 kΩ
8 kΩ
vo
–
Figure P 6.3-9
Solution:
Apply KCL at node a:
v a − ( −18 )
4000
+
va
8000
+ 0 = 0 ⇒ v a = −12 V
The node voltages at the input nodes of
ideal op amps are equal, so v b = v a .
Using voltage division:
vo =
8000
v = −8 V
4000 + 8000 b
(check using LNAP 8/16/02)
R1
P 6.3-10 Determine the current io for the
circuit shown in Figure P 6.3-10.
is
–
Answer: io = 2.5 mA
+
+
R2
R3
vo
–
Figure P 6.3-10
Solution:
Label the circuit as shown. The current in resistor R 3 is
i s . Consequently:
v a = i s R3
Apply KCL at the top node of R 2 to get
⎛ R3 ⎞
va
i=
+ i s = ⎜1 +
i
⎜ R 2 ⎟⎟ s
R2
⎝
⎠
Using Ohm’s law gives
vo − va
R1
⎛ R3 ⎞
= i = ⎜1 +
i
⎜ R 2 ⎟⎟ s
⎝
⎠
⎛
R1 R 3 ⎞
⇒ v o = ⎜ R1 + R 3 +
⎟ is
⎜
R 2 ⎟⎠
⎝
We require
R1 + R 3 +
R1 R 3
R2
= 20
e.g. R1 = 5 kΩ and R 2 = R 3 = 10 kΩ .
(checked: LNAP 6/2/04)
R1
P 6.3-11 Determine the voltage vo for the
circuit shown in Figure P 6.3-11.
+
vo
–
+
–
vs
R2
Answer: vo = – 8 V
R4
R3
Figure P 6.3-11
Solution:
Label the circuit as shown. Apply KCL at the top
node of R 2 to get
vs − va
R1
=
⎛ R2 ⎞
+ 0 ⇒ va = ⎜
⎟⎟ v s
⎜
R2
⎝ R1 + R 2 ⎠
va
Apply KCL at the inverting node of the op amp to
get
R 2 ( R3 + R 4 )
⎛ R3 + R 4 ⎞
⎛ R3 + R 4 ⎞ ⎛ R 2 ⎞
vo − va va
=
+ 0 ⇒ vo = ⎜
va = ⎜
vs =
v
⎟
⎟
⎜
⎟
⎜
⎟
⎜
⎟⎜
⎟
R3
R4
( R1 + R 2 ) R 4 s
⎝ R4 ⎠
⎝ R 4 ⎠ ⎝ R1 + R 2 ⎠
We require
R 2 ( R3 + R 4 )
(R
1
+ R2 ) R4
=5
e.g. R1 = R 2 = 10 kΩ , R 3 = 90 kΩ and R 4 = 10 kΩ .
(checked: LNAP 6/2/04)
5 kΩ
P 6.3-12 The circuit shown in Figure P 6.3-12
has one input, is, and one output, vo. Show that the
output is proportional to the input. Design the
circuit so that the gain is
vo
is
vs
+
–
20 kΩ
–
V
= 20
.
mA
+
1.5 V
+
–
+
10 kΩ
vo
–
Figure P 6.3-12
Solution:
The node voltage at both input nodes of the op amp is
1.5 V. Apply KCL at the inverting input node of the
op amp to get
v s − 1.5
5000
+
v o − 1.5
20000
= 0 ⇒ 4 ( v s − 1.5 ) + v o − 1.5 = 0
⇒ v o = −4 ( v s − 1.5 ) + 1.5 = 0
⇒ v o = −4 v s + 7.5
Comparing this equation to v o = m v s + b , we determine that m = −4 V/V and b = 75. V.
R
P 6.3-13 The circuit shown in Figure P 6.3-13
has one input, vs, and one output, vo. Show that the
output is proportional to the input. Design the
circuit so that vo = 5 vs.
0.2 mA
20 kΩ
ioa
–
+
25 kΩ
1.5 V
+
–
10 kΩ
+
vo
–
Figure P 6.3-13
Solution:
The output of this circuit is v o = 3.5 V.
(a) The current in the 25 kΩ resistor is 0 A
because this current is also the input current of
an ideal op amp. Consequently, the voltage at the
input nodes of the op amp is 1.5 V. Apply KCL
at the inverting input of the op amp to get
1.5
3.5 − 1.5
= 0.2 +
= 0.3
R
20
so
R=
1.5
= 5 kΩ
0.3
(b) The voltage source current is 0A so the voltage source supplies 0 W of power. The voltage
across the current source is equal to the node voltage at the inverting input of the op amp, 1.5 V.
Notice that this voltage and the given current source current do not adhere to the passive
convention so (0.2)(1.5) = 0.3 mW is the power supplied by the current source.
(c) Apply KCL at the output node of the op amp to get
i oa =
3.5 3.5 − 1.5
+
= 0.45 A
10
20
The op amp supplies p oa = i oa × v o = ( 0.45 )( 3.5 ) = 1.575 W
P 6-3.14 Determine the values of the node voltages at nodes a, b, c, and d of the circuit shown in
Figure P 6-3.14.
25 kΩ
5 kΩ
a
20 kΩ c
–
+
b
15 kΩ
–
d
+
2.5 mA
5 kΩ
5 kΩ
10 kΩ
Figure P 6.3-14
Solution:
Label the circuit as shown, noticing that some resistor currents are zero because these currents
are also input currents of op amps:
Now
v a = −2.5 ( 5 ) = −12.5 V , v b = v a = −12.5 V, v c = 0 V
Apply KCL to get
vb − 0
20
= 0+
0 − vd
15
⇒
vd = −
15
3
v b = − ( −12.5 ) = 9.375 V
20
4
P 6-3.15 Determine the values of the node voltages at nodes a, b, c, and d of the circuit shown in
Figure P 6-3.15.
10 kΩ a
40 kΩ
20 kΩ
–
b
10 kΩ c
+
–
+
5V
40 kΩ
–
d
+
10 kΩ
60 kΩ
10 kΩ
40 kΩ
Figure P 6.3-15
Solution:
Label the circuit as shown, noticing that some resistor currents are zero because these currents
are also input currents of op amps:
vb
⎛ 20 ⎞
= 5 V and v d = v c = 5 V.
Now v a = 0 V, v b = − ⎜ ⎟ ( −5 ) = 10 V, v c =
2
⎝ 10 ⎠
Figure P6.3-16
P6.3-16 . Figure P6.3-16 shows four similar circuits. The outputs of the circuits are the voltages
v1, v2, v3 and v4. Determine the values of these four outputs.
Solution:
80
25
80
v1 = − ( 2.1) = −6.72 V , v 2 = − ( −2.1) = 0.65625 V , v 3 = − ( −2.1) = 6.72 V and
25
80
25
25
v 4 = − ( 2.1) = −0.65625 V
80
Figure P6.3-17
P6.3-17 . Figure P6.3-17 shows four similar circuits. The outputs of the circuits are the voltages
v1, v2, v3 and v4. Determine the values of these four outputs.
Solution
⎛ 80 ⎞
⎛ 25 ⎞
⎛ 80 ⎞
v1 = ⎜ 1 + ⎟ ( 2.1) = 8.82 V , v 2 = ⎜1 + ⎟ ( −2.1) = −2.75625 V , v 3 = ⎜ 1 + ⎟ ( −2.1) = −8.82 V a
⎝ 25 ⎠
⎝ 80 ⎠
⎝ 25 ⎠
⎛ 25 ⎞
nd v 4 = ⎜1 + ⎟ ( 2.1) = 2.75625 V .
⎝ 80 ⎠
Section 6-4: Nodal Analysis of Circuits Containing Ideal Operational
Amplifiers
P 6.4-1 Determine the node voltages for the
circuit shown in Figure P 6.4-1.
a
20 kΩ
b
+
d
–
Answer: va = 2 V, vb = – 0.25 V, vc = – 5 V, vd
= – 2.5 V, and ve = – 0.25 V
9 kΩ
2V
+
–
c
40 kΩ
–
+
5V
e
40 kΩ
1 kΩ
Figure P 6.4-1
Solution:
KCL at node b:
vb − 2
vb
v +5
1
+
+ b
= 0 ⇒ vb = − V
20000 40000 40000
4
The node voltages at the input nodes of an ideal op amp are equal so ve = vb = −
KCL at node e:
ve
v −v
10
V
+ e d = 0 ⇒ vd = 10 ve = −
1000 9000
4
1
V.
4
(checked using LNAP 8/16/02)
6 kΩ
P 6.4-2 Find vo and io for the circuit of
Figure P 6.4-2.
Answer: vo = – 4 V and io = 1.33 mA
6 kΩ
6 kΩ
–
+
–
12 V
6 kΩ
io
+
+
6 kΩ
vo
–
Figure P 6.4-2
Solution:
Apply KCL at node a:
0=
va − 12
v
v −0
+ a + a
⇒ va = 4 V
6000 6000 6000
Apply KCL at the inverting input of the op
amp:
⎛v −0⎞
⎛ 0 − vo ⎞
−⎜ a
⎟+0+⎜
⎟ = 0
⎝ 6000 ⎠
⎝ 6000 ⎠
⇒ vo = −va = −4 V
Apply KCL at the output of the op amp:
vo
⎛ 0 − vo ⎞
io − ⎜
= 0
⎟+
⎝ 6000 ⎠ 6000
v
⇒ io = − o = 1.33 mA
3000
(checked using LNAP 8/16/02)
Figure P6.4-3
P6.4-3.
Determine the values of the node voltages, va and vo, of the circuit shown in Figure P6.4-3.
Solution:
Writing node equations:
va
2.25
2.25
+
= 0 ⇒ v a = − 40 × 10 3
= −4.5 V
3
3
20 ×10
40 ×10
20 × 10 3
and
va
va
va − vo
⎛ 8 8 8⎞
+
+
= 0 ⇒ v o = ⎜ + + ⎟ v a = 2 ( −4.5 ) = −9 V
3
3
3
40 ×10 10 ×10
8 × 10
⎝ 40 10 8 ⎠
(
)
+
P 6.4-4 The output of the circuit shown in
Figure P 6.4-4 is vo. The inputs are v1 and v2.
Express the output as a function of the inputs
and the resistor resistances.
–
+
–
v1
+
R1
vo
R2
R3
–
+
+
–
v2
Figure P 6.4-4
Solution:
Ohm’s law:
i=
v1 − v2
R2
KVL:
v0 = ( R1 + R2 + R3 ) i =
R1 + R2 + R3
( v1 − v2 )
R2
–
R3
+
P 6.4-5 The outputs of the circuit shown in
Figure P 6.4-5 are vo and io. The inputs are
v1 and v2. Express the outputs as functions of
the inputs and the resistor resistances.
R5
–
+
–
v1
R1
io
–
R7
+
+
vo
R2
–
–
+
+
–
R4
v2
Figure P 6.4-5
Solution:
⎛ R ⎞
v1 −va v1 −v2
R
+
+ 0 = 0 ⇒ va = ⎜1+ 1 ⎟ v1 − 1 v2
R1
R7
R7
⎝ R7 ⎠
⎛ R ⎞
v2 − vb
v −v
R
− 1 2 + 0 = 0 ⇒ vb = ⎜ 1+ 2 ⎟ v2 − 2 v1
R2
R7
R7
⎝ R7 ⎠
R6
⎛ v −v ⎞ v −0
R6
− ⎜ b c ⎟ + c + 0 = 0 ⇒ vc =
vb
R4 + R6
⎝ R4 ⎠ R6
⎛ v −v ⎞ ⎛ v −v ⎞
−⎜ a c ⎟ + ⎜ c 0 ⎟ + 0 = 0
⎝ R3 ⎠ ⎝ R5 ⎠
⇒ v0 = −
R5
R
va + (1+ 5 )vc
R3
R3
⎡ R R R (R +R )
⎡R
R (R +R ) R ⎤
R ⎤
R
v0 = ⎢ 5 1 + 6 3 5 (1+ 2 ) ⎥ v2 − ⎢ 5 (1+ 1 ) + 6 3 5 2 ⎥ v1
R7 ⎦
R7 R3 ( R4 + R6 ) R7 ⎦
⎣ R3 R7 R3 ( R4 + R6 )
⎣ R3
v −v
i0 = c 0 = "
R5
10 kΩ
P 6.4-6 Determine the node voltages for
the circuit shown in Figure P 6.4-6.
Answer: va = – 0.75 V, vb = 0 V, and vc =
– 0.9375 V
40 kΩ a
–
+
12 V
20 kΩ
40 kΩ
b
25 kΩ
c
–
15 kΩ
+
Figure P 6.4-6
Solution:
KCL at node b:
KCL at node a:
so
va
vc
5
+
= 0 ⇒ vc = − va
3
3
20 × 10 25 × 10
4
5
⎛
⎞
va − ⎜ − va ⎟
va − ( −12 )
va
va + 0
⎝ 4 ⎠ = 0 ⇒ v = − 12 V
+
+
+
a
3
3
3
40 ×10
40 × 10 20 × 10
10 × 103
13
vc = −
5
15
va = −
4
13
(checked using LNAP 6/21/05)
P 6.4-7 Find vo and io for the circuit shown in
Figure P 6.4-7. Assume an ideal operational
amplifier.
10 kΩ
30 kΩ
10 kΩ
6V
+–
30 kΩ
30 kΩ
10 kΩ
–
io
+
+
30 kΩ
vo
–
Figure P 6.4-7
Solution:
Label the circuit to account for the
properties of the ideal op amp and to
identify the supernode corresponding to
the voltage source.
Apply KCL at the inverting input node of
the op amp
⎛ ( va + 6 ) −0 ⎞
⎛ v −0 ⎞
−⎜ a
⎟ = 0
⎟+0−⎜
⎝ 10000 ⎠
⎝ 30000 ⎠
or
va = − 1.5 V
Apply KCL to the super node
corresponding the voltage source:
va − 0 va + 6− 0 va − v b ( va + 6 ) −v b
+
+
+
= 0
10000 30000 30000
10000
Multiply both sides by 30000 to get
3va + ( va + 6 ) + ( va − vb ) + 3 ⎡⎣( va + 6 )− v b ⎤⎦ = 0
Solving gives
vb = 2 va + 6 = 3 V
Apply KCL at node b:
⎛ v −v ⎞ ⎛ ( v a + 6 )−v b ⎞
⎟ = 0
−⎜ a b ⎟−⎜
10000 30000 ⎝ 30000 ⎠ ⎜ 10000 ⎟
⎝
⎠
Multiply both sides by 30000 to get
vb
+
v b −v o
3v b + ( v b − v o ) − ( v a −v b ) − 3 ⎡⎣( v a + 6 )− v b ⎤⎦ = 0
Solving gives
v o = 8 v b − 4 va − 18 = 12 V
Apply KCL at the output node of the op amp:
io +
vo
30000
+
v o −v b
30000
= 0 ⇒ i o = −0.7 mA
10 kΩ
20 kΩ
+
P 6.4-8 Find vo and io for the circuit shown in
Figure P 6.4-8. Assume an ideal operational
amplifier.
–
vo
+
–
io
20 kΩ
10 kΩ
Figure P 6.4-8
Solution:
Apply KVL to the bottom mesh:
−i0 (10000) − i0 (20000) + 5 = 0
⇒ i0 =
1
mA
6
The node voltages at the input nodes of an
ideal op amp are equal. Consequently
va = 10000 i0 =
10
V
6
Apply KCL at node a:
va
v −v
+ a 0 = 0 ⇒
10000 20000
v0 = 3va = 5 V
+
–
5V
P 6.4-9
Determine the node voltages for the circuit shown in Figure P 6.4-9.
Answer: va = – 12 V, vb = – 4 V, vc = – 4 V, vd = – 4 V, ve = – 3.2 V, vf = – 4.8 V, and vg = – 3.2
c
10 kΩ
d
20 kΩ
e
20 kΩ
a
40 kΩ
20 kΩ
–
b
–
+ 12 V
–
g
+
+
f
40 kΩ
20 kΩ
40 kΩ
Figure P 6.4-9
Solution:
vb + 12
vb
+
= 0 ⇒ vb = −4 V
40000 20000
The node voltages at the input nodes of an ideal op amp are equal, so vc = vb = −4 V .
KCL at node b:
The input currents of an ideal op amp are zero, so vd = vc + 0 × 10 4 = −4 V .
KCL at node g:
vg
⎛ v f − vg ⎞
2
−⎜
+
= 0 ⇒ vg = v f
3 ⎟
3
3
⎝ 20 ×10 ⎠ 40 × 10
The node voltages at the input nodes of an ideal op amp are equal, so ve = vg =
2
vd − v f vd − 3 v f
vd − ve
KCL at node d: 0 =
+
=
+
20 ×103 20 ×103 20 ×103 20 ×103
vd − v f
Finally, ve = vg =
2
16
vf = −
V.
3
5
2
vf .
3
6
24
⇒ v f = vd = −
V
5
5
R = Ro + ΔR
P 6.4-10 The circuit shown in Figure P 6.4-10
includes a simple strain gauge. The resistor R changes
its value by ΔR when it is twisted or bent. Derive a
relation for the voltage gain vo/vs and show that it is
proportional to the fractional change in R, namely
ΔR/Ro.
Answer: vo =
Ro ΔR
Ro + R1 Ro
R1
–
+
vs
+
+
R1
Ideal
Ro
–
–
Figure P 6.4-10
Solution:
By voltage division (or by applying KCL at
node a)
R0
va =
vs
R1 + R0
Applying KCL at node b:
vb − vs vb − v0
+
= 0
R1
R0 +ΔR
⇒
vo
R0 +ΔR
( vb −vs )+ vb = v0
R1
The node voltages at the input nodes of an
ideal op amp are equal so vb = va .
⎡⎛ R +ΔR ⎞ R0
⎛
R +ΔR ⎤
R0 ⎞ ΔR
ΔR
v0 = ⎢⎜ 0
+1⎟
− 0
v s = ⎜ − vs
⎥ vs = −
⎟
R1 ⎦
R1 + R0
R1 + R0 ⎠ R0
⎠ R1 + R0
⎝
⎣⎝ R1
+
P 6.4-11 Find vo for the circuit shown in
Figure P 6.4-11. Assume an ideal operational
amplifier.
–
io
+
– 1.5 V
20 kΩ
20 kΩ
8 kΩ
10 kΩ
+
vo
–
Figure P 6.4-11
Solution:
Node equations:
⎛ R2 ⎞
vs vs − va
+
= 0 ⇒ va = ⎜1 +
⎟ vs
⎜
R1
R2
R1 ⎟⎠
⎝
and
vs − va va va − vo
=
+
R2
R3
R4
so
⎛ R4 R4 ⎞
R4
vo = ⎜1 +
+
va −
vs
⎟
⎜ R 2 R3 ⎟
R
2
⎝
⎠
⎛ R4 R4 ⎞ ⎛ R2 ⎞
⎛ R4 R4 R2 R2 R4 ⎞
R4
vo = ⎜1 +
1+
vs −
vs = ⎜ 1 +
v
+
+
+
+
⎟
⎜
⎟
⎜ R 2 R3 ⎟ ⎜
⎜ R 3 R1 R1 R1 R 3 ⎟⎟ s
R1 ⎟⎠
R2
⎝
⎠⎝
⎝
⎠
⎛ ( R1 + R 2 )( R 3 + R 4 ) + R 3 R 4 ⎞
⎟ vs
=⎜
⎜
⎟
R1 R 3
⎝
⎠
with the given values:
⎛ ( 20 + 20 )(10 + 8 ) + 10 × 8 ⎞
⎛ 40 × 18 + 80 ⎞
vo = ⎜
⎟ vs = ⎜
⎟ vs = ( 4 ) vs
20 × 10
200
⎝
⎠
⎝
⎠
(checked: LNAP 5/24/04)
+
P 6.4-12 The circuit shown in Figure P 6.4-12 has
one output, vo, and two inputs, v1 and v2. Show that
when
R1
vo
–
v1 +–
R3
R3 R6
=
, the output is proportional to the
R4 R5
R4
difference of the inputs, v1 – v2. Specify resistance
values to cause vo = 5 (v1 – v2).
+
R2
v2
–
+
–
R5
R6
Figure P 6.4-12
Solution:
Notice that the currents in resistance R1 and R2 are
both zero, as shown. Consequently, the voltages at
the noninverting inputs of the op amps are v1 and v2,
as shown. The voltages at the inverting inputs of the
ideal op amps are also v1 and v2, as shown.
Apply KCL at the top node of R6 to get
va − v2 v2
=
R5
R6
⎛ R5 + R 6 ⎞
⇒ va = ⎜
v
⎜ R 6 ⎟⎟ 2
⎝
⎠
Apply KCL at the top node of R4 to get
vo − v1 v1 − va
=
R3
R4
⎛ R3 ⎞
⎛ R3 ⎞
⇒ vo = ⎜ 1 +
v1 − ⎜ ⎟ va
⎟
⎜ R4 ⎟
⎜ R4 ⎟
⎝
⎠
⎝ ⎠
⎛ R3 ⎞
⎛ R3 ⎞ ⎛ R5 + R6 ⎞
vo = ⎜ 1 +
v1 − ⎜ ⎟ ⎜
v
⎟
⎜ R4 ⎟
⎜ R 4 ⎟ ⎜ R 6 ⎟⎟ 2
⎝
⎠
⎝ ⎠⎝
⎠
When
R3
R4
=
R6
R5
⎛ R3 ⎞
⎛ R3 ⎞ ⎛ R5 ⎞
⎛ R3 ⎞
⎛ R3 ⎞ ⎛ R 4 ⎞
vo = ⎜1 +
v1 − ⎜ ⎟ ⎜1 +
v2 = ⎜1 +
v1 − ⎜ ⎟ ⎜ 1 +
v
⎟
⎟
⎟
⎜ R4 ⎟
⎜ R4 ⎟ ⎜ R6 ⎟
⎜ R4 ⎟
⎜ R 4 ⎟ ⎜ R 3 ⎟⎟ 2
⎝
⎠
⎝ ⎠⎝
⎠
⎝
⎠
⎝ ⎠⎝
⎠
⎛ R3 ⎞
v −v
= ⎜1 +
⎜ R 4 ⎟⎟ ( 1 2 )
⎝
⎠
so vo is proportional to the difference of the inputs, v1 − v2, as required.
R3
Next, to make vo = 5 ( v1 − v2 ) , we choose R3 and R4 so that 5 = 1 +
, e.g. R3 = 40 kΩ and R4 =
R4
10 kΩ. Then with
R3
R4
=
R6
R5
we have
R1 = 50 kΩ, R2 = 50 kΩ, R3 = 40 kΩ, R4 = 10 kΩ, R5 = 10 kΩ and R6 = 40 kΩ.
(checked: LNAP 5/24/04)
R1
P 6.4-13 The circuit shown in Figure P 6.4-13 has
one output, vo, and one input, vi. Show that the
output is proportional to the input. Specify resistance
values to cause vo = 20 vi.
vi
–
vo
+
R2
R3
R4
–
+
Figure P 6.4-13
Solution:
Write a node equation at the inverting input of the
bottom op amp:
vo
R3
+
va
R4
= 0 ⇒ va = −
R4
R3
vo
Write a node equation at the inverting input of the top op
amp:
R4
vo
−
vi va vi
R3
R 2 R3
0=
+
=
+
⇒ vo =
vi
R1 R 2 R 1
R2
R1 R 4
The output is proportional to the input and the constant of proportionality is
vo = 20 vi so
R 2 R3
R1 R 4
R 2 R3
R1 R 4
. We require
= 20 . For example, R1 = R 4 = 10 kΩ, R 2 = 40 kΩ and R 3 = 50 kΩ .
R1
vs
R2
–
+
–
+
P 6.4-14 The circuit shown in
Figure P 6.4-14 has one input, vs, and one
output, vo. Show that the output is
proportional to the input. Design the
circuit so that vo = 20vs.
+
R4
R5
vo
–
R3
Figure P 6.4-14
Solution:
Represent this circuit by node equations.
vo − va
R2
+
vo − vs
R1
vo − va
R4
+
=0
vo
R5
⇒
=0
⎛ R4 ⎞
v a = ⎜1 +
v
⎜ R 5 ⎟⎟ o
⎝
⎠
( R 1 + R 2 ) R 5 − R 1 ( R 4R 5 ) v o
⎛ R1 ⎞
⎛ R1 ⎞⎛ R 4 ⎞
v s = ⎜1 +
vo − ⎜
1+
v =
⎟
⎜ R2 ⎟
⎜ R 2 ⎟⎟ ⎜⎜ R 5 ⎟⎟ o
⎝
⎠
⎝
⎠⎝
⎠
R 2R 5
20 =
Then
R 2 R 5 - R 1R 4
So
For example
R 2 v s = ( R 1 + R 2 ) v o − R 1v a
⇒
R 2R 5
⇒
=
R 2R 5
R 2 R 5 − R 1R 4
vs
19 R 1R 4
=
20 R 2 R 5
R 1 = 19 kΩ, R 4 = 10 kΩ, R 2 = 20 kΩ, R 5 = 10 kΩ, R 3 = 10 kΩ.
(checked: LNAP 5/24/04)
R
P 6.4-15 The circuit shown in Figure P 6.4-15
has one input, vs, and one output, vo. The circuit
contains seven resistors having equal resistance,
R. Express the gain of the circuit, vo/vs, in terms
of the resistance R.
R
R
R
R
R
–
+
–
vs
+
+
R
vo
–
Figure P 6.4-15
Solution:
Writing node equations:
vs
+
v1
=0
⇒
v 1 = −v s
R R
v1 v1 v1 − v 2
+ +
=0
⇒
v 2 = 3v1 = −3v s
R R
R
v 2 − v1 v 2 v 2 − v o
+ +
=0
⇒
v o = 3v 2 − v1 = −8v s
R
R
R
vo
= −8 , does not depend on R.
The gain of this circuit,
vs
(checked: LNAP 6/21/04)
R3
P 6.4-16 The circuit shown in Figure P 6.4-16
has one input, vs, and one output, vo. Express the
gain, vo/vs, in terms of the resistances R1, R2, R3,
R4, and R5. Design the circuit so that vo = – 20 vs.
R1
R4
R2
–
+
+
+
–
vs
R5
vo
–
Figure P 6.4-16
Solution:
Represent this circuit by node equations.
vs − va
R1
+
vo − va
R3
=
va
R2
and
va
R2
+
vo
R4
= 0 ⇒ vo = −
R4
R2
va
So
⎛ 1
⎛ R1 R 2 + R1 R 3 + R 2 R 3 ⎞ ⎛ R 2 ⎞
1
1 ⎞
=⎜ +
+
va = ⎜
vo ⎟
⎟
⎟⎟ ⎜⎜ −
⎜
⎟
R1 R 3 ⎜⎝ R1 R 2 R 2 ⎟⎠
R1 R 2 R 3
⎝
⎠ ⎝ R4 ⎠
⎛ 1 R1 R 2 + R1 R 3 + R 2 R 3 ⎞
⎛ R1 R 4 + R1 R 2 + R1 R 3 + R 2 R 3 ⎞
vs
= −⎜
+
vo = − ⎜
⎟
⎟⎟ v o
⎜ R3
⎟
⎜
R1
R
R
R
R
R
R
1
3
4
1
3
4
⎝
⎠
⎝
⎠
⎛
⎞
R3 R 4
vo = − ⎜
v
⎜ R1 R 4 + R1 R 2 + R1 R 3 + R 2 R 3 ⎟⎟ s
⎝
⎠
R3 R 4
20 =
We require
R1 R 4 + R1 R 2 + R1 R 3 + R 2 R 3
vs
+
vo
Try R1 = R 2 = R and R 3 = R 4 = aR
a2
20 =
3a + 1
2
a − 60a − 20 = 0
Then
So
a=
e.g.
+60 ± 3600 + 4 ( 80 )
= 60.332, − 0.332
2
R1 = R 2 = 10 kΩ and R 3 = R 4 = 603.32 kΩ
(checked: LNAP 6/9/04)
R3
P 6.4-17 The circuit shown in Figure P 6.4-17
has one input, vs, and one output, vo. Express
the gain of the circuit, vo/vs, in terms of the
resistances R1, R2, R3, R4, R5, and R6. Design
the circuit so that
vo = – 20 vs.
R2
R1
–
+
+
+
–
–
vs
R4
R5
+
R6
vo
–
Figure P 6.4-17
Solution:
Represent this circuit by node equations.
vs
R1
+
va
R2
+
vo
R3
=0
and
va
R4
+
va − vo
R5
⎛ R4 ⎞
= 0 ⇒ va = ⎜
v
⎜ R 4 + R 5 ⎟⎟ o
⎝
⎠
vs
So
R1
+
vo
R3
=−
va
R2
=−
R4
R 2 ( R 4 + R5 )
vo
⎛ 1
⎞
R 2 ( R 4 + R5 ) + R 4 R3
R4
⎟ vo = −
= −⎜
+
vo
⎜ R3 R 2 ( R 4 R5 ) ⎟
R1
R
R
R
+
R
(
)
2 3
4
5
⎝
⎠
vs
vo = −
We require
Try
20 =
R 2 R3 ( R 4 + R5 )
R1 ( R 2 R 4 + R 2 R 5 + R 3 R 4 )
vs
R 2 R3 ( R 4 + R5 )
R1 ( R 2 R 4 + R 2 R 5 + R 3 R 4 )
R1 = R 4 = R 5 = R and R 2 = R 3 = aR
Then
e.g.
20 =
2a 2 R 3 2
= a
3aR 3 3
⇒
a = 30
R1 = R 4 = R 5 = 10 kΩ and R 2 = R 3 = 300 kΩ
(checked: LNAP 6/10/04)
R2
R1
P 6.4-18 The circuit shown in Figure P 6.4-18 has one
input, vs, and one output, io. Express the gain of the circuit,
io/vs, in terms of the resistances R1, R2, R3, and Ro. (This
circuit contains a pair of resistors having resistance R1 and
another pair having resistance R2.) Design the circuit so that
io = 0.02 vs.
+
–
–
vs
+
R1
R2
–
+
R3
Ro
io
Figure P 6.4-18
Solution:
Label the node voltages as shown. Represent this
circuit by node equations.
vb − va
R2
io +
vo − va
R2
+
=
va
R1
vo − vb
R3
vs − va
R1
⇒
=0
=0
⇒
⇒
va =
R1
R1 + R 2
vb
v b = R 3i o + v o
⎛ R1 + R2 ⎞
vo
=⎜
va −
⎟
R1 ⎜⎝ R1 R 2 ⎟⎠
R2
vs
So
vs
⎛ R1 + R 2 ⎞ ⎛ R1 ⎞
v o R3
=⎜
R 3i o + v o ) −
=
io
⎟
⎜
⎟
(
R1 ⎜⎝ R1 R 2 ⎟⎠ ⎜⎝ R1 + R 2 ⎟⎠
R2 R2
io
R2
=
v s R1 R 3
We require
R2
R1 R 3
= 0.02, e.g. R 2 = 8 kΩ, R1 = R 3 = 20 kΩ .
(checked: LNAP 6/21/04)
P 6.4-19 The circuit shown in Figure P 6.4-19 has one input, vs, and one output, vo. The circuit
contains one unspecified resistance, R.
(a)
Express the gain of the circuit, vo/vs, in terms of the resistance R.
(b)
Determine the range of values of the gain that can be obtained by specifying a value for
the resistance R.
(c)
Design the circuit so that vo = – 3 vs.
Figure P 6.4-19
Solution:
(a) Use units of volts, mA, and kΩ. Apply KCL at the inverting input of the left op amp to get
vs
10
vo =
(b)
(c) We require
+
va
50
+
50 ⎞
⎛
= 0 ⇒ v a = − ⎜ 5v s + v o ⎟
R
R ⎠
⎝
vo
4
40
⎛ 40 ⎞
v a = −4v s − v o
⇒
⎜ 1 + ⎟ v o = − 4v s
5
R
R⎠
⎝
vo
4
4R
=−
=−
40
vs
R + 40
1+
R
vo
0≤R≤∞
⇒
−4≤
≤0
vs
−3 = −
4R
R + 40
⇒
R = 120 kΩ
(checked: LNAP 6/21/04)
P 6.4-20 The circuit shown in Figure P 6.4-20 has one input, vs, and one output, vo. The circuit
contains one unspecified resistance, R.
(a)
Express the gain of the circuit, vo/vs, in terms of the resistance R.
(b)
Determine the range of values of the gain that can be obtained by specifying a value for
the resistance R.
(c)
Design the circuit so that vo = – 5 vs.
R
30 kΩ
10 kΩ
+
–
+
+
–
–
vs
+
20 kΩ
10 kΩ
vo
–
Figure P 6.4-20
Solution:
(a) Use units of V, mA and kΩ. Apply KCL at the inverting input of the left op amp to get
v1
10
+
va
30
+
30 ⎞
⎛
= 0 ⇒ v a = − ⎜ 3v s + v o ⎟
R
R ⎠
⎝
vo
v o = 3v a = −9v s −
(c) We require
⎛ 90 ⎞
⇒ ⎜ 1 + ⎟ v o = − 9v s
R⎠
⎝
9
9R
=−
90
vs
R + 90
1+
R
vo
0≤R≤∞
⇒
−9 ≤
≤0
vs
vo
(b)
90
vo
R
−5 =
=−
−9 R
R + 90
⇒
R = 112.5 kΩ
(checked: LNAP 7/8/04)
P 6.4-21 The circuit shown in Figure P 6.4-21 has three inputs: v1, v2, and v3. The output of the
circuit is vo. The output is related to the inputs by
vo = av1 + bv2 + cv3
where a, b, and c are constants. Determine the values of a, b, and c.
+
–
20 kΩ
20 kΩ
v1
–
40 kΩ
+
120 kΩ
–
20 kΩ
120 kΩ
+
+
–
+
–
v2
20 kΩ
+
vo
–
20 kΩ
20 kΩ
30 kΩ
–
+
v3
+
–
Figure P 6.4-21
Solution:
Label the node voltages and identify some standard op amp circuits:
Either by recognizing the inverting amplifier or by writing and solving the node equation
corresponding to node a, we obtain
⎛ 20 ⎞
v b = ⎜ − ⎟ v 1 = −v 1
⎝ 20 ⎠
Either by recognizing the voltage divider and voltage follower or by writing and solving the node
equation corresponding to node c, we obtain
1
⎛ 20 ⎞
vd = ⎜
⎟ v2 = v2
2
⎝ 20 + 20 ⎠
Either by recognizing the noninverting amplifier or by writing and solving the node equation
corresponding to node f, we obtain
⎛ 20 ⎞
v g = ⎜1 + ⎟ v 3 = 3 v 3
⎝ 20 ⎠
Either by recognizing the summing amplifier or by writing and solving the node equation
corresponding to node e, we obtain
⎡⎛ 120 ⎞
⎛ 120 ⎞
⎛ 120 ⎞ ⎤
v o = − ⎢⎜
⎟ vb + ⎜
⎟ vd + ⎜
⎟ v g ⎥ = − ⎡⎣3 v b + v d + 4 v g ⎤⎦
⎝ 120 ⎠
⎝ 30 ⎠ ⎦
⎣⎝ 40 ⎠
Substituting for v b , v d and v g gives
⎡
⎤
⎛1 ⎞
v o = − ⎢3 ( −v1 ) + ⎜ v 2 ⎟ + 4 ( 3 v 3 ) ⎥ = 3 v1 − 0.5 v 2 − 12 v 3
⎝2 ⎠
⎣
⎦
so
a = 3, b = −0.5 and c = −12
(checked: LNAP 6/21/04)
P 6.4-22 The circuit shown in Figure P 6.4-22 has two inputs: v1 and v2. The output of the
circuit is vo. The output is related to the inputs by
vo = av1 + bv2
where a and b are constants. Determine the values of a and b.
20 kΩ
20 kΩ
40 kΩ
–
+
+
–
–
v1
+
+
–
20 kΩ
20 kΩ
v2
–
+
vo
20 kΩ
+
20 kΩ
Figure P 6.4-22
Solution:
Label the node voltages as shown. Use units of V, mA and kΩ.
v 3 = v1 and v 4 = −v 2
–
v5 − v3
40
+
v5 − v4
20
=0
⇒
v5 =
1
1
2
v 3 + 2v 2 ) = v 1 − v 2
(
3
3
3
so
a=−
1
2
and b = −
3
3
(checked: LNAP 6/21/04)
P6.4-23
The input to the circuit shown in Figure P6.4-23 is the voltage source voltage v s . The output is
the node voltage v o . The output is related to the input by the equation v o = k v s where k =
is called the gain of the circuit. Determine the value of the gain k.
FigureP6.4-23
Solution:
Label the node voltages as shown. Apply KCL at the
inverting input node of the op amp to get
vs
30000
+
vs − va
80000
= 0 ⇒ 11v s = 3 v a
⇒ va =
11
vs
3
Apply KCL at the right node of the 80 kΩ resistor to get
vs − va
80000
+
vo − va
20000
=
va
20000
⇒ vs − 9 va + 4 vo = 0
⎛ 11 ⎞
⇒ vs − 9 ⎜ vs ⎟ + 4 vo = 0
⎝3 ⎠
⇒ v s − 33 v s + 4 v o = 0
⇒ 4 v o = 32 v s
⇒ vo = 8 vs
Comparing this equation to v o = k v s , we determine that k = 8 V/V.
vo
vs
P6.4-24
The input to the circuit shown in Figure P6.4-23 is the voltage source voltage v s . The output is
the node voltage v o . The output is related to the input by the equation v o = m i s + b where m
and b are constants. Determine the values of m and b.
FigureP6.4-24
Solution:
Label the node voltages as shown. Apply KCL at the
inverting input node of the op amp to get
is =
6 − va
50000
= 0 ⇒ v a = 6 − ( 50 × 10 3 ) i s
Apply KCL at the top node of the 25 kΩ resistor to
get
vo − va
va
is +
=
25000 25000
Solving:
( 25 ×10 3 ) i s − 2 v a + v o = 0
( 25 ×10 ) i s − 2 ( 6 − ( 50 ×10 ) i s ) + v o = 0
(125 ×10 ) i s − 12 + v o = 0
v o = − (125 × 10 ) i s + 12
3
3
3
3
Comparing this equation to v o = m i s + b , we get m = − (125 ×10 3 ) V/A and b =12 V.
P6.4-25
The input to the circuit shown in Figure P6.4-25 is the node voltage v s . The output is the node
voltage v o . The output is related to the input by the equation v o = k v s where k =
the gain of the circuit. Determine the value of the gain k.
FigureP6.4-25
Solution:
Label the node voltages as shown. The
node equations are
vs − va
5000
va − 0
5000
=
vo − vb
5000
=
va − 0
5000
0 − vb
vb
25000
Substituting v a = −
va − vo
50000
⇒ v b = −10 v a
50000
=
+
⇒ v o = 11v b
vo
110
into the first node equation gives:
vo
vs
is called
vs +
vo
−
vo
−
vo
− vo
110 = 110 + 110
5000
5000
50000
vo ⎞
⎛
⎛ vo ⎞ vo
⇒ 10 ⎜ v s +
− vo
⎟ = 10 ⎜ −
⎟−
110 ⎠
⎝
⎝ 110 ⎠ 110
131
⇒ 10 v s = −
vo
110
1100
v s = −8.397 v s
⇒ vo = −
131
Comparing this equation to v o = k v s , we determine that k = −8.397 V/V.
P6.4-26
The values of the node voltages v1, v2 and vo
in Figure P6.4-26 are
v1 = 6.25 V, v2 = 3.75 V and vo = −15 V.
Determine the value of the resistances R1, R2
and R3:
FigureP6.4-26
Solution:
Label the node voltages as shown. Using
units of Volts, kΩ and mAmps, the node
equations are:
2.5 v1 − 2.5 v1 − v 2 v 2
=
=
,
20
R2
20
R1
And
v2 − 0
R3
+
vo − 0
20
=0
Using v1 = 6.25 V gives
2.5 6.25 − 2.5
=
⇒ R1 = 30 kΩ
20
R1
Using v1 = 6.25 V and v 2 = 3.75 V gives
6.25 − 3.75 3.75
=
⇒ R 2 = 30 kΩ
20
R2
Using v 2 = 3.75 V and v o = −15 V gives
3.75 − 0 −15 − 0
+
= 0 ⇒ R 3 = 5 kΩ
R3
20
P6.4-27
The input to the circuit shown in Figure
P6.4-27 is the voltage source voltage, v i . The
output is the node voltage, v o .The output is
related to the input by the equation v o = k v i
where k =
vo
vi
is called the gain of the circuit.
Determine the value of the gain k.
FigureP6.4-27
Solution:
Label the node voltages as shown. The node
voltages at the input nodes of an ideal op amp
are equal so
va = vb
Consequently
i2 =
va − vb
=0
R
The input currents of an ideal op amp are 0 A,
so applying KCL at nodes a and b shows that
i1 = 0 and i 3 = 0 . Consequently, the voltages
across the corresponding resistor are 0 V.
Finally
va = v b = vc = vi
Write an node equation at the right node of the
24 kΩ resistor:
0=
vi
10
+
vi − vo
40
⇒ vo = 5 vi
Finally
k=
vo
vi
=5
Section 6-5: Design Using Operational Amplifier
P 6.5-1 Design the operational amplifier
circuit in Figure P 6.5-1 so that
vout = r · iin
where
r = 20
Operational
amplifier
circuit
iin
+
20 kΩ
vout
–
V
mA
Figure P 6.5-1
Solution:
Use the current-to-voltage converter, entry (g) in Figure 6.6-1.
P 6.5-2 Design the operational amplifier
circuit in Figure P 6.5-2 so that
iout = g · vin
where
g =2
iout
vin
+
–
Operational
amplifier
circuit
mA
V
Figure P 6.5-2
Solution:
Use the voltage –controlled current source, entry (i) in Figure 6.6-1.
5 kΩ
P 6.5-3 Design the operational amplifier
circuit in Figure P 6.5-3 so that
v1
Operational
amplifier
circuit
+
–
vout = 5 · v1 + 2 · v2
+
20 kΩ
v2
vout
–
+
–
Figure P 6.5-3
Solution:
Use the noninverting summing amplifier, entry (e) in Figure 6.6-1.
P 6.5-4
Design the operational amplifier circuit in Figure P 6.5-3 so that vout = 5 · (v1 – v2).
Solution:
Use the difference amplifier, entry (f) in Figure 6.6-1.
P 6.5-5
Design the operational amplifier circuit in Figure P 6.5-3 so that vout = 5 · v1 – 2 · v2
Solution:
Use the inverting amplifier and the summing amplifier, entries (a) and (d) in Figure 6.6-1.
P 6.5-6 The voltage divider shown in Figure P 6.5-6
has a gain of
vout
−10 kΩ
=
=2
vin 5 kΩ + (−10 kΩ)
5 kΩ
+
vin
Design an operational amplifier circuit to implement
the – 10-kΩ resistor.
Solution:
Use the negative resistance converter, entry (h) in Figure 6.6-1.
+
–
–10 kΩ
vout
–
Figure P 6.5-6
P 6.5-7 Design the operational
amplifier circuit in Figure P 6.5-7 so
that
iin = 0 and vout = 3 · vin
iin
+
5V
+
–
10 kΩ
–5 V
+
–
Operational
amplifier
circuit
vin
–
+
20 kΩ
vout
–
Figure P 6.5-7
Solution:
Use the noninverting amplifier, entry (b) in Figure 6.6-1. Notice that the ideal op amp forces the
current iin to be zero.
P 6.5-8 Design an operational amplifier circuit with output vo = 6 v1 + 2 v2, where v1 and v2 are
input voltages.
Solution:
Summing Amplifier: va = − ( 6 v1 + 2 v2 ) ⎫
⎬ ⇒ vo = 6 v1 + 2 v2
Inverting Amplifier:
vo = −va
⎭
8 kΩ
P 6.5-9 Determine the voltage vo for the
circuit shown in Figure P 6.5-9.
Hint: Use superposition.
Answer: vo = (– 3)(3) + (4)(– 4) + (4)(8) = 7 V
4 kΩ
24 kΩ
–
+
+
3V –
4V
–
+
2 mA
+
10 kΩ
vo
–
Figure P 6.5-9
Solution:
Using superposition, vo = v1 + v2 + v3 = −9 − 16 + 32 = 7 V
P 6.5-10 For the op amp circuit shown in
Figure P 6.5-10, find and list all the possible voltage
gains that can be achieved by connecting the resistor
terminals to either the input or the output voltage
terminals.
6
12
12
–
vs
+
+
–
Figure P 6.5-10
Solution:
R1
6
12
24
6||12
6||24
R2
-vo/vs
12||12||24
0.8
6||12||24
0.286
6||12||12
0.125
12||24
2
12||12
1.25
12||24
6||12||12
6||12||24
12||12||24
6||12
0.5
24
8
12
3.5
6
1.25
R1
R2
-vo/vs
12||12
6||24
0.8
24
+
vo
–
P 6.5-11 The circuit shown in Figure P 6.5-11 is
called a Howland current source. It has one input, vin,
and one output, iout. Show that when the resistances
are chosen so that R2 R3 = R1 R4, the output is related
to the input by the equation
iout =
vin
R1
R4
R3
–
+
R1
vin
+
–
R2
RL
iout
Figure P 6.5-11
Solution:
Label the node voltages as shown. Apply KCL at the
inverting input of the op amp to get
va
R3
+
va − vb
R3
⎛ R3 + R 4 ⎞
= 0 ⇒ vb = ⎜
v
⎜ R 3 ⎟⎟ a
⎝
⎠
Apply KCL at the noninverting input of the op amp to get
v a − v in
R1
+
va − vb
R2
+ i out = 0
Solving gives
⎛ R1 + R 2 ⎞ v in v b
⎛ R1 + R 2 R 3 + R 4 ⎞ v in
va ⎜
−
−
+ i out = 0 ⇒ v a ⎜
−
+ i out = 0
⎟
⎟−
⎜ R1 R 2 ⎟ R1 R 2
⎜ R1 R 2
R 2 R 3 ⎟⎠ R1
⎝
⎠
⎝
When R 2 R 3 = R1 R 4 the quantity in parenthesis vanishes leaving
i out =
1
v in
R1
P 6.5-12 The input to the circuit shown in Figure P 6.5-12a is the voltage vs. The output is the
voltage vo. The voltage vb is used to adjust the relationship between the input and output.
(a)
Show that the output of this circuit is related to the input by the equation
vo = avs + b
where a and b are constants that depend on R1, R2, R3, R4, and vb.
Design the circuit so that its input and output have the relationship specified by the graph
shown in Figure P 6.5-12b.
(b)
vo , V
8
+
R1
6
–
+
4
R4
+
–
R2
vs
2
vo
+
–
R3
vb
R5
–4
–6
−
–2
2
4
–2
–4
(a)
(b)
Figure P 6.5-12
Solution:
(a) Label the node voltages as shown. The
node equations are
vs − va
R1
+
vb − va
R2
=
va
R3
and
va
R5
=
vo − va
R4
⎛ R5 ⎞
⇒ va = ⎜
v
⎜ R 4 + R 5 ⎟⎟ o
⎝
⎠
Solving these equations gives
vs ⎛ 1
v b ⎛ R1 R 2 + R 2 R 3 + R1 R 3
R5 ⎞
vb
1
1 ⎞
=⎜ +
+
=⎜
×
va −
vo −
⎟
⎟
R1 ⎝⎜ R1 R 2 R 3 ⎟⎠
R 2 ⎜⎝
R1 R 2 R 3
R 4 + R 5 ⎟⎠
R2
So
⎛
R 2 R3
R 4 + R5 ⎞
R1 R 3
R 4 + R5
vo = ⎜
×
vs +
×
× vb
⎟
⎜ R1 R 2 + R 2 R 3 + R1 R 3
R 5 ⎟⎠
R1 R 2 + R 2 R 3 + R1 R 3
R5
⎝
So
vs , V
⎛
R 2 R3
R 4 + R5 ⎞
R1 R 3
R 4 + R5
a=⎜
×
v s and b =
×
× vb
⎟
⎜ R1 R 2 + R 2 R 3 + R1 R 3
⎟
R
R
R
+
R
R
+
R
R
R
5
1 2
2 3
1 3
5
⎝
⎠
(b) The equation of the straight line is
5
vo = vs + 5
4
We require
R 2 R3
R1 R 2 + R 2 R 3 + R1 R 3
×
R 4 + R5
R5
=
5
4
For example, let R1 = R 2 = R 3 = 10 kΩ, R 4 = 55 kΩ and R 5 = 20 kΩ . Next we require
5=
R1 R 3
R1 R 2 + R 2 R 3 + R1 R 3
×
R 4 + R5
R5
× vb =
5
vb
4
i.e.
vb = 4 V
(checked: LNAP 6/20/04)
R3
R1
P 6.5-13 The input to the circuit shown in
Figure P 6.5-13a is the voltage vs. The output is
the voltage vo. The voltage vb is used to adjust
the relationship between the input and output.
(a)
Show that the output of this circuit is
related to the input by the equation
–
+
–
R2
vs
+
–
+
+
R4
vb
vo
–
vo = avs + b
where a and b are constants that depend
on R1, R2, R3, R4, and vb.
(b)
(a)
Design the circuit so that its input and
output have the relationship specified
by the graph shown in
Figure P 6.5-13b.
vo , V
8
6
4
2
–6
–4
2
–2
4
–2
–4
(b)
P 6.5-13
Solution:
(a) Apply KCL at the inverting input of the op amp to get:
⎛ R3 ⎞
vs vb vo
R3
+
+
= 0 ⇒ vo = ⎜ − ⎟ vs −
vb
⎜
⎟
R1 R 2 R 3
R2
⎝ R1 ⎠
R3
R3
so
a=−
and b = −
vb
R1
R2
(b) The equation of the straight line is
We require
−
R3
5
=−
2
R1
Next, we require
5=−
e.g.
R 3v b
R2
5
vo = − vs + 5
2
R1 = 20 kΩ and R 3 = 50 kΩ .
e.g.
R 2 = R 3 = 50 kΩ and v b = −5 V .
vs, V
P6.5-14 The input to the circuit shown in Figure P6.5-14 is the voltage source voltage v s . The
output is the node voltage v o . the output is related to the input by the equation v o = m v s + b
where m and b are constants. (a) Specify values of R 3 and v a that cause the output to be related
to the input by the equation v o = 4 v s + 7 . (b) Determine the values of m and b when
R 3 = 20 kΩ and v a = 2.5 V .
Figure P6.5-14
Solution: Label the node voltages:
⎛ 30 ⎞
Recognizing an inverting amplifier and a noninverting amplifier we write v b = ⎜ − ⎟ v s .
⎝ 10 ⎠
Applying KCL at the inverting input node of the right op amp gives
⎛ 20 ⎞
⎛ 20 ⎞
⎛ 20 ⎞
⎛ 20 ⎞ ⎛ 30 ⎞
⇒ v o = ⎜1 +
v a − ⎜ ⎟ v b = ⎜1 +
va − ⎜ ⎟ ⎜ − ⎟ vs
⎟
⎟
⎜ R3 ⎟
⎜ R3 ⎟
⎜ R3 ⎟
⎜ R 3 ⎟ ⎝ 10 ⎠
R3
20
⎝
⎠
⎝ ⎠
⎝
⎠
⎝ ⎠
⎛ 20 ⎞
⎛ 60 ⎞
⇒ v o = ⎜1 +
va + ⎜ ⎟ vs
⎟
⎜ R3 ⎟
⎜ R3 ⎟
⎝
⎠
⎝ ⎠
⎛ 60 ⎞
⎛ 20 ⎞
(a) We require ⎜ ⎟ = 4 and ⎜1 +
v = 7 V so R 3 = 15 kΩ and then v a = 3 V.
⎜ R3 ⎟
⎜ R 3 ⎟⎟ a
⎝ ⎠
⎝
⎠
⎛ 60 ⎞ ⎛ 60 ⎞
⎛ 20 ⎞
V
⎛ 20 ⎞
and b = ⎜1 +
v a = ⎜1 + ⎟ 2.5 = 5 V .
(b) m = ⎜ ⎟ = ⎜ ⎟ = 3
⎟
⎜ R 3 ⎟ ⎝ 20 ⎠
⎜ R3 ⎟
V
⎝ 20 ⎠
⎝ ⎠
⎝
⎠
vb − va
=
va − vo
P6.5-15. The circuit shown in Figure 6.5-15 uses a
potentiometer to implement a variable resistor having
a resistance R that varies over the range
0 ≤ R ≤ 200 kΩ
The gain of this circuit is G =
vo
vs
. Varying the
resistance R over it’s range causes the value of the
gain G to vary over the range
vo
≤ G max
G min ≤
vs
Determine the minimum and maximum values of the
gain, G min and G max .
Solution:
Let
R eq = 25 + R || 50
Then, recognizing this circuit as a noninverting amplifier,
we get
R eq
25 + R || 50
G = 1+
= 1+
25
25
G min corresponds to R = 0:
G min = 1 +
25 + 0 || 50
25 + 0
= 1+
= 2 V/V
25
25
G max corresponds to R = 200 kΩ:
G max = 1 +
25 + 200 || 50
25 + 40
= 1+
= 3.6 V/V
25
25
Figure P6.5-15
P6.5-16 The input to the circuit shown in Figure P6.5-16a is the voltage, vs. The output is the
voltage vo. The voltage v b is used to adjust the relationship between the input and output.
Determine values of R 4 and v b that cause the circuit input and output have the relationship
specified by the graph shown in Figure P6.5-21b.
Answers: v b = 1.62 V and R 4 = 62.5 kΩ.
(a)
(b)
Figure P6.5-16
Solution: Recognize the voltage divider, voltage follower and noninverting amplifier to write
R4 ⎞
R4 ⎞
2 R4 ⎞
R4 ⎞
⎛
⎞⎛
⎛
⎛
⎛
20 × 10 3
vo = ⎜
3
3 ⎟⎜ −
3 ⎟ v s + ⎜1 +
3 ⎟ vb = ⎜ −
3 ⎟ v s + ⎜1 +
3 ⎟ vb
⎝ 20 × 10 + 5 × 10 ⎠ ⎝ 30 × 10 ⎠
⎝ 30 × 10 ⎠
⎝ 75 × 10 ⎠
⎝ 30 × 10 ⎠
(Alternately, this equation can be obtained by writing two node equations: one at the
noninverting node of the left op amp and the other at the inverting node of the right op amp.)
The equation of the straight line is
5
vo = − vs + 5
3
Comparing coefficients gives
5
5 75 ×10 3
−
=−
⇒ R4 = ×
= 62.5 × 10 3 = 62.5 kΩ
3
75 ×10
3
3
2
2 R4
and
R4 ⎞
⎛
⎛ 62.5 × 10 3 ⎞
5 = ⎜1 +
v
=
⎜1 +
3 ⎟ b
3 ⎟ v b = 3.08333 v b
30
10
30
10
×
×
⎝
⎠
⎝
⎠
⇒ vb =
5
= 1.62 V
3.08333
Figure P6.5-17
P6.5-17 Figure P6.5-17 shows three similar circuits. The outputs of the circuits are the voltages
v1, v2 and v3. Determine the values of these three outputs.
Solution:
80
80
⎛ 80 ⎞
v1 = ⎜ 1 + ⎟ (1.8 ) = 7.56 V , v 2 =
(1.8) = 1.3714 V and v 3 = − (1.8) = −5.76 V
80 + 25
25
⎝ 25 ⎠
Figure P6.5-18
P6.5-18 . The input to the circuit shown in Figure P6.5-18 is the source voltage vs. The output is
the voltage across the 25 kΩ resistor, vo. The output is related to the input by the equation vo =
(g) vi where g is the gain of the circuit. Determine the value of g.
Solution:
Using equivalent resistance we can draw the circuit as shown.
20 = (15 + 45 ) || 30 and 80 = 64 + ( 80 || 20 )
Recognizing this circuit as an inverting amplifier, we write
g=
vo
vs
=−
80
V
= −4
20
V
Section 6-6: Operational Amplifier Circuits and Linear Algebraic Equations
P 6.6-1
Design a circuit to implement the equation
z = 4w +
x
− 3y
4
The circuit should have one output, corresponding to z, and three inputs, corresponding to w, x,
and y.
Solution:
P 6.6-2
Design a circuit to implement the equation
0 = 4w + x + 10 – (6y + 2z)
The output of the circuit should correspond to z.
Solution:
Section 6-7: Characteristics of the Practical Operational Amplifier
10 kΩ
P 6.7-1 Consider the inverting amplifier shown in
Figure P 6.7-1. The operational amplifier is a typical
OP-07E (Table 6.7-1). Use the offsets model of the
operational amplifier to calculate the output offset voltage.
(Recall that the input, vin, is set to zero when calculating
the output offset voltage.)
100 kΩ
–
+
–
vin
+
+
vo
–
Answer: 0.45 mV
Figure P 6.7-1
Solution:
The node equation at node a is:
vout − vos
vos
=
+ ib1
3
100×10
10×103
Solving for vout:
⎛ 100×103 ⎞
vout = ⎜1+
v + (100 × 103 ) ib1 = 11vos + (100 × 103 ) ib1
3 ⎟ os
10
10
×
⎝
⎠
(
)
= 11 ( 0.03×10−3 )+(100×103 ) 1.2×10−9 = 0.45 mV
+
P 6.7-2 Consider the noninverting amplifier shown in
Figure P 6.7-2. The operational amplifier is a typical
LF351 (Table 6.7-1). Use the offsets model of the
operational amplifier to calculate the output offset voltage.
(Recall that the input, vin, is set to zero when calculating
the output offset voltage.)
–
+
90 kΩ
+
–
vin
vo
10 kΩ
–
Figure P 6.7-2
Solution:
The node equation at node a is:
vos
v −v
+ ib1 = 0 os
10000
90000
Solving for vo:
⎛ 90×103 ⎞
vo = ⎜1+
v + ( 90 × 103 ) ib1 = 10 vos + ( 90 × 103 ) i b1
3 ⎟ os
⎝ 10×10 ⎠
= 10(5 ×10−3 )+ ( 90 ×103 ) (.05 × 10−9 ) = 50.0045 × 10−3 − 50 mV
R1
P 6.7-3 Consider the inverting amplifier shown in
Figure P 6.7-3. Use the finite gain model of the operational
amplifier (Figure 6.7-1c) to calculate the gain of the
inverting amplifier. Show that
vo
Rin ( Ro − AR2 )
=
vin ( R1 + Rin )( Ro − R2 ) + R1 Rin (1 + A)
R2
–
+
–
vin
+
Figure P 6.7-3
Solution:
v1 −vin v1 v1 −v0
⎫
+ +
= 0⎪
R1
Rin
R2
v0
Rin ( R0 − AR2 )
⎪
=
⎬ ⇒
v0 + Av1 v0 − v1
vin
( R1 + Rin )( R0 + R2 ) + R1 Rin (1+ A)
+
=0 ⎪
⎪⎭
R0
R2
+
vo
–
R1
P 6.7-4 Consider the inverting amplifier
shown in Figure P 6.7-3. Suppose the
operational amplifier is ideal, R1 = 5 kΩ, and
R2 = 50 kΩ. The gain of the inverting amplifier
will be
R2
–
+
–
vin
+
vo
= −10
vin
Use the results of Problem P 6.7-3P 6.7-3 to
find the gain of the inverting amplifier in each
of the following cases:
(a)
+
vo
–
Figure P 6.7-3
The operational amplifier is ideal, but 2 percent resistors are used and
R1 = 5.1 kΩ and R2 = 49 kΩ.
(b)
The operational amplifier is represented using the finite gain model with
A = 200,000, Ri = 2 MΩ, and Ro = 75 Ω; R1 = 5 kΩ and R2 = 50 kΩ.
(c)
The operational amplifier is represented using the finite gain model with
A = 200,000, Ri = 2 MΩ, and Ro = 75 Ω; R1 = 5.1 kΩ and R2 = 49 kΩ.
Solution:
a)
v0
R
49×103
= − 2 = −
= −9.6078
vin
R1
5.1×103
(
)
b)
2×106 ) 75−( 200,000 )( 50×103 )
(
v0
=
= −9.9957
vin
(5×103 + 2×106 )(75+50×103 ) + (5×103 )(2×106 )(1+ 200,000)
c)
v0
2×106 (75− (200,000)(49×103 ))
=
= −9.6037
vin
(5.1×103 +2×106 )(75+49×103 )+(5.1×103 )(2×106 )(1+200,000)
R1
P 6.7-5 The circuit in Figure P 6.7-5 is called
a difference amplifier and is used for
instrumentation circuits. The output of a
measuring element is represented by the
common mode signal vcm and the differential
signal (vn + vp). Using an ideal operational
amplifier, show that
R4
(vn + vp )
R1
R4 R3
=
R1 R2
vo = −
When
R4
vn +
–
–
vp +
–
+
+
R2
vo
vcm +
–
R3
–
Figure P 6.7-5
Solution:
Apply KCL at node b:
R3
vb =
(vcm − v p )
R2 + R3
Apply KCL at node a:
va −v0 va −(vcm + vn )
+
= 0
R4
R1
The voltages at the input nodes of
an ideal op amp are equal so
va = vb .
R
R +R
v0 = − 4 (vcm + vn ) + 4 1 va
R1
R1
R
v0 = − 4 (vcm + vn ) +
R1
( R4 + R1 ) R3
(vcm − v p )
R1 ( R2 + R3 )
when
R4
R1
R4
+1
( R4 + R1 ) R3
R3
R
R
R1
then
=
=
× 3 = 4
R3
R2
R1 ( R2 + R3 )
R1
+1 R2
R2
so
v0 = −
R4
R
R
(vcm + vn ) + 4 (vcm − v p ) = − 4 (vn + v p )
R1
R1
R1
Section 6-10 How Can We Check…?
6 kΩ
4 kΩ
P 6.10-1 Analysis of the circuit in Figure P 6.10-1
shows that io = – 1 mA and vo = 7 V. Is this analysis
correct?
io
–
+
–
Hint: Is KCL satisfied at the output node of the op amp?
+
2V
5V
–
+
+
10 kΩ
Figure P 6.10-1
Solution:
Apply KCL at the output node of the op
amp
io =
v − ( −5 )
vo
+ o
=0
10000
4000
Try the given values: io =−1 mA and
vo = 7 V
−1×10−3 ≠ 3.7 × 10−3 =
7 − ( −5 )
7
+
10000
4000
KCL is not satisfied. These cannot be the
correct values of io and vo.
vo
–
P 6.10-2 Your lab partner measured
the output voltage of the circuit shown
in Figure P 6.10-2 to be vo = 9.6 V. Is
this the correct output voltage for this
circuit?
Hint: Ask your lab partner to check
the polarity of the voltage that he or
she measured.
4 kΩ
10 kΩ
+
2 mA
–
–
+
+
vo
–
Figure P 6.10-2
Solution:
va = ( 4 ×103 )( 2 ×10−3 ) = 8 V
12 ×103
va = −1.2 ( 8 ) = −9.6 V
10 ×103
So vo = −9.6 V instead of 9.6 V.
vo = −
12 kΩ
4 kΩ
2 kΩ
+
P 6.10-3 Nodal analysis of the circuit shown
in Figure P 6.10-3 indicates that vo = – 12 V. Is
this analysis correct?
Hint: Redraw the circuit to identify an
inverting amplifier and a noninverting
amplifier.
–
–
+
+
2V
vo
6 kΩ
–
+
3V
+
–
2 kΩ
–
Figure P 6.10-3
Solution: First, redraw the circuit as:
Then using superposition, and recognizing of the inverting and noninverting amplifiers:
⎛ 6 ⎞⎛ 4 ⎞
⎛ 4⎞
vo = ⎜ − ⎟ ⎜ − ⎟ ( −3) + ⎜ 1 + ⎟ ( 2 ) = −18 + 6 = −12 V
⎝ 2 ⎠⎝ 2 ⎠
⎝ 2⎠
The given answer is correct.
P 6.10-4 Computer analysis of the circuit in Figure P 6.10-4 indicates that the node voltages
are va = – 5 V, vb = 0 V, vc = 2 V, vd = 5 V, ve = 2 V, vf = 2 V, and vg = 11 V. Is this analysis
correct? Justify your answer.
Hint: Verify that the resistor currents indicated by these node voltages satisfy KCL at nodes b, c,
d, and f.
a 10 kΩ
–
+
b
4 kΩ
c
6 kΩ
5 kΩ
40 kΩ
5V
e
10 kΩ
+
–
2V
4 kΩ
d
–
f
g
+
10 kΩ
Figure P 6.10-4
Solution:
First notice that ve = v f = vc is required by the ideal op amp. (There is zero current into the input
lead of an ideal op amp so there is zero current in the 10 kΩ connected between nodes e and f,
hence zero volts across this resistor. Also, the node voltages at the input nodes of an ideal op
amp are equal.)
The given voltages satisfy all the node equations at nodes b, c and d:
node b:
0− (−5)
0
0−2
+
+
=0
10000 40000 4000
node c:
0− 2
2 −5
=
+0
4000 6000
node d:
2 −5
5
5−11
=
+
6000 5000 4000
Therefore, the analysis is correct.
a
20 kΩ
b
+
P 6.10-5 Computer analysis of the
d
–
noninverting summing amplifier shown in
Figure P 6.10-5 indicates that the node
9 kΩ
voltages are va = 2 V, vb = – 0.25 V, vc = – 5 V,
c 40 kΩ
+
vd = – 2.5 V, and ve = – 0.25 V.
2V –
e
(a)
Is this analysis correct?
–
(b)
Does this analysis verify that the circuit
+ 5V
1 kΩ
40 kΩ
is a noninverting summing amplifier?
Justify your answers.
1st Hint: Verify that the resistor currents
indicated by these node voltages satisfy KCL
Figure P 6.10-5
at nodes b and e.
2nd Hint: Compare to Figure 6.5-1e to see that Ra = 10 kΩ and Rb = 1 kΩ. Determine K1, K2, and
K4 from the resistance values. Verify that vd = K4(K1va + K2vc).
Solution:
The given voltages satisfy the node equations at nodes b and e:
node b:
−.25− 2 −.25 −.25−( −5 )
+
+
=0
20000 40000
40000
node e:
−2.5−( −0.25 ) −0.25
≠
+0
9000
1000
Therefore, the analysis is not correct.
Notice that
−2.5−( +0.25 ) +0.25
=
+0
9000
1000
So it appears that ve = +0.25 V instead of ve = −0.25 V.
Also, the circuit is an noninverting summer with Ra = 10 kΩ and Rb = 1 kΩ, K1 =1/ 2, K2
= 1/ 4 and K4 = 9. The given node voltages satisfy the equation
−2.5 = vd = K
4
( K v + K v ) = 10 ⎛⎜⎝ 12 ( 2 )+ 14 ( −5) ⎞⎟⎠
None-the-less, the analysis is not correct.
1 a
2 c
PSpice Problems
SP 6-1 The circuit in Figure SP 6-1 has three inputs: vw, vx, and vy. The circuit has one output,
vz. The equation
vz = avw + bvx + cvy
expresses the output as a function of the inputs. The coefficients a, b, and c are real constants.
(a)
Use PSpice and the principle of superposition to determine the values of a, b, and c.
(b)
Suppose vw = 2 V, vx = x, vy = y and we want the output to be vz = z. Express z as a
function of x and y.
Hint: The output is given by vz = a when vw = 1 V, vx = 0 V, and vy = 0 V.
Answer: (a) vz = vw + 4 vx – 5 vy (b) z = 4 x – 5 y + 2
20 kΩ
60 kΩ
–
20 kΩ
+
vx
+
+
–
20 kΩ
+
–
100 kΩ
20 kΩ
–
60 kΩ
vy
–
20 kΩ
+
vw
+
–
20 kΩ
20 kΩ
Figure SP 6-1
Solution:
(a)
vz = a vw + b vx + c v y
The following three PSpice simulations show
1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V
4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V
-5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V
vz
1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V:
4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V:
-5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V:
Therefore
v z = vw + 4 v x − 5 v y
(b) When vw= 2 V:
vz = 4 vx − 5 v y + 2
25 kΩ
SP 6-2 The input to the circuit in Figure SP 6-2 is vs,
and the output is vo.
+
–
(a) Use superposition to express vo as a function of vs.
80 kΩ
–
vs
+
(b) Use the DC Sweep feature of PSpice to plot vo as a
function of vs.
+
vo
2V
–
+
–
(c) Verify that the results of parts (a) and (b) agree with
each other.
Figure SP 6-2
Solution:
a) Using superposition and recognizing the inverting
and noninverting amplifiers:
vo = −
80
⎛ 80 ⎞
vs + ⎜1 + ⎟ ( −2 ) = −3.2 vs − 8.4
25
⎝ 25 ⎠
b) Using the DC Sweep feature of PSpice produces the
plot show below. Two points have been labeled in
anticipation of c).
c) Notice that the equation predicts
( −3.2 ) ( −5) − 8.4 = 7.6
and
( −3.2 ) ( 0 ) − 8.4 = −8.4
Both agree with labeled points on the plot.
1
SP 6-3 A circuit with its nodes identified is
shown in Figure SP 6-3. Determine v34, v23, v50,
and io.
10 kΩ
10 kΩ
30 kΩ
6V
2
3
+–
30 kΩ
10 kΩ
30 kΩ
4
–
+
5
io
30 kΩ
+
vo
–
Figure SP 6-3
Solution:
VOLTAGE SOURCE CURRENTS
NAME
CURRENT
V_V1
V_V2
-3.000E-04
-7.000E-04
v34 = −1.5 − −12 × 10−6 ≅ −1.5 V
v23 = 4.5 − ( −1.5 ) = 6 V
v50 = 12 − 0 = 12 V
io = −7 × 10−4 = −0.7 mA
2 kΩ
SP 6-4 Use PSpice to analyze the VCCS shown in
Figure SP 6-4. Consider two cases:
(a)
The operational amplifier is ideal.
(b)
The operational amplifier is a typical μA741
represented by the offsets and finite gain model.
10 kΩ
–
+
2 kΩ
10 kΩ
iout
20 mV
+
–
50 kΩ
Figure SP 6-4
Solution:
V4 is a short circuit used to measure io.
The input of the VCCS is the voltage of the
left-hand voltage source. (The nominal value
of the input is 20 mV.) The output of the
VCCS is io.
A plot of the output of the VCCS versus the
input is shown below.
The gain of the VCVS is
gain =
50 ×10−6 − ( −50 ×10−6 )
1
A
= × 10−3
V
100 ×10 − ( −100 ×10 ) 2
−3
−3
Design Problems
DP 6-1 Design the operational amplifier circuit in
Figure DP 6-1 so that
iin
iout
1
= ⋅ iin
4
iout
Operational
Amplifier
Circuit
5 kΩ
Figure DP 6-1
Solution:
From Figure 6.6-1g, this circuit
is described by vo = R f i in . Since i out =
Notice that i oa = i in +
i in <
(1250 ) i in
5000
=
Rf
vo
1 i out
, we require =
=
, or Rf = 1250 Ω
5000
4 i in 5000
5
5
i in . To avoid current saturation requires i in < i sat or
4
4
4
i sat . For example, if isat = 2 mA, then iin < 1.6 mA is required to avoid current saturation.
5
DP 6-2 Figure DP 6-2a shows a circuit that has one input, vi, and one output, vo. Figure DP 62b shows a graph that specifies a relationship between vo and vi. Design a circuit having input, vi,
and output, vo, that have the relationship specified by the graph in Figure DP 6-2b.
Hint: A constant input is required. Assume that a 5-V source is available.
vo, V
8
6
4
2
vi
vo
–6
–4
–2
2
4
6
8
–2
–4
–6
–8
(a)
(b)
Figure DP 6-2
Solution:
3
3
3
⎛ 3 ⎞ ⎛ 12 ⎞
⎛ 3 ⎞ ⎛ 12 ⎞
vo = − vi + 3 = − vi + ⎜1 + ⎟ ⎜ ⎟ 5 = − vi + ⎜ 1 + ⎟ ⎜
⎟5
4
4
4
⎝ 4 ⎠ ⎝ 35 ⎠
⎝ 4 ⎠ ⎝ 12 + 23 ⎠
vi, V
DP 6-3 Design a circuit having input, vi, and output, vo, that are related by the equations
(a) vo = 12vi + 6, (b) vo = 12vi – 6, (c) vo = – 12vi + 6, and (d) vo = – 12vi – 6.
Hint: A constant input is required. Assume that a 5-V source is available.
Solution:
1
1
1
⎛1
⎞
(a) 12 v i + 6 = 24 ⎜ v i + ( 5 ) ⎟ ⇒ K 4 = 24, K 1 = , and K 2 = . Take Ra = 18 kΩ and
20
2
20
⎝2
⎠
Rb = 1 kΩ to get
(b)
(c)
(d)
DP 6-4 Design a circuit having three inputs, v1, v2, v3, and two outputs, va, vb, that are related
by the equation
⎡ v1 ⎤
⎡ va ⎤ ⎡12 3 −2 ⎤ ⎢ ⎥ ⎡ 2 ⎤
⎢ ⎥=⎢
⎥ ⎢ v2 ⎥ + ⎢ ⎥
⎣ vb ⎦ ⎣ 8 −6 0 ⎦ ⎢ v ⎥ ⎣ −4 ⎦
⎣ 3⎦
Hint: A constant input is required. Assume that a 5-V source is available.
Solution:
x
DP 6-5 A microphone has an unloaded
voltage vs = 20 mV, as shown in
Figure DP 6-5a. An op amp is available as
shown in Figure DP 6-5b. It is desired to
provide an output voltage of 4 V. Design an
inverting circuit and a noninverting circuit
and contrast the input resistance at terminals
x–y seen by the microphone. Which
configuration would you recommend in order
to achieve good performance in spite of
changes in the microphone resistance Rs?
Hint: We plan to connect terminal a to
terminal x and terminal b to terminal y, or
visa versa.
Rs
vs +–
10 kΩ
Microphone
y
(a)
R2
R1
–
a
b
+
+
vo
–
(b)
Figure DP 6-5
Solution:
We require a gain of
4
= 200 . Using an inverting amplifier:
20 ×10−3
Here we have 200 = −
R2
10 ×103 + R1
noninverting amplifier:
. For example, let R1 = 0 and R2 = 1 MΩ. Next, using the
Here we have 200 = 1 +
R2
R1
. For example, let R1 = 1 kΩ and R2 = 199 kΩ.
The gain of the inverting amplifier circuit does not depend on the resistance of the microphone.
Consequently, the gain does not change when the microphone resistance changes.
Chapter 7 Exercises
Exercise 7.2-1 Determine the current i(t) for t > 0 for the circuit of Figure E 7.2-1b when vs(t) is the
voltage shown in Figure E 7.2-1a.
Hint: Determine iC(t) and iR(t) separately, then use KCL.
vs(t)(V)
5
i(t)
4
iR(t)
iC(t)
3
vs(t)
2
+
–
1Ω
1F
1
1
2
3
4
5
6
7
8
9
t (s)
(a)
(b)
Figure E 7.2-1
Answer:
⎧ 2t − 2 2 < t < 4
⎪
4<t <8
v(t ) = ⎨7 − t
⎪0
otherwise
⎩
Solution:
⎧2 2<t <4
d
⎪
i C ( t ) = 1 v s ( t ) = ⎨ −1 4 < t < 8
dt
⎪ 0 otherwise
⎩
so
and
⎧2 t − 4 2 < t < 4
⎪
i R (t ) = 1 v s (t ) = ⎨ 8 − t
4<t <8
⎪ 0
otherwise
⎩
⎧2 t − 2 2 < t < 4
⎪
i (t ) = i C ( t ) + i R ( t ) = ⎨ 7 − t
4<t <8
⎪ 0
otherwise
⎩
Exercise 7.3-1 A 200-μF capacitor has been charged to 100 V. Find the energy stored by the capacitor.
Find the capacitor voltage at t = 0+ if v(0–) = 100 V.
Answer: w(1) = 1 J and v(0+) = 100 V
Solution:
Cv 2
1
2
= ( 2×10 −4 ) (100 ) = 1 J
2
2
+
−
vc ( 0 ) = vc ( 0 ) = 100 V
W =
Exercise 7.3-2 A constant current i = 2 A flows into a capacitor of 100μF after a switch is closed at t =
0. The voltage of the capacitor was equal to zero at t = 0–. Find the energy stored at (a) t = 1 s and (b) t =
100 s.
Answer: w(1) = 20 kJ and w(100) = 200 MJ
Solution:
(a)
W ( t ) = W ( 0 ) + ∫ 0 vi dt
t
First, W ( 0 ) = 0 since v ( 0 ) = 0
Next, v( t ) = v( 0 ) +
∴ W (t ) =
1 t
4 t
4
i
dt
10
=
∫
∫ 0 2 dt = 2×10 t
C 0
∫ ( 2×10 ) t ( 2 )dt
t
4
0
= 2 × 104 t 2
W (1s ) = 2 ×104 J = 20 kJ
(b)
W (100s ) = 2 × 104 (100 )
2
= 2 × 108 J = 200 MJ
6 mF
Exercise 7.4-1 Find the equivalent capacitance
for the circuit of Figure E 7.4-1
Answer: Ceq = 4 mF
12 mF
9 mF
4 mF
Ceq
Figure E 7.4-1
Solution:
2 mF
Exercise 7.4-2 Determine the equivalent
capacitance Ceq for the circuit shown in
Figure E 7.4-2.
1 3
mF
1 3
1 mF
mF
Answer: 10/19 mF
Ceq
2 mF
1 3
mF
Figure E 7.4-2
Solution:
C eq1 =
1
1
= ,
1
1
1
9
+
+
1/ 3 1/ 3 1/ 3
C eq2 = 1 + C eq1 =
10
1
1
1 10
=
=
=
, C eq =
mF
1
1
1
1
1
9
19
9
19
+ +
+ +
2 2 C eq2 2 2 10 10
Exercise 7.5-1 Determine the voltage v(t) for t > 0 for the circuit of Figure E 7.5-1b when is(t) is the
current shown in Figure E 7.5-1a.
is(t)(V)
5
4
3
1H
1Ω
+ vL(t) –
+ vR(t) –
+ v(t) –
2
1
is(t)
1
2
3
4
5
6
7
8
9
t (s)
(a)
(b)
Figure E 7.5-1b
Hint: Determine vL(t) and vR(t) separately, then use KVL.
⎧2t − 2 2 < t < 4
⎪
Answer:
v(t ) = ⎨7 − t
4<t <8
⎪0
otherwise
⎩
Solution:
⎧ 2 2<t <4
d
⎪
v L ( t ) = 1 i s ( t ) = ⎨ −1 4 < t < 8
dt
⎪ 0 otherwise
⎩
so
and
⎧2 t − 4 2 < t < 4
⎪
v R (t ) = 1 is (t ) = ⎨ 8 − t
4<t <8
⎪ 0
otherwise
⎩
⎧2 t − 2 2 < t < 4
⎪
v(t ) = v L ( t ) + v R ( t ) = ⎨ 7 − t
4<t <8
⎪ 0
otherwise
⎩
Exercise 7.7-1 Find the equivalent inductance of the circuit of Figure E 7.7-1.
Answer: Leq = 14 mH
3 mH
42 mH
5 mH
4 mH
Figure E 7.7-1
Solution:
3 mH
Exercise 7.7-2 Find the equivalent inductance of the circuit of Figure E 7.7-2.
2 mH
20 mH
4 mH
Figure E 7.7-2
Ex. 7.7-2
12 mH
Section 7-2: Capacitors
P 7.2-1 A 15-μF capacitor has a voltage of 5 V across it at t = 0. If a constant current of 25 mA
flows through the capacitor, how long will it take for the capacitor to charge up to 150 μC?
Answer: t = 3 ms
v (t ) = v ( 0) +
Solution:
In our case, the current is constant so
1 t
i (τ ) dτ
C ∫0
and q = Cv
t
∫ i (τ ) dτ .
0
∴ Cv ( t ) = Cv ( 0 ) + i t
−6
−6
q −Cv( 0 ) 150×10 −(15×10 )( 5 )
∴ t=
=
= 3 ms
i
25×10−3
P 7.2-2 The voltage, v(t), across a capacitor and current, i(t), in that capacitor adhere to the
passive convention. Determine the current, i(t), when the capacitance is C = 0.125 F and the
voltage is v(t) = 12 cos(2t + 30°) V.
d
d
Hint:
A cos(ωt + θ ) = − A sin (ωt + θ ) ⋅ (ωt + θ )
dt
dt
= − Aω sin (ωt + θ )
⎛
π ⎞⎞
⎛
= Aω cos ⎜ ωt + ⎜ θ + ⎟ ⎟
2 ⎠⎠
⎝
⎝
Answer: i(t) = 3 cos (2t + 120°) A
Solution:
i (t ) = C
d
1d
1
v (t ) =
12 cos ( 2t + 30° ) = (12 )( −2 ) sin ( 2t + 30° ) = 3cos ( 2t + 120° ) A
dt
8 dt
8
P 7.2-3 The voltage, v(t), across a capacitor and current, i(t), in that capacitor adhere to the
passive convention. Determine the capacitance when the voltage is v(t) = 12 cos(500t–45°) V
and the current is i(t) = 3 cos (500t + 45°) mA.
Answer: C = 0.5 μF
Solution:
( 3×10 ) cos ( 500t + 45 ) = C dtd
−3
so
°
C=
12 cos ( 500t − 45° ) = C (12 )( −500 ) sin ( 500t − 45° )
= C ( 6000 ) cos ( 500t + 45° )
3×10−3 1
1
= ×10−6 = μ F
3
6×10
2
2
P 7.2-4 Determine v(t) for the circuit shown in Figure P 7.2-4a when the is(t) is as shown in
Figure P 7.2-4b and v0(0–) = –1 mV.
is (μA)
4
+
is
2 pF
0
v
–
–2
1
2
3
(a)
4
5
6
t (ns)
(b)
Figure P 7.2-4
Solution:
v (t ) =
0 < t < 2 ×10−9
2 ×10−9 < t < 3 × 10−9
1 t
1
i (τ ) dτ + v ( 0 ) =
∫
C 0
2 × 10−12
is ( t ) = 0 ⇒ v ( t ) =
1
2 × 10−12
t
t
∫ i (τ ) dτ − 10
∫ 0 dτ − 10
0
−3
0
−3
= −10−3
is ( t ) = 4 ×10−6 A
t
1
4 ×10−6 ) dτ − 10−3 = −5 × 10−3 + ( 2 × 106 ) t
−12 ∫2ns (
2 × 10
In particular, v ( 3 × 10−9 ) = −5 ×10−3 + ( 2 × 106 ) ( 3 × 10−9 ) = 10−3
⇒ v (t ) =
3 × 10−9 < t < 5 ×10−9
is ( t ) = −2 × 10−6 A
t
1
−2 ×10−6 ) dτ + 10−3 = 4 × 10−3 − (106 ) t
−12 ∫3ns (
2 × 10
In particular, v ( 5 ×10−9 ) = 4 ×10−3 − (106 ) ( 5 × 10−9 ) = −10−3 V
⇒ v (t ) =
5 ×10−9 < t
is ( t ) = 0 ⇒ v ( t ) =
1
2 ×10−12
∫
t
5ns
0 dτ − 10−3 = −10−3 V
P 7.2-5 The voltage, v(t), and current, i(t), of a 1-F capacitor adhere to the passive convention.
Also, v(0) = 0 V and i(0) = 0 A. (a) Determine v(t) when i(t) = x(t), where x(t) is shown in Figure
P 7.2-5 and i(t) has units of A. (b) Determine i(t) when v(t) = x(t), where x(t) is shown in Figure
P 7.2-5 and v(t) has units of V.
Hint: x(t) = 4t – 4 when 1 < t < 2, and x(t) = –4t + 12 when 2 < t < 3.
x
5
4
3
2
1
0
0
1
2
3
4 t (s)
Figure P 7.2-5
Solution:
(b)
(a)
⎧ 0 0 < t <1
⎪ 4 1< t < 2
d
⎪
i (t ) = C v(t ) = ⎨
dt
⎪ −4 2 < t < 3
3<t
⎩⎪ 0
t
1 t
v ( t ) = ∫ i (τ ) dτ + v ( 0 ) = ∫ i (τ ) dτ
0
C 0
For 0 < t < 1, i(t) = 0 A so
t
v ( t ) = ∫ 0 dτ + 0 = 0 V
0
For 1 < t < 2, i(t) = (4t − 4) A so
t
t
v ( t ) = ∫ ( 4τ − 4 ) dτ + 0 = ( 2τ 2 − 4τ ) =2 t 2 − 4 t + 2 V
1
1
( )
v(2) = 2 22 − 4 ( 2 ) + 2 = 2 V . For 2 < t < 3, i(t) = (−4t + 12) A so
t
t
v ( t ) = ∫ ( −4τ + 12 ) dτ + 2 = ( −2τ 2 + 12τ ) +2= ( −2 t 2 + 12 t − 14 ) V
2
1
v(3) = −2 ( 32 ) + 12 ( 3) − 14 = 4 V
t
For 3 < t, i(t) = 0 A so v ( t ) = ∫ 0 dτ + 4 = 4 V
0
P 7.2-6 The voltage, v(t), and current, i(t), of a 0.5-F capacitor adhere to the passive
convention. Also, v(0) = 0 V and i(0) = 0 A. (a) Determine v(t) when i(t) = x(t), where x(t) is
shown in Figure P 7.2-6 and i(t) has units of A. (b) Determine i(t) when v(t) = x(t), where x(t) is
shown in Figure P 7.2-6 and v(t) has units of V.
Hint: x(t) = 0.2t – 0.4 when 2 < t < 6.
x
1.0
0.8
0.6
0.4
0.2
0.0
0
2
4
6
8 t (s)
Figure P 7.2-6
Solution:
(a)
(b)
⎧ 0 0<t <2
d
⎪
i (t ) = C v(t ) = ⎨0.1 2 < t < 6
dt
⎪
6<t
⎩0
t
1 t
v ( t ) = ∫ i (τ ) dτ + v ( 0 ) = 2 ∫ i (τ ) dτ
0
C 0
t
For 0 < t < 2, i(t) = 0 A so v ( t ) = 2 ∫ 0 dτ + 0 = 0 V
0
For 2 < t < 6, i(t) = 0.2 t − 0.4 V so
t
t
v ( t ) = 2 ∫ ( 0.2τ − 0.4 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) =0.2 t 2 − 0.8 t + 0.8 V
1
2
v(6) = 0.2 ( 62 ) − 0.8 ( 6 ) + 0.8 = 3.2 V .
For 6 < t, i(t) = 0.8 A so
t
v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6 t − 6.4 V
6
P 7.2-7 The voltage across a 40-μF capacitor is 25 V at t0 = 0. If the current through the
capacitor as a function of time is given by i(t) = 6e–6t mA for t < 0, find v(t) for t > 0.
Answer: v(t) = 50 – 25e–6t V
Solution:
v (t ) = v ( 0) +
1 t
4 t
i
τ
d
τ
=
+
×
25
2.5
10
6×10−3 ) e −6τ dτ
(
)
(
∫
∫
0
0
C
t
= 25 + 150∫ 0 e −6τ dτ
t
⎡ 1
⎤
= 2 5 + 150 ⎢ − e −6τ ⎥ = 50 − 25e−6t V
6
⎣
⎦0
P 7.2-8
Find i for the circuit of Figure P 7.2-8 if v = 5(1 – 2e–2t) V.
+
v
i
–
10 μ F
200 kΩ
Figure P 7.2-8
Solution:
v
1
= (1− 2e−2t ) × 10−3 = 25 (1− 2e −2t ) μ A
3
200×10
40
dv
= 10×10−6 ( −2 ) ( −10 e−2t ) = 200 e−2 t μ A
iC = C
dt
i = iR + i C = 200 e−2t + 25 − 50 e −2t
iR =
(
)
= 25 + 150e −2t
μA
is (t)
P 7.2-9 Determine v(t) for t ≥ 0 for the circuit
of Figure P 7.2-9a when is(t) is the current
shown in Figure P 7.2-9b and v(0) = 1 V.
0.5 F
+
v(t)
–
(a)
is (A)
2
4
1
3
2
–2
(b)
Figure P 7.2-9
Solution:
t
v ( t ) = 2 ∫ i ( t ) dt + 1
0
=1
0≤t ≤2
for
t
= 2 ∫ 2 dt + 1 = 4 ( t − 2 ) + 1 = 4t − 7
2
3
t
2
3
3
4
2
3
for
2≤t ≤3
= 2 ∫ 2 dt + 2∫ −2 dt + 1 = 4 − 4 ( t − 3) + 1 = −4t + 17
= 2 ∫ 2 dt + 2∫ −2 dt + 1 = 1
for
t≥4
In summary
0≤t ≤2
⎧ 1
⎪ 4t − 7 2 ≤ t ≤ 3
⎪
v (t ) = ⎨
⎪−4t + 17 3 ≤ t ≤ 4
4≤t
⎩⎪ 1
for
3≤ t ≤ 4
t (s)
+
2Ω
is (t)
v(t)
0.1 F
–
P 7.2-10 Determine v(t) for t ≥ 0 for the circuit of
Figure P 7.2-10a when v(0) = –4 V and is(t) is the
current shown in Figure P 7.2-10b.
(a)
is (A)
2
t (s)
0.25
0.5
(b)
Figure P 7.2-10
Solution:
t
1 t
4
10
i
t
dt
=
−
+
(
)
s
∫ 0 i s ( t ) dt
C ∫0
v (t ) = v ( 0) +
For 0 ≤ t ≤ 0.25 ( i s ( t ) = 8t for 0 ≤ t ≤ 0.25 )
t
⎛τ 2 ⎞
v ( t ) = −4 + 10 ∫ 8τ dτ = −4 + 80 ⎜ ⎟ = −4 + 40t 2
0
⎝ 2 ⎠0
t
⎛1⎞
For example v ( 0 ) = −4, v ⎜ ⎟ = −3.375,
⎝8⎠
For 0.25 ≤ t ≤ 0.5
v ( t ) = −1.5 + 10∫
t
0.25
⎛1⎞
v ⎜ ⎟ = −1.5
⎝4⎠
2 dτ = −1.5 + 20 ( t − 0.25 ) = 20t − 6.5
For example v ( 0.25 ) = −1.5, v ( 0.5 ) = 3.5
For 0.5 ≤ t
In summary
t
v ( t ) = 3.5 + 10∫ 0 dτ = 3.5
0.5
⎧−4 + 40t 2
⎪
v ( t ) = ⎨ 20t − 6.5
⎪ 3.5
⎩
0 ≤ t ≤ 0.25 s
0.25 ≤ t ≤ 0.5 s
t ≥ 0.5 s
i(t)
P 7.2-11 Determine i(t) for t ≥ 0 for the circuit
of Figure P 7.2-11a when vs(t) is the voltage
shown in Figure P 7.2-11b.
vs(t) +
5Ω
5F
–
(a)
vs (V)
20
t (s)
0
0.5
1.0
1.5
(b)
Figure P 7.2-11
Solution:
Representing vs(t) using equations of the straight line segments gives
0
⎧
⎪ 40t − 20
⎪
vs (t ) = ⎨
⎪−20t + 40
⎪⎩
0
t ≤ 0.5
0.5 ≤ t ≤ 1.0
1.0 ≤ t ≤ 2.0
2.0 ≤ t
Use KCL to get
0
⎧
⎪
⎪ 20 + 40t − 20
vs (t ) ⎪
1 d
5
i (t ) =
vs (t ) +
=⎨
2 dt
5
⎪ −10 + 40 − 20t
⎪
5
⎪
0
⎩
⎧ 0
⎪ 8t − 16
⎪
i (t ) = ⎨
⎪−4t − 2
⎪⎩ 0
t ≤ 0.5
0.5 ≤ t ≤ 1.0
1.0 ≤ t ≤ 2.0
t ≥ 2.0
t ≤ 0.5
0.5 ≤ t ≤ 1.0
1.0 ≤ t ≤ 2.0
t ≥ 2.0
2.0
The capacitor voltage in the circuit shown in Figure P 7.2-12 is given by
v(t ) = 12 − 10e −2t V for t ≥ 0
Determine i(t) for t > 0.
P 7.2-12
4Ω
+
v(t)
i(t)
2A
1
F
20
6Ω
–
Figure P 7.2-12
Solution:
1 d
v (t )
20 dt
1
= ( +20e −2t )
20
= e −2t A for t > 0
Apply KCL to get
iC (t ) =
i ( t ) = 2 − i C ( t ) = 2 − e −2t A for t > 0
(checked: LNAP 6/25/04)
P 7.2-13
The capacitor voltage in the circuit shown in Figure P 7.2-13 is given by
v(t ) = 2.4 + 5.6e −5t V for t ≥ 0
Determine i(t) for t > 0.
20 Ω
+
v(t)
400 Ω
2 mF 100 Ω
+
–
i(t)
12 V
–
Figure P 7.2-13
Solution:
We’ll write and solve a node equation. Label
the node voltages as shown. Apply KCL at
node a to get
v (t ) − v a
20
=
+
v a − 12
⇒
100
400
v a = 2.4 + 4.48e −5t V
So
i (t ) =
Then
va
va
100
20v ( t ) + 12
25
for t > 0
va =
= 24 + 44.8e −5t mA for t > 0
(checked: LNAP 6/25/04)
P 7.2-14
The capacitor voltage in the circuit shown in Figure P 7.2-14 is given by
v(t ) = 10 − 8e −5t V for t ≥ 0
Determine i(t) for t > 0.
i(t)
20 mF
+
v(t)
–
60 Ω
12 Ω
+
–
12 V
Figure P 7.2-14
Solution: Apply KVL to the outside loop to get
12 i ( t ) + 12 − v ( t ) = 0 ⇒ i ( t ) =
12 − 10 − 8e −5t
1 2
= − − e −5t A for t > 0
12
6 3
(checked: LNAP 6/25/04)
P 7.2-15
is(t)(A)
4
Determine the voltage v(t) for t > 0
for the circuit of Figure P 7.2-15b when is(t) is
2
the current shown in Figure P 7.2-15a. The
capacitor voltage at time t = 0 is v(0) = –12 V.
–4
–2
2
4
6
8
–2
(a)
+
is(t)
1 3F
v(t)
–
(b)
Figure P 7.2-15
Solution:
v(t ) =
1 t
1 t
(
)
i
τ
d
τ
+
v
t
=
i (τ ) dτ − 12
(
)
s
0
1 ∫0 s
C ∫t 0
3
t
v(t ) = 3∫ 4 dτ − 12 = 12 t − 12 for 0 < t < 4
0
t
v(t ) = 3∫ ( −2 ) dτ + 36 = 60 − 6 t for 4 < t < 10
4
t
v(t ) = 3∫ 0 dτ + 0 = 0 for 10 < t
10
In particular, v(4) = 36 V.
In particular, v(10) = 0 V.
10 t (s)
P7.2-16 The input to the circuit shown in Figure P7.2-16 is
the current
i ( t ) = 3.75 e −1.2 t A for t > 0
The output is the capacitor voltage
v ( t ) = 4 − 1.25 e −1.2 t V for t > 0
Find the value of the capacitance, C.
Figure P7.2-16
Solution: The capacitor voltage is related to the capacitor current by
v (t ) =
1 t
i (τ ) dτ + v ( 0 )
C ∫0
That is:
4 − 1.25 e −1.2 t =
1 t
3.75 −1.2τ
e
3.75 e −1.2τ dτ + v ( 0 ) =
∫
C 0
C ( −1.2 )
t
0
+ v ( 0) =
−3.125 −1.2 t
( e − 1) + v ( 0 )
C
Equating the coefficients of e−1.2 t gives
12.5 =
3.125
3.125
⇒ C=
= 0.25 = 250 mF
C
12.5
P7.2-17 The input to the circuit shown in Figure P7.2-17 is
the current
i ( t ) = 3 e −25 t A for t > 0
The initial capacitor voltage is v C ( 0 ) = −2 V . Determine
the current source voltage, v ( t ) , for t > 0 .
Figure P7.2-17
Solution: Apply KVL to the mesh to get
⎡ 1 t
⎤
v ( t ) = 4 i ( t ) + vC ( t ) = 4 i ( t ) + ⎢
i (τ ) dτ + v ( 0 ) ⎥
∫
⎣ 0.05 0
⎦
That is,
v ( t ) = 4 ( 3 e−25 t ) +
1 t −25τ
3
3 e dτ − 2 = 12 e −25 t +
e−25 t − 1) − 2
(
∫
0
0.05
0.05 ( −25 )
= 12 e −25 t − 2.4 ( e −25 t − 1) − 2 = 9.6 e−25 t + 0.4 V for t > 0
P7.2-18 The input to the circuit shown in Figure P7.2-18 is
the current
i ( t ) = 3 e −25 t A for t > 0
The output is the voltage
v ( t ) = 9.6 e −25 t + 0.4 V for t > 0
The initial capacitor voltage is v C ( 0 ) = −2 V . Determine the
values of the capacitance, C, and resistance, R.
Figure P7.2-18
Solution: Apply KVL to the mesh to get
⎡1 t
⎤
v ( t ) = R i ( t ) + vC ( t ) = R i ( t ) + ⎢ ∫ i (τ ) dτ + v ( 0 ) ⎥
0
⎣C
⎦
That is
⎡1 t
⎤
9.6 e −25 t + 0.4 = R ( 3 e −25 t ) + ⎢ ∫ 3 e−25τ dτ − 2 ⎥
0
⎣C
⎦
= 3 R e −25 t +
⎛
3
1 ⎞ −25 t
3
−2
e−25 t − 1) − 2 = 3 ⎜ R −
(
⎟e +
25 C ⎠
25 C
C ( −25 )
⎝
Equating coefficients gives
0.4 =
and
3
− 2 ⇒ C = 0.05 = 50 mF
25 C
⎛
⎞
⎛
1 ⎞
1
9.6 = 3 ⎜ R −
⎟⎟ = 3 ( R − 0.8 ) R = 4 Ω
⎟ = 3 ⎜⎜ R −
25 C ⎠
25
0.05
(
)
⎝
⎝
⎠
P7.2-19 The input to the circuit shown in Figure P7.2-19 is
the voltage
v ( t ) = 8 + 5 e −10 t V for t > 0
Determine the current, i ( t ) for t > 0 .
Figure P7.2-19
Solution: Apply KCL at either node to get
v (t )
8 + 5 e −10 t
d
d
+ 0.05 v ( t ) =
+ 0.05 ( 8 + 5 e −10 t )
i (t ) =
4
4
dt
dt
−10 t
= 2 + 1.25 e + 0.05 ( 5 )( −10 ) e −10 t
= 2 − 1.25 e −10 t A for t > 0
P7.2-20 The input to the circuit shown in Figure P7.2-20 is the
voltage:
v ( t ) = 3 + 4 e −2 t A for t > 0
The output is the current: i ( t ) = 0.3 − 1.6 e −2 t V for t > 0
Determine the values of the resistance and capacitance
Answer: R = 10 Ω and C = 0.25 F.
Figure P7.2-20
Solution: Apply KCL at either node to get
3 + 4 e −2 t
d
+C
3 + 4 e −2 t
R
dt
−2 t
3+ 4e
3 ⎛4
⎞
=
+ ( −2 ) 4 C e −2 t = + ⎜ − 8 C ⎟ e −2 t
R
R ⎝R
⎠
(
0.3 − 1.6 e −2 t =
)
Equating coefficients:
0.3 =
3
R
⇒ R = 10 Ω
and
−1.6 =
4
− 8 C ⇒ C = 0.25 F
10
P7.2-21 Consider the capacitor shown in Figure P7.2-21. The current
and voltage are given by
⎧0.5 0 < t < 0.5
⎧2 t + 8.6 0 ≤ t ≤ 0.5
⎪
⎪
i ( t ) = ⎨ 2 0.5 < t < 1.5 and v ( t ) = ⎨ a t + b 0.5 ≤ t ≤ 1.5
⎪0
⎪ c
t > 1.5
t ≥ 1.5
⎩
⎩
where a, b and c are real constants. (The current is given in Amps, the
voltage in Volts and the time in seconds.) Determine the values of a, b
and c.
Answer: a = 8 V/s, b = 5.6 V and c = 17.6 V
Solution: At t = 0.5 s
Figure P7.2-21
v ( 0.5 ) = 2 ( 0.5 ) + 8.6 = 9.6 V
For 0.5 ≤ t ≤ 1.5
v (t ) =
1 t
2 dτ + 9.6 = 8τ
0.25 ∫ 0.5
At t = 1.5 s
t
0.5
+ 9.6 = 8 ( t − 0.5 ) + 9.6 = 8 t + 5.6 V
v (1.5 ) = 8 (1.5 ) + 5.6 = 17.6 V
For t ≥ 1.5
v (t ) =
1 t
0 dτ + 17.6 = 17.6
0.25 ∫1.5
Checks:
At t = 1.0 s
At t = 0.5 s
i (t ) =
1 d
1 d
1
v (t ) =
(8 t + 5.6 ) = (8) = 2 A √
4 dt
4 dt
4
v ( 0.5 ) = 8 ( 0.5 ) + 5.6 = 9.6 V
√
P7.2-22At time t =0, the voltage across the capacitor shown in Figure P7.2-22 is v(0) = −20 V.
Determine the values of the capacitor voltage at times 1 ms, 3 ms and 7 ms.
Figure P7.2-22
Solution:
v (t ) = v (0) +
1 t
"area under the curve"
"area under the curve"
i (τ ) dτ = v ( 0 ) +
= −20 +
∫
0
2.5 × 10−6
C
C
1
20 × 10−3 1×10−3
10
v ( 0.001) = −20 + 2
= −20 +
= −16 V
−6
2.5 × 10
2.5
(
)(
)
(When calculating the value of v(0.001), “area under the curve” indicates the area under the
graph of i(t) versus t corresponding to the time interval 0 to 1 ms = 0.001 s.)
1
40 × 10−3 2 ×10−3 + 40 × 10−3 1× 10−3
40 + 40
v ( 0.003) = −20 + 2
= −20 +
= 12 V
−6
2.5 ×10
2.5
(
)(
) (
)(
)
(When calculating the value of v(0.003), “area under the curve” indicates the area under the
graph of i(t) versus t corresponding to the time interval 0 to 3 ms = 0.003 s.)
1
1
40 ×10−3 2 × 10−3 + 40 ×10−3 2 × 10−3 + 40 × 10−3 3 × 10−3
2
v ( 0.007 ) = −20 + 2
= 52 V
−6
2.5 ×10
(
)(
) (
)(
)
(
)(
)
(When calculating the value of v(0.007), “area under the curve” indicates the area under the
graph of i(t) versus t corresponding to the time interval 0 to 7 ms = 0.007 s.)
Section 7-3: Energy Storage in a Capacitor
P 7.3-1 The current, i, through a capacitor is shown in Figure P 7.3-1. When v(0) = 0 and C =
0.5 F, determine and plot v(t), p(t), and w(t) for 0 s < t < 6 s.
i(A)
1.0
0.8
0.6
0.4
0.2
0.0
0
2
4
6
8 t (s)
Figure P 7.3-1
Solution:
0
t<2
⎧
⎪
i ( t ) = ⎨0.2 ( t − 2 ) 2 < t < 6
⎪ 0.8
t >6
⎩
Given
The capacitor voltage is given by
v (t ) =
t
1 t
i (τ ) dτ + v ( 0 ) = 2 ∫ i (τ ) dτ + v ( 0 )
∫
0
0.5 0
t
v ( t ) = 2 ∫ 0 dτ + 0 = 0
For t < 2
0
In particular, v ( 2 ) = 0. For 2 < t < 6
v ( t ) = 2 ∫ 2 (τ − 2 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) = ( 0.2 t 2 − 0.8 t + 0.8 ) V = 0.2 ( t 2 − 4 t + 4 ) V
t
t
2
2
In particular, v ( 6 ) = 3.2 V. For 6 < t
t
v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6τ 2 + 3.2 = (1.6 t − 6.4 ) V = 1.6 ( t − 4 ) V
t
6
Now the power and energy are calculated as
0
⎧
⎪
2
p ( t ) = v ( t ) i ( t ) = ⎨0.04 ( t − 2 )
⎪ 1.28 ( t − 4 )
⎩
t<2
2<t <6
6<t
and
W (t ) = ∫
t
0
0
⎧
t<2
⎪⎪
4
2<t<6
p (τ ) dτ = ⎨ 0.01( t − 2 )
⎪
2
6<t
⎪⎩0.8 ( t − 4 ) − 0.64
These plots were produced using three MATLAB scripts:
capvol.m
function v = CapVol(t)
if t<2
v = 0;
elseif t<6
v = 0.2*t*t - .8*t +.8;
else
v = 1.6*t - 6.4;
end
capcur.m
function i = CapCur(t)
if t<2
i=0;
elseif t<6
i=.2*t - .4;
else
i =.8;
end
c7s4p1.m
t=0:1:8;
for k=1:1:length(t)
i(k)=CapCur(k-1);
v(k)=CapVol(k-1);
p(k)=i(k)*v(k);
w(k)=0.5*v(k)*v(k);
end
plot(t,i,t,v,t,p)
text(5,3.6,'v(t), V')
text(6,1.2,'i(t), A')
text(6.9,3.4,'p(t), W')
title('Capacitor Current, Voltage and Power')
xlabel('time, s')
% plot(t,w)
% title('Energy Stored in the Capacitor, J')
% xlabel('time, s')
In a pulse power circuit the voltage of a 10-μF capacitor is zero for t < 0 and
P 7.3-2
v = 5(1 − e −4000t ) V t ≥ 0
Determine the capacitor current and the energy stored in the capacitor at t = 0 ms and t = 10 ms.
Solution:
⎧⎪
ic ( 0 ) = 0.2 A
dv
= (10×10−6 ) ( −5 )( −4000 ) e −4000t = 0.2e−4000t A ⇒ ⎨
−19
dt
⎪⎩ ic (10ms ) = 8.5×10 A
1
W ( t ) = Cv 2 ( t ) and v ( 0 ) = 5 − 5e0 = 0 ⇒ W ( 0 ) = 0
2
−3
v (10×10 ) = 5 − 5 e −40 = 5 − 21.2 × 10−18 ≅ 5 ⇒ W (10 ) = 1.25×10−4 J
ic = C
P 7.3-3 If vc(t) is given by the waveform shown in Figure P 7.3-3, sketch the capacitor current
for –1 s ≤ t ≤ 2 s. Sketch the power and the energy for the capacitor over the same time interval
when C = 1 mF.
vc (V)
20
–1
0
1
2
t (s)
–20
Figure P 7.3-3
Solution:
dvc
so read off slope of vc (t ) to get i (t )
dt
p (t ) = vc (t ) i (t ) so multiply vc (t ) & i (t ) curves to get p (t )
i (t ) = C
P 7.3-4 The current through a 2-μF capacitor is 50 cos(10t + π/6) μA for all time. The average
voltage across the capacitor is zero. What is the maximum value of the energy stored in the
capacitor? What is the first nonnegative value of t at which the maximum energy is stored?
Solution:
π⎞
1 t
1 t
⎛
i dτ = vc ( 0 ) + ∫0 50 cos⎜10t + ⎟ dτ =
∫
0
2
6⎠
C
⎝
π
5
Now since vc ( t )ave = 0 ⇒ vc ( 0 ) − sin = 0 ⇒ vc ( t )
2
6
vc ( t ) = vc ( 0 ) +
( 2×10 ) ( 2.5) = 6.25 μ J
1
= C v2 =
c max
2
2
−6
∴ Wmax
π⎞
5 π⎤ 5
⎡
⎛
⎢⎣ vc ( 0 ) − 2 sin 6 ⎦⎥ + 2 s in ⎝⎜10t + 6 ⎠⎟
π⎞
5
⎛
= sin ⎜10t + ⎟ V
2
6⎠
⎝
2
First non-negative t for max energy occurs when: 10t +
π
6
=
π
2
⇒t =
π
30
= 0.1047 s
P 7.3-5 A capacitor is used in the electronic flash unit of a camera. A small battery with a
constant voltage of 6 V is used to charge a capacitor with a constant current of 10 μA. How long
does it take to charge the capacitor when C = 10 μF? What is the stored energy?
Solution:
Max. charge on capacitor = C v = (10×10−6 ) ( 6 ) = 60 μ C
Δq
60×10−6
=
= 6 sec to charge
i
10×10−6
1
1
2
stored energy = W = C v 2 = (10×10−6 ) ( 6 ) =180 μ J
2
2
Δt =
P 7.3-6 The initial capacitor voltage of the circuit shown in Figure P 7.3-6 is vc(0–) = 3 V.
Determine (a) the voltage v(t) and (b) the energy stored in the capacitor at t = 0.2 s and t = 0.8 s
when
⎧3e5t A 0 < t < 1
i (t ) = ⎨
t ≥ 1s
⎩0
Answer:
(a)
18e5t V, 0 ≤ t < 1
(b)
w(0.2) = 6.65 J and w(0.8) = 2.68 kJ
t=0
+
5Ω
v
i(t)
+
0.2 F
–
vc
–
Figure P 7.3-6
Solution:
We have v (0+ ) = v (0− ) = 3 V
vc ( t ) =
t
1 t
i (t ) dt + vc (0) = 5 ∫ 0 3 e5t dt + 3 = 3 ( e5t −1)+ 3 = 3 e5t V, 0<t <1
∫
0
C
a)
v(t ) = vR ( t ) + vc ( t ) = 5 i ( t ) + vc ( t ) = 15 e5t + 3 e5t = 18 e5t V, 0 < t < 1
b)
⎧⎪ W (t ) t =0.2 s = 6.65 J
2
W ( t ) = 1 Cvc2 ( t ) = 1 × 0.2 ( 3e5t ) = 0.9e10t J ⇒ ⎨
2
2
⎪⎩W ( t ) t =0.8 s = 2.68 kJ
P7.3-7. (a) Determine the energy stored in the capacitor in the circuit shown in Figure P7.3-7
when the switch is closed and the circuit is at steady state. (b) Determine the energy stored in the
capacitor when the switch is open and the circuit is at steady state.
Figure P7.3-7
Solution:
The capacitor acts like an open circuit when this circuit is at steady state.
(a) When the switch is closed and the circuit is at steady state, v ( t ) = 6 V . The energy stored by
1
( 2.2 ×10 −9 )( 62 ) = 39.6 μJ .
2
(b) When the switch is closed and the circuit is at steady state, v ( t ) = 12 V . The energy stored by
the capacitor is W =
the capacitor is W =
1
2.2 × 10 −9 )(122 ) = 158.4 μJ .
(
2
Section 7-4: Series and Parallel Capacitors
P 7.4-1 Find the current i(t) for the circuit of Figure P 7.4-1.
Answer: i(t) = –1.2 sin 100t mA
i(t)
3 μF
+
–
6 cos 100t V
2 μF
4 μF
Figure P 7.4-1
Solution:
2μ F 4 μ F = 6μ F
6μ F in series with 3μ F =
i (t ) = 2 μ F
6μ F⋅3μ F
= 2μ F
6μ F+3μ F
d
(6 cos100t ) = (2×10−6 ) (6) (100) (− sin100t ) A = −1.2 sin100t mA
dt
P 7.4-2 Find the current i(t) for the circuit of Figure P 7.4-2.
Answer: i(t) = –1.5e–250t mA
i(t)
5 + 3e–250t V
+
–
4 μF
4 μF
2 μF
4 μF
Figure P 7.4-2
Solution:
4 μ F in series with 4 μ F =
4 μ F×4 μ F
= 2 μF
4 μ F+4 μ F
2 μF 2 μF = 4 μF
4 μ F in series with 4 μ F = 2 μ F
i (t ) =(2×10−6 )
d
(5+ 3 e −250t ) = (2×10−6 ) (0+ 3(−250) e −250t ) A = −1.5 e −250t mA
dt
P 7.4-3 The circuit of Figure P 7.4-3 contains five identical capacitors. Find the value of the
capacitance C.
Answer: C = 10 μF
i(t) = 25 cos 250t mA
C
14 sin 250t V
+
–
C
C
C
C
Figure P 7.4-3
Solution:
C in series with C =
C C
C ⋅C C
=
C +C 2
5
C
= C
2
2
C⋅ 5 C
5
5
2
C in series with C =
= C
2
7
C+5 C
2
⎛5 ⎞ d
⎛5 ⎞
(25×10−3 ) cos 250t = ⎜ C ⎟ (14sin 250t ) = ⎜ C ⎟(14)(250) cos 250t
⎝ 7 ⎠ dt
⎝7 ⎠
−3
−6
so 25×10 = 2500 C ⇒ C = 10×10 = 10 μF
P7.4-4 The circuit shown in Figure P 7.4-4 contains seven capacitors, each having capacitance C.
The source voltage is given by
v(t ) = 4 cos(3t ) V
Find the current i(t) when C = 1 F.
i(t)
C
+
–
v(t)
C
C
C
C
C
C
Figure P 7.4-4
Solution: Replacing series and parallel capacitors by equivalent capacitors, the circuit can be
reduced as follows:
Then
i (t ) =
8C d
8C d
8 ×1
32
v (t ) =
4 cos ( 3 t ) =
⎡⎣ −12sin ( 3 t ) ⎤⎦ = − sin ( 3 t ) V
21 d t
21 d t
21
7
(Checked using LNAP 6/25/04)
P7.4-5 Determine the value of the capacitance C in the circuit shown in Figure P 7.4-5, given
that Ceq = 8 F.
Answer: C = 20 F
16 F
A
12 F
C
4F
12 F
10 F
30 F
B
Ceq
Figure P 7.4-5
Solution: The 16 F capacitor is in series with a parallel combination of 4 F and 12 F capacitors.
The capacitance of the equivalent capacitor is
16 ( 4 + 12 )
=8 F
16 + ( 4 + 12 )
The 30 F capacitor is in parallel with a short circuit, which is equivalent to a short circuit. After
making these simplifications, we have
Then
8 = C eq =
10 (12 + C + 8 )
⇒ C = 20 F
10 + (12 + C + 8 )
(Checked using LNAP 6/26/04)
P 7.4-6 Determine the value of the equivalent capacitance, Ceq, in the circuit shown in Figure P
7.4-6.
Answer: Ceq = 10 F
15 F
a
60 F
30 F
10 F
40 F
b
60 F
Ceq
Figure P 7.4-6.
Solution:
C eq =
1
1
1
1
1
+
+ +
60 15 + 10 30 40 + 60
= 10 F
(Checked using LNAP 6/26/04)
P 7.4-7 The circuit shown in Figure P 7.4-7 consists of
nine capacitors having equal capacitance, C. Determine
the value of the capacitance C, given that Ceq = 50 mF.
Answer: C = 90 mF
C
C
C
C
Solution: First
C
Ceq
C
C
C
C
Then
50 = C eq =
1
1
2
2
+
+
C 5C 5C
(Checked using LNAP 6/26/04)
⇒ C = 90 mF
Figure P 7.4-7
P 7.4-8 The circuit shown in Figure P 7.4-8 is at steady state before the switch opens at
time t = 0.
t=0
+
–
20 Ω
5Ω
18 V
+
v(t)
60 mF
20 mF
–
Figure P 7.4-8
The voltage v(t) is given by
⎧ 3.6 V for t ≤ 0
v(t ) = ⎨
−2.5t
for t ≥ 0
⎩3.6e
(a)
(b)
(c)
(d)
Determine the energy stored by each capacitor before the switch opens.
Determine the energy stored by each capacitor 1 s after the switch opens.
The parallel capacitors can be replaced by an equivalent capacitor.
Determine the energy stored by the equivalent capacitor before the switch opens.
Determine the energy stored by the equivalent capacitor 1 s after the switch opens.
Solution:
(a) The energy stored in the 60 mF capacitor is w1 =
1
( 0.060 ) 3.62 = 0.3888 W and the
2
1
( 0.020 ) 3.62 = 0.1296 J .
2
(b) One second after the switch opens, the voltage across the capacitors is
3.6e −2.5 = 0.2955 V . Then w1 = 2.620 mJ and w 2 = 0.873 mJ.
energy stored in the 20 mF capacitor is w 2 =
Next C eq = 0.06 + 0.02 = 80 mF.
1
( 0.08) 3.62 = 0.5184 J = w1 + w 2
2
1
2
(d) w eq = ( 0.08 )( 0.2955 ) = 3.493 mJ = w1 + w 2
2
(c) w eq =
P 7.4-9 The circuit shown in Figure P 7.4-9 is at steady state before the switch closes. The
capacitor voltages are both zero before the switch closes (v1(0) = v2(0) = 0).
t=0
5Ω
+
10 mF
+
– 12 V
v1 (t)
–
25 Ω
i(t)
+
40 mF
v2 (t)
–
Figure P 7.4-9
The current i(t) is given by
for t < 0
⎧ 0A
i (t ) = ⎨
−30 t
⎩2.4e A for t > 0
(a)
(b)
Determine the capacitor voltages, v1(t) and v2(t), for t ≥ 0.
Determine the energy stored by each capacitor 20 ms after the switch closes.
The series capacitors can be replaced by an equivalent capacitor.
(c)
(d)
Determine the voltage across the equivalent capacitor, + on top, for t ≥ 0.
Determine the energy stored by the equivalent capacitor 20 ms after the switch closes.
Solution:
(a)
v1 ( t ) =
1 t
240 −30t
2.4e −30τ dτ + 0 =
( e − 1) = +8 (1 − e−30τ ) V for t ≥ 0
∫
0.01 0
−30
1 t
2.4e −30τ dτ + 0 = 2 (1 − e−30t ) V for t ≥ 0
v 2 (t ) =
∫
0
0.04
(b) When t = 20 ms, v1 ( 0.02 ) = 8 (1 − e −0.6 ) = 3.610 V and v 2 ( 0.02 ) = 2 (1 − e −0.6 ) = 0.902 V so
the energy stored by the 10 mF capacitor is w1 =
by the 40 mF capacitor is w 2 =
Next C eq =
10 × 40
= 8 mF
10 + 40
1
( 0.01) 3.6102 = 65.2 mJ and the energy stored
2
1
( 0.04 ) 0.9022 = 16.3 mJ .
2
1
2.4e −30τ dτ = 10 (1 − e−30t ) V for t ≥ 0
∫
0.08
(d) When t = 20 ms, v ( 0.02 ) = 10 (1 − e −0.6 ) = 4.512 V so the energy stored by the equivalent
(c)
capacitor is w =
v (t ) =
1
( 0.008) 4.5122 = 81.4 mJ = w1 + w 2 .
2
P 7.4-10 Find the relationship for the division of current between two parallel capacitors as
shown in Figure P 7.4-10.
Answer: in = iCn/(C1 + C2), n = 1, 2
i
i1
i2
C1
C2
Figure P 7.4-10
Solution:
v1 = v2 ⇒
dv1 dv2
i
i
C
=
⇒ 1 = 2 ⇒ i1 = 1 i2
dt
dt
C1 C2
C2
⎛C
⎞
KCL: i = i1 + i2 = ⎜ 1 + 1⎟ i2
⎝ C2
⎠
⇒
i2 =
C2
i
C1 +C2
Section 7-5: Inductors
P 7.5-1 Nikola Tesla (1857–1943) was an American electrical engineer who experimented with
electric induction. Tesla built a large coil with a very large inductance. The coil was connected to
a source current
is = 100sin 400t A
so that the inductor current iL = is. Find the voltage across the inductor and explain the discharge
in the air shown in the figure. Assume that L = 200 H and the average discharge distance is 2 m.
Note that the dielectric strength of air is 3 × 106 V/m.
Solution:
di
= 2 0 0 [1 0 0 ( 4 0 0 ) c o s 4 0 0 t ] V
dt
8 ×1 0 6 V
= 4 ×1 0 6 V
∴ v m ax = 8 ×1 0 6 V th u s h a v e a fie ld o f
m
m
2
6
w h ic h e x ce e d s d ie le c tric stre n g th in a ir o f 3 ×1 0 V /m
∴ W e g e t a d isc h a rg e a s th e a ir is io n iz e d .
F in d m a x . v o ltag e a c ro ss c o il:
v (t ) = L
P 7.5-2 The model of an electric motor consists of a series combination of a resistor and
inductor. A current i(t) = 4te–t A flows through the series combination of a 10-Ω resistor and 0.1H inductor. Find the voltage across the combination.
Answer: v(t) = 0.4e–t + 39.6te–t V
Solution:
v=L
di
+ R i = (.1) (4e− t − 4te− t ) + 10(4te− t ) = 0.4 e − t + 39.6t e − t V
dt
P 7.5-3 The voltage, v(t), and current, i(t), of a 1-H inductor adhere to the passive convention.
Also, v(0) = 0 V and i(0) = 0 A.
(a) Determine v(t) when i(t) = x(t), where x(t) is shown in Figure P 7.5-3 and i(t) has units of A.
(b) Determine i(t) when v(t) = x(t), where x(t) is shown in Figure P 7.5-3 and v(t) has units of V.
Hint: x(t) = 4t – 4 when 1 < t < 2, and x(t) = –4t + 12 when 2 < t < 3.
x
5
4
3
2
1
0
0
1
2
3
4 t(s)
Figure P 7.5-3
Solution:
(a)
⎧ 0 0 < t <1
⎪ 4 1< t < 2
d
⎪
v(t ) = L i (t ) = ⎨
dt
⎪ −4 2 < t < 3
3<t
⎩⎪ 0
(b)
i (t ) =
t
1 t
v (τ ) dτ + i ( 0 ) = ∫ v (τ ) dτ
∫
0
L 0
For 0 < t < 1, v(t) = 0 V so
t
i ( t ) = ∫ 0 dτ + 0 = 0 A
0
For 1 < t < 2, v(t) = (4t − 4) V so
t
i ( t ) = ∫ ( 4τ − 4 ) dτ + 0 = 2τ 2 − 4τ =2 t 2 − 4 t + 2 A
0
1
(
t
( )
)
i (2) = 4 22 − 4 ( 2 ) + 2 = 2 A
For 2 < t < 3, v(t) = −4t + 12 V so
(
t
i ( t ) = ∫ ( −4τ + 12 ) dτ + 2 = −2τ 2 + 12τ
2
( )
i (3) = −2 32 + 12 ( 3) − 14 = 4 A
t
For 3 < t, v(t) = 0 V so i ( t ) = ∫ 0 dτ + 4 = 4 A
3
t
) 2 +2= ( −2 t
2
)
+ 12 t − 14 A
P 7.5-4 The voltage, v(t), across an inductor and current, i(t), in that inductor adhere to the
passive convention. Determine the voltage, v(t), when the inductance is L = 250 mH and the
current is i(t) = 120 sin (500t – 30°) mA.
⎛
π ⎞⎞
d
d
⎛
Hint:
A sin(ωt + θ ) = A cos(ωt + θ ) ⋅ (ωt + θ ) = Aω cos(ωt + θ ) = Aω sin ⎜ ωt + ⎜ θ + ⎟ ⎟
dt
dt
2 ⎠⎠
⎝
⎝
Answer: v(t) = 15 sin(500t + 60°) V
Solution:
v (t ) = (250 × 10 −3 )
d
(120 × 10 −3 ) sin(500t − 30° ) = (0.25)(0.12)(500) cos(500 t − 30° )
dt
= 15 cos(500t − 30° )
P 7.5-5 Determine iL(t) for t > 0 when iL(0) = –2 μA for the circuit of Figure P 7.5-5a when
vs(t) is as shown in Figure P 7.5-5b.
vs (mV)
4
iL
vs
+
–
5 mH
–1
1
(a)
2
3
t (μs)
(b)
Figure P 7.5-5
Solution:
iL (t ) =
for
1
5 ×10−3
0< t < 1 μ s
iL (t ) =
1
5×10−3
t
∫ v (τ ) dτ
0
s
− 2 ×10−6
vs (t ) = 4 mV
⎛ 4×10−3 ⎞
−3
−6
d
τ
t − 2×10−6 = 0.8− 2×10−6 A
×
−
×
=
4
10
2
10
⎜
−3 ⎟
∫0
⎝ 5×10 ⎠
t
⎛ 4×10−3
⎞
6
iL (1μs) = ⎜
1×10−6 ) ⎟ − 2×10−6 = − ×10−6 A =−1.2 A
−3 (
5
⎝ 5×10
⎠
for 1μ s <t < 3 μ s vs (t ) = −1 mV
t
1
6
1×10−3
6
−3
−6
d
τ
−
×
−
×
=−
1
10
10
(t −1×10−6 ) − ×10−6 =( −0.2 t −10−6 ) A
(
)
−3 ∫1μ s
−3
5×10
5
5×10
5
−3
⎛ 1×10
⎞
iL ( 3μ s ) = ⎜ −
+ 3×10−6 ⎟ − 1×10−6 = −1.6 μA
−3
⎝ 5×10
⎠
iL (t ) =
for 3μ s < t vs (t ) = 0 so iL (t ) remains − 1.6 μA
P 7.5-6 Determine v(t) for t > 0 for the circuit of Figure P 7.5-6a when iL(0) = 0 and is is as
shown in Figure P 7.5-6b.
is (mA)
1
+
2 kΩ
is
4 mH
v
iL
–
0
–1
t (ms)
0
(a)
5
3
1
7
8
(b)
Figure P 7.5-6
Solution:
v(t ) =
In general
( 2 ×10 ) i
3
s
(t ) +
( 4 ×10 ) dtd
−3
is (t )
⎛ 1×10−3 ⎞
d
For 0<t <1 μ s is (t ) = (1) ⎜
t = 103 t ⇒
is (t ) = 1× 103 . Consequently
−6 ⎟
1
10
dt
×
⎝
⎠
v(t ) = (2 ×103 ) (1×103 ) t + 4 × 10−3 (1×103 ) =
For 1μ s <t < 3μ s is (t ) = 1 mA ⇒
( 2×10
6
t + 4) V
d
is (t ) = 0 . Consequently
dt
v(t ) = (2 ×103 ) (1×10−3 ) + ( 4×10−3 ) × 0 = 2 V
⎛ 1×10−3 ⎞
For 3μ s< t < 5μ s is (t ) = 4 ×10 − ⎜
t ⇒
−6 ⎟
⎝ 1×10 ⎠
−3
d
1×10−3
= −103 . Consequently
is (t ) = −
−6
dt
1×10
v(t ) = ( 2×103 )( 4×10−3 −103 t )+ 4×10−3 ( −103 ) = 4 − ( 2×106 ) t
When 5μ s <t < 7μ s
is (t ) = − 1×10−3 and
d
is (t ) = 0 . Consequently
dt
v(t ) = ( 2×103 )(10−3 ) = − 2 V
⎛ 1×10−3 ⎞
d
t − 8 × 10−3 ⇒
is (t ) = 1× 103
When 7μ s< t < 8μ s is (t ) = ⎜
−6 ⎟
1
10
×
dt
⎝
⎠
v(t ) = ( 2 × 103 )(103 t − 8 ×10−3 ) + ( 4 × 10−3 )(103 ) = −12 + ( 2 × 106 ) t
When 8μ s < t , then is (t ) = 0 ⇒
d
is (t ) = 0 . Consequently v(t ) = 0 .
dt
The voltage, v(t), and current, i(t), of a 0.5-H
P 7.5-7
x
inductor adhere to the passive convention. Also,
1.0
v(0) = 0 V and i(0) = 0 A.
0.8
(a) Determine v(t) when i(t) = x(t), where x(t) is shown
0.6
0.4
in Figure P 7.5-7 and i(t) has units of A.
0.2
(b) Determine i(t) when v(t) = x(t), where x(t) is shown
0.0
in Figure P 7.5-7 and v(t) has units of V.
0
Hint: x(t) = 0.2t – 0.4 when 2 < t < 6.
2
4
6
Figure P 7.5-7
Solution:
(a)
(b)
⎧ 0 0<t <2
d
⎪
v(t ) = L i (t ) = ⎨0.1 2 < t < 6
dt
⎪0
6<t
⎩
t
1 t
i ( t ) = ∫ v (τ ) dτ + i ( 0 ) = 2 ∫ v (τ ) dτ
0
L 0
t
For 0 < t < 2, v(t) = 0 V so i ( t ) = 2 ∫ 0 dτ + 0 = 0 A
0
For 2 < t < 6, v(t) = 0.2 t − 0.4 V so
t
t
i ( t ) = 2∫ ( 0.2τ − 0.4 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) =0.2 t 2 − 0.8 t + 0.8 A
2
2
( )
i (6) = 0.2 62 − 0.8 ( 6 ) + 0.8 = 3.2 A .
For 6 < t, v(t) = 0.8 V so
t
i ( t ) = 2 ∫ 0.8 dτ + 3.2 = (1.6 t − 6.4 ) A
6
8 t (s)
P 7.5-8 Determine i(t) for t ≥ 0 for the current of Figure P 7.5-8a when i(0) = 25 mA and vs(t)
is the voltage shown in Figure P 7.5-8b.
vs (V)
2
1
i(t)
vs (t)
+
–
100 H
2
4
6
8 9
t(s)
−2
−4
(b)
(a)
Figure P 7.5-8
Solution:
i (t ) =
so i (1) = 0.025
i (t ) =
so i ( 3) = −0.055
i (t ) =
so i ( 9 ) = 0.065
1 t
0 dt + 0.025 = 0.025
100 ∫ 0
−4 ( t − 1)
1 t
−
4
d
+
0.025
=
τ
100 ∫1
100
0 < t <1
for
for
2 ( t − 3)
1 t
2 dτ − 0.055 =
− 0.055
∫
100 3
100
i (t ) =
1 t
0 dτ + 0.065 = 0.065
100 ∫9
for
for
1< t < 3
3<t < 9
t >9
P 7.5-9 Determine i(t) for t ≥ 0 for the current of Figure P 7.5-9a when i(0) = –2 A and vs(t) is
the voltage shown in Figure P 7.5-9b.
vs (V)
i(t)
+
vs (t) –
5H
4
2
−1
1
(a)
2
3
t(s)
(b)
Figure P 7.5-9
Solution:
i (t ) = i ( 0) +
For 0 ≤ t ≤ 1 s
1 t
1 t
2
v
τ
d
τ
=
−
+
v s (τ ) dτ
(
)
s
5 ∫0
L ∫0
1 t
4 t
4
4 dτ = −2 + τ ∫ = −2 + t
∫
5 0
5 0
5
6
8
⎛1⎞
For example i ( 0 ) = −2, i (1) = − and i ⎜ ⎟ = −
5
5
⎝2⎠
i ( t ) = −2 +
For 1 ≤ t ≤ 3 s
1 1
1 t
4 1
4 dt + ∫ −1 dτ = −2 + − τ
∫
5 0
5 1
5 5
6
7
8
For example i (1) = − , i ( 2 ) = − , i ( 3) = −
5
5
5
i ( t ) = −2 +
Notice that
i ( t ) = i (1) +
For 3 ≤ t
i (t ) = i ( 0) +
t
0
6 1
t
= − − ( t − 1) = −1 −
5 5
5
1 t
6 1
t
−1 dτ = − − ( t − 1) = −1 −
∫
5 1
5 5
5
1 1
1 3
1 t
8
4 dτ + ∫ −1 dτ + ∫ 0 dτ = −
∫
5 0
5 1
5 3
5
In summary
⎧ −2 + 0.8t
⎪
i ( t ) = ⎨ −1 − 0.2t
⎪
⎩ −1.6
0 ≤ t ≤1
1≤ t ≤ 3
3≤t
P 7.5-10 Determine i(t) for t ≥ 0 for the current of Figure P 7.5-10a when i(0) = 1 A and vs(t) is
the voltage shown in Figure P 7.5-10b.
vs (V)
i(t)
vs(t)
+
–
2
2H
2
4
6
t(s)
−1
(a)
(b)
Figure P 7.5-10
Solution:
1
⎧
⎪ t
⎪ ∫ 2 dτ + 1 = ( t − 2 ) + 1 = t − 1
1 t
⎪
i ( t ) = ∫ v ( t ) dt + 1 = ⎨
0
1 t
1
2
⎪
dτ + 3 = − t + 5
∫
4
2
⎪ 2
⎪⎩
2
t≤2
2≤t ≤4
4≤t ≤6
6≤t
P 7.5-11 Determine i(t) for t ≥ 0 for the circuit of Figure P 7.5-11a when i(0) = 25 mA and vs(t)
is the voltage shown in Figure P 7.5-11b.
vs (V)
1
i(t)
vs(t) +
200 H
–
1 2 3
5
6 7 8
9 t(s)
−1
−2
(a)
(b)
Figure P 7.5-11
Solution:
i (t ) =
i (t ) =
1 t
−t
−dτ + 0.025 =
+ 0.025
∫
0
200
200
for
−2 ( t − 1)
1 t
−
2
d
+
0.02
=
+ 0.02
τ
200 ∫1
200
i (t ) =
1 t
t−4
dτ − 0.01 =
− 0.01
∫
4
200
200
i ( t ) = 0.015 = 15 mA
for
for
t <9
0 < t <1
1< t < 4
4<t<9
The inductor current in the circuit shown in Figure P 7.5-12 is given by
P 7.5-12
i (t ) = 6 + 4e −8t A for t ≥ 0
Determine v(t) for t > 0.
+
v(t)
–
i(t)
2Ω
12 V
+
–
8Ω
0.2 H
Figure P 7.5-12
Solution:
d
i (t )
dt
= −6.4e −8t V for t > 0
v L ( t ) = 0.2
Use KVL to get
v ( t ) = 12 − ( −6.4e −8t ) = 12 + 6.4e −8t V for t > 0
(checked: LNAP 6/25/04)
P 7.5-13
The inductor current in the circuit shown in Figure P 7.5-13 is given by
i (t ) = 5 − 3e−4t A for t ≥ 0
Determine v(t) for t > 0.
+
v(t)
i(t)
–
24 Ω
10 A
24 Ω
24 Ω
4H
Figure P 7.5-13
Solution:
We’ll write and solve a mesh equation. Label the meshes as shown. Apply KVL to the center
mesh to get
i ( t ) + 10
24i a + 24 ( i a − i ( t ) ) + 24 ( i a − 10 ) = 0
⇒ ia =
= 5 − e − 4t A for t > 0
3
v ( t ) = 24i a = 120 − 24e −4t V for t > 0
Then
(checked: LNAP 6/25/04)
P 7.5-14
The inductor current in the circuit shown in Figure P 7.5-14 is given by
i (t ) = 3 + 2e −3t A for t ≥ 0
Determine v(t) for t > 0.
6Ω
5A
+
v(t)
i(t)
9Ω
5H
–
Figure P 7.5-14
Solution: Apply KVL to get
d
d
v ( t ) = 6i ( t ) + 5 i ( t ) = 6 ( 3 + 2e−3t ) + 5 ( 3 + 2e−3t ) = 18 (1 − e−3t ) V for t > 0
dt
dt
(checked: LNAP 6/25/04)
vs(t) (V)
4
P 7.5-15 Determine the current i(t) for t > 0
2
for the circuit of Figure P 7.5-15b when vs(t)
is the voltage shown in Figure P 7.5-15a. The
–4
inductor current at time t = 0 is i(0) = –12 A.
–2
2
4
6
8
–2
(a)
i(t)
+
–
vs(t)
1 3
H
(b)
Figure P 7.5-15
Solution:
i (t ) =
1 t
1 t
v s (τ ) dτ + i (t 0 ) = ∫ v s (τ ) dτ − 12
∫
1 0
L t0
3
t
i (t ) = 3∫ 4 dτ − 12 = 12 t − 12 for 0 < t < 4
0
t
i (t ) = 3∫ ( −2 ) dτ + 36 = 60 − 6 t for 4 < t < 10
4
t
i (t ) = 3∫ 0 dτ + 0 = 0 for 10 < t
10
In particular, i(4) = 36 A.
In particular, i(10) = 0 A.
10 t (s)
2.5 H
i(t)
The initial current in the inductor is i ( 0 ) = 2 A . Determine
+
P7.5-16 The input to the circuit shown in Figure P7.5-16 is
the voltage
v ( t ) = 15 e −4 t V for t > 0
the inductor current, i ( t ) , for t > 0 .
v (t)
–
Figure P7.5-16
Solution: The inductor current is related to the inductor voltage by
i (t ) =
1 t
v (τ ) dτ + i ( 0 )
L ∫0
That is
i (t ) =
1 t
15
e −4τ
15 e −4τ dτ + 2 =
∫
0
2.5
2.5 ( −4 )
t
0
+ 2 = −1.5 ( e−4 t − 1) + 2 = 3.5 − 1.5 e −4 t A for t > 0
P7.5-17 The input to the circuit shown in Figure P7.5-17 is the
voltage
v ( t ) = 4 e −20 t V for t > 0
The output is the current
i ( t ) = −1.2 e −20 t − 1.5 A for t > 0
The initial inductor current is i L ( 0 ) = −3.5 A . Determine the
Figure P7.5-17
values of the inductance, L, and resistance, R.
Solution: Apply KCL at either node to get
i (t ) =
v (t )
v (t ) ⎡ 1 t
⎤
+ i L (t ) =
+ ⎢ ∫ v (τ ) dτ + i ( 0 ) ⎥
0
R
R
⎣L
⎦
That is
−1.2 e
−20 t
4 e −20 t 1 t −20τ
4 e −20 t
4
− 1.5 =
+ ∫ 4e
+
dτ − 3.5 =
e −20 t − 1) − 3.5
(
0
R
L
R
L ( −20 )
⎛ 4 1 ⎞ −20 t 1
=⎜ −
− 3.5
⎟e +
5L
⎝ R 5L ⎠
Equating coefficients gives
−1.5 =
1
− 3.5 ⇒ L = 0.1 H
5L
and
−1.2 =
4 1
4
1
4
−
= −
= −2 ⇒ R =5 Ω
R 5 L R 5 ( 0.1) R
Figure P7.5-18
P7.5-18 . The source voltage the circuit shown in Figure P7.5-18 is v ( t ) = 8 e − 400 t V after time t
= 0. The initial inductor current is i L ( 0 ) = 210 mA . Determine the source current i (t) for t > 0.
Answer: i ( t ) = 360 e − 400 t − 190 mA for t > 0
Solution:
Label the resistor current as shown. The resistor, inductor and voltage
source are connected in parallel so the voltage across each is
v ( t ) = 2.5 e − 400 t V . Notice that the labeled voltage and current of
both the resistor and inductor do not adhere to the passive convention.
8 e − 400 t
= −40 e −400 t mA
200
1 t
−8 e − 400τ dτ A
The inductor current is i L ( t ) = 0.21 +
0.05 ∫0
t
−8
i L ( t ) = 0.21 +
e − 400τ dτ A
∫
0.05 ( −400 ) 0
The resistor current is i R ( t ) = −
(
)
= 0.21 + 0.4 e − 400 t − 1 A
= 400 e − 400 t − 190 mA
Using KCL
i ( t ) = 360 e − 400 t − 190 mA for t > 0
Figure P7.5-19
P7.5-19 . The input to the circuit shown in Figure P7.5-19 is the current
i ( t ) = 5 + 2 e −7 t A for t > 0
v ( t ) = 75 − 82 e −7 t V for t > 0
The output is the voltage:
Determine the values of the resistance and inductance.
Solution:
d
5 + 2 e −7 t )
(
dt
−7 t
= R ( 5 + 2 e ) + L ( ( −7 ) 2 e −7 t ) = 5 R + ( 2 R − 14 L ) e−7 t
v ( t ) = 75 − 82 e −7 t = R ( 5 + 2 e−7 t ) + L
75 = 5 R ⇒ R = 15 Ω and
82 + 30
−82 = 2 R − 14 L = 30 − 14 L ⇒ L =
=8 H
14
Equating coefficients gives
and
P7.5-20 Consider the inductor shown in Figure P7.5-20. The current
and voltage are given by
⎧5 t − 4.6 0 ≤ t ≤ 0.2
⎧12.5 0 < t < 0.2
⎪
⎪
i ( t ) = ⎨ a t + b 0.2 ≤ t ≤ 0.5 and v ( t ) = ⎨ 25 0.2 < t < 0.5
⎪ c
⎪ 0
t ≥ 0.5
t > 0.5
⎩
⎩
where a, b and c are real constants. (The current is given in Amps, the
voltage in Volts and the time in seconds.) Determine the values of a, b
and c.
Answer: a = 10 A/s, b = −5.6 A and c = −0.6 A
Solution: At t = 0.2 s
Figure P7.5-20
i ( 0.2 ) = 5 ( 0.2 ) − 4.6 = −3.6 A
For 0.2 ≤ t ≤ 0.5
i (t ) =
1 t
25 dτ − 3.6 = 10τ
2.5 ∫ 0.2
At t = 0.5 s
t
0.2
− 3.6 = 10 ( t − 0.2 ) − 3.6 = 10 t − 5.6 A
i ( 0.5 ) = 10 ( 0.5 ) − 5.6 = −0.6 A
For t ≥ 0.5
i (t ) =
Checks:
At t = 0.2 s
1 t
0 dτ − 0.6 = −0.6
2.5 ∫ 0.5
i ( 0.2 ) = 10 ( 0.2 ) − 5.6 = −3.6 A
For 0.2 ≤ t ≤ 0.5
v ( t ) = 2.5
√
d
d
i ( t ) = 2.5 (10 t − 5.6 ) = 2.5 (10 ) = 25 V √
dt
dt
−0.6 − ( −3.6 ) = i ( 0.5 ) − i ( 0.2 ) =
1 0.5
25 dτ = 10 ( 0.5 − 0.2 ) = 3 A
2.5 ∫ 0.2
√
P7.5-21 At time t =0, the current in the inductor shown in Figure P7.5-21 is i(0) = 45 mA.
Determine the values of the inductor current at times 1 ms, 4 ms and 6 ms.
Figure P7.5-21
Solution:
i (t ) = i ( 0) +
1 t
"area under the curve"
"area under the curve"
v (τ ) dτ = i ( 0 ) +
= 0.045 +
∫
0.250
L 0
L
i ( 0.001) = 0.045 +
20 ( 0.001)
= 0.125 A = 125 mA ,
0.250
1
20 ( 0.002 ) + 20 ( 0.002 )
2
i ( 0.004 ) = 0.045 +
= 0.285 A = 285 mA
0.250
1
20 ( 0.002 ) + 20 ( 0.002 ) + 0
2
i ( 0.006 ) = 0.045 +
= 0.285 A = 285 mA
0.250
P7.5-22 One of the three elements shown in Figure P7.5-22 is a resistor, one is a capacitor and
one is an inductor. Given
i ( t ) = 0.25cos ( 2 t ) A ,
and
v a ( t ) = −10sin ( 2 t ) V , v b ( t ) = 10sin ( 2 t ) V and v c ( t ) = 10 cos ( 2 t ) V
Determine the resistance of the resistor, the capacitance of the capacitor and the inductance of
the inductor. (We require positive values of resistor, capacitance and inductance.)
Answers: resistance = 64Ω, capacitance = 0.0125 F and inductance = 20 H
Figure P7.5-22
Solution:
d
d
i ( t ) = ( 0.25cos ( 2 t ) ) = − ( 0.25 )( 2 ) sin ( 2 t ) = −0.5sin ( 2 t )
dt
dt
The voltage of an inductor is proportional to the derivative of the current. The constant of
d
proportionality is the inductance. We see that v a ( t ) is proportional to i ( t ) and the constant of
dt
proportionality is positive. Consequently, element a is the inductor. Then
First,
v a (t )
−10sin ( 2 t )
=
= 20 H
d
i ( t ) −0.5sin ( 2 t )
dt
t
t
0.25sin ( 2τ )
Next
= 0.125sin ( 2τ )
∫ −∞ i (τ ) dτ = ∫ −∞ 0.25cos ( 2τ ) dτ =
2
The voltage of a capacitor is proportional to the integral of the current. The constant of
proportionality is the reciprocal of the capacitance. We see that v b ( t ) is proportional to
L=
t
∫ −∞ i (τ ) dτ
and the constant of proportionality is positive. Consequently, element b is the
capacitor. Then
1
=
C
v b (t )
t
∫ −∞ i (τ ) dτ
=
10sin ( 2 t )
1
= 80 ⇒ C =
= 0.0125 F
0.125sin ( 2 t )
80
Finally, the voltage of element c is proportional to the current and the constant of proportionality
v c (t )
is positive. Consequently, element c is the resistor and R =
= 40 Ω .
i (t )
P7.5-23 One of the three elements shown in Figure P7.5-23 is a resistor, one is a capacitor and
one is an inductor. Given
v ( t ) = 24 cos ( 5 t ) V ,
and
i a ( t ) = 3cos ( 5 t ) A , i b ( t ) = 12sin ( 5 t ) A and i c ( t ) = −1.8sin ( 5 t ) A
Determine the resistance of the resistor, the capacitance of the capacitor and the inductance of
the inductor. (We require positive values of resistor, capacitance and inductance.)
Figure P7.5-23
Solution:
First, the current of element a is proportional to the voltage and the constant of proportionality is
v ( t ) 24 cos ( 5 t )
=
=8 Ω.
positive. Consequently, element a is the resistor and R =
i a ( t ) 3cos ( 5 t )
d
d
v ( t ) = ( 24 cos ( 5 t ) ) = − ( 24 )( 5 ) sin ( 5 t ) = −120sin ( 5 t )
dt
dt
Next
The current of a capacitor is proportional to the derivative of the voltage. The constant of
d
proportionality is the capacitance. We see that i c ( t ) is proportional to v ( t ) and the constant
dt
of proportionality is positive. Consequently, element c is the capacitor. Then
ic (t )
−1.8sin ( 5 t )
=
= 0.015 F = 15 mF
d
i ( t ) −120sin ( 5 t )
dt
t
t
24sin ( 5τ )
v
τ
d
τ
=
(
)
∫ −∞
∫ −∞ 24 cos ( 5τ ) dτ = 5 = 4.8sin ( 5τ )
C=
Next
The voltage of an inductor is proportional to the integral of the current. The constant of
proportionality is the reciprocal of the inductance. We see that i b ( t ) is proportional to
t
∫ −∞ v (τ ) dτ
and the constant of proportionality is positive. Consequently, element b is the
inductor. Then
1
=
L
ib (t )
t
∫ −∞ v (τ ) dτ
=
12sin ( 5 t )
1
= 2.5 ⇒ L =
= 0.4 H
4.8sin ( 5 t )
2.5
Section 7-6: Energy Storage in an Inductor
P 7.6-1 The current, i(t), in a 100-mH inductor connected in a telephone circuit changes
according to
t≤0
⎧0
⎪
i (t ) = ⎨4t 0 ≤ t ≤ 1
⎪4
t ≥1
⎩
where the units of time are seconds and the units of current are amperes. Determine the power,
p(t), absorbed by the inductor and the energy, w(t), stored in the inductor.
t≤0
⎧ 0
⎧ 0
⎪
⎪
Answer: p (t ) = ⎨1.6t 0 < t < 1 and w(t ) = ⎨0.8t 2
⎪ 0
⎪ 0.8
t ≥1
⎩
⎩
t≤0
0 < t <1
t ≥1
The units of p(t) are W and the units of w(t) are J.
Solution:
v( t ) =100×10
−3
⎧0
d
⎪
i ( t ) = ⎨0.4
dt
⎪0
⎩
t<0
0≤ t ≤1
t>1
t <0
⎧0
⎪
p ( t ) =v( t ) i ( t ) = ⎨1.6t 0≤t ≤1
⎪0
t >1
⎩
W (t ) = ∫
t
0
⎧ 0
⎪
p(τ ) dτ = ⎨0.8t 2
⎪ 0.8
⎩
t <0
0<t <1
t >1
The current, i(t), in a 5-H inductor is
P 7.6-2
t≤0
⎧ 0
i (t ) = ⎨
⎩4sin 2t t ≥ 0
where the units of time are s and the units of current are A. Determine the power, p(t), absorbed
by the inductor and the energy, w(t), stored in the inductor.
Hint: 2 (cos A)(sin B) = sin (A + B) + sin (A – B)
Solution:
⎡ d
⎤
p (t ) = v (t ) i (t ) = ⎢5 (4sin 2t ) ⎥ (4sin 2t )
⎣ dt
⎦
= 5 (8cos 2t ) (4sin 2t )
= 80 [2 cos 2t sin 2t ]
= 80 [sin(2t + 2t ) + sin(2t − 2t )] = 80 sin 4t W
t
t
80
W (t ) = ∫ p(τ ) dτ = 80∫ sin4τ dτ = −
[cos 4τ |t0 ] = 20 (1 − cos 4t )
0
0
4
The voltage, v(t), across a 25-mH inductor used in a fusion power experiment is
P 7.6-3
0
t≤0
⎧
v(t ) = ⎨
⎩6 cos100t t ≥ 0
where the units of time are s and the units of voltage are V. The current in this inductor is zero
before the voltage changes at t = 0. Determine the power, p(t), absorbed by the inductor and the
energy, w(t), stored in the inductor. Hint: 2(cos A)(sin B) = sin(A + B) + sin(A – B)
Answer: p(t) = 7.2 sin 200t W and w(t) = 3.6[1 – cos 200t] mJ
Solution:
i(t ) =
1
25×10−3
∫
t
0
6 cos 100τ dτ + 0 =
6
[sin 100τ | 0t ] = 2.4sin100 t
(25×10−3 )(100)
p(t ) = v(t ) i(t ) = (6 cos100 t )(2.4 sin100t ) = 7.2 [ 2(cos100 t )(sin100 t ) ]
= 7.2 [sin 200 t + sin 0] = 7.2 sin 200 t
t
t
7.2
⎡⎣cos 200τ |t0 ⎤⎦
W (t ) = ∫0 p (τ ) dτ = 7.2∫0 sin 200τ dτ = −
200
= 0.036[1 − cos 200t ] J = 36 [1 − cos 200t ] mJ
P 7.6-4 The current in an inductor, L = 1/4 H, is i = 4te–t A for t ≥ 0 and i = 0 for t < 0. Find the
voltage, power, and energy in this inductor.
Partial Answer: w = 2t 2 e −2t J
Solution:
v = L
di
⎛1⎞ d
= ⎜ ⎟
4 t e − t ) = (1−t ) e− t V
(
dt
⎝ 4 ⎠ dt
P = vi = ⎡⎣ (1−t ) e − t ⎤⎦ ( 4 t e− t ) = 4 t (1−t ) e −2t W
2
1
1⎛1⎞
W = Li 2 = ⎜ ⎟ ( 4 t e − t ) = 2 t 2 e − 2t J
2
2⎝4⎠
P 7.6-5 The current through the inductor of a
television tube deflection circuit is shown in
Figure P 7.6-5 when L = 1/2 H. Find the
voltage, power, and energy in the inductor.
i (A)
2
Partial Answer: p = 2t for 0 ≤ t <1
= 2(t − 2) for 1< t < 2
= 0 for other t
0
1
2
Figure P 7.6-5
Solution:
v (t ) = L
di 1 di
=
dt 2 dt
t <0
⎧ 0
⎧ 0
⎪ 2t
⎪
0<t <1
⎪
⎪ 1
and i ( t ) = ⎨
⇒ v( t ) = ⎨
⎪−2( t − 2 ) 1<t < 2
⎪ −1
⎪ 0
⎪⎩ 0
t >2
⎩
p (t ) = v (t ) i (t )
t <0
⎧ 0
⎪2t
0<t <1
⎪
= ⎨
⎪2( t − 2 ) 1<t < 2
⎪ 0
t >2
⎩
W ( t ) = W ( t0 ) + ∫ t0 p ( t ) dt
t
i (t ) = 0 for t < 0 ⇒ p ( t ) = 0 for t < 0 ⇒ W ( t0 ) = 0
0 < t < 1: W ( t ) =
∫ 2 t dt = t
1<t < 2 : W ( t ) = W (1)+ ∫ 2 ( t − 2 ) dt = t
t
2
0
t
1
t >2 : W (t ) = W ( 2) = 0
2
− 4t + 4
t <0
0<t <1
1<t < 2
t >2
t (s)
Section 7-7: Series and Parallel Inductors
P 7.7-1 Find the current i(t) for the circuit of Figure P 7.7-1.
Answer: i(t) = 15 sin 100t mA
i(t)
2H
6 cos 100t V
+
–
6H
3H
Figure P 7.7-1
Solution:
6×3
= 2 H and 2 H + 2 H = 4 H
6+3
1 t
6
⎡sin100τ | t0 ⎤⎦ = 0.015sin100 t A = 15sin100 t mA
i (t ) = ∫ 0 6 cos100τ dτ =
4
4×100 ⎣
6H 3H =
P 7.7-2 Find the voltage v(t) for the circuit of Figure P 7.7-2.
Answer: v(t) = –6e–250t mV
5 + 3e–250t A
+
v(t)
–
4 mH
4 mH
8 mH
4 mH
Figure P 7.7-2
Soluton:
(8×10 )×(8×10 )
−3
4 mH + 4 mH = 8 mH , 8mH 8mH =
−3
8×10−3 +8×10−3
= 4 mH
and 4 mH + 4 mH = 8 mH
d
(5+ 3e −250t ) = (8×10−3 ) (0+ 3(−250) e −250t ) =−6 e −250t V
v(t ) = (8×10−3 )
dt
P 7.7-3 The circuit of Figure P 7.7-3 contains four identical inductors. Find the value of the
inductance L.
Answer: L = 2.86 H
i(t) = 14 sin 250t mA
L
25 cos 250t V
+
–
L
L
L
Figure P 7.7-3
Solution:
L⋅ L
L
L
5
=
and L + L +
=
L
2
2
2
L+ L
⎛5 ⎞ d
⎛5 ⎞
25cos 250 t = ⎜ L ⎟
14×10−3 ) sin 250 t = ⎜ L ⎟(14×10−3 )(250) cos 250 t
(
⎝ 2 ⎠ dt
⎝2 ⎠
25
so L =
= 2.86 H
5
−3
(14×10 ) (250)
2
L
L =
(
)
P 7.7-4 The circuit shown in Figure P 7.7-4 contains seven inductors, each having inductance
L. The source voltage is given by
v(t ) = 4 cos(3t ) V
Find the current i(t) when L = 4 H.
i(t)
L
+ v(t)
–
L
L
L
L
L
L
Figure P 7.7-4
⎛ L×2 L
⎞
+ L⎟× L
⎜
L+2L
21
⎠
+2L = L
Solution: The equivalent inductance is: ⎝
8
⎛ L×2 L
⎞
+ L⎟ + L
⎜
⎝ L+2L
⎠
t
1
8
4
Then
4 cos ( 3τ ) dτ =
i (t ) =
× sin ( 3t ) = 127 sin ( 3t ) mA
∫
−∞
21
21× 4 3
L
8
(Checked using LNAP 6/26/04)
P 7.7-5 Determine the value of the inductance L in the circuit shown in Figure P 7.7-5, given
that Leq = 18 H.
Answer: L = 20 H
25 H
A
L
20 H
20 H
60 H
10 H
30 H
B
Leq
Figure P 7.7-5
Solution: The 25 H inductor is in series with a parallel combination of 20 H and 60 H inductors.
The inductance of the equivalent inductor is
25 +
60 × 20
= 40 H
60 + 20
The 30 H inductor is in parallel with a short circuit, which is equivalent to a short circuit. After
making these simplifications, we have
Then
18 = L eq = 10 +
1
1 1 1
+ +
20 L 40
⇒
1 1 1 1
+ +
=
⇒ L = 20 H
20 L 40 8
(Checked using LNAP 6/26/04)
P 7.7-6 Determine the value of the equivalent inductance, Leq, for the circuit shown in Figure P
7.7-6.
Answer: Leq = 120 H
a
60 H
15 H
30 H
10 H
40 H
b
60 H
Leq
Figure P 7.7-6
Solution:
L eq = 60 +
15 × 10
40 × 60
+ 30 +
= 60 + 6 + 30 + 24 = 120 H
15 + 10
40 + 60
(Checked using LNAP 6/26/04)
P 7.7-7 The circuit shown in Figure P 7.7-7 consists of 10 inductors having equal inductance,
L. Determine the value of the inductance L, given that Leq = 12 mH.
Answer: L = 35 mH
L
L
L
L
Leq
L
L
L
L
L
L
Figure P 7.7-7
Solution:
First
Then
12 = L eq
⎛2 ⎞ ⎛2
⎞
⎜ L ⎟ × ⎜ L + 2 L ⎟ 12
5 ⎠ ⎝5
⎠ = L ⇒ L = 35 mH
=⎝
⎛2 ⎞ ⎛2
⎞ 35
⎜ L⎟ + ⎜ L + 2 L⎟
⎝5 ⎠ ⎝5
⎠
(Checked using LNAP 6/26/04)
P 7.7-8 The circuit shown in Figure P 7.7-8 is at steady state before the switch closes. The
inductor currents are both zero before the switch closes (i1(0) = i2(0) = 0).
t=0
24 Ω
i1(t)
+
–
i2(t)
12 V
+
v(t)
−
8H
12 Ω
2H
Figure P 7.7-8
The voltage v(t) is given by
for t < 0
⎧ 0V
v(t ) = ⎨ −5t
⎩ 4e V for t > 0
(a)
Determine the inductor currents, i1(t) and i2(t), for t ≥ 0.
(b)
Determine the energy stored by each inductor 200 ms after the switch closes.
The parallel inductors can be replaced by an equivalent inductor.
(c)
Determine the current in the equivalent inductor, directed downward, for t ≥ 0.
(d)
Determine the energy stored by the equivalent inductor 200 ms after the switch closes.
Solution:
(a)
i1 (t ) =
1 t −5τ
1
e −5t − 1) = 0.1(1 − e −5t ) A for t ≥ 0
4e dτ + 0 =
(
∫
0
−10
8
i2 (t ) =
1 t −5τ
4e dτ = 0.4 (1 − e−5t ) A for t ≥ 0
∫
0
2
(b) When t = 0.25, i1 ( 0.2 ) = 0.1(1 − e −1 ) = 63.2 mA and i 2 ( 0.2 ) = 0.4 (1 − e −1 ) = 252.8 mA so
the energy stored by the 8 H inductor is w1 =
the 2 H indictor is w 2 = 63.9 mJ .
L eq =
(c)
i (t ) =
1
( 8) 0.06322 = 16.0 mJ and the energy stored by
2
8.2
= 1.6 H
8+2
1 t −5τ
4e dτ = 0.5 (1 − e−5t ) A for t ≥ 0
1.6 ∫ 0
(d) When t = 0.2 s, i ( 0.2 ) = 0.5 (1 − e −1 ) = 316 mA so the energy stored by the equivalent
inductor is w =
1
(1.6 ) 0.3162 = 79.9 mJ = w1 + w 2 .
2
P 7.7-9 The circuit shown in Figure P 7.7-9 is at steady state before the switch opens at time
t = 0. The current i(t) is given by
for t ≤ 0
⎧ 0.8 A
i (t ) = ⎨
−2 t
⎩0.8e A for t ≥ 0
t=0
15 Ω
0.5 H
+
–
i(t)
12 V
5Ω
2H
Figure P 7.7-9
(a)
Determine the energy stored by each inductor before the switch opens.
(b)
Determine the energy stored by each inductor 200 ms after the switch opens.
The series inductors can be replaced by an equivalent inductor.
(c)
Determine the energy stored by the equivalent inductor before the switch opens.
(d)
Determine the energy stored by the equivalent inductor 200 ms after the switch opens.
Solution:
(a) The energy stored by the 0.5 H inductor is w1 =
1
( 0.5) ( 0.82 ) = 0.16 J and the energy stored
2
1
( 2 ) ( 0.82 ) = 0.64 J .
2
(b) 200 ms after the switch opens the current in the inductors is 0.8e−0.4 = 0.536 A . Then
1
1
w1 = ( 0.5 ) ( 0.5362 ) = 71.8 mJ and w 2 = ( 2 ) ( 0.5352 ) = 287.3 mJ.
2
2
by the 2 H inductor is w 2 =
Next, L eq = 2 + 0.5 = 2.5 H .
1
( 2.5) ( 0.82 ) = 0.8 J = w1 + w 2
2
1
(d) w eq = ( 2.5 ) ( 0.5362 ) = 359.12 mJ = w1 + w 2
2
(c) w eq =
P 7.7-10 Determine the current ratio i1/i for the circuit shown in Figure P 7.7-10. Assume that
the initial currents are zero at t0.
i
L1
Answer: 1 =
i L1 + L2
L1
i1
i
i2
L2
Figure P 7.7-10
Solution:
i1 =
1 t
∫ v dt + i1 ( t0 ) ,
L1 t 0
i2 =
1 t
∫ v dt + i 2 ( t0 )
L 2 t0
but i1 ( t0 ) = 0 and i 2 ( t0 ) = 0
⎛1
1 t
1 ⎞ t
1 t
t
∫ t v dt + ∫ t v dt = ⎜ +
∫ v dt
⎟ ∫ t v dt =
L1 0
LP t 0
0
⎝ L1 L2 ⎠ 0
1 t
1
∫ t v dt
i
L2
L 0
L1
∴1 = 1
=
=
1 t
1 1
i
L1 + L2
+
∫ t v dt
LP 0
L1 L2
i = i1 + i2 =
P7.7-11
Consider the combination of circuit elements shown in Figure P7.7-11.
(a) Suppose element A is a 20 µF capacitor, element B is a 5 µF capacitor and
element C is a 20 µF capacitor. Determine the equivalent capacitance.
(b) Suppose element A is a 50 mH inductor, element B is a 30 mH inductor
and element C is a 20 mH inductor. Determine the equivalent inductance.
(c) Suppose element A is a 9 kΩ resistor, element B is a 6 kΩ resistor and
element C is a 10 kΩ resistor. Determine the equivalent resistance.
Answers: (a) Ceq = 20 µF, (b) Leq = 16 mH, (c) Req = 6 kΩ
Figure P7.7-11
Solution:
(a) C eq =
20 ( 5 )
( 50 + 30 )( 20 ) = 16 mH (c) R = ( 9 + 6 )(10 ) = 6 kΩ
+ 20 = 24 μF (b) L eq =
eq
20 + 5
( 50 + 30 ) + 20
( 9 + 6 ) + 10
P7.7-12
Consider the combination of circuit elements shown in Figure P7.7-12.
(a) Suppose element A is a 8 µF capacitor, element B is a 16 µF capacitor and
element C is a 12 µF capacitor. Determine the equivalent capacitance.
(b) Suppose element A is a 20 mH inductor, element B is a 5 mH inductor and
element C is a 8 mH inductor. Determine the equivalent inductance.
(c) Suppose element A is a 20 kΩ resistor, element B is a 30 kΩ resistor and
element C is a 16 kΩ resistor. Determine the equivalent resistance.
Answers: (a) Ceq = 8 µF, (b) Leq = 12 mH, (c) Req = 28 kΩ
Figure P7.7-11
Solution:
(a) C eq =
( 8 + 16 )(12 ) = 8 μF
( 8 + 16 ) + 12
(b) L eq =
( 20 )( 5) + 8 = 12 mH
20 + 5
(c) R eq =
( 20 )( 30 ) + 16 = 28 kΩ
20 + 30
Section 7-8: Initial Conditions of Switched Circuits
P 7.8-1
The switch in Figure P 7.8-1 has been open for a long time before closing at time t = 0.
Find vc(0+) and iL(0+), the values of the capacitor voltage and inductor current immediately after
the switch closes. Let vc(∞) and iL(∞) denote the values of the capacitor voltage and inductor
current after the switch has been closed for a long time. Find vc(∞) and iL(∞).
Answer: vc(0+) = 12 V, iL(0+) = 0, vc(∞) = 4 V, and iL(∞) = 1 mA
8 kΩ
t=0
iL(t)
25 mH
12 V
+
–
2 μF
+
vc(t)
–
4 kΩ
Figure P 7.8-1
Solution:
Then
Next
i L ( 0+ ) = i L ( 0 − ) = 0 and v C ( 0+ ) = v C ( 0 − ) = 12 V
P 7.8-2
The switch in Figure P 7.8-2 has been open for a long time before closing at time t = 0.
Find vc(0+) and iL(0+), the values of the capacitor voltage and inductor current immediately after
the switch closes. Let vc(∞) and iL(∞) denote the values of the capacitor voltage and inductor
current after the switch has been closed for a long time. Find vc(∞) and iL(∞).
Answer: vc(0+) = 6 V, iL(0+) = 1 mA, vc(∞) = 3 V, and iL(∞) = 1.5 mA
6 kΩ
12 V
+
–
2 μF
iL(t)
25 mH
+
vc(t)
–
t=0
6 kΩ
3 kΩ
Figure P7.8-2
Solution:
Then
Next
i L ( 0 + ) = i L ( 0− ) = 1 mA and v C ( 0+ ) = v C ( 0− ) = 6 V
P 7.8-3
The switch in Figure P 7.8-3 has been open for a long time before closing at time t = 0.
Find vc(0+) and iL(0+), the values of the capacitor voltage and inductor current immediately after
the switch closes. Let vc(∞) and iL(∞) denote the values of the capacitor voltage and inductor
current after the switch has been closed for a long time. Find vc(∞) and iL(∞).
Answer: vc(0+) = 0 V, iL(0+) = 0, vc(∞) = 8 V, and iL(∞) = 0.5 mA
t=0
8 kΩ
12 V
+
–
2 μF
iL(t)
25 mH
+
vc(t)
–
16 kΩ
Figure P 7.8-3
Solution:
Then
Next
i L ( 0 + ) = i L ( 0− ) = 0 and v C ( 0+ ) = v C ( 0− ) = 0 V
P7.8-4 . The switch in the circuit shown in Figure
P7.8-4 has been closed for a long time before it
opens at time t = 0. Determine the values of vR(0-)
and vL(0-), the voltage across the 4 Ω resistor and
the inductor immediately before the switch opens
and the values of vR(0+) and vL(0+), the voltage
across the 4 Ω resistor and the inductor
immediately after the switch opens.
Figure P7.8-4
Solution:
The circuit is at steady state immediately before the switch opens. We have
The inductor acts like a short circuit so v L ( 0 − ) = 0 .
Noticing that the 80 Ω and 20 Ω are connected in parallel
and using voltage division:
4
4
( 24 ) =
( 24 ) = 4.8 V
4 + ( 80 || 20 )
4 + 16
Using current division:
24
1 ⎛ 24 ⎞
⎛ 20 ⎞
i L ( 0) = ⎜
= ⎜
⎟
⎟ = 0.24 A
⎝ 80 + 20 ⎠ 4 + ( 80 || 20 ) 5 ⎝ 4 + 16 ⎠
vR (0 −) =
The inductor current does not change instantaneously so i L ( 0 + ) = i L ( 0 − ) i L ( 0 ) . Immediately
after the switch opens we have:
v R ( 0 + ) = 4 i L ( 0 ) = 4 ( 0.24 ) = 0.96 V
Using KVL:
v R ( 0 + ) + v L ( 0 + ) + 80 i L ( 0 ) − 24 = 0
0.96 + v L ( 0 + ) + 80 ( 0.24 ) − 24 = 0
v L ( 0 + ) = 3.84 V
v ( t ) = 75 − 82 e −7 t = R ( 5 + 2 e−7 t ) + L
d
5 + 2 e −7 t )
(
dt
−7 t
= R ( 5 + 2 e ) + L ( ( −7 ) 2 e −7 t ) = 5 R + ( 2 R − 14 L ) e−7 t
75 = 5 R ⇒ R = 15 Ω and
82 + 30
−82 = 2 R − 14 L = 30 − 14 L ⇒ L =
=8 H
14
Equation coefficients gives
and
P7.8-5 . The switch in the circuit shown in Figure
P7.8-5 has been open for a long time before it
closes at time t = 0. Determine the values of iR(0-)
and iC(0-), the current in one of the 20 Ω resistors
and in the capacitor immediately before the switch
closes and the values of iR(0+) and iC(0+), the
current in that 20 Ω resistor and in the capacitor
immediately after the switch closes.
Figure P7.8-5
Solution:
The circuit is at steady state immediately before the switch closes. We have
The capacitor acts like an open circuit so i C ( 0 − ) = 0 .
Noticing that two 20 Ω are connected in series and using
current division:
iR (0 −) =
( 20 + 20 ) 120 = 2 120 = 80 mA
( ) ( )
20 + ( 20 + 20 )
3
Using current division and Ohm’s law:
⎡
⎤
20
v C ( 0) = ⎢
(120 )⎥ ( 20 ) = 0.8 V
⎣ 20 + ( 20 + 20 )
⎦
The capacitor does not change instantaneously so v C ( 0 + ) = v C ( 0 − ) v C ( 0 ) . Immediately after
the switch closes we have:
Applying KVL to the loop consisting of the closed
switch, the capacitor and a 20 Ω resistor gives
0 + v C ( 0 ) − 20 i R ( 0 + ) = 0
0 + 0.8 = 20 i R ( 0 + )
i R ( 0 + ) = 40 mA
Applying KCL at the node at the right side of the circuit
gives:
vC (0 +)
+ i C ( 0 + ) + i R ( 0 + ) = 0.120
20
0.8
+ i C ( 0 + ) + 0.04 = 0.120
20
i C ( 0 + ) = 0.04 = 40 mA
Figure P7.8-6
P7.8-6. The switch in the circuit shown in Figure P7.8-6 has been open for a long time before it
closes at time t = 0. Determine the values of vL(0-), the voltage across the inductor immediately
before the switch closes and vL(0+), the voltage across the inductor immediately after the switch
closes.
Solution:
The circuit is at steady state immediately before the switch closes. We have
The inductor acts like a short circuit so
vL (0 −) = 0 .
The inductor current is the negative of the current
source current:
i L ( 0 ) = −120 mA
The inductor current does not change instantaneously so i L ( 0 + ) = i L ( 0 − ) i L ( 0 ) . Immediately
after the switch closes we have:
Applying KVL to the left mesh gives:
v L ( 0 + ) + 20 i L ( 0 ) = 0
v L ( 0 + ) + 20 ( −0.12 ) = 0
v L ( 0 + ) = 2.4
Figure P7.8-7
P7.8-7. The switch in the circuit shown in Figure P7.8-7 has been closed for a long time before it
opens at time t = 0. Determine the values of iC(0-), the current in the capacitor immediately
before the switch opens and iC(0+), the current in the capacitor immediately after the switch
opens.
Solution:
The circuit is at steady state immediately before the switch opens. We have
The capacitor acts like an open circuit so
iC (0 −) = 0 .
The capacitor voltage is equal to the voltage
source voltage:
v C ( 0 ) = 20 V
The capacitor does not change instantaneously so v C ( 0 + ) = v C ( 0 − ) v C ( 0 ) . Immediately after
the switch opens we have:
Applying KCL at the top node of the capacitor,
we see that:
v C ( 0)
iC (0 +) +
=0
20
v C ( 0)
iC (0 +) = −
= −1 A
20
i3(t)
P 7.8-8
3Ω
The circuit shown in Figure P 7.8-8
is at steady state when the switch opens at time
v1(t) −
6H
+
12 V –
6Ω
t=0
i2(t)
Figure P 7.8-8
Solution: The capacitor voltage and inductor current
don’t change instantaneously and so are the keys to
solving this problem.
Label the capacitor voltage and inductor current as
shown.
Before t = 0, with the switch closed and the circuit at
steady state, the inductor acts like a short circuit and
the capacitor acts like an open circuit.
i3 (0 −) = i (0 −) =
+
50 mF
t = 0. Determine v1(0–), v1(0+), i2(0–), i2(0+),
i3(0–), i3(0+), v4(0–), and v4(0+).
+
12
= 1.33 A
9
v 4 (0 −) = v (0 −) = 6 i (0 −) = 8 V
v1 ( 0 − ) = 0 V and i 2 ( 0 − ) = 0 A
The capacitor voltage and inductor current don’t change instantaneously so
v ( 0 + ) = v ( 0 − ) = 8 V and i ( 0 + ) = i ( 0 − ) = 1.33 A
v4(t)
−
After the switch opens the circuit looks like this:
From KCL:
From KVL:
i 3 ( t ) = 0 A and i 2 ( t ) = −i ( t )
v1 ( t ) + 6 i ( t ) = v ( t )
From Ohm’s Law:
At t = 0 +
v 4 (t ) = 6 i (t )
i 3 ( 0 + ) = 0 A and i 2 ( 0 + ) = −i ( 0 + ) = −1.33 A
v1 ( 0 + ) = v ( 0 + ) − 6 i ( 0 + ) = 8 − 6 (1.333) = 0 V
v 4 (0 +) = 6 i (0 +) = 8 V
P 7.8-9
3Ω
The circuit shown in Figure P 7.8-9 is at
v1(t)
steady state when the switch opens at time
t = 0. Determine v1(0–), v1(0+), i2(0–), and i2(0+).
+
6Ω
6H
−
+
–
12 V
50 mF
t=0
Hint: Modeling the open switch as an open circuit
leads us to conclude that the inductor current changes
instantaneously, which would require an infinite
voltage. We can use a more accurate model of the open
switch, a large resistance, to avoid the infinite voltage.
Solution:
The capacitor voltage and inductor current don’t
change instantaneously and so are the keys to
solving this problem.
Label the capacitor voltage and inductor current as
shown.
i2(t)
6Ω
Figure P 7.8-9
Before t = 0, with the switch closed and the circuit at
steady state, the inductor acts like a short circuit and
the capacitor acts like an open circuit.
i 2 (0 −) = 0
i (0 −) =
12
= 1.333 A
9
v1 ( 0 − ) = 0 V
v (0 −) = 6 i (0 −) = 8 V
After the switch opens we model the open switch as
a large resistance, R.
From KVL:
(
)
12 = 3 i ( t ) + i 2 ( t ) + v1 ( t ) + ( R + 6 ) i ( t )
and
v1 ( t ) + ( R + 6 ) i ( t ) = 6 i 2 ( t ) + v ( t )
The capacitor voltage and inductor current don’t
change instantaneously so
v (0 +) = v (0 −) = 8 V
and
i ( 0 + ) = i ( 0 − ) = 1.333 A
At t = 0 +
(
)
12 = 3 i ( 0 + ) + i 2 ( 0 + ) + v 1 ( 0 + ) + ( R + 6 ) i ( 0 + ) ⎫⎪
4
⎬ ⇒ v1 ( 0 + ) = R and i 2 ( 0 + ) = 0
3
v 1 ( 0 + ) + ( R + 6 ) i ( 0 + ) = 6 i 2 ( 0 + ) + v ( 0 + ) ⎪⎭
As expected lim v 1 ( 0 + ) = ∞ .
R →∞
P 7.8-10 The circuit shown in Figure P 7.8-10 is at steady state when the switch closes at time
t = 0. Determine v1(0–), v1(0+), i2(0–), and i2(0+).
30 Ω
i2(t)
t=0
+
+
–
10 Ω
24 V
15 Ω
50 mF
15 Ω
v1(t)
−
3.5 H
Figure P 7.8-10
Solution:
The capacitor voltage and inductor current don’t
change instantaneously and so are the keys to
solving this problem.
Label the capacitor voltage and inductor current as
shown.
Before t = 0, with the switch open and the circuit at
steady state, the inductor acts like a short circuit and
the capacitor acts like an open circuit.
i 2 (0 −) = i (0 −) =
24
= −0.4 A
60
v1 ( 0 − ) = 0 V
v ( 0 − ) − 15 i ( 0 − ) = v 1 ( 0 − ) ⇒ v ( 0 − ) = −6 V
v 3 ( 0 − ) = 15 i ( 0 − ) = −6 V
The capacitor voltage and inductor current don’t change instantaneously so
v ( 0 + ) = v ( 0 − ) = −6 V and i ( 0 + ) = i ( 0 − ) = −0.4 A
After the switch closes the circuit looks like this:
From Ohm’s Law:
i 2 (t ) = −
From KVL:
From KCL:
v (t )
30
v 1 ( t ) = v 3 ( t ) + 24
v1 ( t )
10
+
v 3 (t )
15
= i (t )
At t = 0 +
i 2 (0 +) = −
v (0 +)
= 0.2 A
30
v 1 ( 0 + ) = v 3 ( 0 + ) + 24 ⎫
⎪
⎬ ⇒ v 1 ( 0 + ) = 7.2 V and v 3 ( 0 + ) = −16.8 V
v1 ( 0 + ) v 3 ( 0 + )
+
= i ( 0 + )⎪
10
15
⎭
P7.8-11
The circuit shown in Figure 7.8-11 has reached
steady state before the switch opens at time t = 0.
Determine the values of i L ( t ) , v C ( t ) and v R ( t )
immediately before the switch opens and the value of
v R ( t ) immediately after the switch opens.
Answers: i L ( 0 − ) = 1.25 A, v C ( 0 − ) = 20 V,
v R ( 0 − ) = −5 V and v R ( 0 + ) = −4 V
Figure 7.8-11
Solution: Because
• This circuit has reached steady state before
the switch opens at time t = 0.
• The only source is a constant voltage source.
At t=0−, the capacitor acts like an open circuit and
the inductor acts like a short circuit. From the
circuit
25
25
i1 ( 0 − ) =
=
= 1.25 A ,
4 + ( 20 || 80 ) 4 + 16
⎛ 80 ⎞
iL (0 −) = ⎜
⎟ i1 ( 0 − ) = 1 A ,
⎝ 20 + 80 ⎠
v C ( 0 − ) = 20 i L ( 0 − ) = 20 V
and
v R ( 0 − ) = −4 i 1 ( 0 − ) = −5 V
The capacitor voltage and inductor current don’t
change instantaneously so
v C ( 0 + ) = v C ( 0 − ) = 20 V and
iL (0 +) = iL (0 −) = 1 A
Apply KCL at the top node to see that
i1 ( 0 + ) = i L ( 0 + ) = 1 A
From Ohm’s law
v R ( 0 + ) = −4 i1 ( 0 + ) = −4 V
(Notice that the resistor voltage did change instantaneously.)
P7.8-12. The circuit shown in Figure 7.8-12 has reached steady state before the switch closes at
time t = 0.
Figure 7.8-12
(a) Determine the values of i L ( t ) , v C ( t ) and v R ( t ) immediately before the switch closes.
(b) Determine the value of v R ( t ) immediately after the switch closes.
Solution: Because
• This circuit has reached steady state before the switch closes at time t = 0.
• The only source is a constant voltage source.
At t=0−, the capacitor acts like an open circuit and the inductor acts like a short circuit.
From the circuit
And
iL (0 −) =
35
= 0.5 A , v R ( 0 − ) = 40 i L ( 0 − ) = 20 V ,
30 + 40
v C ( 0 − ) = v R ( 0 − ) = 20 V
The capacitor voltage and inductor current don’t change instantaneously so
v C ( 0 + ) = v C ( 0 − ) = 20 V and i L ( 0 + ) = i L ( 0 − ) = 0.5 A
⎛ 40 ⎞
v R ( 0 + ) = 40 ⎜
⎟ i L ( 0 + ) = 10 V
⎝ 40 + 40 ⎠
(Notice that the resistor voltage did change instantaneously.)
P7.8-13
The circuit shown in Figure 7.8-12 has reached
steady state before the switch opens at time t = 0.
Determine the values of i L ( t ) , v C ( t ) and v R ( t )
immediately before the switch opens and the
value of v R ( t ) immediately after the switch
opens.
Answer: i L ( 0 − ) = 0.4__A, v C ( 0 − ) = 16 V,
v R ( 0 − ) = 0 V and v R ( 0 + ) = −12 V
Solution: Because
• This circuit has reached steady state
before the switch opens at time t = 0.
• The only source is a constant voltage
source.
At t=0−, the capacitor acts like an open circuit
and the inductor acts like a short circuit.
The current in the 30 Ω resistor is zero so
v R ( 0 − ) = 0 V . Next
24
= 0.4 A and
20 + 40
v C ( 0 − ) = 40 i L ( 0 − ) = 16 V
i1 ( 0 − ) =
The capacitor voltage and inductor current
don’t change instantaneously so
v C ( 0 + ) = v C ( 0 − ) = 16 V and
i L ( 0 + ) = i L ( 0 − ) = 0.4 A
Apply KCL at the bottom node and then Ohm’s
law to get
v R ( 0 + ) = −30 i L ( 0 + ) = −12 V
(Notice that the resistor voltage did change instantaneously.)
Figure 7.8-13
Section 7-9: Operational amplifier Circuits and Linear Differential Equations
P 7.9-1 Design a circuit with one input, x(t), and one output, y(t), that are related by this
differential equation:
1 d2
d
5
y (t ) + 4 y (t ) + y (t ) = x(t )
2
2 dt
2
dt
Solution:
P 7.9-2 Design a circuit with one input, x(t), and one output, y(t), that are related by this
differential equation:
1 d2
5
y (t ) + y (t ) = − x(t )
2
2 dt
2
Solution:
P 7.9-3 Design a circuit with one input, x(t), and one output, y(t), that are related by this
differential equation:
d3
d2
d
2 3 y (t ) + 16 2 y (t ) + 8 y (t ) + 10 y (t ) = −4 x(t )
dt
dt
dt
Solution:
P 7.9-4 Design a circuit with one input, x(t), and one output, y(t), that are related by this
differential equation:
d3
d2
d
y (t ) + 16 2 y (t ) + 8 y (t ) + 10 y (t ) = 4 x(t )
3
dt
dt
dt
Solution:
Section 7.11 How Can We Check…?
P 7.11-1
A homework solution indicates that the current and voltage of a 100-H inductor are
t <1
⎧ 0.025
⎪ t
t <1
⎧0
⎪− + 0.065 1 < t < 3
⎪ −4 1 < t < 3
⎪ 25
⎪
i (t ) = ⎨
and v(t ) = ⎨
⎪ t − 0.115 3 < t < 9
⎪ 2 3<t <9
⎪⎩ 0
⎪ 50
t >9
⎪ 0.065
t <9
⎩
where the units of current are A, the units of voltage are V, and the units of time are s. Verify
that the inductor current does not change instantaneously.
Solution: We need to check the values of the inductor current at the ends of the intervals.
at t = 1
at t = 3
? − 1 + 0.065 = 0.025
0.025 =
25
−
3
3
+ 0.065 =?
− 0.115
25
50
−0.055 = −0.055
at t = 9
( Yes!)
( Yes!)
9
− 0.115 =? 0.065
50
0.065 = 0.065
( Yes!)
The given equations for the inductor current describe a current that is continuous; as must be the
case since the given inductor voltage is bounded.
P 7.11-2
A homework solution indicates that the current and voltage of a 100-H inductor are
⎧ t
t <1
⎪− 200 + 0.025
⎪
⎪⎪ − t + 0.03 1 < t < 4
i (t ) = ⎨ 100
⎪ t
− 0.03
4<t <9
⎪
⎪ 100
⎪⎩
0.015
t <9
and
t <1
⎧ −1
⎪ −2 1 < t < 4
⎪
v(t ) = ⎨
⎪ 1 4<t <9
⎪⎩ 0
t >9
where the units of current are A, the units of voltage are V, and the units of time are s. Is this
homework solution correct? Justify your answer.
P7.11-2 We need to check the values of the inductor current at the ends of the intervals.
1
? − 1 + 0.03 ( Yes!)
+ 0.025 =
at t = 1 −
200
100
4
? 4 − 0.03 ( No!)
+ 0.03 =
100
100
The equation for the inductor current indicates that this current changes instantaneously at t = 4s.
This equation cannot be correct.
at t = 4
−
Design Problems
DP 7-1 Consider a single-circuit element, that is, a single resistor, capacitor, or inductor. The
voltage, v(t), and current, i(t), of the circuit element adhere to the passive convention. Consider
the following cases:
(a)
v(t) = 4 + 2e–3t V and i(t) = –3e–3t A for t > 0
(b)
v(t) = –3e–3t V and i(t) = 4 + 2e–3t A for t > 0
(c)
v(t) = 4 + 2e–3t V and i(t) = 2 + e–3t A for t > 0
For each case, specify the circuit element to be a capacitor, resistor, or inductor and give the
value of its capacitance, resistance, or inductance.
Solution:
i (t )
d
= 0.5 F .
v ( t ) = −6 e −3t is proportional to i(t) so the element is a capacitor. C =
d
dt
v (t )
dt
v (t )
d
b) i ( t ) = −6 e −3t is proportional to v(t) so the element is an inductor. L =
= 0.5 H .
d
dt
i (t )
dt
v (t )
c) v(t) is proportional to i(t) so the element is a resistor. R =
= 2 Ω.
i (t )
a)
DP 7-2 Figure DP 7.2 shows a voltage source and unspecified circuit
elements. Each circuit element is a single resistor, capacitor, or
inductor. Consider the following cases:
(a)
i(t) = 1.131 cos (2t + 45°) A
(b)
i(t) = 1.131 cos (2t – 45°) A
For each case, specify each circuit element to be a capacitor, resistor,
or inductor and give the value of its capacitance, resistance, or
inductance.
Hint: cos (θ + φ) = cos θcos φ – sin θsin φ
Solution: (a)
1.131cos ( 2t + 45° ) = 1.131 ⎡⎣ cos ( 45° ) cos ( 2t ) − sin ( 45° ) sin ( 2t ) ⎤⎦
= 0.8 cos 2 t − 0.8 sin 2 t
The first term is proportional to the voltage. Associate it with the resistor.
The noticing that
∫
t
t
−∞
v (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t
−∞
d
d
v ( t ) = 4 cos 2 t = −8 sin 2 t
dt
dt
associate the second term with a capacitor to get the minus sign. Then
4 cos 2 t
4 cos 2 t
R=
=
= 5 Ω and
i1 (t )
0.8 cos 2 t
−0.8 sin 2 t
i2 (t )
=
= 0.1 F
C=
d
8
sin
2
−
t
4 cos 2 t
dt
(b)
1.131cos ( 2t − 45° ) = 1.131 ⎡⎣ cos ( −45° ) cos ( 2t ) − sin ( −45° ) sin ( 2t ) ⎤⎦
= 0.8 cos 2 t + 0.8 sin 2 t
The first term is proportional to the voltage. Associate it with the
resistor. Then noticing that
t
t
−∞
−∞
∫ v (τ ) dτ = ∫
4 cos 2 t dτ = 2 sin 2t
d
d
v ( t ) = 4 cos 2 t = −8 sin 2 t
dt
dt
associate the second term with an inductor to get the plus sign. Then
4 cos 2 t
4 cos 2 t
=
= 5 Ω and
i1 (t )
0.8 cos 2 t
R=
∫
L=
t
−∞
4 cos 2 t dτ
i2 (t )
=
2 sin 2 t
= 2.5 H
0.8 sin 2 t
4 cos 2t V
+–
i(t)
Figure DP 7.2
DP 7-3 Figure DP 7.3 shows a voltage source and
unspecified circuit elements. Each circuit element is a
single resistor, capacitor, or inductor. Consider the
following cases:
(a)
v(t) = 11.31 cos (2t + 45°) V
(b)
v(t) = 11.31 cos (2t – 45°) V
+ v(t) –
4 cos 2t A
Figure DP 7.3
For each case, specify each circuit element to be a capacitor, resistor, or inductor and give the
value of its capacitance, resistance, or inductance.
Hint: cos (θ + φ) = cos θcos φ – sin θsin φ
Solution:
a)
11.31cos ( 2t + 45° ) = 11.31 ⎡⎣ cos ( 45° ) cos ( 2t ) − sin ( 45° ) sin ( 2t ) ⎤⎦
= 8 cos 2 t − 8 sin 2 t
The first term is proportional to the voltage. Associate it with the resistor. The noticing that
t
t
−∞
−∞
∫ i (τ ) dτ = ∫
4 cos 2 t dτ = 2 sin 2t
d
d
i ( t ) = 4 cos 2 t = −8 sin 2 t
dt
dt
associate the second term with an inductor to get the minus sign. Then
v (t )
8 cos 2 t
−8 sin 2 t
v2 (t )
R= 1
=
= 2 Ω and L =
=
=1H
d
4 cos 2 t 4 cos 2 t
−
8
sin
2
t
4 cos 2 t
dt
b)
11.31cos ( 2t + 45° ) = 11.31 ⎡⎣ cos ( −45° ) cos ( 2t ) − sin ( −45° ) sin ( 2t ) ⎤⎦
= 8 cos 2 t + 8 sin 2 t
The first term is proportional to the voltage. Associate it with the resistor. The noticing that
∫
t
−∞
t
i (τ ) dτ = ∫ 4 cos 2 t dτ = 2 sin 2t
−∞
d
d
i ( t ) = 4 cos 2 t = −8 sin 2 t
dt
dt
associate the second term with a capacitor to get the minus sign. Then
8 cos 2 t
v (t )
R= 1
=
= 2 Ω and C =
4 cos 2 t 4 cos 2 t
∫
t
−∞
4 cos 2 t dτ
v2 (t )
=
2 sin 2 t
= 0.25 F
8 sin 2 t
VB
DP 7-4 A high-speed flash unit for sports
photography requires a flash voltage v(0+) =
3 V and
dv(t )
= 24 V/s
dt t =0
3Ω
–+
t=0
Switch 1
Switch 2
1Ω
The flash unit uses the circuit shown in
Figure DP 7.4. Switch 1 has been closed a
long time, and switch 2 has been open a
long time at t = 0. Actually, the long time in
this case is 3 s. Determine the required
battery voltage, VB, when C = 1/8 F.
1 2
C
Flash
voltage
H
+
v
–
3Ω
t=0
+
–
Figure DP 7.4
Solution:
at t=0−
iL ( 0− ) = 0
VB
4
−
We require vC ( 0 ) = 3 V so VB = 12 V
By voltage division: vC ( 0− ) =
at t=0+
Now we will check
vC ( 0+ ) = vC ( 0 − ) = 3 V
and
Apply KCL at node a:
iL ( 0
) + i (0 ) =
+
C
Finally
as required.
dvC
dt
t = 0+
VB − vC ( 0+ )
3
12 − 3
⇒ iC ( 0+ ) = 3 A
3
iC ( 0+ )
3
V
=
=
= 24
C
0.125
s
0 + iC ( 0 + ) =
t = 0+
iL ( 0+ ) = iL ( 0− ) = 0
First:
+
dvC
dt
VB
DP 7-5 For the circuit shown in Figure DP 7.5, select a value of R so that the energy stored in
the inductor is equal to the energy stored in the capacitor at steady state.
20 Ω
10 mH
10 V
+
–
1 μF
R
Figure DP 7.5
1
1
L i L2 = C v C2 where iL and vC are the steady-state inductor current and
2
2
v
capacitor voltage. At steady state, i L = C . Then
R
Solution: We require
2
⎛v ⎞
L ⎜ C ⎟ = C vC2
⎝R⎠
⇒
C=
L
R2
⇒
R =
L
=
C
10 −2
=
10 −6
10 4 = 10 2 Ω
Chapter 8 Exercises
Exercise 8.3-1 The circuit shown in Figure E 8.3-1 is at steady state before the switch closes at
time t = 0. Determine the capacitor voltage, v(t), for t ≥ 0.
Answer: v(t) = 2 + e–2.5t V for t > 0
Figure E 8.3-1
Solution: Before the switch closes:
After the switch closes:
Therefore
Finally,
2
= 8 Ω so τ = 8 ( 0.05 ) = 0.4 s .
0.25
v (t ) = voc + (v (0) − voc ) e− t τ = 2 + e−2.5 t V for t > 0
Rt =
Exercise 8.3-2 The circuit shown in Figure E 8.3-2 is at steady state before the switch closes at
time t = 0. Determine the inductor current, i(t), for t > 0.
1 1
Answer: i (t ) = + e −1.33t A for t > 0
4 12
Figure E 8.3-2
Solution: Before the switch closes:
After the switch closes:
Therefore
Finally,
2
6
= 8 Ω so τ = = 0.75 s .
0.25
8
1 1
i (t ) = isc + (i (0) − isc ) e−t τ = + e−1.33 t A for t > 0
4 12
Rt =
Exercise 8.7-1 The electrical power plant for the orbiting space station shown in Figure E 8.71a uses photovoltaic cells to store energy in batteries. The charging circuit is modeled by the
circuit shown in Figure E 8.7-1b, where vs = 10 sin 20t V. If v(0–) = 0, find v(t) for t > 0.
Answer: v = 4 e–10t – 4 cos 20t + 2 sin 20t V
Figure E 8.7-1
Solution:
Apply KVL:
vs ( t ) = 10sin 20t V
Natural Response:
Forced Response:
⎛ d v( t ) ⎞
−10sin 20t + 10 ⎜ .01
⎟ + v (t ) = 0
dt ⎠
⎝
d v( t )
⇒
+ 10 v ( t ) = 100sin 20t
dt
vn (t ) = Ae −t τ
where τ = Rt C ∴ vn (t ) = Ae−10t
try v f (t ) = B1 cos 20t + B2 sin 20 t
Plugging v f (t ) into the differential equation and equating like terms yields:
−20 B1 sin 20t + 20 B2 cos 20t + 10 B1 cos 20t + 10 B2 sin 20t = 100 sin 20t
20 B2 + 10 B1 = 0 and − 20 B1 + 10 B2 = 100
B1 = −4 and B2 = 2
Complete Response: v(t ) = vn (t ) + v f (t ) = Ae −10t − 4 cos 20t + 2 sin 20t
Now v(0− ) = v(0+ ) = 0 = A − 4 ⇒ A = 4
v(t ) = 4 e−10t − 4 cos 20t + 2sin 20 t V
Section 8.3: The Response of a First Order Circuit to a Constant Input
P 8.3-1 The circuit shown in Figure P 8.3-1 is
at steady state before the switch closes at time
t = 0. The input to the circuit is the voltage of the
voltage source, 12 V. The output of this circuit is
the voltage across the capacitor, v(t). Determine
v(t) for t > 0.
Answer: v(t) = 6 – 2e–1.33t V for t > 0
Figure P 8.3-1
Solution:
Here is the circuit before t = 0, when the switch is
open and the circuit is at steady state. The open
switch is modeled as an open circuit.
A capacitor in a steady-state dc circuit acts like an
open circuit, so an open circuit replaces the
capacitor. The voltage across that open circuit is the
initial capacitor voltage, v (0).
By voltage division
v (0) =
6
(12 ) = 4 V
6+6+6
Next, consider the circuit after the switch closes. The
closed switch is modeled as a short circuit.
We need to find the Thevenin equivalent of the part
of the circuit connected to the capacitor. Here’s the
circuit used to calculate the open circuit voltage, Voc.
Voc =
6
(12 ) = 6 V
6+6
Here is the circuit that is used to determine Rt. A
short circuit has replaced the closed switch.
Independent sources are set to zero when calculating
Rt, so the voltage source has been replaced by a short
circuit.
( 6 )( 6 ) = 3 Ω
Rt =
6+6
Then
τ = R t C = 3 ( 0.25 ) = 0.75 s
Finally,
v ( t ) = Voc + ( v ( 0 ) − Voc ) e−t / τ = 6 − 2 e−1.33 t V for t > 0
P 8.3-2 The circuit shown in Figure P 8.3-2 is at
steady state before the switch opens at time t = 0.
The input to the circuit is the voltage of the voltage
source, 12 V. The output of this circuit is the
current in the inductor, i(t). Determine i(t) for t > 0.
Answer: i(t) = 1 + e–0.5t A for t > 0
Figure P 8.3-2
Solution:
Here is the circuit before t = 0, when the switch is
closed and the circuit is at steady state. The closed
switch is modeled as a short circuit.
An inductor in a steady-state dc circuit acts like an
short circuit, so a short circuit replaces the inductor.
The current in that short circuit is the initial inductor
current, i(0).
12
i ( 0) = = 2 A
6
Next, consider the circuit after the switch opens. The
open switch is modeled as an open circuit.
We need to find the Norton equivalent of the part of
the circuit connected to the inductor. Here’s the
circuit used to calculate the short circuit current, Isc.
12
=1 A
6+6
Here is the circuit that is used to determine Rt. An
open circuit has replaced the open switch.
Independent sources are set to zero when calculating
Rt, so the voltage source has been replaced by an
short circuit.
( 6 + 6 )( 6 ) = 4 Ω
R t = 6 || ( 6 + 6 ) =
( 6 + 6) + 6
L 8
Then
τ=
= =2 s
Rt 4
I sc =
Finally,
i ( t ) = I sc + ( i ( 0 ) − I sc ) e −t / τ = 1 + e−0.5 t A for t > 0
P 8.3-3 The circuit shown in Figure P 8.3-3 is at
steady state before the switch closes at time t = 0.
Determine the capacitor voltage, v(t), for t > 0.
Answer: v(t) = – 6 + 18e–6.67t V for t > 0
Figure P 8.3-3
Solution: Before the switch closes:
After the switch closes:
−6
= 3 Ω so τ = 3 ( 0.05 ) = 0.15 s .
−2
Finally, v (t ) = voc + (v (0) − voc ) e− t τ = −6 + 18 e−6.67 t V for t > 0
Therefore R t =
P 8.3-4 The circuit shown in Figure P 8.3-4 is at steady
state before the switch closes at time t = 0. Determine the
inductor current, i(t), for t > 0.
10
Answer: i (t ) = −2 + e −0.5t A for t > 0
3
Figure P 8.3-4
Solution: Before the switch closes:
After the switch closes:
−6
6
= 3 Ω so τ = = 2 s .
−2
3
t
−
10
Finally, i (t ) = isc + (i (0) − isc ) e τ = −2 + e−0.5 t A for t > 0
3
Therefore R t =
P 8.3-5 The circuit shown in Figure P 8.3-5
is at steady state before the switch opens at
time t = 0. Determine the voltage, vo(t), for
t > 0.
Answer: vo(t) = 10 – 5e–12.5t V for t > 0
Figure P 8.3-5
Solution: Before the switch opens, v o ( t ) = 5 V ⇒ v o ( 0 ) = 5 V . After the switch opens the part of the
circuit connected to the capacitor can be replaced by it's Thevenin equivalent circuit to get:
Therefore τ = ( 20 × 103 )( 4 × 10−6 ) = 0.08 s .
Next, v C (t ) = voc + (v (0) − voc ) e
−
t
τ
= 10 − 5 e−12.5 t V for t > 0
Finally, v 0 (t ) = vC (t ) = 10 − 5 e −12.5 t V for t > 0
P 8.3-6 The circuit shown in Figure P 8.3-6
is at steady state before the switch opens at
time t = 0. Determine the voltage, vo(t),
for t > 0.
Answer: vo(t) = 5e–4000t V for t > 0
Figure P 8.3-6
5
= 0.25 mA ⇒ i o ( 0 ) = 0.25 mA . After the switch
Solution: Before the switch opens, i o ( t ) =
20 × 103
opens the part of the circuit connected to the inductor can be replaced by it's Norton equivalent circuit to
get:
Therefore τ =
5
= 0.25 ms .
20 × 103
Next, i L (t ) = isc + (i L (0) − isc ) e
Finally, vo ( t ) = 5
−
t
τ
= 0.5 − 0.25 e−4000 t mA for t > 0
d
i L ( t ) = 5 e −4000 t V
dt
for t > 0
P 8.3-7
Figure P 8.3-7a shows astronaut Dale Gardner using the manned maneuvering unit to dock
with the spinning Westar VI satellite on November 14, 1984. Gardner used a large tool called the apogee
capture device (ACD) to stabilize the satellite and capture it for recovery, as shown in Figure P 8.3-7a.
The ACD can be modeled by the circuit of Figure P 8.3-7b. Find the inductor current iL for t > 0.
Answer: iL(t) = 6e–20t A
Figure P 8.3-7
Solution: At t = 0− (steady-state)
Since the input to this circuit is constant, the
inductor will act like a short circuit when the
circuit is at steady-state:
for t > 0
iL ( t ) = iL ( 0 ) e− ( R L ) t = 6 e−20t A
P 8.3-8 The circuit shown in Figure P 8.3-8 is at
steady state before the switch opens at time t = 0.
The input to the circuit is the voltage of the voltage
source, Vs. This voltage source is a dc voltage
source; that is, Vs is a constant. The output of this
circuit is the voltage across the capacitor, vo(t). The
output voltage is given by
vo(t) = 2 + 8e–0.5t V for t > 0
Determine the values of the input voltage, Vs, the
capacitance, C, and the resistance, R.
Figure P 8.3-8
Solution: Before the switch opens, the circuit will be at steady state. Because the only input to this
circuit is the constant voltage of the voltage source, all of the element currents and voltages, including
the capacitor voltage, will have constant values. Opening the switch disturbs the circuit. Eventually the
disturbance dies out and the circuit is again at steady state. All the element currents and voltages will
again have constant values, but probably different constant values than they had before the switch
opened.
Here is the circuit before t = 0, when the
switch is closed and the circuit is at steady state.
The closed switch is modeled as a short circuit. The
combination of resistor and a short circuit
connected is equivalent to a short circuit.
Consequently, a short circuit replaces the switch
and the resistor R. A capacitor in a steady-state dc
circuit acts like an open circuit, so an open circuit
replaces the capacitor. The voltage across that open
circuit is the capacitor voltage, vo(t).
Because the circuit is at steady state, the value of the capacitor voltage will be constant. This
constant is the value of the capacitor voltage just before the switch opens. In the absence of unbounded
currents, the voltage of a capacitor must be continuous. The value of the capacitor voltage immediately
after the switch opens is equal to the value immediately before the switch opens. This value is called the
initial condition of the capacitor and has been labeled as vo(0). There is no current in the horizontal
resistor due to the open circuit. Consequently, vo(0) is equal to the voltage across the vertical resistor,
which is equal to the voltage source voltage. Therefore
vo ( 0 ) = Vs
The value of vo(0) can also be obtained by setting t = 0 in the equation for vo(t). Doing so gives
vo ( 0 ) = 2 + 8 e0 = 10 V
Consequently,
Vs = 10 V
Next, consider the circuit after the switch opens.
Eventually (certainly as t →∞) the circuit will again
be at steady state. Here is the circuit at t = ∞, when
the switch is open and the circuit is at steady state.
The open switch is modeled as an open circuit. A
capacitor in a steady-state dc circuit acts like an
open circuit, so an open circuit replaces the
capacitor. The voltage across that open circuit is the
steady-state capacitor voltage, vo(∞). There is no
current in the horizontal resistor and vo(∞) is equal
to the voltage across the vertical resistor. Using
voltage division,
10
vo ( ∞ ) =
(10 )
R + 10
The value of vo(∞) can also be obtained by setting t = ∞ in the equation for vo(t). Doing so gives
vo ( ∞ ) = 2 + 8 e −∞ = 2 V
Consequently,
10
(10 ) ⇒ 2 R + 20 = 100 ⇒ R = 40 Ω
R + 10
−t τ
Finally, the exponential part of vo(t) is known to be of the form e
where τ = R t C and Rt is
2=
the Thevenin resistance of the part of the circuit connected to the capacitor.
Here is the circuit that is used to determine Rt. An
open circuit has replaced the open switch.
Independent sources are set to zero when
calculating Rt, so the voltage source has been
replaced by a short circuit.
R t = 10 +
so
( 40 )(10 ) = 18
40 + 10
Ω
τ = R t C = 18 C
From the equation for vo(t)
t
−0.5 t = −
⇒ τ =2s
τ
Consequently,
2 = 18 C ⇒ C = 0.111 = 111 mF
P 8.3-9
The circuit shown in Figure P 8.3-9 is
at steady state before the switch closes at time
t = 0. The input to the circuit is the voltage of the
voltage source, 24 V. The output of this circuit,
the voltage across the 3-Ω resistor, is given by
vo(t) = 6 – 3e–0.35t V when t > 0
Determine the value of the inductance, L, and of
Figure P 8.3-9
the resistances, R1 and R2.
Solution: Before the switch closes, the circuit will be at steady state. Because the only input to this
circuit is the constant voltage of the voltage source, all of the element currents and voltages, including
the inductor current, will have constant values. Closing the switch disturbs the circuit by shorting out the
resistor R1. Eventually the disturbance dies out and the circuit is again at steady state. All the element
currents and voltages will again have constant values, but probably different constant values than they
had before the switch closed.
The inductor current is equal to the current in the 3 Ω resistor. Consequently,
− 0.35 t
vo (t ) 6 − 3 e
− 0.35 t
=
= 2− e
i (t ) =
A when t > 0
3
3
In the absence of unbounded voltages, the current in any inductor is continuous. Consequently, the value
of the inductor current immediately before t = 0 is equal to the value immediately after t = 0.
Here is the circuit before t = 0, when the switch is
open and the circuit is at steady state. The open
switch is modeled as an open circuit. An inductor in
a steady-state dc circuit acts like a short circuit, so a
short circuit replaces the inductor. The current in that
short circuit is the steady state inductor current, i(0).
Apply KVL to the loop to get
R1 i ( 0 ) + R 2 i ( 0 ) + 3 i ( 0 ) − 24 = 0
24
R1 + R 2 + 3
The value of i(0) can also be obtained by setting t = 0 in the equation for i(t). Do so gives
⇒ i ( 0) =
i ( 0 ) = 2 − e0 = 1 A
Consequently,
1=
24
⇒ R1 + R 2 = 21
R1 + R 2 + 3
Next, consider the circuit after the switch closes.
Here is the circuit at t = ∞, when the switch is closed
and the circuit is at steady state. The closed switch is
modeled as a short circuit. The combination of
resistor and a short circuit connected is equivalent to
a short circuit. Consequently, a short circuit replaces
the switch and the resistor R1.
An inductor in a steady-state dc circuit acts like a short circuit, so a short circuit replaces the inductor.
The current in that short circuit is the steady state inductor current, i(∞). Apply KVL to the loop to get
24
R 2 i ( ∞ ) + 3 i ( ∞ ) − 24 = 0 ⇒ i ( ∞ ) =
R2 + 3
The value of i(∞) can also be obtained by setting t = ∞ in the equation for i(t). Doing so gives
i ( ∞ ) = 2 − e −∞ = 2 A
Consequently
2=
24
⇒ R2 = 9 Ω
R2 + 3
Then
R1 = 12 Ω
Finally, the exponential part of i(t) is known to be of the form e
−t τ
where τ =
L
and Rt is the
Rt
Thevenin resistance of the part of the circuit that is connected to the inductor.
Here is shows the circuit that is used to determine Rt. A
short circuit has replaced combination of resistor R1 and the
closed switch. Independent sources are set to zero when
calculating Rt, so the voltage source has been replaced by an
short circuit.
R t = R 2 + 3 = 9 + 3 = 12 Ω
so
τ=
L
L
=
R t 12
From the equation for i(t)
−0.35 t = −
t
τ
⇒ τ = 2.857 s
Consequently,
2.857 =
L
⇒ L = 34.28 H
12
P 8.3-10 A security alarm for an office building door is modeled by the circuit of Figure P 8.3-10. The
switch represents the door interlock, and v is the alarm indicator voltage. Find v(t) for t > 0 for the circuit
of Figure P 8.3-10. The switch has been closed for a long time at t = 0–.
Figure P 8.3-10
Solution: First, use source transformations to obtain the equivalent circuit
for t < 0:
for t > 0:
So iL ( 0 ) = 2 A, I sc
1
L
1
s
= 0, Rt = 3 + 9 = 12 Ω, τ =
= 2 =
24
Rt 12
and iL ( t ) = 2e−24t
t >0
Finally v ( t ) = 9 iL ( t ) = 18 e−24t
t >0
P 8.3-11 The voltage v(t) in the circuit shown in Figure
P 8.3-11 is given by
v(t) = 8 + 4e–2t V for t > 0
Determine the values of R1, R2, and C.
Figure P 8.3-11
Solution: As t → ∞ the circuit reaches steady state and the
capacitor acts like an open circuit. Also, from the given
equation, v ( t ) → 8 V , as labeled on the drawing to the
right, then
8=
After t = 0
4
24
R2 + 4
⇒
R2 = 8 Ω
v C ( t ) = 24 − v ( t ) = 16 − 4e −2t
Immediately after t = 0
v C ( 0 + ) = 16 − 4 = 12 V
The capacitor voltage cannot change instantaneously so
v ( 0 − ) = 12 V
The circuit is at steady state just before the switch closes so
the capacitor acts like an open circuit. Then
12 =
8
24
R1 + 4 + 8
⇒
R1 = 4 Ω
After t = 0 the Thevenin resistance seen by the capacitor is
8
Rt = 8 & 4 = Ω
3
1
3
2=
⇒ C=
F
so
8
16
C
3
Figure P 8.3-12
P 8.3-12 The circuit shown in Figure P 8.3-12 is at steady state when the switch opens at time
t = 0. Determine i(t) for t ≥ 0.
Solution: Before t = 0, with the switch closed and the circuit at steady state, the inductor acts like a short
circuit so we have
Using superposition
9
− 5 ×10−3 = −2 mA
3000
The inductor current is continuous so i ( 0 + ) = i ( 0 − ) = −2 mA .
i (0 −) =
After t = 0, the switch is open. Determine the Norton equivalent circuit for the part of the circuit
connected to the inductor:
9
= 3 mA
3000
R t = 3000 & 6000 = 2000 Ω
i sc =
The time constant is given by τ =
L
5
1
=
= 0.0025 so = 400 .
R t 2000
τ
The inductor current is given by
i L ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( −0.002 − 0.003) e−400 t + 0.003 = 3 − 5e−400 t mA for t ≥ 0
(checked: LNAP 6/29/04)
Figure P 8.3-13
P 8.3-13 The circuit shown in Figure P 8.3-13 is at steady state when the switch opens at time
t = 0. Determine v(t) for t ≥ 0.
Solution: Before t = 0, with the switch closed and the circuit at the steady state, the capacitor acts like
an open circuit so we have
Using superposition
v (0 −) =
60 & 60
60 & 30
⎛1⎞ ⎛1⎞
6+
36 = ⎜ ⎟ 6 + ⎜ ⎟ 36 = 12 V
30 + ( 60 & 60 )
60 + ( 60 & 30 )
⎝2⎠ ⎝4⎠
The capacitor voltage is continuous so v ( 0 + ) = v ( 0 − ) = 12 V .
After t = 0 the switch is open. Determine the Thevenin equivalent circuit for the part of the circuit
connected to the capacitor:
v oc =
60
6=4V
60 + 30
R t = 30 & 60 = 20 kΩ
The time constant is τ = R t C = ( 20 × 103 )( 5 × 10−6 ) = 0.1 s so
1
τ
= 10
1
.
s
The capacitor voltage is given by
v ( t ) = ( v ( 0 + ) − v oc ) e−t τ + v oc = (12 − 4 ) e−10t + 4 = 4 + 8 e −10t V for t ≥ 0
(checked: LNAP 6/29/04)
Figure P 8.3-14
P 8.3-14 The circuit shown in Figure P 8.3-14 is at steady state when the switch closes at time
t = 0. Determine i(t) for t ≥ 0.
Solution: Before t = 0, with the switch open and the circuit at steady state, the inductor acts like a short
circuit so we have
i (t ) = −
⎤
18 ⎡
5
2 ⎥ = 0.29 A
⎢
4 + 18 ⎣ 5 + 20 + (18 & 4 ) ⎦
After t = 0, we can replace the part of the circuit connected to the inductor by its Norton equivalent
circuit. First, performing a couple of source transformations reduces the circuit to
Next, replace the series voltage sources by an equivalent voltage source, replace the series resistors by
an equivalent resistor and do a couple of source transformations to get
so
The current is given by
2
1
1
= 0.25 ⇒
=5
10
s
τ
−5t
i ( t ) = [ 0.29 − 0.4] e + 0.4 = 0.4 − 0.11e −5t A for t ≥ 0
τ=
P 8.3-15 The circuit in Figure P 8.3-15 is at steady
state before the switch closes. Find the inductor
current after the switch closes.
Hint: i(0) = 0.1 A, Isc = 0.3 A, Rt = 40 Ω
Answer: i(t) = 0.3 – 0.2e–2t A t ≥ 0
Solution: At steady-state, immediately before t = 0:
⎞
12
⎛ 10 ⎞ ⎛
i ( 0) = ⎜
⎟ ⎜
⎟ = 0.1 A
⎝ 10 + 40 ⎠ ⎝ 16 + 40||10 ⎠
After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be
so I sc = 0.3 A, Rt = 40 Ω, τ =
Finally:
L
20
1
=
=
s
Rt
40
2
i (t ) = (0.1 − 0.3)e −2t + 0.3 = 0.3 − 0.2e −2t A
P8.3-16. Consider the circuit shown in Figure P8.316.
a) Determine the time constant, τ, and the steady state
capacitor voltage when the switch is open.
b) Determine the time constant, τ, and the steady state
capacitor voltage when the switch is closed.
Figure P8.3-16.
Solution:
Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit.
⎛ 120 ⎞
a.) When the switch is open, v oc = ⎜
⎟ 24 = 16 V and R t = 120 || 60 = 40 Ω . The steady state
⎝ 60 + 120 ⎠
capacitor voltages is voc =16 V. The time constant is τ = (40)(0.02) = 0.8 s.
⎛ 120 ⎞
b.) When the switch is closed, v oc = ⎜
⎟ 24 = 19.2 V and R t = 120 || 60 || 60 = 24 Ω . The steady
⎝ 30 + 120 ⎠
state capacitor voltages is voc =19.2 V. The time constant is τ = (24)(0.02) = 0.48 s.
LNAPTR, 11/1/07
Figure P 8.3-17
P 8.3-17 The circuit shown in Figure P 8.3-17 is at steady state before the switch closes. The response
of the circuit is the voltage v(t). Find v(t) for t > 0.
Hint: After the switch closes, the inductor current is i(t) = 0.2(1 – e–1.8t) A
Answer: v(t) = 8 + e–1.8t V
Solution: Immediately before t = 0, i (0) = 0.
After t = 0, replace the circuit connected to the inductor by its Norton equivalent to calculate the
inductor current:
I sc = 0.2 A, Rt = 45 Ω, τ =
So i (t ) = 0.2 (1 − e−1.8t ) A
Now that we have the inductor current, we can calculate v(t):
d
i (t )
dt
= 8(1 − e −1.8t ) + 5(1.8)e −1.8t
v(t ) = 40 i (t ) + 25
= 8 + e −1.8t V for t > 0
L 25 5
=
=
Rth 45 9
Figure P 8.3-18
P 8.3-18 The circuit shown in Figure P 8.3-18 is at steady state before the switch closes. The response
of the circuit is the voltage v(t). Find v(t) for t > 0.
Answer: v(t) = 37.5 – 97.5e–6400t V
Solution: At steady-state, immediately before t = 0
so i(0) = 0.5 A.
After t > 0: Replace the circuit connected to the inductor by its Norton equivalent to get
I sc = 93.75 mA, Rt = 640 Ω,
τ =
L
.1
1
=
=
s
Rt 640 6400
i (t ) = 406.25 e−6400t + 93.75 mA
Finally:
v (t ) = 400 i (t ) + 0.1
d
i (t ) = 400 (.40625e −6400t + .09375) + 0.1 (−6400) (0.40625e −6400t )
dt
= 37.5 − 97.5e −6400t V
Figure P 8.3-19
P 8.3-19
≥ 0.
The circuit shown in Figure P 8.3-19 is at steady state before the switch closes. Find v(t) for t
Solution: Before the switch closes v(t) = 0 so v ( 0 + ) = v ( 0 − ) = 0 V .
For t > 0, we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor,
i.e. the part of the circuit to the left of the terminals a – b.
Write mesh equations to find voc:
Mesh equations:
12 i1 + 10 i1 − 6 ( i2 − i1 ) = 0
6 ( i2 − i1 ) + 3 i 2 − 18 = 0
28 i1 = 6 i 2
9 i 2 − 6 i1 = 18
36 i1 = 18 ⇒ i1 =
i2 =
Using KVL,
Find Rt:
⎛7⎞
⎛1⎞
voc = 3 i 2 + 10 i1 = 3 ⎜ ⎟ + 10 ⎜ ⎟ = 12 V
⎝ 3⎠
⎝ 2⎠
14 ⎛ 1 ⎞ 7
⎜ ⎟= A
3 ⎝2⎠ 3
1
A
2
Rt =
Then
and
12 (10 + 2 )
=6Ω
12 + (10 + 2 )
1
1
⎛ 1 ⎞ 1
=4
⎟= s ⇒
s
τ
⎝ 24 ⎠ 4
−4t
+ v oc = ( 0 − 12 ) e + 12 = 12 1 − e−4t
τ = R tC = 6 ⎜
v ( t ) = ( v ( 0 + ) − v oc ) e −t τ
(
)
V for t ≥ 0
(checked: LNAP 7/15/04)
P 8.3-20 The circuit shown in Figure P 8.3-20 is at
steady state before the switch closes. Determine i(t)
for t ≥ 0.
Figure P 8.3-20
Solution: Before the switch closes the circuit is at steady state so the inductor acts like a short circuit.
We have
⎞
1⎛
24
i ( t ) = ⎜⎜
⎟ = 0.8 A
2 ⎝ 5 + ( 20 & 20 ) ⎟⎠
so
i ( 0 + ) = i ( 0 − ) = 0.8 A
After the switch closes, find the Thevenin equivalent circuit for the part of the circuit connected to the
inductor.
Using voltage division twice
⎛ 20 1 ⎞
v oc = ⎜ − ⎟ 24 = 7.2 V
⎝ 25 2 ⎠
R t = ( 5 & 20 ) + ( 20 & 20 ) = 14 Ω
i sc =
v oc
Rt
=
7.2
= 0.514 A
14
Then
τ=
and
L 3.5 1
=
= s
R t 14 4
⇒
1
τ
=4
1
s
i ( t ) = ( i ( 0 + ) − i sc ) e −t τ + i sc = ( 0.8 − 0.514 ) e −4t + 0.514 = 0.286e−4t + 0.514 A for t ≥ 0
(checked: LNAP 7/15/04)
P8.3-21 The circuit in Figure P8.3-21 at steady
state before the switch closes. Determine an
equation that represents the capacitor voltage after
the switch closes.
Figure P8.3-21
Solution:
For t < 0, the switch is open and the capacitor acts like an open circuit because the circuit is at steady
state. Consequently, the current in the 10 Ω resistor is 0A and so the voltage across this resistor is 0 V.
KVL gives v ( t ) = 18 V . Immediately before the switch opens we have v ( 0 − ) = 18 V . The capacitor
voltage does not change instantaneously so v ( 0 + ) = v ( 0 − ) = 18 V .
For t > 0, the Thevenin equivalent of the part of the circuit connected to the capacitor is characterized by
40
R t = 10 || 40 = 8 Ω and, using voltage division, voc =
(18) = 14.4 V .
10 + 40
A = voc = 14.4 V , B = v ( 0 + ) − voc = 18 − 14.4 = 3.6 V and a =
1
τ
=
1
1
1
=
=5
R t C 8 ( 0.025 )
s
LNAPTR, 11/3/07
Figure P 8.3-22
P 8.3-22 The circuit shown in Figure P 8.3-22 is at steady state when the switch closes at time
t = 0. Determine i(t) for t ≥ 0.
Solution: Before t = 0, with the switch open and the circuit at steady state, the inductor acts like a short
circuit so we have
i (0 +) = i (0 −) = 4 A
After t = 0, we can replace the part of the circuit connected to the inductor by its Norton equivalent
circuit.
Using superposition, the short circuit current is
given by
⎛
⎞ ⎛ 3+8 ⎞
8
i sc = ⎜⎜
⎟⎟ 2 + ⎜⎜
⎟⎟ 4 = 3.75 A
⎝ 8 + ( 5 + 3) ⎠ ⎝ ( 3 + 8 ) + 5 ⎠
R t = 8 + 3 + 5 = 16 Ω
so
τ=
2
1
1
= 0.125 s ⇒
=8
16
s
τ
The inductor current is given by
i L ( t ) = ( i ( 0 + ) − i sc ) e −t τ + i sc = ( 4 − 3.75 ) e−8t + 3.75 = 3.75 − 0.25 e −8t A for t ≥ 0
(checked: LNAP 7/15/04)
P8.3-23 The circuit in Figure P8.3-23 at steady
state before the switch closes. Determine an
equation that represents the inductor current after
the switch closes.
Figure P8.3-23
Solution:
For t < 0, the switch is open and the inductor acts like short circuit because the circuit is at steady state.
Consequently, the current in inductor just before the switch closes is i ( 0 − ) = 7 Amps . The inductor
current does not change instantaneously so i ( 0 + ) = i ( 0 − ) = 7 Amps .
For t > 0, the Thevenin equivalent of the part of the circuit connected to the inductor is characterized by
20
R t = 20 + 60 = 80 Ω and, using current division, i sc =
( 7 ) = 1.75 Amps .
20 + 60
1 R t 80
1
=
= 32
A = i sc = 1.75 Amps , B = i ( 0 + ) − i sc = 7 − 1.75 = 5.25 Amps and a = =
τ L 2.5
s
LNAPTR, 11/3/07
P8.3-24 Consider the circuit shown in Figure 8.3-24a and corresponding plot of the inductor current
shown in Figure 8.3-24b. Determine the values of L, R1 and R 2 .
Answer: L = 4.8 H, R1 = 200 Ω and R 2 = 300 Ω.
Hint: Use the plot to determine values of D, E, F and a such that the inductor current can be represented
as
⎧ D for t ≤ 0
i (t ) = ⎨
− at
⎩ E + F e for t ≥ 0
Figure 8.3-24
Solution: From the plot
D = i(t) for t < 0 =120 mA = 0.12 A,
E + F = i(0+) = 120 mA = 0.12 A
and
E = lim i ( t ) = 200 mA = 0.2 A .
t →∞
The point labeled on the plot indicates that i(t)
= 160 mA when t = 27.725 ms = 0.027725 s.
Consequently
160 = 200 − 80 e − a ( 0.027725)
Then
⎧120 mA for t ≤ 0
i (t ) = ⎨
−25 t
mA for t ≥ 0
⎩ 200 − 80 e
⎛ 160 − 200 ⎞
ln ⎜
⎟
80
⎠ = 25 1
⇒ a= ⎝
−0.027725
s
When t < 0, the circuit is at steady state so the inductor
acts like a short circuit.
24
R1 =
= 200 Ω
0.12
As t → ∞, the circuit is again at steady state so the
inductor acts like a short circuit.
R1 || R 2 =
120 = 200 || R 2
24
= 120 Ω
0.2
⇒ R 2 = 300 Ω
Next, the inductance can be determined using the time constant:
25 = a =
1
τ
=
R1 || R 2
L
=
120
120
⇒ L=
= 4.8 H
25
L
P8.3-25 Consider the circuit shown in Figure P8.3-25a and corresponding plot of the voltage across the
40 Ω resistor shown in Figure P8.3-25b. Determine the values of L and R 2 .
Answer:
L = 8 H and R 2 = 10 Ω.
Hint: Use the plot to determine values of D, E, F and a such that the voltage can be represented as
⎧ D for t < 0
v (t ) = ⎨
− at
⎩ E + F e for t > 0
Figure P8.3-25
Solution: From the plot D = v(t) for t < 0 =20 V, E + F = v(0+) = 100 V and E = lim v ( t ) = 20 V . The
t →∞
point labeled on the plot indicates that v(t) = 60 V when t = 0.14 s. Consequently
60 = 20 + 80 e − a ( 0.14)
⎛ 60 − 20 ⎞
ln ⎜
⎟
1
80 ⎠
⇒ a= ⎝
=5
−0.14
s
Then
⎧20 V for t ≤ 0
v (t ) = ⎨
−5t
⎩ 20 + 80 e V for t ≥ 0
At t = 0+,
i (0 +) =
100
= 2.5 A
40
When t < 0, the circuit is at steady state so the inductor
acts like a short circuit.
40 || R 2 =
20
= 8 Ω ⇒ R 2 = 10 Ω
2.5
Next, the inductance can be determined using the time
constant:
1 40
40
5=a= =
⇒ L=
=8 H
5
τ L
Figure P8.3-26
P8.3-26 Determine vo(t) for t > 0 for the circuit shown in Figure P8.3-26.
Solution:
Replace the series inductors with an equivalent inductor and label the current in the inductor:
We will determine the inductor current, i(t) ,first and then use it to determine vo(t) . Determine the initial
condition, i(o), by considering the circuit when t < 0 and the circuit is at steady state. Since an inductor
in a dc circuit acts like a short circuit , we have
Using current division, we have
⎛ 18 ⎞
i ( 0) = ⎜
⎟ 2.4 = 1.2 mA
⎝ 18 + 18 ⎠
Next, consider the circuit when t > 0 and the circuit is not at steady state:
To find the Norton equivalent of the part of the circuit connected to the inductor we determine both the
Thevenin resistance and the short circuit current:
and
τ=
The time constant is:
L 1.5 1
=
= = 0.0556 second
R t 27 18
The inductor current is given by
i ( t ) = ( i ( 0 ) − i sc ) e −t /τ + i sc = (1.2 − 0.8 ) e −18t − 0.8 mA for t ≥ 0
Using KCL
Finally
vo (t )
9
+ ⎡⎣(1.2 − 0.8 ) e −18t − 0.8⎤⎦ ×10−3 = 2.4 × 10−3
v o ( t ) = 14.4 − 3.6 e −18t mV for t > 0
P8.3-27
The circuit shown in Figure P8.3-27 is at steady state before
the switch closes at time t = 0. After the switch closes, the
inductor current is given by
i ( t ) = 0.6 − 0.2 e −5t A
for t ≥ 0
Determine the values of R1 , R 2 and L.
Answers: R1 = 20 Ω , R 2 = 10 Ω and L = 5 H
Figure P8.3-27
Solution:
The steady state current before the switch closes is
equal to i ( 0 ) = 0.6 − 0.2 e−5( 0 ) = 0.4 A .
The inductor will act like a short circuit when this
circuit is at steady state so
0.4 = i ( 0 ) =
12
R1 + R 2
⇒ R1 + R 2 = 30 Ω
After the switch has been open for a long time, the
circuit will again be at steady state. The steady state
inductor current will be i ( ∞ ) = 0.6 − 0.2 e−5( ∞ ) = 0.6 A
The inductor will act like a short circuit when this
circuit is at steady state so
0.6 = i ( ∞ ) =
12
R1
⇒ R1 = 20 Ω
Then R 2 = 10 Ω.
After the switch is closed, the Thevenin resistance of the part of the circuit connected to the inductor is
R t = R1 . Then
5=
1
τ
=
Rt
L
=
R1
L
=
20
⇒ L=4H
L
P8.3-28 After time t = 0, a given circuit is represented
by the circuit diagram shown in FigureP8.3-28.
a.) Suppose that the inductor current is
i ( t ) = 21.6 + 28.4 e −4 t mA
for t ≥ 0
Determine the values of R1 and R 3 .
Figure P8.3-28
b.) Suppose instead that R1 = 16 Ω , R 3 = 20 Ω and the initial condition is i(0) = 10 mA. Determine the
inductor current for t ≥ 0.
Solution: The inductor current is given by i ( t ) = i sc + ( i ( 0 ) − i sc ) e− at
a. Comparing this to the given equation gives 21.6 = i sc =
4=
Rt
2
(
R1
R1 + 4
( 36 )
for t ≥ 0 where a =
⇒ R1 = 6 Ω and
)
⇒ R t = 8 Ω . Next 8 = R t = R1 + 4 || R 3 = 10 || R 3 ⇒ R 3 = 40 Ω .
10
16
= 5 s . also i sc =
( 36 ) = 28.8 mA . Then
τ 2
16 + 4
= 28.8 + (10 − 28.8 ) e−5t = 28.2 − 18.8 e−5t .
b. R t = (16 + 4 ) || 20 = 10 Ω so a =
i ( t ) = i sc + ( i ( 0 ) − i sc ) e − at
1
=
1
τ
=
Rt
L
.
P8.3-29 Consider the circuit shown in Figure P8.3-29.
a.) Determine the time constant, τ, and the steady state
capacitor voltage, v(∞), when the switch is open.
b.) Determine the time constant, τ, and the steady state
capacitor voltage, v(∞), when the switch is closed.
Answers: a.) τ = 3 s and v(∞) = 24 V; b.) τ = 2.25 s
and v(∞) = 2 V;
Figure P8.3-29
Solution: a.) When the switch is open we have
After replacing series and parallel resistors by equivalent resistors, the part of the circuit connected to
the capacitor is a Thevenin equivalent circuit with R t = 33.33 Ω . The time constant is
τ = R t C = 33.33 ( 0.090 ) = 3 s .
Since the input is constant, the capacitor acts like an open circuit when the circuit is at steady state.
Consequently, there is zero current in the 33.33 Ω resistor and KVL gives v(∞) = 24 V.
b.) When the switch is closed we have
This circuit can be redrawn as
Now we find the Thevenin equivalent of the part of the circuit connected to the capacitor:
So R t = 25 Ω and
τ = R t C = 25 ( 0.090 ) = 2.25 s
Since the input is constant, the capacitor acts like an open circuit when the circuit is at steady state.
Consequently, there is zero current in the 25 Ω resistor and KVL gives v(∞) = 12 V.
Section 8-4: Sequential Switching
P 8.4-1 The circuit shown in Figure P 8.4-1 is at steady state before the switch closes at time t
= 0. The switch remains closed for 1.5 s and then opens. Determine the capacitor voltage, v(t),
for t > 0.
Hint: Determine v(t) when the switch is closed. Evaluate v(t) at time t = 1.5 s to get v(1.5). Use
v(1.5) as the initial condition to determine v(t) after the switch opens again.
for 0 < t < 1.5 s
⎧
5 + 5e −0.5t V
Answer: v(t ) = ⎨
−2.5( t −1.5)
V for 1.5 s < t
⎩10 − 2.64e
Figure P 8.4-1
Solution:
Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to
get:
Before the switch closes at t = 0 the circuit is at steady state so v(0) = 10 V. For 0 < t < 1.5s, voc =
5 V and Rt = 4 Ω so τ = 4 × 0.05 = 0.2 s . Therefore
v (t ) = voc + (v (0) − voc ) e −t τ = 5 + 5e−5 t V for 0 < t < 1.5 s
At t =1.5 s, v (1.5) = 5 + 5e (
τ = 8 × 0.05 = 0.4 s . Therefore
−0.05 1.5)
= 5 V . For 1.5 s < t, voc = 10 V and Rt = 8 Ω so
v (t ) = voc + (v (1.5) − voc ) e
−( t −1.5 ) τ
= 10 − 5 e
−2.5 ( t −1.5 )
V for 1.5 s < t
Finally
⎧⎪ 5 + 5 e−5 t V
for 0 < t < 1.5 s
v (t ) = ⎨
−2.5 ( t −1.5)
for 1.5 s < t
V
⎪⎩10 − 5 e
P 8.4-2 The circuit shown in Figure P 8.4-2 is at steady state before the switch closes at time t
= 0. The switch remains closed for 1.5 s and then opens. Determine the inductor current, i(t), for
t > 0.
for 0 < 1 < 1.5 s
⎧
2 + e −0.5t A
Answer: v(t ) = ⎨
−0..667( t −1.5)
A for 1.5 s < t
⎩3 − 0.53e
Figure P 8.4-2
Solution:
Replace the part of the circuit connected to the inductor by its Norton equivalent circuit to get:
Before the switch closes at t = 0 the circuit is at steady state so i(0) = 3 A. For 0 < t < 1.5s, isc = 2
12
A and Rt = 6 Ω so τ = = 2 s . Therefore
6
i (t ) = isc + (i (0) − isc ) e−t τ = 2 + e−0.5 t A for 0 < t < 1.5 s
12
−0.5 1.5
At t =1.5 s, i (1.5) = 2 + e ( ) = 2.47 A . For 1.5 s < t, isc = 3 A and Rt = 8 Ω so τ = = 1.5 s .
8
Therefore
−( t −1.5 ) τ
−0.667 ( t −1.5 )
i (t ) = isc + (i (1.5) − isc ) e
= 3 − 0.53 e
V for 1.5 s < t
Finally
⎧⎪
2 + e−0.5 t A
for 0 < t < 1.5 s
i (t ) = ⎨
−0.667 ( t −1.5 )
for 1.5 s < t
A
⎪⎩3 − 0.53 e
P 8.4-3 Cardiac pacemakers are used by people to maintain regular heart rhythm when they
have a damaged heart. The circuit of a pacemaker can be represented as shown in Figure P 8.4-3.
The resistance of the wires, R, can be neglected since R < 1 mΩ. The heart’s load resistance, RL,
is 1 kΩ. The first switch is activated at t = t0, and the second switch is activated at t1 = t0 + 10 ms.
This cycle is repeated every second. Find v(t) for t0 ≤ t ≤ 1. Note that it is easiest to consider t0 =
0 for this calculation. The cycle repeats by switch 1 returning to position a and switch 2 returning
to its open position.
Hint: Use q = Cv to determine v(0–) for the 100-μF capacitor.
Figure P 8.4-3
Solution:
At t = 0-: Assume that the circuit has reached steady state so that the voltage across the 100 μF
capacitor is 3 V. The charge stored by the capacitor is
q ( 0− ) = (100 × 10−6 ) ( 3) = 300 × 10−6 C
0 < t < 10ms: With R negligibly small, the circuit reaches steady state almost immediately (i.e. at
t = 0+). The voltage across the parallel capacitors is determined by considering charge
conservation:
q ( 0+ ) = (100 μ F) v ( 0+ ) + (400 μ F) v ( 0+ )
v (0
+
) = 100 ×10
( )
q ( 0+ )
−6
+ 400 ×10−6
v 0+ = 0.6 V
10 ms < t < l s: Combine 100 μF & 400 μF in parallel to obtain
v(t ) = v ( 0+ ) e − (t −.01) RC
3
= 0.6e − (t −.01) (10 ) (5 x10
v(t ) = 0.6 e−2(t −.01) V
−4 )
=
q ( 0− )
500 × 10−6
=
300 × 10−6
500 × 10−6
P 8.4-4 An electronic flash on a camera uses the
circuit shown in Figure P 8.4-4. Harold E.
Edgerton invented the electronic flash in 1930. A
capacitor builds a steady-state voltage and then
discharges it as the shutter switch is pressed. The
discharge produces a very brief light discharge.
Determine the elapsed time t1 to reduce the
capacitor voltage to one-half of its initial voltage.
Find the current, i(t), at t = t1.
Solution:
Figure P 8.4-4
v ( 0 ) = 5 V , v ( ∞ ) = 0 and τ = 105 ×10−6 = 0.1 s
∴ v ( t ) = 5 e −10 t V for t > 0
2.5 = 5 e −10 t1
i (t1 ) =
v (t1 )
100 ×10
3
t 1 = 0.0693 s
=
2.5
= 25 μ A
100 × 103
P 8.4-5 The circuit shown in Figure P 8.4-5 is at steady state before the switch opens at t = 0.
The switch remains open for 0.5 second and then closes. Determine v(t) for t ≥ 0.
Figure P 8.4-5
Solution:
The circuit is at steady state before the switch
closes. The capacitor acts like an open circuit.
The initial condition is
⎛ 40 ⎞
v (0 +) = v (0 −) = ⎜
⎟ 24 = 12 V
⎝ 40 + 40 ⎠
After the switch closes, replace the part of the
circuit connected to the capacitor by its
Thevenin equivalent circuit.
R t = 6.67 Ω and v oc = 4 V
Recognize that
The time constant is
τ = R t C = ( 6.67 )( 0.05 ) = 0.335 s
⇒
1
τ
= 2.988 3
1
s
The capacitor voltage is
v ( t ) = ( v ( 0 + ) − v oc ) e−t τ + v oc = (12 − 4 ) e−3t + 4 = 4 + 8e−3t V for 0 ≥ t ≥ 0.5 s
When the switch opens again at time t = 0.5 the capacitor voltage is
−3 0.5
v ( 0.5 + ) = v ( 0.5 − ) = 4 + 8e ( ) = 5.785 V
After time t = 0.5 s, replace the part of the
circuit connected to the capacitor by its
Thevenin equivalent circuit.
Recognize that
R t = 20 Ω and v oc = 12 V
The time constant is
τ = R t C = 20 ( 0.05 ) = 1
1
⇒
τ
=1
1
s
The capacitor voltage is
− t −0.5 τ
−10 t − 0.5
v ( t ) = ( v ( 0.5 + ) − v oc ) e ( ) + v oc = ( 5.785 − 12 ) e ( ) + 12
= 12 − 6.215e
−10( t − 0.5)
V for t ≥ 0.5 s
so
⎧
12 V
⎪
v (t ) = ⎨
4 + 8 e −3t V
⎪
−( t − 0.5)
V
⎩12 − 6.215e
for t ≥ 0
for 0 ≤ t ≤ 0.5 s
for t ≥ 0.5 s
Section 8.5 Stability of First-Order Circuits
P 8.5-1 The circuit in Figure P 8.5-1 contains a current-controlled voltage source. What
restriction must be placed on the gain, R, of this dependent source in order to guarantee stability?
Answer: R < 400 Ω
Figure P 8.5-1
Solution:
This circuit will be stable if the Thèvenin equivalent resistance of the circuit connected to the
inductor is positive. The Thèvenin equivalent resistance of the circuit connected to the inductor
is calculated as
100
⎫
v T (400− R) 100
iT ⎪
=
100+ 400
⎬ ⇒ Rt =
iT
100+ 400
⎪
vT = 400 i (t ) − R i (t ) ⎭
i (t ) =
The circuit is stable when R < 400 Ω.
P 8.5-2 The circuit in Figure P 8.5-2 contains a current-controlled current source. What
restriction must be placed on the gain, B, of this dependent source in order to guarantee stability?
Figure P 8.5-2
Solution:
The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as
Ohm’s law: i ( t ) = −
vT ( t )
6000
KCL: i ( t ) + B i ( t ) + i T ( t ) =
vT ( t )
3000
⎛ v ( t ) ⎞ vT ( t )
∴ i T ( t ) = − ( B + 1) ⎜ − T
⎟+
⎝ 6000 ⎠ 3000
( B + 3) vT ( t )
=
6000
Rt =
vT ( t ) 6000
=
iT ( t ) B + 3
The circuit is stable when B > −3 A/A.
Section 8.6 The Unit Step Source
P 8.6-1
The input to the circuit shown in Figure P 8.6-1
is the voltage of the voltage source, vs(t). The output is the
voltage across the capacitor, vo(t). Determine the output of
this circuit when the input is vs(t) = 8 – 15 u(t) V.
Figure P 8.6-1
Solution:
The value of the input is one constant, 8 V, before time t = 0 and a different constant, −7 V, after
time t = 0. The response of the first order circuit to the change in the value of the input will be
vo ( t ) = A + B e − a t
for t > 0
where the values of the three constants A, B and a are to be determined.
The values of A and B are determined from the steady state responses of this circuit
before and after the input changes value.
Capacitors act like open circuits when the input is constant and the
circuit is at steady state. Consequently, the capacitor is replaced by
an open circuit.
The value of the capacitor voltage at time t = 0, will be
equal to the steady state capacitor voltage before the input
changes. At time t = 0 the output voltage is
The steady-state circuit for
t < 0.
vo ( 0 ) = A + B e
− a ( 0)
= A+ B
Consequently, the capacitor voltage is labeled as A + B. Analysis
of the circuit gives
A+ B = 8 V
Capacitors act like open circuits when the input is constant and the
circuit is at steady state. Consequently, the capacitor is replaced by
an open circuit.
The steady-state circuit for
t > 0.
The value of the capacitor voltage at time t = ∞, will be equal to
the steady state capacitor voltage after the input changes. At time t
= ∞ the output voltage is
vo ( ∞ ) = A + B e
− a (∞)
=A
Consequently, the capacitor voltage is labeled as A.
Analysis of the circuit gives
Therefore
A = -7 V and B = 15 V
The value of the constant a is determined from the time constant, τ, which is in turn
calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit
connected to the capacitor.
1
=τ = Rt C
a
Here is the circuit used to calculate Rt.
Rt = 6 Ω
Therefore
a=
1
1
= 2.5
−3
s
( 6 ) ( 66.7 ×10 )
(The time constant is τ = ( 6 ) ( 66.7 × 10−3 ) = 0.4 s .)
Putting it all together:
8 V
for t ≤ 0
⎧
vo ( t ) = ⎨
− 2.5 t
V for t ≥ 0
⎩−7 + 15 e
P 8.6-2 The input to the circuit shown in
Figure P 8.6-2 is the voltage of the voltage
source, vs(t). The output is the voltage across the
capacitor, vo(t). Determine the output of this
circuit when the input is vs(t) = 3 + 3 u(t) V.
Figure P 8.6-2
Solution:
The value of the input is one constant, 3 V, before time t = 0 and a different constant, 6 V, after
time t = 0. The response of the first order circuit to the change in the value of the input will be
vo ( t ) = A + B e − a t
for t > 0
where the values of the three constants A, B and a are to be determined.
The values of A and B are determined from the steady state responses of this circuit
before and after the input changes value.
Capacitors act like open circuits when the input is constant
and the circuit is at steady state. Consequently, the capacitor is
replaced by an open circuit.
The value of the capacitor voltage at time t = 0, will be
equal to the steady state capacitor voltage before the input
changes. At time t = 0 the output voltage is
The steady-state circuit for
t < 0.
vo ( 0 ) = A + B e
− a ( 0)
= A+ B
Consequently, the capacitor voltage is labeled as A + B.
Analysis of the circuit gives
A+ B =
6
( 3) = 2 V
3+ 6
Capacitors act like open circuits when the input is constant
and the circuit is at steady state. Consequently, the capacitor is
replaced by an open circuit.
The value of the capacitor voltage at time t = ∞, will be equal
to the steady state capacitor voltage after the input changes.
At time t = ∞ the output voltage is
The steady-state circuit for
t > 0.
vo ( ∞ ) = A + B e
− a (∞)
=A
Consequently, the capacitor voltage is labeled as A.
Analysis of the circuit gives
A=
6
( 6) = 4 V
3+ 6
Therefore
B = −2 V
The value of the constant a is determined from the time constant, τ, which is in turn
calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit
connected to the capacitor.
1
=τ = Rt C
a
Here is the circuit used to calculate Rt.
Rt =
( 3)( 6 ) = 2
3+ 6
Ω
Therefore
a=
1
( 2 )(.5)
=1
1
s
(The time constant is τ = ( 2 )( 0.5 ) = 1 s .)
Putting it all together:
for t ≤ 0
⎧ 2 V
vo ( t ) = ⎨
− t
⎩4 − 2 e V for t ≥ 0
P 8.6-3 The input to the circuit shown in Figure P
8.6-3 is the voltage of the voltage source, vs(t). The
output is the current across the inductor, io(t).
Determine the output of this circuit when the input
is vs(t) = – 7 + 13 u(t) V.
Figure P 8.6-3
Solution:
The value of the input is one constant, −7 V, before time t = 0 and a different constant, 6 V, after
time t = 0. The response of the first order circuit to the change in the value of the input will be
vo ( t ) = A + B e − a t
for t > 0
where the values of the three constants A, B and a are to be determined.
The values of A and B are determined from the steady state responses of this circuit
before and after the input changes value.
Inductors act like short circuits when the input is constant and
the circuit is at steady state. Consequently, the inductor is
replaced by a short circuit.
The value of the inductor current at time t = 0, will be
equal to the steady state inductor current before the input
changes. At time t = 0 the output current is
The steady-state circuit for
t < 0.
io ( 0 ) = A + B e
− a ( 0)
= A+ B
Consequently, the inductor current is labeled as A + B.
Analysis of the circuit gives
A+ B =
−7
= −1.4 A
5
Inductors act like short circuits when the input is constant and
the circuit is at steady state. Consequently, the inductor is
replaced by a short circuit.
The steady-state circuit for
t > 0.
The value of the inductor current at time t = ∞, will be equal to
the steady state inductor current after the input changes. At time
t = ∞ the output current is
io ( ∞ ) = A + B e
− a (∞)
=A
Consequently, the inductor current is labeled as A.
Analysis of the circuit gives
A=
6
= 1.2 A
5
Therefore
B = −2.6 V
The value of the constant a is determined from the time constant, τ, which is in turn
calculated from the values of the inductance L and of the Thevenin resistance, Rt, of the circuit
connected to the inductor.
1
L
=τ =
a
Rt
Here is the circuit used to calculate Rt.
Rt =
( 5) ( 4 ) = 2.22
5+ 4
2.22
1
a=
= 1.85
Therefore
1.2
s
1.2
(The time constant is τ =
= 0.54 s .)
2.22
Putting it all together:
−1.4 A
for t ≤ 0
⎧
io ( t ) = ⎨
− 1.85 t
A for t ≥ 0
⎩1.2 − 2.6 e
Ω
P8.6-4 Determine vo(t) for t > 0 for the circuit shown in Figure P8.6-4.
Figure P8.6-4
Solution: Determine the initial condition, vo(o), by considering the circuit when t < 0 and the
circuit is at steady state. Since a capacitor in a dc circuit acts like a open circuit , we have
Recognizing the inverting amplifier, we have
⎛ 45 ⎞
v o ( 0 ) = ⎜ − ⎟ 2.4 = −3.6 V
⎝ 30 ⎠
Next, consider the circuit when t > 0 and the circuit is not at steady state:
To find the Thevenin equivalent of the part of the circuit connected to the capacitor we determine
both the open circuit voltage and the short circuit current:
and
Now we calculate the Thevenin resistance:
v oc
−5.4
Rt =
=
= 20 kΩ
i sc −0.27 × 10−3
and the time constant:
τ = R t C = ( 20 ×103 )( 5 × 10−6 ) = 0.1 second
The capacitor voltage is given by
v o ( t ) = ( v o ( 0 ) − v oc ) e −t /τ + v oc = ( −3.6 − ( −5.4 ) ) e −10 t − 5.4 = 1.8 e −10 t − 5.4 V for t ≥ 0
P 8.6-5 The initial voltage of the capacitor of the circuit
shown in Figure P 8.6-5 is zero. Determine the voltage v(t)
when the source is a pulse, described by
t <1 s
⎧0
⎪
vs = ⎨4 V 1 < t < 2 s
⎪0
t>2s
⎩
Solution
Figure P 8.6-5
τ = R C = ( 5 × 105 )( 2 × 10−6 ) = 1 s
Assume that the circuit is at steady state at t = 1−. Then
v ( t ) = 4 − 4 e − ( t −1) V for 1 ≤ t ≤ 2
so
v ( 2 ) = 4 − 4 e − (2−1) = 2.53 V
and
v ( t ) = 2.53 e − ( t − 2) V for t ≥ 2
Finally
⎧0
⎪
v(t ) = ⎨4− 4e − (t −1)
⎪2.53e − (t − 2)
⎩
t ≤1
1≤t ≤ 2
t ≥2
P 8.6-6 Studies of an artificial insect are being used to understand the nervous system of
animals. A model neuron in the nervous system of the artificial insect is shown in Figure P 8.6-6.
A series of pulses, called synapses, is required. The switch generates a pulse by opening at t = 0
and closing at t = 0.5 s. Assume that the circuit is in steady state and that v(0–) = 10 V.
Determine the voltage v(t) for 0 < t < 2 s.
Figure P 8.6-6
Solution:
The capacitor voltage is v(0−) = 10 V immediately before the switch opens at t = 0.
For 0 < t < 0.5 s the switch is open:
v ( 0 ) = 10 V, v ( ∞ ) = 0 V, τ = 3 ×
1 1
=
s
6 2
so v ( t ) = 10 e − 2 t V
In particular, v ( 0.5 ) = 10 e
For t > 0.5 s the switch is closed:
− 2 ( 0.5)
= 3.679 V
v ( 0 ) = 3.679 V, v ( ∞ ) = 10 V, Rt = 6 || 3 = 2 Ω,
1
6
τ = 2× =
so
1
s
3
v ( t ) = 10 + ( 3.679 − 10 ) e
= 10 − 6.321 e
− 3 ( t − 0.5 )
− 3 ( t − 0.5)
V
V
P8.6-7 Determine the voltage v o ( t ) in the
circuit shown in Figure P8.6-7.
Figure P8.6-7
Solution: This is a first order circuit containing an inductor. First, determine i L ( t ) .
Consider the circuit for time t < 0.
Step 1: Determine the initial inductor current.
The circuit will be at steady state before the
source voltage changes abruptly at time t = 0 .
The source voltage will be 2 V, a constant.
The inductor will act like a short circuit.
i L ( 0) =
2
2
= = 0.25 A
10 || ( 25 + 15 ) 8
Consider the circuit for time t > 0.
Step 2. The circuit will not be at steady state
immediately after the source voltage changes
abruptly at time t = 0 . Determine the Norton
equivalent circuit for the part of the circuit
connected to the inductor.
Replacing the resistors by an equivalent
resistor, we recognize
v oc = −6 V and R t = 8 Ω
Consequently
i sc =
−6
= −0.75 A
8
t < 0 , at steady state:
Step 3. The time constant of a first order circuit
containing an inductor is given by
τ=
Consequently
τ=
L
Rt
L 4
1
1
= = 0.5 s and a = = 2
Rt 8
s
τ
Step 4. The inductor current is given by:
i L ( t ) = i sc + ( i ( 0 ) − i sc ) e − at = −0.75 + ( 0.25 − ( −0.75 ) ) e −2 t = −0.75 + e −2 t for t ≥ 0
Step 5. Express the output voltage as a
function of the source voltage and the inductor
current.
Using current division:
iR =
10
i L = 0.2 i L
10 + ( 25 + 15 )
Then Ohm’s law gives
v o = 15 i R = 3 i L
Step 6. The output voltage is given by
v o ( t ) = −2.25 + 3 e −2 t for t ≥ 0
P 8.6-8
Determine vc(t) for t > 0 for the circuit of Figure P 8.6-8.
Figure P 8.6-8
Solution:
For t < 0, the circuit is:
After t = 0, replace the part of the circuit connected to the capacitor by its Thevenin equivalent
circuit to get:
vc ( t ) = −15 + ( −6 − ( −15 ) ) e
= −15 + 9 e − 5 t V
− t / ( 4000×0.00005 )
P 8.6-9 The voltage source voltage in the circuit shown in
Figure P 8.6-9 is
vs(t) = 7 – 14u (t) V
Determine v(t) for t > 0.
Figure P 8.6-9
Solution:
The input changes abruptly at time t = 0. The voltage v(t) may
not be continuous at t = 0, but the capacitor voltage, vC(t) will
be continuous. We will find vC(t) first and then use KVL to
find v(t).
The circuit will be at steady state before t = 0 so the capacitor
will act like an open circuit.
v (0 +) = v (0 −) =
5
7 = 4.375 V
5+3
After t = 0, we replace the part of the circuit connected to the
capacitor by its Thevenin equivalent circuit.
5
5
v oc = ( 7 − 14 ) = ( −7 ) = −4.375 V
8
8
R t = 5 & 3 = 1.875 Ω
The time constant is
τ = R t C = 0.8625 s
1
So
v C ( t ) = ⎡⎣ 4.375 − ( −4.375 ) ⎤⎦ e −1.16t
1
s
τ
+ ( −4.375 ) = −4.375 + 8.75e−1.16t V for t ≥ 0
= 1.16
Using KVL
v ( t ) = v s ( t ) − vC ( t ) = −7 − ⎡⎣ −4.375 + 8.75e −1.16t ⎤⎦ = −2.625 − 8.75e−1.16t V for t > 0
P 8.6-10 Determine the voltage v(t) for t ≥ 0
for the circuit shown in Figure P 8.6-10.
Figure P 8.6-10
Solution:
For t < 0
Using voltage division twice
v (t ) =
so
32
30
5−
5 = 0.25 V
32 + 96 120 + 30
v ( 0 − ) = 0.25 V
and
v ( 0 + ) = v ( 0 − ) = 0.25 V
For t > 0, find the Thevenin equivalent circuit for the part of the circuit connected to the
capacitor.
Using voltage division twice
v oc =
32
30
20 −
20 = 5 − 4 = 1 V
32 + 96
120 + 30
R t = ( 96 || 32 ) + (120 || 30 ) = 24 + 24 = 48 Ω
then
τ = 48 × 0.0125 = 0.6 s
so
1
τ
Now
= 1.67
1
s
v ( t ) = [ 0.25 − 1] e −1.67t + 1 = 1 − 0.75e −1.67t V for t ≥ 0
(checked: LNAP 7/1/04)
P 8.6-11
The voltage source voltage in the circuit shown in Figure P 8.6-11 is
vs(t) = 5 + 20u (t)
Determine i(t) for t ≥ 0.
Figure P 8.6-11
Solution:
For t > 0 the circuit is at steady state so the inductor acts like a short circuit:
Apply KVL to the supermesh corresponding to
the dependent source to get
−5000i b + 1000 ( 3i b ) − 5 = 0
⇒
i b = 0.2 mA
Apply KVL to get
i ( t ) = 3i b = 0.6 mA
i ( 0 − ) = 0.6 mA
so
i ( 0 + ) = i ( 0 − ) = 0.6 mA
and
For t > 0, find the Norton equivalent circuit for the part of the circuit that is connected to the
inductor.
Apply KCL at the top node of the dependent
source to see that i b = 0 A . Then
( )
v oc = 25 − 5000 i b = 25 V
Apply KVL to the supermesh corresponding to
the dependent source to get
( )
−5000 i b + 10000 3 i b − 25 = 0 ⇒ i b = 1 mA
Apply KCL to get
i sc = 3 i b = 3 mA
Then
Rt =
v oc
i sc
= 8.33 kΩ
τ=
Then
So
Now
25
= 3 ms
8333
1
1
= 333
τ
s
i ( t ) = [ 0.6 − 3] e −333t + 3 = 3 − 2.4e −333t mA for t ≥ 0
(checked: LNAP 7/2/04)
P 8.6-12
The voltage source voltage in the
circuit shown in Figure P 8.6-12 is
vs(t) = 12 – 6u (t) V
Determine v(t) for t ≥ 0.
Figure P 8.6-12
Solution:
For t > 0, the circuit is at steady state so the capacitor acts like an open circuit. We have the
following situation.
Notice that v(t) is the node voltage at node a.
Express the controlling voltage of the dependent
source as a function of the node voltage:
va = −v(t)
Apply KCL at node a:
⎛ 12 − v ( t ) ⎞ v ( t ) ⎛ 3
⎞
−⎜
+ ⎜ − v (t ) ⎟ = 0
⎟+
8
4
⎝ 4
⎠
⎝
⎠
−12 + v ( t ) + 2 v ( t ) − 6 v ( t ) = 0 ⇒ v ( t ) = −4 V
So
v ( 0 + ) = v ( 0 − ) = −4 V
For t > 0, we find the Thevenin equivalent circuit for the part of the circuit connected to the
capacitor, i.e. the part of the circuit to the left of terminals a – b.
Notice that voc is the node voltage at node a. Express
the controlling voltage of the dependent source as a
function of the node voltage:
va = −voc
Apply KCL at node a:
⎛ 6 − voc ⎞ voc ⎛ 3 ⎞
−⎜
+ ⎜ − voc ⎟ = 0
⎟+
⎝ 8 ⎠ 4 ⎝ 4 ⎠
−6 + voc + 2 voc − 6 voc = 0 ⇒ voc = −2 V
Find Rt:
We’ll find isc and use it to calculate Rt. Notice that
the short circuit forces
va = 0
Apply KCL at node a:
⎛ 6−0⎞ 0 ⎛ 3 ⎞
−⎜
⎟ + + ⎜ − 0 ⎟ + i sc = 0
⎝ 8 ⎠ 4 ⎝ 4 ⎠
i sc =
Rt =
6 3
= A
8 4
voc −2
8
=
=− Ω
i sc 3 4
3
Then
⎛ 8 ⎞⎛ 3 ⎞
1
τ = R tC = ⎜ − ⎟ ⎜ ⎟ = − s
5
⎝ 3 ⎠ ⎝ 40 ⎠
and
⇒
1
τ
= −5
1
s
v ( t ) = ( v ( 0 + ) − v oc ) e −t τ + v oc = ( −4 − ( −2 ) ) e5t + ( −2 ) = −2 (1 + e5t ) V for t ≥ 0
Notice that v(t) grows exponentially as t increases.
(checked: LNAP 7/8/04)
P 8.6-13
Determine i(t) for t ≥ 0 for the circuit shown in Figure P 8.6-13.
Figure P 8.6-13
Solution: When t < 0 and the circuit is at steady state, the inductor acts like a short circuit.
The mesh equations are
(
)
(
)
2 i x + 4 i x + i (0 −) + 3i x + 6 i x + i (0 −) = 0
1i ( 0 − ) − 3 i x = 0
so
i (0 +) = i (0 −) = 0
For t ≥ 0 , we find the Norton equivalent circuit for the part of the circuit connected to the
inductor. First, simplify the circuit using a source transformation:
Identify the open circuit voltage and short circuit current.
Apply KVL to the mesh to get:
(10 + 2 + 3) i x − 15 = 0
⇒ ix = 1 A
Then
v oc = 3 i x = 3 V
Express the controlling current of the
CCVS in terms of the mesh currents:
i x = i1 − i sc
The mesh equations are
10 i1 + 2 ( i1 − i sc ) + 3 ( i1 − i sc ) − 15 = 0 ⇒ 15 i1 − 5 i sc = 15
and
i sc − 3 ( i1 − i sc ) = 0 ⇒ i1 =
4
i sc
3
so
⎛4 ⎞
15 ⎜ i sc ⎟ − 5 i sc = 15 ⇒ i sc = 1 A
⎝3 ⎠
The Thevenin resistance is
Rt =
The time constant is given by τ =
3
=3Ω
1
L 5
1
1
= = 1.67 s so = 0.6 .
Rt 3
τ
s
The inductor current is given by
i L ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( 0 − 1) e−0.6 t + 1 = 1 − e −0.6t A for t ≥ 0
P 8.6-14
Determine i(t) for t ≥ 0 for the circuit shown in Figure P 8.6-14.
Figure P 8.6-14
Solution:
When t < 0 and the circuit is at steady state, the
inductor acts like a short circuit. The initial condition
is
0
2
i (0 +) = i (0 −) =
+
= 0.02 A
150 100
For t ≥ 0 , we find the Norton equivalent circuit for the part of the circuit connected to the
inductor. First, simplify the circuit using source transformations:
i sc = 20 + 40 = 60 mA
R t = 100 ||150 =
The time constant is given by τ =
100 × 150
= 60 Ω
100 + 150
L
2
1
1
=
= 0.0333 s so = 30 .
R t 60
τ
s
The inductor current is given by
i L ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( 20 − 60 ) e −30 t + 60 = 60 − 40 e −30 t mA for t ≥ 0
(checked: LNAP 7/15/04)
P 8.6-15 Determine v(t) for t ≥ 0 for the
circuit shown in Figure P 8.6-15.
Figure P 8.6-15
Solution:
When t < 0 and the circuit is at steady state, the
capacitor acts like an open circuit. The 0 A current
source also acts like an open circuit. The initial
condition is
v (0 + ) = v (0 −) = 0 V
For t ≥ 0 , we find the Thevenin equivalent circuit for the part of the circuit connected to the
capacitor.
30
⎡ 170
voc = ⎢
( 20 ) ⎤⎥ 10 − ⎡⎢
( 20 ) ⎤⎥ 50
⎣170 + 30
⎦
⎣170 + 30
⎦
=
170(20)(10) − 30(20)(50) 4000
=
= 20 V
200
200
Rt = 8 +
( 20 + 120 ) (10 + 50 ) = 50 Ω
( 20 + 120 ) + (10 + 50 )
The time constant is τ = R t C = ( 50 ) (10−3 ) = 0.05 s s so
1
τ
= 20
1
.
s
The capacitor voltage is given by
(
v ( t ) = ( v ( 0 + ) − v oc ) e −t τ + v oc = ( 0 − 20 ) e −20t + 20 = 20 1 − e−20t
)
V for t ≥ 0
P 8.6-16
Determine v(t) for t ≥ 0 for the circuit shown in Figure P 8.6-16.
Figure P 8.6-16
Solution:
When t < 0 and the circuit is at steady
state, the capacitor acts like an open circuit.
The 0 A current source also acts like an
open circuit.
After a couple of source transformations,
the initial condition is calculated as
v (0 +) = v (0 −) =
18
16 = 10.667 V
9 + 18
For t ≥ 0 , we find the Thevenin equivalent circuit for the part of the circuit connected to the
capacitor. Using source transformations, reduce the circuit as follows.
Now recognize R t = 10 Ω and v oc = 4 V .
The time constant is τ = R t C = (10 ) ( 20 ×10−3 ) = 0.2 s s so
1
τ
=5
1
.
s
The capacitor voltage is given by
v ( t ) = ( v ( 0 + ) − v oc ) e −t τ + v oc = (10.667 − 4 ) e−5t + 4 = 4 + 6.667 e −5t V for t ≥ 0
(checked: LNAP 7/15/04)
P 8.6-17
Determine i(t) for t ≥ 0 for the circuit shown in Figure P 8.6-17.
Figure P 8.6-17
Solution:
When t < 0 and the circuit is at steady state, the
inductor acts like a short circuit. The 0 V
voltage source also acts like a short circuit.
After a replacing series and parallel resistors by
equivalent resistors, the equivalent resistors,
current source and short circuit are all
connected in parallel. Consequently
i (0 +) = i (0 −) = 2 A
For t ≥ 0 , we find the Thevenin equivalent circuit for the part of the circuit connected to the
capacitor.
Replace series and parallel resistors by an
equivalent resistor.
18 & (12 + 24 ) = 12 Ω
Do a source transformation, then replace series
voltage sources by an equivalent voltage
source.
Do two more source transformations.
Now recognize R t = 8 Ω and i sc = 3 A .
The time constant is given by
τ=
L 2
1
1
= = 0.25 s so = 4 .
τ
Rt 8
s
The inductor current is given by
i L ( t ) = ( i ( 0 + ) − i sc ) e −t τ + i sc = ( 2 − 3) e −4t + 3 = 3 − e −4t A for t ≥ 0
(checked: LNAP 7/15/04)
P 8.6-18
The voltage source voltage in the circuit shown in Figure P 8.6-18 is
vs(t) = 8 + 12u(t) V
Determine v(t) for t ≥ 0.
Figure P 8.6-18
Solution: Simplify the circuit by replacing the parallel capacitors by an equivalent capacitor and
by doing a couple of source transformations:
For t < 0 the circuit is at steady state so the
capacitor acts like an open circuit. The voltage
source voltage is 6.4 V, so
v ( 0 + ) = v ( 0 − ) = 6.4 V
For t > 0 we find the Thevenin equivalent circuit of the part of the circuit connected to the
capacitor. In this case we recognize the voc = 16 V and Rt = 50 Ω.
The time constant is
τ = R t C = ( 50 ) ( 8 × 10−3 ) = 0.4 s
⇒
1
τ
= 2.5
1
s
Then
−t
v ( t ) = ( v ( 0 + ) − v oc ) e τ + v oc = ( 6.4 − 16 ) e −2.5t + 16 = 16 − 9.6e −2.5t V
for t ≥ 0
(checked: LNAP 7/12/04)
P8.6-19 Determine the current i o ( t ) in the
circuit shown in Figure P8.6-19.
Figure P8.6-19
Solution: This is a first order circuit containing a capacitor. First, determine v C ( t ) .
Consider the circuit for time t < 0.
Step 1: Determine the initial capacitor voltage.
t < 0 , at steady state:
The circuit will be at steady state before the
source voltage changes abruptly at time t = 0 .
The source voltage will be 5 V, a constant.
The capacitor will act like an open circuit.
Apply KVL to the mesh to get:
(10 + 2 + 3) i x − 5 = 0
Then
⇒ ix =
1
A
3
v C (0) = 3 i x = 1 V
Consider the circuit for time t > 0.
Step 2. The circuit will not be at steady state immediately after the source voltage changes
abruptly at time t = 0 . Determine the Thevenin equivalent circuit for the part of the circuit
connected to the capacitor. First, determine the open circuit voltage, v oc :
Apply KVL to the mesh to get:
(10 + 2 + 3) i x − 15 = 0
⇒ ix = 1 A
Then
v oc = 3 i x = 3 V
Next, determine the short circuit current, i sc :
Express the controlling current of the
CCVS in terms of the mesh currents:
i x = i1 − i sc
The mesh equations are
10 i1 + 2 ( i1 − i sc ) + 3 ( i1 − i sc ) − 15 = 0 ⇒ 15 i1 − 5 i sc = 15
i sc − 3 ( i1 − i sc ) = 0 ⇒ i1 =
And
4
i sc
3
⎛4 ⎞
15 ⎜ i sc ⎟ − 5 i sc = 15 ⇒ i sc = 1 A
⎝3 ⎠
so
The Thevenin resistance is
Rt =
3
=3Ω
1
Step 3. The time constant of a first order circuit containing
an capacitor is given by
τ = Rt C
Consequently
1
1
⎛1⎞
⎟ = 0.25 s and a = = 4
τ
s
⎝ 12 ⎠
τ = Rt C = 3⎜
Step 4. The capacitor voltage is given by:
(
)
v C ( t ) = v oc + v C ( 0 ) − v oc e − at = 3 + (1 − 3) e−4 t = 3 − 2 e−4 t for t ≥ 0
Step 5. Express the output current as a function of the source voltage and the capacitor voltage.
io (t ) = C
d
1 d
v C (t ) =
vC (t )
dt
12 dt
Step 6. The output current is given by
io (t ) =
1 d
1
2
3 − 2 e −4 t = ( −2 )( −4 ) e −4 t = e −4 t for t ≥ 0
12 dt
12
3
(
)
P 8.6-20
The voltage source voltage in the circuit shown in Figure P 8.6-20 is
vs(t) = 25u(t) – 10 V
Determine i(t) for t ≥ 0.
Figure P 8.6-20
Solution: Simplify the circuit by replacing the series inductors by an equivalent inductor. Then,
after a couple of source transformations, we have
For t < 0 the circuit is at steady state and so the
inductor acts like a short circuit. The voltage
source voltage is −6 V so
i ( 0 + ) = i ( 0 − ) = −60 mA
For t > 0 we find the Norton equivalent circuit for the part of the circuit connected to the
inductor. In this case we recognize voc = 9V and Rt = 100 Ω so isc = 90 mA.
The time constant is
Then
τ=
L
20
=
= 0.2 s
R t 100
⇒
1
τ
=5
1
s
i ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( −60 − 90 ) e −5t + 90 = 90 − 150e−5t mA for t ≥ 0
(checked: LNAP 7/12/04)
P 8.6-21
The voltage source voltage in the circuit shown in Figure P 8.6-21 is
vs(t) = 30 – 24u(t) V
Determine i(t) for t ≥ 0.
Figure P 8.6-21
Solution: Simplify the circuit by replacing the parallel inductors by an equivalent inductor.
Then, after doing a couple of source transformations, we have
For t < 0 the circuit is at steady state and the
inductor acts like a short circuit. The voltage
source voltage is 12 V so
i ( 0 + ) = i ( 0 − ) = 0.2 A
For t > 0 we find the Norton equivalent circuit for the part of the circuit connected to the
inductor. In this case, we recognize voc = 2.4 V and Rt = 60 Ω so isc = 0.04 A.
The time constant is
τ=
Then
L
4
1
=
=
s
R t 60 15
⇒
1
τ
= 15
1
s
i ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( 0.2 − 0.04 ) e−15t + 0.04 = 40 + 160e −15t mA
(checked: LNAP 7/13/04)
P 8.6-22
The voltage source voltage in the circuit shown in Figure P 8.6-22 is
vs(t) = 10 + 40u(t) V
Determine v(t) for t ≥ 0.
Figure P 8.6-22
Solution: Simplify the circuit by replacing the series capacitors by an equivalent capacitor.
Then, after doing some source transformations, we have
For t < 0 the circuit is at steady state so the
capacitor acts like an open circuit. The voltage
source voltage is 8 V so
v (0 +) = v (0 −) = 8 V
For t > 0 we find the Thevenin equivalent circuit of the part of the circuit connected to the
capacitor. In this case we recognize voc = 40 V and Rt = 8 Ω.
The time constant is
τ = R t C = ( 8 ) ( 60 ×10−3 ) = 0.48
Then
⇒
1
τ
= 2.08
1
s
v ( t ) = ( v ( 0 + ) − v oc ) e −t τ + v oc = ( 8 − 40 ) e −2.08t + 40 = 40 − 32e−2.08t V for t ≥ 0
(checked: LNAP 7/13/04)
P 8.6-23
Determine v(t) for t > 0 for the circuit shown in Figure P 8.6-23.
Figure P 8.6-23
Solution: The resistor voltage, v(t), may not be
continuous at time t = 0. The inductor will be
continuous. We will find the inductor current
first and then find v(t). Label the inductor
current as i(t).
For t < 0 the circuit is at steady state and the
inductor acts like a short circuit. The initial
condition is
i (0 +) = i (0 −) = 0 A
For t > 0 use source transformations to
simplify the part of the circuit connected to the
inductor until is a Norton equivalent circuit.
Recognize that
R t = 2 Ω and i sc = 1.333 A
The time constant is
τ=
L 3
=
Rt 2
⇒
1
τ
= 0.667
1
s
Then
(
i ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = 1.333 1 − e−0.667t
)
A for t ≥ 0
Returning to the original circuit we see that
d
3 i (t )
v (t )
d
= i ( t ) + dt
= i (t ) + i (t )
dt
2
3
= 1.333 (1 − e −0.667 t ) + ( −0.667 )(1.333) ( −e −0.667 t ) = 1.333 − 0.4439e−0.667
Finally
v ( t ) = 2.667 − 0.889e0.667 t V for t > 0
(checked: LNAP 7/14/04)
P 8.6-24
The input to the circuit shown in Figure P 8.6-24 is the current source current
is(t) = 2 + 4u(t) A
The output is the voltage v(t). Determine v(t) for t > 0.
Figure P 8.6-24
Solution:
Label the inductor current, i(t).
We will find i(t) first, then find
v(t).
For t < 0 the circuit is at steady
state and the inductor acts like a
short circuit. The initial condition
is
⎛ 3 ⎞
i (0 +) = i (0 −) = ⎜
⎟2 =1 A
⎝ 3+3⎠
For t > 0 use source transformations to
simplify the part of the circuit connected to
the inductor until it is a Norton equivalent
circuit.
Recognize that R t = 2 Ω and i sc = 3 A
The time constant is
Then
τ=
L 0.25
=
= 0.125 s
Rt
2
⇒
1
τ
=8
1
s
i ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = (1 − 3) e−8t + 3 = 3 − 2e −8t A for t ≥ 0
Returning to the original circuit
d
⎛
⎞
0.25 i ( t ) ⎟
⎜
d
d
−8t
−8t
dt
v (t ) = 3⎜ i (t ) +
⎟ + 0.25 i ( t ) = 3i ( t ) + 0.5 i ( t ) = 3 ( 3 − 2e ) + 0.5 (16e )
3
dt
dt
⎜⎜
⎟⎟
⎝
⎠
−8t
= 9 + 2 e V for t > 0
(checked: LNAP 7/26/04)
P 8.6-25
The input to the circuit shown in Figure P 8.6-25 is the voltage source voltage
vs = 6 + 6u (t)
The output is the voltage vo(t). Determine vo(t) for t > 0.
Figure P 8.6-25
Solution:
Label the capacitor voltage, v(t). W will
find v(t) first then find vo(t).
For t < 0 the circuit is at steady state and
the capacitor acts like an open circuit. The
initial condition is
v (0 +) = v (0 −) = 6 V
For t > 0 we replace series and then parallel
resistors by equivalent resistors in order to replace
the part of the circuit connected to the capacitor by
its Thevenin equivalent circuit.
We recognize R t = 4 Ω and v oc = 12 V
The time constant is
τ = R t C = 4 ( 0.125 ) = 0.5 s
⇒
1
τ
=2
1
s
The capacitor voltage is given by
−t
v ( t ) = ( v ( 0 + ) − v oc ) e τ + v oc = ( 6 − 12 ) e −2t + 12 = 12 − 6e−2t V for t ≥ 0
Returning to the original circuit and applying KCL we see
0.125
so
v o ( t ) = 0.25
12 − v ( t ) v o ( t )
d
v (t ) =
+
6
2
dt
v (t )
d
= 0.25 (12e −2t ) − 4 + 4 − 2e−2t = e −2t V for t > 0
v (t ) − 4 +
3
dt
(checked: LNAP 7/26/04)
P 8.6-26 Determine v(t) for t > 0 for the circuit shown
in Figure P 8.6-26.
Figure P 8.6-26
Solution:
Label the inductor current as i(t). We will find i(t)
first then use it to find v(t).
For t < 0 the circuit is at steady state and the inductor
acts like a short circuit. The initial condition is
i (0 +) = i (0 −) = 0 A
For t > 0 we replace series and parallel resistors by
equivalent resistors. Then the part of the circuit
connected to the inductor will be a Thevenin equivalent
circuit.
We recognize
R t = 2 Ω and v oc = 12 V
so
i sc =
v oc
Rt
=6A
The time constant is
τ=
L 0.5
=
= 0.25
Rt
2
⇒
1
τ
=4
1
s
The inductor current is given by
(
i ( t ) = ( i ( 0 + ) − i sc ) e −t τ + i sc = ( 0 − 6 ) e−4t + 6 = 6 1 − e−4t
)
A for t ≥ 0
Returning to the original circuit and applying KCL we see
i (t ) +
0.5
d
i ( t ) − 12 v t
()
dt
=
3
4
so
v ( t ) = 4i ( t ) +
2d
⎛ 2⎞
i ( t ) − 16 = 24 (1 − e −4t ) + ⎜ ⎟ ( 24e −4t ) − 16 = 8 − 8e −4t V for t > 0
3 dt
⎝ 3⎠
(checked: LNAP 7/26/04)
P 8.6-27 When the input to the circuit shown
in Figure P 8.6-27 is the voltage source voltage
vs(t) = 3 – u(t) V
the output is the voltage
vo(t) = 10 + 5 e–50t V for t ≥ 0
Determine the values of R1 and R2.
Figure P 8.6-27
Solution: Apply KCL at the inverting input of the op amp to get
v o (t ) − v (t )
R2
=
R2 ⎞
⎛
v (t )
⇒ v o ( t ) = ⎜1 +
⎟ v (t )
1000
⎝ 1000 ⎠
We will determine the capacitor voltage first and then use it to determine the output voltage.
When t < 0 and the circuit is at steady state,
the capacitor acts like an open circuit. Apply
KCL at the noninverting input of the op amp to
get
3 − v (0 −)
= 0 ⇒ v (0 −) = 3 V
R1
The initial condition is
v (0 +) = v (0 −) = 3 V
For t ≥ 0 , we find the Thevenin equivalent circuit for the part of the circuit connected to the
capacitor.
2 − voc
= 0 ⇒ voc = 2 V
R1
2
= i sc
R1
⇒ Rt =
voc
= R1
i sc
The time constant is τ = R t C = R t (10−6 ) . From the given equation for v o ( t ) ,
R t (10−6 ) =
1
τ
= 50
1
, so
s
6
1
10
⇒ R1 = R t =
= 20 kΩ
50
50
The capacitor voltage is given by
v ( t ) = ( v ( 0 ) − v oc ) e −t τ + v oc = ( 3 − 2 ) e −50t + 2 = 2 + e−50t V for t ≥ 0
So
v o (t ) = 5 v (t ) ⇒ 5 = 1 +
R2
1000
⇒ R 2 = 4 kΩ
(checked LNAPTR 7/31/04)
P8.6-28
The time constant of a particular circuit is τ = 0.25 s. In response to a step input, a capacitor
voltage changes from −2.5 V to 4.2 V. How long did it take for the capacitor voltage to increase
from −2.0 V to +2.0 V?
Solution:
The capacitor voltage can be represented by the equation v ( t ) = A + B e − 4t for t ≥ 0. Given that
A + B = v(0) = −2.5 V and A = v(∞) = 4.2 V we determine v ( t ) = 4.2 − 6.7 e − 4 t .
⎛ −2 − 4.2 ⎞
ln ⎜
⎟
−6.7 ⎠
Let t1 be the time at which v(t1) = −2.0 V. Then t 1 = ⎝
= 0.01939 s .
−4
⎛ 2 − 4.2 ⎞
ln ⎜
⎟
−6.7 ⎠
Let t2 be the time at which v(t2) = 2.0 V. Then t 2 = ⎝
= 0.27841 s .
−4
The transition requires 0.27841 − 0.01939 = 0.25902 s.
Section 8.7 The Response of a First-Order Circuit to a Nonconstant Source
P 8.7-1 Find vc(t) for t > 0 for the circuit shown in Figure P 8.7-1 when v1 = 8e–5tu(t) V.
Assume the circuit is in steady state at t = 0–.
Answer: vc(t) = 4e–9t + 18e–5t V
Figure P 8.7-1
Solution: Assume that the circuit is at steady state before t = 0:
KVL : 12ix + 3(3 ix ) + 38.5 = 0 ⇒ ix = −1.83 A
Then vc (0− ) = −12 ix = 22 V = vc (0+ )
After t = 0:
KVL : 12i (t ) − 8e −5t + v ( t ) = 0
x
c
dv ( t )
1 dvc ( t )
KCL : −ix ( t )−2ix ( t ) + (1 36) c
= 0 ⇒ ix ( t ) =
dt
108 dt
⎡ 1 dvc ( t ) ⎤
−5t
∴ 12 ⎢
+ v (t ) = 0
⎥ − 8e
c
dt
108
⎣
⎦
dv (t )
c + 9v (t ) = 72e −5t ⇒ v (t ) = Ae−9t
c
cn
dt
Try v (t ) = Be−5t & substitute into the differential equation ⇒ B = 18
cf
∴ v (t ) = Ae −9t + 18 e−5t
c
v (0) = 22 = A + 18 ⇒ A = 4
c
∴ v (t ) = 4e−9t + 18e−5t V
c
P 8.7-2 Find v(t) for t > 0 for the circuit shown in Figure P 8.7-2. Assume steady state at t = 0–.
Answer: v(t) = 20e–10t/3 – 12e–2t V
Figure P 8.7-2
Solution: Assume that the circuit is at steady state before t = 0:
iL (0+ ) = iL (0− ) =
12
= 3A
4
After t = 0:
v( t ) −12
v( t )
+ iL ( t ) +
= 6 e−2t
4
2
di ( t )
also : v ( t ) = (2 / 5) L
dt
di ( t ) ⎤
3⎡
iL ( t ) + ⎢(2 / 5) L ⎥ = 3 + 6 e−2t
4⎣
dt ⎦
KCL :
diL ( t ) 10
+ iL ( t ) = 10 + 20 e−2t
3
dt
∴ in (t ) = Ae
− (10 / 3)t
, try i f (t ) = B + Ce−2t , substitute into the differential equation,
and then equating like terms ⇒ B =3, C =15 ⇒ i f (t ) =3+15 e−2t
∴iL (t ) =in (t ) + i f (t ) = Ae −(10 / 3)t + 3+15e−2t , iL (0) = 3 = A + 3 + 15 ⇒ A = −15
∴ iL (t ) = −15e− (10 / 3) t + 3 + 15e−2t
Finally, v(t ) =( 2 / 5 )
diL
= 20 e− (10 / 3) t −12 e −2t V
dt
P 8.7-3 Find vc(t) for t > 0 for the circuit shown
in Figure P 8.7-3 when is = [2 cos 2t] u(t) mA.
Figure P 8.7-3
Solution: Assume that the circuit is at steady state before t = 0:
Replace the circuit connected to the capacitor by its Thevenin equivalent (after t=0) to get:
dvc ( t ) ⎞
⎛
+v t =0 ⇒
KVL: − 10 cos 2t + 15 ⎜ 1
30 dt ⎟ c ( )
⎝
⎠
dvc ( t )
+ 2vc ( t ) = 20 cos 2t
dt
vn (t ) = Ae−2t , Try v f (t ) = B cos 2t + C sin 2t & substitute into the differential equation to get
B = C = 5 ⇒ v f (t ) = 5cos 2t + 5sin 2t. ∴ vc (t ) = vn (t ) + v f (t ) = Ae−2t + 5cos 2t + 5sin 2t
Now vc (0) = 0 = A + 5 ⇒ A = −5 ⇒ vc (t ) = −5e −2t + 5cos 2t + 5sin 2t V
P 8.7-4 Many have witnessed
the use of an electrical
megaphone for amplification of
speech to a crowd. A model of a
microphone and speaker is shown
in Figure P 8.7-4a, and the circuit
model is shown in Figure P 8.74b. Find v(t) for
vs = 10(sin 100t)u(t)
which could represent a person
whistling or singing a pure tone.
Figure P 8.7-4
Solution: Assume that the circuit is at steady state before t = 0. There are no sources in the
circuit so i(0) = 0 A. After t = 0, we have:
di ( t )
KVL : − 10sin100t + i ( t ) + 5
+ v (t ) = 0
dt
v( t )
Ohm's law : i ( t ) =
8
dv( t )
∴
+18 v( t ) = 160sin100t
dt
∴ vn (t ) = Ae−18t , try v f (t ) = B cos100t + C sin100t , substitute into the differential equation
and equate like terms ⇒ B = −1.55 & C = 0.279 ⇒ v f (t ) = −1.55cos100t + 0.279sin100t
∴ v(t ) = vn (t ) + v f (t ) = Ae−18t −1.55 cos100 t + 0.279 sin100 t
v(0) = 8 i (0) = 0 ⇒ v (0) = 0 = A−1.55 ⇒ A = 1.55
so v(t ) = 1.55e−18t −1.55cos100t + 0.279 sin100t V
P 8.7-5 A lossy integrator is shown in
Figure P 8.7-5. The lossless capacitor of
the ideal integrator circuit has been
replaced with a model for the lossy
capacitor, namely, a lossless capacitor in
parallel with a 1-kΩ resistor. If
and
vs = 15e–2tu(t) V
vo(0) = 10 V,
find vo(t) for t > 0.
Figure P 8.7-5
Solution: Assume that the circuit is at
steady state before t = 0.
vo ( t ) = −vc ( t )
vC (0+ ) = vC (0− ) = −10 V
After t = 0, we have
i (t ) =
vs ( t )
8 e−5 t
=
= 0.533 e−5 t mA
15000
15000
The circuit is represented by the differential
dv ( t ) vC ( t )
+
. Then
equation: i ( t ) = C C
dt
R
( 0.533 ×10 ) e
−3
−5 t
= ( 0.25 ×10−6 )
dvc ( t )
+ (10−3 ) vc ( t ) ⇒
dt
dvc ( t )
+ 4000 vc ( t ) = 4000 e−5t
dt
Then vn ( t ) = Ae−4000t . Try v f ( t ) = Be−5t . Substitute into the differential equation to get
(
d B e−5t
dt
) + 4000 ( B e ) = 4000 e
−5t
−5t
⇒ B=
4000
= −1.00125 ≅ −1
−3995
vC (t ) = v f ( t ) + vn ( t ) = e −5t + Ae−4000t
vC (0) = −10 = 1 + A ⇒ A = −11 ⇒ vC (t ) = 1 e −2t − 11 e−4000t V
Finally
vo (t ) = − vC (t ) = 11e−4000t −1e −5t V , t ≥ 0
P 8.7-6 Determine v(t) for the circuit
shown in Figure P 8.7-6.
Figure P 8.7-6
Solution: Assume that the circuit is at steady
state before t = 0.
2
v (0+ ) = v (0− ) =
30 = 10 V
4+ 2
After t = 0 we have
KVL :
⎛ 1 d v( t ) ⎞
5 d v( t )
+ v (t ) + 4 ⎜
−i ⎟ = 30
2
dt
2 dt
⎝
⎠
⎛
1 d v( t ) ⎞
−3t
2 i ( t ) + 4⎜ i ( t ) −
⎟ + 30 = e
2
dt
⎝
⎠
The circuit is represented by the differential equation
d v( t )
6
6
2
(10 + e −3t )
v (t ) =
+
19
19
3
dt
Take vn ( t ) = Ae
−( 6 /19 ) t
. Try , v f ( t ) = B + Ce −3t , substitute into the differential equation to get
−3Ce −3t +
6
60 4 −3t
( B + Ce−3t ) =
+ e
19
19 19
Equate coefficients to get
4
4
⇒ v f ( t ) = e−3t + Ae− (6 /19) t
51
51
4
v ( t ) = vn ( t ) + v f ( t ) = 10 − e−3t + Ae− (6 /19) t
51
4
4
vc (0+ ) = 10 V, ⇒ 10 = 10 − + A ⇒ A =
51
51
4 − (6 /19) t −3t
∴ vc (t ) = 10 + (e
−e ) V
51
B = 10 , C = −
Then
Finally
P 8.7-7 Determine v(t) for the circuit shown in Figure P 8.7-7a when vs varies as shown in
Figure P 8.7-7b. The initial capacitor voltage is vc(0) = 0.
Figure P 8.7-7
Solution: We are given v(0) = 0. From part b
of the figure:
⎧5t 0 ≤ t ≤ 2 s
vs ( t ) = ⎨
t > 2s
⎩10
Find the Thevenin equivalent of the part of the
circuit that is connected to the capacitor:
The open circuit voltage:
The short circuit current:
(ix=0 because of the short across the right 2 Ω
resistor)
Replace the part of the circuit connected to the capacitor by its Thevenin equivalent:
dv( t )
+ v ( t ) − vs ( t ) = 0
dt
dv( t ) v ( t ) vs ( t )
+
=
dt
2
2
vn ( t ) = Ae−0.5 t
2
For 0 < t < 2 s, vs ( t ) = 5 t . Try v f ( t ) = B + C t . Substituting into the differential equation and
equating coefficients gives B = −10 and C =5. Therefore v ( t ) = 5t − 10 + A e − t / 2 . Using v(0) = 0,
we determine that A =10. Consequently, v ( t ) = 5t + 10(e−t / 2 − 1) .
At t = 2 s, v( 2 ) = 10e−1 = 3.68 .
Next, for t > 2 s, vs ( t ) = 10 V . Try v f ( t ) = B . Substituting into the differential equation and
equating coefficients gives B = 10. Therefore v ( t ) = 10 + Ae
determine that A = −6.32. Consequently, v ( t ) = 10 − 6.32 e
− (t −2) / 2
− (t − 2) / 2
.
. Using v ( 2 ) = 3.68 , we
P 8.7-8 The electron beam, which is used to “draw” signals on an oscilloscope, is moved
across the face of a cathode-ray tube (CRT) by a force exerted on electrons in the beam. The
basic system is shown in Figure P 8.7-8a. The force is created from a time-varying, ramp-type
voltage applied across the vertical or the horizontal plates. As an example, consider the simple
circuit of Figure P 8.7-8b for horizontal deflection where the capacitance between the plates is C.
Derive an expression for the voltage across the capacitance. If v(t) = kt and Rs = 625 kΩ,
k = 1000, and C = 2000 pF, compute vc as a function of time. Sketch v(t) and vc(t) on the same
graph for time less than 10 ms. Does the voltage across the plates track the input voltage?
Figure P 8.7-8
Solution:
⎡ d v (t ) ⎤
KVL: − kt + Rs ⎢C C ⎥ + vC ( t ) = 0
dt ⎦
⎣
d vC ( t )
k
1
vC ( t ) =
t
⇒
+
Rs C
Rs C
dt
vc ( t ) = vn ( t ) + v f ( t ) , where vc ( t ) = Ae− t / Rs C . Try v f ( t ) = B0 + B1 t
& plug into D.E. ⇒ B1 +
1
k
t thus B0 = − kRs C , B1 = k .
[ B0 + B1t ] =
Rs C
Rs C
Now we have vc (t ) = Ae− t / Rs C + k (t − Rs C ). Use vc (0) = 0 to get 0 = A − kRs C ⇒ A = kRs C.
∴ vc (t ) = k[t − Rs C (1− e −t / Rs C )]. Plugging in k =1000 , Rs = 625 kΩ & C = 2000 pF get
vc (t ) = 1000[t − 1.25 × 10−3 (1 − e −800 t )]
v(t) and vC(t) track well on a millisecond time scale.
Section 8.10 How Can We Check…?
P 8.10-1 Figure P 8.10-1 shows the transient response of a first-order circuit. This transient
response was obtained using the computer program PSpice. A point on this transient response
has been labeled. The label indicates a time and the capacitor voltage at that time. Placing the
circuit diagram on the plot suggests that the plot corresponds to the circuit. Verify that the plot
does indeed represent the voltage of the capacitor in this circuit.
Figure P 8.10-1
Solution: First look at the circuit. The initial capacitor voltage is vc(0) = 8 V. The steady-state
capacitor voltage is vc = 4 V.
We expect an exponential transition from 8 volts to 4 volts. That’s consistent with the plot.
Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of
( 2000 )( 4000 ) = 4 kΩ so the time constant is
the circuit connected to the capacitor is R t =
2000 + 4000
3
2
⎛4
⎞
τ = R t C = ⎜ × 103 ⎟ ( 0.5 × 10−6 ) = ms . Thus the capacitor voltage is
3
⎝3
⎠
vc (t ) = 4 e− t
0.67
+4 V
where t has units of ms. To check the point labeled on the plot, let t1 = 1.33 ms. Then
vc (t1 ) =
So the plot is correct.
⎛ 1.33 ⎞
−⎜
⎟
4 e ⎝ .67 ⎠
+ 4 = 4.541 ~ 4.5398 V
P 8.10-2 Figure P 8.10-2 shows the transient response of a first-order circuit. This transient
response was obtained using the computer program PSpice. A point on this transient response
has been labeled. The label indicates a time and the inductor current at that time. Placing the
circuit diagram on the plot suggests that the plot corresponds to the circuit. Verify that the plot
does indeed represent the current of the inductor in this circuit.
Figure P 8.10-2
Solution: The initial and steady-state inductor currents shown on the plot agree with the values
obtained from the circuit.
Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of
( 2000 )( 4000 ) = 4 kΩ so the time constant is
the circuit connected to the inductor is R t =
2000 + 4000
3
5
15
L
ms . Thus inductor current is
τ=
=
=
R t 4 ×103
4
3
iL (t ) − 2 e − t 3.75 + 5 mA
where t has units of ms. To check the point labeled on the plot, let t1 = 3.75 ms. Then
iL (t1 ) =
⎛ 3.75 ⎞
−⎜
⎟
−2 e ⎝ 3.75 ⎠
+ 5 = 4.264 mA ≠ 4.7294 mA
so the plot does not correspond to this circuit.
P 8.10-3 Figure P 8.10-3 shows the transient response of a first-order circuit. This transient
response was obtained using the computer program PSpice. A point on this transient response
has been labeled. The label indicates a time and the inductor current at that time. Placing the
circuit diagram on the plot suggests that the plot corresponds to the circuit. Specify that value of
the inductance, L, required to cause the current of the inductor in this circuit to be accurately
represented by this plot.
Figure P 8.10-3
Solution: Notice that the steady-state inductor current does not depend on the inductance, L.
The initial and steady-state inductor currents shown on the plot agree with the values obtained
from the circuit.
After t = 0
So I sc = 5 mA and τ =
The inductor current is given by iL (t ) = −2e −1333t
has units of Henries. Let t 1 = 3.75 ms, then
L
+ 5 mA , where t has units of seconds and L
4.836 = iL (t1 ) = −2 e−(1333)⋅(0.00375) L + 5 = −2e−5 L + 5
4.836−5
= e −5 L
−2
so
and
is the required inductance.
L=
L
1333
−5
=2 H
⎛ 4.836−5 ⎞
ln ⎜
⎟
⎝ −2 ⎠
P 8.10-4 Figure P 8.10-4 shows the transient response of a first-order circuit. This transient
response was obtained using the computer program PSpice. A point on this transient response
has been labeled. The label indicates a time and the capacitor voltage at that time. Assume that
this circuit has reached steady state before time t = 0. Placing the circuit diagram on the plot
suggests that the plot corresponds to the circuit. Specify values of A, B, R1, R2, and C that cause
the voltage across the capacitor in this circuit to be accurately represented by this plot.
Figure P 8.10-4
Solution: First consider the circuit. When t < 0 and the circuit is at steady-state:
For t > 0
So
Voc =
R2
R1 R2
RRC
( A + B) , Rt =
and τ = 1 2
R1 + R2
R1 + R2
R1 + R2
Next, consider the plot. The initial capacitor voltage is (vc (0)=) –2 and the steady-state capacitor
voltage is (Voc =) 4 V, so
vC (t ) = − 6e−t τ + 4
At t 1 = 1.333 ms
3.1874 = vC (t1 ) = − 6 e −0.001333 τ + 4
−0.001333
= 0.67 ms
⎛ −4+ 3.1874 ⎞
ln ⎜
⎟
−6
⎝
⎠
τ =
so
Combining the information obtained from the circuit with the information obtained from the plot
gives
R2
R2
R1 R2C
A = −2,
( A + B ) = 4,
= 0.67 ms
R1 + R2
R1 + R2
R1 + R2
There are many ways that A, B, R , R , and C can be chosen to satisfy these equations. Here is
one convenient way. Pick R = 3000 and R = 6000. Then
1
1
2
2
2A
= −2 ⇒ A = − 3
3
2( A+ B)
= 4 ⇒ B −3 = 6 ⇒ B = 9
3
2
1
μF = C
2000 ⋅ C = ms ⇒
3
3
Spice Problems
SP 8-1 The input to the circuit shown in Figure SP 8.1 is the voltage of the voltage source,
vi(t). The output is the voltage across the capacitor, vo(t). The input is the pulse signal specified
graphically by the plot. Use PSpice to plot the output, vo(t), as a function of t.
Hint: Represent the voltage source using the PSpice part named VPULSE.
Figure SP 8.1
Solution:
SP 8-2 The input to the circuit shown in Figure SP 8.2 is the voltage of the voltage source,
vi(t). The output is the current in the inductor, io(t). The input is the pulse signal specified
graphically by the plot. Use PSpice to plot the output, io(t), as a function of t.
Hint: Represent the voltage source using the PSpice part named VPULSE.
Figure SP 8.2
Solution:
SP 8-3 The circuit shown in Figure SP 8.3 is at
steady state before the switch closes at time t = 0.
The input to the circuit is the voltage of the
voltage source, 12 V. The output of this circuit is
the voltage across the capacitor, v(t). Use PSpice
to plot the output, v(t), as a function of t. Use the
plot to obtain an analytic representation of v(t) for
t > 0.
Hint: We expect v(t) = A + Be–t/τ for t > 0, where
A, B, and τ are constants to be determined.
Figure SP 8.3
Solution:
v(t ) = A + B e −t / τ
7.2 = v(0) = A + B e 0
8.0 = v(∞) = A + B e −∞
for t > 0
⇒ 7.2 = A + B ⎫⎪
⎬ ⇒ B = −0.8 V
⇒ A = 8.0 V ⎪⎭
7.7728 = v(0.05) = 8 − 0.8 e −0.05 / τ
⎛ 8 − 7.7728 ⎞
= ln ⎜
⎟ = −1.25878
τ
0.8
⎝
⎠
0.05
⇒ τ=
= 39.72 ms
1.25878
⇒ −
0.05
Therefore
v(t ) = 8 − 0.8 e −t / 0.03972 V for t > 0
SP 8-4 The circuit shown in Figure SP 8.4 is at steady
state before the switch closes at time t = 0. The input to
the circuit is the current of the current source, 4 mA.
The output of this circuit is the current in the inductor,
i(t). Use PSpice to plot the output, i(t), as a function of
t. Use the plot to obtain an analytic representation of
i(t) for t > 0.
Hint: We expect i(t) = A + Be–t/τ for t > 0, where A, B,
and τ are constants to be determined.
Solution:
Figure SP 8.4
i (t ) = A + B e−t / τ
0 = i (0) = A + B e 0
⇒ 0 = A+ B
4 × 10−3 = i (∞) = A + B e −∞
⇒
A = 4 × 10−3
2.4514 ×10 = v(5 ×10 ) = ( 4 ×10
−3
⇒ −
−6
5 × 10−6
⇒ τ=
Therefore
for t > 0
τ
−3
⎫⎪
−3
⎬ ⇒ B = −4 × 10 A
A ⎪⎭
) − ( 4 ×10 ) e (
−3
)
− 5×10−6 / τ
⎛ ( 4 − 2.4514 ) × 10−3 ⎞
= ln ⎜
⎟ = −0.94894
−3
4
×
10
⎝
⎠
5 × 10−6
= 5.269 μ s
0.94894
i (t ) = 4 − 4 e−t / 5.269×10
−6
mA for t > 0
DESIGN PROBLEMS
DP 8-1 Design the circuit in Figure DP
8.1 so that v(t) makes the transition from
v(t) = 6 V to v(t) = 10 V in 10 ms after the
switch is closed. Assume that the circuit is
at steady state before the switch is closed.
Also assume that the transition will be
complete after 5 time constants.
Figure DP 8.1
Solution:
Steady-state response when the switch is open: 6 =
R3
R1 + R 2 + R 3
Steady-state response when the switch is closed: 10 =
10 ms = 5 τ = ( R 1 || R 3 ) C =
R3
6
R3
R1 + R 3
12 ⇒ R1 + R 2 = R 3 .
12
⇒ R1 =
R3
5
.
C
Let C = 1 μF. Then R 3 = 60 kΩ, R 1 = 30 kΩ and R 2 = 30 kΩ.
DP 8-2 Design the circuit in Figure DP 8.2 so that
i(t) makes the transition from i(t) = 1 mA to
i(t) = 4 mA in 10 ms after the switch is closed.
Assume that the circuit is at steady state before the
switch is closed. Also assume that the transition will
be complete after 5 time constants.
Figure DP 8.2
Solution:
steady state response when the switch is open: 0.001 =
12
R1+R 2
steady state response when the switch is closed: 0.004 =
12
R1
⇒ R + R = 12 kΩ .
1
2
⇒ R1 = 3 kΩ .
Therefore, R 2 = 9 kΩ.
⎛ L ⎞
L
10 ms = 5 τ = 5 ⎜
⇒ L = 240 H
⎟=
⎜ R1 + R 2 ⎟ 2400
⎝
⎠
DP 8-3 The switch in Figure DP 8.3 closes at time
0, 2Δt, 4Δt, …2kΔt and opens at times Δt, 3Δt,
5Δt….(2k + 1)Δt. When the switch closes, v(t)
makes the transition from v(t) = 0 V to v(t) = 5 V.
Conversely, when the switch opens, v(t) makes the
transition from v(t) = 5 V to v(t) = 0 V. Suppose we
require that Δt = 5τ so that one transition is complete
before the next one begins. (a) Determine the value
of C required so that Δt = 1 μs. (b) How large must
Δt be when C = 2 μF?
Answer: (a) C = 4 pF; (b) Δt = 0.5s
Figure DP 8.3
Solution:
Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use
Rt = 50 kΩ.
10−6
(a) Δt = 5 Rt C ⇒ C =
= 4 pF
5 50×103
(
(b) Δ t = 5 50×10
3
)( 2×10
−6
(
)
) = 0.5 s
DP 8-4 The switch in Figure DP 8.3 closes at time 0, 2Δt, 4Δt, …2kΔt and opens at times Δt,
3Δt, 5Δt….(2k + 1)Δt. When the switch closes, v(t) makes the transition from v(t) = 0 V to v(t) =
5 V. Conversely, when the switch opens, v(t) makes the transition from v(t) = 5 V to v(t) = 0 V.
Suppose we require that one transition be 95 percent complete before the next one begins. (a)
Determine the value of C required so that Δt = 1 μs. (b) How large must Δt be when C = 2 μF?
Hint: Show that Δt = – τ ln(1 – k) is required for the transition to be 100 k percent complete.
Answer: (a) C = 6.67 pF; (b) Δt = 0.3 s
Solution: Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed
so use Rt = 50 kΩ.
Δt
−Δt τ
When the switch is open: 5 e
= (1 − k ) 5 ⇒ ln (1− k ) = −
⇒ Δ t = −τ ln (1− k )
When the switch is open: 5 − 5 e
(a) C =
−Δt τ
10−6
= 6.67 pF
− ln (1−.95 ) ( 50×103 )
τ
= k 5 ⇒ Δ t = −τ ln (1− k )
(b) Δ t = − ln(1 − .95) ( 50×103 )( 2×10−6 ) = 0.3 s
DP 8-5 A laser trigger circuit is shown in Figure DP 8.5. In order to trigger the laser, we
require 60 mA < |i| < 180 mA for 0 < t < 200 μs. Determine a suitable value for R1 and R2.
Figure DP 8.5
Solution:
i (0) =
R1
20
×
40 R1 R1 + 40
40 +
40 + R1
For t > 0:
i (t ) = i (0) e
−t
τ
where τ =
L
10−2
=
R t 40+ R 2
At t < 200μ s we need i ( t ) > 60 mA and i ( t ) <180 mA
First let's find a value of R 2 to cause i (0) < 180 mA.
1
t
A = 166.7 mA so i (t ) = 0.1667 e− τ .
6
Next, we find a value of R 2 to cause i (0.0002) > 60 mA.
Try R 2 = 40 Ω . Then i (0) =
10−2
1
s.
= 0.2 ms =
50
5000
i (0.0002) = 166.7×10−3 e −5000×0.0002 = 166.7×10−3 e −1 = 61.3 mA
Try R 2 = 10Ω, then τ =
DP 8-6 Fuses are used to open a circuit when
excessive current flows (Wright, 1990). One
fuse is designed to open when the power
absorbed by R exceeds 10 W for 0.5 s.
Consider the circuit shown in Figure DP 8.6.
The input is given by
vs = A[u(t) – u(t – 0.75)] V.
Assume that iL(0–) = 0. Determine the largest
value of A that will not cause the fuse to open.
Solution:
The current waveform will look like this:
Figure DP 8.6
We only need to consider the rise time:
iL (t ) =
−t
−t
Vs
A
τ
(1 − e τ ) =
(1 − e )
R+2
R+2
where
L 0.2 1
s
=
=
Rt
3 15
A
∴ iL (t ) = (1 − e −15t )
3
τ =
Now find A so that iL2 R fuse ≥ 10 W during 0.25 ≤ t ≤ 0.75 s
∴ we want [iL2 (0.25)]R fuse = 10 W ⇒
A2
(1−e −15(.25) ) 2 (1) =10 ⇒ A = 9.715 V
9
Figure DP 8-7
DP 8-7 Design the circuit in Figure DP 8-7(a) to have the response in Figure DP 8-7 (b) by
specifying the values of C, R1 and R2.
Solution:
D
⎧
The voltage v(t) is represented by an equation of the form v ( t ) = ⎨
− at
⎩E + F e
for t < 0
for t > 0
where D, E, F and a are unknown constants. The constants D, E and F are described by
D = v ( t ) when t < 0 , E = lim v ( t ) , E + F = lim v ( t )
t →∞
t →0+
From the plot, we see that
D = 10, E = 20, and E + F = 10 V
Consequently,
10
⎧
v (t ) = ⎨
− at
⎩ 20 − 10 e
for t < 0
for t > 0
To determine the value of a , we pick a time when the circuit is not at steady state. One such
point is labeled on the plot. We see v ( 3.22 ) = 18 V , that is, the value of the voltage is 18 volts at
time 3.22 seconds. Substituting these into the equation for v ( t ) gives
− a 3.22
18 = 20 − 10 e ( )
⇒ a=
ln ( 0.2 )
= 0.5
−3.22
Consequently
10
⎧
v (t ) = ⎨
−0.5 t
⎩20 − 10 e
for t < 0
for t > 0
Now let’s turn our attention to the circuit. When the circuit is at steady state, the capacitor acts
like an open circuit.
After t = 0, the switch is closed and the steady state
voltage is determined from the plot to be v(t) = E = 20
V. On the right we see the circuit that results from
replacing the capacitor by an open circuit and the switch
by a short circuit. Using voltage division gives
80
20 =
( 25) ⇒ R 2 = 20 Ω
80 + R 2
Before t = 0, the switch is open and the steady state
voltage is determined from the plot to be v(t) = E + F =
10 V. On the right we see the circuit that results from
replacing the capacitor by an open circuit and the switch
by an open circuit. Using voltage division gives
10 =
80
80
( 25 ) =
( 25) ⇒ R1 = 100 Ω
80 + R1 + R 2
100 + R1
Recalling that a = 0.5 from the plot, consider the time
1
constant 2 = = τ = C R t . After t = 0, the Thevenin
a
resistance of the part of the circuit connected to the capacitor
is R t = 80 || R 2 = 80 || 20 = 16 Ω .
Then C =
2
2
= = 0.125 F .
R t 16
Figure DP 8-8
DP 8-8 Design the circuit in Figure DP 8-8(a) to have the response in Figure DP 8-8 (b) by
specifying the values of L, R1 and R2.
Solution:
D
⎧
The voltage v(t) is represented by an equation of the form v ( t ) = ⎨
− at
⎩E + F e
for t < 0
for t > 0
where D, E, F and a are unknown constants. The constants D, E and F are described by
D = v ( t ) when t < 0 , E = lim v ( t ) , E + F = lim v ( t )
t →∞
t →0+
From the plot, we see that
D = 2, E = 8, and E + F = 2 V
Consequently,
2
⎧
v (t ) = ⎨
− at
⎩8 − 6 e
for t < 0
for t > 0
To determine the value of a , we pick a time when the circuit is not at steady state. One such
point is labeled on the plot. We see v ( 0.358 ) = 7 V , that is, the value of the voltage is 7 volts at
time 0.358 seconds. Substituting these into the equation for v ( t ) gives
7 = 8− 6e
− a ( 0.358)
⇒ a=
( 6) = 5
ln 1
−0.358
Consequently
2
⎧
v (t ) = ⎨
−5 t
⎩8 − 6 e
for t < 0
for t > 0
Now let’s turn our attention to the circuit. When the circuit is at steady state, the inductor acts
like a short circuit.
After t = 0, the switch is closed and the steady state
voltage is determined from the plot to be v(t) = E = 8 V.
On the right we see the circuit that results from
replacing the inductor by a short circuit and the switch
by a short circuit. Using voltage division gives
8
8=
(12 ) ⇒ R1 = 4 Ω
8 + R1
Before t = 0, the switch is open and the steady state
voltage is determined from the plot to be v(t) = E + F = 2
V. On the right we see the circuit that results from
replacing the inductor by a short circuit and the switch by
an open circuit. Using voltage division gives
8
8
2=
(12 ) =
(12 ) ⇒ R 2 = 36 Ω
8 + R1 + R 2
12 + R 2
Recalling that a = 5 from the plot, consider the time constant
1
L
0.2 = = τ =
. After t = 0, the Thevenin resistance of the
a
Rt
part of the circuit connected to the capacitor is
R t = 8 + R 2 = 8 + 4 = 12 Ω .
Then L = 0.2 R t = 2.4 H .
Figure DP 8-9
DP 8-9 Design the circuit in Figure DP 8-9(a) to have the response in Figure DP 8-9 (b) by
specifying the values of C, R1 and R2.
Solution:
D
⎧
The voltage v(t) is represented by an equation of the form v ( t ) = ⎨
− at
⎩E + F e
for t < 0
for t > 0
where D, E, F and a are unknown constants. The constants D, E and F are described by
D = v ( t ) when t < 0 , E = lim v ( t ) , E + F = lim v ( t )
t →∞
From the plot, we see that
Consequently,
t →0 +
D = 9, E = 5, and E + F = 9 V
9
⎧
v (t ) = ⎨
− at
⎩5 + 4 e
for t < 0
for t > 0
To determine the value of a , we pick a time when the circuit is not at steady state. One such
point is labeled on the plot. We see v ( 0.173) = 6 V , that is, the value of the voltage is 6 volts at
time 0.173 seconds. Substituting these into the equation for v ( t ) gives
− a 0.173)
6 = 5+ 4e (
⇒ a=
ln ( 0.25 )
=8
−0.173
Consequently
9
⎧
v (t ) = ⎨
−8 t
⎩5 + 4 e
for t < 0
for t > 0
Now let’s turn our attention to the circuit. When the circuit is at steady state, the capacitor acts
like an open circuit.
Before t = 0, the switch is open and the steady
state voltage is determined from the plot to be
v(t) = E + F = 9 V. On the right we see the
circuit that results from replacing the capacitor
by an open circuit and the switch by an open
circuit. Using voltage division gives
9=
R2
240 + R 2
(15 )
⇒ R 2 = 360 Ω
After t = 0, the switch is closed and the steady
state voltage is determined from the plot to be
v(t) = E = 5 V. On the right we see the circuit
that results from replacing the capacitor by an
open circuit and the switch by a short circuit.
Using voltage division gives
5=
R1 || R 2
240 + R1 || R 2
(15)
⇒ R1 || R 2 = 120 Ω
R1 || R 2 = R1 || 360 = 120 Ω ⇒ R1 = 180 Ω
Recalling that a = 8 from the plot, consider the
1
time constant 0.125 = = τ = C R t . After t = 0,
a
the Thevenin resistance of the part of the circuit
connected to the capacitor is
R t = 240 || R1 || R 2 = 240 ||180 || 360 = 80 Ω .
Then
0.125 0.125
C=
=
= 0.0015625 F = 1.5625 mF .
Rt
80
Figure DP 8-10
DP 8-10 Design the circuit in Figure DP 8-10(a) to have the response in Figure DP 8-10 (b) by
specifying the values of L, R1 and R2.
Solution:
D
⎧
The current i(t) is represented by an equation of the form i ( t ) = ⎨
− at
⎩E + F e
for t < 0
for t > 0
where D, E, F and a are unknown constants. The constants D, E and F are described by
D = i ( t ) when t < 0 , E = lim i ( t ) , E + F = lim i ( t )
t →∞
t →0 +
From the plot, we see that
D = 120, E = 60, and E + F = 120 mA
Consequently,
120 mA
for t < 0
⎧
i (t ) = ⎨
− at
⎩60 + 60 e mA for t > 0
To determine the value of a , we pick a time when the circuit is not at steady state. One such
point is labeled on the plot. We see i ( 0.1195 ) = 70 mA , that is, the value of the current is 70
mA at time 0.1195 seconds. Substituting these into the equation for i ( t ) gives
70 = 60 + 60 e
− a ( 0.1195)
⇒ a=
( 6 ) = 15
ln 1
−0.1195
Consequently
120 mA
for t < 0
⎧
i (t ) = ⎨
−15 t
mA for t > 0
⎩60 + 60 e
Now let’s turn our attention to the circuit. When the circuit is at steady state, the inductor acts
like a short circuit.
Before t = 0, the switch is closed and the steady state current
is determined from the plot to be i(t) = E + F = 120 mA. On
the right we see the circuit that results from replacing the
inductor by a short circuit and the switch by a short circuit.
Using current division gives
28
120 =
(180 ) ⇒ R1 = 14 Ω
28 + R1
After t = 0, the switch is open and the steady state voltage is
determined from the plot to be v(t) = E + F = 2 V. On the
right we see the circuit that results from replacing the
inductor by a short circuit and the switch by an open circuit.
Using current division gives
28
28
60 =
(180 ) =
(180 ) ⇒ R 2 = 42 Ω
28 + R1 + R 2
28 + 14 + R 2
Recalling that a = 15 from the plot, consider the time constant
1 1
L
= =τ =
. After t = 0, the Thevenin resistance of the part of the
15 a
Rt
circuit connected to the capacitor is
R t = 28 + R1 + R 2 = 28 + 14 + 42 = 84 Ω .
Then L =
Rt
15
=
84
= 5.6 H .
15
Chapter 9 - Complete Response of Circuits with Two Energy
Storage Elements
Exercises
2H
Exercise 9.2-1 Find the second-order differential
equation for the circuit shown in Figure E 9.2-1 in terms
of i using the direct method.
Answer:
is
1Ω
1 2F
i
d 2i 1 di
1 dis
+
+i =
2
dt
2 dt
2 dt
Figure E 9.2-1
Solution:
KVL a : 2 di
dt
+ v + 1(i − is ) = 0
⇒ v = − 2 di
i = 1
= 1
∴
2
2
dv
dis
dt
dt
= 1
−1
2
2
di
d
dt
dt
dt
−
d 2i
1 di
1 dis
+
+i=
2
2 dt
2 dt
dt
− i + is
( −2di dt −i +i )
s
d 2i
dt 2
Exercise 9.2-2 Find the second-order differential
equation for the circuit shown in Figure E 9.2-2 in terms
of v using the operator method.
v
is
1Ω
di
d 2v
di
Answer: 2 + 2 + 2v = 2 s
dt
dt
dt
1H
Ground
Figure E 9.2-2
Solution:
KCL at v : using s = d
v
+ i + 1 sv = is
2
1
also v = si
(2) Solving for i in (1) & plugging into
(2)
di
d 2v
dv
+2
+ 2v = 2 s
yields s 2v + 2 sv + 2v = 2 sis or
dt
dt
dt
dt
(1)
1 2F
v
Exercise 9.3-1 Find the characteristic equation and the
natural frequencies for the circuit shown in Figure E 9.3-1.
Answer:
6Ω
is
s2 + 7s + 10 = 0; s1 = –2, s2 = –5
4Ω
1 4
F
1H
Ground
Figure E 9.3-1
Solution:
v
+ iL + 1 dv
= i
4
dt
s
4
di
KVL for the right mesh : v = 6iL + L
dt
KCL at the top node :
(1)
(2)
Plugging (2) into (1) yields
d 2i
di
L + 7 L + 10i = 4i
L
s
dt
dt 2
∴ characteristic equation ⇒ s 2 + 7 s + 10 = 0 &
natural frequencies ⇒ s = − 2, − 5
v
Exercise 9.4-1 Find the natural response of the RLC circuit of
Figure 9.4-1 when R = 6 Ω, L = 7 H, and C = 1/42 F. The initial
conditions are v(0) = 0 and i(0) = 10 A.
–t
L
R
i
–6t
Answer: vn(t) = – 84(e – e ) V
Figure 9.4-1
Solution:
This is a parallel RLC circuit with
α =
1
1
=
= 7
2
2 RC
2(6)(1/ 42)
1
1
=
= 6
LC
(7)(1/ 42)
and ω o2 =
The roots of the characteristic equation are
s = −α ±
α 2 −ω o2 = − 7 2 ±
( 7 2 ) −6
2
= − 1, − 6
so the natural response is
vn (t ) = A 1e− t + A2 e−6t
Need vn (0) and
dvn
dt
t =0
to evaluate A1 & A2 . We are given vn (0) = 0.
Apply KCL at the top node to get
i (t ) +
v (t ) 1 d
+
v (t ) = 0 ⇒
6
42 dt
d
v ( t ) = − ( 7 v ( t ) + 42 i ( t ) )
dt
At time t=0,
d
V
v ( t ) = − ( 7 v ( 0 ) + 42 i ( 0 ) ) = − ( 7 × 0 + 42 × 10 ) = −420
dt
s
t =0
So
vn (0) = 0 = A1 + A2
⎫
⎪
dvn
⎬ A = − 84 , A2 = 84
=−420= − A1 − 6 A2 ⎪ 1
dt t =0
⎭
and
vn (t ) = −84e− t + 84e −6t V
C
Exercise 9.5-1 A parallel RLC circuit has R = 10 Ω, C = 1 mF, L = 0.4 H, v(0) = 8 V, and
i(0) = 0. Find the natural response vn(t) for t < 0.
Answer: vn(t) = e –50t(8 – 400t) V
Solution:
For parallel RLC
1
1
1
1
α =
=
= 50, ω o2 =
=
= 2500
−3
LC
2 RC
2(10)(10 )
(0.4)(10−3 )
∴ s = −50 ±
(50)2 − 2500 = −50, −50
∴ vn (t ) = A1e −50t + A2te−50t
+ ) = 0 & v(0+ ) = 8 V
−v(0+ )
i (0+ ) =
= −0.8 V
C
10Ω
i (0+ )
dv
V
= C
= − 800
dt t = 0+
C
s
with i (0
At t = 0+
vn (0) = 8 = A1 ⇒ vn (t ) = 8e −50t + A2te−50t
dv(0)
= −800 = − 400 + A2 ⇒ A2 = − 400
dt
∴ vn (t ) = 8e −50t − 400t e−50t V
Exercise 9.6-1 A parallel RLC circuit has R = 62.5 Ω, L = 10 mH, C = 1 μF, v(0) = 10 V, and
i(0) = 80 mA. Find the natural response vn(t) for t > 0.
Answer: vn(t) = e–8000t [10 cos 6000t – 26.7 sin 6000t] V
Solution:
1
1
=
= 8000
2 RC 2(62.5)(10−6 )
1
1
=
= 108
ω2 =
0 LC
−
6
(.01)(10 )
α=
∴ s = − α ± α 2 − ω 02 = − 8000 ±
(8000) 2 − 108 = − 8000 ± j 6000
∴ vn (t ) = e−8000t [ A1 cos 6000 t + A2 sin 6000t ]
KCL at top : 0.08 +
10
+ iC = 0
62.5
⇒ iC (0+ ) = − 0.24 A
∴
dv(0+ ) iC (0+ )
=
= −2.4 ×10+5 V/s
dt
C
vn (0) = 10 = A1
dvn (0)
= − 2.4×105 = 6000 A2 − 8000(10) ⇒ A2 = − 26.7
dt
∴ vn (t ) = e −8000t [10 cos 6000 t − 26.7 sin 6000t ] V
Exercise 9.7-1 A circuit is described for t > 0 by the equation
d 2i
di
+ 9 + 20i = 6is
2
dt
di
where is = 6 + 2t A. Find the forced response if for t > 0.
Answer: if = 1.53 + 0.6t A
Solution:
i "+ 9i '+ 20i = 36 + 12t
Try i f = A + Bt
& plug into above
0 + 9 B + 20( A + Bt ) = 36 + 12t
Equating the constant coefficients and the coefficients of t gives
20 Bt = 12t ⇒ B = 0.6 and 9 B + 20 A = 36 ⇒ A = 1.53
∴ if = 1.53 + 0.6t A
i
Exercise 9.9-1 Find v2(t) for t > 0 for the
circuit of Figure E 9.9-1. Assume there is no
initial stored energy.
3 10
+
1Ω
v1
+
v2
–
1 12 F
–
Answer: v2(t) = – 15e–2t + 6e– 4t – e–6t + 10 V
H
5 6
F
10u(t) A
Figure E 9.9-1
Solution:
no initial stored energy ⇒ v1 (0+ ) = v2 (0+ ) = i (0+ ) = 0
t = 0+
KVL : − 0 + 3
di(0+ )
di(0+ )
+0=0 ⇒
=0
10 dt
dt
0V
KCL at A :
+ i1 (0+ )+0 = 0
1Ω
dv1 (0+ )
=0
⇒
dt
KCL at B : − 0 + i 2 (0+ ) − 10 = 0 ⇒ i 2 (0+ ) = 5 6
dv 2 (0+ )
dv 2 (0+ )
= 12 V s
= 10 ⇒
dt
dt
t>0
v1
1
+
v1 ' + i = 0
1
12
KCL at B : − i + (5 / 6)v '2 =10
KCL at A :
(1)
KVL : − v1 + ( 3 /10 ) i '+ v2 = 0
( 2)
( 3)
Eliminating i from (1) & (3) yields
1
v '1 + (5 / 6)v2 '− 10 = 0
12
3 ⎛5 ⎞
−v1 + ⎜ v2" ⎟ + v2 = 0
10 ⎝ 6 ⎠
v1 +
(4)
( 5)
From (5)
1
v1 = v2 + v2" ⇒ v1' = v2" + (1/ 4)v2"
4
Now substituting into (4) yields
1
1⎛
1 ⎞ 5
v2 ' + v2" + ⎜ v2 ' + v2"' ⎟ + v2 ' = 10
4
12 ⎝
4 ⎠ 6
v2"' + 12v2" + 44v2 ' + 48v2 = 480
Natural Response:
Forced Response:
Complete Response:
v2 n : s 3 + 12 s 2 + 44s + 48 = 0
⇒ s = − 2, −4, −6
∴ v2 n = A1e −2t + A2 e −4t + A3e−6t
v2 f : try v2 f = B and plug into Diff. Eq. ⇒ B = 10
v2 (t ) = A1e −2t + A2e −4t + A3e −6t + 10
dv2 (0+ )
d 2 v2 (0+ )
Recall v2 (0 ) = 0,
= 12 V/s, then from (5)
= 4[v1 (0+ ) − v2 (0+ )] = 0 .
2
dt
dt
+
v2 (0 ) = 0 = A1 + A2 + A3 + 10
(6)
+
dv2 (0+ )
= 12 = − 2 A1 − 4 A2 − 6 A3
(7)
dt
d 2 v2 (0+ )
= 0 = 4 A1 + 16 A2 + 36 A3
(8)
dt 2
Solving (6)-(80 simultaneously gives A1 = −15, A2 = 6, A3 = −1
Then
v2 (t ) = − 15e −2t + 6e −4t − e−6t + 10 V
jω
Exercise 9.10-1 A parallel RLC circuit has
L = 0.1 H and C = 100 mF. Determine the
roots of the characteristic equation and plot
them on the s-plane when (a) R = 0.4 Ω and
(b) R = 1.0 Ω.
×
–20
×
–5
0
Answer: (a) s = – 5, – 20 (Figure E 9.10-1)
Figure E 9.10-1
Solution:
s2 +
1
1
s+
= 0 and L = 0.1, C = 0.1 ⇒
RC
LC
a)
R = 0.4 Ω ⇒ s 2 + 25s + 100 = 0
s = − 5, − 20
b)
R = 1Ω
⇒ s 2 + 10s + 100 = 0
s = − 5 ± j5 3
s2 +
10
s + 100 = 0
R
σ
Section 9-2: Differential Equations for Circuits with Two Energy Storage
Elements
2Ω
P 9.2-1 Find the differential equation for the
circuit shown in Figure P 9.2-1 using the direct
method.
1 mH
+
–
vs
100 Ω
Figure P 9.2-1
Solution:
KCL: i L =
v
dv
+C
R2
dt
KVL: Vs = R 1i L + L
L v + C dv OP + L dv + LC d v + v
= R M
N R dt Q R dt dt
L R + 1OPv + LMR C + L OP dv + [LC] d v
= M
dt
N R Q N R Q dt
2
vs
1
2
2
2
2
vs
1
1
2
2
2
R 1 = 2Ω, R 2 = 100Ω, L = 1mH, C = 10μF
2
dv
−8 d v
. v +.00003 + 1 × 10
v s = 102
dt
dt 2
dv d 2 v
8
8
1 × 10 v s = 102
. × 10 v + 3000 + 2
dt dt
di L
+v
dt
10 μ F
P 9.2-2 Find the differential equation for the circuit
shown in Figure P 9.2-2 using the operator method.
Answer:
d2
d
i (t ) + 11, 000 iL (t ) + 1.1 × 108 iL (t ) = 108 is (t )
2 L
dt
dt
10 Ω
is
10 μ F
100 Ω
1 mH
iL
Figure P 9.2-2
Solution:
v
+ i L + Csv
R1
KVL: v = R 2 i L + Lsi L
KCL: i s =
Solving Cramer's rule for i L :
iL =
is
R 2 Ls
+
+ R 2 Cs + LCs2 +1
R1 R1
LM1 + R OPi + LM L + R COPsi
N R Q NR
Q
2
L
1
2
L
+ LC s2 i L = i s
1
R 1 = 100Ω, R 2 = 10Ω, L = 1mH, C = 10μF
1.1i L +.00011si L + 1 × 10 −8 s2 i L = i s
11
. × 108 i L + 11000si L + s2 i L = 1 × 108 i s
P 9.2-3 Find the differential equation for iL(t) for t > 0
for the circuit of Figure P 9.2-3.
iL
is
R1
L
t=0
+
–
vs
C
+
vc
–
Figure P 9.2-3
Solution:
t>0
KCL: i L + C
dv c v s + v c
+
= 0
dt
R2
KVL: R 1i s + R 1i L + L
di L
− vc − vs = 0
dt
Solving for iL :
− R1
d 2iL ⎡ R1
1 ⎤ diL ⎡ R1
1 ⎤
R1 dis 1 dvs
+
+
+
+
=
−
+
i
i
s
⎢
⎥
⎢
⎥L
dt 2 ⎣ L R2C ⎦ dt ⎣ LR2C LC ⎦
LCR2
L dt L dt
R2
P 9.2-4 The input to the circuit shown in Figure P 9.2-4 is
the voltage of the voltage source, Vs. The output is the
inductor current i(t). Represent the circuit by a secondorder differential equation that shows how the output of
this circuit is related to the input, for t > 0.
t=0
+
–
Hint: Use the direct method.
R1
i(t)
+
L
Vs
v(t)
C
–
Figure P 9.2-4
Solution:
After the switch opens, apply KCL and KVL to get
d
⎛
⎞
R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) = Vs
dt
⎝
⎠
Apply KVL to get
v (t ) = L
Substituting v ( t ) into the first equation gives
d
i (t ) + R2 i (t )
dt
⎛
d⎛ d
d
⎞⎞
R1 ⎜ i ( t ) + C ⎜ L i ( t ) + R 2 i ( t ) ⎟ ⎟ + L i ( t ) + R 2 i ( t ) = Vs
dt ⎝ dt
dt
⎠⎠
⎝
then
R1 C L
d2
dt
2
(
i ( t ) + R1 C R 2 + L
) dtd i ( t ) + ( R1 + R 2 ) i ( t ) = Vs
Dividing by R1 C L :
⎛ R1 C R 2 + L ⎞ d
⎛ R1 + R 2 ⎞
Vs
+
+
=
i
t
i
t
i
t
⎜
⎟
⎜
⎟
(
)
(
)
(
)
⎜ R1 C L ⎟ dt
⎜ R1 C L ⎟
R1 C L
dt 2
⎝
⎠
⎝
⎠
d2
R3
R2
P 9.2-5 The input to the circuit shown in Figure P 9.2-5 is the voltage of the voltage source, vs.
The output is the capacitor voltage v(t). Represent the circuit by a second-order differential
equation that shows how the output of this circuit is related to the input, for t > 0.
Hint: Use the direct method.
i(t)
t=0
R1
L
+
–
vs
+
R2
C
v(t)
–
Figure P 9.2-5
Solution:
After the switch closes, use KCL to get
i (t ) =
v (t )
d
+ C v (t )
R2
dt
Use KVL to get
v s = R1 i ( t ) + L
d
i (t ) + v (t )
dt
Substitute to get
vs =
R1
R2
= CL
v ( t ) + R1C
d
L d
d2
v (t ) +
v ( t ) + CL 2 v ( t ) + v ( t )
dt
R 2 dt
dt
⎛
R1 + R 2
d2
L ⎞d
v t + R1C +
v (t ) +
v (t )
⎟
2 ( ) ⎜
⎜
⎟ dt
dt
R
R
CL
2
2
⎝
⎠
Finally,
vs
CL
=
⎛ R1
R1 + R 2
d2
1 ⎞d
+
+
v
t
v (t ) +
v (t )
(
)
⎜
⎟
2
⎜ L R 2C ⎟ dt
dt
R 2CL
⎝
⎠
t=0
P 9.2-6 The input to the circuit shown in
Figure P 9.2-6 is the voltage of the voltage
source, vs. The output is the inductor current
i(t). Represent the circuit by a second-order
differential equation that shows how the output
of this circuit is related to the input, for t > 0.
R1
R2
+
–
+
C
vs
L
i(t)
Hint: Use the direct method.
Figure P 9.2-6
Solution:
After the switch closes use KVL to get
R2 i (t ) + L
d
i (t ) = v (t )
dt
Use KCL and KVL to get
d
⎛
⎞
v s = R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t )
dt
⎝
⎠
Substitute to get
d
d2
d
i ( t ) + R1CL 2 i ( t ) + R 2i ( t ) + L i ( t )
dt
dt
dt
2
d
d
= R1CL 2 i ( t ) + ( R1 R 2C + L ) i ( t ) + ( R1 + R 2 ) i ( t )
dt
dt
v s = R1i ( t ) + R1CR 2
Finally
vs
R1CL
=
⎛ R2
R1 + R 2
d2
1 ⎞d
+
i t +
i (t ) +
i (t )
⎟
2 ( ) ⎜
⎜ L R1C ⎟ dt
dt
R
CL
1
⎝
⎠
v(t)
–
P 9.2-7 The input to the circuit shown in Figure P 9.2-7 is the voltage of the voltage source, vs.
The output is the inductor current i2(t). Represent the circuit by a second-order differential
equation that shows how the output of this circuit is related to the input, for t > 0.
Hint: Use the operator method.
R2
t=0
L1
R1
+
–
i1(t)
L2
i2(t)
vs
R3
Figure P 9.2-7
Solution:
After the switch opens, KVL gives
L1
d
d
i1 ( t ) = R 2 i 2 ( t ) + L 2 i 2 ( t )
dt
dt
L1
d
i1 ( t ) + R1 ( i1 ( t ) + i 2 ( t ) ) = 0
dt
KVL and KCL give
Use the operator method to get
L1s i1 = R 2 i 2 + L 2 s i 2
L1s i1 + R1 ( i1 + i 2 ) = 0
L1s 2i1 + R1s i1 + R1s i 2 = 0
s ( R 2i 2 + L 2 s i 2 ) +
R1
L1
(R i
2 2
+ L 2 s i 2 ) + R1s i 2 = 0
⎛
⎞
L2
R1 R 2
L 2 s 2 i 2 + ⎜ R 2 + R1
+ R1 ⎟ s i 2 +
i2 = 0
⎜
⎟
L
L
1
1
⎝
⎠
⎛ R 2 R1 R1 ⎞
R1 R 2
s 2i 2 + ⎜
+
+ ⎟ s i2 +
i2 = 0
⎜ L 2 L 2 L1 ⎟
L1 L 2
⎝
⎠
so
⎛ R 2 R1 R1 ⎞ d
R1 R 2
d2
i t +
+
+ ⎟ i 2 (t ) +
i 2 (t ) = 0
2 2( ) ⎜
⎜ L 2 L 2 L1 ⎟ dt
dt
L
L
1
2
⎝
⎠
t=0
P 9.2-8 The input to the circuit shown in
Figure P 9.2-8 is the voltage of the voltage
source, vs. The output is the capacitor voltage
v2(t). Represent the circuit by a second-order
differential equation that shows how the output
of this circuit is related to the input, for t > 0.
R2
R1
+
+
+
–
Hint: Use the operator method.
C1
vs
v1(t)
C2
–
R3
Figure P 9.2-8
Solution:
After the switch closes, KVL and KCL give
d
⎛ d
⎞
v1 ( t ) + R 3 ⎜ C 1 v1 ( t ) + C 2 v 2 ( t ) ⎟ = v s
dt
⎝ dt
⎠
KVL gives
v 1 ( t ) = R 2 C2
Using the operator method
d
v 2 (t ) + v 2 (t )
dt
v1 + R 3 ( C 1sv1 + C 2 sv 2 ) = v s
v1 = R 2C 2 sv 2 + v 2
so
(1 + R C s ) v
2
Then
2
v1 = (1 + R 2C2 s ) v 2
2
+ R 3C 1s (1 + R 2C 2 s ) v 2 + R 3C 2 sv 2 = v s
R 2 R 3C 1C 2 s 2 v 2 + ( R 2C 2 + R 3C 1 + R 3C 2 ) sv 2 + v 2 = v s
s 2v 2 +
R 2C 2 + R 3C 1 + R 3C 2
R 2 R 3C 1C 2
sv 2 +
vs
1
v2 =
R 3 R 2C 1C 2
R 2 R 3C 1C 2
⎛ 1
vs
1
1 ⎞
1
+
+
s 2v 2 + ⎜
sv 2 +
v2 =
⎟
⎜ R 3C 1 R 2 C 2 R 2 C 1 ⎟
R 2 R 3C 1C 2
R 2 R 3C1C 2
⎝
⎠
so
⎛ 1
1
1 ⎞d
1
d2
= 2 v 2 (t ) + ⎜
+
+
v (t )
⎟⎟ v ( t ) +
⎜
R 2 R 3C 1C 2 dt
R 2 R 3C 1C 2
⎝ R 3C1 R 2C 2 R 2C 1 ⎠ dt
vs
v2(t)
–
t=0
P 9.2-9 The input to the circuit shown in
Figure P 9.2-9 is the voltage of the voltage source,
vs. The output is the capacitor voltage v(t).
Represent the circuit by a second-order differential
equation that shows how the output of this circuit
is related to the input, for t > 0.
L
+
–
vs
R1
R2
Hint: Use the direct method.
C
Figure P 9.2-9
Solution:
After the switch closes
i (t ) = C
d
v (t )
dt
KCL and KVL give
⎛
1 ⎛ d
d
⎞⎞
vs = R2 ⎜ i (t ) + ⎜ L i (t ) + v (t ) ⎟ ⎟ + L i (t ) + v (t )
⎜
R1 ⎝ dt
dt
⎠ ⎟⎠
⎝
Substituting gives
⎛ R2 ⎞
⎛ R2 ⎞
d2
d
v s = ⎜1 +
LC
v ( t ) + R 2C v ( t ) + ⎜ 1 +
⎟
⎟⎟ v ( t )
2
⎜
⎟
⎜
R
dt
dt
R
1 ⎠
1 ⎠
⎝
⎝
⎛ R2 ⎞
⎛ R2 ⎞
d2
d
= ⎜1 +
LC 2 v ( t ) + R 2C v ( t ) + ⎜ 1 +
⎟
⎟ v (t )
⎜
⎜
R1 ⎟⎠
dt
dt
R1 ⎟⎠
⎝
⎝
Finally
R1v s
LC ( R1 + R 2 )
=
R1 R 2
1
d2
d
v (t ) +
v (t ) +
v (t )
dt
LC
L ( R1 + R 2 ) dt
i(t)
+
v(t)
–
P 9.2-10 The input to the circuit shown in Figure P 9.2-10 is the voltage of the voltage source,
vs. The output is the capacitor voltage v(t). Represent the circuit by a second-order differential
equation that shows how the output of this circuit is related to the input, for t > 0.
Hint: Find a Thévenin equivalent circuit.
ia
i(t)
t=0
L
R1
+
–
vs
bia
+
R2
C
v(t)
–
Figure P 9.2-10
Solution:
Find the Thevenin equivalent circuit for the part of the circuit to the left of the inductor.
v s − v oc ⎫
⎪
R1 ⎪
v s R 2 (1 + b )
⎬ ⇒ v oc =
v oc ⎪
R1 + R 2 (1 + b )
ia + bia =
R 2 ⎪⎭
ia =
i sc = i a (1 + b ) =
vs
R1
(1 + b )
v s R 2 (1 + b )
Rt =
v oc
i sc
=
R1 + R 2 (1 + b )
R1 R 2
=
vs
R1 + R 2 (1 + b )
(1 + b )
R1
Rt i (t ) + L
d i (t )
+ v ( t ) − v oc = 0
dt
i (t ) = C
d v (t )
d 2 v (t )
Rt C
+ LC
+ v ( t ) = v oc
dt
d t2
Finally,
⇒
d v (t )
dt
d 2 v (t ) Rt d v (t )
v (t )
1
+
+
v
t
=
(
)
d t2
L dt
LC
LC
R1 R 2
d 2 v (t )
d v (t )
v (t )
1
+
+
v (t ) =
2
dt
LC
LC
L ( R1 + R 2 (1 + b ) ) d t
P 9.2-11 The input to the circuit shown in
Figure P 9.2-11 is the voltage of the voltage
source, vs(t). The output is the voltage v2(t).
Derive the second-order differential equation
that shows how the output of this circuit is
related to the input.
–
R1
R2
+
+
+
–
vs(t)
v1(t)
C1
–
Hint: Use the direct method.
Figure P 9.2-11
Solution:
KCL gives
v s ( t ) − v1 ( t )
= C1
d
v1 ( t )
dt
⇒
v s ( t ) = R1C 1
d
v1 ( t ) + v1 ( t )
dt
= C2
d
v 2 (t )
dt
⇒
v1 ( t ) = R 2C 2
d
v 2 (t ) + v 2 (t )
dt
R1
and
v1 ( t ) − v 2 ( t )
R2
Substituting gives
v s ( t ) = R1C 1
d ⎡
d
d
⎤
R 2C 2 v 2 ( t ) + v 2 ( t ) ⎥ + R 2 C 2 v 2 ( t ) + v 2 ( t )
⎢
dt ⎣
dt
dt
⎦
so
⎛ 1
1
d2
1 ⎞
1
vs (t ) = 2 v 2 (t ) + ⎜
+
v 2 (t )
⎟⎟ v 2 ( t ) +
⎜
R1 R 2C 1C 2
dt
R1 R 2C 1C 2
⎝ R1C 1 R 2C 2 ⎠
+
C2
v2(t)
–
C2
P 9.2-12 The input to the circuit shown in
Figure P 9.2-12 is the voltage of the voltage
source, vs(t). The output is the voltage vo(t).
Derive the second-order differential equation
that shows how the output of this circuit is
related to the input.
– v2(t) +
R1
R2
C1
–
Hint: Use the operator method.
+
–
+ v1(t) –
+
vs(t)
+
vo(t)
–
Figure P 9.2-12
Solution:
d
v1 ( t ) + v1 ( t )
dt
v 2 (t )
d
d
C 1 v1 ( t ) + C 2 v 2 ( t ) +
=0
dt
dt
R2
v s ( t ) = R1C 1
KVL gives
KCL gives
v o (t ) = v 2 (t )
KVL gives
Using the operator method
v s = R1C 1sv1 + v1
C 1sv1 + C 2 sv 2 +
v2
R2
=0
Solving
⎛ C2
⎞
1
v1 = − ⎜
v2 +
v2 ⎟
⎜ C1
R 2C 1s ⎟⎠
⎝
⎛ C2
1 ⎞
sv s = ( sR1C 1 + 1) ⎜
s+
⎟ vo
⎜ C1
R 2C 1 ⎟⎠
⎝
⎛ 1
1
1 ⎞
1
sv s = s 2 v o + ⎜
vo
+
⎟⎟ sv o +
⎜
R1C 2
R
C
R
C
R
R
C
C
2 2 ⎠
1 2 1 2
⎝ 1 1
The corresponding differential equation is
⎛ 1
1 d
d2
1 ⎞d
1
vs (t ) = 2 v o (t ) + ⎜
+
v o (t )
⎟⎟ v o ( t ) +
⎜
R1C 2 dt
dt
R1 R 2C 1C 2
⎝ R1C 1 R 2C 2 ⎠ dt
t=0
P 9.2-13 The input to the circuit shown in
Figure P 9.2-13 is the voltage of the voltage
source, vs(t). The output is the voltage vo(t).
Derive the second-order differential equation
that shows how the output of this circuit is
related to the input.
Hint: Use the direct method.
– v(t) +
C
R1
i(t)
–
+
–
+
vs(t)
+
L
R2
vo(t)
–
Figure P 9.2-13
Solution:
After the switch opens, KCL gives
vs (t )
R1
+C
d
v (t ) = 0
dt
KVL gives
v (t ) − v o (t ) = L
and Ohm’s law gives
d
i (t )
dt
v o (t ) = R2 i (t )
so
d
1
v (t ) = −
vs (t )
dt
R1 C
and
d
d
d2
v (t ) − v o (t ) = L 2 i (t )
dt
dt
dt
Then
−
1
d
d2
d
v s (t ) = v (t ) = L 2 i (t ) + R2 i (t )
R1C
dt
dt
dt
or
R2 d
1
d2
−
vs (t ) = 2 i (t ) +
i (t )
R1CL
dt
L dt
R2
P 9.2-14 The input to the circuit shown in
Figure P 9.2-14 is the voltage of the voltage
source, vs(t). The output is the voltage v2(t).
Derive the second-order differential equation
that shows how the output of this circuit is
related to the input.
+ v1(t) –
C1
R1
–
+
–
Hint: Use the direct method.
R3
+
+
vs(t)
C2
v2(t)
–
Figure P 9.2-14
Solution:
KCL gives
vs (t )
R1
and
=
v1 ( t )
R2
v 2 ( t ) + v1 ( t )
R3
+ C1
+ C2
d
v1 ( t )
dt
d
v 2 (t ) = 0
dt
so
v1 ( t ) + R 2C 1
R2
d
v1 ( t ) =
vs (t )
dt
R1
and
d
⎛
⎞
v 1 ( t ) = − ⎜ v 2 ( t ) + R 3C 2 v 2 ( t ) ⎟
dt
⎝
⎠
Substituting gives
R2
⎡
d
d ⎡
d
⎤⎤
⎢ v 2 ( t ) + R 3C 2 dt v 2 ( t ) + R 2C 1 dt ⎢v 2 ( t ) + R 3C 2 dt v 2 ( t ) ⎥ ⎥ = − R v s ( t )
⎣
⎦⎦
⎣
1
or
⎛ 1
d2
1 ⎞d
1
1
v t +
+
v 2 (t ) +
v 2 (t ) = −
v s (t )
⎟
2 2( ) ⎜
⎜ R 2C 1 R 3C 2 ⎟ dt
dt
R
R
C
C
R
R
C
C
2
3
1
2
1
3
1
2
⎝
⎠
P 9.2-15 Find the second-order differential
equation for i2 for the circuit of Figure P 9.2-15
using the operator method. Recall that the operator
for the integral is 1/s.
Answer:
3
1Ω
vs
+
–
i1
d 2 vs
d 2i2
di2
+
+
i
=
4
2
2
dt 2
dt
dt 2
Apply KVL to the left mesh : i1 + s(i1 −i2 ) = vs
(1)
dt
⎛1⎞
Apply KVL to the right mesh : 2i2 + 2 ⎜ ⎟ i2
⎝s⎠
⎛1⎞
⇒ i1 = 2⎜ ⎟i2 +
⎝s⎠
+ s(i2 − i1 ) = 0
⎛1⎞
2⎜ 2 ⎟i2 + i2
⎝s ⎠
(2)
Plugging (2) into (1) yields
3s 2i + 4si + 2i = s 2v
2
2
2
s
or
1H
i2
Figure P 9.2-15
Solution:
where s = d
2Ω
d 2v
d 2i
di
s
2
2
+ 4
+ 2i =
3
2
2
2
dt
dt
dt
1 2
F
Section 9-3: Solution of the Second Order Differential Equation - The Natural
Response
P 9.3-1 Find the characteristic equation and its roots
for the circuit of Figure P 9.2-2.
10 Ω
is
10 μ F
100 Ω
1 mH
iL
Figure P 9.2-2
Solution:
From Problem P 9.2-2 the characteristic equation is
−11000 ± (11000)2 − 4(1.1×108 )
1.1× 108 + 11000s + s 2 = 0 ⇒ s , s =
= −5500 ± j8930
1 2
2
100 mH
P 9.3-2 Find the characteristic equation and its
roots for the circuit of Figure P 9.3-2.
2
4
Answer: s + 400s + 3 × 10 = 0
roots: s = – 300, – 100
iL
is
40 Ω
+
vc
–
1 3 mF
Figure P 9.3-2
Solution:
KVL: 40 (is − iL ) = 100 ×10−3
diL
+ vc
dt
⎛1
⎞ dv
i L = i c = ⎜ × 10−3 ⎟ c
⎝3
⎠ dt
di 40
di 100
d 2iL
40
× 10−3 s − ×10−3 L −
× 10−6
dt 3
dt
3
3
dt 2
di
d 2iL
di
+ 400 L +30000i L = 400 s
2
dt
dt
dt
2
s + 400s + 30000 = 0 ⇒ ( s + 100)( s + 300) = 0 ⇒ s1 = −100,
iL =
s2 = −300
P 9.3-3 Find the characteristic equation and
its roots for the circuit shown in Figure P 9.3-3.
2Ω
1Ω
vs
+
vc
–
+
–
1 mH
iL
Figure P 9.3-3
Solution:
v − vs
dv
+ i L + 10 × 10−6
= 0
1
dt
di
KVL: v = 2i L +10−3 L
dt
KCL:
2
diL
diL
−6
−6
−3 d iL
0 = 2iL + 10
− vs + iL + 10 × 10 ⋅ 2
+ 10 × 10 × 10
dt
dt
dt
2
di
d i
vs = 3iL + .00102 L + 1× 10−8 2L
dt
dt
2
d iL
di
+ 102000 L + 3 × 10−8 iL = 1× 108 vs
dt
dt
2
s + 102000s + 3 ×108 = 0, ∴ s1 = 3031, s2 = − 98969
−3
10 μ F
P 9.3-4 German automaker Volkswagen, in
its bid to make more efficient cars, has come
+
10u(t) V –
1 2H
up with an auto whose engine saves energy by
5 mF
shutting itself off at stoplights. The “stop–
start” system springs from a campaign to
10 Ω
+
– 7u(t) V
develop cars in all its world markets that use
less fuel and pollute less than vehicles now on
the road. The stop–start transmission control
Figure P 9.3-4
has a mechanism that senses when the car does
not need fuel: coasting downhill and idling at an intersection. The engine shuts off, but a small
starter flywheel keeps turning so that power can be quickly restored when the driver touches the
accelerator.
A model of the stop–start circuit is shown in Figure P 9.3-4. Determine the characteristic
equation and the natural frequencies for the circuit.
Answer: s2 + 20s + 400 = 0
s = – 10 ± j17.3
Solution:
Assume zero initial conditions
1 di1
1 di2
−
= 10 − 7
2 dt
2 dt
1 di1
1 di2
+
+ 200 ∫ i2 dt = 7
loop 2 : −
2 dt
2 dt
⎡⎛
⎤
1 ⎞
1
− s
⎢⎜ 10 + 2 s ⎟
⎥
2
⎝
⎠
⎥
determinant : ⎢
⎢
1
1
200
⎛
⎞⎥
⎜ s+
⎟
⎢ −2s
s ⎠ ⎥⎦
⎝2
⎣
s 2 + 20s + 400 = 0, ∴ s = − 10 ± j 17.3
loop 1 : 10i1 +
Section 9.4: Natural Response of the Unforced Parallel RLC Circuit
P 9.4-1 Determine v(t) for the circuit of
Figure P 9.4-1 when L = 1 H and vs = 0 for t ≥
0. The initial conditions are v(0) = 6 V and
dv/dt(0) = – 3000 V/s.
L
vs(t)
+
–
+
80 Ω
v(t)
–
Answer: v(t) = – 2e–100t + 8e–400t V
Figure P 9.4-1
Solution:
v ( 0 ) = 6,
dv ( 0 )
= −3000
dt
Using operators, the node equation is: Csv +
v
+
R
So the characteristic equation is: s 2 +
So v ( t ) = Ae −100t + Be−400t
( v −vs )
sL
L
⎛
⎞
= 0 or ⎜ LCs 2 + s + 1⎟ v = v
s
⎝
R
⎠
1
1
s+
= 0
RC
LC
⇒ s1,2 = − 250 ± 2502 − 40, 000 = − 100, − 400
v ( 0) = 6 = A + B
dv ( 0 )
⎫ A = −2
= − 3000 = − 100 A − 400 B ⎬
dt
⎭ B = 8
∴ v ( t ) = − 2e −100t + 8e−400t
t>0
25μ F
P 9.4-2 An RLC circuit is shown in
Figure P 9.4-2, where v(0) = 2 V. The switch
has been open for a long time before closing at
t = 0. Determine and plot v(t).
t=0
+
1 3F
v(t)
–
3 4
Ω
Figure P 9.4-2
Solution:
v ( 0 ) = 2, i ( 0 ) = 0
Characteristic equation s 2 +
v ( t ) = Ae − t + Be−3t
1
1
s+
= 0 ⇒ s 2 + 4s + 3 = 0 ⇒ s = − 1, − 3
RC
LC
Use eq. 9.5 − 12 ⇒ s1 A + s2 B = −
−1A − 3B = −
also have v ( 0 ) = 2 = A + B
2
−0 = − 8
1
4
From (1) & ( 2 ) get A = −1, B = 3
∴ v ( t ) = −e − t + 3 e −3t V
v ( 0) i ( 0)
−
RC
C
(1)
( 2)
1H
2H
P 9.4-3 Determine i1(t) and i2(t) for the circuit of
Figure P 9.4-3 when i1(0) = i2(0) = 11 A.
1 Ω
i2
i1
3H
2 Ω
Figure P 9.4-3
Solution:
di1
di
−3 2 = 0
dt
dt
di
di
KVL : − 3 1 + 3 2 + 2i2 = 0
dt
dt
KVL : i1 + 5
(1)
( 2)
in operator form
(1 + 5s ) i1 + ( −3s ) i2 = 0⎫⎪
⎬
( −3s ) i1 + ( 3s + 2 ) i2 = 0 ⎪⎭
Thus i1 ( t ) = Ae
i 2 ( t ) = Ce
-t
−t
6
6
thus Δ =
(1 + 5s )( 3s + 2 ) − 9s 2 = 6s 2 + 13s + 2 = 0
+ Be −2t
+ De-2t
Now i1 ( 0 ) = 11 = A + B; i2 ( 0 ) = 11 = C + D
from (1) & ( 2 ) get
di1 ( 0 )
di2 ( 0 )
A
C
33
143
= −
= − − 2B ;
= −
= − − 20
dt
2
6
dt
6
6
which yields A = 3, B = 8, C = − 1, 0 = 12
i1 (t ) = 3e − t /6 + 8e−2t A
&
i2 (t ) = − e− t /6 + 12e−2t A
⇒ s = − 1 −2
6,
P 9.4-4 The circuit shown in Figure P 9.4-4 contains a switch that is sometimes open and
sometimes closed. Determine the damping factor, α, the resonant frequency, ω0, and the damped
resonant frequency, ωd, of the circuit when (a) the switch is open and (b) the switch is closed.
10 Ω
40 Ω
i(t)
2H
+
–
+
50 Ω
20 V
v(t)
5 mF
–
Figure P 9.4-4
Solution:
Represent this circuit by a differential equation.
(R1 = 50 Ω when the switch is open and R1 = 10
Ω when the switch is closed.)
Use KCL to get
i (t ) =
v (t )
d
+ C v (t )
R2
dt
Use KVL to get
v s = R1 i ( t ) + L
d
i (t ) + v (t )
dt
Substitute to get
vs =
R1
R2
v ( t ) + R1C
d
L d
d2
v (t ) +
v ( t ) + CL 2 v ( t ) + v ( t )
dt
R 2 dt
dt
⎛
R1 + R 2
d2
L ⎞d
= CL 2 v ( t ) + ⎜ R1C +
v (t ) +
v (t )
⎟
⎜
dt
R 2 ⎟⎠ dt
R2
⎝
Finally,
Compare to
to get
vs
CL
=
⎛ R1
R1 + R 2
d2
1 ⎞d
v t +
+
v (t ) +
v (t )
⎟
2 ( ) ⎜
⎜ L R 2C ⎟ dt
dt
R
CL
2
⎝
⎠
d2
d
v ( t ) + ω 02 v ( t ) = f (t )
dt
dt
R1
R1 + R 2
1
+
2α =
and ω 0 2 =
L R 2C
R 2CL
v t + 2α
2 ( )
(a) When the switch is open α = 14.5 , ω 0 = 14.14 rad/s and ω d = j3.2 (the circuit is
overdamped).
(b) When the switch is closed α = 4.5 , ω 0 = 10.954 rad/s and ω d = 9.987 (the circuit is
underdamped).
P 9.4-5 The circuit shown in Figure P 9.4-5 is used to detect smokers in airplanes who
surreptitiously light up before they can take a single puff. The sensor activates the switch, and
the change in the voltage v(t) activates a light at the flight attendant’s station. Determine the
natural response v(t).
Answer: v(t) = – 1.16e–2.7t + 1.16e–37.3t V
Sensor
t=0
Light
bulb 0.4 H
1Ω
1A
+
1 40
F
v(t)
–
Figure P 9.4-5
Solution:
1
1
s +
= 0
RC
LC
s 2 + 40 s + 100 = 0
s2 +
s = − 2.7 , − 37.3
The initial conditions are v(0) = 0 , i (0) 1 A .
vn = A1e −2.7 t + A2 e−37.3t , v(0) = 0 = A1 + A2
KCL at t = 0+ yields :
(1)
v(0+ )
1 dv(0+ )
+ i (0+ ) +
=0
1
40 dt
dv(0+ )
= − 40v(0+ ) − 40i (0+ ) = − 40(1) = − 2.7 A1 − 37.3 A2
dt
from (1) and (2) ⇒ A1 = −1.16 , A2 = 1.16
∴
So v(t ) = vn (t ) = − 1.16e −2.7 t + 1.16e −37.3t
(2)
Section 9.5: Natural Response of the Critically Damped Unforced Parallel
RLC Circuit
25 mH
P 9.5-1 Find vc(t) for t > 0 for the circuit shown
in Figure P 9.5-1.
Answer: vc(t) = (3 + 6000t)e
–2000t
30u(–t) mA
100 Ω
V
vc
Figure P 9.5-1
Solution:
t >0
di c
dv
+ v c = 0, ic = 10−5 c
dt
dt
2
d vc
dv
∴
+ 4000 c + 4 × 106 vc = 0
2
dt
dt
2
6
s + 4000 s + 4 × 10 = 0 ⇒ s = −2000, − 2000 ∴ vc ( t ) = A1e −2000t + A2te −2000t
KVL a : 100ic + .025
t = 0− (Steady − State)
iL = ic ( 0
−
)=0
= ic ( 0
+
vc ( 0− ) = 3 V = vc ( 0+ )
so vc ( 0+ ) = 3 = A1
dvc ( 0+ )
dt
vc ( t ) =
= 0 = −2000 A1 + A2 ⇒ A2 = 6000
( 3 + 6000t ) e−2000t
V
)
⇒
dvc ( 0+ )
dt
= 0
+
–
10 mF
P 9.5-2 Find vc(t) for t > 0 for the circuit of
Figure P 9.5-2. Assume steady-state
20 V
conditions exist at t = 0–.
t=0
10 Ω
+
–
1 Ω
1H
+
1 4
Answer: vc(t) = – 8te–2t V
F
vc
–
Figure P 9.5-2
Solution:
( )
dvc
t
=0
KCL at vc : ∫−∞ vc dt + vc + 1
4 dt
d 2 vc
dv
⇒
+ 4 c + 4vc = 0
dt
dt
t>0
s 2 + 4 s + 4 = 0, s = −2, − 2
t = 0− (Steady-State)
⇒ vc ( t ) = A1 e −2t + A2 t e −2t
vc ( 0− ) = 0 = vc ( 0+ ) & iL ( 0− ) =
20 V
= 2 A = iL ( 0+ )
10 Ω
Since vc ( 0+ ) = 0 then ic ( 0+ ) = −iL ( 0+ ) = −2 A
∴
So vc ( 0
)
dv ( 0 )
+
= 0 = A1
+
= − 8 = A2
dt
∴ vc ( t ) = − 8te −2t V
c
dvc ( 0+ )
dt
ic ( 0+ )
=
= −8 V
S
1
4
10 mH
P 9.5-3 Police often use stun guns to
incapacitate potentially dangerous felons. The
+
t=0
hand-held device provides a series of high+
4
v
R = 106 Ω
10 V –
voltage, low-current pulses. The power of the
C
–
pulses is far below lethal levels, but it is
enough to cause muscles to contract and put
Figure P 9.5-3
the person out of action. The device provides a
pulse of up to 50,000 V, and a current of 1 mA
flows through an arc. A model of the circuit for one period is shown in Figure P 9.5-3. Find v(t)
for 0 < t < 1 ms. The resistor R represents the spark gap. Select C so that the response is critically
damped.
Solution:
Assume steady − state at t = 0− ∴ vc ( 0− ) = 10 4 V & iL ( 0− ) = 0
t>0
Also
∴ 0.01C
diL
+ 106 iL = 0
(1)
dt
⎡ d 2iL
di ⎤
dvc
: iL = −C
= −C ⎢.01 2 + 106 L ⎥
dt
dt
dt ⎦
⎣
KVL a : − vc + .01
( 2)
d 2iL
di
+ 106 C L + iL = 0
2
dt
dt
Characteristic eq. ⇒ 0.01C s 2 + 106 s + 1 = 0 ⇒ s =
−106 C ±
(10 C )
6
2 (.01C )
for critically damped: 1012 C2 − .04C = 0
⇒ C = 0.04 pF ∴ s = −5 × 107 , −5 × 107
So iL ( t ) = A1e −5×10 t + A2te−5×10
7
7t
diL +
0 ) = 100 ⎡⎣vc ( 0+ ) − 106 iL ( 0+ ) ⎤⎦ = 106 A
(
s
dt
di ( 0 )
7
So i L ( 0 ) = 0 = A1 and L
= 106 = A2 ∴ iL ( t ) = 106 te−5×10 t A
dt
Now from (1) ⇒
Now v ( t ) = 106 iL ( t ) = 1012 te−5×10 t V
7
2
− 4 (.01C )
P 9.5-4 Reconsider Problem P 9.4-1 when L = 640
mH and the other parameters and conditions remain the
same.
vs(t)
– 250t
Answer: v(t) = (6 – 1500t)e
V
L
+
–
+
80 Ω
v(t)
–
Figure P 9.4-1
Solution:
s2 +
1
1
1
1
s+
= 0 with
= 500 and
= 62.5 × 103 yields s = −250, −250
RC
LC
RC
LC
v ( t ) = Ae −250t + Bte −250t
v ( 0) = 6 = A
dv ( 0 )
= −3000 = − 250 A + B ⇒ B = − 1500
dt
∴ v ( t ) = 6e −250t − 1500te −250t
25μ F
P 9.5-5 An automobile ignition uses an electromagnetic
trigger. The RLC trigger circuit shown in Figure P 9.5-5
has a step input of 6 V, and v(0) = 2 V and i(0) = 0. The
resistance R must be selected from 2 Ω < R < 7 Ω so that
the current i(t) exceeds 0.6 A for greater than 0.5 s in
order to activate the trigger. A critically damped response
i(t) is required to avoid oscillations in the trigger current.
Select R and determine and plot i(t).
1H
Trigger
di
t
+ Ri + 2 + 4∫0 idt = 6
dt
v (t )
Figure P 9.5-5
(1)
taking the derivative with respect to t :
d 2i
di
+ R + 4i = 0
2
dt
dt
Characteristic equataion: s 2 + Rs + 4 = 0
Let R = 4 for critical damping ⇒
( s + 2)
2
=0
So i ( t ) = Ate −2t + Be −2t
i ( 0) = 0 ⇒ B = 0
di ( 0 )
= 4 − R (i ( 0)) = 4 − R ( 0) = 4 = A
dt
= 4te −2t A
from (1)
∴ i (t )
F
+ –
v(t)
6 u(t) V +–
Solution:
KVL :
1 4
i
R
Section 9-6: Natural Response of an Underdamped Unforced Parallel RLC
Circuit
P 9.6-1 A communication system from a
space station uses short pulses to control a
robot operating in space. The transmitter
circuit is modeled in Figure P 9.6-1. Find the
output voltage vc(t) for t > 0. Assume steadystate conditions at t = 0–.
0.8 H
vc
250 Ω
+
–
5 × 10-6 F
250 Ω
t=0
Answer:
vc(t) = e– 400t [3 cos 300t + 4 sin 300t] V
+
–
6V
Figure P 9.6-1
Solution:
t>0
KCL at vc :
also :
Solving for i
L
vc
250
+ iL + 5 ×10−6
vc = 0.8
dvc
= 0
dt
diL
dt
in (1) & plugging into ( 2 )
d 2 vc
dv
+800 c + 2.5×105 vc = 0 ⇒ s 2 +800s+ 250,000 = 0, s = −400± j 300
2
dt
dt
∴ v (t ) = e
c
t = 0−
−400t ⎡
⎤
⎣ A1 cos300t + A 2 sin 300t ⎦
(Steady − State)
−6 V −6
=
A = iL ( 0+ )
500
500 Ω
−
+ 6 = 3 V = vc ( 0+ )
vc ( 0 ) = 250 −6
500
iL ( 0− ) =
(
Now from (1) :
So vc ( 0
+
)
dvc ( 0+ )
)
dvc ( 0+ )
dt
= 3 = A1
= − 2 × 105 iL ( 0+ ) − 800vc ( 0+ ) = 0
= 0 = − 400 A1 + 300 A2 ⇒ A2 = 4
dt
∴ vc ( t ) = e −400t [3cos 300t + 4sin 300t ] V
( 2)
(1)
P 9.6-2 The switch of the circuit shown in Figure P 9.6-2 is opened at t = 0. Determine and plot
v(t) when C = 1/4 F. Assume steady state at t = 0–.
Answer: v(t) = – 4e–2t sin 2t V
3Ω
6V
t=0
+
–
1Ω
+
1 2
H
C
–
v(t)
Figure P 9.6-2
Solution:
t = 0−
i (0) = 2 A
v (0) = 0
t = 0−
KCL at node a:
t
v
dv 1
+ C + ∫ vdt + i ( 0 ) = 0 (1)
dt L 0
1
in operator form have v + Csv +
with s 2 + 4s + 8 = 0
1
v + i ( 0 ) = 0 or
Ls
1 ⎞
⎛ 2 1
⎜s + s+
⎟v = 0
C
LC ⎠
⎝
⇒ s = −2 ± j 2
v ( t ) = e −2t [ B1 cos 2t + B2 sin 2t ]
v ( 0 ) = 0 = B1
dv ( 0 )
1
⎡ −i ( 0 ) − v ( 0 ) ⎤⎦ = −4 [ 2] = −8 = 2 B2 or B2 = −4
dt
C ⎣
So v ( t ) = −4e −2t sin 2 t V
From (1) ,
=
P 9.6-3 A 240-W power supply circuit is shown in Figure P 9.6-3a. This circuit employs a
large inductor and a large capacitor. The model of the circuit is shown in Figure P 9.6-3b. Find
iL(t) for t > 0 for the circuit of Figure P 9.6-3b. Assume steady-state conditions exist at t = 0–.
Answer: iL(t) = e–2t(– 4 cos t + 2 sin t) A
4H
iL
t=0
14
2Ω
F
8Ω
4Ω
7A
(b)
(a)
Figure P 9.6-3
Solution:
t>0
1 dvc vc
+ + iL = 0
4 dt 2
4diL
KVL : vc =
+ 8 iL
( 2)
dt
KCL at vc :
d 2iL
di
+ 4 L + 5iL = 0 ⇒ s 2 + 4s + 5 = 0 ⇒ s = − 2 ± i
2
dt
dt
−2 t
= e [ A1 cos t + A2 sin t ]
( 2 ) into (1) yields
∴ iL ( t )
t = 0−
( Steady − State )
vc ( 0− )
2
⎛ 48 ⎞
= 7 ⎜⎜
⎟⎟
⎝ 4 8+ 2⎠
⇒ vc ( 0− ) = 8 V = vc ( 0+ )
−8 V
= −4 A = iL ( 0+ )
2Ω
diL ( 0+ ) vc ( 0+ )
8V
A
+
=
− 2iL ( 0 ) =
− 2 ( −4 ) = 10
dt
4
4
s
iL ( 0− ) =
∴ from ( 2 )
So i L ( 0
∴ iL ( t )
+
) = -4 = A
1
and
diL ( 0+ )
dt
−2 t
= e [ −4 cos t + 2 sin t
= 10 = −2 A1 + A2 ⇒ A2 = 2
]A
(1)
600
P 9.6-4 The natural response of a
parallel RLC circuit is measured and
plotted as shown in Figure P 9.6-4.
Using this chart, determine an
expression for v(t).
500
400
300
200
Hint: Notice that v(t) = 260 mV at
t = 5 ms and that v(t) = – 200 mV at
t = 7.5 ms. Also, notice that the time
between the first and third zerocrossings is 5 ms.
v(t)
(mV)
100
0
– 100
Answer:
v(t) = 544e– 276t sin 1257t V
– 200
– 300
– 400
0
5
10
15
20
25
Time (ms)
Figure P 9.6-4
Solution:
The response is underdamped so
∴ v ( t ) = e − α t [ k1 cos ωt + k2 sin ωt ] + k3
v ( ∞ ) = 0 ⇒ k3 = 0, v ( 0 ) = 0 ⇒ k1 = 0
∴ v ( t ) = k2 e − α t sin ωt
From Fig. P 9.6-4
t ≈ 5ms ↔ v ≈ 260mV (max)
t ≈ 7.5ms ↔ v ≈ −200 mV (min)
∴ distance between adjacent maxima is ≈ ω =
so
0.26 = k2 e
−0.2 = k2 e
−α (.005 )
−α
2π
= 1257 rad
s
T
sin (1257 (.005 ) )
(.0075) sin
(1)
( 1257 (.0075) ) ( 2 )
Dividing (1) by (2) gives
⎛ sin ( 6.29 rad ) ⎞
− 1.3 = eα ( 0.0025) ⎜⎜
⎟⎟ ⇒
⎝ sin ( 9.43 rad ) ⎠
From (1) k 2 = 544 so
v ( t ) = 544e−267t sin1257t
e0.0025 α = 1.95 ⇒ α = 267
( approx. answer )
30
P 9.6-5 The photovoltaic cells of the
proposed space station shown in
Figure P 9.6-5a provide the voltage v(t) of the
circuit shown in Figure P 9.6-5b. The space
station passes behind the shadow of earth (at
t = 0) with v(0) = 2 V and i(0) = 1/10 A.
Determine and sketch v(t) for t > 0.
Photocells
(a)
i
5Ω
+
v
1 10 F
2H
–
Space
station
electric motors
The photovoltaic
cells connected
in parallel
(b)
Figure P 9.6-5
Solution:
v ( 0 ) = 2 V and i ( 0 ) = 1 A
10
Char. eq. ⇒ s 2 +
1
1
s+
= 0 or s 2 + 2 s + 5 = 0 thus the roots are s = −1 ± j 2
RC
LC
So have v(t ) = e − t ⎡ B cos 2t + B2 sin 2t ⎤ now v(0+ ) = 2 = B1
⎣ 1
⎦
Need
So
dv ( 0+ )
dv ( 0
dt
dt
+
v ( 0+ )
1
1 V
+
+
= ic ( 0 ) . KCL yields ic ( 0 ) = −
− i ( 0+ ) = −
C
5
2 s
) = 10 ⎛ − 1 ⎞ = − B + 2 B
⎜
⎟
⎝ 2⎠
1
2
⇒ B2 = − 3
2
3
Finally, we have v ( t ) = 2e − t cos 2t − e− t sin 2t V
2
t>0
Section 9-7: Forced Response of an RLC Circuit
P 9.7-1 Determine the forced response for
the inductor current if when (a) is = 1 A, (b) is =
is u(t) A
0.5t A, and (c) is = 2e–250t A for the circuit of
Figure P 9.7-1.
100 65 Ω
10 mH
i
Figure P 9.7-1
Solution:
v
dv
+ iL + C
R
dt
diL
KVL : v = L
dt
d 2iL
L diL
+ iL + LC
is =
R dt
dt 2
KCL : is =
(a)
is = l u (t ) ∴ assume i f = A
d 2iL
1 diL
1
iL = is
+
+
2
dt
RC dt LC
1
to get: 0 + 0 + A
= 1 ⇒ A = 1× 10−5 = i f
−3
.01
1
10
×
( )(
)
Let iL = i f = A in
(b)
is = 0.5t u (t ) ∴ assume i f = At + B
0+ A
65
1
+ ( At + B )
= 0.5 t
(.01)(.001)
(100 ) (.001)
⇒ 650 A + 100000 B = 0 and 100000 At = 0.5t
A = 5 ×10−6
B = 3.25 × 10−8
i f = 5 × 10−6 t − 3.25 × 10−8 A
(c)
is = 2e −250t Assumming i f = Ae −250t does not work
because i f cannot have the same form as is ∴ we choose i f = Bte−250t
Be−250t −250 Bte−250t Bte−250t
+
+
= 2 e −250t
RC
RC
LC
150 B = 2
B = 0.0133
if
= 0.0133 te −250t A
1 mF
7Ω
P 9.7-2 Determine the forced response for the
capacitor voltage, vf, for the circuit of Figure P 9.7-2
when
vs u(t) V –+
0.1 H
+
v
–
833.3 μ F
(a) vs = 2 V, (b) vs = 0.2t V, and (c) vs = 1e–30t V.
Figure P 9.7-2
Solution:
Represent the circuit by the differential equation:
(a)
vs = 2 ∴ assume v f = A
Then 0 + 0 + 12000 A = 2 so A = 1
(b)
d 2 v R dv 1
v = vs
+
+
dt L dt LC
6000
= vf
vs = 0.2t ∴ assume v f = At + B
70 A + 12000 At + 12000 B = 0.2t ⇒ 70 A + 12000 B = 0 and 12000 At = 0.2t
1
70 A
, B=
60000
12000
t
∴v f =
+ 350 V
60000
A=
(c)
vs = e −30t
⇒
B = 350
∴ assume v f = Ae−30t
900 A − 2100 Ae −30t + 12000 Ae−30t = e−30t
vf =
e −30t
V
10800
⇒ 10800 Ae−30t = e −30t
⇒
A =
1
10800
P 9.7-3
A circuit is described for t > 0 by the equation
d 2v
dv
+ 5 + 6v = vs
2
dt
dt
Find the forced response vf for t > 0 when (a) vs = 8 V, (b) vs = 3e–4t V, and (c) vs = 2e–2t V.
3
Answer: (a) vf = 8/6 V (b) vf = e– 4t V (c) vf = 2te–2t V
2
Solution:
(a)
(b)
d 2v
dv
+ 5 + 6v = 8 so we try v f = B.
2
dt
dt
Substituting v f = B into the differential equation gives 6 B = 8 ∴ vf = 8 / 6 V
The differential equation is
d 2v
dv
+ 5 + 6v = 3e −4t so we try v f = Be −4t .
2
dt
dt
into the differential equation gives
The differential equation is
Substituting v f = Be −4t
(−4) 2 B + 5(−4) B + 6 B = 3 ⇒ B = 3 / 2
∴ vf = 3 / 2e −4t
(c)
d 2v
dv
+ 5 + 6v = 2e −2t so we try vf = Bte−2t
2
dt
dt
(since − 2 is a natural frequency). Substituting vf = Bte −2t into the differential equation gives
The differential equation is
⇒ (4t − 4) B + 5B (1 − 2t ) + 6 Bt = 2 ⇒ B = 2
∴ vf = 2te −2t
Section 9-8: Complete Response of an RLC Circuit
P 9.8-1
Determine i(t) for t > 0 for the circuit shown in Figure P 9.8-1.
iL
i
2 kΩ +
vc
1 kΩ
–
11 mA
t=0
6.25 H
1 μF
+
–
4V
Figure P 9.8-1
Solution:
First, find the steady state response for t < 0, when the switch is open. Both inputs are constant
so the capacitor will act like an open circuit at steady state, and the inductor will act like a short
circuit. After a source transformation at the left of the circuit:
i L ( 0) =
and
22 − 4
= 6 mA
3000
v C ( 0) = 4 V
After the switch closes
Apply KCL at node a:
vC
R
+C
d
vC + iL = 0
dt
Apply KVL to the right mesh:
L
d
d
i L + Vs − vC = 0 ⇒ vC = L i L + Vs
dt
dt
After some algebra:
Vs
d2
1 d
1
d2
d
16
⎛ 4
⎞
i
+
i
+
i
=
−
⇒
i + (103 ) i L + ⎜ × 106 ⎟ i L = − × 103
L
L
2 L
2 L
dt
R C dt
LC
R LC
dt
dt
25
⎝ 25
⎠
The characteristic equation is
⎛ 4
⎞
s 2 + (103 ) s + ⎜ ×106 ⎟ = 0 ⇒ s1,2 = −200, − 800 rad/s
⎝ 25
⎠
After the switch closes the steady-state inductor current is iL(∞) = -4 mA so
i L ( t ) = −0.004 + A1 e −200 t + A2 e −800 t
4
⎛ 4 ⎞d
⎡⎣( −200 ) A1 e −200 t + ( −800 ) A2 e−800 t ⎤⎦ + 4
vC (t ) = ⎜ ⎟ i L (t ) + 4 =
25
⎝ 25 ⎠ dt
= ( −32 ) A1 e −200 t + ( −128 ) A2 e −800 t + 4
Let t = 0 and use the initial conditions:
0.006 = −0.004 + A1 + A2
⇒ 0.01 = A1 + A2
4 = ( −32 ) A1 + ( −128 ) A2 + 4 ⇒
A1 = ( −4 ) A2
So A1 = 8.01 and A2 = 2.00 and
i L ( t ) = −0.004 + 8.01 e −200 t + 2.00 e −800 t A
v C ( t ) = ( −104 ) e −200 t + (104 ) e −800 t + 4 V
i (t ) =
vC (t )
1000
= ( −10 ) e −200 t + (10 ) e−800 t + 0.004 A
Determine i(t) for t > 0 for the circuit shown in Figure P 9.8-2.
d2
d
Hint: Show that 1 = 2 i (t ) + 5 i (t ) + 5i (t ) for t > 0
dt
dt
P 9.8-2
1Ω
2u(t) – 1 V
+
–
4Ω
+
v(t)
–
0.25 F
i(t)
4H
Figure P 9.8-2
Answer:
–3.62t
i(t) = 0.2 + 0.246 e
– 0.646 e–1.38t A
for t > 0.
Solution:
First, find the steady state response for t < 0. The input is constant so the capacitor will act like
an open circuit at steady state, and the inductor will act like a short circuit.
i ( 0) =
−1
= 0.2 A
1+ 4
and
v (0) =
4
( −1) = −0.8 V
1+ 4
For t > 0
Apply KCL at node a:
v − Vs
d
+C v+i = 0
R1
dt
Apply KVL to the right mesh:
R2 i + L
d
d
i − v = 0 ⇒ v = R2 i + L iL
dt
dt
After some algebra:
L + R1 R 2C d
R1 + R 2
d2
Vs
i
+
i
+
i=
2
dt
R1 L C dt
R1 L C
R1 L C
The forced response will be a constant, if = B so 1 =
⇒
d2
d
i +5 i +5i =1
2
dt
dt
d2
d
B + 5 B + 5B ⇒ B = 0.2 A .
2
dt
dt
To find the natural response, consider the characteristic equation:
0 = s 2 + 5 s + 5 = ( s + 3.62 )( s + 1.38 )
The natural response is
in = A1 e−3.62 t + A2 e−1.38 t
so
i ( t ) = A1 e−3.62 t + A2 e−1.38 t + 0.2
Then
d
⎛
⎞
v ( t ) = ⎜ 4 i ( t ) + 4 i ( t ) ⎟ = −10.48 A1 e−3.62 t − 1.52 A2 e−1.38 t + 0.8
dt
⎝
⎠
At t=0+
−0.2 = i ( 0 + ) = A1 + A2 + 0.2
−0.8 = v ( 0 + ) = −10.48 A1 − 1.52 A2 + 0.8
so A1 = 0.246 and A2 = -0.646. Finally
i ( t ) = 0.2 + 0.246 e−3.62 t − 0.646 e−1.38 t A
P 9.8-3
Answer:
Determine v1(t) for t > 0 for the circuit shown in Figure P 9.8-3.
4
3
v1 (t ) = 10 + e−2.4 × 10 t − 6e−4 × 10 t V for t > 0
1 kΩ
10 V
+
–
1 kΩ
+
v1(t)
–
1/6 μ F
+
v2(t)
–
t=0
1/16 μ F
Figure P 9.8-3
Solution:
First, find the steady state response for t < 0. The input is constant so the capacitors will act like
an open circuits at steady state.
v1 ( 0 ) =
and
1000
(10 ) = 5 V
1000 + 1000
v2 ( 0 ) = 0 V
For t > 0,
Node equations:
v1 − 10 ⎛ 1
v −v
⎞ d
+ ⎜ × 10−6 ⎟ v1 + 1 2 = 0
1000 ⎝ 6
1000
⎠ dt
⎛1
⎞ d
⇒ 2 v1 + ⎜ × 10−3 ⎟ v1 − 10 = v2
⎝6
⎠ dt
v1 − v2 ⎛ 1
⎞ d
= ⎜ × 10−6 ⎟ v2
1000 ⎝ 16
⎠ dt
⎛1
⎞ d
⇒ v1 − v2 = ⎜ × 10−3 ⎟ v2
⎝ 16
⎠ dt
After some algebra:
d2
d
v + ( 2.8 × 104 ) v1 + ( 9.6 × 107 ) v1 = 9.6 × 108
2 1
dt
dt
The forced response will be a constant, vf = B so
d2
d
B + ( 2.8 × 104 ) B + ( 9.6 ×107 ) B = 9.6 × 108 ⇒ B = 10 V .
2
dt
dt
To find the natural response, consider the characteristic equation:
s 2 + ( 2.8 ×104 ) s + ( 9.6 × 107 ) = 0 ⇒ s1,2 = −4 × 103 , −2.4 × 104
The natural response is
3
vn = A1 e −4×10 t + A2 e−2.4×10
4t
so
3
v1 ( t ) = A1 e−4×10 t + A2 e−2.4×10
4t
+ 10
At t = 0
5 = v1 ( 0 ) = A1 e
−4×103 ( 0 )
+ A2 e
−2.4×104 ( 0 )
+ 10 = A1 + A2 + 10
(1)
Next
⎛1
⎞ d
2 v1 + ⎜ ×10−3 ⎟ v1 − 10 = v2
⎝6
⎠ dt
⇒
d
v1 = 12000v1 + 6000 v2 − 6 × 104
dt
At t = 0
d
v1 ( 0 ) = 12000v1 ( 0 ) + 6000 v2 ( 0 ) − 6 × 104 = 12000 ( 5 ) + 6000 ( 0 ) − 6 × 104 = 0
dt
so
3
4
d
v1 ( t ) = A1 −4 ×103 e−4×10 t + A2 −2.4 × 104 e−2.4×10 t
dt
(
)
(
)
At t = 0+
0=
d
−4×103 ( 0 )
−2.4×104 ( 0 )
v1 ( 0 ) = A1 −4 × 103 e
+ A2 −2.4 × 104 e
= A1 −4 × 103 + A2 −2.4 × 104
dt
(
)
(
)
(
so A1 = -6 and A2 = 1. Finally
v1 ( t ) = 10 + e −2.4 ×10
4t
3t
− 6 e −4 ×10
V for t > 0
)
(
)
1Ω
P 9.8-4 Find v(t) for t > 0 for the circuit shown in
Figure P 9.8-4 when v(0) = 1 V and iL(0) = 0.
+
5 cos t V –
Answer:
v = 25e −3t −
1
⎡ 429e−4t − 21cos t + 33 sin t ⎤⎦ V
17 ⎣
0.5 H
iL
1Ω
1 12 F
+
–
v
Figure P 9.8-4
Solution:
t > 0
di
⎛
⎞
KCL at top node : ⎜ 0.5 L − 5cos t ⎟ + iL + 1 dv
= 0
12 dt
dt
⎝
⎠
di
dv
KVL at right loop : 0.5 L = 1
+v
( 2)
12 dt
dt
d
d
d 2iL diL 1 d 2 v
+
+
= −5sin t (3)
dt 2
dt 12 dt 2
d 2iL
d 2 v dv
1
⇒ 0.5 2 =
+
( 4)
12 dt 2 dt
dt
dt
of (1) ⇒ 0.5
dt
of ( 2 )
Solving for
d 2iL
di
in ( 4 ) and L in ( 2 ) & plugging into ( 3)
2
dt
dt
d 2v
dv
+ 7 + 12v = − 30sin t
2
dt
dt
⇒ s 2 + 7 s + 12 = 0 ⇒ s = − 3, − 4
so v(t ) = A1e−3t + A2 e−4t + v f
Try v f = B1 cos t + B2 sin t & plug
into D.E., equating like terms
yields B1 = 21 , B2 = − 33
17
17
t = 0+
5 −1
dv(0+ )
ic ( 0 ) =
= 2A ∴
= 2
= 24 V
1
s
11
dt
12
⎫
So v(0+ ) = 1 = A1 + A2 + 21
A1 = 25
17
⎪
⎬
429
dv(0+ )
⎪ A2 = −
= 24 = − 3 A1 − 4 A2 − 33
17
17
dt
⎭
∴ v(t ) = 25e −3t − 1 ( 429e −4t − 21cos t + 33sin t ) V
17
+
(1)
Find v(t) for t > 0 for the circuit of Figure P 9.8-5.
P 9.8-5
Answer: v(t) = [ – 16e–t + 16e – 3t + 8]u(t) + [16e–(t–2) – 16e–3(t–2) – 8]u(t – 2) V
1 3
F
+
2[u(t) – u(t – 2)] A
4Ω
v
1H
–
Figure P 9.8-5
Solution: Use superposition − first consider the 2u (t ) source.
KVL at right mesh : vc + siL + 4(iL − 2) = 0
also : iL = (1/ 3) svc ⇒ vc = (3 / s ) iL
(1)
(2)
Plugging (2) into (1) yields ( s 2 + 4s + 3) iL = 0 , roots : s = −1 , −3
So iL (t ) = A1e − t + A2 e−3t
−
t = 0 ⇒ circuit is dead ∴ vc (0) = iL (0) = 0
Now from (1)
diL (0+ )
= 8 − 4iL (0+ ) − vC (0+ ) = 8 A/s
dt
So iL (0) = 0 = A1 + A2 ⎫
⎪
⎬ A1 = 4 , A2 = −4
diL (0)
= 8 = − A1 − 3 A2 ⎪
dt
⎭
−t
−3t
∴ iL (t ) = 4e − 4e
∴ v1 (t ) = 8 − 4 iL (t ) = 8 − 16e − t + 16e −3t V
Now for 2u (t − 2) source, just take above expression and replace t → t − 2 and flip signs
∴ v2 (t ) = −8 + 16e − (t − 2) − 16 e −3(t − 2) V
∴ v(t ) = v1 (t ) + v2 (t )
v(t ) = ⎡⎣8 − 16e − t + 16e −3t ⎤⎦ u (t ) + ⎡⎣ −8 + 16e− (t − 2) − 16 e−3( t − 2) ⎤⎦ u (t − 2) V
P 9.8-6 An experimental space station power supply system is modeled by the circuit shown in
Figure P 9.8-6. Find v(t) for t > 0. Assume steady-state conditions at t = 0–.
t=0
(10 cos t)u(t) V
– +
0.125 F
4Ω
4H
+
v(t)
–
+
–
2Ω
5V
i(t)
Figure P 9.8-6
Solution:
First, find the steady state response for t < 0, when the switch is closed. The input is constant so
the capacitor will act like an open circuit at steady state, and the inductor will act like a short
circuit.
i ( 0) = −
and
5
= −1.25 mA
4
v (0) = 5 V
After the switch closes
Apply KCL at node a:
v
d
+ 0.125 v = i
dt
2
Apply KVL to the right mesh:
−10 cos t + v + 4
After some algebra:
d
i+4i =0
dt
d2
d
v + 5 v + 6 v = 20 cos t
2
dt
dt
The characteristic equation is
s 2 + 5 s + 6 = 0 ⇒ s1,2 = −2, − 3 rad/s
Try
vf = A cos t + B sin t
d2
d
A cos t + B sin t ) + 5 ( A cos t + B sin t ) + 6 ( A cos t + B sin t ) = 20 cos t
2 (
dt
dt
( − A cos t − B sin t ) + 5 ( − A sin t + B cos t ) + 6 ( A cos t + B sin t ) = 20 cos t
( − A + 5 B + 6 A) cos t + ( − B − 5 A + 6 B ) sin t = 20 cos t
vf = 2 cos t + 2 sin t
So A =2 and B =2. Then
v ( t ) = 2 cos t + 2 sin t + A1 e −2 t + A2 e−3 t
v (t )
d
+ 0.125 v ( t ) = i ( t ) ⇒
dt
2
Next
d
v (t ) = 8 i (t ) − 4 v (t )
dt
d
V
⎛ 5⎞
v ( 0 ) = 8 i ( 0 ) − 4 v ( 0 ) = 8 ⎜ − ⎟ − 4 ( 5 ) = −30
dt
s
⎝ 4⎠
Let t = 0 and use the initial conditions:
5 = v ( 0 ) = 2 cos 0 + 2 sin 0 + A1 e −0 + A2 e−0 = 2 + A1 + A2
d
v ( t ) = −2 sin t + 2 cos t − 2 A1 e −2 t − 3 A2 e−3 t
dt
−30 =
d
v ( 0 ) = −2 sin 0 + 2 cos 0 − 2 A1 e −0 − 3 A2 e −0 = 2 − 2 A1 − 3 A2
dt
So A1 = -23 and A2 = 26 and
v ( t ) = 2 cos t + 2 sin t − 23 e −2 t + 26 e −3 t
8Ω
P 9.8-7 Find vc(t) for t > 0 in the circuit of Figure P 9.8-7
when (a) C = 1/18 F, (b) C = 1/10 F, and (c) C = 1/20 F.
4Ω
2u(t) A
Answer:
(a)
vc(t) = 8e–3t + 24te – 3t – 8 V
(b)
vc(t) = 10e–t – 2e–5t – 8 V
(c)
vc(t) = e – 3t(8 cos t + 24 sin t) – 8 V
C
2H
a
+
v(t)
–
i(t)
Figure P 9.8-7
Solution:
First, find the steady state response for t < 0, when the switch is closed. The input is constant so
the capacitor will act like an open circuit at steady state, and the inductor will act like a short
circuit.
i ( 0) = 0 A
and
v ( 0) = 0 V
After the switch closes
d
v=i
dt
Apply KVL to the right mesh:
d
8 i + v + 2 i + 4 (2 + i) = 0
dt
d
12 i + v + 2 i = −8
dt
⎛ 1 ⎞
d2
d
4
After some algebra:
v + ( 6) v + ⎜
⎟v = −
2
dt
dt
C
⎝2C⎠
The forced response will be a constant, vf = B so
Apply KCL at node a:
C
⎛ 1 ⎞
d2
d
4
B + ( 6) B + ⎜
⎟B = −
2
dt
dt
C
⎝2C⎠
(a)
⇒ B = −8 V
d2
d
v + ( 6 ) v + ( 9 ) v = −72
2
dt
dt
2
The characteristic equation is s + 6 s + 9 = 0 ⇒ s1,2 = −3, −3
When C = 1/18 F the differential equation is
Then v ( t ) = ( A1 + A2 t ) e−3t − 8 .
Using the initial conditions:
0 = v ( 0 ) = ( A1 + A2 ( 0 ) ) e0 − 8 ⇒
0=
So
(b)
d
v ( 0 ) = −3 ( A1 + A2 ( 0 ) ) e0 + A2 e0
dt
A1 = 8
⇒
A2 = 24
v ( t ) = ( 8 + 24 t ) e −3t − 8 V for t > 0
d2
d
When C = 1/10 F the differential equation is 2 v + ( 6 ) v + ( 5 ) v = −40
dt
dt
2
The characteristic equation is s + 6 s + 5 = 0 ⇒ s1,2 = −1, −5
Then v ( t ) = A1 e − t + A2 e−5 t − 8 .
Using the initial conditions:
0 = v ( 0 ) = A1 e0 + A2 e0 − 8 ⇒ A1 + A2 = 8 ⎫
⎪
⎬ ⇒ A1 = 10 and A2 = −2
d
0
0
0 = v ( 0 ) = − A1 e − 5 A2 e ⇒ − A1 − 5 A2 = 0 ⎪
dt
⎭
So
v ( t ) = 10 e − t − 2 e −5 t − 8 V for t > 0
(c)
d2
d
v + ( 6 ) v + (10 ) v = −80
2
dt
dt
2
The characteristic equation is s + 6 s + 10 = 0 ⇒ s1,2 = −3 ± j
When C = 1/20 F the differential equation is
Then v ( t ) = e −3 t ( A1 cos t + A2 sin t ) − 8 .
Using the initial conditions:
0 = v ( 0 ) = e0 ( A1 cos 0 + A2 sin 0 ) − 8 ⇒
0=
So
A1 = 8
d
v ( 0 ) = −3 e0 ( A1 cos 0 + A2 sin 0 ) + e0 ( − A1 sin 0 + A2 cos 0 ) ⇒
dt
v ( t ) = e −3 t ( 8cos t + 24 sin t ) − 8 V for t > 0
A2 = 24
P 9.8-8 Find vc(t) for t > 0 for the circuit
shown in Figure P 9.8-8.
4Ω
8Ω
Hint:
d2
d
2 = 2 vc (t ) + 6 vc (t ) + 2vc (t ) for t > 0
dt
dt
Answer:
vc(t) = 0.123e – 5.65t + 0.877e – 0.35t + 1 V
for t > 0.
– 1 4 u(t) + 1 2 A
+
1 4
2H
F
iL(t)
Figure P 9.8-8
Solution:
The circuit will be at steady state for t<0:
so iL(0+) = iL(0-) = 0.5 A and vC(0+) = vC(0-) = 2 V.
For t>0:
Apply KCL at node b to get:
1
1 d
1 1 d
= i L (t ) +
v C (t ) ⇒ i L (t ) = −
v C (t )
4
4 dt
4 4 dt
Apply KVL to the right-most mesh to get:
4 i L (t ) + 2
d
⎛1 d
⎞
i L (t ) = 8 ⎜
v C (t ) ⎟ + v C (t )
dt
⎝ 4 dt
⎠
Use the substitution method to get
d ⎛1 1 d
⎛1 1 d
⎞
⎞
⎛1 d
⎞
4⎜ −
vC ( t ) ⎟ + 2 ⎜ −
vC ( t ) ⎟ = 8 ⎜
vc ( t ) ⎟ + v ( t )
dt ⎝ 4 4 dt
⎝ 4 4 dt
⎠
⎠
⎝ 4 dt
⎠ c
d2
d
2=
v C (t ) + 6 v C (t ) + 2 v C (t )
or
dt
dt 2
d
d2
The forced response will be a constant, vC= B so 2 = 2 B + 6 B + 2 B ⇒ B = 1 V .
dt
dt
To find the natural response, consider the characteristic equation:
0 = s 2 + 6 s + 2 = ( s + 5.65 )( s + 0.35 )
vC(t)
–
The natural response is
vn = A1 e −5.65t + A e −0.35t
2
vC ( t ) = A e
so
1
−5.65 t
+A e
−0.35t
2
+1
Then
i L (t ) =
At t=0+
1 1 d
1
+
vC ( t ) = + 1.41A e −5.65t + 0.0875 A e−0.35t
1
2
4 4 dt
4
2 = vC ( 0 + ) = A + A + 1
1
2
1
1
= i L ( 0 + ) = + 1.41A1 + 0.0875 A
2
2
4
so A1 = 0.123 and A2 = 0.877. Finally
vC ( t ) = 0.123 e −5.65 t + 0.877 e −0.35 t + 1 V
P 9.8-9 In Figure P 9.8-9, determine the
inductor current i(t) when is = 5u(t) A. Assume
that i(0) = 0, vc(0) = 0.
is
2Ω
Answer: i(t) = 5 + e–2t [ – 5 cos 5t – 2 sin 5t] A
8 29 H
1 8F
i
Figure P 9.8-9
Solution:
dv
v
+i+ =i
2 s
dt
d 2i
⎛ L ⎞ di
LC
+ i + ⎜ ⎟ = 5 u (t )
dt
⎝ 2 ⎠ dt
KCL : C
1 d 2i
⎛ 4 ⎞ di
+ i + ⎜ ⎟ = 5 u (t )
29 dt
⎝ 29 ⎠ d t
d 2i
di
+ 4 + i = 145 u ( t )
dt
dt
Characteristic eqn: s 2 + 4s + 29 = 0 ⇒
roots : s = − 2 ± j5
∴ in = e −2t [ A cos 5t + B sin 5t ] and i f = 145 = 5
29
−2t
So i (t ) = 5 + e [ A cos 5t + B sin 5t ]
Now i (0) = 0 = A + 5 ⇒ A = −5
di (0)
= 0 = −2 A + 5 B ⇒ B = − 2
dt
P 9.8-10 Railroads widely use
automatic identification of
railcars. When a train passes a
tracking station, a wheel detector
activates a radio-frequency
module. The module’s antenna, as
shown in Figure P 9.8-10a,
transmits and receives a signal
that bounces off a transponder on
the locomotive. A trackside
processor turns the received signal
into useful information consisting
of the train’s location, speed, and
direction of travel. The railroad
uses this information to schedule
locomotives, trains, crews, and
equipment more efficiently.
One proposed transponder
circuit is shown in Figure P 9.810b with a large transponder coil
of L = 5 H. Determine i(t) and
v(t). The received signal is
is = 9 + 3e–2tu(t) A.
(a)
Vehicle-mounted
transponder tag
Wheel detector
input
Antenna
+
–
(b)
0.5 F
L
1Ω
v
1.5 Ω
i
is
0.5 Ω
Figure P 9.8-10
Solution:
t = 0−
t > 0
2
× 9 = 6 A = i (0+ )
2 +1
1
& v (0− ) =
× 9× 1.5 = 4.5 V = v(0+ )
2 +1
i (0− ) =
dv v
+
=i
dt 1.5 s
5di
dv v
KVL : v + (0.5
+ ) (0.5) =
+ i
dt 1.5
dt
KCL at middle node: i + 0.5
(1)
(2)
Solving for i in (1) and plugging into (2) yields
d
d 2 v ⎛ 49 ⎞ dv ⎛ 4 ⎞
⎛2⎞
is where i = 9 + 3e−2t A
2
+
+
=
+
v
i
⎜ ⎟
⎜ ⎟
⎜ ⎟s
s
5
dt 2 ⎝ 30 ⎠ dt ⎝ 5 ⎠
dt
⎝ ⎠
49
4
So the characteristic equation is s 2 + s + = 0 and its roots are s = − 0.817 ± j 0.365
30
5
−.817 t
vn (t ) = e
[ A1 cos(0.365 t ) + A2 sin(0.365 t )]
Try vf (t ) = B0 + B1e −2t and substitute vf (t ) into the differential equation and equate like terms
to get B0 = 4.5, B1 = −7.04
Then v(t ) = e −.817 t [ A1 cos(0.365t ) + A2 sin(0.365t ) ] + 4.5 − 7.04e−2t
Now using the initial conditions gives v(0) = 4.5 = A1 + 4.5 − 7.04 ⇒ A1 = 7.04
4
dv(0)
4
= 2is (0) − 2i (0) − v(0) = 2(9 + 3) − 2(6) − (4.5) = 6
3
dt
3
∴ 6 = −0.817 A1 + 0.365 A2 + 14.08 ⇒ A2 = −22.82
and
v(t )
dv(t )
− 0.5
so i (t ) = i (t ) −
s
dt
1.5
−0.817 t
i (t ) =e
[ 2.37 cos(0.365t ) + 7.14sin(0.365t ) ] + 6 + 0.65e−2t A
i(t)
P 9.8-11 Determine v(t) for t > 0 for the
circuit shown in Figure P 9.8-11.
0.1 H
+
–
Answer:
– va(t) +
4Ω
6u(t) + 10 V
0.625 F
2va(t)
vc(t) = 0.75 e–4t – 6.75 e–36t + 16 V for t > 0
+
v(t)
–
Figure P 9.8-11
Solution:
First, find the steady state response for t < 0, when the switch is closed. The input is constant so
the capacitor will act like an open circuit at steady state, and the inductor will act like a short
circuit.
va ( 0 ) = −4 i ( 0 )
i ( 0 ) = 2 ( −4 i ( 0 ) ) ⇒ i ( 0 ) = 0 A
and
v ( 0 ) = 10 V
For t > 0
Apply KCL at node 2:
va
d
+ K va + C v = 0
R
dt
KCL at node 1 and Ohm’s Law:
va = − R i
so
d
1+ K R
v=
i
dt
CR
Apply KVL to the outside loop:
L
d
i + R i + v − Vs = 0
dt
After some algebra:
d2
R d
1+ K R
1+ K R
v+
v+
v=
Vs
2
dt
L dt
LC
LC
⇒
d2
d
v + 40 v + 144 v = 2304
2
dt
dt
The forced response will be a constant, vf = B so
d2
d
B + ( 40 ) B + (144 ) B = 2304 ⇒ B = 16 V
2
dt
dt
The characteristic equation is s 2 + 40 s + 144 = 0 ⇒ s1,2 = −4, −36 .
v ( t ) = A1 e − 4 t + A2 e −36 t + 16 .
Then
Using the initial conditions:
⎫
10 = v ( 0 ) = A1 e0 + A2 e0 + 16 ⇒ A1 + A2 = −6
⎪
⎬ ⇒
d
0
0
0 = v ( 0 ) = −4 A1 e − 36 A2 e ⇒ − 4 A1 − 36 A2 = 0 ⎪
dt
⎭
A1 = 0.75 and A2 = −6.75
So
v ( t ) = 0.75 e−4 t − 6.75 e−36 t + 16 V for t > 0
P 9.8-12 The circuit shown in Figure P 9.8-12 is at
steady state before the switch opens. The inductor
current is given to be
i(t) = 240 + 193e–6.25tcos(9.27t – 102°) mA for t ≥ 0.
Determine the values of R1, R3, C, and L.
t=0
R1
+
–
i(t)
24 V
L
+
v(t)
C
20 Ω
–
Figure P 9.8-12
Solution:
Two steady state responses are of interest, before and
after the switch opens. At steady state, the capacitor acts
like an open circuit and the inductor acts like a short
circuit.
For t > 0, the switch is open. At steady state, inductor
24
current is i ( ∞ ) =
. From the given equation,
R1 + 20
i ( ∞ ) = lim i ( t ) = 0.24 . Thus,
t →∞
0.24 =
24
⇒ R1 = 80 Ω .
R1 + 20
For t < 0, the switch is closed and the circuit is at steady
state.
24
= 0.24 + 0.193 cos ( −102° ) = 0.2
80 || R 3 + 20
(
)
Consequently, R 3 = 80 Ω
After the switch opens, apply KCL and KVL to get
d
⎛
⎞
R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) = Vs
dt
⎝
⎠
Apply KVL to get
v (t ) = L
d
i (t ) + R2 i (t )
dt
Substituting v ( t ) into the first equation gives
R3
d⎛ d
d
⎛
⎞⎞
R1 ⎜ i ( t ) + C ⎜ L i ( t ) + R 2 i ( t ) ⎟ ⎟ + L i ( t ) + R 2 i ( t ) = Vs
dt ⎝ dt
dt
⎠⎠
⎝
then
R1 C L
d2
dt
(
i ( t ) + R1 C R 2 + L
2
) dtd i ( t ) + ( R1 + R 2 ) i ( t ) = Vs
Dividing by R1 C L :
⎛ R1 C R 2 + L ⎞ d
⎛ R1 + R 2 ⎞
Vs
i
t
i
t
+
+
⎜
⎟
⎜
⎟ i (t ) =
(
)
(
)
2
⎜
⎟
⎜
⎟
R1 C L
dt
⎝ R1 C L ⎠ dt
⎝ R1 C L ⎠
d2
Compare to
d2
dt
to get
2α =
2
i ( t ) + 2α
R1 C R 2 + L
R1 C L
d
i ( t ) + ω 0 2 i ( t ) = f (t )
dt
, ω 02 =
R1 + R 2
R1 C L
and
f (t ) =
Vs
R1 C L
From the given equation, we have α = 6.25 and ω d = 9.27 rad/s . Consequently,
ω 0 = ω d 2 + α 2 = 11.18 rad/s . Next
12.5 =
R1 C R 2 + L
R1 C L
=
20
1
+
L 80 C
and 125 =
R1 + R 2
R1 C L
=
1.25
1
⇒ 100 =
CL
CL
So
12.5 =
20
1
+
1
80 C
100 C
⇒ 0 = 2000 C 2 − 12.5 C + 0.0125 ⇒ C = 1.25, 5 mF
The corresponding values of the inductance are L = 8, 2 H .
There are two solutions:
R1 = 80 Ω, R 3 = 80 Ω, C = 1.25 mF and L = 8 H
and
R1 = 80 Ω, R 3 = 80 Ω, C = 5 mF and L = 2 H
We have used the initial condition i ( 0 ) = 0.2 A but we have not yet used the initial condition
v (t ) = L
d
i (t ) + R2 i (t ) ⇒
dt
from the given equation,
v (0) R2 i ( 0) 8 4 4
d
i ( 0) =
−
= − =
dt
L
L
L L L
i ( t ) = 0.24 + e −6.25 t ( −0.04 cos ( 9.27 t ) + 0.1888sin ( 9.27 t ) ) A for t ≥ 0
d
i ( t ) = ( −6.25 ) e −6.25 t ( −0.04 cos ( 9.27 t ) + 0.1888sin ( 9.27 t ) )
dt
+ ( 9.27 ) e −6.25 t ( 0.04sin ( 9.27 t ) + 0.1888cos ( 9.27 t ) ) for t ≥ 0
d
i ( 0 ) = ( −6.25 )( −0.04 ) + ( 9.27 )(1.888 ) = 2
dt
Consequently,
2=
d
4
i ( 0) =
dt
L
⇒ L=2H
and we choose
R1 = 80 Ω, R 3 = 80 Ω, C = 5 mF and L = 2 H
P 9.8-13 The circuit shown in Figure P 9.8-13 is
at steady state before the switch opens. Determine
the inductor current, i(t), for t > 0.
t=0
8Ω
+
–
i(t)
18 V
+
v(t)
0.4 H
25 mF
–
Figure P 9.8-13
Solution:
First, we find the initial conditions;
For t < 0, the switch is closed and the circuit is at steady
state. At steady state, the capacitor acts like an open
circuit and the inductor acts like a short circuit.
v (0 −) =
12
×18 = 12 V
( 8 || 24 ) + 12
and
i (0 −) =
24
18
×
= 0.75 A
8 + 24 ( 8 || 24 ) + 12
Next, represent the circuit by a differential equation.
After the switch opens, apply KCL and KVL to get
d
⎛
⎞
R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) = Vs
dt
⎝
⎠
Apply KVL to get
v (t ) = L
24 Ω
d
i (t ) + R2 i (t )
dt
Substituting v ( t ) into the first equation gives
d⎛ d
d
⎛
⎞⎞
R1 ⎜ i ( t ) + C ⎜ L i ( t ) + R 2 i ( t ) ⎟ ⎟ + L i ( t ) + R 2 i ( t ) = Vs
dt ⎝ dt
dt
⎠⎠
⎝
12 Ω
then
R1 C L
d2
dt
2
(
i ( t ) + R1 C R 2 + L
) dtd i ( t ) + ( R1 + R 2 ) i ( t ) = Vs
Dividing by R1 C L :
⎛ R1 C R 2 + L ⎞ d
⎛ R1 + R 2 ⎞
Vs
i
t
i
t
+
+
⎜
⎟
⎜
⎟ i (t ) =
(
)
(
)
2
⎜ R1 C L ⎟ dt
⎜ R1 C L ⎟
R1 C L
dt
⎝
⎠
⎝
⎠
d2
Compare to
d2
dt
to get
2α =
i t + 2α
2 ( )
R1 C R 2 + L
R1 C L
d
i ( t ) + ω 02 i ( t ) = f (t )
dt
, ω 02 =
R1 + R 2
R1 C L
and
f (t ) =
Vs
R1 C L
With the given element values, we have α = 17.5 and ω 0 2 = 250 . Consequently, the roots of the
characteristic equation are s 1 = −α − α 2 − ω 0 2 = −25 and s 2 = −α + α 2 − ω 02 = −10 . The
natural response is
i n ( t ) = A1 e−10 t + A 2 e−25 t
Next, determine the forced response.
The steady state response after the switch opens will be
used as the forced response. At steady state, the
capacitor acts like an open circuit and the inductor acts
like a short circuit.
if =
18
= 0.9 A
8 + 12
So
i ( t ) = i n ( t ) + i f ( t ) = A1 e−10 t + A2 e−25 t + 0.9
It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have
0.75 = i ( 0 ) = A1 + A 2 + 0.9
The other initial condition comes from
v (t ) R 2
d
i (t ) =
−
i (t ) ⇒
dt
L
L
then
12 12
d
i ( 0) =
−
× 0.75 = 7.5
0.4 0.4
dt
7.5 =
d
i ( 0 ) = −10 A1 − 25 A 2
dt
Solving these equations gives A1 = 0.25 and A2 = −0.4 so
i ( t ) = 0.25 e−10 t − 0.4 e−25 t + 0.9 A for t > 0
(checked using LNAPTR 7/21/04)
*P 9.8-14 The circuit shown in Figure P 9.8-14 is at steady state before the switch closes.
Determine the capacitor voltage, v(t), for t > 0.
ia
+
–
i(t)
t=0
0.4 H
10 Ω
3ia
20 V
10 Ω
25 mF
+
v(t)
–
Figure P 9.8-14
Solution:
First, we find the initial conditions;
For t < 0, the switch is open and the circuit
is at steady state. At steady state, the
capacitor acts like an open circuit and the
inductor acts like a short circuit.
v ( 0 − ) = 0 V and i ( 0 − ) = 0 A
also
i ( 0)
d
v ( 0) =
=0
0.025
dt
Next, represent the circuit after the switch closes by a differential equation. To do so, we find the
Thevenin equivalent circuit for the part of the circuit to the left of the inductor.
v s − v oc ⎫
⎪
R1 ⎪
v s R 2 (1 + b )
⎬ ⇒ v oc =
v oc ⎪
R1 + R 2 (1 + b )
ia + bia =
R 2 ⎪⎭
ia =
i sc = i a (1 + b ) =
v s R 2 (1 + b )
Rt =
With the given values, v oc = 16 V and R t = 2 Ω .
After the switch closes, apply KVL to get
v oc
i sc
=
vs
R1
(1 + b )
R1 + R 2 (1 + b )
R1 R 2
=
vs
R1 + R 2 (1 + b )
(1 + b )
R1
R t i (t ) + L
d
i ( t ) + v ( t ) = voc
dt
d
v (t )
dt
Substituting i ( t ) into the first equation gives
Apply KCL to get
i (t ) = C
d2
⎛ 1 ⎞
voc
⎛R⎞ d
v (t ) + ⎜ ⎟ v (t ) + ⎜
⎟ v (t ) =
CL
dt
⎝ L ⎠ dt
⎝CL⎠
2
d2
d
v ( t ) + ω 02 v ( t ) = f (t )
dt
dt
Rt
v
1
2α =
, ω 02 =
and f (t ) = oc
L
CL
CL
Compare to
2
to get
v (t ) + 2α
With the given element values, we have α = 2.5 and ω 0 2 = 100 . Consequently, the roots of the
characteristic equation are s 1,2 = −α ± α 2 − ω 02 = −2.5 ± j 9.682 and the circuit is
underdamped. The damped resonant frequency is ω d = ω 0 2 − α 2 = 9.682 rad/s . The natural
response is
(
v n ( t ) = e−2.5 t A1 cos 9.682 t + A 2 sin 9.682 t
)
Next, determine the forced response.
The steady state response after the switch closes will be
used as the forced response. At steady state, the
capacitor acts like an open circuit and the inductor acts
like a short circuit.
v f = v oc = 16 V
(
v ( t ) = 16 + e−2.5 t A1 cos 9.682 t + A 2 sin 9.682 t
so
)
It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have
0 = v ( 0 ) = 16 + A1 ⇒
And
0=
d
v ( 0 ) = −2.5 A1 + 9.682 A 2
dt
⇒
A1 = −16
A2 = −
2.5 × 16
= −4.131
9.682
Finally,
v ( t ) = 16 + e −2.5 t ( −16 cos 9.682 t − 4.131sin 9.682 t )
= 16 + 16.525 e −2.5 t cos ( 9.682 t + 165.5° ) V for t ≥ 0
(checked using LNAPTR 7/22/04)
P 9.8-15 The circuit shown in Figure P 9.8-15 is at steady state before the switch closes.
Determine the capacitor voltage, v(t), for t > 0.
t=0
50 Ω
i(t)
2H
+
–
+
50 Ω
20 V
v(t)
5 mF
–
Figure P 9.8-15
Solution:
First, we find the initial conditions;
For t < 0, the switch is open and the circuit is
at steady state. At steady state, the capacitor
acts like an open circuit and the inductor acts
like a short circuit.
v ( 0 − ) = 0 V and i ( 0 − ) = 0 A
also
i ( 0)
v ( 0)
d
v ( 0) =
−
=0
0.005 50 × 0.005
dt
Next, represent the circuit after the switch closes by a differential equation.
After the switch closes, use KCL to get
i (t ) =
v (t )
d
+ C v (t )
R2
dt
Use KVL to get
v s = R1 i ( t ) + L
d
i (t ) + v (t )
dt
Substitute to get
vs =
R1
R2
= CL
Finally,
v ( t ) + R1C
d
L d
d2
v (t ) +
v ( t ) + CL 2 v ( t ) + v ( t )
dt
R 2 dt
dt
⎛
R1 + R 2
d2
L ⎞d
v t + R1C +
v (t ) +
v (t )
⎟
2 ( ) ⎜
⎜
⎟ dt
dt
R
R
2
2
⎝
⎠
⎛ R1
R1 + R 2
d2
1 ⎞d
= 2 v (t ) + ⎜ +
v (t ) +
v (t )
⎟
⎜ L R 2C ⎟ dt
CL dt
R 2CL
⎝
⎠
vs
Compare to
d2
dt
to get
2α =
R1
L
2
+
i ( t ) + 2α
d
i ( t ) + ω 02 i ( t ) = f (t )
dt
R1 + R 2
1
, ω 02 =
R 2C
R 2CL
f (t ) =
and
vs
CL
With the given element values, we have α = 14.5 and ω 0 2 = 200 . Consequently, the roots of the
characteristic equation are s 1 = −11.3 and s 2 = −17.7 so the circuit is overdamped. The natural
response is
v n ( t ) = A1 e −11.3 t + A 2 e −17.7 t
Next, determine the forced response.
The steady state response after the switch opens will be
used as the forced response. At steady state, the
capacitor acts like an open circuit and the inductor acts
like a short circuit.
vf =
1
v s = 10 V
2
So
v n ( t ) = 10 + A1 e −11.3 t + A 2 e−17.7 t
It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have
0 = v ( 0 ) = 10 + A1 + A 2
and
0=
d
v ( 0 ) = −11.3 A1 − 17.7 A 2
dt
Solving these equations gives
A1 = −27.6 and A 2 = 17.6
Finally,
v ( t ) = 10 − 27.6 e−11.3 t + 17.6 e−17.7 t
(checked using LNAPTR 7/26/04)
t=0
P 9.8-16 The circuit shown in Figure P 9.8-16 is
at steady state before the switch closes. Determine
the inductor current, i(t), for t > 0.
16 Ω
9Ω
+
–
+
25 mF
20 V
0.4 H
i(t)
Figure P 9.8-16
Solution:
First, we find the initial conditions;
For t < 0, the switch is closed and the circuit is at
steady state. At steady state, the capacitor acts like an
open circuit and the inductor acts like a short circuit.
v ( 0 − ) = 0 V and i ( 0 − ) = 0 A
Also
9 i ( 0 ) + 0.4
d
i ( 0) = v ( 0) ⇒
dt
d
i (0) = 0
dt
Next, represent the circuit by a differential equation.
After the switch closes use KVL to get
R2 i (t ) + L
d
i (t ) = v (t )
dt
Use KCL and KVL to get
d
⎛
⎞
v s = R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t )
dt
⎝
⎠
Substitute to get
d
d2
d
v s = R1i ( t ) + R1CR 2 i ( t ) + R1C L 2 i ( t ) + R 2i ( t ) + L i ( t )
dt
dt
dt
2
d
d
= R1CL 2 i ( t ) + ( R1 R 2C + L ) i ( t ) + ( R1 + R 2 ) i ( t )
dt
dt
then
⎛ R2
vs
R1 + R 2
d2
1 ⎞d
= 2 i (t ) + ⎜
+
i (t ) +
i (t )
⎟
⎜ L R1C ⎟ dt
R1CL dt
R
CL
1
⎝
⎠
v(t)
–
Compare to
d2
2
dt
to get
2α =
R2
L
+
i ( t ) + 2α
d
i ( t ) + ω 02 i ( t ) = f (t )
dt
R1 + R 2
1
, ω 02 =
R1C
R1 C L
and
f (t ) =
Vs
R1 C L
With the given element values, we have α = 12.5 and ω 0 2 = 156.25 . Consequently, the roots of
the characteristic equation are s 1,2 = −α ± α 2 − ω 0 2 = −12.5, − 12.5 so the circuit is critically
damped. The natural response is
(
)
i n ( t ) = A1 + A 2 t e−12.5 t
Next, determine the forced response.
The steady state response after the switch opens will be
used as the forced response. At steady state, the
capacitor acts like an open circuit and the inductor acts
like a short circuit.
20
if =
= 0.8 A
16 + 9
So
i ( t ) = i n ( t ) + i f ( t ) = A1 + A 2 t e−12.5 t + 0.8
(
)
It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have
And
0=
0 = i ( 0 ) = A1 + 0.8 ⇒
A1 = −0.8
d
i ( 0 ) = −12.5 A1 − A 2
dt
⇒
A 2 = 10
Thus
i ( t ) = ( −0.8 + 10 t ) e−12.5 t + 0.8 for t > 0
(checked using LNAPTR 7/27/04)
75 Ω
P 9.8-17 The circuit shown in
Figure P 9.8-17 is at steady state before
the switch opens. Determine the inductor
current, i2(t), for t > 0.
t=0
24 Ω
+
–
4H
i1(t)
20 V
15 Ω
Figure P 9.8-17
Solution:
First, we find the initial conditions;
For t < 0, the switch is closed and the
circuit is at steady state. At steady state, the
inductors act like short circuits.
i1 ( 0 − ) =
and
20
= 1.333 A
15
i 2 (0 −) = 0 A
Next, represent the circuit by a differential equation.
After the switch opens, KVL gives
L1
d
d
i1 ( t ) = R 2 i 2 ( t ) + L 2 i 2 ( t )
dt
dt
KVL and KCL give
L1
d
i 1 ( t ) + R1 ( i 1 ( t ) + i 2 ( t ) ) = 0
dt
Use the operator method to get
L1s i1 = R 2 i 2 + L 2 s i 2
L1s i1 + R1 ( i1 + i 2 ) = 0
1.6 H
i2(t)
L1s 2i1 + R1s i1 + R1s i 2 = 0
s ( R 2i 2 + L 2 s i 2 ) +
R1
L1
(R i
2 2
+ L 2 s i 2 ) + R1s i 2 = 0
⎛
⎞
L2
R1 R 2
+ R1 ⎟ s i 2 +
L 2 s 2 i 2 + ⎜ R 2 + R1
i2 = 0
⎜
⎟
L
L
1
1
⎝
⎠
⎛ R 2 R 1 R1 ⎞
R1 R 2
+
+ ⎟ s i2 +
s 2i 2 + ⎜
i2 = 0
⎜ L 2 L 2 L1 ⎟
L1 L 2
⎝
⎠
so
⎛ R 2 R1 R1 ⎞ d
R1 R 2
d2
i t +
+
+ ⎟ i 2 (t ) +
i 2 (t ) = 0
2 2( ) ⎜
⎜ L 2 L 2 L1 ⎟ dt
dt
L
L
1
2
⎝
⎠
Compare to
d2
dt
to get
2α =
R2
L2
2
+
i ( t ) + 2α
R1
L2
+
R1
L1
d
i ( t ) + ω 02 i ( t ) = f (t )
dt
, ω 02 =
R1R 2
L1L 2
and
f (t ) = 0
With the given element values, we have α = 33.9 and ω 02 = 281.25 . Consequently, the roots of
the characteristic equation are s 1,2 = −α ± α 2 − ω 02 = −4.4, − 63.4 so the circuit is
overdamped. The natural response is
i n ( t ) = A1 e−4.4 t + A 2 e−63.4 t
Next, determine the forced response.
The steady state response after the switch
opens will be used as the forced response.
At steady state the inductors act like short
circuits.
if = 0 A
So
i 2 ( t ) = i n ( t ) + i f ( t ) = A1 e−4.4 t + A2 e−63.4 t
It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have
0 = i 2 ( 0 ) = A1 + A 2
L2
d
i 2 ( 0 ) + R 2 i 2 ( 0 ) + R1 i 1 ( 0 ) + R 1 i 2 ( 0 ) ⇒
dt
d
i 2 ( 0 ) = −20
dt
and
−20 =
d
i ( 0 ) = −4.4 A1 − 63.4 A 2
dt
Solving these equations gives A1 = −0.339 and A2 = 0.339 so
i 2 ( t ) = −0.339 e−4.4 t + 0.339 e−63.4 t for t ≥ 0
(checked using LNAPTR 7/27/04)
t=0
P 9.8-18 The circuit shown in Figure P 9.8-18 is at
steady state before the switch closes. Determine the
capacitor voltage, v(t), for t > 0.
2H
+
–
20 V
50 Ω
5 mF
50 Ω
i(t)
+
v(t)
–
Figure P 9.8-18
Solution:
First, we find the initial conditions;
For t < 0, the switch is open and the circuit is at steady
state. At steady state, the capacitor acts like an open
circuit and the inductor acts like a short circuit.
v ( 0 − ) = 0 V and i ( 0 − ) = 0 A
also
i (0)
d
v ( 0) =
=0
0.005
dt
Next, represent the circuit after the switch closes by a differential equation.
After the switch closes
i (t ) = C
d
v (t )
dt
KCL and KVL give
⎛
1 ⎛ d
d
⎞⎞
vs = R2 ⎜ i (t ) + ⎜ L i (t ) + v (t ) ⎟ ⎟ + L i (t ) + v (t )
⎜
R1 ⎝ dt
dt
⎠ ⎟⎠
⎝
Substituting gives
⎛ R2 ⎞
⎛ R2 ⎞
⎛ R2 ⎞
R2
d2
d
d2
d
vs =
LC 2 v ( t ) + R 2C v ( t ) + ⎜ 1 +
v
t
=
1
+
LC
v ( t ) + R 2C v ( t ) + ⎜1 +
(
)
⎟
⎜
⎟
⎟⎟ v ( t )
2
⎜
⎟
⎜
⎟
⎜
R1
dt
dt
R
R
dt
dt
R
1
1
1
⎝
⎠
⎝
⎠
⎝
⎠
So the differential equation is
R1v s
LC ( R1 + R 2 )
=
R1 R 2
d2
d
1
v (t ) +
v (t ) +
v (t )
dt
LC
L ( R1 + R 2 ) dt
Compare to
d2
dt
to get
2α =
(
2
i ( t ) + 2α
R1R 2
L R1 + R 2
)
d
i ( t ) + ω 02 i ( t ) = f (t )
dt
, ω 02 =
1
and
CL
f (t ) =
(
R1v s
LC R1 + R 2
)
With the given element values, we have α = 6.25 and ω 0 2 = 100 . Consequently, the roots of the
characteristic equation are s 1,2 = −α ± α 2 − ω 02 = −6.25 ± j 7.806 and the circuit is
underdamped. The damped resonant frequency is ω d = ω 0 2 − α 2 = 7.806 rad/s . The natural
response is
(
v n ( t ) = e−6.25 t A1 cos 7.806 t + A 2 sin 7.806 t
)
Next, determine the forced response.
The steady state response after the switch opens will be
used as the forced response. At steady state, the
capacitor acts like an open circuit and the inductor acts
like a short circuit.
vf =
So
50
× 20 = 10 V
50 + 50
(
v ( t ) = 10 + e−6.25 t A1 cos 7.806 t + A 2 sin 7.806 t
)
It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have
0 = v ( 0 ) = 10 + A1 ⇒
And
0=
d
v ( 0 ) = −6.25 A1 + 7.806 A 2
dt
⇒
A1 = −10
A2 = −
6.25 × 10
= −8.006
7.806
Finally,
v ( t ) = 10 + e −6.25 t ( −10 cos 7.806 t − 8.006sin 7.806 t )
= 10 + 12.81 e −6.25 t cos ( 7.806 t + 141.3° ) V for t ≥ 0
(checked using LNAPTR 7/26/04)
iL(t)
P 9.8-19 Find the differential equation
for vc(t) in the circuit of Figure P 9.8-19
using the direct method. Find vc(t) for
time t > 0 for each of the following sets
of component values:
L
R1
vs(t) = u(t) +–
R2
C
+
vc(t)
–
Figure P 9.8-19
(a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω
(b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω
(c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω
Answer:
(a) vc(t) =
1
2
– e–2t +
–2t
1
2
e–4t V; (b) vc(t) =
1
4
–(
1
4
1
+
2
t)e–2t V;
(c) vc(t) = 0.8 – e (0.8 cos 4t + 0.4 sin 4t) V
Solution:
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) =
R2
R1 + R 2
1
Next, represent the circuit by a 2nd order differential equation:
KCL at the top node of R2 gives:
KVL around the outside loop gives:
vC ( t )
R2
vs ( t ) = L
+C
d
vC ( t ) = iL ( t )
dt
d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt
Use the substitution method to get
vs ( t ) = L
⎞
⎛ v (t )
⎞
d ⎛ vC ( t )
d
d
+ C vC ( t ) ⎟ + R1 ⎜ C
+ C vC ( t ) ⎟ + vC ( t )
⎜⎜
⎟
⎜ R2
⎟
dt ⎝ R 2
dt
dt
⎠
⎝
⎠
⎛ L
⎞d
⎛
R1 ⎞
d2
= LC 2 vC ( t ) + ⎜
+ R1 C ⎟ vC ( t ) + ⎜1 +
v t
⎜ R2
⎟ dt
⎜ R 2 ⎟⎟ C ( )
dt
⎝
⎠
⎝
⎠
(a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω
Use the steady state response as the forced response:
R2
1
v f = vC ( ∞ ) =
1=
R1 + R 2
2
The characteristic equation is
R ⎞
⎛
1+ 1 ⎟
⎜
⎛ 1
R1 ⎞
R2 ⎟ 2
s2 + ⎜
+ ⎟s+⎜
= s + 6s + 8 = ( s + 2 )( s + 4 )
⎜ R 2 C L ⎟ ⎜ LC ⎟
⎝
⎠ ⎜
⎟⎟
⎜
⎝
⎠
so the natural response is
vn = A1 e −2 t + A2 e−4 t V
The complete response is
vc ( t ) =
iL ( t ) =
vC ( t )
1.309
At t = 0+
+
1
+ A1 e −2 t + A2 e−4 t V
2
d
vC ( t ) = −1.236 A1 e −2 t − 3.236 A2 e−4 t + 0.3819
dt
0 = vc ( 0 + ) = A1 + A2 + 0.5
0 = iL ( 0 + ) = −1.236 A1 − 3.236 A2 + 0.3819
Solving these equations gives A1 = -1 and A2 = 0.5, so
vc ( t ) =
1 −2 t 1 −4 t
−e + e V
2
2
(b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω
Use the steady state response as the forced response:
R2
1
1=
v f = vC ( ∞ ) =
4
R1 + R 2
The characteristic equation is
R ⎞
⎛
1+ 1 ⎟
⎜
⎛ 1
R1 ⎞
R2 ⎟ 2
2
s2 + ⎜
+ ⎟s+⎜
= s + 4s + 4 = ( s + 2 )
⎜ R 2 C L ⎟ ⎜ LC ⎟
⎝
⎠ ⎜
⎟⎟
⎜
⎝
⎠
so the natural response is
v f = ( A1 + A2 t ) e −2 t V
The complete response is
vc ( t ) =
iL ( t ) = vC ( t ) +
At t = 0+
1
+ ( A1 + A2 t ) e −2 t V
4
d
1
vC ( t ) = +
dt
4
(( A
2
)
− A1 ) − A2 t e −2 t
0 = vc ( 0 + ) = A1 +
0 = iL ( 0 + ) =
1
4
1
+ A2 − A1
4
Solving these equations gives A1 = -0.25 and A2 = -0.5, so
1 ⎛ 1 1 ⎞ −2 t
− ⎜ + t ⎟e V
4 ⎝4 2 ⎠
(c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω
Use the steady state response as the forced response:
R2
4
v f = vC ( ∞ ) =
1=
R1 + R 2
5
The characteristic equation is
R ⎞
⎛
1+ 1 ⎟
⎜
⎛ 1
R1 ⎞
R2 ⎟ 2
s2 + ⎜
+ ⎟s+⎜
= s + 4 s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 )
⎜ R 2 C L ⎟ ⎜ LC ⎟
⎝
⎠ ⎜
⎟⎟
⎜
⎝
⎠
so the natural response is
v f = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
vc ( t ) =
The complete response is
iL ( t ) =
At t = 0+
vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
vC ( t )
4
+
A2 −2 t
A1
1d
vC ( t ) = 0.2 +
e cos 4 t − e−2 t sin 4 t
8 dt
2
2
0 = vc ( 0 + ) = 0.8 + A1
0 = iL ( 0 + ) = 0.2 +
A2
2
Solving these equations gives A1 = -0.8 and A2 = -0.4, so
vc ( t ) = 0.8 − e −2 t ( 0.8cos 4 t + 0.4sin 4 t ) V
P 9.8-20 Find the differential equation for
vo(t) in the circuit of Figure P 9.8-20 using
the direct method. Find vo(t) for time t > 0
for each of the following sets of component
values:
iL(t)
R1
vs(t) = u(t)
+
–
C
L
+
+
vc(t)
–
R2
vo(t)
–
Figure P 9.8-20
(a)
C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω
(b)
C = 1 F, L = 1 H, R1 = 1 Ω, R2 = 3 Ω
(c)
C = 0.125 F, L = 0.5 H, R1 = 4 Ω, R2 = 1 Ω
Answer:
1
– e–2t +
1
(a)
vo(t) =
(b)
vo(t) =
(c)
vo(t) = 0.2 – e–2t(0.2 cos 4t + 0.1 sin 4t) V
2
3
4
–(
3
4
+
2
3
2
e – 4t
V
t)e–2t V
Solution:
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) =
R2
R1 + R 2
1, iL ( ∞ ) =
1
R1 + R 2
and
vo ( ∞ ) =
R2
R1 + R 2
1
Next, represent the circuit by a 2nd order differential equation:
KVL around the right-hand mesh gives:
KCL at the top node of the capacitor gives:
d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
vC ( t ) = L
Use the substitution method to get
vs ( t ) = R1 C
= R1 LC
Using iL ( t ) =
vo ( t )
gives
R2
d ⎛ d
⎞ ⎛ d
⎞
⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t )
dt ⎝ dt
⎠ ⎝ dt
⎠
d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt
⎛ L
⎞d
⎛ R1 + R 2 ⎞
d2
vs (t ) =
LC 2 v o ( t ) + ⎜
+ R1 C ⎟ v o ( t ) + ⎜
v t
⎜ R2
⎟ dt
⎜ R 2 ⎟⎟ o ( )
R2
dt
⎝
⎠
⎝
⎠
R1
(a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω
Use the steady state response as the forced response:
R2
1
v f = vo ( ∞ ) =
1=
R1 + R 2
2
The characteristic equation is
R ⎞
⎛
1+ 2 ⎟
⎜
⎛ 1
R2 ⎞ ⎜
R1 ⎟ 2
s2 + ⎜
+
s+
= s + 6 s + 8 = ( s + 2 )( s + 4 )
⎟
⎜ R1 C L ⎟ ⎜ LC ⎟
⎝
⎠ ⎜
⎟⎟
⎜
⎝
⎠
so the natural response is
vn = A1 e −2 t + A2 e −4 t V
The complete response is
1
+ A1 e −2 t + A2 e−4 t V
2
A1 −2 t
A 2 −4 t
vo ( t )
1
iL ( t ) =
e +
e V
=
+
1.309 2.618 1.309
1.309
vo ( t ) =
vC ( t ) = 1.309 iL ( t ) +
1 d
1
iL ( t ) = + 0.6167 A1 e−2 t + 0.2361 A2 e−4 t
4 dt
2
At t = 0+
0 = iL ( 0 + ) =
0 = vC ( 0 + ) =
A1
A2
1
+
+
2.618 1.309 1.309
1
+ 0.6167 A1 + 0.2361 A2
2
Solving these equations gives A1 = -1 and A2 = 0.5, so
vo ( t ) =
1 −2 t 1 −4 t
−e + e V
2
2
(b) C = 1 F, L = 1 H, R1 = 1 Ω, R2 = 3 Ω
Use the steady state response as the forced response:
R2
3
v f = vo ( ∞ ) =
1=
R1 + R 2
4
The characteristic equation is
R ⎞
⎛
1+ 2 ⎟
⎜
⎛ 1
R2 ⎞ ⎜
R1 ⎟ 2
2
s2 + ⎜
+
s+
= s + 4s + 4 = ( s + 2 )
⎟
⎜ R1 C L ⎟ ⎜ LC ⎟
⎝
⎠ ⎜
⎟⎟
⎜
⎝
⎠
so the natural response is
v f = ( A1 + A2 t ) e −2 t V
The complete response is
vo ( t ) =
iL ( t ) =
3
+ ( A1 + A2 t ) e −2 t V
4
vo ( t )
vC ( t ) = 3 iL ( t ) +
3
=
1 ⎛ A1 A2 ⎞ −2 t
t⎟ e V
+⎜ +
4 ⎝ 3
3 ⎠
3 ⎛ ⎛ A1 A2 ⎞ A2 ⎞ −2 t
d
iL ( t ) = + ⎜⎜ ⎜ +
t ⎟e
⎟+
4 ⎝⎝ 3
3 ⎠ 3 ⎟⎠
dt
At t = 0+
0 = iL ( 0 + ) =
A1
+
1
4
3
3 A1 A2
0 = vC ( 0 + ) = + +
4 3
3
Solving these equations gives A1 = -0.75 and A2 = -1.5, so
vo ( t ) =
3 ⎛ 3 3 ⎞ −2 t
− ⎜ + t ⎟e V
4 ⎝4 2 ⎠
(c) C = 0.125 F, L = 0.5 H, R1 = 4 Ω, R2 = 1 Ω
Use the steady state response as the forced response:
R2
1
1=
v f = vo ( ∞ ) =
5
R1 + R 2
The characteristic equation is
R ⎞
⎛
1+ 2 ⎟
⎜
⎛ 1
R2 ⎞ ⎜
R1 ⎟ 2
s2 + ⎜
+
s+
= s + 4 s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 )
⎟
⎜ R1 C L ⎟ ⎜ LC ⎟
⎝
⎠ ⎜
⎟⎟
⎜
⎝
⎠
so the natural response is
v f = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
The complete response is
vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
iL ( t ) =
vo ( t )
vC ( t ) = iL ( t ) +
At t = 0+
1
= 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
1 d
iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e −2 t sin 4 t
2 dt
0 = iL ( 0 + ) = 0.2 + A1
0 = vC ( 0 + ) = 0.2 + 2 A2
Solving these equations gives A1 = -0.8 and A2 = -0.4, so
vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V
Section 9-9: State Variable Approach to Circuit Analysis
P 9.9-1 Find v(t) for t > 0 using the state variable method of
Section 9.9 when C = 1/5 F in the circuit of Figure P 9.9-1.
Sketch the response for v(t) for 0 < t < 10 s.
Answer: v(t) = – 25e–t + e– 5t + 24 V
6Ω
+
C
4u(t) A
1H
Figure P 9.9-1
Solution:
t = 0− circuit is source free ∴ iL (0) = 0 & v(0) = 0
t>0
⎛1⎞
= 4
KCL at top node: i L + ⎜ ⎟ dv
(1)
⎝ 5 ⎠ dt
di
− 6i L = 0
KVL at right loop: ( v − 1) L
dt
Solving for i in (1) & plugging into (2) ⇒
1
d 2v
dv
+ 6 + 5v = 120
2
dt
dt
The characteristic equation is: s 2 + 6s + 5 = 0,
The roots of the characteristic equation are s = −1, −5
∴ Tthe natural response is: vn (t ) = A1 e −t + A2 e −5t
Try vf = B & plug into D.E. ⇒
B = 24 = vf
dv(0)
= 20 − 5iL (0) = 20 V
s
dt
So v(0) = 0 = A1 + A2 + 24 ⎫
A1 = −25, A2 = 1
⎪
⎬ ∴ v(t ) = −25 e −t + e−5t + 24 V
dv(0)
= 20 = − A1 − 5 A2 ⎪
dt
⎭
From (1)
–
v
P 9.9-2 Repeat Problem P9.9-1when C = 1/10 F. Sketch the
response for v(t) for 0 < t < 3 s.
6Ω
+
Answer: v(t) = e – 3t(– 24 cos t – 32 sin t) + 24 V
C
4u(t) A
–
v
1H
Figure P 9.9-1
Solution:
At t = 0− the circuit is source free ∴ iL (0) = 0, & v(0) = 0
At t > 0
( )
i = 4 − 1 dv dt
10
L
KCL at top node :
KVL at right node : v −
(1) into (2) yields
2
⇒ s + 6 s + 10 = 0, s = −3 ± j
∴ vn ( t ) = e
−3t
d 2v
dt
2
diL
+6
dt
dv
+10v = 240
dt
⎡⎣ A1 cos t + A2 sin t ⎤⎦
Try vf = B & plug into D.E. ⇒ vf = B = 24
dv (0)
= 40−10 i L (0) = 40 V
s
dt
dv (0)
= 40 = −3 A1 + A2
So v (0) = 0 = A1 + 24 ⇒ A1 = −24 &
dt
From (1)
⇒
∴ v (t ) = e
−3 t
[ −24cos t
− 32sin t ] + 24 V
− 6i L = 0
A2 = −32
(1)
(2)
Determine the current i(t) and the voltage v(t) for the circuit of Figure P 9.9-3.
P 9.9-3
Answer: i(t) = (3.08e–2.57t – 0.08e–97.4t – 6) A
v
0.2 H
–3u(t) A
i
+
–
20 mF
0.5 Ω
3A
Figure P 9.9-3
Solution:
i (0) = −3, v(0) = 0
t >0
dv v
+ +6=0
dt R
di
KVL: v = L
dt
KCL: i + C
d 2i
dt
2
+ 100
di
+ 250i = −1500
dt
s = − 2.57, − 97.4
i f (t ) =
−1500
= −6
250
i (t ) = A1e −2.57 t + A2 e−97.4 t −6
⎫
⎪ A1 = 3 .081
⎬
di (0)
= 0 = − 2.57 A1 −97.4 A2 ⎪ A2 =−0.081
dt
⎭
i (0) = A1 + A2 − 6 = −3
i (t ) = 3.081 e
v (t ) = 0.2
−2.57 t
− 0.081e
di
= −1.58e
dt
−2.57 t
−97.4 t
−6 A
+ 1.58e−97.4 t
V
P 9.9-4 Clean-air laws are pushing the auto industry toward the development of electric cars.
One proposed vehicle using an ac motor is shown in Figure P 9.9-4a. The motor-controller
circuit is shown in Figure P 9.9-4b with L = 100 mH and C = 10 mF. Using the state equation
approach, determine i(t) and v(t) where i(t) is the motor-control current. The initial conditions are
v(0) = 10 V and i(0) = 0.
+
+ –
2vx
Transistorized
dc to ac inverter
+
Integrated interior
permanent magnet Sodium-sulfur
battery
ac motor and
automatic transaxle
2Ω
2ix
Electric power
steering
System
controller
–
C
–
vx
v
L
ix
i
1Ω
(a)
(b)
Figure P 9.9-4
Apply KCL to the supernode corresponding to the VCVS to get
ix +
vx
2
= 2ix + C
dv
dt
⇒ − ix +
Apply KCL to the right node of the CCCS to get i +
vx
2
= 0.01
dv
dt
vx
= 2ix .
2
Apply KVL to the mesh consisting of the 2-Ω resistor, inductor and capacitor to get
vx + v − L
Apply KVL to the outside loop to get
di
di
= 0 ⇒ v x + v = 0.1
dt
dt
v + ix + 2vx = 0
(1)
(2)
(3)
(4)
Combine equations (2) and (4) to get
4 1
2
4
i − v and v x = − i − v
9
9
9
9
Use (5) to eliminate ix and vx from (1) and (3) to get
dv
5 1
di
2
5
0.01
= − i − v and 0.1 = − i + v
dt
9 9
dt
9
9
Use operators to write
500 100
20
50
sv = −
i−
v and s i = − i +
v
9
9
9
9
The characteristic equation is : s 2 + 13.33s + 333.33 = 0 ⇒ s1 , s2 = −6.67 ± j 17
ix =
(5)
(6)
(7)
The natural response is v ( t ) = e −6.67 t ⎡⎣ A cos (17 t ) + B sin (17 t ) ⎤⎦ and there is no forced because
there is no forcing function. The constants A and B are evaluated using the initial conditions:
dv(0)
v(0) = 10 = A and
= −111 = −6.67 A + 17 B ⇒ B = −2.6
dt
Then
v ( t ) = e −6.67 t ⎡⎣10 cos (17 t ) − 2.6 sin (17 t ) ⎤⎦ V
Similarly, the natural response is i ( t ) = e −6.67 t ⎡⎣ A cos (17 t ) + B sin (17 t ) ⎤⎦ and again there is no
forced because there is no forcing function. The constants A and B are evaluated using the initial
conditions:
di (0)
i (0) = 0 = A and
= 55.6 = −6.67 A + 17 B ⇒ B = −3.27
dt
Then
i ( t ) = e −6.67 t ⎡⎣3.27 sin (17 t ) ⎤⎦ V
P 9.9-5 Studies of an artificial insect are being used
to understand the nervous system of animals. A
model neuron in the nervous system of the artificial
insect is shown in Figure P 9.9-5. The input signal, vs,
is used to generate a series of pulses, called synapses.
The switch generates a pulse by opening at t = 0 and
closing at t = 0.5 s. Assume that the circuit is at
steady state and that v(0 –) = 10 V.
Switch
6Ω
3Ω
vs +
–
+
v
1 6F
30 V
1 2H
–
Determine the voltage v(t) for 0 < t < 2 s.
Figure P 9.9-5
Solution
First consider t < 0 :
v(0) = 10 V, i (0) =
L
10
A
3
Next consider 0 < t < 0.5s
α=
R 3
1
and ω02 = LC =
=
L 2
12
s =−α ±
α 2 − ω02
⇒
s1 = −.028 and s2 = −2.97
The natural response is v(t ) = Ae −0.028t + Be−2.97 t and the forced response is vf = 0.
The constants are evaluated using the initial conditions:
v(0) = 10 = A + B
⎫
⎪ A = 16. 89
dv(0)
⎬
= 20 = −0.028 A − 2.97 B ⎪ B = −6.89
dt
⎭
−0.028t
−2.97 t
so v(t ) = 16. 89 e
− 6.89e
Similarly i (t ) = − .079e −0.028t + 3.41e−2.97 t
At t = 0.5 s, v(0.5) = 15.1 V and i (0.5) = 0.7 A
For t > 0.55 s:
v − 30
1 dv
+ iL +
=0
6
6 dt
di L
KVL: v = 3 i L + 1
2 dt
KCL:
Characteristic equation: 0 = s 2 − 7 s − 18 ⇒ s = −1,9
vf = 10 V
v(t ) = Ae9t + Be − t + 10
v(0.5) = 15.1 = 90 A + 0.61 B + 10
dv(0.5)
= 10.7 = 810 A − 0.61 B
dt
t
0
→ .5
.5
→2
⎫
⎪
⎬
⎪⎭
A = 17.6 × 10−3
B = 5.77
v(t)
16.89e−0.28τ − 6.89e−2.97 τ V
17.6 × 10-3e9t + 5.77e-t + 10 V
Section 9-10: Roots in the Complex Plane
2 kΩ
P 9.10-1 For the circuit of Figure P 9.10-1,
determine the roots of the characteristic
equation, and plot the roots on the s-plane.
12 – 6u(t) V
3 kΩ
+
–
2 mH
2 mH
i1
i2
Figure P 9.10-1
Solution:
s 2 + 3.5 × 106 s + 1.5 × 1012 = 0
s = −5 × 105
1
s2 = −3 × 106
P 9.10-2 For the circuit of Figure P 9.6-1,
determine the roots of the characteristic
equation and plot the roots on the s-plane.
0.8 H
vc
250 Ω
t=0
250 Ω
+
–
6V
Figure P 9.6-1
Solution:
s 2 + 800s + 250000 = 0
s = 400 ± j 300
+
–
5 × 10-6 F
4H
P 9.10-3 For the circuit of Figure P 9.10-3,
determine the roots of the characteristic
equation and plot the roots on the s-plane.
vs +
1 4
–
μF
4 kΩ
Figure P 9.10-3
Solution:
KCL:
KVL:
Characteristic equation: s 2 + 1×103 s + 1× 106 = 0
s = −500 ± j 866
dv
v
1
× 10−6
+
dt
4
4000
di
vs = 4 + v
dt
i=
P 9.10-4 An RLC circuit is shown in
Figure P 9.10-4.
(a) Obtain the two-node voltage equations using
operators.
(b) Obtain the characteristic equation for the
circuit.
(c) Show the location of the roots of the
characteristic equation in the s-plane.
(d) Determine v(t) for t > 0.
1H
a
36u(t) V
+
–
12 Ω
b
6 Ω
1 18 F
+
–
v(t)
Figure P 9.10-4
Solution:
at t = 0 the initial conditions are v(0) = v (0) = 0,
b
dvb vb − va
i (0) = 0 and C
+
+ i =0 (1)
dt
6
t=0
Node a:
va (0) − 36
v (0) − vb (0)
− i (0) + a
= 0 then va (0) + 2va (0) = 36 so va (0) = 12 V
12
6
t ≥ 0
va − vs
va − vb
1
+
= 0
∫ (va − vb ) dt +
L
12
6
v −v
dv
1
Node b :
C b + b a +
∫ (va − vb )dt =0
dt
6
L
v
1 1⎞
⎛1
⎛ 1 1⎞
Using operators ⎜
+ + ⎟ va + ⎜ − − ⎟ vb = s
6
s⎠
12
⎝ 12
⎝ 6 s⎠
1 1
1
1 1
( − − ) va + (
s + + ) vb = 0
6 s
18
6 s
2
Cramers rule: ( s + 5s + 6)vb = ( s + 6)vs
Node a :
Then
vb = 36 + A1 e −2t + A2 e−3t
vb (0) = 36 + A1 + A2
need
(2)
dvb
(0) = −2 A1 − 3 A2
dt
v ( 0 ) − vb (0)
dvb (0)
1
12
=
(−2 A1 − 3 A2 ) = a
− i (0) =
=2
dt
18
6
6
Use (2) and (3) to get
Use 1 above: C
A1 = −72 and A2 = 36 so v ( t ) = vb ( t ) = 36 − 72 e −2t + 36 e −3t , t ≥ 0
(3)
Section 9-11 How Can We Check…?
P 9.11-1 Figure P 9.11-1a shows an RLC circuit. The voltage, vs(t), of the voltage source is the
square wave shown in Figure P 9.11-1a. Figure P 9.11-1c shows a plot of the inductor current,
i(t), which was obtained by simulating this circuit using PSpice. Verify that the plot of i(t) is
correct.
Answer: The plot is correct.
25
i(t)
100 Ω
vs, V
vs
0
4
8
12
+
–
2 μF
12 mH
16
t, ms
(a)
(b)
400 mA
(550.562u, 321.886m)
(1.6405m, 256.950m)
(3.6854m, 250.035m)
200 mA
(1.0787m, 228.510m)
I (L1)
0A
–200 mA
0s
I (L1)
2.0 ms
4.0 ms
Time
6.0 ms
8.0 ms
(c)
Figure P 9.11-1a
Solution:
This problem is similar to the verification example in this chapter. First, check the steady-state
inductor current
v
25
i (t ) = s =
= 250 mA
100 100
This agrees with the value of 250.035 mA shown on the plot. Next, the plot shows an
underdamped response. That requires
12 ⋅10−3 = L < 4 R 2C = 4(100) 2 (2 ⋅10−6 ) = 8 ⋅10−2
This inequality is satisfied, which also agrees with the plot. The damped resonant frequency is
given by
2
2
⎛
⎞
1
1
1
⎛ 1 ⎞
⎟ = 5.95 ⋅103
ω =
−⎜
− ⎜
⎟ =
d
6
−
⎜
⎟
LC ⎝ 2RC ⎠
2 ⋅10−6 12 ⋅10−3
⎝ 2(100) (2 ⋅10 ) ⎠
(
)(
)
The plot indicates a maxima at 550.6μs and a minima at 1078.7μs. The period of the damped
oscillation is
T = 2 (1078.7 μs − 550.6μs) = 1056.2μs
d
2π
2π
Finally, check that 5.95 ⋅103 = ω =
=
= 5.949 ⋅103
d
T
1056.2 ⋅10−6
d
The value of ωd determined from the plot agrees with the value obtained from the circuit.
The plot is correct
P 9.11-2 Figure P 9.11-2b shows an RLC circuit. The voltage, vs(t), of the voltage source is the
square wave shown in Figure P 9.11-2a. Figure P 9.11-2c shows a plot of the inductor current,
i(t), which was obtained by simulating this circuit using PSpice. Verify that the plot of i(t) is
correct.
Answer: The plot is not correct.
15
i(t)
100 Ω
vs, V
vs
0
2
4
6
+
–
0.2 μ F
8 mH
8
t, ms
(b)
(a)
300 mA
(426.966u, 172.191m)
200 mA
(1.7753m, 149.952m)
100 mA
I (L1)
0A
(831.461u, 146.570m)
–100 mA
0s
2.0 ms
4.0 ms
Time
6.0 ms
8.0 ms
(c)
Figure P 9.11-2
Solution:
This problem is similar to the verification example in this chapter. First, check the steady-state
inductor current.
v
15
i (t ) = s =
= 150 mA
100 100
This agrees with the value of 149.952 mA shown on the plot. Next, the plot shows an under
damped response. This requires
8 ⋅10−3 = L < 4 R 2C = 4 (100) 2 (0.2 ⋅10−6 ) = 8 ⋅10−3
This inequality is not satisfied. The values in the circuit would produce a critically damped, not
underdamped, response. This plot is not correct.
PSpice Problems
SP 9-1 The input to the circuit shown in Figure
SP 9-1 is the voltage of the voltage source, vi(t).
The output is the voltage across the capacitor,
vo(t). The input is the pulse signal specified
graphically by the plot. Use PSpice to plot the
output, vo(t), as a function of t for each of the
following cases:
vi(V)
5
0
(a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω
(b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω
(c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω
Plot the output for these three cases on the same
axis.
Hint: Represent the voltage source using the
PSpice part named VPULSE.
10
5
L
+
vi(t) –
15 t (s)
R1
R2
C
+
vo(t)
–
Figure SP 9-1
Solution:
Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure
to edit the labels of the parts so, for example, there is only one R1.)
V(C1:2), V(C2:2) and V(C3:2) are the capacitor voltages, listed from top to bottom.
SP 9-2 The input to the circuit shown in Figure
SP 9-2 is the voltage of the voltage source, vi(t).
The output is the voltage, vo(t), across resistor R2.
The input is the pulse signal specified graphically
by the plot. Use PSpice to plot the output, vo(t), as
a function of t for each of the following cases:
(a)
(b)
(c)
vi(V)
5
0
Plot the output for these three cases on the same
axis.
Hint: Represent the voltage source using the
PSpice part named VPULSE.
R1
+
vi(t) –
15 t (s)
10
5
C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω
C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω
C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω
L
C
R2
+
vo(t)
–
Figure SP 9-2
Solution:
Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure
to edit the labels of the parts so, for example, there is only one R1.)
V(R2:2), V(R4:2) and V(R6:2) are the output voltages, listed from top to bottom.
10 Ω
SP 9-3 Determine and plot the
capacitor voltage
50 Ω
ig
v(t) for 0 < t < 300 μs
for the circuit shown in
Figure SP 9-3a. The sources
are pulses as shown in
Figures SP 9-3b, c.
1 mH
+
v
–
0.1 μ F
+
–
vg
(a)
0.2 A
5V
vg
ig
0
0
100
t ( μ s)
200
0
0
100
t ( μ s)
(c)
(b)
Figures SP 9-3
Solution:
50 Ω
200
3 kΩ
SP 9-4 Determine and plot v(t) for the circuit
of Figure SP 9-4 when vs(t) = 5u(t) V. Plot v(t)
for 0 < t < 0.25 s.
vs(t)
+
–
6 kΩ
+
v(t)
–
2 kΩ
3 kΩ
2 μF
3 μF
Figure SP 9-4
Solution:
Design Problems
DP 9-1 Design the circuit shown in Figure DP 9-1 so that
vc(t) =
1
2
+ A1e–2t + A2e–4t V for t > 0
Determine the values of the unspecified constants, A1 and A2.
Hint: The circuit is overdamped, and the natural frequencies are 2 and 4 rad/sec.
iL(t)
vs(t) = u(t)
R1
L
+
–
R2
C
+
vc(t)
–
Figure DP 9-1
Solution:
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
R2
vC ( ∞ ) =
1
R1 + R 2
1
The specifications require that vC ( ∞ ) = so
2
R2
1
=
⇒ R1 = R 2
2 R1 + R 2
Next, represent the circuit by a 2nd order differential equation:
KCL at the top node of R2 gives:
KVL around the outside loop gives:
vC ( t )
R2
vs ( t ) = L
+C
d
vC ( t ) = iL ( t )
dt
d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt
Use the substitution method to get
vs ( t ) = L
⎞
⎛ v (t )
⎞
d ⎛ vC ( t )
d
d
+ C vC ( t ) ⎟ + R1 ⎜ C
+ C vC ( t ) ⎟ + vC ( t )
⎜⎜
⎟
⎜ R2
⎟
dt ⎝ R 2
dt
dt
⎠
⎝
⎠
= LC
⎛ L
⎞d
⎛
R1 ⎞
d2
v t +
v t
+ R1 C ⎟ vC ( t ) + ⎜1 +
2 C ( ) ⎜
⎜ R2
⎟ dt
⎜ R 2 ⎟⎟ C ( )
dt
⎝
⎠
⎝
⎠
The characteristic equation is
R
⎛
1+ 1
⎜
⎛ 1
R1 ⎞
R2
s2 + ⎜
+ ⎟s+⎜
⎜ R 2 C L ⎟ ⎜ LC
⎝
⎠ ⎜
⎜
⎝
Equating coefficients of like powers of s:
⎞
⎟
⎟ = s 2 + 6s + 8 = ( s + 2 )( s + 4 )
⎟
⎟⎟
⎠
R1
1
+
= 6 and
R2 C L
1+
R1
R2
LC
=8
Using R1 = R 2 = R gives
1
R
1
+ = 6 and
=4
RC L
LC
These equations do not have a unique solution. Try C = 1 F. Then L =
1
H and
4
1
3
1
+ 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω
R
2
4
Pick R = 1.309 Ω. Then
vc ( t ) =
iL ( t ) =
At t = 0+
vC ( t )
1.309
+
1
+ A1 e−2 t + A2 e−4 t V
2
d
vC ( t ) = −1.236 A1 e −2 t − 3.236 A2 e−4 t + 0.3819
dt
0 = vc ( 0 + ) = A1 + A2 + 0.5
0 = iL ( 0 + ) = −1.236 A1 − 3.236 A2 + 0.3819
Solving these equations gives A1 = -1 and A2 = 0.5, so
vc ( t ) =
1 −2 t 1 −4 t
−e + e V
2
2
DP 9-2 Design the circuit shown in Figure DP 9-1 so that
vc(t) =
1
4
+ (A1 + A2t)e–2t V for t > 0
Determine the values of the unspecified constants, A1 and A2.
Hint: The circuit is critically damped, and the natural frequencies are both 2 rad/sec.
iL(t)
vs(t) = u(t)
L
R1
+
–
R2
C
+
vc(t)
–
Figure DP 9-1
Solution:
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
R2
vC ( ∞ ) =
1
R1 + R 2
1
The specifications require that vC ( ∞ ) = so
4
R2
1
=
⇒ 3 R 2 = R1
4 R1 + R 2
Next, represent the circuit by a 2nd order differential equation:
vC ( t )
d
KCL at the top node of R2 gives:
+ C vC ( t ) = iL ( t )
R2
dt
d
KVL around the outside loop gives:
vs ( t ) = L iL ( t ) + R1 iL ( t ) + vC ( t )
dt
Use the substitution method to get
vs ( t ) = L
⎞
⎛ v (t )
⎞
d ⎛ vC ( t )
d
d
+ C vC ( t ) ⎟ + R1 ⎜ C
+ C vC ( t ) ⎟ + vC ( t )
⎜⎜
⎟
⎜ R2
⎟
dt ⎝ R 2
dt
dt
⎠
⎝
⎠
⎛ L
⎞d
⎛
R1 ⎞
d2
= LC 2 vC ( t ) + ⎜
+ R1 C ⎟ vC ( t ) + ⎜1 +
v t
⎜ R2
⎟ dt
⎜ R 2 ⎟⎟ C ( )
dt
⎝
⎠
⎝
⎠
The characteristic equation is
R ⎞
⎛
1+ 1 ⎟
⎜
⎛ 1
R1 ⎞
R2 ⎟ 2
2
s2 + ⎜
+ ⎟s+⎜
= s + 4s + 4 = ( s + 2 )
⎜ R 2 C L ⎟ ⎜ LC ⎟
⎝
⎠ ⎜
⎟⎟
⎜
⎝
⎠
Equating coefficients of like powers of s:
1
R2 C
+
R1
L
1+
= 4 and
R1
R2
LC
=4
Using R 2 = R and R1 = 3R gives
1
3R
1
+
= 4 and
=1
LC
RC L
These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and
1
4
1
1
+ 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω
R
3
3
3
Pick R = 1 Ω. Then R1 = 3 Ω and R 2 = 1 Ω .
vc ( t ) =
iL ( t ) = vC ( t ) +
1
+ ( A1 + A2 t ) e −2 t V
4
d
1
vC ( t ) = +
dt
4
(( A
2
At t = 0+
0 = vc ( 0 + ) = A1 +
0 = iL ( 0 + ) =
1
4
1
+ A2 − A1
4
Solving these equations gives A1 = -0.25 and A2 = -0.5, so
vc ( t ) =
)
− A1 ) − A2 t e −2 t
1 ⎛ 1 1 ⎞ −2 t
− ⎜ + t ⎟e V
4 ⎝4 2 ⎠
DP 9-3 Design the circuit shown in Figure DP 9-1 so that
vc(t) = 0.8 + e–2t(A1 cos 4t + A2 sin 4t) V for t > 0
Determine the values of the unspecified constants, A1 and A2.
Hint: The circuit is underdamped and the damped resonant frequency is 4 rad/sec and the
damping coefficient is 2.
iL(t)
vs(t) = u(t)
R1
L
+
–
R2
C
+
vc(t)
–
Figure DP 9-1
Solution:
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
R2
1
vC ( ∞ ) =
R1 + R 2
4
The specifications require that vC ( ∞ ) = so
5
R2
4
=
⇒ 4 R1 = R 2
5 R1 + R 2
Next, represent the circuit by a 2nd order differential equation:
KCL at the top node of R2 gives:
KVL around the outside loop gives:
vC ( t )
R2
vs ( t ) = L
+C
d
vC ( t ) = iL ( t )
dt
d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt
Use the substitution method to get
vs ( t ) = L
⎞
⎛ v (t )
⎞
d ⎛ vC ( t )
d
d
+ C vC ( t ) ⎟ + R1 ⎜ C
+ C vC ( t ) ⎟ + vC ( t )
⎜⎜
⎟
⎜ R2
⎟
dt ⎝ R 2
dt
dt
⎠
⎝
⎠
⎛ L
⎞d
⎛
R1 ⎞
d2
+ R1 C ⎟ vC ( t ) + ⎜1 +
v t +
v t
2 C ( ) ⎜
⎜ R2
⎟ dt
⎜ R 2 ⎟⎟ C ( )
dt
⎝
⎠
⎝
⎠
The characteristic equation is
R ⎞
⎛
1+ 1 ⎟
⎜
⎛ 1
R1 ⎞
R2 ⎟ 2
s2 + ⎜
+ ⎟s+⎜
= s + 4 s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 )
⎜ R 2 C L ⎟ ⎜ LC ⎟
⎝
⎠ ⎜
⎟⎟
⎜
⎝
⎠
Equating coefficients of like powers of s:
= LC
R1
1
+
= 4 and
R2 C L
1+
R1
R2
LC
= 20
Using R1 = R and R 2 = 4 R gives
R
1
1
+ = 4 and
= 16
4R C L
LC
These equations do not have a unique solution. Try C =
1
1
F . Then L = H and
8
2
2
+ 2 R = 4 ⇒ R2 − 2R + 2 = 0 ⇒ R = 1 Ω
R
Then R1 = 1 Ω and R 2 = 4 Ω . Next
vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
iL ( t ) =
At t = 0+
vC ( t )
4
+
A2 −2 t
A1
1d
vC ( t ) = 0.2 +
e cos 4 t − e−2 t sin 4 t
8 dt
2
2
0 = vc ( 0 + ) = 0.8 + A1
0 = iL ( 0 + ) = 0.2 +
A2
2
Solving these equations gives A1 = -0.8 and A2 = -0.4, so
vc ( t ) = 0.8 − e −2 t ( 0.8cos 4 t + 0.4sin 4 t ) V
DP 9-4 Show that the circuit shown in Figure DP 9-1 cannot be designed so that
vc(t) = 0.5 + e–2t(A1 cos 4t + A2 sin 4t) V for t > 0
Hint: Show that such a design would require 1/RC + 10RC = 4 where R = R1 = R2. Next, show
that 1/RC + 10 RC = 4 would require the value of RC to be complex.
iL(t)
vs(t) = u(t)
R1
L
+
–
R2
C
+
vc(t)
–
Figure DP 9-1
Solution:
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
R2
vC ( ∞ ) =
1
R1 + R 2
1
The specifications require that vC ( ∞ ) = so
2
R2
1
=
⇒ R1 = R 2
2 R1 + R 2
Next, represent the circuit by a 2nd order differential equation:
KCL at the top node of R2 gives:
KVL around the outside loop gives:
vC ( t )
R2
vs ( t ) = L
+C
d
vC ( t ) = iL ( t )
dt
d
iL ( t ) + R1 iL ( t ) + vC ( t )
dt
Use the substitution method to get
⎞
⎛ v (t )
⎞
d ⎛ v (t )
d
d
vs ( t ) = L ⎜ C
+ C vC ( t ) ⎟ + R1 ⎜ C
+ C vC ( t ) ⎟ + vC ( t )
⎟
⎜ R2
⎟
dt ⎜⎝ R 2
dt
dt
⎠
⎝
⎠
⎛ L
⎞d
⎛
R1 ⎞
d2
v t +
+ R1 C ⎟ vC ( t ) + ⎜1 +
v t
2 C ( ) ⎜
⎜ R2
⎟ dt
⎜ R 2 ⎟⎟ C ( )
dt
⎝
⎠
⎝
⎠
The characteristic equation is
R ⎞
⎛
1+ 1 ⎟
⎜
⎛ 1
R1 ⎞
R2 ⎟ 2
s2 + ⎜
+ ⎟s+⎜
= s + 4 s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 )
⎜ R 2 C L ⎟ ⎜ LC ⎟
⎝
⎠ ⎜
⎟⎟
⎜
⎝
⎠
Equating coefficients of like powers of s:
= LC
R1
1
+
= 4 and
R2 C L
1+
R1
R2
LC
= 20
Using R1 = R 2 = R gives
1
R
1
+ = 4 and
= 10
RC L
LC
Substituting L =
1
into the first equation gives
10 C
0.4 ± 0.42 − 4 ( 0.1)
4
1
( RC ) + = 0 ⇒ RC =
10
10
2
Since RC cannot have a complex value, the specification cannot be satisfied.
( RC )
2
−
DP 9-5 Design the circuit shown in Figure DP 9-5 so that
vo(t) =
1
2
+ A1 e–2t + A2 e– 4t V for t > 0
Determine the values of the unspecified constants, A1 and A2.
Hint: The circuit is overdamped, and the natural frequencies are 2 and 4 rad/sec.
iL(t)
R1
L
+
vs(t) = u(t)
+
–
C
+
R2
vc(t)
–
vo(t)
–
Figure DP 9-5
Solution:
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) =
R2
R1 + R 2
1, iL ( ∞ ) =
The specifications require that vo ( ∞ ) =
1
R1 + R 2
and
vo ( ∞ ) =
R2
R1 + R 2
1
1
so
2
R2
1
=
⇒ R1 = R 2
2 R1 + R 2
Next, represent the circuit by a 2nd order differential equation:
KVL around the right-hand mesh gives:
KCL at the top node of the capacitor gives:
d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
vC ( t ) = L
Use the substitution method to get
vs ( t ) = R1 C
= R1 LC
Using vo ( t ) =
iL ( t )
gives
R2
d ⎛ d
⎞ ⎛ d
⎞
⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t )
dt ⎝ dt
⎠ ⎝ dt
⎠
d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt
⎛ L
⎞d
⎛ R1 + R 2 ⎞
d2
vo ( t ) =
LC 2 iL ( t ) + ⎜
+ R1 C ⎟ iL ( t ) + ⎜
i t
⎜ R2
⎟ dt
⎜ R 2 ⎟⎟ L ( )
R2
dt
⎝
⎠
⎝
⎠
R1
The characteristic equation is
R
⎛
1+ 2
⎜
⎛ 1
R2 ⎞ ⎜
R1
s2 + ⎜
+
s+
⎟
⎜ R1 C L ⎟ ⎜ LC
⎝
⎠ ⎜
⎜
⎝
Equating coefficients of like powers of s:
⎞
⎟
⎟ = s 2 + 6 s + 8 = ( s + 2 )( s + 4 )
⎟
⎟⎟
⎠
R2
1
+
= 6 and
R1 C L
1+
R2
R1
LC
=8
Using R1 = R 2 = R gives
1
R
1
+ = 6 and
=4
RC L
LC
These equations do not have a unique solution. Try C = 1 F. Then L =
1
H and
4
1
3
1
+ 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω
2
4
R
Pick R = 1.309 Ω. Then
1
+ A1 e −2 t + A2 e−4 t V
2
A1 −2 t
A 2 −4 t
v (t )
1
iL ( t ) = o
e +
e V
=
+
1.309 2.618 1.309
1.309
vo ( t ) =
vC ( t ) = 1.309 iL ( t ) +
1 d
1
iL ( t ) = + 0.6167 A1 e−2 t + 0.2361 A2 e−4 t
4 dt
2
At t = 0+
0 = iL ( 0 + ) =
0 = vC ( 0 + ) =
A1
A2
1
+
+
2.618 1.309 1.309
1
+ 0.6167 A1 + 0.2361 A2
2
Solving these equations gives A1 = -1 and A2 = 0.5, so
vo ( t ) =
1 −2 t 1 −4 t
−e + e V
2
2
DP 9-6 Design the circuit shown in Figure DP 9-5 so that
3
vo(t) =
4
+ (A1 + A2t)e–2t V for t > 0
Determine the values of the unspecified constants, A1 and A2.
Hint: The circuit is critically damped, and the natural frequencies are both 2 rad/sec.
iL(t)
R1
L
+
+
–
vs(t) = u(t)
C
+
R2
vc(t)
–
vo(t)
–
Figure DP 9-5
Solution:
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
vC ( ∞ ) =
R2
R1 + R 2
1, iL ( ∞ ) =
The specifications require that vo ( ∞ ) =
1
R1 + R 2
and
vo ( ∞ ) =
R2
R1 + R 2
1
3
so
4
R2
3
=
⇒ 3R1 = R 2
4 R1 + R 2
Next, represent the circuit by a 2nd order differential equation:
KVL around the right-hand mesh gives:
KCL at the top node of the capacitor gives:
d
iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
R1
dt
vC ( t ) = L
Use the substitution method to get
vs ( t ) = R1 C
= R1 LC
Using vo ( t ) =
iL ( t )
gives
R2
vo ( t ) =
d ⎛ d
⎞ ⎛ d
⎞
⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t )
dt ⎝ dt
⎠ ⎝ dt
⎠
d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt
R1
R2
LC
The characteristic equation is
⎛ L
⎞d
⎛ R1 + R 2 ⎞
d2
i t +
+ R1 C ⎟ iL ( t ) + ⎜
i t
2 L( ) ⎜
⎜ R2
⎟ dt
⎜ R 2 ⎟⎟ L ( )
dt
⎝
⎠
⎝
⎠
R
⎛
1+ 2
⎜
⎛ 1
R2 ⎞ ⎜
R1
s2 + ⎜
+
s+
⎟
⎜ R1 C L ⎟ ⎜ LC
⎝
⎠ ⎜
⎜
⎝
Equating coefficients of like powers of s:
⎞
⎟
⎟ = s 2 + 4s + 4 = ( s + 2 )2
⎟
⎟⎟
⎠
R2
1
+
= 4 and
R1 C L
1+
R2
R1
LC
=4
Using R1 = R and R 2 = 3R gives
1
3R
1
+
= 4 and
=1
RC L
LC
These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and
1
4
1
1
+ 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω
3
3
3
R
Pick R = 1 Ω. Then R1 = 1 Ω and R 2 = 3 Ω .
vo ( t ) =
iL ( t ) =
3
+ ( A1 + A2 t ) e −2 t V
4
vo ( t )
vC ( t ) = 3 iL ( t ) +
3
=
1 ⎛ A1 A2 ⎞ −2 t
t⎟ e V
+⎜ +
4 ⎝ 3
3 ⎠
d
3 ⎛ ⎛ A1 A2 ⎞ A2 ⎞ −2 t
iL ( t ) = + ⎜⎜ ⎜ +
t ⎟e
⎟+
dt
4 ⎝⎝ 3
3 ⎠ 3 ⎟⎠
At t = 0+
0 = iL ( 0 + ) =
A1
+
1
4
3
3 A1 A2
0 = vC ( 0 + ) = + +
4 3
3
Solving these equations gives A1 = -0.75 and A2 = -1.5, so
vo ( t ) =
3 ⎛ 3 3 ⎞ −2 t
− ⎜ + t ⎟e V
4 ⎝4 2 ⎠
DP 9-7 Design the circuit shown in Figure DP 9-5 so that
vc(t) = 0.2 + e–2t(A1 cos 4t + A2 sin 4t) V for t > 0
Determine the values of the unspecified constants, A1 and A2.
Hint: The circuit is underdamped, the damped resonant frequency is 4 rad/sec, and the damping
coefficient is 2.
iL(t)
R1
L
+
vs(t) = u(t)
+
–
C
vc(t)
–
+
R2
vo(t)
–
Figure DP 9-5
Solution:
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
R2
R2
1
vC ( ∞ ) =
and vo ( ∞ ) =
1, iL ( ∞ ) =
1
R1 + R 2
R1 + R 2
R1 + R 2
1
The specifications require that vo ( ∞ ) = so
5
R2
1
=
⇒ R1 = 4 R 2
5 R1 + R 2
Next, represent the circuit by a 2nd order differential equation:
d
KVL around the right-hand mesh gives:
vC ( t ) = L iL ( t ) + R 2 iL ( t )
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
KCL at the top node of the capacitor gives:
R1
dt
Use the substitution method to get
vs ( t ) = R1 C
= R1 LC
Using vo ( t ) =
iL ( t )
gives
R2
vo ( t ) =
d ⎛ d
⎞ ⎛ d
⎞
⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t )
dt ⎝ dt
⎠ ⎝ dt
⎠
d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt
R1
R2
LC
The characteristic equation is
⎛ L
⎞d
⎛ R1 + R 2 ⎞
d2
i
t
+
+
R
C
i
t
+
i t
(
)
(
)
⎜
⎟
⎜
L
1
⎜ R2
⎟ dt L
⎜ R 2 ⎟⎟ L ( )
dt 2
⎝
⎠
⎝
⎠
R
⎛
1+ 2
⎜
⎛ 1
R2 ⎞ ⎜
R1
+
s2 + ⎜
s+
⎟
⎜ R1 C L ⎟ ⎜ LC
⎝
⎠ ⎜
⎜
⎝
Equating coefficients of like powers of s:
⎞
⎟
⎟ = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 )
⎟
⎟⎟
⎠
R2
1
+
= 4 and
R1 C L
1+
R2
R1
LC
= 20
Using R 2 = R and R1 = 4 R gives
1
R
1
+ = 4 and
= 16
4R C L
LC
1
1
These equations do not have a unique solution. Try C = F . Then L = H and
8
2
2
+ 2 R = 4 ⇒ R2 − 2R + 2 = 0 ⇒ R = 1 Ω
R
Then R1 = 4 Ω and R 2 = 1 Ω . Next
vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V
vo ( t )
= 0.2 + e−2 t ( A1 cos 4 t + A2 sin 4 t ) V
1
1 d
vC ( t ) = iL ( t ) +
iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e −2 t sin 4 t
2 dt
iL ( t ) =
At t = 0+
0 = iL ( 0 + ) = 0.2 + A1
0 = vC ( 0 + ) = 0.2 + 2 A2
Solving these equations gives A1 = -0.8 and A2 = -0.4, so
vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V
DP 9-8 Show that the circuit shown in Figure DP 9-5 cannot be designed so that
vc(t) = 0.5 + e–2t(A1 cos 4t + A2 sin 4t) V for t > 0
Hint: Show that such a design would require 1/RC + 10RC = 4 where R = R1 = R2. Next, show
that 1/RC + 10 RC = 4 would require the value of RC to be complex.
iL(t)
R1
L
+
vs(t) = u(t)
+
–
C
vc(t)
–
+
R2
vo(t)
–
Figure DP 9-5
Solution:
When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the
inductor acts like a short circuit. Under these conditions
R2
R2
1
1, iL ( ∞ ) =
1
vC ( ∞ ) =
and vo ( ∞ ) =
R1 + R 2
R1 + R 2
R1 + R 2
1
The specifications require that vC ( ∞ ) = so
2
R2
1
=
⇒ R1 = R 2
2 R1 + R 2
Next, represent the circuit by a 2nd order differential equation:
d
vC ( t ) = L iL ( t ) + R 2 iL ( t )
KVL around the right-hand mesh gives:
dt
vs ( t ) − vC ( t )
d
− C vC ( t ) = iL ( t )
KCL at the top node of the capacitor gives:
R1
dt
Use the substitution method to get
vs ( t ) = R1 C
= R1 LC
Using vo ( t ) =
iL ( t )
gives
R2
vo ( t ) =
d ⎛ d
⎞ ⎛ d
⎞
⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t )
dt ⎝ dt
⎠ ⎝ dt
⎠
d2
d
i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t )
2 L( ) (
dt
dt
R1
R2
LC
The characteristic equation is
⎛ L
⎞d
⎛ R1 + R 2 ⎞
d2
i t +
+ R1 C ⎟ iL ( t ) + ⎜
i t
2 L( ) ⎜
⎜ R2
⎟ dt
⎜ R 2 ⎟⎟ L ( )
dt
⎝
⎠
⎝
⎠
R
⎛
1+ 2
⎜
⎛ 1
R2 ⎞ ⎜
R1
+
s2 + ⎜
s+
⎟
⎜ R1 C L ⎟ ⎜ LC
⎝
⎠ ⎜
⎜
⎝
Equating coefficients of like powers of s:
⎞
⎟
⎟ = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 )
⎟
⎟⎟
⎠
R2
1
+
= 4 and
R1 C L
1+
R2
R1
LC
= 20
Using R1 = R 2 = R gives
1
R
1
+ = 4 and
= 10
LC
RC L
Substituting L =
1
into the first equation gives
10 C
( RC )
2
0.4 ± 0.42 − 4 ( 0.1)
4
1
− ( RC ) + = 0 ⇒ RC =
10
10
2
Since RC cannot have a complex value, the specification cannot be satisfied.
L
DP 9-9 A fluorescent light uses cathodes (coiled
1
2
tungsten filaments coated with an electron-emitting
i
t=0
substance) at each end that send current through
mercury vapors sealed in the tube. Ultraviolet
+
4 Ω
radiation is produced as electrons from the cathodes 10 V –
1 3F
knock mercury electrons out of their natural orbits.
Some of the displaced electrons settle back into
orbit, throwing off the excess energy absorbed in
Figure DP 9-9
the collision. Almost all of this energy is in the
form of ultraviolet radiation. The ultraviolet rays, which are invisible, strike a phosphor coating
on the inside of the tube. The rays energize the electrons in the phosphor atoms, and the atoms
emit white light. The conversion of one kind of light into another is known as fluorescence.
One form of a fluorescent lamp is represented by the RLC circuit shown in Figure DP 99. Select L so that the current i(t) reaches a maximum at approximately t = 0.5 s. Determine the
maximum value of i(t). Assume that the switch was in position 1 for a long time before switching
to position 2 at t = 0.
Hint: Use PSpice to plot the response for several values of L.
Solution:
R
1
s+
=0
L
LC
Select L so fast response and i achieve maximum at t = 0.5 s
Characteristic equation: s 2 +
4
3
s + = 0 Try L = 1 H ⇒ s 2 + 4s + 3 = 0 or s = −3, −1
L
L
i (t ) = A1 e − t + A2 e −3t
s2 +
i (0) = 0 = A1 + A2
⎫
⎪
⎬ A1 = 5, A2 = −5
di (0) v(0) 10
=
=
= − A1 − 3 A2 ⎪
dt
L
1
⎭
−t
−3t
i (t ) = 5e − 5e
at t = 0.5 s i − 1.92
Section 10.2 Sinusoidal Sources
P10.2‐1 Given the sinusoids v1 ( t ) = 8cos ( 250 t + 15° ) V and v 2 ( t ) = 6 cos ( 250 t − 45° ) V ,
determine the time by which v 2 ( t ) is advanced or delayed with respect to v1 ( t ) .
Solution:
T=
The period of both sinusoids is
2π
= 25.1327 ms
250
The difference in the phase angles is
θ 2 − θ1 = −45° − 15° = −60°
The delay time is
td =
−60 ( 25.1327 )
= −4.188 ms
2π
(The minus sign indicates a delay.) The voltage v 2 ( t ) is delayed by 4.188 ms with respect to
v1 ( t ) .
P10.2‐2 Given the sinusoids v1 ( t ) = 8cos (100 t − 54° ) V and v 2 ( t ) = 8cos (100 t − 102° ) V ,
determine the time by which v 2 ( t ) is advanced or delayed with respect to v1 ( t ) .
Solution:
T=
The period of both sinusoids is
2π
= 62.8319 ms
100
The difference in the phase angles is
θ 2 − θ1 = −102° − ( −54° ) = −48°
−48° ( 62.8319 )
= −8.3776 ms
360°
(The minus sign indicates a delay.) The voltage v 2 ( t ) is delayed by 48.3776 ms with respect to
The delay time is
v1 ( t ) .
td =
P10.2‐3 A sinusoidal current is given as
i ( t ) = 125cos ( 5000 π t − 135° ) mA
Determine the period, T, and the time, t1, at which the first positive peak occurs.
Answer: T = 0.4 ms and t1 = 0.15 ms.
Solution: The frequency is 5000 π rad/s or 2500 Hertz so the period is
1
T=
= 0.0004 = 0.4 ms
2500
⎛ π ⎞
Converting the angle from degrees to radians, we get −135° ⎜
⎟ = −0.75 π radians . The
⎝ 180° ⎠
−0.75 π
= −15 ms . The minus sign
current sinusoid is shifted from 125cos ( 5000 π t ) mA by
5000 π
indicates a delay. A positive peak occurs at t 1 = 0.15 ms . Since 15 ms is less than the period of
i ( t 1 ) , the positive peak at t 1 = 0.15 ms is the first positive peak.
P10.2‐4 Express the voltage shown in Figure P10.2‐7 in the general form
v ( t ) = A cos (ω t + θ ) V
where A ≥ 0 and −180° < θ ≤ 180° .
Figure P10.2‐4
Solution: The amplitude is A = 45 mv and the period is given by
period is T = 80 ms. The frequency is given by ω =
T
= 60 − 20 = 40 ms so the
2
2π
= 78.54 rad/s . Noticing that v(t) is 0
80 ×10−3
at time 0 and is increasing at time 0, we can write
v ( t ) = 45sin ( 78.54 t ) = 45cos ( 78.54 t − 90° ) mV
20 V
P 10.2-5 Figure P 10.2-5 shows a sinusoidal
voltage, v(t), plotted as a function of time, t.
Represent v(t) by a function of the form
A cos (ω + θ).
v(t)
0V
Answer: v(t) = 18 cos (393t–27°)
–20 V
0s
20 ms
t
40 ms
Figure P 10.2-5
Solution:
A = 18 V
T = 18 − 2 = 16 ms
ω=
2π
2π
=
= 393 rad/s
0.016
T
⎛ 16 ⎞
⎟ = −27°
⎝ 18 ⎠
θ = − cos −1 ⎜
v ( t ) = 18 cos ( 393 t − 27° ) V
20 V
P 10.2-6 Figure P 10.2-6 shows a sinusoidal
voltage, v(t), plotted as a function of time, t.
Represent v(t) by a function of the form
A cos (ωt+θ).
v(t)
0V
–20 V
0s
20 ms
40 ms
60 ms
t
Figure P 10.2-6
Solution:
A = 15 V
T = 43 − 11 = 32 ms
ω=
2π
2π
=
= 196 rad/s
T
0.032
⎛8⎞
⎟ = −58°
⎝ 15 ⎠
θ = − cos −1 ⎜
v ( t ) = 15 cos (196 t − 58° ) V
Section 10.3 Phasors and Sinusoids
P10.3‐1 Express the current
i ( t ) = 2 cos ( 6 t + 120° ) + 2sin ( 6 t − 60° ) mA
In the general form
i ( t ) = A cos (ω t + θ ) mA
where A ≥ 0 and −180° < θ ≤ 180° .
Solution:
i ( t ) = 2 cos ( 6 t + 120° ) + 2sin ( 6 t − 60° ) mA
= 2 cos ( 6 t + 120° ) + 2 cos ( 6 t − 60° − 90° ) mA
Representing the sinusoids using phasors gives:
I (ω ) = 2∠120° + 4∠ − 150° = ( −1 + j 1.732 ) + ( −3.464 − j 2 )
= −4.464 − j 0.268 = 4.472∠183.4° = 4.472∠ − 176.6° mA
The corresponding sinusoid is:
P10.3‐2 Express the voltage
i ( t ) = 4.472 cos ( 6 t − 176.6° ) mA
v ( t ) = 5 2 cos ( 8 t ) + 2sin ( 8 t + 45° ) V
In the general form
v ( t ) = A cos (ω t + θ ) V
where A ≥ 0 and −180° < θ ≤ 180° .
Solution:
v ( t ) = 5 2 cos ( 8 t ) + 2sin ( 8 t + 45° )
= 5 2 cos ( 8 t ) + 2 cos ( 8 t + 45° − 90° ) = 5 2 cos ( 8 t ) + 2 cos ( 8 t − 45° ) V
Representing the sinusoids using phasors gives:
V (ω ) = 7.0711 + 10∠ − 45° = 7.0711 + ( 7.0711 − j 7.0711)
= 14.1422 − j 7.0711 = 15.811∠ − 26.6° V
The corresponding sinusoid is:
v ( t ) = 15.811cos ( 8 t − 26.6° ) V
P 10.3-3
Determine the polar form of the quantity
(25∠36.9°) ( 80 ∠ − 53.1° )
(4 + j 8) + (6 − j 8)
Answer: 200∠ − 16.2°
Solution:
(25∠36.9°) ( 80 ∠ − 53.1° ) 25 ⋅ 80∠ ( 36.9° − 53.1° ) 2000∠ − 16.2°
=
=
= 200∠ − 16.2°
(4 + j 8) + (6 − j 8)
10
( 4 + 6 ) + j (8 − 8)
P 10.3-4
Determine the polar and rectangular form of the expression
⎛
3 2 ∠ − 45° ⎞
5 ∠ + 81.87° ⎜⎜ 4 − j 3 +
⎟
7 − j1 ⎟⎠
⎝
Answer: 88.162 ∠30.127° = 76.2520 + j 44.2506
Solution:
⎛
⎛
⎞
40 ∠ − 45° ⎞
40 ∠ − 45°
8 ∠42° ⎜ 8 − j 3 +
⎟ = 8 ∠42° ⎜ 8 − j 3 +
⎟
7 − j 12 ⎠
13.892 ∠ − 59.744° ⎠
⎝
⎝
= 8 ∠42° ( 8 − j 3 + 2.8793 ∠14.744° )
= 8 ∠42° ( 8 − j 3 + 2.7845 + j 0.7328 )
= 8 ∠42° (10.7845 − j 2.2672 )
= 8 ∠42° (11.02∠ − 11.873° ) = 88.162 ∠30.127° = 76.2520 + j 44.2506
Determine the polar and rectangular form of the expression
P 10.3-5
(60 ∠120°)(−16 + j12 + 20∠15°)
5∠ − 75°
Solution:
(60 ∠120°)(−16 + j12 + 20∠15°) (60 ∠120°)(−16 + j12 + 19.3185 + j 5.1764)
=
5∠ − 75°
5∠ − 75°
(60 ∠120°)(3.3185 + j 17.1764)
=
5∠ − 75°
(60 ∠120°)(17.494∠79.065°)
=
5∠ − 75°
1049.6∠ − 160.93°
=
= 139.95∠109.07° = 45.714 + j 132.28
5∠ − 75°
P10.3‐6 The circuit shown in Figure 10.3‐6 is at steady state. The input currents are
i1 ( t ) = 10 cos ( 25 t ) mA and i 3 ( t ) = 10 cos ( 25 t + 135° ) mA
Determine the voltage v2(t).
Figure 10.3‐6
Solution:
Using first Ohm’s law and then KCL
v2 ( t ) = 250 i 2 ( t )
i 2 ( t ) = i1 ( t ) − i 3 ( t ) = 10 cos ( 25 t ) − 10 cos ( 25 t + 135° ) mA
and
Using phasors
I 2 (ω ) = I 1 (ω ) − I 3 (ω ) = 10 − 10 ∠135 = 10 − ( −7.071 + j 7.071) mA
= 17.071 − j 7.071 = 18.478∠ − 22.5° mA
The corresponding sinusoid is i 2 ( t ) = 18.478cos ( 25 t − 22.5° ) mA
Finally
v2 ( t ) = 250 i 2 ( t ) = 4.6195cos ( 25 t − 22.5° ) V
P10.3‐7 The circuit shown in Figure 10.3‐7 is at steady state. The inputs to this circuit are the
current source current
i1 ( t ) = 0.12 cos (100 t + 45° ) A
and the voltage source voltage
v 2 ( t ) = 24 cos (100 t − 60° ) V
Determine the current i2(t).
Figure P10.3‐7
Solution: Using Ohm’s and Kirchhoff’s laws
v 2 (t )
24 cos (100 t − 60° )
i 2 ( t ) = i1 ( t ) −
= 0.12 cos (100 t + 45° ) −
96
96
= 0.12 cos (100 t + 45° ) − 0.25cos (100 t − 60° )
Using phasors
I 2 (ω ) = 0.12 ∠45° − 0.25 ∠60° = ( 0.0849 + j 0.0849 ) − ( 0.1250 − j 0.2165 )
= −0.0401 + j 0.3014 = 0.3040∠97.6° A
The corresponding sinusoid is
i 2 ( t ) = 0.3040 cos (100 t + 97.6° ) A
Checked using LNAPAC 8/16/11
P 10.3-8
i1 (t)
Given that
i2 (t)
i1(t) = 30 cos (4t + 45°) mA
and
i2(t) = – 40 cos (4t) mA
Determine v(t) for the circuit shown in
100 Ω
15 H
+
120 Ω
v(t)
–
Figure P 10.3-8.
Figure P 10.3-8
Solution:
V = j ( 4 )(15 ) ( I 1 + I 2 ) = j 60 ( 0.03∠45° − 0.04∠0° ) = j 60 ( 0.0212 + j 0.0212 − 0.04 )
= −1.273 − j1.127
So
= 1.7∠ − 138.5° V
v ( t ) = 1.7 cos ( 4t − 138.5° ) V
(checked: LNAP 8/7/04)
P 10.3-9 For the circuit shown in Figure P 10.3-9,
find (a) the impedances Z1 and Z2 in polar form,
(b) the total combined impedance in polar form, and
(c) the steady-state current i(t).
3Ω
Z1
i
Answer:
(a) Z1 = 5 ∠53.1°; Z 2 = 8 2 ∠ − 45°
(b) Z1 + Z 2 = 11.7 ∠ − 20°
(c) i (t ) = (8.55) cos (1250t + 20°) A
100 μF
3.2 mH
8Ω
Z2
+
–
100 cos (1250t) V
Figure P 10.3-9
Solution:
(a)
(b)
(c)
Z1 =3+ j 4 = 5∠53.1° Ω
and
Z 2 =8− j8 = 8 2 ∠− 45° Ω
Total impedance = Z1 + Z 2 = 3 + j 4 + 8 − j8 = 11 − j 4 = 11.7∠− 20.0° Ω
I=
100∠0°
100
100
=
=
∠20.0° ⇒
Z1 +Z 2 11.7 ∠− 20° 11.7
i (t ) = 8.55 cos (1250 t + 20.0°) A
P 10.3-10 The circuit shown in Figure P 10.3-10 is at
steady state. The voltages vs(t) and v2(t) are given by
+ v1(t) –
vs(t) = 7.68 cos (2t + 47°) V
and
vs(t)
+
–
C
R
v2(t) = 1.59 cos (2t + 125°) V
Find the steady-state voltage v1(t).
+
v2(t)
–
Figure P 10.3-10
Answer: v1(t) = 7.51 cos (2t + 35°) V
Solution:
V1 (ω ) = Vs (ω ) − V2 (ω ) = 7.68∠47° − 1.59∠125°
= ( 5.23 + j 5.62 ) − ( −0.91 + 1.30 )
= ( 5.23 + 0.91) + j ( 5.62 − 1.30 )
= 6.14 + j 4.32
= 7.51∠35°
v1 ( t ) = 7.51 cos ( 2 t + 35° ) V
P 10.3-11 The circuit shown in Figure P 10.3-11 is at steady state. The currents i1(t) and i2(t)
are given by
i1(t) = 744 cos (2t – 118°) mA
and
i2(t) = 540.5 cos (2t + 100°) mA
Find the steady-state current i(t).
Answer: i(t) = 460 cos (2t + 196°) mA
10 Ω
v1(t)
+
–
i1(t)
i2(t)
6H
10 Ω
0.05 F
+
–
10 Ω
v2(t)
i(t)
Figure P 10.3-11
Solution:
I = I 1 + I 2 = 0.744∠ − 118° + 0.5405∠100 = ( −0.349 − j 0.657 ) + ( −0.094 + j 0.532 )
= ( −0.349 − 0.094 ) + j ( −0.657 + 0.532 )
= −0.443 − j 0.125
= 0.460∠196°
i ( t ) = 460 cos (2 t + 196°) mA
i(t)
P 10.3-12 Determine i(t) of the RLC circuit shown in
Figure P 10.3-12 when
3H
6Ω
vs
vs = 2 cos (4t + 30°) V.
+
–
1 12
Answer: i(t) = 0.185 cos (4t – 26.3°) A
Figure P 10.3-12
Solution:
Vs = 2 ∠30° V
and I =
2 ∠30°
= 0.185 ∠ − 26.3° A
6+ j12+ 3 / j
i (t ) = 0.185 cos (4 t − 26.3°) A
F
Section 10.4 Impedances
P10.4-1 Figure P10.4-1a shows a circuit represented in the time domain. Figure P10.4-1b shows the
same circuit represented in the frequency domain, using phasors and impedances. ZR, ZC, ZL1, and
ZL2 are the impedances corresponding to the resistor, capacitor, and two inductors in Figure P10.41a. Vs is the phasor corresponding to the voltage of the voltage source. Determine ZR, ZC, ZL1, ZL2,
and Vs.
+
8Ω
+
Z L1
ZR
2H
Vs
5 sin 5t V
vo
+–
1
F
12
4H
Vo
+–
Z L2
ZC
–
(a)
–
(b)
Figure P10.4-1
Hint: 5 sin 5t = 5 cos (5t – 90°)
Answer: Z R = 8Ω, Z C =
1
2.4 j 2.4
=
=
= − j 2.4Ω, Z L1 = j 5(2) = j10Ω,
j
j× j
⎛ 1⎞
j5 ⎜ ⎟
⎝ 12 ⎠
ZL2 = j5(4) = j20 Ω, and Vs = 5 ∠–90° V
Solution:
ZR = 8 Ω, ZC =
1
=
2.4 j 2.4
=
= − j 2.4 Ω, ZL1 = j 5 (2) = j 10 Ω,
j
j× j
1
12
ZL2 = j 5 (4) = j 20 Ω and VS = 5 ∠-90° V.
j5
1
P10.4-2 Figure P10.4-2a shows a circuit represented in the time domain. Figure P10.4-2b shows the
same circuit represented in the frequency domain, using phasors and impedances. ZR, ZC, ZL1, and
ZL2 are the impedances corresponding to the resistor, capacitor, and two inductors in Figure P10.4-
2a. Is is the phasor corresponding to the current of the current source. Determine ZR, ZC, ZL1, ZL2,
and Is.
+
8Ω
+
ZR
2H
vo
1 12
Z L1
Vo
4 cos (3t + 15°) A
F 4H
ZC
Is
Z L2
–
–
(a)
(b)
Figure P10.4-2
Answer: Z R = 8 Ω, Z C =
1
4
j4
= =
= − j 4 Ω, Z L1 = j 3(2) = j 6 Ω,
⎛ 1 ⎞ j j× j
j3 ⎜ ⎟
⎝ 12 ⎠
ZL2 = j3(4) = j12 Ω, and Is = 4 ∠15° A
Solution:
ZR = 8 Ω, ZC =
1
=
4
j4
=
= − j 4 Ω, ZL1 = j 3 (2) = j 6 Ω,
j j× j
1
12
ZL2 = j 3 (4) = j 12 Ω and IS = 4 ∠15° A.
j3
2
P10.4-3 Represent the circuit shown in Figure P10.4-3 in the frequency domain using impedances
and phasors.
Figure P10.4-3
Solution:
P10.4-4 Represent the circuit shown in Figure P10.4-4 in the frequency domain using impedances
and phasors.
Figure P10.4-4
Solution:
3
P10.4-5 Determine the current, i(t), for the circuit shown in Figure P10.4-5.
Figure P10.4-5
Solution: Represent the circuit in the frequency domain using phasors and impedances:
j 96 I − j 312.5 I + 250 I + 7.4∠ − 24° = 0
Using KVL:
Solving:
I=
7.4∠ − 24°
7.4∠ − 24°
7.4∠ − 24°
=
=
= 0.0224∠16.9 A
j 96 − j 312.5 + 250 − j 216.5 + 250 330.715∠ − 40.9°
In the time domain:
i ( t ) = 22.4 cos ( 200 t + 16.9 ) mA
(Checked using LNAP)
4
P10.4-6 The input to the circuit shown in Figure P10.4-6 is the current
i ( t ) = 120 cos ( 4000 t ) mA
Determine the voltage, v(t), across the 40 Ω resistor.
Figure P10.4-6
Solution: Represent the circuit in the frequency domain using phasors and impedances:
Using KCL
0.120∠0° = I 1 + I 2
Using KVL
j 60 I 2 + 40 I 2 − ( − j 12.5 ) I 1 = 0
(
)
j 60 I 2 + 40 I 2 − ( − j 12.5 ) 0.12∠0° − I 2 = 0
j 60 I 2 + 40 I 2 + ( − j 12.5 ) I 2 = ( − j 12.5 )( 0.12∠0° )
I2 =
1.5∠ − 90°
1.5∠ − 90° 1.5∠ − 90°
=
=
= 0.024155∠ − 140° A
j 60 + 40 + ( − j 12.5 ) 40 + j 47.5 62.1∠50°
V = 40 I 2 = 0.9662∠ − 140° V
In the time domain
v ( t ) = 0.9662 cos ( 4000 t − 140° ) V
(Checked using LNAP)
5
P10.4-7 The input to the circuit shown in Figure P10.4-7 is the current
i ( t ) = 82 cos (10000 t ) μA
Determine the voltage, v(t), across the 50 kΩ resistor.
Figure P10.4-7
Solution: Represent the circuit in the frequency domain using phasors and impedances:
Using KCL
Using KVL
In the time domain
82 ×10−6 ∠0° = 0 + I 1
( 20 ×103 ) I 1 + ( − j 20 ×103 ) I 1 + V = 0
V = − ( 20 × 103 − j 20 ×103 )( 82 × 10−6 ) = 2.3193∠135°
V
v ( t ) = 2.3193cos (10000 t + 135° ) V
(Checked using LNAP)
6
P 10.4-8 Each of the following pairs of element voltage and element current adheres to the passive
convention. Indicate whether the element is capacitive, inductive, or resistive and find the element
value.
(a)
v(t) = 15 cos (400t + 30°); i = 3 sin (400t + 30°)
(b)
v(t) = 8 sin (900t + 50°); i = 2 sin (900t + 140°)
(c)
v(t) = 20 cos (250t + 60°); i = 5 sin (250t + 150°)
Answer: (a) L = 12.5 mH
(b) C = 277.77 μF
(c) R = 4 Ω
Solution:
(a)
v = 15cos (400 t + 30°) V
i = 3 sin(400 t+30°) = 3 cos (400 t − 60°) V
v leads i by 90° ⇒ element is an inductor
v
15
Z L = peak =
= 5 = ω L = 400 L ⇒ L = 0.0125 H = 12.5 mH
ipeak
3
(b)
i leads v by 90° ⇒ the element is a capacitor
(c)
8
1
1
=4=
=
⇒ C = 277.77 μ F
ω C 900 C
ipeak 2
v = 20 cos (250 t + 60°) V
Zc =
vpeak
=
i = 5sin (250 t +150°) =5cos (250 t + 60°) A
Since v & i are in phase ⇒ element is a resistor
v
20
∴ R = peak =
=4Ω
ipeak
5
7
Figure P10.4-9
P10.4-9 This voltage and current for the circuit shown in Figure P10.4-9 are given by
v ( t ) = 20 cos (20 t + 15°) V and i ( t ) = 1.49 cos (20 t + 63°) A
Determine the values of the resistance, R, and capacitance, C.
Solution:
1
V
20∠15°
20
=Z= =
=
∠ (15° − 63° ) = 13.42∠ − 48° = 8.98 − j9.97 Ω
20 C
I 1.49∠63° 1.49
1
Equating real and imaginary parts gives R = 9 Ω and C =
= 5 mF .
20 × 9.97
R− j
8
P10.4-10
Figure P10.4-10 shows an ac circuit represented in both the time domain and the frequency domain.
Determine the values of A, B, a and b.
Figure P10.4-10
Solution:
The impedance between nodes a and b is given by
18 + j (10 )( 2.5 ) = 18 + j 25 = 30.8∠54.2°
To find the impedance between nodes b and c we first find the impedance of the capacitor:
−j
then
1
1
=−j
= − j 25
0.04
(10 )( 0.004 )
9 ( − j 25 )
− j 225
225∠ − 90°
=
=
= 8.47∠ − 19.8° Ω
9 − j 25 26.57∠ − 70.2° 26.57∠ − 70.2°
The impedance between nodes c and d is given by
( 5 ) ( j (10 )( 0.88) )
j 40
j 40 ⎛ 5 − j 8 ⎞
1
1
−j
=
−j
=
− j 20
5 + j (10 )( 0.8 )
(10 )( 0.005 ) 5 + j 8 0.05 5 + j 8 ⎜⎝ 5 − j 8 ⎟⎠
320 + j 200
− j 20
25 + 64
= 3.60 + j 2.25 − j 20 = 3.60 − j 17.75 Ω
=
So
A = 30.8 V, B = 8.47 Ω, , a = 3.57 Ω and b = −17.75 Ω.
9
P 10.4-11 Represent the circuit shown in Figure P 10.4-11 in the frequency domain using
impedances and phasors.
6Ω
15 cos 4t V
+
–
2H
5i(t)
0.125 F
i(t)
+
v(t)
–
Figure P 10.4-11
Solution:
P 10.4-12 Represent the circuit shown in Figure P 10.4-12 in the frequency domain using
impedances and phasors.
12 cos (5t – 30°) V
+ –
2H
0.25 F
0.05 F
6Ω
+
v(t)
–
+ –
i(t)
15 cos (5t + 60°) V
Figure P 10.4-12
Solution:
10
P 10.4-13 Find R and L of the circuit of Figure P 10.4-13 when
v(t) = 10 cos (ωt + 40°) V; i(t) = 2 cos (ωt + 15°) mA,
+
v
6
and ω = 2 × 10 rad/s.
–
Answer: R = 4.532 kω, L = 1.057 mH
R
L
i
Figure P 10.4-13
Solution:
Z=
V
10 ∠40°
=
= − 5000∠205°Ω = 4532 + j 2113 = R + j ω L
−I −2×10−3 ∠−165°
so R = 4532 Ω and L =
2113
ω
=
2113
= 1.057 mH
2×106
11
Section 10.5 Series and Parallel Impedances
P10.5-1 Determine the steady state voltage v(t) in the circuit shown in Figure P10.5-1.
Answer: v ( t ) = 32 cos ( 250 t − 57.9° ) V
Figure P10.5-1
Solution: Represent the circuits in the frequency domain using phasors and impedances:
Using voltage division: V = −
In the time domain
16
16∠180°
( 40∠ − 15° ) =
( 40∠ − 15° ) = 32∠128.1° V
16 + j 12
20∠36.9°
v ( t ) = 32 cos ( 250 t − 57.9° ) V
1
P10.5-2 Determine the voltage v(t) in the circuit shown in Figure P10.5-2.
Answer: v ( t ) = 14.57 cos ( 800 t + 111.7° ) V
Figure P10.5-2
Solution:
Represent the circuit in the frequency domain:
Replace the series impedances at the right of the
circuit by an equivalent impedance
Z S = j 112 + ( − j 250 ) + 80 = 80 − j 138 Ω
Replace the parallel impedances at right of the
circuit by an equivalent impedance
ZP =
(80 − j 138)120 = (80 − j 138 )120
80 − j 138 + 120
200 − j 138
=
(159.51∠ − 59.9 )120
242.99∠ − 34.6
= 78.77∠25.3° Ω
Using voltage division
V=−
78.77∠ − 25.3°
78.77∠ − 25.3°
18∠0° = −
18∠0° = 14.57∠111.73° V
j 100 + 78.77∠ − 25.3°
97.325∠42.97°
In the time domain
v ( t ) = 14.57 cos ( 800 t + 111.7° ) V
(checked using LNAP)
2
P10.5-3 Determine the voltage v(t) in the circuit shown in Figure P10.5-3.
Answer: v ( t ) = 14.1cos ( 2500 t − 35.2° ) V
Figure P10.5-3
Solution:
Represent the circuit in the frequency
domain:
Replace the parallel impedances at right of
the circuit by an equivalent impedance:
( − j 100 )( j 250 ) = − j166.67 Ω
− j 100 + j 250
Using voltage division
V=−
In the time domain
− j 166.67
166.67∠ − 90°
22∠15° = −
22∠15° = 14.1∠ − 35.2° V
200 − j 166.67
260.3∠ − 39.8°
v ( t ) = 14.1cos ( 2500 t − 35.2° ) V
3
P10.5-4 The input to the circuit shown in Figure P10.5-4 is the current i s ( t ) = 48cos ( 25 t ) mA .
Determine the current i1(t).
Answer: i1 ( t ) = 144 cos ( 25 t + 180° ) mA
Figure P10.5-4
Solution: Represent the circuits in the frequency domain using phasors and impedances:
Using current division
I1 =
In the time domain
j 15
15
( 48∠0° ) = ( 48∠0° ) = 144∠180° mA
−5
j 15 − j 20
i1 ( t ) = 144 cos ( 25 t + 180° ) mA
4
P10.5-5 The input to the circuit shown in Figure P10.5-5 is the current i s ( t ) = 48cos ( 25 t ) mA .
Determine the current i2(t).
Figure P10.5-5
Solution: Represent the circuits in the frequency domain using phasors and impedances:
Using current division
I 2 ( ω) =
In the time domain
j15
15∠90°
( 48∠0° ) =
( 48∠0° ) = 28.8∠53.1° mA
20 + j15
25∠36.9°
i 2 ( t ) = 28.8cos ( 25 t + 53.1° ) mA
5
P10.5-6 The input to the circuit shown in Figure P10.5-6 is the current i s ( t ) = 48cos ( 25 t ) mA .
Determine the current i3(t).
Answer: i 3 ( t ) = 16.85cos ( 25 t + 69.4° ) mA
Figure P10.5-6
Solution: Represent the circuits in the frequency domain using phasors and impedances:
Using current division
7.5∠90°
j15 || j15
j 7.5
I3 =
( 48∠0° ) =
( 48∠0° ) =
( 48∠0° ) = 16.8539∠69.44° V
j 7.5 + 20
21.36∠69.56°
( j15 || j15 ) + 20
In the time domain
i 3 ( t ) = 16.85cos ( 25 t + 69.4° ) mA
6
P10.5-7 Figure P10.5-7 shows a circuit represented in the frequency domain. Determine the voltage
phasor V1.
Answer: V1 = 14.59∠ − 13.15° V
Figure P10.5-7
Solution:
V1 =
j 80 || − j 50
− j 133.33
133.33∠ − 90°
( 20∠30° ) =
( 20∠30° ) =
( 20∠30° )
125 + ( j 80 || − j 50 )
125 − j 133.33
182.8∠ − 46.85°
= 14.59∠ − 13.15° V
7
P10.5-8 Figure P10.5-8 shows a circuit represented in the frequency domain. Determine the current
phasor I2.
Answer: I 2 = 18.48∠ − 93.7° mA
Solution:
I2 =
Figure P10.5-8
− j 50
− j 50
( 20∠30° ) =
( 20∠30° )
− j 50 + 45 + j 80
45 + j 30
=
50∠ − 90
( 20∠30° ) = 18.49∠ − 93.7° mA
54.08∠ 33.7
8
P10.5-9 Here’s an ac circuit represented both in the time domain and frequency domain:
R1
+
–
15 cos 20t V
C
Z1
+
R2
v(t)
+
–
L
15 0° V
+
Z2
–
V(ω)
–
(b)
(a)
Z 1 = 15.3∠ − 24.1° Ω and Z 2 = 14.4∠53.1° Ω
Suppose
Determine the node voltage v(t) and the values of R1, R2, L and C.
Solution:
Consider Z1:
1
1
R1 − j
= 15.3∠ − 24.1° = 14 − j 6.25 ⇒ R1 = 14 Ω and C =
= 0.008 F = 8 mF
20 C
20 ( 6.25 )
Next consider Z2:
1
= 14.4∠53.1° ⇒
1
1
+
R 2 j 20 L
1
1
1
1
+
=
=
∠53.1 = 0.05556 − j 0.04167
R 2 j 20 L 14.4∠53.1° 14.4
Equating coefficients gives
R2 =
1
1
= 18 Ω and L =
= 1.2 H
0.05556
20 ( 0.04167 )
Next, consider the voltage divider:
A∠31.5° =
In the time domain,
(15)(14.4 ) ∠36.9°
14.4∠36.9°
(15∠0° ) =
15.3∠ − 24.1° + 14.4∠36.9°
(14 − j 6.25)(11.52 + j 8.64 )
=
216∠36.9°
25.52 + j 2.39
=
216∠36.9°
= 8.43∠31.5° V
25.63∠5.4°
v ( t ) = 8.43cos ( 20 t + 31.5° ) V .
9
P 10.5-10 Find Z and Y for the circuit of
1 μF
160 μH
36 Ω
Figure P 10.5-10 operating at 10 kHz.
Z,Y
Figure P 10.5-10
Solution:
ZR
ω = 2π f = 2π (10 ×103 ) = 62832 rad sec
1
1
= R = 36 Ω ⇔ YR =
=
= 0.0278 S
36
ZR
Z L = jω L = j (62830)(160 ×10−6 ) = j10.053 ≈ j10 Ω ⇔ YL =
ZC =
1
= − 0.1 j S
ZL
−j
−j
1
=
= − j15.915 ≈ − j 16 Ω ⇔ YC =
= j 0.0625 S
−6
ZC
ω C (62830)(1×10 )
Yeq = YR + YL + YC = 0.0278 − j0.0375 = 0.0467 ∠ − 53.4° S
Z eq =
1
= 21.43∠ 53.4° = 12.75 + j17.22 Ω
Yeq
10
300 Ω
P 10.5-11 For the circuit of Figure P 10.5-11, find the value
of C required so that Z = 590.7Ω when f = 1 MHz.
C
Answer: C = 0.27 nF
47 μ H
Z
Figure P 10.5-11
Solution:
Z L = j ω L = j (6.28 ×106 ) (47 × 10−6 ) = j 295 Ω
⎛ 1 ⎞
⎜
⎟( 300 + j 295 )
jω C ⎠
⎝
Z eq = Z c || ( Z R +Z L ) =
= 590.7 Ω
1
+300+ j 295
jω C
300+300 j
590.7 =
⇒ 590.7 − (590.7)(295 ω C ) + j (590.7)(300ω C ) = 300 + j 295
1+300 j ω C −300 ω C
(
Equating imaginary terms ω =2π f = 6.28×106 rad sec
)
(590.7) (300ω C ) = 295 ⇒ C = 0.27 nF
11
P 10.5-12 Determine the impedance Z for the circuit shown in Figure P 10.5-12.
2.5 H
2 mF
2 mF
100 Ω
2 mF
Z
1.5 H
Figure P 10.5-12
Solution:
Replace series and parallel capacitors by an
equivalent capacitor and series inductors by an
equivalent inductor:
Then
100
Z = jω 4 +
100 +
4
−j
1
(
jω 5 ×10−3
(
)
1
jω 5 ×10
−3
)
⎛ 200 ⎞
200
100 ⎜ − j
1+
−j
⎟
ω
⎝
⎠ = jω 4 +
ω ×
= jω 4 +
2
⎛ 200 ⎞
1− j
1+
100 + ⎜ − j
⎟
ω
ω ⎠
⎝
j
j
2
ω
2
ω
2
ω = jω 4 + 100 4 − j 2 ω = 400 + j ⎛ 4 ω − 200 ω ⎞
Z = jω 4 + 100 ω
⎜
⎟
4
4 +ω2
4 +ω2
4 +ω2 ⎠
⎝
1+ 2
2
ω
12
P 10.5-13 The big toy from the hit movie
Speaker
Big is a child’s musical fantasy come true—a
sidewalk-sized piano. Like a hopscotch grid,
20 Ω
this once-hot Christmas toy invites anyone v = 12 cos ω t V
+
–
C
i
3 mH
who passes to jump on, move about, and
make music. The developer of the “toy” piano
Figure P 10.5-13
used a tone synthesizer and stereo speakers as
shown in Figure P 10.5-13 (Gardner, 1988). Determine the current i(t) for a tone at 796 Hz when
C = 10 μF.
Solution:
j (2π ⋅ 796) (3 ⋅10 −3 ) = j15 Ω
12
= 0.48 ∠ − 37° A
20 + j15
i (t ) = 0.48 cos (2π ⋅ 796 t − 37°) A
I=
13
P 10.5-14 Determine i(t), v(t), and L for the circuit shown in Figure P 10.5-14.
Answer: i(t) = 1.34 cos (2t – 87°) A, v(t) = 7.29 cos (2t – 24°) V, and L = 4 H
i(t)
2 cos(3t – 15°) A
8Ω
L
Figure P 10.5-14
Solution:
Z1 = R = 8 Ω, Z 2 = j 3 L, I = B ∠ − 51.87° and I s = 2 ∠ − 15° A
8
I B ∠−51.87°
Z1
=
=
=
=
2 ∠−15°
Is
Z1 + Z 2 8+ j 3L
8 ∠0°
⎛ 3L ⎞
82 + (3L) 2 ∠ tan −1 ⎜ ⎟
⎝ 8 ⎠
Equate the magnitudes and the angles.
⎛ 3L ⎞
angles: + 36.87 = + tan −1 ⎜ ⎟ ⇒ L = 2 H
⎝ 8 ⎠
B
8
magnitudes:
=
⇒ B =1.6
2
2
64+ 9 L
14
P 10.5-15 Spinal cord injuries result in paralysis of the lower body and can cause loss of bladder
control. Numerous electrical devices have been proposed to replace the normal nerve pathway
stimulus for bladder control. Figure P 10.5-15 shows the model of a bladder control system where vs
= 20 cos ωt V and ω = 100 rad/s. Find the steady-state voltage across the 10-Ω load resistor.
Answer: v(t) = 10 2 cos (100t + 45°) V
1 mF
vs +
–
50 Ω
100 μF 10 Ω
Regular nerve
pathway load
Figure P 10.5-15
Solution:
⎛ 10 ⎞
V10 = Vs ⎜
⎟
⎝ 10− j10 ⎠
10
⎛
⎞
= 20∠0° ⎜
⎟
⎝ 10 2∠− 45° ⎠
= 10 2∠45°
v10 (t ) = 10 2 cos (100 t + 45°) V
15
P 10.5-16 There are 500 to 1000 deaths each year in
Person's body
the United States from electric shock. If a person makes a
i
good contact with his hands, the circuit can be
represented by Figure P 10.5-16, where vs = 160 cos ωt
V and ω = 2πf. Find the steady-state current i flowing
Source
vs +–
300 Ω
2 μF
100 mH
through the body when (a) f = 60 Hz and (b) f = 400 Hz.
Answer: (a)
(b)
i(t) = 0.53 cos (120πt + 5.9°)
i(t) = 0.625 cos (800πt + 59.9°) A
Figure P 10.5-16
Solution:
(a)
(b)
160 ∠0°
160 ∠0°
=
(− j1326) (300+ j 37.7) 303 ∠−5.9°
− j1326 + 300+ j 37.7
= 0.53 ∠5.9° A
i(t ) = 0.53cos (120π t +5.9°) A
I=
I=
160∠0°
160∠0°
=
(− j199)(300+ j 251) 256∠−59.9°
− j199 +300+ j 251
= 0.625∠59.9° A
i(t ) =0.625 cos (800π t +59.9°) A
16
P 10.5-17 Determine the steady-state voltage, v(t), and current, i(t), for each of the circuits shown
in Figure P 10.5-17.
– v(t)
+
4Ω
+
24 V –
40 Ω
10 Ω
i(t)
(a)
– v(t)
+
4H
24 cos (4t + 15°) V
+
–
40 Ω
10 mF
i(t)
(b)
Figure P 10.5-17
Solution:
(a)
v (t ) = −
i (t ) =
4
× 24 = −8 V
4 + ( 40 & 10 )
40
24
8
×
= = 1.6 A
40 + 10 4 + ( 40 & 10 ) 5
(b) Represent the circuit in the frequency domain using impedances and phasors.
17
V=−
(16∠ − 90° )( 24∠15° ) = 33.66∠ − 65° V
j16
× 24∠15° =
40 ( − j 25 )
j16 + ( 40 & − j 25 )
j16 +
40 − j 25
I=
so
and
40
×
40 − j 25
24∠15°
= 1.78∠57° A
40 ( − j 25 )
j16 +
40 − j 25
v ( t ) = 33.66 cos ( 4t − 65° ) V
i ( t ) = 1.78cos ( 4t + 57° ) A
(checked: LNAP 8/1/04)
18
P 10.5-18
Determine the steady-state current, i(t), for the circuit shown in Figure P 10.5-18.
5H
5 mF
20 Ω
2 mF
30 Ω
4H
i(t)
+–
5 cos (10t + 30°) V
Figure P 10.5-18
Solution:
I=
so
5∠30°
5∠30°
5∠30°
+
+
= 0.100∠ − 23.1° + 0.0923∠98.2° + 0.1667∠ − 60°
30 + j 40 20 − j 50 j 50 − j 20
= 0.186∠ − 29.5° A
i ( t ) = 0.186 cos (10t − 29.5° ) A
(checked: LNAP 8/1/04)
19
P10.5-19 Determine the steady state voltage, v(t), for this circuit:
20 Ω
4H
+
v(t)
10 cos(5t + 45°) mA
–
30 Ω
5 mF
2H
4 mF
Solution:
V = 0.01∠45° ⎡⎣( 20 & j 20 ) + ( 30 & ( − j 40 ) ) + ( j10 & ( − j 50 ) ) ⎤⎦
⎡ 20 ( j 20 ) 30 ( − j 40 ) j10 ( − j 50 ) ⎤
= 0.01∠45° ⎢
+
+
⎥
j10 − j 50 ⎦
⎣ 20 + j 20 30 − j 40
= 0.01∠45° [14.14∠45° + 24∠ − 36.9° + 12.5∠90°]
= 0.01∠45° [10 + j10 + 19.2 − j14.4 + j12.5]
= 0.303∠60.5° V
so
v ( t ) = 0.303cos ( 5t + 60.5° ) V
(checked: LNAP 8/1/04)
20
P10.5-20 Determine the steady state voltage, v(t), for this circuit:
2.5H
20 Ω 10 mF
+
–
Solution:
Let
and
10 cos(4t + 60°) V
5H
5 mF
+
v(t) 40 Ω
–
⎛
( 20 − j 25) j10 = 250 − j 200 = 12.81∠75.5° Ω
1 ⎞
Z 1 = ⎜⎜ 20 − j
⎟⎟ & j10 =
4 ( 0.01) ⎠
20 − j 25 + j10
20 − j15
⎝
⎛
⎞ j 20 ( 40 − j 50 ) 1000 + j800
1
Z 2 = j 20 & ⎜⎜ − j
+ 40 ⎟⎟ =
=
= 25.61∠75.5° Ω
40 − j 30
⎝ 4 ( 0.005 )
⎠ j 20 + 40 − j 50
Then
V=
so
Z2
Z1 + Z 2
× 10∠60° =
25.61∠75.5°
× 10∠60° = 6.67∠60° V
12.81∠75.5° + 25.6∠75.5°
v ( t ) = 6.67 cos ( 4t + 60° ) V
(checked: LNAP 8/1/04)
21
P 10.5-21 The input to the circuit shown in Figure P 10.5-21 is the current source current
is(t) = 25 cos (10t + 15°) mA
The output is the current i1(t). Determine the steady-state response, i1(t).
i1(t)
5H
is(t)
5 mF
40 Ω
2H
2 mF
25 Ω
Figure P 10.5-21
Solution:
Represent the circuit in the frequency domain using impedances and phasors. Let
40 ( − j 50 )
⎛
⎞
1
= j 50 +
= 39.0∠51.3° Ω
Z 1 = j 50 + ⎜ 40 &
−3 ⎟
40 − j 50
j10 × 2 × 10 ⎠
⎝
and
Z2 = − j
j 20 ( 25 )
1
+ j 20 & 25 = − j 20 +
= 12.5∠ − 38.7 Ω
−3
25 + j 20
10 ( 5 ×10 )
Z1 and Z2 are connected in parallel. Current division gives
I1 =
so
Z1
Z1 + Z 2
× 0.025∠15° = 0.024∠32.7° A
i1 ( t ) = 0.024 cos (10t + 32.7° ) A
(checked: LNAP 8/1/04)
22
P10.5-22 Determine the steady state voltage, v(t), and current i(t) for each of these circuits:
– v(t)
+
4Ω
+
24 V –
40 Ω
10 Ω
i(t)
(a)
– v(t)
+
4H
24 cos (4t + 15°) V
+
–
40 Ω
10 mF
i(t)
(b)
Solution:
i (t ) =
(a)
v (t ) =
80 + 80
0.024 = 19.2 mA
40 + ( 80 + 80 )
80
1
× ( 40 & ( 80 + 80 ) ) 0.024 = ( 32 )( 0.024 ) = 0.384 V
80 + 80
2
(b) Represent the circuit in the frequency domain using impedances and phasors.
I=
80 + j80
× 0.024∠15° = 0.028∠25.5° A
− j 25 + ( 80 + j80 )
V=
80
× ⎡ − j 25 & ( 80 + j80 ) ⎤⎦ × 0.024∠15° = 0.494∠ − 109.5° V
80 + j80 ⎣
So
i ( t ) = 28cos (10t + 25.5° ) mA
and
v ( t ) = 0.494 cos (10t − 109.5° ) V
(checked: LNAP 8/1/04)
23
P10.5-23 Determine the steady state current i(t) for this circuit:
25 Ω
2H
2 mF
20 Ω
+
–
40 Ω
i(t)
16 cos(20t + 75°) V
5 mF
2H
Solution:
Represent the circuit in the frequency domain using impedances and phasors. Let
Z 1 = 25 + j ( 20 ) 2 +
Z 2 = 20 +
1
= 25 + j15 = 29.2∠31° Ω
j ( 20 )( 0.002 )
1
= 20 − j10 = 22.36∠ − 26.6° Ω
j ( 20 )( 0.005 )
Z 3 = 40 + j ( 20 ) 2 = 40 + j 40 = 56.57∠45° Ω
and let
Z p = Z 2 & Z 3 = 18.86∠ − 8° = 18.67 − j 2.67 Ω
Then
I=
so
Z2
16∠75°
×
= 0.118∠6.1° A
Z1 + Z p Z 2 + Z 3
i ( t ) = 0.118cos ( 20t + 6.1° ) A
(checked: LNAP 8/2/04)
24
P 10.5-24 When the switch in the circuit shown in
Figure P 10.5-24 is open and the circuit is at steady
R2
R1
state, the capacitor voltage is
v(t) = 14.14 cos (100t –45°) V
When the switch is closed and the circuit is at steady
+
–
20 cos (100t) V
state, the capacitor voltage is
0.5 μF
+
v(t)
–
Figure P 10.5-24
v(t) = 17.89 cos (100t –26.6°) V
Determine the values of the resistances R1 and R2.
Solution:
Represent the circuit in the frequency domain using phasors and impedances. The impedance
1
capacitor is
= − j 20, 000 . When the switch is closed
j (100 ) ( 0.5 ×10−6 )
17.89∠ − 26.6° = V =
− j 20, 000
× 20∠0°
R 2 − j 20, 000
Equating angels gives
⎛ −20, 000 ⎞
−26.6° = −90° − tan −1 ⎜
⎜ R 2 ⎟⎟
⎝
⎠
When the switch is open
14.14∠ − 45° = V =
⇒
R2 =
−20, 000
= 10015 Ω
tan ( −63.4 )
− j 20, 000
× 20∠0°
R1 + R 2 − j 20, 000
Equating angles gives
⎛ −20, 000 ⎞
−45° = −90° − tan −1 ⎜
⎜ R1 + R 2 ⎟⎟
⎝
⎠
⇒
R1 + R 2 =
−20, 000
= 20, 000
tan ( −45° )
So
R1 = 20, 000 − 10015 = 9985 Ω
(checked: LNAP 8/2/04)
25
P10.5-25 Determine the steady state current i(t) for this circuit:
20 cos (5t + 30°) mA
4H
20 Ω
10 mF
15 Ω
2H
8H
40 Ω
8 mF
i(t)
Solution:
Represent the circuit in the frequency domain using phasors and impedances. Let
Z 1 = ( j 20 & 20 ) +
1
= 10 − j10 = 14.14∠ − 45° Ω
j 0.05
⎛
1 ⎞
Z 2 = j 40 + 40 + ⎜ j10 &
⎟ + 15 = 55 + j 56.67 = 79∠46.3° Ω
j 0.04 ⎠
⎝
Z1
I=−
× 20∠30° = 3.535∠129.3° mA
Z1 + Z 2
so
i ( t ) = 3.535cos ( 5t + 129.3° ) mA
(checked: LNAP 8/2/04)
26
P10.5-26 Determine the steady state voltage, v(t), and current i(t) for each of these circuits:
i(t)
+
–
50 Ω
4i(t)
20 V
40 Ω
+
v(t)
–
2H
10 Ω
(a)
i(t)
+
–
5 mF
40 Ω
3H
20 cos (10t + 15°) V
4i(t)
+
v(t)
–
(b)
Solution:
(a) Using KCL and then KVL gives
20
= 80 mA
250
v ( t ) = 40 ( 5i ( t ) ) = 200 ( 0.08 ) = 16 V
20 = 50 i ( t ) + 40 ( 5 i ( t ) ) ⇒ i ( t ) =
Then
(b) Represent the circuit in the frequency domain using phasors and impedances.
Where
And
Z 1 = 40 + j (10 ) 3 +
1
= 40 + j10 = 41.23∠26.6° Ω
j (10 )( 0.005 )
Z 2 = j (10 ) 2 & 10 = 8 + j 4 = 8.944∠26.6° Ω
Using KCL and then KVL gives
20∠15° = Z 1I + 5Z 2 I
Then
⇒
I = 0.234∠ − 5.6° A
V = Z 2 ( 5I ) = 10.47∠21° A
so
i ( t ) = 0.234 cos (10t − 5.6° ) A
and
v ( t ) = 10.47 cos (10t + 21° ) V
(checked: 8/3/04)
27
P10.5-27 Determine the steady state voltage, v(t), for each of these circuits:
20 Ω
80 Ω
24 V
+
+
–
v(t)
40 Ω
−
100 Ω
(a)
3H
5 mF
25 Ω
20 Ω
+
+
–
v(t)
–
24 cos (20t + 45°) V
4 mF
4H
20 Ω
2 mF
15 Ω
(b)
Solution:
(a) Using voltage division twice
v (t ) =
40
100
× 24 −
× 24 = −12 V
40 + 80
20 + 100
28
(b) Represent the circuit in the frequency domain using phasors and impedances.
Where
Z 1 = 20 Ω
⎛
⎞
1
& 20 ⎟⎟ = 12.2 + j 70.2 = 71.30∠80.2° Ω
Z 2 = j ( 20 ) 4 + ⎜⎜
⎝ j ( 20 )( 0.002 )
⎠
1
+ 25 = 25 + j 50 = 55.90∠63.4° Ω
Z 3 = j ( 20 ) 3 +
j ( 20 )( 0.005 )
Z4 =
1
+ 15 = 15 − j12.5 = 19.53∠ − 39.8° Ω
j ( 20 )( 0.004 )
Using voltage division twice
V=
so
Z2
Z1 + Z 2
× 24∠45° −
Z4
Z3 + Z4
× 24∠45° = 24.8∠80° V
v ( t ) = 24.8cos ( 20t + 80° ) V
(Checked using LNAP 10/5/04)
29
P 10.5-28 The input to the circuit shown in
Figure P 10.5-28 is the voltage of the voltage
source
vs(t) = 5 cos (2t + 45°) V
0.1 F
+
–
4Ω
vs(t)
The output is the inductor voltage, v(t).
Determine the steady-state output voltage.
3H
+
v(t)
–
Figure P 10.5-28
Solution:
Represent the circuit in the frequency domain using phasors and impedances.
4 & j6 =
4 ( j 6 ) 24∠90°
=
= 3.33∠34° = 2.76 + j1.86 Ω
4 + j 6 7.2∠56°
Using voltage division
V=
3.33∠34°
3.33∠34°
3.33∠34°
× 5∠45° =
× 5∠45° =
× 5∠45° = 3.98∠127° V
− j 5 + 2.76 + j1.86
2.76 − j 3.14
4.18∠ − 48°
The corresponding voltage in the time domain is
v ( t ) = 3.98cos ( 2t + 127° ) V
30
P10.5-29 Determine the steady state voltage, v(t), for this circuit:
+
8Ω
2H
5 sin 5t V
v(t)
+–
1 12
F
4H
_
Solution:
V1 (ω ) =
V 2 (ω ) =
j10
5 e − j 90 = 3.9 e − j 51 V
8 + j10
j 20
5 e − j 90 = 5.68 e − j 90 V
j 20 − j 2.4
V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90
= 3.58 e j 47 V
31
P10.5-30 Determine the steady state voltage, v(t), for this circuit:
+
8Ω
2H
v(t)
1 12
4 cos (3t + 15°) A
F
4H
–
Solution:
V1 (ω ) =
V 2 (ω ) =
8 ( j 6 ) j15
4 e = 19.2 e j 68 V
8 + j6
j12 ( − j 4 ) j15
4 e = 24 e − j 75 V
j12 − j 4
V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22 V
32
P10.5-31
The input to the circuit in Figure P10.5-31 is the voltage source voltage, v s ( t ) . The output is the
voltage v o ( t ) . When the input is v s ( t ) = 8cos ( 40 t ) V , the output is v o ( t ) = 2.5cos ( 40 t + 14° ) V .
Determine the values of the resistances R1 and R 2 .
Figure P10.5-31
Solution:
Using voltage division in the frequency domain:
jω L R 2
L
R 2 + jω L
R1
Vo (ω )
=
=
jω L R 2
L
Vi (ω )
1 + jω
R1 +
Rp
R 2 + jω L
jω
R1 R 2
where R p =
R1 + R 2
Representing the given input and output in the frequency domain:
2.5∠14° Vo (ω )
=
=
8∠0°
Vi (ω )
In this case the angle of
and the magnitude of
ω
L
R1
⎛
j ⎜ 90 − tan −1 ω
⎛ L ⎞
1+ ⎜ω
⎜ R p ⎟⎟
⎝
⎠
2
e
⎜
⎝
L ⎞⎟
R p ⎟⎠
Vo (ω )
L L ( R1 + R 2 ) tan ( 90° − 14° )
=
=
= 0.1
is specified to be 14° so
40
Rp
R1 R 2
Vi (ω )
Vo (ω )
2.5
is specified to be
so
8
Vi (ω )
40
L
R1
=
2.5
⇒
8
1 + 16
values that satisfies these two equations is L = 1 H, R1 = 31 Ω, R 2 = 14.76 Ω .
L
= 0.0322 . One set of
R1
33
Section 10.6 Mesh and Node Equations
P10.6‐1 The input to the circuit shown in Figure P10.6‐1 is the voltage
v s ( t ) = 48cos ( 2500 t + 45° ) V
Write and solve node equations to determine the steady state output voltage v o ( t ) .
Figure P10.6‐1
Solution: Represent the circuit in the frequency domain as
The node voltages are 48∠45° = V1 , V2, V3 and Vo. Express the dependent source voltage in
terms of the node voltages:
⎛ V2 ⎞
V3 − V 2 = 25 I = 25 ⎜
⎟ ⇒ V3 = 2.25 V 2
⎝ 20 ⎠
Apply KCL to the supernode corresponding to the CCVS to get
48∠45° − V 2
j 50
=
V2
20
+
V3 − V o
30
+
V3
− j 40
V3
48∠45° V 2 V 2 V3 − Vo
=
+
+
+
j 50
j 50 20
30
− j 40
⎛ 1
48∠45° ⎛ 1
1 ⎞
1 ⎞
1
=⎜
+ ⎟ V2 + ⎜ +
⎟ V3 − V o
j 50
30
⎝ j 50 20 ⎠
⎝ 30 − j 40 ⎠
⎛ 1
48∠45° ⎛ 1
1 ⎞
1 ⎞
1
=⎜
+ ⎟ V2 + ⎜ +
⎟ 2.25V 2 − Vo
j 50
30
⎝ j 50 20 ⎠
⎝ 30 − j 40 ⎠
48∠45° ⎛ 1
1 2.25 2.25 ⎞
1
=⎜
+ +
+
⎟ V 2 − Vo
j 50
30
⎝ j 50 20 30 − j 40 ⎠
Apply KCL at the right node of the 30 Ω resistor to get
V3 − V o
30
=
⎛ 1
1 ⎞
⎛ 1 ⎞
⇒ 0 = ⎜ − ⎟ 2.25 V 2 + ⎜ +
⎟ Vo
j 25
⎝ 30 ⎠
⎝ 30 j 25 ⎠
Vo
1 2.25 2.25
⎡ 1
⎢ j 50 + 20 + 30 + − j 40
⎢
In matrix form
2.25
⎢
−
⎢
30
⎣
Solving, perhaps using MATLAB,
1 ⎤
⎡ 48∠45° ⎤
30 ⎥ ⎡ V 2 ⎤ ⎢
⎥⎢ ⎥ =
j 50 ⎥
⎢
⎥
1
1 ⎥ ⎣ Vo ⎦
+
0
⎢
⎥⎦
⎣
30 j 25 ⎥⎦
−
⎡ V 2 ⎤ ⎡10.18∠ − 44.6°⎤
⎢ ⎥=⎢
⎥ V
⎣⎢ Vo ⎦⎥ ⎣ 14.67∠5.6° ⎦
P10.6‐2 Figure P10.6‐2 shows an ac circuit represented in the frequency domain. Determine the
values of the phasor node voltages, Vb and Vc.
Figure P10.6‐2
Solution:
12∠45° − V b
Writing Node equations:
=
Vb
+
Vb − Vc
20 − j 25 15 − j 30
j 30
V b − Vc 12∠45° − Vc
Vc
+
=
15 − j 30
40 + j 20
j 40
Rearranging:
⎞
⎛
⎞
12∠45° ⎛ 1
1
1
1
=⎜
+
+
⎟ Vb − ⎜
⎟ Vc
j 30
⎝ j 30 20 − j 25 15 − j 30 ⎠
⎝ 15 − j 30 ⎠
⎛
⎞
⎛
12∠45°
1
1
1
1 ⎞
= −⎜
+
+
⎟ Vb + ⎜
⎟ Vc
40 + j 20
⎝ 15 − j 30 ⎠
⎝ 15 − j 30 40 + j 20 j 40 ⎠
In matrix from:
1
1
1
⎡ 1
⎤
⎡ 12∠45° ⎤
+
+
−
⎢ j 30 20 − j 25 15 − j 30
⎥ ⎡ V b ⎤ ⎢ j 30 ⎥
15 − j 30
⎢
⎥⎢ ⎥ = ⎢
⎥
1
1
1
1 ⎥ ⎣⎢ Vc ⎦⎥ ⎢ 12∠45° ⎥
⎢
−
+
+
⎢
⎢ 40 + j 20 ⎥
15 − j 30
15 − j 30 40 + j 20 j 40 ⎥⎦
⎣
⎣
⎦
Solving using MATLAB:
V b = 7.69∠ − 19.8° and Vc = 10.18∠7.7° V
Checked using LNAPAC
P10.6‐3 Figure P10.6‐3 shows an ac circuit represented in the frequency domain. Determine the
value of the phasor node voltage V.
Answer: V = 71.0346∠ − 39.627° V
Figure P10.6‐3
Solution:
Writing a node equation:
Rearranging
Solving
Finally
20∠30° − V −20∠30° − V
+
= 0.25∠15°
j 40
25 − j 50
⎛ 1
⎞
1
20∠30° −20∠30°
+
+
− 0.25∠15°
⎜
⎟V =
25 − j 50
j 40
⎝ j 40 25 − j 50 ⎠
( 0.012042∠ − 48.366° ) V = 0.85537∠87.993°
V=
0.85537∠87.993°
= 71.0346∠ − 39.627° V
0.012042∠ − 48.366°
P10.6‐4 Figure P10.6‐4 shows an ac circuit represented in the frequency domain. Determine the
values of the phasor mesh currents.
Figure P10.6‐4
Solution:
( 40 + j 15) I 1 + ( 25 − j 50 ) ( I 1 − I 3 ) − 48∠75° = 0
( 65 − j 35 ) I 1 − ( 25 − j 50 ) I 3 = 48∠75°
Mesh 1:
48∠75° + ( − j 50 ) ( I 2 − I 3 ) + ( 32 + j 16 ) I 2 = 0
Mesh 2:
( 32 − j 34 ) I 2 + j 50I 3 = −48∠75°
j 40I 3 − ( − j 50 ) ( I 2 − I 3 ) − ( 25 − j 50 ) ( I 1 − I 3 ) = 0
Mesh 3:
In matrix form:
( −25 + j 50 ) I 1 + j 50 I 2 + ( 25 − j 160 ) I 3 = 0
0
−25 + j 50 ⎤ ⎡ I 1 ⎤ ⎡ 48∠75° ⎤
⎡ 65 − j 35
⎢ ⎥
⎢
+ j 50 ⎥⎥ ⎢I 2 ⎥ = ⎢⎢ −48∠75° ⎥⎥
0
32 − j 34
⎢
⎢⎣ −25 + j 50
⎥⎦
+ j 50
25 − j 60 ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣
0
Solving using MATLAB:
I 1 = 0.794∠111°, I 2 = 0.790∠ − 61.7° and I 3 = 0.229∠176° A
P 10.6-5 A commercial airliner has sensing
devices to indicate to the cockpit crew that each
R
door and baggage hatch is closed. A device called
vs
a search coil magnetometer, also known as a
b
+
–
proximity sensor, provides a signal indicative of
R
a
LR
Door
Ls
the proximity of metal or other conducting
material to an inductive sense coil. The inductance
of the sense coil changes as the metal gets closer to
Figure P 10.6-5
the sense coil. The sense coil inductance is compared to a reference coil inductance with a circuit
called a balanced inductance bridge (see Figure P 10.6-5). In the inductance bridge, a signal
indicative of proximity is observed between terminals a and b by subtracting the voltage at b, vb,
from the voltage at a, va (Lenz, 1990).
The bridge circuit is excited by a sinusoidal voltage source vs = sin (800 πt) V. The two
resistors, R = 100 ω, are of equal resistance. When the door is open (no metal is present), the
sense coil inductance, LS, is equal to the reference coil inductance, LR = 40 mH. In this case,
what is the magnitude of the signal Va – Vb?
When the airliner door is completely closed, LS = 60 mH. With the door closed, what is
the phasor representation of the signal Va – Vb?
Solution:
vs = sin (2π ⋅ 400 t ) V
R = 100 Ω
LR = 40 mH
⎧ 40 mH
LS = ⎨
⎩ 60 mH
door opened
door closed
With the door open VA − VB = 0 since the bridge circuit is balanced.
With the door closed Z LR = j (800π )(0.04) = j100.5 Ω and Z LS = j (800π )(0.06) = j150.8 Ω.
The node equations are:
KCL at node B:
KCL at node A :
Since VC = Vs =1 V
VB − VC VB
j100.5
+
= 0 ⇒ VB =
VC
Z LR
R
j100.5+100
VA − VC VA
+
=0
R
Z LS
VB =0.709∠44.86° V and VA = 0.833∠33.55 V
Therefore
VA − VB = 0.833∠33.55° − 0.709∠44.86° = (0.694 + j.460) − (0.503 + j 0.500) = 0.191 − j 0.040
= 0.195∠ − 11.83° V
P 10.6-6 Using a tiny diamond-studded burr operating at 190,000 rpm, cardiologists can
remove life-threatening plaque deposits is coronary arteries. The procedure is fast,
uncomplicated, and relatively painless (McCarty, 1991). The Rotablator, an angioplasty system,
consists of an advancer/catheter, a guide wire, a console, and a power source. The
advancer/catheter contains a tiny turbine that drives the flexible shaft that rotates the catheter
burr. The model of the operational and control circuit is shown in Figure P 10.6-6. Determine
v(t), the voltage that drives the tip, when vs =
2 cos (40t – 135°) V.
Answer: v(t) = 1.414 cos (40t + 135°) V
1 20
vs
+
–
1 80
H
F
1 80
F
+
2 Ω
1 80
2i
i
F
–
v(t)
Figure P 10.6-6
Solution:
Represent the circuit in the frequency domain
V1 − (−1+ j ) V1 V1 − V2
+ +
=0
−j2
j2
2
V2 − V1
V2
+
− IC = 0
− j2
− j2
Also, expressing the controlling signal of the dependent source in terms of the node voltages
⎡ −1+ j ⎤
−1 + j
yields
⇒ IC = 2 Ix = 2 ⎢
Ix =
⎥ = −1 − j A
-2 j
⎣ -2 j ⎦
Solving these equations yields
The node equations are:
V2 =
−3− j
= 2 ∠ − 135° V ⇒ v(t ) = v2 (t ) = 2 cos (40 t − 135°) V
1+ j 2
(checked: LNAP 7/19/04
Figure P 10.6-7, it is known that
1H
v2(t) = 0.7571 cos (2t + 66.7°) V
v3(t) = 0.6064 cos (2t–69.8°) V
i1
5Ω
P 10.6-7 For the circuit of
A cos 2t V
+
–
1F
5H
+
v2
–
3i1
1 4
F
+
v3
–
Determine i1(t).
Figure P 10.6-7
Solution:
V2 = 0.7571∠66.7° V
V3 = 0.6064∠ − 69.8° V
⎫
I1 = I 2 + I 3 ⎪
⎪
⎧ I 3 =0.3032 ∠20.2° A
V3 − V2 ⎪
⎪
I2 =
⎬ yields ⎨ I 2 =0.1267∠−184° A
j 10 ⎪
⎪ I =0.195∠36° A
⎩ 1
⎪
V3
I3 =
⎪
−j2 ⎭
therefore
i1 (t ) =0.195cos (2 t +36°) A
(checked: MATLAB 7/18/04)
P10.6‐8 The input to the circuit shown in Figure P10.6‐8 is the voltage
vs = 25cos(40t + 45°) V
Determine the mesh currents i1 and i2 and the voltage vo.
Figure P10.6‐8
Solution
Represent the circuit in the frequency domain:
Apply KVL to mesh 1:
j 320 I 1 + 400 ( I 1 − I 2 ) − 25∠45° = 0
Apply KVL to mesh 2:
50 I 2 + j 240 I 2 + j 120 I 2 − 400 ( I 1 − I 2 ) = 0
In matrix form:
−400 ⎤ ⎡ I 1 ⎤ ⎡ 25∠45° ⎤
⎡ 400 + j 320
⎢ ⎥=
⎢ −400
450 + j 360 ⎥⎦ ⎣ I 2 ⎦ ⎢⎣ 0 ⎥⎦
⎣
Solving using MATLAB:
Using Ohm’s Law
⎡ I 1 ⎤ ⎡ 47.5∠ − 24.6°⎤
⎢I ⎥ = ⎢
⎥ mA
⎣ 2 ⎦ ⎣ 33.0∠ − 63.3° ⎦
Vo = 400 ( I 1 − I 2 ) = 12∠18.8° V
In the time domain i1 ( t ) = 47.5cos ( 40 t − 24.6° ) mA , i 2 ( t ) = 33cos ( 40 t − 63.3° ) mA
and
v o ( t ) = 12 cos ( 40 t + 18.8° ) V
P10.6‐9 The input to the circuit shown in Figure P10.6‐9 is the voltage
vs(t) = 42cos(800t + 60°) mV
Determine the output voltage vo(t).
Answer: vo(t) = 823.5 cos(800t − 55.6°) mV
Figure P10.6‐9
Solution
Represent the circuit in the frequency domain:
Apply KCL at the top node of the inductor, node a:
0.042∠60° − Va
=
Va
+
Va
⇒ Va =
4
( 0.042∠60° )
5− j
500
j 2000 2000
Apply KCL at the inverting input node of the op amp:
Vb
Va
+
= 0 ⇒ V b = −5 V a
2000 10, 000
Apply KCL at the top node of the capacitor, node b:
Vb
Vb
V b − Vo
+
+
= 0 ⇒ Vo = ( 3 + j 4 ) V b
10, 000 − j 5000 20, 000
Combining these results we get:
( 5∠53.1° )( 20∠ − 180° ) 0.042∠60° = 0.8235∠ − 55.6°
4
V o = ( 3 + j 4 )( −5 )
( 0.042∠60° ) =
(
)
5− j
5.1∠ − 11.3°
In the time domain
v o ( t ) = 832.5cos ( 800 t − 55.6° ) mV
P 10.6-10 The idea of using an induction coil in a
C
lamp isn’t new, but applying it in a commercially
available product is. An induction coil in a bulb
1Ω
1Ω
induces a high-frequency energy flow in mercury
+
1Ω
L
vapor to produce light. The lamp uses about the same
2Ω
vs +–
amount of energy as a fluorescent bulb but lasts six
times longer, with 60 times the life of a conventional
incandescent bulb. The circuit model of the bulb and
v
–
Induction bulb
Figure P 10.6-10
its associated circuit are shown in Figure P 10.6-10.
Determine the voltage v(t) across the 2-Ω resistor when C = 40 μF, L = 40 μH, vs = 10 cos (ω0t +
30°), and ω0 = 105 rad/s.
Answer: v(t) = 6.45 cos (105t + 44°) V
Solution:
Represent the circuit in the frequency domain:
The mesh equations are:
−1
− j 4 ⎤ ⎡ I1 ⎤ ⎡10∠30°⎤
⎡ (2 + j 4)
⎢ −1
−1 ⎥⎥ ⎢⎢I 2 ⎥⎥ = ⎢⎢ 0 ⎥⎥
(2+1/ j 4)
⎢
⎢⎣ − j 4
−1
(3+ j 4) ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ 0 ⎥⎦
Using Cramer’s rule yields
I3 =
Then
2 + j8
(10∠30° ) = 3.225∠44° A
12+ j 22.5
V = 2 I 3 = 2 ( 3.225∠44° ) = 6.45∠44° V ⇒ v(t ) = 6.45cos (105 t + 44° ) V
(checked: LNAP 7/19/04)
P 10.6-11 The development of coastal hotels in various parts of the world is a rapidly growing
enterprise. The need for environmentally acceptable shark protection is manifest where these
developments take place alongside shark-infested waters (Smith, 1991). One concept is to use an
electrified line submerged in the water in order to deter the sharks, as shown in Figure P 10.611a. The circuit model of the electric fence is shown in Figure P 10.6-11b, where the shark is
represented by an equivalent resistance of 100Ω. Determine the current flowing through the
shark’s body, i(t), when vs = 375 cos 400t V.
Electric
fence
(a)
100 μ F
25 μF
i
vs
+
–
100 Ω
250 mH
Source
Electric fence
Shark
(b)
Figure P 10.6-11
Solution: Represent the circuit in the frequency domain:
Mesh Equations:
j 75 I1 − j 100 I 2 = 375
− j 100 I1 + (100+ j 100) I 2 = 0
Solving for I 2 yields I 2 = 4.5 + j 1.5 = 3 ∠53.1° A
⇒
i 2 ( t ) 3cos ( 400t + 53.1° ) A
(checked: LNAP 7/19/04)
P 10.6‐12 Determine the node voltage at nodes a and b in each of these circuits:
Solution
(a)
24 − v a
40
24 − v b
The node equations are
25
or
1 1
⎡1
+
⎢ 40 20 + 15
⎢
1
⎢
−
⎢⎣
20
Solving using MATLAB gives
=
+
va − vb
20
va − vb
20
+
=
va
15
vb
50
1
⎤ ⎡ v ⎤ ⎡ 24 ⎤
⎥ ⎢ a ⎥ ⎢ 40 ⎥
20
=⎢ ⎥
⎥
1
1
1 ⎥ ⎢ ⎥ ⎢ 24 ⎥
⎢v ⎥
+ +
25 20 50 ⎥⎦ ⎣ b ⎦ ⎢⎣ 25 ⎥⎦
−
v a = 8.713 V and v b = 12.69 V
(b) Use phasors and impedances to represent the circuit in the frequency domain as
where
Z 1 = 25 + j ( 20 ) 4 = 25 + j80 = 83.82∠72.7° Ω
⎛
⎞
1
Z 2 = ⎜⎜ 40 &
⎟ + j ( 20 ) 5 = 3.56 + j88.6 = 88.68∠87.7° Ω
j ( 20 )( 0.004 ) ⎟⎠
⎝
Z 3 = 20 Ω
Z 4 = 15 + j ( 20 ) 2 = 15 + j 40 = 42.72∠69.4°
Z 5 = j ( 20 ) 3 +
1
= j 50 = 50∠90° Ω
j ( 20 )( 0.005 )
24∠45° − Va
The node equations are
Z2
24∠45° − V b
Z1
1
1
⎡ 1
⎢Z + Z + Z
3
4
⎢ 2
⎢
1
−
⎢
Z3
⎢⎣
Solving using MATLAB gives
So
=
+
Va
Z4
+
Va − V b
Va − Vb
Z3
Z3
=
Vb
Z5
⎤
⎡ 24∠45° ⎤
⎥ ⎡ Va ⎤ ⎢ Z
⎥
2
⎥⎢ ⎥ = ⎢
⎥
1
1
1 ⎥ ⎢ ⎥ ⎢ 24∠45° ⎥
+
+
⎥ ⎢V ⎥ ⎢
⎥
Z 1 Z 3 Z 5 ⎥⎦ ⎣ b ⎦ ⎢⎣ Z 1 ⎥⎦
−
1
Z3
Va = 7.89∠44.0°
V b = 8.45∠45.1°
v a ( t ) = 7.89 cos ( 20t + 44° ) V
v a ( t ) = 8.45cos ( 20t + 45.1) V
(checked: LNAP 8/3/04)
P 10.6‐13 Determine the voltage v(t):
Solution:
Represent the circuit in the frequency domain using impedances and phasors
The mesh currents are I and 0.05∠ − 30° A . Apply KVL to the top mesh to get
15∠45° + ( − j 25 ) I + (15 + j 32 )( I − 0.05∠ − 30° ) + 25I = 0
So
Then
So
I=
−15∠45° + (15 + j 32 )( 0.05∠ − 30° )
= 0.3266∠ − 143.6° = −0.2629 − j 0.1939 A
25 − j 25 + 15 + j 32
V = ( − j 25 ) I = 8.166∠126.4° = −4.8475 + j 6.5715 V
v ( t ) = 8.166 cos ( 8t + 126.4° ) V
(checked: LNAP 8/3/04)
P 10.6‐14 Determine the voltage vo(t) when vs(t) = 25 cos (100t‐15°) V.
Solution:
Represent the circuit in the frequency domain using impedances and phasors.
The mesh currents are I and 10I. Apply KVL to the supermesh corresponding to the dependant
current source to get
( j500 ) I + ( − j5)(10I ) + 40 (10I ) − 25∠ − 15° = 0
So
I=
25∠ − 15°
= 0.04152∠ − 63.37° A
400 + j 450
The output voltage is
V = 40 (10I ) = 16.61∠ − 63.37° V
So
v ( t ) = 16.61cos (100t − 63.37° ) V
(checked: LNAP 8/3/04)
P 10.6‐15 Determine the mesh current is when i(t) = 0.8394 cos (10t‐138.5°) A. Determine the
values of L and R.
Solution:
Represent the circuit in the frequency domain using phasors and impedances. Apply KVL to the
center mesh to get
0.8394∠138.5° = I =
So
8∠210° − 30∠ − 15°
R + j10 L
R = 35 Ω
⇒
R + j10 L = 35 + j 25 = 35 + j (10 ) 2.5
and L = 2.5 H
(checked: LNAP 8/3/04)
P 10.6-16 The circuit shown in Figure P 10.6-16
has two inputs:
50 Ω
5 mF
v1(t) = 50 cos (20t – 75°) V
6H
40 Ω
2H
v2(t) = 35 cos (20t + 110°) V
When the circuit is at steady state, the node voltage
is
+
–
v1(t)
R
+
v(t)
–
L
+
v2(t) –
v(t) = 21.25 cos (20t – 168.8°) V
Determine the values of R and L.
Figure P 10.6-16
Solution:
Represent the circuit in the frequency domain using phasors and impedances. Apply KCL at the
top node of R and L to get
( 50∠ − 75° ) − V + 35∠100° − V =
j 40
40
⇒
V
R & jω L
50∠ − 75° 35∠110° ⎛ 1
1 1
1 ⎞
+
=⎜
+
+ −j
⎟V
40∠90°
40
20 L ⎠
⎝ j 40 40 R
Using the given equation for v(t) we get
21.25∠ − 168.8° = V =
Then
Finally
1.587∠161.7°
1
1
0.025 (1 − j ) + − j
R
20 L
1
1
1.587∠161.7°
−j
=
− 0.025 (1 − j ) = 0.04 − j 0.01176
R
20 L 21.25∠ − 168.8°
R=
1
1
= 25 Ω and L =
= 4.25 H
0.04
20 ( 0.01176 )
(checked: LNAP 8/3/04)
P 10.6‐17 Determine the steady state current i(t):
Solution: Represent the circuit in the frequency domain using phasors and impedances.
The node equations are
50∠0° − Va
15
50∠0° − V b
− j 20
or
1
1
⎡1
⎢15 + j100 + 25
⎢
1
⎢
−
⎢
j100
⎣
+
=
Vb − Va
j100
V b − Va
j100
=
+
Va
25
Vb
j 50
1
⎤
⎡ Va ⎤ ⎡ 50∠0° ⎤
⎥
⎢
⎥
j100
⎥ ⎢ ⎥ = ⎢ 15 ⎥
1
1
1 ⎥ ⎢ ⎥ ⎢ 50∠0° ⎥
⎢V ⎥
+
+
j 50 j100 − j 20 ⎥⎦ ⎣ b ⎦ ⎢⎣ − j 20 ⎥⎦
−
⎡ 0.1067 − j 0.010
⎢
j 0.010
⎣
j 0.010 ⎤ ⎡ Va ⎤ ⎡3.333⎤
⎢ ⎥=
j 0.020 ⎥⎦ ⎣ V b ⎦ ⎢⎣ j 2.5 ⎥⎦
Solving, e.g. using MATLAB, gives
Va = 33.05∠ − 12.6° V and V b = 108.9∠1.9° V
Then
So
I=
Va
= 1.322∠ − 12.6° A
25
i ( t ) = 1.322 cos ( 25t − 12.6° ) A
(checked: LNAP 8/3/04)
P 10.6‐18 Determine the steady state current i(t):
Solution:
Represent the circuit in the frequency domain using phasors and impedances. Label the node
voltages.
The node equations are
24∠15° − Va
25
24∠15° − V b
− j 6.25
or
Then
So
Va
+
Va − Vb
j 40
10
Va − Vb Vb
+
=
10
45
1
⎤ ⎡ V ⎤ ⎡ 24∠15° ⎤
⎥⎢ a⎥ ⎢
⎥
10
25
⎥⎢ ⎥ = ⎢
⎥
1
1 1⎥
24∠15° ⎥
⎢
⎢V ⎥
+ +
j
6.25 45 10 ⎥⎦ ⎣ b ⎦ ⎢⎣ 6.25∠ − 90° ⎥⎦
1 1
⎡1
⎢ 25 − j 40 + 10
⎢
1
⎢
−
⎢⎣
10
−
⎡ 0.140 − j 0.025
⎢
−0.10
⎣
Solving gives
=
− 0.10
⎤ ⎡ Va ⎤ ⎡ 0.960∠15° ⎤
⎢ ⎥=
0.1222 + j 0.160 ⎥⎦ ⎣ V b ⎦ ⎢⎣3.840∠105°⎥⎦
Va = 24.67∠32.6° V and V b = 25.59∠25.2° V
I=
Va − Vb
10
= 0.3347∠134.9° A
i ( t ) = 0.3347 cos (10t + 134.9° ) A
(checked: LANP 8/4/04)
P 10.6‐19 Determine the steady state voltage vo(t):
Solution:
Represent the circuit in the frequency domain using phasors and impedances.
20∠0° − V V V − 5V
=
+
j 40
25 − j 20
5V − Vo
Vo
=
10
− j10
The node equations are
⎡1
⎢ 25 −
⎢
⎢
⎢⎣
1
1
j −j
5
40
1
−
2
⎡ 0.04 − j 0.225
⎢
−0.50
⎣
Solving gives
So
⎤ ⎡ V ⎤ ⎡ − j 0.5⎤
⎥⎢ ⎥ ⎢
⎥
⎥⎢ ⎥ = ⎢
⎥
1
1
+ j ⎥ ⎢⎣ Vo ⎥⎦ ⎢⎣ 0 ⎥⎦
10
10 ⎥⎦
0
⎤ ⎡ V ⎤ ⎡ − j 0.5⎤
⎢ ⎥=
0.10 + j 0.10 ⎥⎦ ⎣ Vo ⎦ ⎢⎣ 0 ⎥⎦
0
V = 2.188∠ − 10.1° V and Vo = 7.736∠ − 55.1° V
v o ( t ) = 7.736 cos ( 5t − 55.1° ) V
(checked: LNAP 8/4/04)
P 10.6‐20 Determine the steady state current i(t) in each of these circuits:
4 i(t)
+ –
20 Ω
36 mA
i(t)
8Ω
(a)
4 i(t)
+ –
20 Ω
2H
i(t)
36 cos (25t) mA
2 mF
15 Ω
4 mF
(b)
Solution:
(a) Use KVL to see that the voltage across the 8 Ω resistor is 20i ( t ) − 4i ( t ) = 16i ( t ) .
Apply KCL to the supernode corresponding to the dependent voltage source to get
16i ( t )
0.036 = i ( t ) +
= 3i ( t )
8
so
i ( t ) = 12 mA
(b) Represent the circuit in the frequency domain using phasors and impedances.
Z 1 = 20 +
where
1
= 20 − j 20 Ω
j ( 25 )( 0.002 )
⎛
⎞
1
Z 2 = j 50 + ⎜⎜ 15 &
⎟⎟ = 43.3∠83.9° Ω
j
25
0.004
(
)(
)
⎝
⎠
V = Z 1I − 4I = ( Z 1 − 4 ) I
Use KVL to get
Then apply KCL to the supernode corresponding to the dependent source to get
0.036∠0° = I +
so
so
I=
(Z
1
− 4) I
Z2
Z 2 ( 0.036∠0° )
Z1 + Z 2 − 4
⎛ Z1 + Z 2 − 4 ⎞
=⎜
⎟⎟ I
⎜
Z
2
⎝
⎠
= 50.4∠35.7° mA
i ( t ) = 50.4 cos ( 25t + 35.7° ) mA
(checked: LNAP 8/4/04)
10.6‐21 The input to the circuit show in Figure
10.9‐24 is the current
i s ( t ) = 50 cos ( 200 t ) mA
Determine the steady‐state mesh current i 2 .
Figure P10.6‐21
Solution:
Represent the circuit in the
frequency domain using phasors
and impedances:
Apply KVL to the right mesh to
get:
( j 20 − j 62.5) I 2 + 60 ( I 2 − 0.050∠0° ) = 0
In the time domain
⇒ I2 =
3∠0°
= 0.0408∠35.3° A
60 − j 42.5
i 2 ( t ) = 40.8cos ( 200 t + 35.3° ) mA
MATLAB, 11/20/09
10.6‐22 The input to the circuit show in Figure
10.9‐24 is the current
i s ( t ) = 80 cos ( 250 t ) mA
The steady‐state mesh current in the right
mesh is
i s ( t ) = 66.56 cos ( 250 t + 33.7° ) mA
Figure P10.6‐22
Determine the value of the resistance R.
Solution:
Represent the circuit in the
frequency domain using phasors
and impedances. The mesh
currents are
I 1 = 0.080∠0° A
and
I 2 = 0.06656∠33.7° A
Apply KVL to the right to get
( j 150 − j 200 )( 0.06656∠33.7° ) + R ( 0.06656∠33.7° − 0.080∠0° ) = 0
( − j 50 )( 0.06656∠33.7° ) + R ( 0.044376∠123.7° ) = 0
R=
( 50∠90° )( 0.06656∠33.7° ) = 74.9955 75 Ω
0.044376∠123.7°
MATLAB, 11/20/09
P10.6-23
This circuit shown in Figure P10.6-23 is at steady state. The voltage source voltages are given by
v1(t) = 12 cos (2t – 90°) V and v2(t) = 5 cos (2t + 90°) V
The currents are given by
i1(t) = 744 cos (2t – 118°) mA , i2(t) = 540.5 cos (2t + 100°) mA
Determine the values of R1 , R 2 , L and C.
Figure P10.6-23
Solution: Represent the circuit in the frequency domain using impedances and phasors:
I = I 1 + I 2 = 0.744∠ − 118° + 0.5405∠100 = ( −0.349 − j 0.657 ) + ( −0.094 + j 0.532 )
= ( −0.349 − 0.094 ) + j ( −0.657 + 0.532 )
= −0.443 − j 0.125
= 0.460∠ − 164°
In the time domain
i ( t ) = 460 cos (2 t − 164°) mA
Replacing series impedances by equivalent impedances gives
Z 1 = R1 + j ω L
and
Z 2 = R2 − j
From KVL
1
ωC
Z 1 I 1 + 10 I − V1 = 0 ⇒ Z 1 =
V1 − 10 I
I1
=
12∠ − 90° − 10 ( 0.460∠ − 164° )
0.744∠ − 118°
− j 12 − 10 ( −0.443 − j 0.125 )
0.744∠ − 118°
4.43 − j 10.75 11.63∠ − 67.6°
=
=
0.744∠ − 118° 0.744∠ − 118°
= 15.63∠50.4°
=
= 10 + j 12 Ω
and
− Z 2 I 2 + V 2 − 10 I = 0 ⇒ Z 2 =
V 2 − 10 I
I2
=
=
5∠90° − 10 ( 0.460∠ − 164° )
0.5405∠100°
j 5 − 10 ( −0.443 − j 0.125 )
0.5405∠100°
4.43 + j 6.25
7.66∠54.7°
=
=
0.5405∠100° 0.5405∠100°
= 14.14∠ − 55.3°
= 10 − j 10 Ω
Next
and
12
=6H
2
1
⇒ R 2 = 10 Ω and C =
= 0.05 F
2 (10 )
10 + j 12 = R1 + j ω L = R1 + j 2 L ⇒ R1 = 10 Ω and L =
10 − j 10 = R 2 − j
1
1
= R2 − j
2C
ωC
10.7 Thevenin and Norton Equivalent Circuits
P10.7‐1 Determine the Thevenin equivalent circuit of the circuit shown in Figure P10.7‐1 when
(a) ω = 1000 rad/s, (b) ω = 2000 rad/s and (c) ω = 4000 rad/s.
Figure P10.7‐1
Solution: Represent the circuit in the frequency domain as
Determine the open circuit voltage and Thevenin impedance:
Voc =
ZC
2500 + Z C
(12∠0° )
Zt =
2500 Z C
2500 + Z C
+ ZL
a. Voc = 2.3534∠ − 78.69° V and Z t = 775.22∠82.875° Ω
b. Voc = 1.194∠ − 84.29° V and Z t = 2252.6∠89.37° Ω
c. Voc = 0.59925∠ − 87.14° V and Z t = 4875.3∠89.93° Ω
The Thevenin Equivalent Circuit changes whenever the input frequency changes.
P10.7‐2 Determine the Thevenin equivalent of this circuit when vs(t) = 5 cos (4000t‐30°) V.
1 80
mF
20 mH
vs
+
–
80 Ω
Solution:
Find Voc :
⎛
80 + j80 ⎞
Voc = ( 5 ∠−30° ) ⎜
⎟
⎝ 80 + j80− j 20 ⎠
⎛ 80 2∠− 45° ⎞
= ( 5 ∠−30° ) ⎜
⎟
⎝ 100∠36.90° ⎠
= 4 2∠ − 21.9° V
Find Z t :
Zt =
The Thevenin equivalent is
( − j 20 )( 80 + j80 ) = 23 ∠ − 81.9°
− j 20 + 80 + j80
Ω
P10.7‐3 Determine the Thevenin equivalent of this circuit
2v
600 Ω
+
9 cos 500t –
v
+
–
+
–
1 150
mF
300 Ω
Solution:
First, determine Voc :
The mesh equations are
600 I1 − j 300 (I1 − I 2 ) = 9 ⇒ (600 − j 300) I1 + j 300 I 2 = 9∠0°
−2 V + 300 I 2 − j 300 (I1 − I 2 ) = 0 and V = j 300 (I1 − I 2 ) ⇒
j 3 I1 + (1 − j 3) I 2 = 0
I 2 = 0.0124∠ − 16° A
Using Cramer’s rule:
Voc = 300 I 2 = 3.71∠ − 16° V
Then
Next, determine I sc :
−2 V − V = 0 ⇒ V = 0 ⇒ I sc =
The Thevenin impedance is
ZT =
The Thevenin equivalent is
9∠0°
= 0.015∠0° A
600
Voc 3.545∠−16°
=
= 247∠ − 16° Ω
I sc
0.015∠0°
P10.7‐4 Determine the Thevenin equivalent of this circuit when vs(t) = 10 cos (10,000t−53.1°) V.
3i / 2
2Ω
200 μH
vs
+
–
a
i
25 μ F
b
Solution:
First, determine Voc :
The node equation is:
Voc Voc − (6 + j8) 3 ⎛ Voc − (6+ j8) ⎞
+
− ⎜
⎟=0
− j4
j2
2⎝
j2
⎠
Voc =3+ j 4 =5∠53.1° V
Vs = 10∠53° = 6 + j 8 V
Next, determine I sc :
The node equation is:
V
V
V − (6 + j8) 3 ⎡ V − (6 + j8) ⎤
+
+
− ⎢
⎥=0
2 − j4
j2
2⎣
j2
⎦
V=
I sc =
3 + j4
1− j
V 3+ j 4
=
2 2− j 2
The Thevenin impedance is
ZT =
Vs = 10∠53° = 6 + j 8 V
⎛ 2− j 2 ⎞
Voc
= 3 + j4 ⎜
⎟ = 2 − j2 Ω
I sc
⎝ 3+ j 4 ⎠
The Thevenin equivalent is
(checked: LNAP 7/18/04)
P10.7‐5 Determine the frequency at which Y is a pure conductance
20 k Ω
Y
Solution:
Y = G + YL + YC
1
Y = G when YL + YC = 0 or
+ jω C = 0
jω L
1
1
1
, fO =
ωO =
=
2π LC 2π 39.6×10−15
LC
= 0.07998×107 Hz =800 kHz
(80 on the dial of the radio)
1 nF
39.6 μH
P 10.7-6 Consider the circuit of Figure P 10.7-6, where we wish to determine the current I. Use
a series of source transformations to reduce the part of the circuit connected to the 2-Ω resistor to
a Norton equivalent circuit, and then find the current in the 2-Ω resistor by current division.
j4 Ω
3 30° A
–j3 Ω
4Ω
–j2 Ω
2Ω
I
Figure P 10.7-6
Solution:
Z1 =
(− j 3)(4)
= 2.4∠ − 53.1° Ω
− j 3+ 4
=1.44 − j1.92 Ω
Z 2 = Z1 + j 4
= 1.44 + j 2.08
= 2.53∠55.3° Ω
Z3 = 3.51∠ − 37.9° Ω
= 2.77 − j 2.16 Ω
⎛ 3.51∠−37.9° ⎞
( 3.51∠−37.9° ) = 1.9∠ − 92° A
I = ( 2.85∠− 78.4° ) ⎜
⎟ = ( 2.85∠− 78.4° )
( 5.24∠− 24.4° )
⎝ 2.77 − j 2.16 + 2 ⎠
(checked: LNAP 7/18/04)
P 10.7-7 For the circuit of Figure P 10.7-7,
200 Ω
determine the current I using a series of source
transformations.
20 45° V
+
–
100 Ω
160 μ H
10 μ F
The source has ω = 25 × 103 rad/s.
I
Answer: i(t) = 4 cos (25,000t – 44°) mA
Figure P 10.7-7
Solution:
Z2 =
I=
(200)(− j 4)
= 4∠ − 88.8° Ω
200− j 4
0.4∠− 44°
= 4∠ − 44° mA
−4 j +100+ j 4
i (t ) = 4 cos (25000 t − 44°) mA
P10.7‐8 Determine the Thevenin equivalent of the circuit in (a):
a
8Ω
Zt
j10 Ω
–j5
a
+
–
+–
Vt
b
j20 Ω
–j2.4 Ω
b
(a)
(b)
Solution:
V1 =
V2 =
j10
5 e − j 90 = 3.9 e− j 51
8 + j10
j 20
5 e − j 90 = 5.68 e− j 90
j 20 − j 2.4
Vt = V1 − V 2 = 3.9 e − j 51 − 5.68 e − j 90
= 3.58 e j 47
Zt =
8 ( j10 ) − j 2.4 ( j 20 )
+
= 4.9 + j 1.2
8 + j10 − j 2.4 + j 20
P10.7‐9 Determine the voltage v(t) for this circuit:
+
8Ω
2H
5 sin 5t V
v(t)
+–
1 12 F
4 cos (3t + 15°) A
4H
–
Solution:
V1 (ω ) =
V 2 (ω ) =
j10
5 e − j 90 = 3.9 e − j 51
8 + j10
j 20
5 e − j 90 = 5.68 e − j 90
j 20 − j 2.4
V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90
= 3.58 e j 47
V1 (ω ) =
V 2 (ω ) =
8 ( j6)
4 e j15 = 19.2 e j 68
8 + j6
j12 ( − j 4 )
4 e j15 = 24 e − j 75
j12 − j 4
V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22
Using superposition: v(t) = 3.58 cos ( 5t + 47° ) + 14.4 cos ( 3t ‐ 22° ) V.
10.8 Superposition
P10.8‐1 Determine the steady state current i(t) in the circuit shown in Figure P10.8‐1 when the
voltage source voltages are
vs1 (t) = 12 cos (2500 t) V and vs2 (t) = 12 cos (4000 t) V
Figure 10.8‐1
Solution:
Use superposition in the time domain. These circuits can be used to find the part of io caused by
vs1 and the part of io caused by vs2.
In the frequency domain:
12∠0°
= 0.03578∠26.6° V
300 + j 250 − j 400
12∠0°
I2 =
= 0.03578∠ − 26.6° V
300 − j 250 + j 400
I1 =
In the time domain
i o1 ( t ) = 35.78cos ( 2500 t + 26.6° ) mA and i o2 ( t ) = 35.78cos ( 4000 t − 26.6° ) mA
and
i o ( t ) = i o1 ( t ) + i o2 ( t ) = 35.78cos ( 2500 t + 26.6° ) + 35.78cos ( 4000 t − 26.6° ) mA V
P10.8‐2 Determine the steady state voltage v(t) in the circuit shown in Figure P10.8‐2 when the
current source current is (a) 400 rad/s and (b) 200 rad/s.
Figure 10.8‐2
Solution:
(a) Represent the circuit in the frequency domain as
Use superposition in the frequency domain to write
Vo = −
=
100
j 150
(12∠0° ) + 100
( 0.1∠0° )
100 + j 150
100 + j 150
−1200 + j 1500
= 10.66∠72.35°
100 + j 150
In the time domain v o ( t ) = 10.66cos ( 400 t + 72.35 ) V
(b) Use superposition in the time domain. These circuits can be used to find the part of vo
caused by the current source and the part of vo caused by the voltage source.
In the frequency domain:
Vo1 = 100
Vo2 = −
j 75
( 0.1∠0° ) = 6∠53.1° V
100 + j 75
100
(12∠0° ) = 6.656∠123.7° V
100 + j 150
In the time domain
v o1 ( t ) = 6cos ( 200 t + 53.1° ) V and v o2 ( t ) = 6.656cos ( 400 t + 123.7° ) V
and
v o ( t ) = v o1 ( t ) + v o2 ( t ) = 6cos ( 200 t + 53.1° ) + 6.656cos ( 400 t + 123.7° ) V
P10.8‐3 Determine the steady state current i(t) in the circuit shown in Figure P10.8‐3 when the
voltage source voltage is
vs (t) = 8 + 8 cos (400t − 135° ) V
Figure 10.8‐3
Solution:
Use superposition in the time domain:
An inductor in a dc circuit acts like a short circuit so:
i1 ( t ) =
8
= 0.533 A
15
Represent the right circuit the frequency domain:
8∠ − 135°
= 0.32∠ − 188° A
15 + j 20
i 2 ( t ) = 0.32cos ( 400 t − 188° ) A
I2 =
In the time domain
and
i ( t ) = i1 ( t ) + i 2 ( t ) = 0.533 + 0.32cos ( 400 t − 188° ) A
P10.8‐4 Determine the steady state current i(t) in the circuit shown in Figure P10.8‐4 when the
voltage source voltages are
vs1 (t) = 10 cos (800t + 30° ) V and vs2 (t) = 15 sin (200t − 30° ) V
Figure 10.8‐4
Solution:
Use superposition in the time domain:
Represent these circuits the frequency domain:
I1 =
10∠30°
15∠ − 120°
= 0.1342∠33.4° A
= 0.0447∠ − 33.4° A and I 2 = −
100 + j 200
100 + j 50
In the time domain
i ( t ) = i1 ( t ) + i 2 ( t ) = 44.7 cos ( 800 t − 33.4° ) + 134.2cos ( 200 t + 33.4° ) mA
4i(t)
P 10.8-5
The input to the circuit shown in
+ –
Figure P 10.8-5 is the current source current
20 Ω
is(t) = 36 cos (25t) + 48 cos (50t + 45°) mA
Determine the steady-state current, i(t).
2H
i(t)
is(t)
2 mF
15 Ω
4 mF
Figure P 10.8-5
Solution:
Use superposition in the time domain. Let
i s1 ( t ) = 36 cos ( 25 t ) mA and i s 2 ( t ) = 48cos ( 50 t + 45° ) mA
We will find the response to each of these inputs separately. Let ii(t) denote the response to
isi(t) for i = 1,2. The sum of the two responses will be i(t), i.e.
i ( t ) = i1 ( t ) + i 2 ( t )
Represent the circuit in the frequency domain as
Use KVL to get
Vi = Z i I i − 4I i
Apply KCL to the supernode corresponding to the dependent voltage source.
I si = I i +
Vi
Z2
=
Z1 + Z 2 − 4
Z2
or
Ii =
Z 2I s i
Z1 + Z 2 − 4
Ii
Consider the case i = 1 : is1(t) = 26cos(25t) mA.
Here ω = 25 rad/s and
I si = 36∠0° mA
Z 1 = 20 +
1
= 20 − j 20 Ω
j ( 25 )( 0.002 )
⎛
⎞
1
Z 2 = j 50 + ⎜⎜15 &
⎟⎟ = 43.3∠83.9° Ω
25
0.004
j
(
)(
)
⎝
⎠
and
I 1 = 50.4∠35.7° mA
so
i ( t ) = 50.4 cos ( 25t + 35.7° ) mA
Next consider i = 2 : is2 = 48cos(50t + 45°) mA.
Here ω = 50 rad/s and
I s2 = 48∠45° mA
Z 1 = 20 +
1
= 20 − j10 Ω
j ( 50 )( 0.002 )
⎛
⎞
1
Z 2 = j100 + ⎜⎜15 &
⎟ = 95.5∠89.1° Ω
j ( 50 )( 0.004 ) ⎟⎠
⎝
(Notice that Z1 and Z2 change when ω changes.)
I 2 = 52.5∠55.7° mA
so
i 2 ( t ) = 52.5cos ( 50t + 55.7° ) mA
Finally, using superposition in the time domain gives
i ( t ) = 50.4 cos ( 25t + 35.7° ) + 52.5cos ( 50t + 55.7° ) mA
(checked: LNAP 8/7/04)
20 Ω
20 Ω
4H
5 mF
P 10.8-6 The inputs to the circuit shown in
10 Ω
Figure P 10.8-6 are
i(t)
vs1(t) = 30 cos (20t + 70°) V
and
vs2(t) = 18 cos (10t – 15°) V
+
–
vs1(t)
+
15 Ω
vs2(t) –
2H
Figure P 10.8-6
Solution:
Use superposition in the time domain. Let i1(t) be the part of i(t) due to vs1(t) and i2(t) be the
part of i(t) due to vs2(t). To determine i1(t), set vs2(t) = 0. Represent the resulting circuit in the
frequency domain to get
where
Z 1 = 20 + j80 = 82.46∠76° Ω
Z 2 = 10 + ( j 40 & 15 ) = 23.15 + j 4.93 = 23.67∠12° Ω
Z 3 = 20 +
1
= 20 − j10 = 22.36∠ − 26.6° Ω
j ( 20 )( 0.005 )
Next, using Ohm’s law and current division gives
I1 =
so
Z3
Z 3 ( 30∠70° )
30∠70°
×
=
= 0.182∠ − 17.6° A
Z 1 + ( Z 2 & Z 3 ) Z 2 + Z 3 Z 1Z 2 + Z 2 Z 3 + Z 1Z 3
i ( t ) = 0.182 cos ( 20t − 17.6° ) A
To determine i2(t), set vs1(t) = 0. Represent the resulting circuit in the frequency domain to get
where
Z 4 = 20 + j 40 = 44.72∠63.4° Ω
Z 5 = 10 + ( j 20 & 15 ) = 19.6 + j 7.2 = 20.88∠20.2° Ω
Z 6 = 20 +
1
= 20 − j 20 = 28.28∠ − 45° Ω
j (10 )( 0.005 )
Next, using Ohm’s law and current division gives
I2 =
Z4
Z 1 (18∠ − 15° )
18∠ − 15°
×
=
= 0.377∠18° A
Z 6 + ( Z 4 & Z 5 ) Z 4 + Z 5 Z 1Z 2 + Z 2 Z 3 + Z 1Z 3
so
i 2 ( t ) = 0.377 cos (10t + 18° ) A
Using superposition,
i ( t ) = i1 ( t ) + i 2 ( t ) = 0.182 cos ( 20t − 17.6° ) + 0.377 cos (10t + 18° ) A
(checked: LNAP 8/8/04)
P 10.8-7 The input to the circuit shown in Figure
5Ω
25 Ω
P 10.8-7 is the voltage source voltage
vs(t) = 5 + 30 cos (100t) V
+
–
i(t)
vs(t)
Determine the steady-state current, i(t).
20 μF
50 mH
Figure P 10.8-7
Solution:
Use superposition in the time domain. Let vs1(t) = 5 V and vs2(t) = 30cos(100t) V.
Find the steady state response to vs1(t).
When the input is constant and the circuit is at
steady state, the capacitor acts like an open
circuit and the inductor acts like a short
circuit.
So
5
i1 ( t ) = = 1 A
5
Find the steady state response to vs2(t).
Represent the circuit in the frequency domain
using impedances and phasors.
I2 =
So
30∠0°
= 4.243∠ − 45 A
5 + j5
i 2 ( t ) = 4.243cos (100t − 45° ) A
Using superposition
i ( t ) = i1 ( t ) + i 2 ( t ) = 1 + 4.243cos (100t − 45° )
P10.8‐8 Determine the voltage v(t) for the circuit
+
8Ω
2H
5 sin 5t V
v(t)
+–
1 12 F
4 cos (3t + 15°) A
4H
–
Solution:
V1 (ω ) =
V 2 (ω ) =
j10
5 e − j 90 = 3.9 e − j 51
8 + j10
j 20
5 e − j 90 = 5.68 e − j 90
j 20 − j 2.4
V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90
= 3.58 e j 47
V1 (ω ) =
V 2 (ω ) =
8 ( j6)
4 e j15 = 19.2 e j 68
8 + j6
j12 ( − j 4 )
4 e j15 = 24 e − j 75
j12 − j 4
V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22
Using superposition: v(t) = 3.58 cos ( 5t + 47° ) + 14.4 cos ( 3t ‐ 22° ) V.
P10.8‐9 Determine the current i(t) for this circuit when v1(t)=10cos(10t) V
5Ω
v1 +
–
i(t)
1.5 H
10 mF
10 Ω
3A
Solution:
Use superposition. First, find the response to the voltage source acting alone:
Z eq =
− j10⋅10
= 5(1 − j ) Ω
10− j10
Replacing the parallel elements by the equivalent impedance. The write a mesh equation :
−10 + 5 I1 + j15 I1 + 5(1 − j ) I1 = 0 ⇒ I1 =
10
= 0.707∠ − 45° A
10+ j10
Therefore:
i1 (t ) = 0.707 cos(10 t − 45° ) A
Next, find the response to the dc current source acting alone:
Current division:
I2 = −
Using superposition:
i (t ) = 0.707 cos(10 t − 45°) − 2 A
10
× 3 = −2 A
15
Section 10-9: Phasor Diagrams
P 10.9-1
Using a phasor diagram, determine V when
V = V1 – V2 + V3* and V1 = 3 + j3, V2 = 4 + j2, and V3 = –3 – j2. (Units are volts.)
Answer: V = 5 ∠143.1° V
.
Solution:
V = V1 − V2 + V3 = ( 3+ j 3) − ( 4 + j 2 ) + ( −3− j 2 ) = −4 + j 3
*
*
P 10.9-2 Consider the series RLC circuit of Figure P 10.9-2 when
R = 10 Ω, L = 1 mH, C = 100 μF, and ω = 103 rad/s.
Find I and plot the phasor diagram
j ωL
R
10 0° V
+
–
I
1
j ωC
Figure P 10.9-2
Solution:
I=
VR = R I = 7.4∠42° V
VL = Z L I = (1∠90°)(0.74∠42°) = 0.74∠132° V
VC = Z C I = (10∠−90°)(0.74∠42°) = 7.4∠− 48° V
VS = 10∠0° V
10∠0°
= 0.74∠42° A
10+ j1− j10
10.10 Op Amps in AC Circuits
P10.10‐1 The input to the circuit shown in Figure P10.10‐1 is the voltage
v s ( t ) = 2.4 cos ( 500 t ) V .
Determine the output voltage vo(t).
Answer: v o ( t ) = 6.788cos ( 500 t + 135° ) V
Figure P10.10‐1
Solution:
Represent the circuit in the frequency domain as
Recognizing this circuit as an inverting amplifier, we can write
14.14∠ − 45° ⎞
⎛ 20 || − j 20 ⎞
⎛
Vo = ⎜ −
⎟ ( 2.4∠0 ) = ⎜ (1∠180° )
⎟ ( 2.4∠0 ) = 6.788∠135° V
5
5
⎝
⎠
⎝
⎠
In the time domain
v o ( t ) = 6.788cos ( 500 t + 135° ) V
(Checked using LNAPAC 3/15/12)
P10.10‐2 The input of the circuit shown in Figure P10.10‐2 is the voltage
v s ( t ) = 1.2 cos ( 400 t + 20° ) V .
Determine the output voltage vo(t).
Figure P10.10‐2
Solution:
Represent the circuit in the frequency domain as
Recognizing this circuit as a noninverting amplifier, we can write
⎛
48 ⎞
V0 = ⎜1 +
⎟ (1.2∠20° ) = (1 + j 4.8 )(1.2∠20° ) = 5.88∠98° V
⎝ − j 10 ⎠
In the time domain
v o ( t ) = 5.88cos ( 400 t + 98° ) V
(Checked using LNAPAC 3/15/12)
P10.10‐3 The input of the circuit shown in Figure P10.10‐3 is the voltage
v s ( t ) = 3.2 cos ( 200 t ) V .
Determine the output voltage vo(t).
Figure P10.10‐3
Solution:
Represent the circuit in the frequency domain as
Recognizing this circuit as a voltage divider followed by a noninverting amplifier, we can write
⎞
20
20
⎛ 40 ⎞ ⎛
⎛
⎞
Vo = ⎜1 + ⎟ ⎜
⎟ ( 3.2∠0° ) = ⎜
⎟ ( 9.6∠0° ) = 4.293∠63.4° V
⎝ 20 ⎠ ⎝ − j 40 + 20 ⎠
⎝ 44.72∠ − 63.4° ⎠
In the time domain
v o ( t ) = 4.293cos ( 200 t + 63.4° ) V
(Checked using LNAPAC 3/15/12)
P10.10‐4 The input of the circuit shown in Figure P10.10‐4 is the voltage
v s ( t ) = 1.2 cos ( 2000 t ) V .
Determine the output voltage vo(t).
Figure P10.10‐4
Solution:
Represent the circuit in the frequency domain as
Recognizing this circuit as a voltage divider followed by a voltage follower, we can write
⎛ − j 2.5 ⎞
⎛ 2.5∠ − 90° ⎞
Vo = ⎜
⎟ (1.2∠0° ) = ⎜
⎟ (1.2∠0° ) = 0.1974∠ − 80.54° V
⎝ 15.2∠ − 9.46° ⎠
⎝ 15 − j 2.5 ⎠
v o ( t ) = 0.1974 cos ( 400 t − 80.54° ) V
In the time domain
(Checked using LNAPAC 3/15/12)
P10.10‐5 The input of the circuit shown in Figure P10.10‐5 is the voltage
v s ( t ) = 1.2 cos ( 2000 t ) V .
Determine the output voltage vo(t).
Figure P10.10‐5
Solution:
Represent the circuit in the frequency domain as
Recognizing this circuit as a inverting amplifier, we can write
In the time domain
⎛ − j 50 ⎞
Vo = ⎜ −
⎟ (1.2∠0 ) = 12∠90° V
5 ⎠
⎝
v o ( t ) = 12 cos ( 400 t + 90° ) V
(Checked using LNAPAC 3/15/12)
P 10.10-6
Determine the ratio Vo/Vs for the circuit shown in Figure P 10.10-6.
Z2
Z1
+
–
Vs
Z4
Z3
–
–
+
+
+
Vo
–
Figure P 10.10-6
Solution:
Label the nodes:
The ideal op amps force Va = 0 and Vc = 0.
Apply KCL at node a to get
Vb =
Apply KCL at node c to get
Vo =
Z2
Z1 + Z 2
Z4
Z3 + Z4
Vs
Vb
Therefore
Z4
Z2
Vo
=
×
Vs Z 3 + Z 4 Z 1 + Z 2
P 10.10‐7 Determine the ratio Vo/Vs for both of the circuits shown in Figure P 10.10‐7.
Z1
–
Z1
Z3
Z3
+
+
+
+
–
Vs
Z4
Z2
+
–
Vo
Vs
Z4
Z2
–
–
(a)
(b)
Figure P 10.10-7
Solution:
Label a node voltage as Va in each of the
circuits.
In both circuits, we can apply KCL at the node
between Z3 and Z4 to get
Vo =
In (a)
Va =
=
(
Z4
Z3 + Z4
Z 2 || Z 3 + Z 4
(
)
Z 1 + Z 2 || Z 3 + Z 4
(
)
Va
Vs
Z2 Z3 +Z4
(
)
)
(
Z1 Z 2 + Z 3 + Z 4 + Z 2 Z 3 + Z 4
)
Vs
so
Z2 Z4
Va
=
Vs Z 1 Z 2 + Z 3 + Z 4 + Z 2 Z 3 + Z 4
(
)
(
In (b)
Va =
Z2
Z1 + Z 2
Vs
so
Z4
Z2
Vo
=
×
Vs Z 3 + Z 4 Z 1 + Z 2
Vo
)
P 10.10‐8 Determine the ratio Vo/Vs for the circuit shown in Figure P 10.10‐8.
Z4
Z3
–
Z1
Z5
+
+
–
Vs
+
Z6
Z2
Vo
–
Figure P 10.10‐8
Solution:
Label the node voltages Va and Vb as shown:
Apply KCL at the node between Z1 and Z2 to
get
Z2
Va =
Vs
Z1 + Z 2
Apply KCL at the node between Z1 and Z2 to
get
Z3 + Z4
Vb =
Va
Z3
Apply KCL at the node between Z5 and Z6 to
get
Z6
Vo =
Vb
Z5 + Z6
so
Z6
Z3 + Z4
Z2
Vo
=
×
×
Vs Z 5 + Z 6
Z3
Z1 + Z 2
+
P 10.10-9 When the input to the circuit shown in
Figure P 10.10-9 is the voltage source voltage
R1
vs(t) = 2 cos (1000t) V
+
–
the output is the voltage
–
+
R2
vs(t)
vo(t)
+
vo(t) = 5 cos (1000t – 71.6°) V
C=1 mF
Determine the values of the resistances R1 and R2.
v(t)
1 kΩ
–
Figure P 10.10-9
Solution:
The network function of the circuit is
1
1+
R2
1+
R2
R2 ⎞ jω C
⎛
1000 =
1000
= ⎜1 +
=
⎟
1
V s ⎝ 1000 ⎠ R +
1 + j ω C R1 1 + j 10−3 R1
1
jω C
Vo
Converting the given input and output sinusoids to phasors gives
Vo
Vs
=
5 ∠71.6°
2
Consequently
1+
R2
5 ∠71.6°
1000
=
2
1 + j 10−3 R1
Equating angles gives
(
71.6° = − tan −1 10−3 R1
)
⇒ R1 = tan ( 71.6° ) × 10 3 = 3006 Ω
Equating magnitudes gives
R2
R2
1+
1+
5
1000
1000
=
=
2
2
1 + 10−3 R1
1 + 10−3 × 3006
(
)
(
)
2
⎛5
⎞
⇒ R 2 = ⎜ 10 − 1⎟ × 10 3 = 6906 Ω
⎝2
⎠
–
P 10.10-10
When the input to the circuit
10 kΩ
source voltage
vs(t)
+
–
C
R
10 kΩ
shown in Figure P 10.10-10 is the voltage
+
–
v(t)
+
–
vs(t) = 4 cos (100t) V
+
10 kΩ
–
the output is the voltage
vo(t) = 8 cos (100t + 135°) V
Figure P 10.10‐10
Determine the values of C and R.
Solution:
Represent the circuit in the frequency domain as
Apply KCL at the top node of the impedance of the capacitor to get
Vs − V
1
V
V
=
+ 4
⇒
Vs = 1 + j ( 5 × 105 ) C V
4
1
10
10
2
j100C
(
)
Apply KCL at the inverting node of the op amp to get
V Vo
+
=0
104 R
⇒
Vo = −
R
V
104
R
2 ×104
=
Vs 1 + j ( 5 ×105 ) C
Vo
so
−
Converting the input and output sinusoids to phasors gives
Vo
Vs
=
8∠135°
= 2∠135°
4∠0°
R
R
4
2 ×10
2 ×104
2∠135° =
=
∠180° − tan −1 ( 5 × 105 ) C
5
2
1 + j ( 5 ×10 ) C
1 + ⎡⎣( 5 × 105 ) C ⎤⎦
−
so
vo(t)
(
)
Equating angles gives
(
135° = 180° − tan −1 ( 5 × 105 ) C
)
⇒
C=
tan ( 45° )
= 2 × 10−6 = 2 μ F
5 ×105
Next, equating magnitudes gives
R
R
4
4
2 ×10
2=
= 2 ×10
2
1 + ( 5 × 105 )( 2 × 10−6 )
⇒
R = 104 = 10 kΩ
P10.10‐11 The input to the circuit shown in
Figure P10.10‐11 is the voltage source voltage,
v s ( t ) . The output is the voltage v o ( t ) . The input
v s ( t ) = 2.5cos (1000 t ) V
causes the output to be
v o ( t ) = 8cos (1000 t + 104° ) V .
Determine the values of the resistances R1 and
R2 .
Answers: R1 = 1515 Ω and R 2 = 20 kΩ.
Figure P10.10-11
Solution:
R2
1
jω C
=
R2
1 + jω CR 2
R2
R2
R1
1 + jω CR 2
Vo (ω )
=−
=−
R1
Vi (ω )
1 + jω CR 2
R2
R1
Vo (ω )
j (180− tan −1 ω CR 2 )
e
=
2
Vi (ω )
1 + (ω CR 2 )
In this case the angle of
the magnitude of
Vo (ω )
tan (180° − 104° )
is specified to be 104° so CR 2 =
= 0.004 and
Vi (ω )
1000
Vo (ω )
8
is specified to be
so
2.5
Vi (ω )
R2
R1
1 + 16
=
8
⇒
2.5
R2
R1
= 13.2 . One set of
values that satisfies these two equations is C = 0.2 μ F, R1 = 1515 Ω, R 2 = 20 kΩ .
Section 10.11 The Complete Response
Figure P10.11‐1
P10.11‐1 The input to the circuit shown in Figure P10.11‐1 is the voltage v s = 12 cos ( 4000 t ) V .
The output is the capacitor voltage, vo. Determine vo.
Solution:
Before the switch closes the circuit is represented in the frequency domain as
Express the dependent source voltage in
terms of the node voltages:
150 Ι a = Va − Vo
Using Ohm’s law
150
so
Va
200
= Va − Vo
Vo =
1
Va
4
Apply KCL to the supernode corresponding to the dependent source to get
Vs − V a
100
=
Va
200
+
Vo
− j 12.5
⇒ Vo =
− j 12.5
⎛ − j 12.5 − j 12.5 ⎞
Vs − ⎜
+
⎟ Va
100
200 ⎠
⎝ 100
Vo = − j 0.125 (12∠0° ) + j 0.1875 ( 4 Vo )
− j 0.125 (12∠0° )
= 1.2∠ − 53.1° V
1 − j 0.75
The corresponding sinusoid is 1.2 cos ( 4000 t − 53.1° ) V . The initial capacitor voltage is
Vo (1 − j 0.75 ) = − j 0.125 (12∠0° ) ⇒ Vo =
v o ( 0 ) = 1.2 cos ( −53.1° ) = 0.7205 V .
The steady state response after the switch closes is the forced response. The circuit is
represented in the frequency domain as
Express the dependent source voltage in
terms of the node voltages:
150 Ι a = Va − Vo
Using Ohm’s law
Va
150
= Va − Vo
200
1
Vo = Va
so
4
Apply KCL to the supernode corresponding to the dependent source to get
Vs − V a
100
=
Va
200
+
Vo
− j 12.5
+
⎛ 1
1 ⎞
1
1 ⎞
⎛ 1
⇒ Vo ⎜
+ ⎟=
+
Vs − ⎜
⎟ Va
25
⎝ 100 200 ⎠
⎝ − j 12.5 25 ⎠ 100
Vo
Multiply by 200 to get
Vo ( 8 + j 16 ) = 2 (12∠0° ) − 3 ( 4 Vo ) ⇒ Vo =
24
= 0.937∠ − 38.7° V
20 + j 16
The corresponding sinusoid is the forced response:
v f ( t ) = 0.937 cos ( 4000 t − 38.7° ) V
The natural response is
v n ( t ) = k e− t τ V
To determine the time constant τ we need find to find the Thevenin resistance of the part of
the circuit connected to the capacitor after the switch closes. Here’s the circuit:
The terminals separate the capacitor from the part of the circuit connected to the capacitor.
Now (1) remove the capacitor, (2) replace the voltage source by a short circuit to set the input
to zero and (3) connect a current source to the terminals to get
The Thevenin resistance is given by
Rt =
vt
it
Express the dependent source voltage in terms of the node voltages to get
v a − v t = 150 i a = 150
va
⇒ va = 4 vt
200
Apply KCL to the supernode corresponding to the dependent source to get
it =
va
100
The time constant is
The natural response is
+
va
200
+
vt
25
=
4vt
100
+
4vt
200
+
vt
25
=
vt
10
⇒ Rt =
vt
it
= 10 Ω
τ = R t C = 10 ( 20 × 10−6 ) = 0.2 × 10−3 = 0.2 ms
v n ( t ) = k e − t τ = k e − 5000 t V
The complete response is
v o ( t ) = 0.937 cos ( 4000 t − 38.7° ) + k e −5000 t V for t ≥ 0
Using the initial condition we calculate
0.7205 = v o ( 0 ) = 0.937 cos ( −38.7° ) + k ⇒ k = −0.0108
Finally
v o ( t ) = 0.937 cos ( 4000 t − 38.8° ) − 0.0108 e −5000t V for t ≥ 0
Figure P10.11‐2
P10.11‐2 The input to the circuit shown in Figure P10.11‐2 is the voltage v s = 12 cos ( 4000 t ) V .
The output is the capacitor voltage, vo. Determine vo.
Solution:
Before the switch opens the circuit is represented in the frequency domain as
Express the dependent source voltage in
terms of the node voltages:
150 Ι a = Va − Vo
Using Ohm’s law
Va
150
= Va − Vo
200
1
so
Vo = Va
4
Apply KCL to the supernode corresponding to the dependent source to get
Vs − V a
100
=
Va
200
+
Vo
− j 12.5
+
⎛ 1
1 ⎞
1
1 ⎞
⎛ 1
Vs − ⎜
⇒ Vo ⎜
+ ⎟=
+
⎟ Va
25
⎝ 100 200 ⎠
⎝ − j 12.5 25 ⎠ 100
Vo
Multiply by 200 to get
Vo ( 8 + j 16 ) = 2 (12∠0° ) − 3 ( 4 Vo ) ⇒ Vo =
The corresponding sinusoid is
The initial condition is
24
= 0.937∠ − 38.7° V
20 + j 16
0.937 cos ( 4000 t − 38.7° ) V
v o ( 0 ) = 0.937 cos ( −38.7° ) = 0.7313 V
The steady state response after the switch closes is the forced response. The circuit is
represented in the frequency domain as
Express the dependent source voltage in
terms of the node voltages:
150 Ι a = Va − Vo
Using Ohm’s law
150
Va
200
= Va − Vo
Vo =
so
1
Va
4
Apply KCL to the supernode corresponding to the dependent source to get
Vs − V a
100
=
Va
200
+
Vo
− j 12.5
⇒ Vo =
− j 12.5
⎛ − j 12.5 − j 12.5 ⎞
+
Vs − ⎜
⎟ Va
100
200 ⎠
⎝ 100
Vo = − j 0.125 (12∠0° ) + j 0.1875 ( 4 Vo )
Vo (1 − j 0.75 ) = − j 0.125 (12∠0° ) ⇒ Vo =
− j 0.125 (12∠0° )
= 1.2∠ − 53.1° V
1 − j 0.75
The corresponding sinusoid is the forced response:
v f ( t ) = 1.2 cos ( 4000 t − 53.1° ) V
The natural response is
v n ( t ) = k e− t τ V
To determine the time constant τ we need find to find the Thevenin resistance of the part of
the circuit connected to the capacitor after the switch closes. Here’s the circuit:
The terminals separate the capacitor from the part of the circuit connected to the capacitor.
Now (1) remove the capacitor, (2) replace the voltage source by a short circuit to set the input
to zero and (3) connect a current source to the terminals to get
The Thevenin resistance is given by
Rt =
vt
it
Express the dependent source voltage in terms of the node voltages to get
v a − v t = 150 i a = 150
va
200
⇒ va = 4 vt
Apply KCL to the supernode corresponding to the dependent source to get
it =
va
100
+
va
200
=
4vt
100
+
4vt
200
= 0.06 v t
⇒ Rt =
vt
it
= 16.67 Ω
The time constant is τ = R t C = 16.67 ( 20 ×10−6 ) = 0.333 × 10−3 = 0.333 ms
The natural response is
v n ( t ) = k e − t τ = k e − 3000 t V
The complete response is
v o ( t ) = 1.2 cos ( 4000 t − 53.1° ) + k e −3000 t V for t ≥ 0
Using the initial condition we calculate
0.7313 = v o ( 0 ) = 1.2 cos ( −53.1° ) + k ⇒ k = 0.0108
Finally
v o ( t ) = 1.2 cos ( 4000 t − 53.1° ) + 0.0108 e −3000 t V for t ≥ 0
Section 10.12 Using MATLAB to Analyze Electric Circuits
10.12-1 Determine the mesh currents for the circuit shown in Figure P10.12-1 when
vs(t) = 12 cos(2500t + 60°) V and is(t) = 2 cos(2500t − 15°) mA
Figure P10.12‐1
Solution:
Represent the circuit in the frequency domain:
Represent the source current in terms of the mesh currents:
I 2 − I 1 = I s = 0.002∠ − 15° A
Apply KVL to the supermesh corresponding to the current source:
( 2000 − j 2000 ) I 1 + j 1200 ( I 1 − I 3 ) + 1500 ( I 2 − I 3 ) + ( 4000 − j 1000 ) I 2 = 0
Apply KVL to mesh 3:
( 2400 + j 1875) I 3 + 1500 ( I 3 − I 2 ) + j 1200 ( I 3 − I 1 ) = −Vs = −12∠60°
−1
1
0
⎡
⎤ ⎡ I 1 ⎤ ⎡0.002∠ − 15° ⎤
⎢ ⎥
⎢
⎥
In matrix form: ⎢ 2000 − j 800 5500 − j 1000 −1500 − j 1200 ⎥⎥ ⎢I 2 ⎥ = ⎢⎢
0
⎥
⎢⎣ − j 1200
3900 + j 3075 ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ −12∠60° ⎥⎦
−1500
Solving, using MATLAB, gives
In the time domain:
⎡ I 1 ⎤ ⎡1.549∠ − 164° ⎤
⎢ ⎥ ⎢
⎥
⎢ I 2 ⎥ = ⎢ 1.039∠ − 65° ⎥ mA
⎢ I 3 ⎥ ⎢⎣ 2.904∠ − 148° ⎥⎦
⎣ ⎦
⎡ i1 ⎤ ⎡1.549 cos ( 2500 t − 164° ) ⎤
⎢ ⎥ ⎢
⎥
⎢i 2 ⎥ = ⎢ 1.039 cos ( 2500 t − 65° ) ⎥ mA
⎢ i 3 ⎥ ⎢ 2.904 cos ( 2500 t − 148° ) ⎥
⎦
⎣ ⎦ ⎣
10.12-2 Determine the node voltages for the circuit shown in Figure P10.12-2 when
vs(t) = 12 cos(400t + 45°) V.
Figure P10.12‐2
Solution:
Represent the circuit in the frequency domain:
Represent the dependent source voltage in terms of the node voltages currents:
V2
V3 − V 2 = 10000
⇒ V3 = 3.5 V 2
4000
Apply KCL to the supernode corresponding to the dependent voltage source:
Vs − V 2
j 2000
=
V2
4000
+
V3
j 3200
+
V3 − V 4
4000
⎛ 1
⎛ 1
1 ⎞
1 ⎞
⎛ 1 ⎞
=⎜
+
+
⎟ V2 + ⎜
⎟ V3 − ⎜
⎟ V4
j 2000 ⎝ 4000 j 2000 ⎠
⎝ 4000 ⎠
⎝ 4000 j 3200 ⎠
Vs − V 4 V3 − V 4
V4
Apply KCL at node 3:
+
=
10, 000
4000
− j 20, 000
Rearranging:
Vs
Rearranging:
Vs
10, 000
=−
⎡
⎢
3.5
⎢
⎢ 1
1
+
In matrix form: ⎢
⎢ 4000 j 2000
⎢
0
⎢
⎣
Solving, using MATLAB, gives
In the time domain:
⎛ 1
⎞
1
1
+⎜
+
+
⎟ V4
4000 ⎝ 4000 10, 000 − j 20, 000 ⎠
V3
⎡
⎤
⎤
⎢ 0 ⎥
⎥
−1
0
⎥
⎥ ⎡ V2 ⎤ ⎢
⎥ ⎢ ⎥ ⎢ Vs ⎥
1
1
1
+
−
⎥
⎥ ⎢ V3 ⎥ = ⎢
4000 j 3200
4000
⎥ ⎢ V ⎥ ⎢ j 2000 ⎥
⎥ ⎣ 4 ⎦ ⎢ Vs ⎥
1
1
1
1
−
+
+
⎢
⎥
⎥
4000
4000 10, 000 − j 20, 000 ⎦
⎣10, 000 ⎦
⎡ V 2 ⎤ ⎡ 3.236∠34° ⎤
⎢ ⎥ ⎢
⎥
⎢ V3 ⎥ = ⎢11.324∠34° ⎥ V
⎢ V 4 ⎥ ⎢⎣10.798∠29°⎥⎦
⎣ ⎦
⎡ v 2 ⎤ ⎡ 3.236 cos ( 400 t + 34° ) ⎤
⎢ ⎥ ⎢
⎥
⎢ v 3 ⎥ = ⎢11.324 cos ( 400 t + 34° ) ⎥ V
⎢ v 4 ⎥ ⎢10.798cos ( 400 t + 29° ) ⎥
⎦
⎣ ⎦ ⎣
10.12-3 Determine the mesh currents for the circuit shown in Figure P10.12-3
Figure P10.12‐3
Solution:
Represent the source current in terms of the mesh currents:
I 1 − I 2 = 4.2∠30° A
Apply KVL to the supermesh corresponding to the current source:
j 8 ( I 1 − I o ) + 5 ⎡⎣ j 8 ( I 1 − I o ) ⎤⎦ + ( 4 + j 5 ) I 2 + ( 3 − j 8 ) I 1 = 0
5 I o + 6 ( j 8) ( I 1 − I o ) = 0
Apply KVL to mesh 3:
In matrix form:
−1
0 ⎤ ⎡ I 1 ⎤ ⎡ 4.2∠30°⎤
⎡ 1
⎢
⎥⎢ ⎥ ⎢
⎥
⎢3 + j 40 4 + j 5 − j 48 ⎥ ⎢ I 2 ⎥ = ⎢ 0 ⎥
⎢⎣ − j 48
0
5 + j 48⎥⎦ ⎢⎣I o ⎥⎦ ⎢⎣ 0 ⎥⎦
Solving, using MATLAB, gives
⎡ I 1 ⎤ ⎡ 2.204∠93.1°⎤
⎢ ⎥ ⎢
⎥
⎢ I 2 ⎥ = ⎢ 3.758∠178° ⎥ A
⎢ I o ⎥ ⎢⎣ 2.192∠99° ⎥⎦
⎣ ⎦
10.12-4 Determine the node voltages for the circuit shown in Figure P10.12-4.
Figure P10.12‐4
Solution:
⎛ 24∠45° − V 2 ⎞ V 2 V 2 − Vo
+ 10 ⎜
+
⎟=
2+ j4
2+ j4
j6
⎝
⎠ 4
24∠45° − V3 V 2 − Vo Vo
+
=
Apply KCL at node 3:
j6
5− j2
4
11
1
⎡1 1
⎤
⎡ 11
⎤
−
24∠45° ) ⎥
(
⎢4 + j 6 + 2+ j 4
⎥
⎢
j6
⎡V ⎤
2+ j4
⎥⎢ 2⎥ = ⎢
⎥
In matrix form: ⎢
1
1
1 1 ⎥ ⎣ Vo ⎦ ⎢ 1
⎢
⎥
−
+
+ ⎥
( 24∠45° ) ⎥
⎢
⎢
j6
5− j 2 j 6 4⎦
⎣
⎣5− j 2
⎦
V
⎡ 2 ⎤ ⎡ 22.13∠40.8°⎤
Solving, using MATLAB, gives
⎢V ⎥ = ⎢
⎥ V
⎣ o ⎦ ⎣10.07∠30.4° ⎦
Apply KCL at node 2:
24∠45° − V 2
10.12-5 The input to the circuit shown in Figure P10.12-5 is the voltage source voltage
vs(t) = 12 cos(20,000t + 60°) V .
and the output is the steady state voltage vo(t). Use MATLAB to plot the input and output
sinusoids.
Figure P10.12‐5
Solution:
Represent the circuit in the frequency domain:
Write a node equation:
12∠60° − Vo
200, 000
+
12∠60° − Vo
− j 25, 000
+
12∠60°
=0
20, 000
⎛
⎛
⎞
1
1
1 ⎞
1
1
Rearrange: ⎜
+
+
+
⎟12∠60° = ⎜
⎟ Vo
⎝ 200, 000 − j 25, 000 20, 000 ⎠
⎝ 200, 000 − j 25, 000 ⎠
Modify the MATLAB script given in the textbook (and posted on the Student Companion Site for
Introduction to Electric Circuits):
%----------------------------------------------%
Describe the input voltage source.
%----------------------------------------------w = 20000;
A = 12;
theta = (pi/180)*60;
Vs = A*exp(j*theta);
%-----------------------------------------------%
Describe the impedances.
%-----------------------------------------------R1=50e3; R2=200e3; ZC=-j*25e3;
%-----------------------------------------------%
Calculate the phasor corresponding to the
%
output voltage.
%-----------------------------------------------Vo=(1/R2 + 1/ZC + 1/R1)*Vs/(1/R2 + 1/ZC);
B = abs(Vo)
phi = angle(Vo)
%-----------------------------------------------%
%-----------------------------------------------T = 2*pi/w;
tf = 2*T; N = 100; dt = tf/N;
t = 0 : dt : tf;
%-----------------------------------------------%
Plot the input and output voltages.
%-----------------------------------------------for k = 1 : 101
vs(k) = A * cos(w * t(k) + theta);
vo(k) = B * cos(w * t(k) + phi);
end
plot (t, vs, t, vo)
to get the plot:
10.12-6 The input to the circuit shown in Figure P10.12-6 is the voltage source voltage
vs(t) = 3 cos(4000t + 30°) V .
and the output is the steady state voltage vo(t). Use MATLAB to plot the input and output
sinusoids.
Figure P10.12‐6
Solution:
Represent the circuit in the frequency domain:
Recognize this circuit as an inverting amplifier to write:
20 || − j 12.5
⎛ 20 || − j 12.5 ⎞
Vo = ⎜ −
(12∠60° )
⎟ Vs = −
5
5
⎝
⎠
Modify the MATLAB script given in the textbook (and posted on the Student Companion Site for
Introduction to Electric Circuits):
%----------------------------------------------%
Describe the input voltage source.
%----------------------------------------------w = 4000;
A = 3;
theta = (pi/180)*30;
Vs = A*exp(j*theta);
%-----------------------------------------------%
Describe the impedances.
%-----------------------------------------------R1=5e3; R2=20e3; ZC=-j*12.5e3;
Zp=R2*ZC/(R2+ZC);
%-----------------------------------------------%
Calculate the phasor corresponding to the
%
output voltage.
%-----------------------------------------------Vo=(-Zp/R1)*Vs;
B = abs(Vo)
phi = angle(Vo)
%-----------------------------------------------%
%-----------------------------------------------T = 2*pi/w;
tf = 2*T; N = 100; dt = tf/N;
t = 0 : dt : tf;
%-----------------------------------------------%
Plot the input and output voltages.
%-----------------------------------------------for k = 1 : 101
vs(k) = A * cos(w * t(k) + theta);
vo(k) = B * cos(w * t(k) + phi);
end
plot (t, vs, t, vo)
to get the plot:
Section 10.14 How Can We Check…?
P 10.14-1
Computer analysis of the circuit in Figure P 10.14-1 indicates that the values of the
node voltages are V1 = 20 ∠–90° and V2 = 44.7 ∠–63.4°. Are the values correct?
Hint: Calculate the current in each circuit element using the values of V1 and V2. Check to see
whether KCL is satisfied at each node of the circuit.
j10 Ω
V1
10 Ω
2A
V2
10 Ω
3Ix
Ix
Figure P 10.14-1
Solution:
Generally, it is more convenient to divide complex numbers in polar form. Sometimes, as in this
case, it is more convenient to do the division in rectangular form.
Express V1 and V2 as: V1 = − j 20 and V 2 = 20 − j 40
KCL at node 1:
2−
V1
10
−
V1 − V 2
j 10
= 2−
− j 20 − j 20 − ( 20 − j 40 )
−
= 2+ j2−2− j2 = 0
10
j 10
KCL at node 2:
V1 − V 2
j 10
−
⎛ V1 ⎞ − j 20 − ( 20 − j 40 ) 20 − j 40
⎛ − j 20 ⎞
+ 3⎜ ⎟ =
−
+ 3⎜
⎟ = ( 2 + j 2) − ( 2 − j4) − j 6 = 0
j 10
10
10
⎝ 10 ⎠
⎝ 10 ⎠
V2
The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1
and V2 are correct.
P 10.14-2 Computer analysis of the circuit in
Figure P 10.14-2 indicates that the mesh currents are
i1(t) = 0.39 cos (5t + 39°) A and i2(t) = 0.28 cos (5t + 180°) A.
i1
8Ω
2H
5 sin 5t V
Is this analysis correct?
+–
Hint: Represent the circuit in the frequency domain using
impedances and phasors. Calculate the voltage across each
circuit element using the values of I1 and I2. Check to see
whether KVL is satisfied for each mesh of the circuit.
1 12
F
i2
4H
Figure P 10.14-2
Solution:
I 1 = 0.390 ∠ 39° and I 2 = 0.284 ∠ 180°
Generally, it is more convenient to multiply complex
numbers in polar form. Sometimes, as in this case, it is more
convenient to do the multiplication in rectangular form.
Express I1 and I2 as: I 1 = 0.305 + j 0.244 and I 2 = −0.284
KVL for mesh 1:
8 ( 0.305 + j 0.244 ) + j 10 ( 0.305 + j 0.244 ) − (− j 5) = j 10 ≠ 0
Since KVL is not satisfied for mesh 1, the mesh currents are
not correct.
Here is a MATLAB file for this problem:
Vs = -j*5;
Z1 = 8;
Z2 = j*10;
Z3 = -j*2.4;
Z4 = j*20;
% Mesh equations in matrix form
Z = [ Z1+Z2
0;
0
Z3+Z4 ];
V = [ Vs;
-Vs ];
I = Z\V
abs(I)
angle(I)*180/3.14159
% Verify solution by obtaining the algebraic sum of voltages for
% each mesh. KVL requires that both M1 and M2 be zero.
M1 = -Vs + Z1*I(1) +Z2*I(1)
M2 = Vs + Z3*I(2) + Z4*I(2)
P 10.14-3 Computer analysis of the circuit in
Figure P 10.14-3 indicates that the values of
the node voltages are
v1(t) = 19.2 cos (3t + 68°) V
v1
8Ω
2H
and
4 cos (3t + 15°) A
v2(t) = 2.4 cos (3t + 105°) V.
Is this analysis correct?
1 12
F 4H
Hint: Represent the circuit in the frequency
domain using impedances and phasors.
Calculate the current in each circuit element
using the values of V1 and V2. Check to see
whether KCL is satisfied at each node of the
circuit.
v2
Figure P 10.14-3
Solution:
V1 = 19.2 ∠ 68° and V 2 = 24 ∠ 105° V
KCL at node 1 :
19.2 ∠ 68° 19.2 ∠ 68°
+
− 4∠15 = 0
8
j6
KCL at node 2:
24 ∠105° 24 ∠105°
+
+ 4∠15 = 0
j12
− j4
The currents calculated from V1 and V2 satisfy
KCL at both nodes, so it is very likely that the V1
and V2 are correct.
Here is a MATLAB file for this problem:
Is = 4*exp(j*15*3.14159/180);
Z1 = 8;
Z2 = j*6;
Z3 = -j*4;
Z4 = j*12;
Y = [ 1/Z1 + 1/Z2
0;
0
1/Z3 + 1/Z4 ];
I = [ Is;
-Is ];
V = Y\I
abs(V)
angle(V)*180/3.14159
% Verify solution by obtaining the algebraic sum of currents for
% each node. KCL requires that both M1 and M2 be zero.
M1 = -Is + V(1)/Z1 + V(1)/Z2
M2 = Is + V(2)/Z3 + V(2)/Z4
P 10.14-4 A computer program reports that the currents of the circuit of Figure P 10.14-4 are
I = 0.2 ∠53.1° A, I1 = 632 ∠–18.4° mA, and I2 = 190 ∠71.6° mA.
Verify this result.
j500 Ω
I
100 0° V
+
–
3000 Ω
–j1000 Ω
I1
I2
Figure P 10.14-4
Solution: First, replace the parallel resistor and capacitor by an equivalent impedance
ZP =
(3000)(− j 1000)
= 949 ∠ − 72° = 300 − j 900 Ω
3000− j 1000
The current is given by
VS
100 ∠0°
=
= 0.2∠53° A
I=
j 500+ Z P j 500+ 300− j 900
Current division yields
⎛ − j 1000 ⎞
I1 = ⎜
⎟ ( 0.2 ∠53° ) = 63.3 ∠ − 18.5° mA
⎝ 3000 − j 1000 ⎠
⎛
⎞
3000
I2 = ⎜
⎟ ( 0.2 ∠53° ) = 190∠71.4° mA
⎝ 3000 − j 1000 ⎠
The reported value of I1 is off by an order of magnitude.
P 10.14-5 The circuit shown in Figure P 10.14-5 was built using a 2 percent resistor having a
nominal resistance of 500 Ω and a 10 percent capacitor with a nominal capacitance of 5 μF. The
steady-state capacitor voltage was measured to be
v(t) = 18.3 cos (200t – 24°) V
The voltage source represents a “signal generator.” Suppose that the signal generator was
adjusted so carefully that errors in the amplitude, frequency, and angle of the voltage source
voltage are all negligible. Is the measured response explained by the component tolerances? That
is, could the measured v(t) have been produced by this circuit with a resistance R that is within 2
percent of 500 Ω and a capacitance C that is within 5 percent of 5 μF?
R
+
20 cos (200t) V
+
–
C
v(t)
–
Figure P 10.14-5
Solution:
Represent the circuit in the frequency domain using phasors and impedances. Use voltage
division to get
1
j 200C
18.3∠ − 24° =
× 20∠0°
1
R+
j 200C
1
1
0.915∠ − 24° =
=
∠ − tan −1 ( 200CR )
So
2
1 + j 200CR
1 + ( 200CR )
Equating angles gives
−24° = − tan −1 ( 200CR )
⇒
200CR = tan ( 24° ) = 0.4452
The nominal component values cause 200CR = 0.5. So we expect that the actual component
values are smaller than the nominal values.
Try
C = 5 (1 − 0.10 ) × 10−6 = 4.5 μ F
Then
R=
0.4452
= 494.67 Ω
200 × 4.5 × 10−6
500 − 494.67
= 0.01066 = 1.066% this resistance is within 2% of 500 Ω. We conclude
500
that the measured angle could have been caused by a capacitance that is within 10% of 5 μF and
the resistance is within 2% of 500 Ω. Let’s check the amplitude. We require
Since
1
1 + ( 0.4452 )
2
= 0.9136 0.915
So the measured amplitude could also have been caused by the given circuit with C = 4.5 μF and
R = 494.67 Ω.
We conclude that he measured capacitor voltage could indeed have been produced by the given
circuit with a resistance that is within 2 % of 500 Ω and a capacitance that is within 10% of 5 μF.
PSpice Problems
SP 10-1 The circuit shown in
Figure SP 10.1 has two inputs, vs(t) and is(t),
and one output, v(t). The inputs are given by
vs(t) = 10 sin (6t + 45°) V
and
is(t) = 2 sin (6t + 60°) A
1H
vs(t)
+
–
3Ω
1 12
F
+
v(t)
–
is(t)
Figure SP 10.1
Use PSpice to demonstrate superposition. Simulate three versions of the circuit simultaneously.
(Draw the circuit in the PSpice workspace. “Cut and paste” to make two copies. Edit the part
names in the copies to avoid duplicate names. For example, the resistor will be R1 in the original
circuit. Change R1 to R2 and R3 in the two copies.) Use the given vs(t) and is(t) in the first
version. Set is(t) = 0 in the second version and vs(t) = 0 in the third version. Plot the capacitor
voltage, v(t), for all three versions of the circuit. Show that the capacitor voltage in the first
version of the circuit is equal to the sum of the capacitor voltages in the second and third
versions.
Hint: Use PSpice parts VSIN and ISIN for the voltage and current source. PSpice uses hertz
rather than rad/s as the unit for frequency.
Remark: Notice that v(t) is sinusoidal and has the same frequency as vs(t) and is(t).
Solution:
SP 10-2 The circuit shown in
Figure SP 10.1 has two inputs, vs(t) and is(t),
and one output, v(t). The inputs are given by
vs(t) = 10 sin (6t + 45°) V
and
is(t) = 2 sin (18t + 60°) A
1H
vs(t)
+
–
3Ω
1 12
F
+
v(t)
–
is(t)
Figure SP 10.1
Use PSpice to demonstrate superposition. Simulate three versions of the circuit simultaneously.
(Draw the circuit in the PSpice workspace. “Cut and paste” to make two copies. Edit the part
names in the copies to avoid duplicate names. For example, the resistor will be R1 in the original
circuit. Change R1 to R2 and R3 in the two copies.) Use the given vs(t) and is(t) in the first
version. Set is(t) = 0 in the second version and vs(t) = 0 in the third version. Plot the capacitor
voltage, v(t), for all three versions of the circuit. Show that the capacitor voltage in the first
version of the circuit is equal to the sum of the capacitor voltages in the second and third
versions.
Hint: Use PSpice parts VSIN and ISIN for the voltage and current source. PSpice uses hertz
rather than rad/s as the unit for frequency.
Remark: Notice that v(t) is not sinusoidal.
Solution:
SP 10-3 The circuit shown in
Figure SP 10-1 has two inputs, vs(t) and is(t),
and one output, v(t). The inputs are given by
vs(t) = 10 sin (6t + 45°) V
and
is(t) = 0.8 A
1H
vs(t)
+
–
3Ω
1 12
F
+
v(t)
–
is(t)
Figure SP 10.1
Use PSpice to demonstrate superposition. Simulate three versions of the circuit simultaneously.
(Draw the circuit in the PSpice workspace. “Cut and paste” to make two copies. Edit the part
names in the copies to avoid duplicate names. For example, the resistor will be R1 in the original
circuit. Change R1 to R2 and R3 in the two copies.) Use the given vs(t) and is(t) in the first
version. Set is(t) = 0 in the second version and vs(t) = 0 in the third version. Plot the capacitor
voltage, v(t), for all three versions of the circuit. Show that the capacitor voltage in the first
version of the circuit is equal to the sum of the capacitor voltages in the second and third
versions.
Hint: Use PSpice part VSIN and IDC for the voltage and current source. PSpice uses hertz rather
than rad/s as the unit for frequency.
Remark: Notice that v(t) looks sinusoidal, but it’s not sinusoidal because of the dc offset.
Solution:
SP 10-4
The circuit shown in Figure SP 10-
1H
3Ω
1 has two inputs, vs(t) and is(t), and one
output, v(t). When inputs are given by
vs(t)
+
–
1 12
F
vs(t) = Vm sin 6t V
and
is(t) = Im A
+
v(t)
–
is(t)
Figure SP 10.1
the output will be
vo(t) = A sin (6t + θ) + B V
Linearity requires that A be proportional to Vm and that B be proportional to Im.
Consequently, we can write A = k1Vm and B = k2Im, where k1 and k2 are constants yet to be
determined.
(a) Use PSpice to determine the value of k1 by simulating the circuit using Vm = 1 V and Im = 0.
(b) Use PSpice to determine the value of k2 by simulating the circuit using Vm = 0 V and Im = 1.
(c) Knowing k1 and k2, specify the values of Vm and Im that are required to cause
vo(t) = 5 sin (6t + θ) + 5 V
Simulate the circuit using PSpice to verify the specified values of Vm and Im.
Solution:
The following simulation shows that k1 = 0.4and k2 = -3 V/A. The required values of Vm and Im
are Vm = 12.5 V and Im = -1.667 A.
Design Problems
DP 10-1
Design the circuit shown in Figure DP 10-1 to produce the specified output voltage
vo(t) = 8cos(1000t + 104°) V when provided with the input voltage vi(t) = 2.5cos(1000t) V.
Figure DP 10-1
Solution:
R2
1
jω C
=
R2
1 + jω CR 2
R2
R2
1 + jω CR 2
R1
Vo (ω )
=−
=−
Vi (ω )
R1
1 + jω CR 2
R2
R1
Vo (ω )
j (180 − tan −1 ω CR 2 )
e
=
2
Vi (ω )
1 + (ω CR 2 )
In this case the angle of
the magnitude of
Vo (ω )
tan (180° − 104° )
is specified to be 104° so CR 2 =
= 0.004 and
1000
Vi (ω )
Vo (ω )
8
is specified to be
so
Vi (ω )
2.5
R2
R1
1 + 16
=
8
⇒
2.5
R2
R1
= 13.2 . One set of
values that satisfies these two equations is C = 0.2 μ F, R1 = 1515 Ω, R 2 = 20 kΩ .
DP 10-2
Design the circuit shown in Figure DP 10-2 to produce the specified output voltage
vo(t) = 2.5cos(1000t − 76°) V when provided with the input voltage vi(t) = 12cos(1000t) V.
Figure DP 10-2
Solution:
R2
Vo (ω )
=
Vi (ω )
where K =
1
jω C
=
R2
1 + jω CR 2
R2
1 + jω CR 2
K
=
R2
1 + jω CR p
R1 +
1 + jω CR 2
R1
R1 + R 2
and R p =
R1 R 2
R1 + R 2
Vo (ω )
K
− j tan −1 ω CR p
=
e
2
Vi (ω )
1 + (ω CR p )
In this case the angle of
C Rp = C
R1 R 2
R1 + R 2
=−
Vo (ω )
is specified to be -76° so
Vi (ω )
tan ( −76 )
V (ω )
2.5
is specified to be
= 0.004 and the magnitude of o
so
1000
Vi (ω )
12
R2
K
2.5
=
⇒ 0.859 = K =
. One set of values that satisfies these two equations is
R1 + R 2
1 + 16 12
C = 0.2 μ F, R1 = 23.3 kΩ, R 2 = 142 kΩ .
DP 10-3
Design the circuit shown in Figure DP 10-3 to produce the specified output voltage
vo(t) = 2.5cos(40t + 14°) V when provided with the input voltage vi(t) = 8cos(40t) V.
Figure DP 10-3
Solution:
jω L R 2
L
R 2 + jω L
R1
Vo (ω )
=
=
jω L R 2
L
Vi (ω )
1 + jω
R1 +
Rp
R 2 + jω L
where R p =
R1 R 2
R1 + R 2
Vo (ω )
Vi (ω )
In this case the angle of
jω
ω
=
L
R1
⎛ L ⎞
1+ ⎜ω
⎜ R p ⎟⎟
⎝
⎠
2
e
⎛
L ⎞
j ⎜ 90 − tan −1 ω
⎟
⎜
⎟
R
p ⎠
⎝
Vo (ω )
is specified to be 14° so
Vi (ω )
V (ω )
L L ( R1 + R 2 ) tan ( 90° − 14° )
2.5
is specified to be
so
=
=
= 0.1 and the magnitude of o
40
Rp
R1 R 2
Vi (ω )
8
40
L
R1
=
2.5
⇒
8
L
= 0.0322 . One set of values that satisfies these two equations is
R1
1 + 16
L = 1 H, R1 = 31 Ω, R 2 = 14.76 Ω .
DP 10-4 Show that it is not possible to design the circuit shown in Figure DP 10-4 to produce
the specified output voltage vo(t) = 2.5cos(40t − 14°) when provided with the input voltage
vi(t) = 8cos(40t) V.
Figure DP 10-4
Solution:
jω L R 2
L
R 2 + jω L
R1
Vo (ω )
=
=
jω L R 2
L
Vi (ω )
1 + jω
R1 +
Rp
R 2 + jω L
where R p =
R1 R 2
R1 + R 2
Vo (ω )
Vi (ω )
In this case the angle of
jω
ω
=
L
R1
⎛ L ⎞
1+ ⎜ω
⎜ R p ⎟⎟
⎝
⎠
Vo (ω )
is specified to be −14°. This requires
Vi (ω )
L L ( R1 + R 2 ) tan ( 90 + 14 )
=
=
= −0.1
Rp
R1 R 2
40
This condition cannot be satisfied with positive element waves.
2
e
⎛
L ⎞
j ⎜ 90 − tan −1 ω
⎟
⎜
R p ⎟⎠
⎝
v
DP 10-5 A circuit with an unspecified R, L,
and C is shown in Figure DP 10-5. The input
source is
is = 10 cos 1000t A,
R
is
10 Ω
C
L
and the goal is to select the R, L, and C so that
the node voltage is
v = 80 cos 1000t V.
Figure DP 10-5
Solution:
Z1 =10 Ω
1
Z2 =
jω C
Z3 = R + jω L
1
S
10
Y2 = jω C
1
Y3 =
R + jω L
Y1 =
v(t ) = 80 cos (1000 t − θ ) V ⇒ V = 80∠ − θ V
iS (t ) = 10 cos 1000 t A ⇒ I s =10∠0° A
try θ = 0° . Then
⎡1
⎤
1
+ jωC ⎥ = 10∠0° ⇒ R + 10 − 10 ω 2 LC + j (ω L + 10 ω RC ) = 1.25 R + j1.25 ω L
(80∠−∞ ) ⎢ +
⎣10 R + jω L
⎦
2
Equate real part: 40 − 40ω LC = R where ω = 1000 rad sec
Equate imaginary part: 40 RC = L
Solving yields R = 40(1− 4×107 RC 2 )
Now try R = 20 Ω ⇒ 1 − 2(1 − 4 × 107 (20)C 2 )
which yields C = 2.5×10−5 F= 25 μ F so L = 40 RC = 0.02 H=20 mH
Now check the angle of the voltage. First
Y1 = 1/10 = 0.1 S
Y2 = j 0.25 S
Y3 = 1/(20+ j 20) = .025− j.025 S
then
Y = Y1 + Y2 + Y3 = 0.125 , so V =YI s = (0.125∠0°)(10∠0°) = 1.25∠0° V
So the angle of the voltage is θ =0° , which satisfies the specifications.
R
DP 10-6 The input to the circuit shown in Figure DP 10-6 is
the voltage source voltage
+
+
–
vs(t) = 10 cos (1000t) V
vs(t)
C
vo(t)
–
The output is the steady-state capacitor voltage
vo(t) = A cos (1000t + θ) V
Figure DP 10-6
(a)
Specify values for R and C such that θ = –30°. Determine the resulting value of A.
(b)
Specify values for R and C such that A = 5 V. Determine the resulting values of θ.
(c)
Is it possible to specify values for R and C such that A = 4 and θ = –60°? (If not, justify
your answer. If so, specify R and C.)
(d)
Is it possible to specify values of R and C such that A = 7.07 V and θ = –45°? (If not,
justify your answer. If so, specify R and C.)
Solution:
Represent the circuit in the frequency domain using phasors and impedances. Using voltage
division gives
1
10
j1000C
×10∠0° =
A∠θ =
1
1 + j103 RC
R+
j1000C
Equating magnitudes and angles gives
2
A=
10
⇒
1 + 106 R 2C 2
⎛ 10 ⎞
⎜ ⎟ −1
⎝ A⎠
RC =
1 + j103 RC
and
θ = − tan −1 (103 RC )
(a)
Pick C = 1 μF, then R =
θ = −30°
⇒
⇒
RC =
RC =
tan ( 30° ) 0.577
=
.
103
103
0.577
= 577 Ω and A = 8.66 V.
106 ×103
2
(b)
tan ( −θ )
103
A=5 V
⇒
⎛ 10 ⎞
⎜ ⎟ −1
3
⎝5⎠
RC =
= 3.
3
10
10
Pick C = 1 μF, then R =
3
= 1732 Ω and θ = −60° .
10 × 103
−6
2
(c)
A=4
θ = −60°
⎛ 10 ⎞
⎜ ⎟ −1
2.29
⎝ 4⎠
RC =
= 3
3
10
10
⇒
⇒
RC =
tan ( 60° ) 1.73
= 3
103
10
Since RC cannot be both 0.00229 and 0.00173 simultaneously, the specifications cannot be
satisfied using this circuit.
2
(d)
A = 7.07
θ = −45°
⇒
⇒
⎛ 10 ⎞
⎜
⎟ −1
⎝ 7.07 ⎠
RC =
= 10−3
3
10
RC =
tan ( 45° )
= 10−3
103
Both specifications can be satisfied by taking R = 1000 Ω and C = 1 μF.
Chapter 11: AC Steady State Power
Exercises
Exercise 11.3-1
Determine the instantaneous power delivered to an element and sketch p(t) when the element is
(a) a resistance R and (b) an inductor L. The voltage across the element is v(t) = Vm cos (ωt + θ)
V.
V2
V2
Answer: (a) PR = m [1 + cos (2ωt + 2θ)] W (b) PL = m cos (2ωt + 2θ – 90°) W
2R
2ω L
Solution:
(a) When the element is a resistor, the current has the same phase angle as the voltage:
i (t ) =
v(t ) Vm
cos (ω t + θ ) A
=
R
R
The instantaneous power delivered to the resistor is given by
2
2
2
V
V
Vm
V
pR (t ) = v(t ) ⋅ i (t ) = Vm cos (ω t + θ ) ⋅ cos (ω t + θ ) = m cos 2 (ω t + θ ) = m + m cos (2ω t + θ )
R
R
2R 2R
(b) When the element is an inductor, the current will lag the voltage by 90°.
Z L = jω L = ω L∠90° Ω
⇒
I=
V
V ∠θ
V
= m
= m ∠ (θ −90° )
Z ω L∠90° ω L
The instantaneous power delivered to the inductor is given by
2
V
V
pL (t ) = i (t ) ⋅ v(t ) = m cos (ω t + θ − 90° ) ⋅Vm cos (ω t + θ ) = m cos ( 2ω t + 2θ −90° ) W
ωL
2ω L
Exercise 11.4-1
Find the effective value of the following currents: (a) cos 3t + cos 3t; (b) sin 3t + cos(3t + 60°);
(c) 2 cos 3t + 3 cos 5t
Answer: (a) 2 (b) 0.366 (c) 2.55
Ex. 11.4-1
(a)
I
2
i (t ) = 2 cos 3 t A ⇒ I eff max =
= 2A
2
2
(b)
i (t ) = cos (3 t − 90° ) + cos (3 t + 60° ) A
1
3
+j
= 0.518∠ − 15° A
2
2
0.518
i (t ) = 0.518 cos (3t − 15°) A ⇒ I eff =
= 0.366 A
2
I = (1∠−90° ) + (1∠60° ) = − j +
(c)
2
I eff
2
⎛ 2 ⎞ ⎛ 3 ⎞
=⎜
⎟ +⎜
⎟
⎝ 2⎠ ⎝ 2⎠
2
⇒ I eff = 2.55 A
Exercise 11.5-1 Determine the average power delivered to
each element of the circuit shown in Figure E 11.5-1. Verify
that average power is conserved.
Answer: 4.39 + 0 = 4.39 W
Figure E 11.5-1
Solution:
Analysis using Mathcad (ex11_5_1.mcd):
A := 12
Enter the parameters of the voltage source:
Enter the values of R and L
R := 10
L := 4
The impedance seen by the voltage source is:
The mesh current is: I :=
ω := 2
Z := R + j ⋅ω ⋅L
A
Z
⎯
I ⋅( I ⋅Z)
The complex power delivered by the source is: Sv :=
Sv = 4.39 + 3.512i
2
⎯
I ⋅( I ⋅R)
The complex power delivered to the resistor is: Sr :=
Sr = 4.39
2
⎯
I ⋅( I ⋅j ⋅ω ⋅L)
The complex power delivered to the inductor is: Sl :=
Sl = 3.512i
2
Verify Sv = Sr + Sl :
Sr + Sl = 4.39 + 3.512i
Sv = 4.39 + 3.512i
Exercise 11.5-2 Determine the complex power
delivered to each element of the circuit shown in Figure
E 11.5-2. Verify that complex power is conserved.
Answer: 6.606 + j5.248 – j3.303 + 6.606 = j1.982 VA
Figure E 11.5-2
Solution:
Analysis using Mathcad (ex11_5_2.mcd):
Enter the parameters of the voltage source: A := 12
R := 10
Enter the values of R, L an
dC
ω := 2
L := 4
C := 0.1
The impedance seen by the voltage source is: Z := R + j ⋅ω ⋅L +
The mesh current is:
I :=
A
Z
The complex power delivered by the sourceis:
Sv :=
1
j ⋅ω ⋅C
⎯
I ⋅( I ⋅Z)
2
⎯
I ⋅( I ⋅R)
The complex power delivered to the resistor is: Sr :=
2
⎯
I ⋅( I ⋅j ⋅ω ⋅L)
The complex power delivered to the inductor is: Sl :=
2
⎯⎛
1 ⎞
I ⋅⎜ I ⋅
⎟
j ⋅ω ⋅C ⎠
⎝
The complex power delivered to the capacitor is: Sc :=
2
Verify Sv = Sr + Sl + Sc :
Sr + Sl + Sc = 6.606 + 1.982i
Sv = 6.606 + 1.982i
Sr = 6.606
Sl = 5.284i
Sc = −3.303i
Sv = 6.606 + 1.982i
Exercise 11.6-1 A circuit has a large motor connected to the ac power lines [ω = (2π)60 = 377
rad/s]. The model of the motor is a resistor of 100 ω in series with an inductor of 5 H. Find the
power factor of the motor.
Answer: pf = 0.053 lagging
Solution:
( )
(377) (5) ⎤
⎡
pf = cos (∠ Z) = cos ⎡ tan −1 ω L ⎤ = cos ⎢ tan −1
= 0.053 lagging
R ⎥⎦
100 ⎥⎦
⎣⎢
⎣
Exercise 11.6-2 A circuit has a load impedance Z = 50 + j80 Ω, as shown in Figure 11.6-5.
Determine the power factor of the uncorrected circuit. Determine the impedance ZC required to
obtain a corrected power factor of 1.0.
Answer : ZC = –j111.25 Ω
Figure 11.6-5
Solution:
( )
( )
pf = cos (∠ Z) = cos ⎡ tan −1 X ⎤ = cos ⎡ tan −1 80 ⎤ = 0.53 lagging
⎢⎣
⎢⎣
50 ⎥⎦
R ⎥⎦
(50)2 + (80)2
XC =
= −111.25 Ω ⇒ ZC = − j 111.25 Ω
50 tan (cos −1 1) −80
Exercise 11.6-3 Determine the power factor for the total plant of Example 11.6-1 when the
resistive heating load is decreased to 30 kW. The motor load and the supply voltage remain as
described in Example 11.6-1.
Answer: pf = 0.915
Figure from Example 11.6-1
Solution:
PT = 30 + 86 = 116 kW and QT = 51 kVAR
S T = PT + j QT = 116 + j 51 = 126.7 ∠23.7° kVA
pf plant = cos 23.7° = 0.915
Exercise 11.6-4 A 4-kW, 110-Vrms load, as shown in Figure 11.6-5, has a power factor of 0.82
lagging. Find the value of the parallel capacitor that will correct the power factor to 0.95 lagging
when ω = 377 rad/s.
Answer: C = 0.324 mF
Figure 11.6-5
Solution:
P = V rms I rms cos θ
Z=
V rms
I rms
⇒ I rms =
P
4000
=
= 44.3 A
V rms cos θ
(110)(.82)
∠ cos −1 (0.82) = 2.48 ∠34.9° = 2.03+ j 1.42 = R + j X
To correct power factor to 0.95 requires
(2.03) 2 + (1.42) 2
R2 + X 2
=
= − 8.16 Ω
(2.03) tan (18.19° ) −1.42
R tan (cos −1 pfc) − X
−1
C=
=325 μ F
ω X1
X1 =
Exercise 11.7-1 Determine the average power absorbed by the resistor in Figure 11.7-2a for
these two cases:
(a)
vA(t) = 12 cos 3t V and vB(t) = 4 cos 3t V;
(b)
vA(t) = 12 cos 4t V and vB(t) = 4 cos 3t V
Answer: (a) 2.66 W (b) 4.99 W
Figure 11.7-2a
Solution:
(a) I = I1 + I 2 = (1.414∠ − 45° ) + ( 0.4714∠135° ) = 0.9428∠ − 45° A
0.94282
( 6 ) = 2.66 W
2
(b)
1.22
I1 = 1.2∠53° A ⇒ p1 =
( 6 ) = 4.32 W
2
0.47142
I 2 = 0.4714∠135° A ⇒ p2 =
( 6 ) = 0.666 W
2
∴ p = p1 + p2 = 4.99 W
⇒
p=
Exercise 11.8-1 For the circuit of Figure 11.8-1, find ZL
to obtain the maximum power transferred when the
Thévenin equivalent circuit has Vt = 100 ∠0°V and Zt = 10
+ j14 Ω. Also determine the maximum power transferred to
the load.
Answer: ZL = 10–j14 Ω and P = 125 W
Figure 11.8-1
Solution:
For maximum power transfer
Z L = Z*t = 10 − j14 Ω
100
I=
=5 A
(10+ j14) + (10− j14)
2
⎛ 5 ⎞
PL = ⎜
⎟ Re {10− j14} = 125 W
⎝ 2⎠
Exercise 11.8-2 A television receiver uses a cable to connect the antenna to the TV, as shown
in Figure E 11.8-2, with vs = 4 cos ωt mV. The TV station is received at 52 MHz. Determine the
average power delivered to each TV set if (a) the load impedance is Z = 300 Ω; (b) two identical
TV sets are connected in parallel with Z = 300 Ω for each set; (c) two identical sets are
connected in parallel and Z is to be selected so that maximum power is delivered at each set.
Answer: (a) 9.6 nW (b) 4.9 nW (c) 5 nW
Figure E 11.8-2
Solution:
If the station transmits a signal at 52 MHz
then
ω = 2π f = 104π ×106 rad/sec
so the received signal is
vs (t ) = 4 cos (104π × 106 t ) mV
(a) If the receiver has an input impedance of Zin =300 Ω then
Zin
300
1 ⎛ 1 ⎞ ( 2.4 ×10
Vin =
Vs =
× 4 × 10−3 = 2.4 mV ⇒ P = Vin2 ⎜
⎟=
R + Zin
200+300
2 ⎝ Zin ⎠
2(300)
)
−3 2
(b) If two receivers are connected in parallel then Z in = 300||300 = 150 Ω and
Vin =
total P =
Zin
150
VS =
(4 ×10−3 ) = 1.71× 10−3 V
R + Zin
200+150
2
Vin ⎛ 1 ⎞ (1.71×10−3 ) 2
= 9.7 nW or 4.85 nW to each set
⎜
⎟=
2 ⎝ Zin ⎠
2(150)
(c) In this case, we need Zin = R || R = 200 Ω ⇒ R = 400 Ω , where R is the input
impedance of each television receiver. Then
Ptotal =
Vin 2 (2×10−3 ) 2
=
= 10 nW ⇒ 5 nW to each set
2 Zin
2(200)
= 9.6 nW
Exercise 11.9-1 Determine the voltage vo for
the circuit of Figure E 11.9-1.
Hint: Write a single mesh equation. The
currents in the two coils are equal to each other
and equal to the mesh current.
Answer: vo = 14 cos 4t V
Figure E 11.9-1
Solution:
Coil voltages:
V1 = j 24 I 1 + j 16 I 2 = j 40 I
V 2 = j 16 I 1 + j 40 I 2 = j 56 I
Mesh equation:
24 = V1 + V 2 = j 40 I + j 56 I = j 96 I
24
1
=−j
4
j 96
⎛ 1⎞
Vo = V2 = ( j 56 ) ⎜ − j ⎟ = 14
⎝ 4⎠
vo = 14 cos 4t V
I=
Exercise 11.9-2 Determine the voltage vo for
the circuit of Figure E 11.9-2.
Hint: This exercise is the same as Exercise
11.9-1, except for the position of the dot on the
vertical coil.
Answer: vo = 18 cos 4t V
Figure E 11.9-2
Solution:
Coil voltages:
V1 = j 24 I 1 − j 16 I 2 = j 8 I
V 2 = − j 16 I 1 + j 40 I 2 = j 24 I
Mesh equation:
24 = V1 + V 2 = j 8 I + j 24 I = j 32 I
24
3
=−j
4
j 32
⎛ 3⎞
Vo = V 2 = ( j 24 ) ⎜ − j ⎟ = 18
⎝ 4⎠
vo = 18 cos 4t V
I=
Exercise 11.9-3 Determine the current io for
the circuit of Figure E 11.9-3.
Hint: The voltage across the vertical coil is
zero because of the short circuit. The voltage
across the horizontal coil induces a current in
the vertical coil. Consequently, the current in
the vertical coil is not zero.
Answer: io = 1.909 cos (4t – 90°) A
Solution:
Figure E 11.9-3
0 = V 2 = j 16 I 1 + j 40 I 2
40
I 2 = −2.5 I 2
16
V s = V1 = j 24 I 1 + j 16 I 2
⇒ I1 = −
= j (24(−2.5) + 16) I 2
= − j 44 I 2
24
6
= j
11
− j 44
I o = I 1 − I 2 = (−2.5 − 1) I 2
I2 =
= −3.5 I 2
⎛ 6⎞
= −3.5 ⎜ j ⎟ = − j 1.909
⎝ 11 ⎠
io = 1.909 cos ( 4t - 90° ) A
Exercise 11.9-4 Determine the current io for
the circuit of Figure E 11.9-4.
Hint: This exercise is the same as Exercise
11.9-3, except for the position of the dot on the
vertical coil.
Answer: io = 0.818 cos (4t – 90°) A
Figure E 11.9-4
Solution:
0 = V 2 = − j 16 I 1 + j 40 I 2
40
I 2 = 2.5 I 2
16
V s = V1 = j 24 I 1 − j 16 I 2
⇒ I1 =
= j (24(2.5) − 16) I 2
= j 44 I 2
24
6
=−j
j 44
11
I o = I 1 − I 2 = (2.5 − 1) I 2
I2 =
= 1.5 I 2
6⎞
⎛
= 1.5 ⎜ − j ⎟ = − j 0.818
⎝ 11 ⎠
io = 0.818 cos ( 4t - 90° ) A
Exercise 11.10-1 Determine the impedance Zab for the circuit of Figure E 11.10-1. All the
transformers are ideal.
Answer: Zab = 4.063Z
Figure E 11.10-1
Ex. 11.10-1
Z1 =
1 ⎛
Z⎞
Z⎞
1
Z Z
⎛
=
,
=
= 9 ⎜ Z + ⎟ and Z 3 = 2 (Z + Z 2 )
Z
Z
+
⎜
2
2
2 ⎜
2 ⎟
⎟
n3
4
n1
n2 ⎝
n3 ⎠
4⎠
⎝
then
1⎛
Z ⎞⎞
⎛
Z ab = Zin = Z + Z3 = Z + ⎜ Z +9⎜ Z+ ⎟ ⎟ = 4.0625 Z
4⎝
4 ⎠⎠
⎝
PROBLEMS
Section 11.3 Instantaneous Power and Average Power
P 11.3-1 An RLC circuit is shown in Figure P 11.3-1. Find the instantaneous power delivered
to the inductor when is = 1 cos ωt A and ω = 6283 rad/s.
Figure P 11.3-1
Solution:
1∠0° =
V
V
V
+
+
⇒ V = 14.6∠ − 43° V
20 j 63 − j16
V
= 0.23∠ − 133° A
j 63
p(t ) = i (t )v(t ) = 0.23cos (2π ⋅103 t − 133° ) × 14.6 cos (2π ⋅103 t − 43° )
I=
= 3.36 cos (2π ⋅103 t −133° ) cos (2π ⋅103 t − 43° )
= 1.68 (cos (90° ) + cos (4π ⋅103 t −176° ))
=1.68 cos (4π ⋅103 t −176° )
P 11.3-2 Find the average power absorbed by the 0.6-kΩ resistor and the average power
supplied by the current source for the circuit of Figure P 11.3-2.
Figure P 11.3-2
Solution:
Current division:
⎡ 1800 − j 2400 ⎤
I=4 5⎢
⎥
⎣1800− j 2400+ 600 ⎦
=5
5
∠ − 8.1° mA
2
2
P600Ω
I 600
⎛5⎞
=
= 300(25) ⎜ ⎟ = 1.875 ×104 μW = 18.75 mW
2
⎝ 2⎠
Psource =
⎛ 5⎞
V I cos θ 1
= (600) ⎜ 5 ⎟ 4 5 cos(−8.1° ) = 2.1× 104 μW = 21 mW
2
2
⎝ 2⎠
(
)
P 11.3-3 Use nodal analysis to find the average power absorbed by the 20-Ω resistor in the
circuit of Figure P 11.3-3.
Answer: P = 200 W
Figure P 11.3-3
Solution:
Node equations:
20I X −100
20I X − V
+ IX +
= 0 ⇒ I X (20 − j15) − V = − j 50
10
− j5
V − 20I X
V
− 3I X + = 0 ⇒ I X (−40 + j 30) + V (2 − j ) = 0
10
− j5
Solving the node equations using Cramer’s rule yields
IX =
j 50(2 − j )
50 5∠63.4°
=
= 2 5∠10.3° A
(40 − j 30) − (20 − j15)(2 − j )
25∠53.1°
Then
2
PAVE
(
I
= X (20) = 10 2 5
2
)
2
= 200 W
P 11.3-4 Nuclear power stations
have become very complex to operate,
as illustrated by the training simulator
for the operating room of the Pilgrim
Power Station shown in Figure P 11.34a. One control circuit has the model
shown in Figure P 11.3-4b. Find the
average power delivered to each
element.
Answer: Psource current = −12.8 W
P8 Ω = 6.4 W
PL = 0 W
Pvoltage source = 6.4 W
P 11.3-4
Solution:
A node equation:
(V − 16) V
+ − (2 2∠45°) = 0
j4
8
⎛
2⎞
⇒ V = ⎜⎜16
⎟ ∠18.4° V
5 ⎟⎠
⎝
Then
I=
16 − V
= 3.2 ∠ − 116.6° A
j4
2
PAVE 8Ω
PAVE current source = −
1 V
= ×
2 8
2
⎛
2⎞
⎜16
⎟
5⎠
1 ⎝
= ×
= 6.4 W absorbed
2
8
1
1⎛
2⎞
V 2 2 cos θ = − ⎜ 16
⎟ 2 2 cos ( 26.6° ) = −12.8 W absorbed
2
2⎝
5⎠
(
)
(
)
PAVE inductor = 0
1
1
PAVE voltage source = − (16) I cosθ = − (16)( 3.2)cos( − 116.6°) = 6.4 W absorbed
2
2
P 11.3-5 Find the average power delivered to each element for the circuit of Figure P 11.3-5.
Figure P 11.3-5
Solution:
A node equation:
−20 +
V1 V1 + (3 / 2)V1
+
= 0 ⇒ V1 = 50 5 ∠ − 26.6° V
10
15 − j 20
Then
I=
V1 + ( 3/2 ) V1
( 5 / 2 ) V1 = 5 5 ∠26.6° A
=
15 − j 20
25∠ − 53.1°
Now the various powers can be calculated:
2
2
PAVE 10Ω
PAVE current source = −
1
1
V (20) cosθ = − (50 5)(20) cos (−26.6°) = −1000 W absorbed
2
2
PAVE 15Ω
PAVE voltage source = −
1 V1
1 (50 5)
=
=
= 625 W absorbed
2 10
2 10
(
)
(
)(
2
5 5
I
=
(15 ) = −
2
2
2
(15 ) = 937.5 W absorbed
)
1 3
1
I V1 cos θ = − 5 5 75 5 cos ( −53.1° ) = −562.5 W absorbed
2 2
2
PAVE capacitor = 0 W
P 11.3-6 A student experimenter in
the laboratory encounters all types of
electrical equipment. Some pieces of
test equipment are battery-operated or
operate at low voltage so that any
hazard is minimal. Other types of
equipment are isolated from electrical
ground so that there is no problem if a
grounded object makes contact with
the circuit. Some types of test
equipment, however, are supplied by
voltages that can be hazardous or have
dangerous voltage outputs. The
standard power supply used in the
United States for power and lighting in
laboratories is the 120, grounded, 60Hz sinusoidal supply. This supply
provides power for much of the
laboratory
equipment,
so
an
understanding of its operation is
essential in its safe use (Bernstein,
Figure P 11.3-6
1991).
Consider the case where the experimenter has one hand on a piece of electrical equipment
and the other hand on a ground connection, as shown in the circuit diagram of Figure P 11.3-6a.
The hand-to-hand resistance is 200 Ω. Shocks with an energy of 30 J are hazardous to
humans. Consider the model shown in Figure P 11.3-6b, which represents the human with R.
Determine the energy delivered to the human in 1 s.
Solution:
Z=
I =
200 ( j 200 ) 200 ∠90° 200
=
=
∠45° Ω
200 (1 + j )
2 ∠45°
2
⎛
120∠0D
j 200 ⎞
= 0.85 ∠ − 45° A, I R = ⎜
⎟ I = 0.6∠0° A
200
200
+
j
200
D
⎝
⎠
∠45
2
P = I R = ( 0.6 ) ( 200 ) = 72 W and w = ( 72 )(1) = 72 J
2
2
P 11.3-7 An RLC circuit is shown in Figure P 11.3-7 with a voltage source vs = 7 cos 10t V.
(a)
Determine the instantaneous power delivered to the circuit by the voltage source.
(b)
Find the instantaneous power delivered to the inductor.
Answer: (a) p = 7.54 + 15.2 cos (20t – 60.3°) W
(b) p = 28.3 cos (20t – 30.6°) W
Figure P 11.3-7
Solution:
Z = j3 +
4(− j 2)
= 0.8 + j1.4
4− j 2
= 1.6 ∠60.3° Ω
∴I =
7∠0°
V
=
= 4.38 ∠ − 60.3° A
Z 1.6∠60.3°
i (t ) = 4.38cos (10 t − 60.3° ) A
The instantaneous power delivered by the source is given by
(7)(4.38)
[cos (60.3°) + cos (20 t −60.3°)]
2
= 7.6 + 15.3cos (20 t − 60.3°) W
p(t ) = v(t ) ⋅ i (t ) = (7 cos 10 t )(4.38cos (10 t − 60.3°)) =
The inductor voltage is calculated as
VL = I ⋅ Z L = (4.38 ∠ − 60.3°)(j 3) = 13.12 ∠29.69° V
vL (t ) = 13.12 cos (10 t + 29.69°) V
The instantaneous power delivered to the inductor is given by
pL (t ) = vL (t ) ⋅ i (t ) = ⎡⎣ (13.12 cos (10t + 29.69°)(4.38cos (l0t − 60.3° ) ⎤⎦
57.47
[ cos (29.69°+ 60.3°)+ cos (20 t + 29.69°−60.3°)]
2
= 28.7 cos (20t − 30.6°) W
=
P 11.3-8
(a)
Find the average power delivered by the source to the circuit shown in Figure P 11.3-8.
(b)
Find the power absorbed by resistor R1.
Answer: (a) 30 W
(b) 20 W
Figure P 11.3-8
Solution:
The equivalent impedance of the parallel resistor
(1)( j ) = 1 1+ j Ω . Then
and inductor is Z =
( )
1+ j
2
I=
(a) Psource =
I V
cos θ =
2
10∠0°
20
20
=
=
∠ − 18.4° A
1
j
3
+
10
1+ (1+ j )
2
⎛ 20 ⎞
⎟
⎝ 10 ⎠ cos −18.4° = 30.0 W
(
)
2
(10 ) ⎜
2
⎛ 20 ⎞
2
(1)
I R1 ⎜⎝ 10 ⎟⎠
(b) PR 1 =
=
= 20 W
2
2
Section 11.4 Effective Value of a Periodic Waveform
P 11.4-1 Find the rms value of the current i for (a) i = 2 – 4 cos 2t A, (b) i = 3 sin π t +
π t A, and (c)i = 2 cos 2t + 4 2 cos (2t + 45°) + 12 sin 2t A.
Answer: (a) 2 3 (b) 2.35 A (c) 5 2 A
Solution:
( Treat i as two sources of different frequencies.)
i = 2 − 4 cos 2t = i1 + i 2
(a)
2A source: I eff = lim
T →∞
1
T
∫
T
o
2
(2) dt = 2 A
and
4 cos 2t source:
Ieff =
4
A
2
The total is calculated as
2
I eff
(b)
2
⎛ 4 ⎞
= ( 2) + ⎜
⎟ = 12 A ⇒ I rms = I eff = 12 = 2 3 A
⎝ 2⎠
2
i ( t ) = 3cos (π t − 90° ) + 2 cos π t ⇒ I = ( 3∠ − 90° ) +
(
2 ∠0°
)
= 2 − j 3 = 3.32∠ − 64.8° A
I rms =
(c)
3.32
= 2.35 A
2
i ( t ) = 2 cos 2t + 4 2 cos ( 2t + 45° ) + 12 cos ( 2 t − 90° )
(
)
I = ( 2∠0° ) + 4 2 ∠45° + (12∠ − 90° ) = ( 2 + 4 ) + ( j 4 − j12 ) =10∠ − 53.1° A
I rms =
10
=5 2 A
2
2 cos
P 11.4-2 Determine the rms value for each of the waveforms shown in Figure P 11.4-2.
Answer: (a) 4.10 V (b) 4.81 V (c) 4.10
v (V)
6
v (V)
v (V)
6
6
2
2
2
2
5
7
10 t (s)
2
(a)
5
7
10 t (s)
3
5
8
10 t (s)
(c)
(b)
Figure P 11.4-2
Solution:
(a)
(b)
(c)
)
1
5
(∫
Vrms =
1
5
( ∫ 2 dt +∫ 6 dt ) = 15 ( ∫ 4 dt +∫ 36 dt ) =
1
116
= 4.81 V
(8 + 108) =
5
5
Vrms =
1
5
( ∫ 2 dt +∫ 6 dt ) = 15 ( ∫ 4 dt +∫ 36 dt ) =
1
84
= 4.10 V
(12 + 72 ) =
5
5
2
0
2
5
62 dt + ∫ 22 dt =
2
2
0
3
0
5
2
2
2
5
3
2
1
5
(∫
)
Vrms =
2
0
5
36 dt + ∫ 4 dt =
2
2
5
0
2
3
5
0
3
1
84
= 4.10 V
( 72 + 12 ) =
5
5
P 11.4-3 Determine the rms value for each of the
waveforms shown in Figure P 11.4-3.
Answer: (a) 4.16 V (b) 4.16 V (c) 4.16
Solution:
(a)
2
2
Vrms
1 4⎛ 4 2 ⎞
=
⎜ t + ⎟ dt
3 ∫1 ⎝ 3 3 ⎠
=
1
4 4
2
2 t + 1) dt
(
∫
27 1
4
1
4t 2
+
2
4
1
4
7
10 t (s)
(a)
4 4 2
=
( 4 t + 4 t + 1) dt
27 ∫1
4 ⎛ 4 t3
⎜
=
27 ⎜ 3
⎝
v (V)
6
⎞
4
+t 1 ⎟
⎟
⎠
=
4
( (85.33 − 1.33) + ( 2 )(16 − 1) + 3)
27
=
4
(117 ) = 4.16 V
27
(b)
v (V)
6
2
1
4
7
10 t (s)
(b)
v (V)
6
2
2
Vrms =
1 4 ⎛ 4 22 ⎞
⎜ − t + ⎟ dt
3 ∫1 ⎝ 3
3 ⎠
=
4 4
2
( −2 t + 11) dt
∫
1
27
4 4 2
=
( 4 t − 44 t + 121) dt
27 ∫1
4 ⎛ 4 t3
⎜
=
27 ⎜ 3
⎝
4
1
44t 2
−
2
4
1
⎞
4
+ 121 t 1 ⎟
⎟
⎠
=
4
(84 + ( −22 )15 + (121) 3)
27
=
4
(117 ) = 4.16 V
27
3
6
9
(c)
Figure P 11.4-3
t (s)
2
(c)
Vrms
1 3⎛ 4
4 3
4 3 2
2
⎞
=
t + 2 ⎟ dt =
2 t + 3) dt =
(
( 4 t + 12 t + 9 ) dt
⎜
∫
∫
3 0⎝3
27 0
27 ∫ 0
⎠
4 ⎛ 4 t3
⎜
=
27 ⎜ 3
⎝
3
12t 2
+
2
0
⎞
3
+ 9t 0 ⎟
⎟
0
⎠
3
=
4
( 36 + 54 + 27 )
27
=
4
(117 ) = 4.16 V
27
P 11.4-4 Find the rms value for each of the waveforms of Figure P 11.4-4.
Answer: Vrms = 1.225 V, Irms = 5 mA
v(t) V
2
Sinusoid +
Constant
–T/ 2
0
T/ 2
T
t
(a)
i(t) mA
10
Sinusoidal
–5
0
5
15 t ( ms)
10
(b)
Figure P 11.4-4
Solution:
(a)
⎛ 2π
v ( t ) = 1 + cos ⎜
⎝ T
vdc eff
2
t
⎛1 T ⎞
= ⎜ ∫ 0 1dt ⎟ =
⎝T
⎠ T
T
0
⎞
t ⎟ = vdc + vac
⎠
⎛T
⎞
2
= ⎜ − 0 ⎟ = 1 V and vac eff =
T
⎝
⎠
1
V
2
2
veff = vdc eff + vac eff
2
(b)
2
2
=
⎛ 1 ⎞
1 +⎜
⎟ = 1.225 V
⎝ 2⎠
2
The period of the sinusoid is half the period of i(t). Let T be the period of i(t) and T/2 be
the period of the sinusoid. Then
I RMS
4π
T
⎧
0<t <
⎪ 10sin T t
4
⎪
4π
3T
T
⎪
<t <
i ( t ) = ⎨−10sin
t
2
4
T
⎪
0
otherwise
⎪
⎪
⎩
3T / 4
100 ⎡ T / 4 2 4π
4π
⎤
=
t dt + ∫
t dt ⎥
sin
sin 2
∫
⎢
T /4
T ⎣ 0
T
T
⎦
⎡⎛
8π
sin
100 ⎢⎜ t
T
⎢⎜ −
=
18
π
T ⎢⎜ 2
⎢⎣⎝
T
I rms = 25
T /4
8π
⎞
⎛
t⎟
⎜ t sin T
⎟ + ⎜ − 18π
⎟
⎜2
T
⎠0
⎝
⇒
I rms = 5 mA
2T / 4
⎞
t⎟
⎟
⎟
⎠T /2
⎤
⎥
⎥ = 25
⎥
⎥⎦
P 11.4-5 Find the rms value of the voltage v(t) shown in Figure P 11.4-5.
Answer:
Vrms = 4.24 V
v(t) (V)
9
–0.3
–0.2
–0.1
0
0.1
0.2
0.3
0.4
0.5 t (s)
Figure P 11.4-5
Solution:
⎧90 t
⎪
v ( t ) = ⎨90 ( 0.2−t )
⎪0
⎩
2
=
Vrms
0 ≤ t ≤ 0.1
0.1 ≤ t ≤ 0.2
0.2 ≤ t ≤ 0.3
2
0.2
2
1 ⎡ 0.1
2
⎤ = 90
+
⎡
−
⎤
90
t
dt
90
0.2
t
dt
(
)
(
)
⎦ ⎥⎦ .3
∫0.1 ⎣
.3 ⎢⎣ ∫ 0
902
=
.3
V
rms
⎡ 0.1 t 2 dt + 0.2 0.2−t 2 dt ⎤
) ⎥
∫
∫0.1 (
⎣⎢ 0
⎦
⎡ .001 .001 ⎤
⎢⎣ 3 + 3 ⎥⎦ = 18 V
= 18 = 4.24 V
P 11.4-6 Find the effective value of the current waveform shown in Figure P 11.4-6.
Answer: Ieff = 8.66
10
5
–1
0
1
2
3
4
5
t (s)
Figure P 11.4-6
Solution:
I eff =
3
1 t 2
1⎡ 2
2
(
)
(10)
(5) 2 dt ⎤ = 8.66
i
t
dt
=
dt
+
∫
∫
∫
0
0
2
⎢
⎥⎦
3⎣
T
P 11.4-7 Calculate the effective value of the voltage across the resistance R of the circuit
shown in Figure P 11.4-7 when ω = 100 rad/s.
Hint: Use superposition.
Answer: Veff = 4.82 V
10 cos ω t A
5A
6 sin ω t A
R
12
Ω
Figure P 11.4-7
Solution:
Use superposition:
V1 = 5∠0° V
V2 = 2.5 V (dc)
V3 = 3∠−90° V
V1 and V3 are phasors having the same frequency, so we can add them:
V1 + V3 = ( 5∠0° ) + ( 3∠ − 90° ) = 5 − j3 = 5.83∠ − 31.0° V
Then
vR (t ) = v1 (t ) + ( v2 (t ) + v3 (t ) ) = 2.5 + 5.83cos (100 t − 31.0°) V
Finally
2
V
2
R eff
⎛ 5.83 ⎞
= (2.5) + ⎜
⎟ = 23.24 V ⇒ VR eff = 4.82 V
⎝ 2 ⎠
2
Section 11.5 Complex Power
i(t)
P 11.5-1 The complex power delivered by the voltage source in
Figure P 11.5-1 is S = 3.6 + j7.2 V A. Determine the values of the
resistance, R, and inductance, L.
+
–
R
12 cos 4t V
L
Answer: R = 4 Ω and L = 2 H
Figure P 11.5-1
Solution:
I* =
R+ j4 L =
2 ( 3.6 + j 7.2 )
2S
=
= 0.6 + j 1.2 = 1.342∠63.43° A
12∠0°
12∠0°
12∠0°
= 8.94∠63.43 = 4 + j 8 ⇒ R = 4 Ω and L = 2 H
1.342∠ − 63.43
i(t)
P 11.5-2 The complex power delivered by the voltage
source in Figure P 11.5-2 is S = 18 + j9 VA. Determine
the values of the resistance, R, and inductance, L.
+
–
12 cos 4t V
R
Answer: R = 4 Ω and L = 2 H
Figure P 11.5-2
Solution:
I* =
2 (18 + j 9 )
2S
=
= 3 + j 1.5 = 3.35∠26.56° A
12∠0°
12∠0°
1
1
1
1
3.35∠ − 26.56°
+
= −j
=
= 0.2791∠ − 26.56 = 0.250 − j 0.125
R j4 L R
4L
12∠0°
⇒ R = 4 Ω and L = 2 H
L
i(t)
P 11.5-3 Determine the complex power delivered by the
voltage source in the circuit shown in Figure P 11.5-3.
Answer: S = 7.2 + j3.6 VA
+
–
4Ω
12 cos 4t V
8Ω
2H
Figure P 11.5-3
Solution:
Let
Zp =
Next
I=
Finally
S=
8 ( j 8)
j 8 1− j 8 + j 8
=
×
=
= 4+ j4 V
8 + j 8 1+ j 1− j
2
12∠0°
12∠0°
=
= 1.342∠ − 26.6° A
4 + Z p 4 + (8 + j 8)
(12∠0° )(1.342∠ − 26.6° ) * = 7.2 + j 3.6
2
VA
P 11.5-4 Many engineers are working to develop photovoltaic power plants that provide ac
power. An example of an experimental photovoltaic system is shown in Figure P 11.5-4a. A
model of one portion of the energy conversion circuit is shown in Figure P 11.5-4b. Find the
average, reactive, and complex power delivered by the dependent source.
Answer: S = + j8/9 VA
(a)
5 mH
1Ω
+
5 cos 400t A
2Ω
v1
1 800
F
+
–
–( 1 8)v1
–
(b)
Figure P 11.5-4
Soution:
Represent the circuit in the frequency domain and label the node voltages:
The node equations are:
V1 V1 − V2
+
= 0 ⇒ V1 (1 − j ) + j V2 = 10
2
j2
⎛1 ⎞
V2 − ⎜ V1 ⎟
V2 − V1 V2
⎝ 8 ⎠ = 0 ⇒ V 0.25 + j + V 2 = 0
=
+
) 2( )
1(
j2
− j2
1
5∠0 =
Using MATLAB:
V1 = 5.33∠36.9° V
j ⎤ ⎡ V1 ⎤ ⎡10 ⎤
⎡ 1− j
⎢ 0.25 + j 2 ⎥ ⎢ V ⎥ = ⎢ 0 ⎥ ⇒ V = 2.75∠ − 67.2° V
⎣
⎦⎣ 2⎦ ⎣ ⎦
2
then
1
− V1 − V2
I = 8
= 2.66∠126.9° A
1
Now the complex power can be calculated as
(
)
* ⎛ 5.33
⎛ ⎛1⎞ ⎞
⎞
I* ⎜ −⎜ ⎟ V1 ⎟
2.667∠−126.9° ⎜ −
∠36.9° ⎟
⎝ ⎝8⎠ ⎠ =
⎝ 8
⎠ = j 0.8889 = j 8 VA
S=
2
9
2
Finally
S = P + jQ = j
8
8
⇒ P = 0, Q = VAR
9
9
(Checked using MATLAB and LNAP)
12 Ω
P 11.5-5 For the circuit shown in Figure P 11.5-5,
determine I and the complex power S delivered by
the source when V = 50 ∠120° V rms.
Answer: S = 100 + j75 VA
+
V –
I
j20 Ω
–j10 Ω
20 Ω
Figure P 11.5-5
Solution:
I=
S = VI* =
50∠120°
50∠120°
=
= 2.5∠83.13° A
16+ j12 20∠36.87°
( 50∠120° )( 2.5∠−83.13° ) =
125 ∠36.87° = 100 + j 75 VA
50 mF
P 11.5-6 For the circuit of Figure P 11.5-6,
determine the complex power of the R, L, and C
elements and show that the complex power
delivered by the sources is equal to the complex
power absorbed by the R, L, and C elements.
10 Ω
6 cos 10t A
+
–
5 cos 10t V
2H
Figure P 11.5-6
Solution:
KVL:
(10+ j 20 ) I1 = 5∠0° − j 2 I 2
⇒ (10 + j 20 ) I1 + j 2 I 2 = 5∠0°
KCL:
I1 + I 2 = 6∠0°
Solving these equations using Cramer’s rule:
Δ=
I1 =
10 + j 20
1
j2
= 10 + j18
1
1 5 j2
5 − j12
=
= 0.63∠232° A = −0.39 − j 0.5 A
Δ 6 1
10 + j18
I 2 = 6 − I1 = 6 + 0.39 + j.5 = 6.39 + j.5 = 6.41∠4.47° A
Now we are ready to calculate the powers. First, the powers delivered:
1
( 5∠0° )( −I 2* ) = 2.5 ( 6.41∠(180− 4.47) ) = −16.0 + j1.25 VA
2
1
S 6∠ 0° = [5− j 2 I 2 ]( 6∠0° ) = ⎡⎣5− j 2( 6.39+ j.5 ) ⎤⎦ 3 = 18.0− j 38.3 VA
2
S Total = S5∠ 0° + S 6∠ 0° = 2.0− j 37.2 VA
S5∠ 0° =
delivered
Next, the powers absorbed:
1
10
2
2
10 I1 = (.63) = 2.0 VA
2
2
j 20 2
S j20Ω =
I1 = j 4.0 VA
2
1
2
2
S − j2Ω = ( − j 2 ) I 2 = − j ( 6.41) = − j 41.1 VA
2
S Total = 2.0 − j 37.1 VA
S10Ω =
absorbed
To our numerical accuracy, the total complex power delivered is equal to the total complex
power absorbed.
P 11.5-7 A circuit is shown in Figure P 11.5-7 with an unknown
impedance Z. However, it is known that v(t) = 100 cos (100t + 20°) V and
i(t) = 25 cos (100t–10°) A. (a) Find Z. (b) Find the power absorbed by the
impedance. (c) Determine the type of element and its magnitude that should
be placed across the impedance Z (connected to terminals a–b) so that the
voltage v(t) and the current entering the parallel elements are in phase.
Answer: (a) 4 ∠30° Ω
(b) 1082.5 W (c) 1.25 mF
a
i(t)
+
v(t)
Z
–
b
Figure P 11.5-7
Solution:
(a)
(b)
Z=
V 100∠20°
=
= 4∠30° Ω
I 25∠ − 10°
P=
I V cos θ (100 )( 25 ) cos 30°
=
= 1082.5 W
2
2
1
= 0.25∠ − 30° = 0.2165 − j 0.125 S . To cancel the phase angle we add a capacitor
Z
having an admittance of YC = j 0.125 S . That requires ω C =0.125 ⇒ C = 1.25 mF .
(c) Y =
P 11.5-8 Find the complex power delivered
by the voltage source and the power factor
seen by the voltage source for the circuit of
Figure P 11.5-8.
+
v1
–
1Ω
4Ω
10 cos 2t V
+
–
(3 4) v1
13F
Figure P 11.5-8.
Solution:
Apply KCL at the top node to get
10 − V1
V1 3
− V1 +
= 0 ⇒ V1 = 4∠36.9° V
3
4 4
1− j
2
V
I1 = 1 = 1∠36.9° A
Then
4
The complex power delivered by the source is calculated as
−
S=
Finally
(1∠36.9° )* (10∠0° ) =
2
5∠ − 36.9° VA
pf = cos ( −36.9° ) = .8 leading
R
P 11.5-9 The circuit in Figure P 11.5-9 consists of a
source connected to a load.
(a)
Suppose R = 9 Ω and L = 5 H. Determine the
+
L
– 24 cos (3t +75°) V
average, complex, and reactive powers delivered
by the source to the load.
Source
Load
(b)
Suppose R = 15 Ω and L = 3 H. Determine the
average, complex, and reactive powers delivered
Figure P 11.5-9
by the source to the load.
(c) Suppose the source delivers 8.47 + j14.12 VA to the load. Determine the values of the
resistance, R, and the inductance, L.
(d) Suppose the source delivers 14.12 + j8.47 VA to the load. Determine the values of the
resistance, R, and the inductance, L.
Solution:
Represent the circuit in the frequency domain as
(a)
24∠75° 24∠75°
=
= 1.37∠16° A
9 + j15 17.5∠59°
24 (1.37 )
1
∠ ( 75 − 16 ) ° = 16.44∠59° = 8.47 + j14.1 VA
S = ( 24∠75° )(1.37∠16° )∗ =
2
2
I=
so P = 8.47 W and Q = 14.1 VAR
(b)
(c)
24∠75° 24∠75°
=
= 1.37∠44° A
15 + j 9 17.5∠31°
1
S = ( 24∠75° )(1.37∠44° )∗ = 16.44∠31° = 14.1 + j8.47 VA
2
I=
∗
∗
⎛ 2 ( 8.47 + j14.12 ) ⎞
⎛ 2 (16.44∠59° ) ⎞
∗
I=⎜
⎟ =⎜
⎟ = (1.37∠ − 16° ) = 1.37∠16° A
24∠75°
⎝
⎠
⎝ 24∠75° ⎠
R + j 3L =
24∠75°
= 17.5∠59° = 9 + j15 Ω
1.37∠16°
R = 9 Ω and L =
(d)
15
=5 H
3
∗
⎛ 2 (14.12 + j8.47 ) ⎞
∗
I=⎜
⎟ = (1.37∠ − 44° ) = 1.37∠44° A
24∠75°
⎝
⎠
R + j 3L =
24∠75°
= 17.5∠31° = 15 + j 9 Ω
1.37∠44°
R = 15 Ω and L =
9
=3H
3
P 11.5-10 The circuit in Figure P 11.5-10 consists of a
source connected to a load. Suppose the amplitude of the
source voltage is doubled so that
vi (t) = 48 cos (3t + 75°)V.
How will each of the following change?
(a)
The impedance of the load
(b)
The complex power delivered to the load
(c)
The load current
Solution:
Represent the circuit in the frequency domain as
Doubling the amplitude of vi(t):
(c) doubles the amplitude of the load current.
(a) does not change the impedance.
(b) multiplies the complex power by 22 = 4.
R
+
–
24 cos (3t +75°) V
Source
Load
Figure P 11.5-10
L
P 11.5-11 The circuit in Figure P 11.5-11 consists of a
source connected to a load. Suppose the phase angle of
the source voltage is doubled so that vi (t) = 24 cos (3t +
150°) V. How will the following change?
(a)
The impedance of the load
(b)
The complex power delivered to the load
(c)
The load current
Solution:
Represent the circuit in the frequency domain as
Doubling the angles of vi(t) increases the angle of vi(t) by 75°
(c) increases the angle of the load current by 75°.
(a) does not change the impedance.
(b) does not change the complex power.
R
+
–
24 cos (3t +75°) V
Source
Load
Figure P 11.5-11
L
R
P 11.5-12 The circuit in Figure P 11.5-12 consists of a
source connected to a load. The complex power delivered
by the source to the load is S = 6.61 + j1.98 VA.
Determine the values of R and C.
+
–
C
12 cos 2t V
Source
4H
Load
Figure P 11.5-12
Solution:
Represent the circuit in the frequency domain as
1
S = VI∗
2
so
⇒
∗
⎛ 2S ⎞
I=⎜ ⎟
⎝V⎠
∗
⎛ 2 ( 6.61 + j1.98 ) ⎞
∗
− j1.67
A
I (ω ) = ⎜
⎟ = (1.10 + j 0.33) = (1.10 − j 0.33) = 1.15e
j0
12
e
⎝
⎠
Using KVL gives
1
⎛
⎞
+ j8 ⎟ (1.15e − j16.7 ) = 12e j 0
⎜R− j
2C
⎝
⎠
1 ⎞
12e j 0
⎛
= 10.4e j16.7 = 10 + j 3 Ω
R + j ⎜8 −
⎟=
− j16.7
2C ⎠ 1.15e
⎝
R = 10 Ω
So
And
8−
1
=3 ⇒
2C
1
1
=5 ⇒ C =
F
2C
10
R
P 11.5-13 Design the circuit shown in Figure P 11.5-13, that is,
specify values for R and L so that the complex power delivered to
the RL circuit is 8 + j6 VA.
Answer: R = 5.76 Ω and L = 2.16 H
12 cos 2t
+
–
L
Figure P 11.5-13
Solution:
Analysis using Mathcad (ex11_5_3.mcd):
Enter the parameters of the voltage source:
A := 12
ω := 2
Enter the Average and Reactive Power delivered to the RL circuit: P := 8
The complex power delivered to the RL circuit is:
Q := 6
S := P + j ⋅Q
2
The impedance seen by the voltage source is:
Z :=
Calculate the required values of R and L R := Re( Z)
The mesh current is: I :=
A
Z
A
⎯
2 ⋅S
L :=
Im( Z)
ω
⎯
I ⋅( I ⋅Z)
The complex power delivered by the source is: Sv :=
2
⎯
I ⋅( I ⋅R)
The complex power delivered to the resistor is: Sr :=
2
⎯
I ⋅( I ⋅j ⋅ω ⋅L)
The complex power delivered to the inductor is: Sl :=
2
Verify Sv = Sr + Sl :
Sr + Sl = 8 + 6i
Sv = 8 + 6i
R = 5.76
Sv = 8 + 6i
Sr = 8
Sl = 6i
L = 2.16
P11.5-14
The source voltage in the circuit shown in Figure P11.5-14 is V s = 24 ∠ 30° V. Consequently
I 1 = 3.13 ∠ 25.4° A, I 2 = 1.99 ∠ 52.9° A and V 4 = 8.88 ∠ −10.6° V
Determine (a) the average power absorbed by Z 4, (b) the average power absorbed by Z 1, and (c)
the complex power delivered by the voltage source. (All phasors are given using peak, not RMS,
values.)
Figure P11.5-14
Solution:
a. The average power absorbed by Z 4 is P4 =
( 8.88)(1.99 ) cos
2
( −10.6° − 52.9° ) = 3.942 W .
⎛ 3.13 2 ⎞
b. The average power absorbed by Z 1 is P1 = ⎜
⎟ 4 = 19.59 W .
⎝ 2 ⎠
c. The complex power delivered by the voltage source is
S=
( 24∠30° )( 3.13∠25.4° ) * = 37.56∠4.6° = 37.44 + j 3.01 VA
2
Section 11.6 Power Factor
P 11.6-1 An industrial firm has two electrical loads connected in parallel across the power
source. Power is supplied to the firm at 4000 V rms. One load is 30 kW of heating use, and the
other load is a set of motors that together operate as a load at 0.6 lagging power factor and at 150
kVA. Determine the total current and the plant power factor.
Answer: I = 42.5 A rms and pf = 1/ 2
Solution:
Heating:
P = 30 kW
Motor:
θ = cos −1 ( 0.6 ) = 53.1° ⎪⎫
S = 150 kVA
Total (plant):
⎧ P = 150 cos 53.1° = 90 kW
⎬ ⇒ ⎨
⎩Q = 150sin 53.1° = 120 kVAR
⎭⎪
P = 30 + 90 = 120 kW ⎫
⎬ ⇒ S = 120 + j120 = 170∠45° VA
Q = 0 + 120 = 120 kVAR ⎭
The power factor is pf = cos 45° = 0.707 lagging.
The rms current required by the plant is I =
S 170 kVA
=
= 42.5 A rms .
V
4 kV
P 11.6-2 Two electrical loads are connected in parallel to a 400-V rms, 60-Hz supply. The first
load is 12 kVA at 0.7 lagging power factor; the second load is 10 kVA at 0.8 lagging power
factor. Find the average power, the apparent power, and the power factor of the two combined
loads.
Answer: Total power factor = 0.75 lagging
Solution:
Load 1:
P1 = S cosθ = (12 kVA )( 0.7 ) = 8.4 kW
Q1 = S sin ( cos −1 (.7 ) ) = (12 kVA ) sin ( 45.6° ) = 8.57 kVAR
Load 2:
P2 = (10 kVA )( 0.8 ) = 8 kW
Q2 = 10sin ( cos −1 ( 0.8 ) ) = 10sin ( 36.9° ) = 6.0 kVAR
Total:
S = P + jQ = 8.4 + 8 + j ( 8.57 + 6.0 ) = 16.4 + j14.57 = 21.9∠41.6° kVA
The power factor is pf = cos ( 41.6° ) = 0.75 . The average power is P = 16.4 kW. The
apparent power is |S| = 21.9 kVA.
P 11.6-3 The source of Figure P 11.6-3 delivers
50 VA with a power factor of 0.8 lagging. Find the
unknown impedance Z.
6Ω
20 0° V
Answer: Z = 6.39 ∠26.6° Ω
+
–
Z
j8 Ω
Figure P 11.6-3
Solution:
The source current can be calculated from the apparent
power:
2( 50∠ cos −1 0.8 )
Vs I s∗
2S
∗
⇒I =
=
= 5∠36.9° A
S=
s
20∠0°
Vs
2
I s = 5∠ − 36.9° = 4 − j3 A
Next
I1 =
Vs
20∠0°
=
= 2∠ − 53.1° = 1.2 − j1.6 A
6 + j8 10∠53.1°
I 2 = I s − I1 = 4 − j 3 − 1.2 + j1.6 = 2.8 − j1.4
= 3.13∠ − 26.6° A
Finally,
Z=
Vs
20∠0°
=
= 6.39∠26.6° Ω
I 2 3.13∠ − 26.6°
P 11.6-4 Manned space stations require several continuously available ac power sources. Also,
it is desired to keep the power factor close to 1. Consider the model of one communication
circuit shown in Figure P 11.6-4. If an average power of 500 W is dissipated in the 20Ω resistor,
find (a) Vrms, (b) Is rms, (c) the power factor seen by the source, and (d) |Vs|.
Is
Vs
+
–
I
–j20 Ω
+
j20 Ω
20 Ω
V = ⎜V ⎜ 0° V
–
IL
Figure P 11.6-4
Solution:
(Using all rms values.)
2
(a)
V
2
2
P = I R=
⇒ V = P ⋅ R = ( 500 )( 20 ) ⇒
R
(b)
Is = I + I L =
V = 100 Vrms
V
V 100∠0° 100∠0°
+
=
+
= 5 − j 5 = 5 2∠ − 45° A
20 j 20
20
j 20
(c)
Zs = − j 20 +
( 20 )( j 20 ) = 10
20 + j 20
pf = cos ( −45° ) =
(d)
2∠ − 45° Ω
1
leading
2
No average power is dissipated in the capacitor or inductor. Therefore,
PAVE = PAVE = 500 W ⇒
source
20Ω
Vs I s cos θ = 500 ⇒ Vs =
500
=
I s cos θ
(
500
= 100 V
⎛ 1 ⎞
5 2 ⎜
⎟
⎝ 2⎠
)
I
P 11.6-5 Two impedances are supplied by
V = 100 ∠160° Vrms, as shown in Figure P 11.6-5,
where I = 2 ∠190° A rms. The first load draws P1 =
23.2 W and Q1 = 50 VAR. Calculate I1, I2, the power
factor of each impedance, and the total power factor
of the circuit.
I1
+
V –
Z1
I2
Z2
Figure P 11.6-5
Solution:
Load 1:
V = 100∠160° Vrms
I = 2∠190° Arms = −1.97 − j 0.348 Arms
P1 = 23.2 W, Q1 = 50 VAR
S1 = P1 + jQ1 = 23.2 + j50 = 55.12∠65.1° VA
pf1 = cos 65.1° = 0.422 lagging
I1∗ =
Load 2:
S1 55.12∠65.1°
=
= 0.551∠ − 94.9°, so I1 = 0.551∠94.9° Arms
Vs
100∠160°
I 2 = I − I 1 = −1.97 − j 0.348 + 0.047 − j 0.549 = 2.12∠ − 155° Arms
S 2 = V I 2 * = (100∠160° )( 2.12∠155° ) * = 212∠315° = 212∠ − 45° = 150 − j 150 VA
pf 2 = cos ( −45° ) = 0.707 leading
Total:
S = S1 + S 2 = ( 23.2 + j50 ) + (150 − j150 ) = 173.2 − j100 = 200∠ − 30° VA
pf = cos ( −30° ) = 0.866 leading
P 11.6-6 A residential electric
supply three-wire circuit from a
transformer is shown in Figure P
11.6-6a. The circuit model is shown
in Figure P 11.6-6b. From its 7.2 kV
nameplate, the refrigerator motor is
known to have a rated current of 8.5
A rms. It is reasonable to assume an
inductive impedance angle of 45°
for a small motor at rated load.
Lamp and range loads are 100 W
and 12 kW, respectively.
(a)
(b)
Calculate the currents in line
1, line 2, and the neutral
wire.
Calculate: (i) Prefrig, Qrefrig,
(ii) Plamp, Qlamp, and (iii)
Ptotal, Qtotal, Stotal, and overall
power factor.
Line 1
120 Vrms
Refrig
Neutral
Kitchen
range
Lamp
120 Vrms
Line 2
+
–
(a)
Refrig
120 0° Vrms
Range
+
–
Lamp
120 0° Vrms
(b)
(c)
The neutral connection resistance increases, because of corrosion and looseness, to 20 Ω.
(This must be included as part of the neutral wire.) Use mesh analysis and calculate the voltage
across the lamp.
Solution:
(Using rms values)
Z refrig =
120
= 14.12 Ω
8.5
Z refrig = 14.12∠45° = 10 + j10 Ω
V 2 (120 )
=
=
= 144 Ω
100
P
2
R lamp
R range
(a)
Ι refrig =
and
( 240 )
=
2
12, 000
= 4.8 Ω
120∠0°
120∠0°
= 8.5∠ − 45° Arms , I lamp =
= 0.83∠0° Arms
10 + j10
144
I range =
From KCL:
240∠0°
= 50∠0° Arms
4.8
I1 = I refrig + I range = 56 − j 6 = 56.3∠ − 6.1° Arms
I 2 = −I lamp − I range = 50.83∠180° Arms
I N = −I1 − I 2 = 7.92∠131° Arms
(b)
Prefrig = I refrig
2
2
Rrefrig = 722.5 W and Qrefrig = I refrig X refrig = 722.5 VAR
Plamp = 100 W and Q lamp = 0
Ptotal = 722 + 100 + 12, 000 = 12.82 kW ⎫
⎬ ⇒ S = 12,822 + j 722 = 12.84∠3.2° kVA
Qtotal = 722 + 0 + 0 = 722 VAR
⎭
The overall power factor is pf = cos ( 3.2° ) = 0.998 ,
(c)
Mesh equations:
− 20
⎡ 30 + j10
⎢ − 20
164
⎢
⎢⎣ −10 − j10 − 144
Solve to get:
− 10 − j10 ⎤ ⎡ I A ⎤ ⎡120∠0° ⎤
− 144 ⎥⎥ ⎢⎢ I B ⎥⎥ = ⎢⎢120∠0° ⎥⎥
⎥⎦
158.8 + j10 ⎥⎦ ⎢⎣ I C ⎥⎦ ⎢⎣ 0
I A = 64.3 − j1.57 = 64.3∠ − 1.7° Arms
I B = 61.3 − j 0.19 = 61.3∠ − 0.5°Arms
I C = 60 + j 0 = 60∠0° Arms
The voltage across the lamp is
Vlamp = Rlamp I B − IC = 144 1.27∠ − 8.6° = 183.2 Vrms
(checked: MATLAB 7/20/04)
P 11.6-7 A motor connected to a 220-V supply line from the power company has a current of
7.6 A. Both the current and the voltage are rms values. The average power delivered to the motor
is 1317 W.
(a)
Find the apparent power, the reactive power, and the power factor when ω = 377 rad/s.
(b)
Find the capacitance of a parallel capacitor that will result in a unity power factor of the
combination.
(c)
Find the current in the utility lines after the capacitor is installed.
Answer: (a) pf = 0.788 (b) C = 56.5 μF (c) I = 6.0 A rms
Solution:
(Using all rms values)
(a)
VI = 220 ( 7.6 ) = 1672 VA
P 1317
=
= .788
VI 1672
θ =cos −1pf = 38.0° ⇒ Q = VIsinθ =1030 VAR
pf =
(b)
To restore the pf to 1.0, a capacitor is required to eliminate Q by introducing –Q, then
V2
(220)2
⇒ X c = 47Ω
1030 =
=
Xc
Xc
∴C= 1
(c)
1
ω X = (377)(47) = 56.5μ F
P = VI cosθ where θ = 0°
then 1317 = 220I
∴ I = 6.0Arms for corrected pf
*
Note I = 7.6Arms for uncorrected pf .
P 11.6-8 Two loads are connected in parallel across a 1000-V rms, 60-Hz source. One load
absorbs 500 kW at 0.6 power factor lagging, and the second load absorbs 400 kW and 600
kVAR. Determine the value of the capacitor that should be added in parallel with the two loads
to improve the overall power factor to 0.9 lagging.
Answer: C = 2.2 μF
Solution:
First load:
(
)
S 1 = P + jQ = P 1 + j tan ( cos −1 (.6 ) ) = 500 (1 + j tan 53.1° ) = 500 + j 667 kVA
S 2 = 400 + j 600 kVA
Second load:
Total:
S = S1 + S 2 = 900 + j1267 kVA
S desired = P + jP tan (cos −1 (.90)) = 900 + j 436 VA
From the vector diagram: S desired = S + jQ . Therefore
900 + j 436 = 900 + j1277 + j Q ⇒ Q = −841 VAR
2
2
V
V
(1000) 2
j
*
841
=
−
⇒
=
=
= j1189 ⇒ Z = − j1189 = −
j
Z
*
377 C
− j841 − j841
Z
Finally,
C=
1
= 2.20 μ F
(1189)(377)
6.4 Ω
P 11.6-9 A voltage source with a complex internal
impedance is connected to a load, as shown in Figure V
P 11.6-9. The load absorbs 1 kW of average power at 1
100 V rms with a power factor of 0.80 lagging. The
source frequency is 200 rad/s.
24 mH
+
+
–
ZL
VL
–
Figure P 11.6-9
(a)
Determine the source voltage V1.
(b)
Find the type and value of the element to be placed in parallel with the load so that
maximum power is transferred to the load.
Solution:
(a)
S = P + jQ = P + jP tan (cos −1 pf ) = 1000 + j1000 tan (cos −1 0.8) = 1000 + j 750 VA
S 1000+ j750
=
= 10 + j7.5 ⇒ I = 10 − j 7.5 Arms
Let VL = 100∠0° Vrms. Then I* =
VL
100∠0°
ZL =
(b)
VL
100∠0°
=
= 8∠36.9° = 6.4 + j 4.8 Vrms
I 12.5∠−36.9°
V1 =[6.4 + j (200)(.024) + Z L )(I ) = (12.8 + j 9.6)(10 − j 7.5) = 200∠0° Vrms
1
For maximum power transfer, we require ( 6.4 + j 4.8 )* = Z L || Z new =
.
YL +Ynew
1
1
1
= YL + Ynew ⇒ Ynew =
−
= j 0.15 S
(6.4 − j4.8)
6.4 − j 4.8 6.4 + j 4.8
Then Z new = − j 6.67 Ω so we need a capacitor given by
1
1
= 6.67 ⇒ C =
= 750 μ F
(6.67)(200)
ωC
P 11.6-10 The circuit shown in Figure P 11.6-10a can be represented in the frequency domain
as shown in Figure P 11.6-10b. In the frequency domain, the value of the mesh current is I =
1.076 ∠–38.3° A.
(a)
Determine the complex power supplied by the voltage source.
(b)
Given that the complex power received by Z1, is 6.945 + j 13.89 VA, determine the
values of R1 and L1.
(c)
Given that the real power received by Z3 is 4.63 W at a power factor of 0.56 lagging,
determine the values of R3 and L3.
Z1 = R1 + jωL1
R1
L1
R2
+
–
48 cos (6t) V
R3
48 0° V
+
–
I
Z2 = R2 − j
C2
L3
Z3 = R3 + jωL3
(a)
(b)
Figure P 11.6-10
Solution:
( 48∠0° )(1.076∠ − 38.3° )
(a) S =
*
2
(b) V1 =
Z1 =
2 S1
I
*
V1
I
=
=
2 ( 6.945 + j 13.89 )
(1.076∠ − 38.3° )
*
= 25.82∠38.3° = 20.27 + j 16 VA
=
2 (15.53∠63.4° )
= 28.87∠25.1° V
(1.076∠38.3° )
28.87∠25.1°
= 26.83∠63.4° = 12 + j 24 = 12 + j 6 ( 4 ) Ω
1.076∠ − 38.3°
R1 = 12 Ω and L1 = 4 H.
(c) θ = cos −1 ( 0.56 ) = 56° , S =
V3 =
Z3 =
2 ( 8.268∠56° )
(1.076∠ − 38.3° )
*
V3
I
=
P 4.63
=
= 8.268 VA
pf 0.56
= 15.37∠17.7° V
15.37∠17.7°
= 14.28∠56° = 8 + j 11.83 = 8 + j 6 (1.97 ) Ω
1.076∠ − 38.3°
R3 = 8 Ω and L3 = 2 H.
1
ωC2
R
P 11.6-11 The circuit in Figure P 11.6-11 consists
of a source connected to a load. The source delivers
14.12 W to the load at a power factor of 0.857
lagging. What are the values of the resistance, R, and
the inductance, L?
+
–
24 cos (3t +75°) V
Source
L
Load
Figure P 11.6-11
Solution:
Represent the circuit in the frequency domain as
⎧0.857 = cos (θ )
⎪
pf = 0.857 lagging ⇒ ⎨
and
⎪
θ >0
⎩
so
θ = 31°.
Next
14.12 = P = S cos θ = S ( 0.857 )
so
S =
Then
and
so
14.12
= 16.48 VA
0.857
S = 16.48∠31° = 14.12 + j8.49
∗
∗
⎡ 2 (16.48∠31° ) ⎤
⎛ 2S ⎞
I=⎜ ⎟ =⎢
⎥ = 1.37∠44°
24
75
∠
°
⎝V⎠
⎣
⎦
24∠75°
R + j 3L =
= 17.5∠31° = 15 + j 9 Ω
1.37∠44°
R = 15 Ω
L=3H
P 11.6-12 The circuit in Figure P 11.6-12
consists of a source connected to a load.
Determine the impedance of the load and the
complex power delivered by the source to the
load under each of the following conditions:
+
–
A
24 cos (3t + 75°) V
Source
B
Load
Figure P 11.6-12
(a)
(b)
(c)
(d)
The source delivers 14.12 + j8.47 VA to load A and 8.47 + j14.12 VA to load B.
The source delivers 8.47 + j14.12 VA to load A, and the impedance of load B is 15 + j9
Ω.
The source delivers 14.12 W to load A at a power factor of 0.857 lagging, and the
impedance of load B is 9 + j15 Ω.
The impedance of load A is 15 + j9 Ω, and the impedance of load B is 9 + j15 Ω.
Solution:
Represent the circuit in the frequency domain as
(a)
∗
⎛ 2 (14.12 + j8.47 ) ⎞
IA = ⎜
⎟ = 1.37∠44° A
24∠75°
⎝
⎠
∗
⎛ 2 ( 8.47 + j14.12 ) ⎞
IB = ⎜
⎟ = 1.37∠16° A
24∠75°
⎝
⎠
I = I A + I B = (1.37∠44° ) + (1.37∠16° ) = ( 0.986 + j 0.954 ) + (1.319 + j 0.377 )
= 2.305 + j1.331 = 2.662∠30° A
Z=
S=
24∠75°
= 9.016∠45°
2.662∠30°
1
∗
( 24∠75° )( 2.662∠30° ) = 31.9∠45° = 22.59 + j 22.59 VA
2
(b)
∗
⎛ 2 ( 8.47 + j14.12 ) ⎞
IA = ⎜
⎟ = 1.37∠16° A
24∠75°
⎝
⎠
IB =
24∠75°
= 1.37∠44° A
15 + j 9
I = I A + I B = 2.662∠30° A
Z=
24∠75°
= 9.016∠45° Ω
2.662∠30°
S = 22.59 + j 22.59 VA
(c)
P = 14.12 W =
0.857 = cos ( 75 − θ A ) ⎫⎪
⎬
75 − θ A > 0
⎪⎭
Then
IA =
24 I A
2
⇒
cos ( 75 − θ A )
θ A = 75° − 31° = 44°
2 (14.12 )
= 1.37
24 cos ( 31° )
so
I A = 1.37∠44° A
Also
24∠75°
= 137∠16° A
9 + j15
I = I A + I B = 2.662∠30° A
IB =
(d)
24∠75°
= 1.37∠44°
15 + j 9
24∠75°
IB =
= 1.37∠16°
9 + j15
I = I A + I B = 2.662∠30° A
IA =
Z=
24∠75°
= 9.016∠45° Ω
2.662∠30°
S = 22.59 + j 22.59 VA
P 11.6-13 Figure P 11.6-13 shows two possible
representations of an electrical load. One of these
representations is used when the power factor of the load is
lagging, and the other is used when the power factor is
leading. Consider two cases:
(a)
At the frequency ω = 4 rad/s, the load has the power
factor pf = 0.8 lagging.
(b)
At the frequency ω = 4 rad/s, the load has the power
factor pf = 0.8 leading.
R
R
L
C
Figure P 11.6-13
In each case, choose one of the two representations of the load. Let R = 6 Ω and determine the
value of the capacitance, C, or the inductance, L.
Solution:
⎛X⎞
Let Z = R + j X = R 2 + X 2 ∠ tan −1 ⎜ ⎟ be the impedance of the load. Further, let V = A∠θ
⎝R⎠
and I = B∠φ be the voltage across and current through the load. Then
A∠θ A
⎛X⎞
= ∠ (θ − φ )
R 2 + X 2 ∠ tan −1 ⎜ ⎟ = R + j X = Z =
B∠φ B
⎝R⎠
⎛X⎞
Equating angles gives
tan −1 ⎜ ⎟ = θ − φ
⎝R⎠
Also, the complex power delivered to the load is
S = ( A∠θ )( B∠φ ) * =
So
AB
AB
cos (θ − φ ) + j
sin (θ − φ )
2
2
pf = cos (θ − φ ) ⇒ θ − φ = cos −1 ( pf )
(Remark: cos −1 ( pf ) is positive when the power factor is lagging and negative when the power
factor is leading.)
a)
pf = 0.8 lagging ⇒ cos −1 ( 0.8 ) = 36.9°
X
= tan ( 36.9 ) = 0.75 ⇒ X = 6 ( 0.75 ) = 4.5 Ω
R
Choose the inductor to implement the positive reactance. Then
We require
4.5 = ω L = 4 L ⇒ L = 1.125 H
b)
pf = 0.8 leading ⇒ cos −1 ( 0.8 ) = −36.9°
X
= tan ( −36.9 ) = −0.75 ⇒ X = 6 ( −0.75 ) = −4.5 Ω
R
Choose the capacitor to implement the negative reactance. Then
We require
−4.5 = −
1
1
=−
4C
ωC
⇒ C=
8
= 0.889 F
9
P 11.6-14 Figure P 11.6-14 shows two possible
representations of an electrical load. One of these
representations is used when the power factor of the load
L
R
C
R
is lagging, and the other is used when the power factor is
leading. Consider two cases:
(a)
At the frequency ω = 4 rad/s, the load has the
Figure P 11.6-14
power factor pf = 0.8 lagging.
(b)
At the frequency ω = 4 rad/s, the load has the
power factor pf = 0.8 leading.
In each case, choose one of the two representations of the load. Let R = 6 Ω and determine the
value of the capacitance, C, or the inductance, L.
Solution:
1
1
1
1
1
1
⎛ R⎞
= −j =
+ 2 ∠ tan −1 ⎜ − ⎟ be the admittance of the load. Further,
Let Y = +
2
R jX R
X
R
X
⎝ X⎠
let V = A∠θ and I = B∠φ be the voltage across and current through the load. Then
1
1
B∠φ B
⎛ R⎞
+ 2 ∠ tan −1 ⎜ − ⎟ = Y =
= ∠ (φ − θ )
2
R
X
A∠θ A
⎝ X⎠
⎛ R ⎞
tan −1 ⎜ −
⎟ = (φ − θ )
⎝ ω L⎠
Also, the complex power delivered to the load is
Equating angles gives
S = ( A∠θ )( B∠φ ) * =
AB
AB
cos (θ − φ ) + j
sin (θ − φ )
2
2
pf = cos (θ − φ ) ⇒ θ − φ = cos −1 ( pf )
So
(Remark: cos −1 ( pf ) is positive when the power factor is lagging and negative when the power
factor is leading.)
Therefore
R
⎛ R⎞
tan −1 ⎜ − ⎟ = (φ − θ ) = − cos −1 ( pf ) ⇒ − = tan ( − cos −1 ( pf ) ) = − tan ( cos −1 ( pf ) )
X
⎝ X⎠
R
⇒ = tan ( cos −1 ( pf ) )
X
a)
pf = 0.8 lagging ⇒ cos −1 ( 0.8 ) = 36.9°
R
6
= tan ( 36.9 ) = 0.75 ⇒ X =
=8 Ω
0.75
X
Choose the inductor to implement the positive reactance. Then
We require
8=ω L = 4L ⇒ L = 2 H
pf = 0.8 leading ⇒ cos −1 ( 0.8 ) = −36.9°
b)
R
6
= tan ( −36.9 ) = −0.75 ⇒ X =
= −8 Ω
−0.75
X
Choose the capacitor to implement the negative reactance. Then
We require
−8 = −
1
1
=−
4C
ωC
⇒ C=
1
= 31.25 mF
32
R
P 11.6-15 Figure P 11.6-15 shows two electrical loads.
Express the power factor of each load in terms of ω, R,
and L.
R
L
L
(a)
(b)
Figure P 11.6-15
Solution:
Let V = A∠θ and I = B∠φ be the voltage across and current through the load. The complex
power delivered to the load is
S = ( A∠θ )( B∠φ ) * =
So
AB
AB
cos (θ − φ ) + j
sin (θ − φ )
2
2
pf = cos (θ − φ ) ⇒ θ − φ = cos −1 ( pf )
(Remark: cos −1 ( pf ) is positive when the power factor is lagging and negative when the power
factor is leading.)
a.) Let
Also
Equating angles gives
⎛ω L ⎞
Z = R + j ω L = R 2 + ω 2 L2 ∠ tan −1 ⎜
⎟
⎝ R ⎠
A∠θ A
Z=
= ∠ (θ − φ )
B∠φ B
⎛ω L ⎞
tan −1 ⎜
⎟ = θ −φ
⎝ R ⎠
⎛ −1 ⎛ ω L ⎞ ⎞
⎛ω L⎞
−1
tan −1 ⎜
⎟ = θ − φ = cos ( pf ) ⇒ pf = cos ⎜ tan ⎜
⎟ ⎟ lagging
⎝ R ⎠
⎝ R ⎠⎠
⎝
⎛ω L ⎞
(The power factor is lagging because tan −1 ⎜
⎟ is an angle in the first quadrant.)
⎝ R ⎠
so
b.) Let
Y=
1
1
1
1
+
= −j
=
R jω L R
ωL
Also
⎛ R ⎞
1
1
+ 2 2 ∠ tan −1 ⎜ −
⎟
2
R ω L
⎝ ω L⎠
B∠φ B
= ∠ (φ − θ )
A∠θ A
⎛ R ⎞
tan −1 ⎜ −
⎟ = (φ − θ )
⎝ ω L⎠
Y=
Equating angles gives
⎛ R ⎞
R
−1
= tan ( − cos −1 ( pf ) ) = − tan ( cos −1 ( pf ) )
tan −1 ⎜ −
⎟ = (φ − θ ) = − cos ( pf ) ⇒ −
ωL
⎝ ω L⎠
⎛
⎛ R ⎞⎞
so
pf = cos ⎜ tan −1 ⎜
⎟ ⎟ lagging
⎝ ω L ⎠⎠
⎝
⎛ R ⎞
(The power factor is lagging because tan −1 ⎜
⎟ is an angle in the first quadrant.)
⎝ω L ⎠
R
P 11.6-16 Figure P 11.6-16 shows two electrical loads.
Express the power factor of each load in terms of ω, R,
and C.
R
C
C
(b)
(a)
Figure P 11.6-16
Solution:
Let V = A∠θ and I = B∠φ be the voltage across and current through the load. The complex
power delivered to the load is
AB
AB
S = ( A∠θ )( B∠φ ) * =
cos (θ − φ ) + j
sin (θ − φ )
2
2
So
pf = cos (θ − φ ) ⇒ θ − φ = cos −1 ( pf )
(Remark: cos −1 ( pf ) is positive when the power factor is lagging and negative when the power
factor is leading.)
a.) Let
Also
Z = R+
1
jω C
= R− j
⎛
1
1
1 ⎞
= R 2 + 2 2 ∠ tan −1 ⎜ −
⎟
ωC
ω C
⎝ ωCR⎠
Z=
A∠θ A
= ∠ (θ − φ )
B∠φ B
⎛
1 ⎞
tan −1 ⎜ −
⎟ =θ −φ
⎝ ωCR⎠
Equating angles gives
So
⎛
1 ⎞
−1
tan −1 ⎜ −
⎟ = θ − φ = cos ( pf ) ⇒
⎝ ωCR⎠
⎛
⎛
1 ⎞⎞
pf = cos ⎜ tan −1 ⎜ −
⎟ ⎟ leading
⎝ ω C R ⎠⎠
⎝
⎛ −1 ⎞
(The power factor is leading because tan −1 ⎜
⎟ is an angle in the third quadrant.)
ω
C
R
⎝
⎠
Y=
b.) Let
Also
Equating angles gives
So
1
1
+ jω C =
+ ω 2 C 2 ∠ tan −1 (ω C R )
2
R
R
B∠φ B
Y=
= ∠ (φ − θ )
A∠θ A
tan −1 (ω C R ) = (φ − θ )
tan −1 (ω C R ) = − cos −1 ( pf ) ⇒
pf = cos ( − tan −1 (ω C R ) ) leading
(The power factor is leading because − tan −1 (ω C R ) is an angle in the third quadrant.)
P11.6-17
The source voltage in the circuit shown in Figure P11.6-17 is V s = 24 ∠ 30° V. Consequently
I 1 = 3.13 ∠ 25.4° A, I 2 = 1.99 ∠ 52.9° A and V 4 = 8.88 ∠ −10.6° V
Determine (a) the average power absorbed by Z 4, (b) the average power absorbed by Z 1, and (c)
the complex power delivered by the voltage source. (All phasors are given using peak, not RMS,
values.)
Figure P11.6-17
Solution:
⎛
⎛ −2 ⎞ ⎞
a. The power factor of Z 1 is pf1 = cos ⎜ tan −1 ⎜ ⎟ ⎟ = cos ( −26.6° ) = 0.894 leading .
⎝ 4 ⎠⎠
⎝
⎛
⎛ 8 ⎞⎞
b. The power factor of Z 3 is pf 3 = cos ⎜ tan −1 ⎜ ⎟ ⎟ = cos ( 69.4° ) = 0.352 lagging .
⎝ 3 ⎠⎠
⎝
c. The power factor of Z 4 is pf 4 = cos ( −10.6° − 52.9° ) = cos ( −63.5° ) = 0.45 leading .
P11.6-18 The current source the circuit shown in Figure P11.6-18 supplies 131.16 – j36.048 VA
and the voltage source supplies 64.2275 – 87.8481VA. Determine the values of the impedances
Z1 and Z2.
Figure P11.6-18
Solution:
The current source this circuit supplies 131.16 – j36.048 VA and the voltage source supplies
64.2275 – 87.8481VA. Determine the values of the impedances Z1 and Z2.
The voltage V1 is calculated from the current source current and complex power:
2S 2 (131.16 − j 36.048 )
=
= 136.02∠ − 0.368° V
I*
( 2∠15° ) *
The current I2 is calculated from the voltage source voltage and complex power:
V1 =
*
*
⎛ 2S ⎞ ⎛ 2 ( 64.2275 + j87.8481) ⎞
I2 = ⎜ ⎟ = ⎜
⎟ = 2.1765∠ − 83.829° A
100∠ − 30°
⎝V⎠ ⎝
⎠
The impedances are calculated as
Z1 =
Z2 =
V1
2∠15° + I 2
=
V1 − 100∠ − 30°
−I 2
136.02∠ − 0.368°
= 40 + j 30 Ω
2∠15° + ( 2.1765∠ − 83.829° )
=
136.02∠ − 0.368° − 100∠ − 30°
= 20 − j 25 Ω
−2.1765∠ − 83.829°
Section 11.7 The Power Superposition Principle
14 A
P 11.7-1 Find the average power absorbed by
the 2-Ω resistor in the circuit of Figure P 11.7-1.
Answer: P = 413 W
+
110 cos 20t V
2Ω
–
12 Ω
1 100 F
Figure P 11.7-1
Solution:
Use superposition since we have two different
frequency sources. First consider the dc source
(ω = 0):
⎛ 12 ⎞
I1 = 14 ⎜
⎟ = 12 A
⎝ 12 + 2 ⎠
P1 = I12 R = (12) 2 (2) = 288 W
Next, consider the ac source (ω = 20 rad/s):
After a source transformation, current division gives
− j 60
⎡
⎤
⎢
⎥
25
(12 − j 5)
I 2 = −9.167 ⎢
∠116.6° A
⎥ =
5
⎢ − j 60 + 2+ j 4 ⎥
⎢⎣ (12 − j 5)
⎥⎦
2
Then
I
(125)(2)
P2 = 2 (2) =
=125 W
2
2
Now using power superposition
P = P1 + P2 = 288 + 125 = 413 W
0.2 H
+
P 11.7-2 Find the average power absorbed by the 8-Ω resistor in the circuit of Figure P 11.7-2.
Answer: P = 22 W
5 cos 2000t A
18
–
40 cos 8000t V
mF
1 mH
8Ω
Figure P 11.7-2
Solution:
Use superposition since we have two different frequency sources. First consider ω = 2000 rad/s
source:
Current division yields
⎡ 8 ⎤
⎢ −j2 ⎥ 5
I1 = 5 ⎢
∠63.4° A
⎥=
5
⎢ 8 +8 ⎥
⎢⎣ − j 2 ⎥⎦
2
Then
I 8
P1 = 1
= 20 W
2
Next consider ω = 8000 rad/s source.
Current division yields
⎡ 8 ⎤
⎢ j7 ⎥
5
I 2 = − j5 ⎢
∠ − 171.9° A
⎥=
8
50
⎢ +8 ⎥
⎢⎣ j 7 ⎥⎦
2
Then
I 8
P2 = 2
=2W
2
Now using power superposition
P = P1 + P2 = 22 W
P 11.7-3 For the circuit shown in Figure P 11.7-3, determine the average power absorbed by
each resistor, R1 and R2. The voltage source is vs = 10 + 10 cos (5t + 40°) V, and the current
source is is = 4 cos (5t – 30°) A.
2H
10 Ω
6i2
R1
R2
5 Ω
+
–
is
1 10
vs
F
i2
Figure P 11.7-3
P11.7-3
Use superposition since we have two different frequencies. First consider the dc source (ω = 0):
⎛1⎞
i 2 (t ) = 0 and i1 (t ) = 10 ⎜ ⎟ = 1 A
⎝ 10 ⎠
PR1 = i12 R1 = 12 (10) = 10 W
PR 2 = 0 W
Next consider ω = 5 rad/s sources.
Apply KCL at the top node to get
(10I1 −10∠40°)
=0
j10
−10 I1 + (5− j 2) I 2 = 0
−6 I 2 + I1 + I 2 − ( 4 ∠−30° ) +
Apply KVL to get
Solving these equations gives
I1 = −0.56 ∠ − 64.3° A and I 2 = −1.04∠ − 42.5° A
Then
PR1 =
I12 R1
2
( 0.56 )
=
2
2
(10)
= 1.57 W and PR 2 =
I 22 R 2
2
(1.04 )
=
2
Now using power superposition
PR = 10 + 1.57 = 11.57 W and PR 2 = 0 + 2.7 = 2.7 W
1
2
(5)
= 2.7 W
P 11.7-4 For the circuit shown in Figure P 11.7-4,
determine the effective value of the resistor voltage
vR and the capacitor voltage vC.
+
4 cos 10t V
+
–
2 Ω
6 sin 5t V
+
–
20 mF
Figure P 11.7-4
Solution:
Use superposition since we have two different frequencies. First consider the ω = 10 rad/s
source:
V1 4∠0°
=
= 0.28 + j 0.7 A
Z 2− j5
V = 2 I1 = 2(0.28 + j.7) = 0.56 + j1.4 = 1.51 ∠68.2° V
I1 =
R1
V = − j 5 I1 = 3.77 ∠ − 21.8° V
C1
Next consider ω = 5 rad/s source.
V2 6∠−90°
=
= 0.577 − j 0.12 A
Z
2 − j10
VR = 2 I 2 = 2(.577 − j 0.12) =1.15− j 0.24
I2 =
2
V
C2
=1.17∠−11.8° V
= − j10I 2 =5.9∠258.3° V
Now using superposition
vR (t) = 1.51cos (10 t + 68.2°) + 1.17 cos (5 t − 11.8°) V
vC (t) = 3.77 cos (10 t − 21.8°) + 5.9 cos (5 t − 258.3°) V
Then
2
2
2
Reff
⎛ 1.51 ⎞ ⎛ 1.17 ⎞
=⎜
⎟ +⎜
⎟ = 1.82 ⇒ VReff =1.35 V
⎝ 2⎠ ⎝ 2 ⎠
2
Ceff
⎛ 3.77 ⎞ ⎛ 5.9 ⎞
=⎜
⎟ +⎜
⎟ = 24.52 ⇒ VCeff = 4.95 V
⎝ 2 ⎠ ⎝ 2⎠
V
2
V
2
vR
–
+
vC
–
Section 11.8 The Maximum Power Transfer Theorem
P 11.8-1 Determine values of R and L for the circuit shown in Figure P 11.8-1 that cause
maximum power transfer to the load.
Answer: R = 800 Ω and L = 1.6 H
4000 Ω
R
5 cos (1000t + 60°) V
+
–
0.5 μF
L
Source
Load
Figure P 11.8-1
Solution:
Z t = 4000 || − j 2000 = 800 − j1600 Ω
Z L = Z*t =800+ j1600 Ω
⎧ R =800 Ω
R + j1000 L = 800 + j1600 ⇒ ⎨
⎩ L =1.6 H
P 11.8-2 Is it possible to choose R and L for the circuit shown in Figure P 11.8-2 so that the
average power delivered to the load is 12 mW?
Answer: Yes
R
2.5 cos 100t mA
0.2 μF
25 kΩ
Source
L
Load
Figure P 11.8-2
Solution:
Z t = 25, 000 || − j 50, 000 = 20, 000 − j10, 000 Ω
Z L =Z*t = 20,000+ j10,000 Ω
⎧ R = 20 kΩ
⎪
R + jω L = 20, 000 + j10, 000 ⇒ ⎨100 L =10,000
⎪
L =100 H
⎩
After selecting these values of R and L,
2
Since Pmax
⎛ 0.14×10−2 ⎞
3
I = 1.4 mA and Pmax = ⎜
⎟ ( 20×10 ) = 19.5 mW
2
⎝
⎠
> 12 mW , yes, we can deliver 12 mW to the load.
P 11.8-3 The capacitor has been added to the load in the circuit shown in Figure P 11.8-3 in
order to maximize the power absorbed by the 4000-Ω resistor. What value of capacitance should
be used to accomplish that objective?
Answer: 0.1 μF
800 Ω
+
–
0.32 H
5 cos (5000t + 45°) V
Source
4000 Ω
C
Load
Figure P 11.8-3
Solution:
⎛ −j ⎞
R⎜
ω C ⎟⎠ R − jω R 2C
Z t = 800 + j1600 Ω and Z L = ⎝
=
j
1 + (ω RC ) 2
R−
ωC
⎛ −j ⎞
R⎜
ω C ⎟⎠ R − jω R 2C
*
⎝
ZL = Zt ⇒
=
= 800 − j1600 Ω
j
1+ (ω RC ) 2
R−
ωC
Equating the real parts gives
800 =
R
4000
=
2
1+ (ω RC ) 1+[(5000)(4000)C ]2
⇒ C = 0.1 μ F
P 11.8-4 What is the value of the average power delivered to the 2000-Ω resistor in the circuit
shown in Figure P 11.8-4? Can the average power delivered to the 2000-Ω resistor be increased
by adjusting the value of the capacitance?
Answer: 8 mW. No.
400 Ω
5 cos 1000t V
0.8 H
+
–
2000 Ω
Source
1 μF
Load
Figure P 11.8-4
Solution:
Z t = 400 + j 800 Ω and Z L = 2000 || − j1000 = 400 − j 800 Ω
Since Z L = Z*t the average power delivered to the load is maximum and cannot be increased by
adjusting the value of the capacitance. The voltage across the 2000 Ω resistor is
VR = 5
ZL
= 2.5 − j 5 = 5.59e − j 63.4 V
Zt +ZL
2
⎛ 5.59 ⎞ 1
P=⎜
= 7.8 mW
⎟
⎝ 2 ⎠ 2000
So
is the average power delivered to the 2000 Ω resistor.
P 11.8-5 What is the value of the resistance R in Figure P 11.8-5 that maximizes the average
power delivered to the load?
R
5 cos 1000t V
0.8 H
+
–
2000 Ω
Source
1 μF
Load
Figure P 11.8-5
Solution:
Notice that Zt,not ZL, is being adjusted .When Zt is fixed, then the average power delivered to
the load is maximized by choosing ZL = Zt*. In contrast, when ZL is fixed, then the average
power delivered to the load is maximized by minimizing the real part of Zt. In this case, choose
R = 0. Since no average power is dissipated by capacitors or inductors, all of the average power
provided by source is delivered to the load.
Section 11.9 Coupled Inductors
M
a
P 11.9-1 Two magnetically coupled coils are connected as
shown in Figure P 11.9-1. Show that an equivalent inductance at
terminals a–b is Lab = L1 + L2 – 2M.
L1
L2
b
Figure P 11.9-1
Solution:
Vs = I jω L1 − I jω M + I jω L2 − I jω M
⇒ jω ( L1 + L2 − 2 M ) =
Vs
I
Therefore
Lab = L1 + L2 − 2M
a
P 11.9-2 Two magnetically coupled coils are shown connected in
Figure P 11.9-2. Find the equivalent inductance Lab.
M
L1
b
Figure P 11.9-2
Solution:
The coil voltages are given by:
V1 = I1 jω L1 + I 2 jω M
V2 = I 2 jω L2 + I1 jω M
From KVL the coil voltages are equal so
I1 jω L1 + I 2 jω M = I 2 jω L2 + I1 jω M
⇒ I1 jω ( L1 − M ) = I 2 jω ( L2 − M )
⎛ L −M
⇒ I1 = ⎜ 2
⎝ L1 − M
⎞
⎟ I2
⎠
From KCL
⎛ L −M
⎞
⎛ L + L −2 M ⎞
I s = I1 + I 2 = ⎜ 2
+ 1⎟ I 2 = ⎜ 2 1
⎟ I2
⎝ L1 − M
⎠
⎝ L1 − M
⎠
Next
⎛ L1 − M
I2 = ⎜
⎜ L2 + L1 − 2 M
⎝
⎞
⎛ L −M
⎟⎟ I s and I1 = ⎜ 2
⎝ L1 − M
⎠
⎞
⎛ L2 − M
⎟ I2 = ⎜
⎠
⎝ L1 − M
⎞ ⎛ L1 − M
⎟ ⎜⎜
⎠ ⎝ L2 + L1 − 2 M
⎞
⎟⎟ I s
⎠
Substituting gives
⎛ L2 − M ⎞
⎛ L1 − M
⎞
V1 = jω L1 I1 + jω M I 2 = jω M ⎜
I + jω M ⎜
I
⎜ L2 + L1 − 2 M ⎟⎟ s
⎜ L2 + L1 − 2 M ⎟⎟ s
⎝
⎠
⎝
⎠
⎛ M ( L2 − M ) + M ( L1 − M ) ⎞
⎛ L1 L 2 − M 2
j
= jω ⎜
I
=
ω
⎟⎟ s
⎜⎜
⎜
L2 + L1 − 2 M
⎝
⎠
⎝ L1 + L 2 − 2 M
⎛ L1 L 2 − M 2 ⎞
V1
Z=
= jω ⎜
Finally
⎜ L1 + L 2 − 2 M ⎟⎟
Is
⎝
⎠
⎞
⎟⎟ I s
⎠
L2
P 11.9-3 The source voltage of the circuit
shown in Figure P 11.9-3 is vs = 141.4 cos 100t
V. Determine i1(t) and i2(t).
2Ω
vs +
–
M = 0.6 H
i1
0.4 H
i2
1.6 H
200 Ω
Figure P 11.9-3
Solution:
Mesh equations:
−141.4∠0° + 2 I1 + j 40 I1 − j 60 I 2 = 0
200 I 2 + j160 I 2 − j 60 I1 = 0 ⇒ I 2 = (0.23∠51°) I1
Solving yields
I1 = 4.17∠ − 68° A and I 2 = 0.96∠ − 17°A
Finally
i1 ( t ) = 4.17 cos(100t −68°) A and i 2 ( t ) = 0.96 cos(100t −17°) A
(checked: LNAPAC 7/21/04)
P 11.9-4 A circuit with a mutual inductance is shown in Figure P 11.9-4. Find the voltage V2
when ω = 5000.
10 Ω
V1 = 10 0° V
+
–
M = 10 mH
1 mH
100 mH
400 Ω
+
V2
–
Figure P 11.9-4
Solution:
Mesh equations:
(10+ j5 ) I 1 − j50 I 2 = 10
− j 50 I1 + ( 400+ j 500 ) I 2 = 0
Solving the mesh equations using Cramer’s rule:
I2 =
Then
(10+ j5)( 0 )−( − j50 )(10 )
2
(10+ j5) ( 400+ j500 ) − ( − j50 )
= 0.062 ∠29.7° A
V2 = 400 I 2 = 400 ( 0.062 ∠29.7° ) = 24.8 ∠29.7°
1 150
P 11.9-5 Determine v(t) for the circuit of Figure
P 11.9-5 when vs = 10 cos 30t V.
Answer: v(t) = 23 cos (30t + 9°) V
28 Ω
F
+
vs
+
–
v(t)
0.1 H
0.3 H
0.2 H
–
Figure P 11.9-5
Solution:
Mesh equations:
10 = − j 5I 1 + ⎡⎣ j 9 ( I 1 − I 2 ) + j 3I 2 ⎤⎦
0 = 28I 2 + ⎡⎣ j 6I 2 + j 3 ( I 1 − I 2 ) ⎤⎦ − ⎡⎣ j 9 ( I 1 − I 2 ) + j 3I 2 ⎤⎦
or
− j 6 ⎤ ⎡ I 1 ⎤ ⎡10 ⎤
⎡ j4
⎢ − j 6 28 + j 9 ⎥ ⎢I ⎥ = ⎢ 0 ⎥
⎣
⎦⎣ 2⎦ ⎣ ⎦
Solving the mesh equations, e.g. using MATLAB, yields
I1 = 2.62∠ − 72° A and I 2 = 0.53∠0° A
then
V = j 9(I1 − I 2 ) + j 3I 2 = j 9I1 − j 6I 2 = 23 ∠10° V
Finally
v (t ) = 23cos (30t + 10°) V
(checked: LNAPAC 7/21/04)
P 11.9-6 Find the total energy stored in the
circuit shown in Figure P 11.9-6 at t = 0 if the
secondary winding is (a) open-circuited, (b)
short-circuited, (c) connected to the terminals
of a 7-Ω resistor.
Answer: (a) 15 J
12 Ω
M = 0.6 H
0.3 H
10 cos 5t A
(b) 0 J (c) 5 J
1.2 H
Figure P 11.9-6
Solution:
(a)
I 2 = 0 ⇒ I1 = 10 ∠0° A ⇒ i1 (0) = 10 A
2
w=
(b)
L1 i1 (0)
2
=
(0.3) (10) 2
= 15 J
2
Mesh equations:
j 6 I 2 - j 3 I1 = 0 ⇒ I1 = 2 I 2
I1 = 10∠0° A ⇒ I 2 = 5∠0° A
Then
w=
1
1
1
1
L1i 2 2 (0) + L 2i12 (0) − M i1 (0) i 2 (0) = (0.3)(10) 2 + (1.2)(5) 2 − (0.6) (10)(5) = 0
2
2
2
2
(c)
(7 + j 6) I 2 − j 3 I1 = 0
I 2 = 3.25 ∠49.4° A
i 2 (t ) = 3.25cos(5t + 49.4°) A
i 2 (0) = 2.12 A
Finally
1
1
w = (0.3) (10) 2 + (1.2) (2.12) 2 − (0.6) (10) (2.12) = 5.0 J
2
2
8 mH
P 11.9-7 Find the input impedance, Z, of the
circuit of Figure P 11.9-7 when ω = 1000 rad/s.
Answer: Z = 8.4 ∠14° Ω
1 6
mF
5 mH
6 mH
3Ω
Z
Figure P 11.9-7
Solution:
Mesh equations:
− VT + j8 I1 + j 5(I1 − I 2 ) − j 6 I1 + j 6 (I1 − I 2 ) + j 5 I1 = 0
3 I 2 + j 6 (I 2 − I1 ) − j 5 I1 = 0
Solving yields
I2 =
j 11
I1
3+ j 6
⎛
j 11 ⎞
VT = I1 ( j 18) + I 2 (− j 11) = ⎜ j 18+
⎟I
3+ j 6 ⎠ 1
⎝
Then
Z =
j 11
VT
= j 18 +
= 8.28 ∠13° = 8.0667 + j 1.8667 Ω
I1
3+ j 6
(checked: LNAPAC 7/21/04)
P 11.9-8 A circuit with three mutual inductances is shown in Figure P 11.9-8. When vs = 10
cos 2t V, M1 = 2 H, and M2 = M3 = 1 H, determine the capacitor voltage v(t).
M1
5Ω
4H
3H
M2
vs
2Ω
6Ω
M3
+
–
+
1 10
v
F
2H
–
Figure P 11.9-8
Solution:
The coil voltages are given by
V1 = j 6 I1 − j 2 (I1 − I 2 ) − j 4 I 2 = j 4 I1 − j 2 I 2
V2 = j 4 ( I1 − I 2 ) − j 2 I1 + j 2 I 2 = j 2 I1 − j 2 I 2
V3 = j8 I 2 − j 4 I1 + j 2 ( I1 − I 2 ) = − j 2 I1 + j 6 I 2
The mesh equations are
5 I1 + V1 + 6 (I1 − I 2 ) + V2 = 10∠0°
− V2 + 6 (I 2 − I1 ) + 2 I 2 + V3 − j 5 I 2 = 0
Combining and solving yields
11 + j 6 10
−6 − j 4 0
60 + j 40
I2 =
=
= 1.2 ∠0.28° A
11 + j 6 −6 − j 4
50 + j 33
−6 − j 4 8 + j 3
Finally
V = − j 5 I 2 = 6.0 ∠ − 89.72° A ⇒ v(t ) = 6 cos(2 t − 89.7°) V
0.25 mF
P 11.9-9 The currents i1(t) and i2(t) in Figure,
P 11.9-9 are mesh currents. Represent the
circuit in the frequency domain and write the
mesh equations.
6H
5H
40 Ω
i1(t)
i2(t)
8H
+ –
15 cos (25t + 30°) V
80 Ω
P 11.9-9
Solution:
The equations describing the coupled coils give:
V1 = j 200 ( I 1 − I 2 ) + j125 I 2 = j 200I 1 − j 75 I 2
V 2 = j150 I 2 + j125 ( I 1 − I 2 ) = j125 I 1 + j 25 I 2
The mesh equation for the left mesh is
− j 160 I 1 + V1 − 15∠30° + 40 I 1 = 0
− j 160 I 1 + j 200I 1 − j 75 I 2 − 15∠30° + 40 I 1 = 0
( 40 + j 40 ) I 1 − j 75 I 2 = 15∠30°
The mesh equation for the right mesh is
V 2 + 80 I 2 − V1 = 0
j125 I 1 + j 25 I 2 + 80 I 2 − ( j 200I 1 − j 75 I 2 ) = 0
− j 75 I 1 + ( 80 + j100 ) I 2 = 0
P 11.9-10 Determine the mesh currents for the circuit shown in Figure P 11.9-10.
3H
10 Ω
+
–
–
+
4H
v1
12 cos 5t V
6H
+
–
i1
v2
20 Ω
50 Ω
i2
P 11.9-10
Solution:
In the frequency domain, the coil voltages are
given by
V1 = j ω L1 I 1 − j ω M I 2
V2 = j ω L 2 I 2 − j ω M I 1
The mesh equations are
R1 I 1 + V1 + R 2 ( I 1 − I 2 ) = Vs
V2 + R3 I 2 − R 2 ( I 1 − I 2 ) = 0
Substituting for the coil voltages gives
⎡ R1 + R 2 + j ω L1
⎢
⎢
⎣ − ( R2 + j ω M )
− ( R 2 + j ω M ) ⎤ ⎡ I 1 ⎤ ⎡ Vs ⎤
⎥⎢ ⎥ = ⎢ ⎥
R 2 + R 3 + j ω L 2 ⎦⎥ ⎣I 2 ⎦ ⎣ 0 ⎦
Using the given values
⎡ 30 + j 20
− ( 20 + j 15 ) ⎤ ⎡ I 1 ⎤ ⎡12∠0 ⎤
⎢
⎥⎢ ⎥ =
70 + j 30 ⎦ ⎣I 2 ⎦ ⎢⎣ 0 ⎥⎦
⎣ − ( 20 + j 15 )
Solving, for example using MATLAB, gives
⎡ I 1 ⎤ ⎡0.4240∠ − 28.9° ⎤
⎢ ⎥=⎢
⎥
⎣ I 2 ⎦ ⎣ 0.1392∠ − 15.2° ⎦
Back in the time domain, the mesh currents are
⎡ i1 ⎤ ⎡0.4240 cos ( 5t − 28.9° ) ⎤
⎥ A
⎢i ⎥ = ⎢
−
°
t
0.1392
cos
5
15.2
(
)
2
⎣ ⎦ ⎣
⎦
(checked using LNAP, 9/9/04)
P 11.9-11 Determine the coil voltages, v1, v2, v3, and v4, for the circuit shown in Figure P 11.911.
v1
–
2 cos (5t + 45°) A
+
4H
10 Ω
3H
+
10 Ω
v2
v3
+
6H
–
5H
20 Ω
4H
–
–
5H
v4
+
1.25 cos (5t − 45°) A
2.75 cos (5t) A
Figure P 11.9-11
Solution:
In the frequency domain, the coil voltages are given by
V1 = − j ω L1 I a + j ω M 1 ( I a − I b )
= j ω ( M 1 − L1 ) I a − j ω M 1 I b
= ( − j 5 )( 2∠45° ) + ( − j 15 )(1.25∠ − 45° )
= 21.25∠ − 106.9° V
V2 = j ω L 2 ( I b − I a ) + j ω M 1 I a
= j ω ( M 1 − L2 ) I a + j ω L2 I b
= ( − j 15 )( 2∠45° ) + ( j 30 )(1.25∠ − 45° )
= 48.02∠6.4° V
V3 = j ω L 3 ( I c − I a ) − j ω M 2 I c = − j ω L 2 I a + j ω ( L 3 − M 2 ) I c
= ( − j 25 )( 2∠45° ) + ( j 5 )( 2.75∠0° ) = 41.43∠ − 31.4° V
V4 = − j ω L 4 I c + j ω M 2 ( I c − I a ) = − j ω M 2 I a + j ω ( M 2 − L 4 ) I c
= ( − j 20 )( 2∠45° ) + ( − j 5 )( 2.75∠0° ) = 50.66∠ − 56.1° V
Back in the time domain, the coil voltages are
v1 = 21.25 cos ( 5t − 106.9° ) V , v 2 = 48.02 cos ( 5t + 6.3° ) V ,
v 3 = 41.43 cos ( 5t − 31.4° ) V and v 4 = 50.66 cos ( 5t − 56.1° ) V
(checked using LNAP, 9/9/04)
4H
Solution:
(a) In (b) and (c) the coils are coupled by mutual
inductance, but not in (a). This circuit can be
represented in the frequency domain, using phasors and
impedances.
+
5Ω
5H
vo(t )
–
+
P 11.9-12 Figure P 11.9-12 shows three similar
circuits. In each, the input to the circuit is the voltage of
the voltage source, vs(t). The output is the voltage across
the right-hand coil, vo(t). Determine the steady-state
output voltage, vo(t), for each of the three circuits.
–
vs(t ) = 5.94 cos (3t + 140°) V
(a)
4H
+
2H
5Ω
5H
vo(t )
+
–
–
vs(t ) = 5.94 cos (3t + 140°) V
Voltage division gives
Vo =
(b)
j 15
5.94∠140°
5 + j 12 + j 15
4H
= ( 0.5463∠10.5° )( 5.94∠140° ) = 3.245∠150.5° V
vo ( t ) = 3.245 cos ( 3 t + 150.5° ) V
(b) The input voltage is a sinusoid and the circuit is at
steady state. The output voltage is also a sinusoid and
has the same frequency as the input voltage.
Consequently, the circuit can be represented in the
frequency domain, using phasors and impedances.
5Ω
5H
vo(t )
–
+
In the time domain, the output voltage is given by
+
2H
–
vs(t ) = 5.94 cos (3t + 140°) V
(c)
Figure P 11.9-12
This circuit of a single mesh. Notice that the mesh current, I(ω), enters the undotted ends of both
coils. Apply KVL to the mesh to get
5 I (ω ) + ( j12 I (ω ) + j 6 I (ω ) ) + ( j 6 I (ω ) + j15 I (ω ) ) − 5.94∠140° = 0
5 I (ω ) + ( j12 + j 6 + j 6 + j15 ) I (ω ) − 5.94∠140° = 0
I (ω ) =
5.94∠140°
5.94∠140° 5.94∠140°
=
=
= 0.151∠57° A
5 + j (12 + 6 + 6 + 15 )
5 + j 39
39.3∠83
Notice that the voltage, Vo(ω), across the right-hand coil and the mesh current, I(ω), adhere to
the passive convention. The voltage across the right-hand coil is given by
Vo (ω ) = j 15 I (ω ) + j 6 I (ω ) = j 21 I (ω ) = j 21 ( 0.151∠57° )
= (21∠90°) ( 0.151∠57° )
= 3.17∠147° V
In the time domain, the output voltage is given by
vo ( t ) = 3.17 cos ( 3 t + 147° ) V
(c) Circuit (c) is very similar to the circuit (b). There is only one difference: the dot of the lefthand coil is located at the right of the coil in (c) and at the left of the coil in (b). As before, our
first step is to represent the circuit in the frequency domain, using phasors and impedances.
This circuit consists of a single mesh. Notice that the mesh current, I(ω), enters the dotted end of
the left-hand coil and the undotted end of the right-hand coil. Apply KVL to the mesh to get
5 I (ω ) + ( j12 I (ω ) − j 6 I (ω ) ) + ( − j 6 I (ω ) + j15 I (ω ) ) − 5.94∠140° = 0
5 I (ω ) + ( j12 − j 6 − j 6 + j15 ) I (ω ) − 5.94∠140° = 0
I (ω ) =
5.94∠140°
5.94∠140° 5.94∠140°
=
=
= 0.376∠68.4° A
5 + j (12 − 6 − 6 + 15 )
5 + j 15
15.8∠71.6
Notice that the voltage, Vo(ω), across the right-hand coil and the mesh current, I(ω), adhere to
the passive convention. The voltage across the right-hand coil is given by
Vo (ω ) = j 15 I (ω ) − j 6 I (ω ) = j 9 I (ω ) = j 9 ( 0.376∠68.4° )
= (9∠90°) ( 0.376∠68.4° ) = 3.38∠158.4° V
In the time domain, the output voltage is given by
vo ( t ) = 3.38 cos ( 3 t + 158.4° ) V
P 11.9-13 Figure P 11.9-13 shows three similar circuits. In each, the input to the circuit is the
voltage of the voltage source,
vs(t) = 5.7 cos (4t + 158°) V
The output in each circuit is the voltage across the right-hand coil, vo(t). Determine the steadystate output voltage, vo(t), for each of the three circuits.
+
4Ω
+
–
vs(t)
5H
4H
2H
4Ω
vo(t)
+
–
vs(t)
+
5H
4H
–
vo(t)
–
(a)
(b)
2H
4Ω
+
–
vs(t)
+
5H
4H
vo(t)
–
(c)
Figure P 11.9-13
Solution:
(a) In (b) and (c) the coils are coupled by mutual inductance, but not in (a). This circuit can be
represented in the frequency domain, using phasors and impedances.
Voltage division gives
j 16 || j 20
5.7∠158° = ( 0.9119∠24° )( 5.7∠158° ) = 5.198∠182° V
4 + ( j 16 || j 20 )
In the time domain, the output voltage is given by
Vo =
vo ( t ) = 5.2 cos ( 4 t + 182° ) V
(b) The input voltage is a sinusoid and the circuit is at steady state. The output voltage is also a
sinusoid and has the same frequency as the input voltage. Consequently, the circuit can be
represented in the frequency domain, using phasors and impedances.
The coil currents, I1(ω) and I2(ω), and the coil voltages, V1(ω) and V2(ω), are labeled. Reference
directions for these currents and voltages have been selected so that the current and voltage of
each coil adhere to the passive convention. Notice that both coil currents, I1(ω) and I2(ω), enter
the undotted ends of their respective coils. The device equations for coupled coils are:
and
V1 (ω ) = j 16 I1 (ω ) + j 8 I 2 (ω )
(1)
V2 (ω ) = j 8 I1 (ω ) + j 20 I 2 (ω )
(2)
The coils are connected in parallel, consequently V1 (ω ) = V2 (ω ) . Equating the expressions for
V1(ω) and V2(ω) gives
j 16 I1 (ω ) + j 8 I 2 (ω ) = j 8 I1 (ω ) + j 20 I 2 (ω )
j 8 I1 (ω ) = j 12 I 2 (ω )
I 1 (ω ) =
3
I 2 (ω )
2
Apply Kirchhoff’s Current Law (KCL) to the top node of the coils to get
I ( ω ) = I 1 ( ω ) + I 2 (ω ) =
I 1 (ω ) =
Therefore
3
5
I 2 ( ω ) + I 2 (ω ) = I 2 (ω )
2
2
3
2
I (ω ) and I 2 (ω ) = I (ω )
5
5
(3)
Substituting the expressions for I1(ω) and I2(ω) from Equation 3 into Equation 1 gives
16 ( 3) + 8 ( 2 )
⎛3
⎞
⎛2
⎞
V1 (ω ) = j 16 ⎜ I (ω ) ⎟ + j 8 ⎜ I (ω ) ⎟ = j
I (ω ) = j 12.8 I (ω )
5
⎝5
⎠
⎝5
⎠
Apply KVL to the mesh consisting of the voltage source, resistor and left-hand coil to get
4 I (ω ) + V1 (ω ) − 5.7∠158° = 0
Using Equation 4 gives
Solving for I (ω) gives
4 I (ω ) + j 12.8 I (ω ) − 5.7∠158° = 0
I (ω ) =
5.7∠158° 5.7∠158°
=
= 0.425∠85° A
4 + j 12.8 13.41∠73°
Now the output voltage can be calculated using Equation 4:
(4)
Vo (ω ) = V1 (ω ) = j 12.8 I (ω ) = j 12.8 ( 0.425∠85° )
= (12.8∠90° )( 0.425∠85° ) = 5.44∠175° V
In the time domain, the output voltage is given by
vo ( t ) = 5.44 cos ( 4 t + 175° ) V
(c) Circuit (c) is very similar to the circuit (b). There is only one difference: the dot of the righthand coil is located at the bottom of the coil in (b) and at the top of the coil in (c). Here is the
circuit from (c) represented in the frequency domain, using impedances and phasors.
The coil currents, I1(ω) and I2(ω), and the coil voltages, V1(ω) and V2(ω), are labeled. Reference
directions for these currents and voltages have been selected so that the current and voltage of
each coil adhere to the passive convention. Notice that one of coil currents, I1(ω), enters the
undotted end of the coil while the other coil current, I2(ω), enters the dotted end of the coil.
The device equations for coupled coils are:
and
V1 (ω ) = j 16 I1 (ω ) − j 8 I 2 (ω )
(5)
V2 (ω ) = − j 8 I1 (ω ) + j 20 I 2 (ω )
(6)
The coils are connected in parallel, consequently V1 (ω ) = V2 (ω ) . Equating the expressions for
V1(ω) and V2(ω)gives
j 16 I1 (ω ) − j 8 I 2 (ω ) = − j 8 I1 (ω ) + j 20 I 2 (ω )
j 24 I1 (ω ) = j 28 I 2 (ω )
I 1 (ω ) =
28
7
I 2 (ω ) = I 2 (ω )
24
6
Apply Kirchhoff’s Current Law (KCL) to the top node of the coils to get
7
13
I 2 (ω ) + I 2 (ω ) = I 2 (ω )
6
6
7
6
I1 (ω ) = I (ω ) and I 2 (ω ) = I (ω )
Therefore
13
13
Substituting the expressions for I1(ω) and I2(ω) from Equation 7 into Equation 5 gives
I (ω ) = I 1 (ω ) + I 2 (ω ) =
(7)
16 ( 7 ) − 8 ( 6 )
⎛7
⎞
⎛6
⎞
V1 (ω ) = j 16 ⎜ I (ω ) ⎟ − j 8 ⎜ I (ω ) ⎟ = j
I (ω ) = j 4.9 I (ω )
13
⎝ 13
⎠
⎝ 13
⎠
Apply KVL to the mesh consisting of the voltage source, resistor and left-hand coil to get
4 I (ω ) + V1 (ω ) − 5.7∠158° = 0
Using Equation 8 gives
4 I (ω ) + j 4.9 I (ω ) − 5.7∠158° = 0
Solving for I (ω) gives
I (ω ) =
5.7∠158° 5.7∠158°
=
= 0.901∠107° A
4 + j 4.9 6.325∠51°
Now the output voltage can be calculated using Equation 8:
Vo (ω ) = V1 (ω ) = j 4.9 I (ω ) = j 4.9 ( 0.901∠107° )
= ( 4.9∠90° )( 0.901∠107° ) = 4.41∠197° V
In the time domain, the output voltage is given by
vo ( t ) = 4.41 cos ( 4 t + 197° ) V
(8)
P11.9-14
The circuit shown in Figure 11.9-14 is represented in the
time domain. Determine coil voltages v1 and v2.
Answer: v1 = 104.0 cos (6t + 46.17 °) V
and
v2 = 100.6 cos (6t + 63.43 °) V.
Figure 11.9-14
Solution:
Represent the circuit in the frequency domain as
shown. Then
V1 = j 48 (1.5∠ − 90° ) + j 30 ( 2.5∠0° )
= j 48 ( − j 1.5 ) + j 30 ( 2.5 ) = 72 + j 75
= 103.97∠46.17° V
and
V 2 = j 36 ( 2.5∠0° ) + j 30 (1.5∠ − 90° )
= j 36 ( 2.5 ) + j 30 ( − j 1.5 ) = 45 + j 90
= 100.62∠63.43° V
P11.9-15
The circuit shown in Figure P11.9-15 is represented in the frequency domain (e.g. –j30 Ω is the
impedance due to the mutual inductance of the coupled coils). Suppose V ( ω ) = 70∠0° V. Then
I 1 ( ω ) = B∠θ A and I 2 ( ω ) = 0.875∠ − 90° A. Determine the values of B and θ.
Answer: B = 1.75 A and θ = −90 °
Figure 11.9-15
Solution:
Suppose V ( ω ) = 70∠0° V. Then I 1 ( ω ) = B∠θ A and I 2 ( ω ) = 0.875∠ − 90° A.
V = − j 25 I 1 + j 50 I 1 + j 30 I 2 = j 25 I 1 + j 30 I 2 = j 25 I 1 + j 30 ( − j 0.875 )
I1 =
V − j 30 ( − j 0.875 ) V − 26.25 70 − 26.25 43.75
=
=
=
= − j 1.75
j 25
j 25
j 25
j 25
P11.9-16
Determine the values of the inductances L1
and L2 in the circuit shown in Figure
P11.9-16, given that
i ( t ) = 0.319 cos ( 4 t − 82.23° ) A
and
v ( t ) = 0.9285cos ( 4 t − 62.20° ) V .
Figure P11.9-16
Solution:
Represent the circuit in the frequency domain as
Apply KVL to the left mesh to get
5∠0° + j 3 ( 0.9285∠ − 62.20° )
⎛ 0.9285∠ − 62.20° ⎞
5∠0° = j 4 L1 ( 0.319∠ − 82.23° ) + j 12 ⎜ −
⎟ ⇒ L1 =
4
j 4 ( 0.319∠ − 82.23° )
⎝
⎠
= 6.0004 − j 0.0005
≅6H
Apply KVL to the left mesh to get
⎛ 0.9285∠ − 62.20° ⎞
0.9285∠ − 62.20° = j 4 L 2 ⎜ −
⎟ + j 12 ( 0.319∠ − 82.23° )
4
⎝
⎠
L2 =
( 0.9285∠ − 62.20° ) − j 12 ( 0.319∠ − 82.23° )
⎛ 0.9285∠ − 62.20° ⎞
j 4⎜ −
⎟
4
⎝
⎠
= 3.9998 + j 0.0005
≅4H
P11.9-17 Determine the complex power supplied by the source in the circuit shown in Figure
P11.9-17.
Figure P11.9-17
Solution: Represent the circuit in the frequency domain as
The coil voltages are given by
V1 = j 100 I 1 − j 80 I 2 and V 2 = j 120 I 2 − j 80 I 1
V1 + 40 ( I 1 − I 2 ) + 20 I 1 − 180∠0° = 0
Using KVL
V 2 + 50 I 2 − j 50 I 2 − 40 ( I 1 − I 2 ) = 0
Substituting the coil voltages:
( j 100 I 1 − j 80 I 2 ) + 40 ( I 1 − I 2 ) + 20 I 1 = 180∠0°
( j 120 I
2
− j 80 I 1 ) + 50 I 2 − j 50 I 2 − 40 ( I 1 − I 2 ) = 0
( 60 + j 100 ) I 1 − ( 40 + j 80 ) I 2 = 180∠0°
− ( 40 + j 80 ) I 1 + ( 90 + j 70 I 1 ) I 2 = 0
Simplifying
I 1 = 2.731∠ − 26.9° A and I 2 = 2.142∠ − 1.36° A
Solving gives
The complex power delivered by the source is
S=
VI * (180∠0° )( 2.731∠ − 26.9° ) *
=
= 219.19 + j111.20 VA
2
2
P11.9-18 The input to the circuit shown in Figure P11.9-18 is
v s ( t ) = 12 cos ( 5 t ) V
The impedance of the load is 20 + j 15 Ω. Determine the complex power (a) supplied by the
source, (b) received by the 20 Ω resistor, (c) received by the coupled inductors and (d) received
by the load.
.
Figure P11.9-18
Solution:
Represent the circuit in the frequency domain as
The coil voltages are given by
V1 = j 20 I 1 − j 25 I 2 and V 2 = j 40 I 2 − j 25 I 1
Using KVL
20 I 1 + V1 − 12∠45° = 0 and 20 I 2 + j15 I 2 + V 2 = 0
Substituting the coil voltages:
20 I 1 + j 20 I 1 − j 25 I 2 = 12∠0°
20 I 2 + j15 I 2 + j 40 I 2 − j 25 I 1 = 0
Writhing these equations in matrix form
− j 25 ⎤ ⎡ I 1 ⎤ ⎡12∠0° ⎤
⎡ 20 + j 20
⎢ ⎥=
⎢ − j 25
20 + j 55⎥⎦ ⎣ I 2 ⎦ ⎢⎣ 0 ⎥⎦
⎣
I 1 = 0.4676∠ − 22.8° A and I 2 = 0.1998∠ − 2.86° A
Solving gives
(a) The complex power delivered by the source is
S=
(12∠0° ) I1 * = (12∠0° )( 0.4676∠ − 22.8° ) * = 2.5855 + j1.0893 VA
2
2
(b) The complex power received by the 20 Ω resistor is
I
S= 1
2
2
( 20 ) =
( 0.4676 )
2
2
( 20 ) = 2.1865 + j 0 VA
(c) The complex power received by the coupled inductors is
S=
V1I1 * V2 I 2 *
+
= 0 + j 0.79 VA
2
2
(d) The complex power received by the load is
I
S= 2
2
2
( 20 + j 15)
( 0.1998)
=
2
2
( 20 + j 15) = 0.399 + j 0.299 VA
(checked using LNAP 2/6/09)
P11.9-19 Figure P11.9-18a shows a source connected to a 160 Ω load. In Figure P11.9-18a an
ideal transformer and capacitor have been inserted between the source and the load.
(a) Determine the value of the average power delivered to the 160 Ω load in Figure P11.918a.
(b) Determine the values of n and C Figure P11.9-18b that maximize the average power in
the load.
(c) Using the values of n and C from part (b), determine the value of the average power
delivered to the 160 Ω load in Figure P11.9-18b.
(a)
(b)
Figure P11.9-19
Solution:
(a)
Represent the circuit in the frequency domain. Using
KVL
I=
and
120∠0°
= 0.7058∠ − 0.674° A
170 + j 2
P=
0.7058 2
(160 ) = 39.85 W
2
(b)
For maximum power transfer to the 160 Ω load
1
1
⎛ 1 ⎞
160 ) = 10 ⇒ n = 4 and − ( j 2 ) = ⎜ 2 ⎟ − j
2 (
n
250 C
⎝4 ⎠
⇒ C = 0.125 mF
(c)
Represent the circuit in the frequency domain
I1 =
and
120∠0°
2
⎛1⎞
10 + j 2 + ⎜ ⎟ ( − j 32 + 160 )
⎝4⎠
=
120∠0°
1
= 6∠0° A and I 2 = I1 = 1.5∠0° A
20
4
P=
1.5 2
(160 ) = 180 W
2
Section 11.10 The Ideal Transformer
P 11.10-1
Find V1, V2, I1, and I2 for the circuit of Figure P 11.10-1, when n = 5.
I1
2 + 3j
12 0° V
+
–
1:n
I2
+
+
V1
V2
–
100 – j75
–
Ideal
Figure P 11.10-1
Solution:
Z = (2 + j 3) +
I1 =
12∠0° 12∠0°
=
=2A
Z
6
⎛ 100− j 75 ⎞
⎛ 100− j 75 ⎞
V1 = I1 ⎜
⎟ = (2) ⎜
⎟ = 10∠ − 36.9° V
2
n
⎝
⎠
⎝ 25 ⎠
V2 = nV1 = 5 (10∠−36.9°) = 50 ∠−36.9° V
I2 =
I1 2
=
A
n 5
(100− j 75)
=6Ω
52
P 11.10-2 A circuit with a transformer is shown in Figure P 11.10-2.
(a)
Determine the turns ratio.
(b)
Determine the value of Rab.
(c)
Determine the current supplied by the voltage source.
Answer: (a) n = 5
(b) Rab = 400 Ω
i
a
5 mA rms
1:n
+
10 V rms
+
–
V0
–
Rab
b
Ideal
Figure P 11.10-2
Solution:
(a)
V0 = (5 ×10−3 )(10, 000) = 50 V
n=
(b)
(c)
V
N2
50
= 0 =
= 5
N1
V1
10
Rab =
Is =
1
1
R2 = (10 × 103 ) = 400 Ω
2
25
n
10
10
=
= 0.025 A = 25 mA
Rab 400
10 kΩ
P 11.10-3 Find the voltage Vc in the circuit shown in Figure P 11.10-3. Assume an ideal
transformer. The turns ratio is n = 1/3.
Answer: Vc = 21.0 ∠ − 105.3°
30 Ω
j20 Ω
1:n
+
5Ω
–j8 Ω
80 –50° V –
+
–
Vc
Ideal
Figure P 11.10-3
Solution:
1
Z 2 = 9 Z 2 = 9 (5 − j8) = 45 − j 72 Ω
n2
Using voltage division, the voltage across Z 1 is
Z1 =
⎛
⎞
45− j 72
V1 = ( 80∠ − 50° ) ⎜
⎟ = 74.4 ∠ − 73.3° V
⎝ 45− j 72+ 30 + j 20 ⎠
Then
V2 = nV1 =
74.4 ∠ − 73.3°
= 24.8 ∠ − 73.3° V
3
Using voltage division again yields
⎛ − j8 ⎞
⎛ 8∠−90° ⎞
Vc = V2 ⎜
= ( 24.8∠−73.3° ) ⎜
⎟
⎟ = 21.0∠ − 105.3° V
⎝ 89∠−58° ⎠
⎝ 5− j 8 ⎠
P 11.10-4 An ideal transformer is connected in the circuit shown in Figure P 11.10-4, where
vs = 50 cos 1000t V and n = N2/N1 = 5. Calculate V1 and V2.
2Ω
vs
+
–
1:5
+
+
v1
v2
–
200 Ω
–
Ideal
Figure P 11.10-4
Solution:
n = 5, Z1 =
200
( 5)
2
= 8 Ω ⇒ V1 =
8
( 50∠0° ) = 40∠0° V ⇒ V2 = n V1 = 200∠0° V
8+ 2
P11.10-5 Figure P11.10-5 shows a load connected to a source through an ideal transformer. The
input to the circuit is
v s ( t ) = 12 cos ( 5 t ) V
Determine
(a) The values of the turns ration, n, and load inductance, L, required for maximum power
transfer to the load.
(b) The complex power delivered to the transformer by the source.
(c) The complex power delivered to the load by the transformer.
Figure P11.10-5
Solution:
*
(a) For maximum power transfer:
Equating real parts gives n =
1
1
⎛
⎞
288 + j 20 L ) = ⎜ 18 − j
⎟ = 18 + j 5
2 (
20 × 0.01 ⎠
n
⎝
( )
5 42
288
= 4 . Equating imaginary parts gives L =
=4 H.
20
18
(b) Represent the circuit in the frequency domain as
Replace the transformer and load by an equivalent impedance
Z equiv =
1
( 288 + j 80 ) = 18 + j 5 Ω
42
I1 =
and
12∠0°
12∠0° 1
=
= ∠0° A
(18 − j 5) + (18 + j 5) 36 3
⎛1
⎞
V1 = (18 + j 5 ) I 1 = (18 + j 5 ) ⎜ ∠0° ⎟ = 6.227∠15.5° V
⎝3
⎠
The secondary coil current and voltages
and
1
1⎛1
1
⎞
I 2 = − I 1 = − ⎜ ∠0° ⎟ = − ∠0° = −0.0833∠0° A
4
4⎝3
12
⎠
4
V 2 = V1 = 24.91∠15.5° V
1
The complex power delivered to the transformer by the source.
2
I
V1I1 *
= 1 + j 0.277 VA = 1 (18 + j 5 )
2
2
(c) The complex power delivered to the load by the transformer is
I
VI *
− 2 2 = 1 + j 0.277 VA = 2
2
2
2
( 288 + j80 )
(checked using LNAP and MATLAB 2/6/09)
a
Ideal
P 11.10-6 Find the Thévenin equivalent at
terminals a–b for the circuit of Figure P 11.10-6 v
when v = 16 cos 3t V.
Answer: Voc = 12 and Zt = 3.75 Ω
2Ω
2Ω
+
–
6Ω
b
1:2
Figure P 11.10-6
Solution:
Z=
1
( 2 + 6) = 2 Ω
22
⎞
⎛ 6 ⎞ ⎛⎛ 2 ⎞
Voc = ⎜
⎟ ( 2) ⎜ ⎜
⎟16∠0° ⎟ = 12∠0° V
⎝ 6 + 2 ⎠ ⎝⎝ 2 + 2 ⎠
⎠
Z=
1
1
2 = Ω
2 ( )
2
2
⎛
⎞
1
1 ⎜ 16∠0° ⎟
I sc = −I 2 = I1 = ⎜
⎟ = 3.25∠0° A
2
2⎜ 2+ 1 ⎟
⎝
2 ⎠
Then
Zt =
12∠0°
= 3.75∠0° Ω
3.2∠0°
2Ω
P 11.10-7 Find the input impedance Z for the
circuit of Figure P 11.10-7.
Answer: Z = 6 Ω
2:1
6Ω
Z
Ideal
Figure P 11.10-7
Solution:
1
V1
2
V − V2 V1
=
I3 = 1
2
4
V
V
I 2 = I3 − 2 = 1
6
6
V
1
I1 = − I 2 = − 1
2
12
V
I T = I 3 + I1 = 1
6
V
Z= 1 =6
IT
V2 =
P 11.10-8 In less developed regions in mountainous areas, small hydroelectric generators are
used to serve several residences (Mackay, 1990). Assume each house uses an electric range and
an electric refrigerator, as shown in Figure P 11.10-8. The generator is represented as Vs
operating at 60 Hz and V2 = 230 ∠0° V. Calculate the power consumed by each home connected
to the hydroelectric generator when n = 5.
Source and line
1:n
2 + j3
Vs
+
+
–
10 Ω
20 Ω
V2
20 mH
–
Ideal
Range
Refrigerator
Figure P 11.10-8
Solution:
ZL =
1 ⎛ 20 (10+ j 7.54) ⎞ 8.1∠23°
=
= 0.3 + j 0.13 Ω
52 ⎜⎝ 20+10+ j 7.54 ⎟⎠
25
V
V
( 230 ) = 88 kW/home
PL = L = 2 =
2 R 2 2 R L 2( 0.3)
2
2
Therefore, 529 kW are required for six homes.
2
P 11.10-9 Three similar circuits are
shown in Figure P 11.10-9. In each of
these circuits
8Ω
vs(t)
vs(t) = 5 cos (4t + 45°) V.
+
–
i2(t)
+
+
v1(t)
4H
3H
–
Determine v2 (t) for each of the three
circuits.
Answer:
v2(t) = 0 V
(a)
v2(t) = 1.656 cos (4t + 39°) V
(b)
(c)
v2(t) = 2.88 cos (4t + 45°) V
i1(t)
v2(t)
12 Ω
–
(a)
8Ω
i1(t)
i2(t)
2H
+
vs(t)
+
–
v1(t)
+
4H
3H
–
v2(t)
12 Ω
–
(b)
8Ω
vs(t)
+
–
i1(t)
i2(t)
10:8.66
+
+
v1(t)
v2(t)
–
–
Ideal
(c)
Figure P 11.10-9
Solution:
(a)
Coil voltages:
V1 = j16 I1
V2 = j12 I 2
Mesh equations:
8 I1 + V1 − 5∠45° = 0
−12 I 2 − V2 = 0
12 Ω
Substitute the coil voltages into the mesh equations and do some algebra:
8 I1 + j16 I1 = 5∠45° ⇒ I1 = 0.28∠ − 18.4°
12 I 2 + j12 I 2 = 0 ⇒ I 2 = 0
V2 = −12 I 2 = 0
(b)
Coil voltages:
V1 = j16 I1 + j8 I 2
V2 = j12 I 2 + j8 I1
Mesh equations:
8 I1 + V1 − 5∠45° = 0
−12 I 2 − V2 = 0
Substitute the coil voltages into the mesh equations and do some algebra:
8 I1 + ( j16 I1 + j8 I 2 ) = 5∠45°
12 I 2 + ( j12 I 2 + j8 I1 ) = 0
I1 = −
12 + j12
3
I 2 = ( j − 1) I 2
j8
2
⎡
⎤
⎛3⎞
⎢( 8 + j16 ) ⎜ 2 ⎟ ( j − 1) + j8⎥ I 2 = 5∠45° ⇒ I 2 = 0.138∠ − 141°
⎝ ⎠
⎣
⎦
V2 = −12 I 2 = 1.656∠39°
(c )
Coil voltages and currents:
10
V2
8.66
8.66
I1 = −
I2
10
V1 =
Mesh equations:
8 I1 + V1 − 5∠45° = 0
−12 I 2 − V2 = 0
Substitute into the second mesh equation and do some algebra:
2
⎛ 10 ⎞ 8.66
⎛ 10 ⎞
−12 ⎜ −
I1 ⎟ =
V1 ⇒ V1 = 12 ⎜
⎟ I1
⎝ 8.66 ⎠ 10
⎝ 8.66 ⎠
2
⎛ 10 ⎞
8 I1 + 12 ⎜
⎟ I1 = 5∠45° ⇒ I1 = 0.208∠45°
⎝ 8.66 ⎠
⎛ 10 ⎞ 12 (10 )
0.208∠45° = 2.88∠45°
V2 = −12 I 2 = −12 ⎜ −
I1 ⎟ =
8.66
⎝ 8.66 ⎠
P 11.10-10
Find V1 and I1 for the circuit of Figure P 11.10-10 when n = 5.
I1
1 + j3
10 0° V
+
–
1:n
I2
+
V1
100 – j75
–
Ideal
Figure P 11.10-10
Solution:
I1 =
10∠0°
10∠0°
=
= 2∠0° A
100 − j 75 (1 + j 3) + ( 4 − j 3)
1
j
3
+
+
(
)
52
V1 = ( 4 − j 3) 2∠0° = 10∠ − 36.9° V
P 11.10-11 Determine v2 and i2 for the circuit shown in Figure P 11.10-11 when n = 2. Note
that i2 does not enter the dotted terminal.
Answer: v2 = 0.68 cos (10t + 47.7°) V
i2 = 0.34 cos (10t + 42°) A
20 mF
5 cos 10t V
+
–
i1
5Ω
2Ω
1:n
+
v1
–
v2
–
+
20 mH
i2
Ideal
Figure P 11.10-11
Solution:
I1 =
5∠0°
5∠0°
5∠0°
=
=
= 0.68∠42° A
2 + j 0.2 5.5 + j 4.95 7.4∠− 42°
−
+
5
j
5
(
)
22
1
I 2 = I1 = 0.34∠42° A
2
V2 = ( 2 + j 0.2 ) I 2 = ( 2.01∠5.7° )( 0.34∠42° ) = 0.68∠47.7° V
so
v2 (t ) = 0.68cos (10 t + 47.7°) V and i 2 (t ) = 0.34 cos (10 t + 42°) A
P11.10-12
The circuit shown in Figure P11.10-12 is represented in the frequency domain. Given the line
current is I Line = 0.5761∠ − 75.88° A , determine PSource , the average power supplied by the
source, PLine , the average power delivered to the line and PLoad , the average power delivered to
the load.
Hint: Use conservation of (average) power to check your answers.
Answer: PSource = 42.15 W, PLine = 0.6638 W and PLoad = 41.49 W.
Figure P11.10-12
Solution:
⎛5⎞
⎜ ⎟120∠0°
⎝1⎠
= 0.5761∠ − 75.88° A
I Line =
2
⎛5⎞
4 + j 10 + ⎜ ⎟ (10 + j 40 )
⎝1⎠
⎛5⎞
I Source = ⎜ ⎟ I Line = 2.8805∠ − 75.88° A
⎝1⎠
⎛5⎞
I Load = ⎜ ⎟ I Line = 2.8805∠ − 75.88° A
⎝1⎠
(120 )( 2.8805) cos 0 − −75.88° = 42.16 W
PSource =
))
( (
2
0.5761 2
PLine =
( 4 ) = 0.6638 W
2
2.8805 2
PLoad =
(10 ) = 41.49 W
2
P11.10-13
The circuit shown in Figure P11.10-13 is represented in the frequency domain. Determine R and
X, the real and imaginary parts of the equivalent impedance, Z eq.
Answer: R = 180 Ω and X =110 Ω
Figure P11.10-13
Solution:
2
Z eq
⎛6⎞
= − j 250 + ⎜ ⎟ ( 5 + j 10 ) = 180 + j 110 Ω
⎝1⎠
P11.10-14 Figure P11.10-14 shows a load connected to a source through an ideal transformer.
Determine the complex power delivered to the transformer by the source.
Answer: S = 698.3 + j1745.7 VA
Figure P11.10-14
Solution:
Represent the circuit in the frequency domain.
Using the element equation of the ideal
transformer:
15
V 2 = (120∠15° ) = 450∠15° V
4
Using Ohm’s law:
V2
450∠15°
=
= 8.3563∠ − 53.2°
I2 =
20 + j 50 53.852∠68.2°
The power supplied by the source is equal to the power received by the load. The power received
by the load is calculated as
2
8.35632
S=
( 20 + j 50 ) =
( 20 + j 50 ) = 698.3 + j1745.7 VA
2
2
I2
(checked using LNAP and MATLAB 2/17/12)
Section 11.11 How Can We Check …?
P 11.11-1
Computer analysis of the circuit shown in Figure P 11.11-1 indicates that when
vs(t) = 12 cos (4t + 30°) V
the mesh currents are given by
i1(t) = 2.327 cos (4t – 25.22°) A
and
i2(t) = 1.229 cos (4t – 11.19°) A
Check the results of this analysis by checking that the average power supplied by the voltage
source is equal to the sum of the average powers received by the other circuit elements.
4Ω
v1(t)
+
–
i1(t)
2Ω
2H
i2(t)
2H
Figure P 11.11-1
P11.11-1
The average power supplied by the source is
Ps =
(12 )( 2.327 ) cos
2
( 30° − ( −25.22° ) ) = 7.96
W
Capacitors and inductors receive zero average power, so the average power supplied by the
voltage source should be equal to the sum of the average powers received by the resistors:
PR =
2.327 2
1.1292
4
+
( )
( 2 ) = 10.83 + 1.27 = 12.10 W
2
2
The average power supplied by the voltage source is not equal to the sum of the average powers
received by the other circuit elements. The mesh currents cannot be correct.
(What went wrong? It appears that the resistances of the two resistors were interchanged when
the data was entered for the computer analysis. Notice that
PR =
2.327 2
1.1292
( 2) +
( 4 ) = 5.41 + 2.55 = 7.96 W
2
2
The mesh currents would be correct if the resistances of the two resistors were interchanged. The
computer was used to analyze the wrong circuit.)
P 11.11-2
Computer analysis of the circuit shown in Figure P 11.11-2 indicates that when
vs(t) = 12 cos (4t + 30°) V
the mesh currents are given by
i1(t) = 1.647 cos (4t – 17.92°) A
and
i2(t) = 1.094 cos (4t – 13.15°) A
Check the results of this analysis by checking that the complex power supplied by the voltage
source is equal to the sum of the complex powers received by the other circuit elements.
4Ω
vs(t)
+
–
i1(t)
2Ω
2H
i2(t)
4H
Figure P 11.11-2
Solution:
The average complex supplied by the source is
Ss =
(12∠30° )(1.647∠ − 17.92° ) * = (12∠30° )(1.647∠17.92° ) = 9.88∠47.92° = 6.62 + j 7.33
2
2
The complex power received by the 4 Ω resistor is
S4Ω =
( 4 ×1.647∠ − 17.92° )(1.647∠ − 17.92° ) * = 5.43 + j 0
2
VA
The complex power received by the 2 Ω resistor is
S2Ω =
( 2 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 1.20 + j 0
2
VA
The current in the 2 H inductor is
(1.647∠ − 17.92° ) − (1.094∠ − 13.15° ) = 0.5640∠ − 27.19°
The complex power received by the 2 H inductor is
S 2H =
( j 8 × 0.5640∠ − 27.19° )( 0.5640∠ − 27.19° ) * = 0 + j 1.27
2
VA
The complex power received by the 4 H inductor is
S 4H =
( j 16 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 0 + j 9.57
2
VA
W
S 4 Ω + S 2 Ω + S 2H + S 4H = ( 5.43 + j 0 ) + (1.20 + j 0 ) + ( 0 + j 1.27 ) + ( 0 + j 9.57 ) = 6.63 + j 10.84 ≠ Ss
The complex power supplied by the voltage source is not equal to the sum of the complex
powers received by the other circuit elements. The mesh currents cannot be correct.
(Suppose the inductances of the inductors were interchanged. Then the complex power received
by the 4 H inductor would be
S 4H =
( j 16 × 0.5640∠ − 27.19° )( 0.5640∠ − 27.19° ) * = 0 + j 2.54
2
VA
The complex power received by the 2 H inductor would be
S 2H =
( j 8 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 0 + j 4.79
2
VA
S 4 Ω + S 2 Ω + S 2H + S 4H = ( 5.43 + j 0 ) + (1.20 + j 0 ) + ( 0 + j 2.54 ) + ( 0 + j 4.79 ) = 6.63 + j 7.33 ≈ S s
The mesh currents would be correct if the inductances of the two inductors were interchanged.
The computer was used to analyze the wrong circuit.)
P 11.11-3
Computer analysis of the circuit shown in Figure P 11.11-3 indicates that when
vs(t) = 12 cos (4t + 30°) V
the mesh currents are given by
i1(t) = 1.001 cos (4t – 47.01°) A
and
i2(t) = 0.4243 cos (4t – 15.00°) A
Check the results of this analysis by checking that the equations describing currents and voltages
of coupled coils are satisfied.
vs(t)
+
–
i1(t)
4H
3H
6H
i2(t)
15 Ω
Figure P 11.11-3
Solution:
The voltage across the left coil must be equal to the voltage source voltage. Notice that the mesh
currents both enter the undotted ends of the coils. In the frequency domain, the voltage across the
left coil is
( j 16 )(1.001∠ − 47.01° ) + ( j12 )( 0.4243∠ − 15° ) = 16.016∠42.99° + 5.092∠75°
= (11.715 + j 10.923) + (1.318 + j 4.918 )
= 13.033 + j 15.841 = 20.513∠50.55°
The voltage across the left coil isn’t equal to the voltage source voltage so the computer analysis
isn’t correct.
What happened? A data entry error was made while doing the computer analysis. Both coils
were described as having the dotted end at the top. If both coils had the dot at the top, the
equation for the voltage across the right coil would be
( j 16 )(1.001∠ − 47.01° ) − ( j12 )( 0.4243∠ − 15° ) = 16.016∠42.99° − 5.092∠75°
= (11.715 + j 10.923) − (1.318 + j 4.918 )
= 10.397 + j 6.005 = 12.007∠30.01°
This is equal to the voltage source voltage. The computer was used to analyze the wrong circuit.
Computer analysis of the circuit shown in Figure P 11.11-4 indicates that when
vs(t) = 12 cos (4t + 30°) V
the mesh currents are given by
i1(t) = 25.6 cos (4t + 30°) mA
and
i2(t) = 64 cos (4t + 30°) mA
Check the results of this analysis by checking that the equations describing currents and voltages
of ideal transformers are satisfied.
P 11.11-4
2:5
vs(t)
+
–
i1(t)
i2(t)
75
Figure P 11.11-4
Solution:
First check the ratio of the voltages across the coils.
n1 2
12∠30°
= 2.5 ≠
=
n2 5
( 75)( 0.064∠30° )
The transformer voltages don’t satisfy the equations describing the ideal transformer. The given
mesh currents are not correct.
That’ enough but let’s also check the ratio of coil currents. (Notice that the reference direction of
the i2(t) is different from the reference direction that we used when discussing transformers.)
n1 2
0.064∠30°
= 2.5 ≠
=
0.0256∠30°
n2 5
The transformer currents don’t satisfy the equations describing the ideal transformer.
n1
1
to be 2.5 instead of 0.4 =
In both case, we calculated
. This suggests that a data entry
2.5
n2
error was made while doing the computer analysis. The numbers of turns for the two coils was
interchanged.
PSpice Problems
The input to the circuit shown in Figure SP 11-1 is the voltage of the voltage source,
vs(t) = 7.5 sin (5t + 15°) V
The output is the voltage across the 4-Ω resistor, vo(t). Use PSpice to plot the input and output
voltages.
Hint: Represent the voltage source using the PSpice part called VSIN.
SP 11-1
8Ω
3H
vs(t)
+
–
5H
2H
+
vo(t)
–
Figure SP 11-1
Solution:
The coupling coefficient is k =
3
= 0.94868 .
2×5
4Ω
8Ω
3H
vs(t)
+
–
5H
2H
+
vo(t)
4Ω
–
Figure SP 11-1
The input to the circuit shown in Figure SP 11-1 is the voltage of the voltage source,
vs(t) = 7.5 sin (5t + 15°) = 7.5 cos (5t – 75°) V
The output is the voltage across the 4-Ω resistor, vo(t). Use PSpice to determine the average
power delivered to the coupled inductors.
Hint: Represent the voltage source using the PSpice part called VAC. Use printers (PSpice parts
called IPRINT and VPRINT) to measure the ac current and voltage of each coil.
SP 11-2
Solution: Here is the circuit with printers inserted to measure the coil voltages and currents:
Here is the output from the printers, giving the voltage of coil 2 as 2.498∠107.2°, the current of
coil 1 as 0.4484∠-94.57°, the current of coil 2 as 0.6245∠-72.77° and the voltage of coil 1 as
4.292∠-58.74°:
FREQ
7.958E-01
VM(N00984)
2.498E+00
VP(N00984)
1.072E+02
FREQ
IM(V_PRINT1)IP(V_PRINT1)
7.958E-01
4.484E-01 -9.457E+01
FREQ
IM(V_PRINT2)IP(V_PRINT2)
7.958E-01
6.245E-01 -7.277E+01
FREQ
7.958E-01
VM(N00959) VP(N00959)
4.292E+00 -5.874E+01
The power received by the coupled inductors is
p=
( 4.292 )( 0.4484 ) cos
2
= 0.78016 − .78000 ≈ 0
( −58.74 − ( −94.57 ) ) +
( 2.498)( 0.6245) cos
2
(107.2 − ( −72.77 ) )
The input to the circuit shown in Figure SP 11-3 is the voltage of the voltage source,
vs(t) = 48 cos (4t + 114°) V
The output is the voltage across the 9-Ω resistor, vo(t). Use PSpice to determine the average
power delivered to the transformer.
Hint: Represent the voltage source using the PSpice part called VAC.
SP 11-3
i(t)
8Ω
2:3
+
–
+
9Ω
vs(t)
vo(t)
–
Figure SP 11-3
Solution:
The inductance are selected so that
L2
L1
=
N2
N1
=
3
and the impedance of these inductors are
2
much larger that other impedance in the circuit. The 1 GΩ resistor simulates an open circuit
while providing a connected circuit.
Here is the output from the printers, giving the voltage of coil 2 as 24.00∠114.1°, the current of
coil 1 as 4.000∠114.0°, the current of coil 2 as 2.667∠-65.90° and the voltage of coil 1 as
16.00∠114.1°:
FREQ
6.366E-01
VM(N00984)
2.400E+01
VP(N00984)
1.141E+02
FREQ
IM(V_PRINT1)IP(V_PRINT1)
6.366E-01
4.000E+00
1.140E+02
FREQ
IM(V_PRINT2)IP(V_PRINT2)
6.366E-01
2.667E+00 -6.590E+01
FREQ
6.366E-01
VM(N00959)
1.600E+01
VP(N00959)
1.141E+02
The power received by the transformer is
p=
(16 )( 4 ) cos
(114 − 114 ) +
2
= 32 − 32.004 ≈ 0
( 24 )( 2.667 ) cos
2
(114 − ( −66 ) )
SP 11-4 Determine the value of the input impedance, Zt, of the circuit shown in Figure SP 11-4
at the frequency ω = 4 rad/s.
Hint: Connect a current source across the terminals of the circuit. Measure the voltage across the
current source. The value of impedance will be equal to the ratio of the voltage to the current.
8Ω
2Ω
5:2
4H
Zt
Figure SP 11-4
Solution:
The inductance are selected so that
L2
L1
=
N2
N1
=
2
and the impedance of these inductors are
5
much larger that other impedance in the circuit. The 1 GΩ resistor simulates an open circuit
while providing a connected circuit.
FREQ
6.366E-01
VM(N00921)
1.011E+02
VP(N00921)
7.844E+01
VR(N00921)
2.025E+01
VI(N00921)
9.903E+01
The printer output gives the voltage across the current source as
20.25 + j 99.03 = 101.1∠78.44° V
The input impedance is
Zt =
20.25 + j 99.03
= 20.25 + j 99.03 Ω = 101.1∠78.44° Ω
1
⎛ 52 ⎞
(We expected Z t = 8 + ⎜ 2 ⎟ ( 2 + j ( 4 )( 4 ) ) = 20.5 + j 100 Ω . That’s about 1% error.)
⎝2 ⎠
Design Problems
DP 11-1 A 100-kW induction motor, shown in Figure DP 11-1, is receiving 100 kW at 0.8
power factor lagging. Determine the additional apparent power in kVA that is made available by
improving the power factor to (a) 0.95 lagging and (b) 1.0. (c) Find the required reactive power
in kVAR provided by a set of parallel capacitors for parts (a) and (b). (d) Determine the ratio of
kVA released to the kVAR of capacitors required for parts (a) and (b) alone. Set up a table
recording the results of this problem for the two values of power factor attained.
Distribution
line
Capacitor
Induction motor
Figure DP 11-1
Solution:
(a)
P 100
⎧
=
= 125 kVA
| S |=
P = 100 kW ⎫
⎪
pf
0.8
⇒
⎬
⎨
pf=0.8 ⎭
⎪Q =| S |sin ( cos −1 0.8 ) = 125sin ( 36.9° ) = 75 kVAR
⎩
Now pf = 0.95 so
P 100
S= =
=105.3 kVA
pf 0.95
Q = S sin (cos −1 0.95) =105.3sin (18.2°) =32.9 kVAR
so an additional 125 − 105.3 = 19.7 kVA is available.
(b)
Now pf = 1 so
P 100
=
=100 kVA
pf 1
Q = S sin (cos −1 1) = 0
S=
and an additional 125−100= 25 kVA is available.
(c)
In part (a), the capacitors are required to reduce Q by 75 – 32.9 = 42.1 kVAR. In
part (b), the capacitors are required to reduce Q by 75 – 0 = 75 kVAR.
(d)
Corrected power factor
Additional available
apparent power
Reduction in reactive power
0.95
19.7 kVA
1.0
25 kVA
42.1 kVAR
75 kVAR
DP 11-2 Two loads are connected in parallel and supplied from a 7.2-kV rms 60-Hz source.
The first load is 50-kVA at 0.9 lagging power factor, and the second load is 45 kW at 0.91
lagging power factor. Determine the kVAR rating and capacitance required to correct the overall
power factor to 0.97 lagging.
Answer: C = 1.01 μF
Solution:
This example demonstrates that loads can be specified either by kW or kVA. The procedure is
as follows:
First load:
Second load:
Total load:
⎧⎪
P1 = S1 pf =( 50 )( 0.9 ) = 45 W
S1 =50 VA ⎫
⎬ ⇒ ⎨
−1
pf =0.9 ⎭
⎪⎩Q1 = S1 sin (cos 0.9) =50sin (25.8°) = 21.8 kVAR
P 45
⎧
S2 = 2 =
= 49.45 kVA
P2 = 45 W ⎫
⎪
pf 0.91
⎬ ⇒ ⎨
pf = 0.91 ⎭
⎪Q = S sin (cos −1 0.91) = 49.45sin (24.5°) = 20.5 kVAR
⎩ 2 2
S L = S1 + S 2 = (45 + 45) + j(21.8+20.5) = 90 + j 42.3 kVA
Specified load:
P 90
⎧
Ss = s =
=92.8 kVA
Ps =90 W ⎫
⎪
pf 0.97
⎬ ⇒ ⎨
pf = 0.97 ⎭
⎪Q = S sin (cos −1 0.97) =92.8sin (14.1°) = 22.6 kVAR
⎩ s s
The compensating capacitive load is Qc = 42.3 − 22.6 = 19.7 kVAR .
The required capacitor is calculated as
2
V
(7.2 × 103 ) 2
1
Xc = c =
= 2626 Ω ⇒ C =
= 1.01 μ F
3
Qc
19.7 × 10
377 (2626)
DP 11-3
(a)
Determine the load impedance Zab that will absorb maximum power if it is connected to
terminals a–b of the circuit shown in Figure DP 11-3.
(b)
Determine the maximum power absorbed by this load.
(c)
Determine a model of the load and indicate the element values.
5Ω
a
100 mH
+
10 cos 100 t V
+
–
–
+
vab
0.5 vab
–
b
Figure DP 11-3
Solution:
Find the open circuit voltage:
−10 + 5I + j10I − 0.5 Voc = 0
and
10 − Voc
5
Voc = 8∠36.9° = 6.4 + j 4.8 V
I=
so
Find the short circuit current:
I sc =
10∠0°
= 2∠0° A
5
The the Thevenin impedance is:
Zt =
Voc
= 3.2 + j 2.4 Ω
I sc
The short circuit forces the controlling voltage to be
zero. Then the controlled voltage is also zero.
Consequently the dependent source has been replaced
by a short circuit.
(a)
Maximum power transfer requires Z L = Z t * = 3.2 − j 2.4 Ω .
(c)
ZL can be implemented as the series combination of a resistor and a capacitor with
1
R = 3.2 Ω and C =
= 4.17 mF .
(100 ) (2.4)
(b)
Pmax =
| Voc |2
64
=
= 2.5 W
8R
8 ( 3.2 )
DP 11-4 Select the turns ratio n necessary to provide maximum power to the resistor R of the
circuit shown in Figure DP 11-4. Assume an ideal transformer. Select n when R = 4 and 8 Ω.
3 Ω
j4 Ω
1:n
R
Vs
+
–
j3 Ω
Ideal
Figure DP 11-4
Solution:
When n is selected to deliver maximum power to Z3, the value of the maximum power is given
as
2
⎛ R ⎞ Vs
⎜ 2⎟
⎝n ⎠ 2
P=
2
2
R⎞ ⎛ 3
⎛
⎞
3
+
+
+
4
⎜
⎟ ⎜
⎟
n2 ⎠ ⎝ n2
⎝
⎠
When R = 4 Ω,
2
n 2 R Vs
P=
25n 4 + 48n 2 + 25
4
2
2
3
dP
2 ⎡ 2 n(25n + 48n + 25) − n (100n + 96n ) ⎤
= R Vs ⎢
⎥
dn
(25n 4 + 48n 2 + 25) 2
⎣
⎦
5
4
⇒ − 50n + 50n = 0 ⇒ n = 1 ⇒ n = 1
0=
When R = 8 Ω, a similar calculation gives n = 1.31.
DP 11-5 An amplifier in a short-wave radio operates at 100 kHz. The load Z2 is connected to a
source through an ideal transformer, as shown in Figure DP 11-5. The load is a series connection
of a 10-Ω resistance and 10-μH inductance. The Zs consists of a 1-Ω resistance and a 1-μH
inductance.
(a)
Select an integer n in order to maximize the energy delivered to the load. Calculate I2 and
the energy to the load.
(b)
Add a capacitance C in series with Z2 in order to improve the energy delivered to the
load.
Zs
Vs
I2
1:n
+
–
Z2
Ideal
Figure DP 11-5
Solution:
Maximum power transfer requires
Equating real parts gives
10 + j 6.28
= Z1 = (1 + j 0.628 ) *
n2
10
= 1 ⇒ n = 3.16
n2
Equating imaginary parts requires
jX
= − j 0.628 ⇒
3.162
X = −6.28
This reactance can be realized by adding a capacitance C in series with the resistor and inductor
that comprise Z2. Then
−6.28 = X = −
1
1
+ 6.28 ⇒ C =
= 0.1267 μ F
5
5
( 2π ×10 ) C
( 2π ×10 ) (12.56 )
DP 11-6 A new electronic lamp (e-lamp) has been developed that uses a radio-frequency
sinusoidal oscillator and a coil to transmit energy to a surrounding cloud of mercury gas as
shown in Figure DP 11-6a. The mercury gas emits ultraviolet light that is transmitted to the
phosphor coating, which, in turn, emits visible light. A circuit model of the e-lamp is shown in
Figure DP 11.6b. The capacitance C and the resistance R are dependent on the lamp’s spacing
design and the type of phosphor. Select R and C so that maximum power is delivered to R, which
relates to the phosphor coating (Adler, 1992). The circuit operates at ω0 = 107 rad/s.
Phosphorus
coating
100 Ω
Mercury
vapor
1 mH
Coil
Coil
V0 sin ω 0 t V
+
–
C
R
(b)
(a)
Figure DP 11-6
Solution:
Maximum power transfer requires
1
|| R = (100 + j107 ×10−6 ) *
7
j10 C
R
= 100 − j10
1 + j107 R C
R = (100 − j10) (1 + j107 R C ) = 100 + 108 R C + j (109 R C − 10 )
Equating real and imaginary parts yields
R = 100 + 108 R C and 109 R C − 10 = 0
then
RC = 10−8
⎛ 10−8 ⎞
10−8
⇒ R = 100 + 108 R ⎜
= 0.101 nF
⎟ = 99 Ω ⇒ C =
99
⎝ R ⎠
Chapter 12: Three-Phase Circuits
Exercises
Exercise 12.2-1 The Y-connected three-phase voltage source has Vc = 120∠ − 240° V rms .
Find the line-to-line voltage Vbc.
Answer: 207.8 ∠90° V rms
Solution:
VC = 120∠ − 240° so VA = 120∠0° and VB = 120∠ − 120°
Vbc = 3 (120 ) ∠ − 90°
1
Exercise 12.3-1 Determine complex power delivered to the three-phase load of a four-wire Yto-Y circuit such as the one shown in Figure 12.3-1. The phase voltages of the Y-connected
source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. The load
impedances are ZA = 80 + j50 Ω, ZB = 80 + j80 Ω, and ZC = 100 – j25 Ω.
Answer:
SA = 129 + j81 VA, SB = 90 + j90 VA, SC = 136 – j34 VA, and S = 355 + j137 VA
Ia A
B
b
a
A
Ib B
–
–
+
+
Va
ZB
Vb
ZA
n
N
IN n
ZC
+
–
Vc
c
C
Ic C
Figure 12.3-1
Solution:
Mathcad analysis (12x4_1.mcd):
Vp := 120
Describe the three-phase source:
j⋅
Va := Vp⋅ e
π
180
⋅0
j⋅
Vb := Va⋅ e
Describe the three-phase load:
Calculate the line currents:
IaA = 1.079 − 0.674i
IaA = 1.272
180
π
⋅ arg( IaA ) = −32.005
IaA :=
π
180
⋅ − 120
Vc := Va⋅ e
ZA := 80 + j⋅ 50
Va
IbB :=
ZA
j⋅
Vb
ZB
IbB = 1.061
π
⋅ 120
ZB := 80 + j⋅ 80
IbB = −1.025 − 0.275i
180
π
180
⋅ arg( IbB) = −165
IcC :=
ZC := 100 − j⋅ 25
Vc
ZC
IcC = −0.809 + 0.837i
IcC = 1.164
180
π
⋅ arg( IcC) = 134.036
2
3
Exercise 12.3-2 Determine complex power delivered to the three-phase load of a four-wire Yto-Y circuit such as the one shown in Figure 12.3-1. The phase voltages of the Y-connected
source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. The load
impedances are ZA = ZB = ZC = 40 + j30 Ω.
Answer: SA = SB = SC = 230 + j173 VA and S = 691 + j518 VA
Ia A
B
b
a
A
Ib B
–
–
+
+
Va
ZB
Vb
ZA
n
N
IN n
ZC
+
–
Vc
c
C
Ic C
Figure 12.3-1
Solution:
Mathcad analysis (12x4_2.mcd):
Vp := 120
Describe the three-phase source:
j⋅
Va := Vp⋅ e
π
180
⋅0
j⋅
Vb := Va⋅ e
Describe the three-phase load:
Calculate the line currents:
IaA = 1.92 − 1.44i
IaA = 2.4
180
π
⋅ arg( IaA ) = −36.87
IaA :=
π
180
⋅ − 120
Vc := Va⋅ e
ZA := 40 + j⋅ 30
Va
IbB :=
ZA
j⋅
Vb
ZB
IbB = 2.4
π
180
⋅ 120
ZB := ZA
IbB = −2.207 − 0.943i
180
π
⋅ arg( IbB) = −156.87
IcC :=
ZC := ZA
Vc
ZC
IcC = 0.287 + 2.383i
IcC = 2.4
180
π
⋅ arg( IcC) = 83.13
4
5
Exercise 12.3-3 Determine complex power delivered to the three-phase load of a three-wire Yto-Y circuit such as the one shown in Figure 12.3-2. The phase voltages of the Y-connected
source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. The load
impedances are ZA = 80 + j50 Ω, ZB = 80 + j80 Ω, and ZC = 100 – j25 Ω.
Intermediate Answer: VnN = 28.89 ∠–150.5 V rms
Answer: S = 392 + j142 VA
IaA
b
B
a
A
IbB
–
–
+
+
ZB
Va
ZA
Vb
n
–
N
+
VNn
ZC
Vc
+
–
c
C
IcC
Figure 12.3-2
Solution:
Mathcad analysis (12x4_3.mcd):
Vp := 120
Describe the three-phase source:
j⋅
Va := Vp⋅ e
π
180
⋅0
j⋅
Vb := Va⋅ e
Describe the three-phase load:
π
180
⋅ − 120
ZA := 80 + j⋅ 50
j⋅
Vc := Va⋅ e
π
180
⋅ 120
ZB := 80 + j⋅ 80
ZC := 100 − j⋅ 25
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN :=
ZA ⋅ ZC⋅ e
4
j⋅ ⋅ π
3
+ ZA ⋅ ZB⋅ e
2
j⋅ ⋅ π
3
+ ZB⋅ ZC
ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC
VnN = −25.137 − 14.236i
VnN = 28.888
⋅ Vp
180
π
⋅ arg( VnN) = −150.475
6
Calculate the line currents:
IaA :=
IaA = 1.385 − 0.687i
IaA = 1.546
180
π
Check:
⋅ arg( IaA ) = −26.403
Va − VnN
ZA
IbB :=
Vb − VnN
ZB
IbB = −0.778 − 0.343i
IbB = 0.851
180
π
IcC :=
⋅ arg( IbB) = −156.242
Vc − VnN
ZC
IcC = −0.606 + 1.03i
IcC = 1.195
180
π
⋅ arg( IcC) = 120.475
IaA + IbB + IcC = 0
Calculate the power delivered to the load:
⎯
⎯
SA := IaA ⋅ IaA ⋅ ZA
SB := IbB⋅ IbB⋅ ZB
SA = 191.168 + 119.48i
Total power delivered to the load:
SB = 57.87 + 57.87i
⎯
SC := IcC⋅ IcC⋅ ZC
SC = 142.843 − 35.711i
SA + SB + SC = 391.88 + 141.639i
7
Exercise 12.3-4 Determine complex power delivered to the three-phase load of a three-wire Yto-Y circuit such as the one shown in Figure 12.3-2. The phase voltages of the Y-connected
source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. The load
impedances are ZA = ZB = ZC = 40 + j30 Ω.
Answer: SA = SB = SC = 230 + j173 VA and S = 691 + j518 VA
IaA
b
B
a
A
IbB
ZB
–
–
+
+
Va
ZA
Vb
n
–
N
+
VNn
ZC
Vc
+
–
c
C
IcC
Figure 12.3-2
Solution:
Mathcad analysis (12x4_4.mcd):
Vp := 120
Describe the three-phase source:
j⋅
Va := Vp⋅ e
π
180
⋅0
Describe the three-phase load:
j⋅
Vb := Va⋅ e
π
180
⋅ − 120
ZA := 40 + j⋅ 30
j⋅
Vc := Va⋅ e
ZB := ZA
π
180
⋅ 120
ZC := ZA
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN :=
ZA ⋅ ZC⋅ e
4
j⋅ ⋅ π
3
+ ZA ⋅ ZB⋅ e
2
j⋅ ⋅ π
3
+ ZB⋅ ZC
ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC
⋅ Vp
8
9
Exercise 12.5-1
Consider the three-phase circuit shown in Figure 12.5-1. The voltages of the Y-connected source
are
360
360
360
∠–30° V rms, Vb =
∠–150° V rms, and Vc =
∠90° V rms
Va =
3
3
3
The Δ-connected load is balanced. The impedance of each phase is ZΔ = 180 ∠45° Ω. Determine
the phase and line currents when the line-to-line voltage is 360 V rms.
Partial Answer: IAB = 2 ∠45° A rms and IaA = 3.46 ∠15° A rms
I aA
b
a
–
I bB
+
+
B
Z3
–
–
Va
Vb
n
I AB
+
–
+
VAB
Z1
VBC
Vc
+
A
I CA
VCA
I BC
–
c
–
Z2
+
C
I cC
Figure 12.5-1
Solution:
Balanced delta
load:
phase currents:
line currents:
(See Table 12.5-1)
Z Δ = 180∠ − 45°
I AB
VAB
360∠0°
=
=
= 2∠45° A
D
Z Δ 180∠− 45
I BC =
VBC 360∠−120°
=
= 2∠− 75° A
Z Δ 18∠− 45°
I CA =
VCA 360∠120°
=
= 2∠165° A
°
Z Δ 180∠− 45
I A = I AB − I CA = 2∠45° − 2∠165° = 2 3∠15° A
I B = 2 3∠−105° A
I C = 2 3∠135° A
10
Exercise 12.6-1 Figure 12.6-1a shows a balanced Y-to-Δ three-phase circuit. The phase
voltages of the Y-connected source are Va = 110 ∠0° V rms, Vb = 110 ∠–120° V rms, and Vc =
110 ∠120° V rms. The line impedances are each ZL = 10 + j25 Ω. The impedances of the Δconnected load are each ZΔ = 150 + j270 Ω. Determine the phase currents in the Δ-connected
load.
Answer: IAB = 0.49 ∠–32.5° A rms, IBC = 0.49 ∠–152.5° A rms, ICA = 0.49 ∠87.5° A rms
Figure 12.6-1a
Solution:
Three-wire Y-to-Delta Circuit with line impedances
Mathcad analysis (12x4_4.mcd):
Vp := 110
Describe the three-phase source:
j⋅
Va := Vp⋅ e
π
180
⋅0
j⋅
Vb := Va⋅ e
π
180
⋅ − 120
Z1 := 150 + j⋅ 270
Describe the delta connected load:
j⋅
Vc := Va⋅ e
π
180
Z2 := Z1
⋅ 120
Z3 := Z1
Convert the delta connected load to the equivalent Y connected load:
ZA :=
Z1⋅ Z3
ZB :=
Z1 + Z2 + Z3
ZA = 50 + 90i
Z2⋅ Z3
Z1 + Z2 + Z3
ZB = 50 + 90i
Z1 + Z2 + Z3
ZC = 50 + 90i
ZaA := 10 + j⋅ 25
Describe the three-phase line:
Z1⋅ Z2
ZC :=
ZbB := ZaA
ZcC := ZaA
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN :=
( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e
4
j⋅ ⋅ π
3
+ ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e
2
j⋅ ⋅ π
3
+ ( ZbB + ZB) ⋅ ( ZcC + ZC)
( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC)
− 14
VnN = −1.172 × 10
− 14
+ 1.784i× 10
− 14
VnN = 2.135 × 10
180
π
⋅ Vp
⋅ arg( VnN) = 123.304
11
Calculate the line currents:
IaA :=
IaA = 0.392 − 0.752i
IaA = 0.848
180
π
Check:
⋅ arg( IaA ) = −62.447
Va − VnN
IbB :=
ZA + ZaA
Vb − VnN
ZB + ZbB
IbB = −0.847 + 0.036i
IbB = 0.848
180
π
IcC :=
⋅ arg( IbB) = 177.553
Vc − VnN
ZC + ZcC
IcC = 0.455 + 0.716i
IcC = 0.848
180
π
⋅ arg( IcC) = 57.553
IaA + IbB + IcC = 0
Calculate the phase voltages of the Y-connected load:
VAN := IaA ⋅ ZA
VAN = 87.311
180
π
⋅ arg( VAN) = −1.502
VBN := IbB⋅ ZB
VBN = 87.311
180
π
⋅ arg( VBN) = −121.502
VCN := IcC⋅ ZC
VCN = 87.311
180
π
⋅ arg( VCN) = 118.498
Calculate the line-to-line voltages at the load:
VAB := VAN − VBN
VAB = 151.227
180
π
⋅ arg( VAB) = 28.498
VBC:= VBN − VCN
VBC = 151.227
180
π
⋅ arg( VBC) = −91.502
VCA := VCN − VAN
VCA = 151.227
180
π
⋅ arg( VCA) = 148.498
Calculate the phase currents of the Δ -connected load:
IAB :=
VAB
Z3
IAB = 0.49
180
π
⋅ arg( IAB) = −32.447
IBC :=
VBC
Z1
IBC = 0.49
180
π
⋅ arg( IBC) = −152.447
ICA :=
VCA
Z2
ICA = 0.49
180
π
⋅ arg( ICA) = 87.553
12
Exercise 12.7-1 Figure 12.6-1a shows a balanced Y-to-Δ three-phase circuit. The phase
voltages of the Y-connected source are Va = 110 ∠0° V rms, Vb = 110 ∠–120° V rms, and Vc =
110 ∠120° V rms. The line impedances are each ZL = 10 + j25 Ω. The impedances of the Δconnected load are each ZΔ = 150 + j 270 Ω. Determine the average power delivered to the Δconnected load.
Intermediate Answer: IaA = 0.848 ∠–62.5° A rms and VAN = 87.3 ∠–1.5° V rms
Answer: P = 107.9 W
Figure 12.6-1a
Solution:
Continuing Ex. 12.6-1:
Calculate the power delivered to the load:
⎯
⎯
SA := IaA ⋅ IaA ⋅ ZA
SB := IbB⋅ IbB⋅ ZB
SA = 35.958 + 64.725i
Total power delivered to the load:
SB = 35.958 + 64.725i
⎯
SC := IcC⋅ IcC⋅ ZC
SC = 35.958 + 64.725i
SA + SB + SC = 107.875 + 194.175i
13
Exercise 12.8-1 The line current to a balanced three-phase load is 24 A rms. The line-to-line
voltage is 450 V rms, and the power factor of the load is 0.47 lagging. If two wattmeters are
connected as shown in Figure 12.8-2, determine the reading of each meter and the total power to
the load.
Answer: P1 = –371 W, P2 = 9162 W, and P = 8791 W
IB
W2
B
IA
A
+
W1
B
A
Z
Z
VAC
Z
C
–
C
FIGURE 12.8-2
Solution:
P1 = VAB I A cos(θ +30° ) + VCB I C cos(θ − 30° )= P + P2
1
pf = .4 lagging ⇒ θ = 61.97
°
So P T = 450(24) ⎡⎣ cos 91.97° + cos 31.97° ⎤⎦ = 8791 W
∴ P 1 = − 371 W P2 = 9162 W
14
Exercise 12.8-2 The two wattmeters are connected as shown in Figure 12.8-2 with P1 = 60 kW
and P2 = 40 kW, respectively. Determine (a) the total power and (b) the power factor.
Answers: (a) 100 kW (b) 0.945 leading
IB
W2
B
IA
A
W1
+
A
B
Z
Z
VAC
Z
C
–
C
Figure 12.8-2
Ex. 12.8-2
Consider Fig. 12.8-1 with P1 = 60 kW P2 = 40 kW .
(a.) P = P1 + P2 = 100 kW
(b.) use equation 12.9-7 to get
tan θ = 3
P2 − P1
40−60
= 3
= − .346 ⇒ θ = − 19.11°
PL + P2
100
then
pf = cos ( − 19.110°) = 0.945 leading
15
Problems
Section 12-2: Three Phase Voltages
P 12.2-1
A balanced three-phase Y-connected load has one phase voltage:
Vc = 277 ∠45° V rms
The phase sequence is abc. Find the line-to-line voltages VAB, VBC, and VCA. Draw a phasor
diagram showing the phase and line voltages.
Solution:
Given VC = 277 ∠45° and an abc phase sequence:
VA = 277 ∠ ( 45−120 ) ° = 277 ∠ − 75°
VB = 277 ∠( 45° +120 )° = 277 ∠165°
VAB = VA − VB =( 277 ∠− 75° )−( 277 ∠165° )
=( 71.69 − j 267.56 ) −( −267.56+ j 71.69 )
=339.25− j 339.25 = 479.77 ∠− 45° 480 ∠− 45°
Similarly:
VBC = 480 ∠ − 165° and VCA = 480 ∠75°
P 12.2-2
A three-phase system has a line-to-line voltage
VBA = 12,470 ∠–35° V rms
with a Y load. Find the phase voltages when the phase sequence is abc.
Solution:
VAB
3∠30°
= 12470 ∠145° V
VAB = VA × 3∠30° ⇒ VA =
In our case:
So
VAB = −VBA = − (12470 ∠−35° )
VA =
12470 ∠145°
= 7200∠115°
3∠30°
Then, for an abc phase sequence:
VC = 7200 ∠ (115 + 120 ) ° = 7200 ∠235° = 7200 ∠ − 125°
VB = 7200 ∠ (115 − 120 ) ° = 7200 ∠ − 5° V
1
P 12.2-3
A three-phase system has a line-to-line voltage
Vab = 1500 ∠30° V rms
with a Y load. Determine the phase voltage.
Solution:
Vab = Va × 3∠30° ⇒ Va =
Vab
3∠30°
In our case, the line-to-line voltage is
So the phase voltage is
Vab = 1500 ∠30° V
1500 ∠30°
Va =
= 866∠0° V
3∠30°
2
Section 12.3
P 12.3-1
The Y-to-Y Circuit
Consider a three-wire Y-to-Y circuit. The voltages of the Y-connected source are
Va = (208/ 3 ) ∠0° V rms, Vb = (208/ 3 ) ∠–120° V rms, and Vc = (208/ 3 )∠120° V rms.
The Y-connected load is balanced. The impedance of each phase is Z = 12 ∠30° Ω.
(a)
(b)
(c)
(d)
Find the phase voltages.
Find the line currents and phase currents.
Show the line currents and phase currents on a phasor diagram.
Determine the power dissipated in the load.
Solution:
Balanced, three-wire, Y-Y circuit:
where Z A = Z B = Z C = 12∠30 = 10.4 + j 6
MathCAD analysis (12p4_1.mcd):
Vp :=
Describe the three-phase source:
208
3
π
j⋅
Va := Vp⋅ e
⋅0
180
j⋅
⋅ − 120
j⋅
180
Vb := Va⋅ e
Describe the balanced three-phase load:
π
Vc := Va⋅ e
ZA := 10.4 + j⋅ 6
π
⋅ 120
180
ZB := ZA
ZC := ZB
Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero:
VnN :=
ZA ⋅ ZC⋅ e
4
j⋅ ⋅ π
3
+ ZA ⋅ ZB⋅ e
IaA :=
Va − VnN
ZA
IaA = 8.663 − 4.998i
Check:
− 14
⋅ Vp
IbB :=
VnN = 2.762 × 10
Vb − VnN
IbB = −8.66 − 5.004i
IaA = 10.002
π
+ ZB⋅ ZC
ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC
Calculate the line currents:
180
2
j⋅ ⋅ π
3
IbB = 10.002
180
⋅ arg( IaA ) = −29.982
π
− 15
IaA + IbB + IcC = 4.696 × 10
IcC :=
ZB
Vc − VnN
ZC
−3
IcC = −3.205 × 10
+ 10.002i
IcC = 10.002
⋅ arg( IbB) = −149.982
180
π
⋅ arg( IcC) = 90.018
− 14
− 1.066i× 10
1
Calculate the power delivered to the load:
⎯
⎯
SA := IaA ⋅ IaA ⋅ ZA
SB := IbB⋅ IbB⋅ ZB
3
SA = 1.04 × 10 + 600.222i
Total power delivered to the load:
⎯
SC := IcC⋅ IcC⋅ ZC
3
SB = 1.04 × 10 + 600.222i
3
3
SC = 1.04 × 10 + 600.222i
3
SA + SB + SC = 3.121 × 10 + 1.801i× 10
Consequently:
(a) The phase voltages are
208
∠0° = 120∠0° V rms, Vb = 120∠ − 120° V rms and Vc = 120∠120° V rms
3
(b) The currents are equal the line currents
(c)
Va =
I a = I aA = 10∠ − 30° A rms, I b = I bB = 10∠ − 150° A rms
and
I c = I cC = 10∠90° A rms
(d) The power delivered to the load is S = 3.121 + j1.801 kVA .
2
P 12.3-2 A balanced three-phase Y-connected supply delivers power through a three-wire plus
neutral-wire circuit in a large office building to a three-phase Y-connected load. The circuit
operates at 60 Hz. The phase voltages of the Y-connected source are
Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms.
Each transmission wire, including the neutral wire, has a 2-Ω resistance, and the balanced Y load
has a 10-Ω resistance in series with 100 mH. Find the line voltage and the phase current at the
load.
Solution:
Balanced, three-wire, Y-Y circuit:
where
Va = 120∠0° Vrms, Vb = 120∠ − 120° Vrms
and Vc = 120∠120° Vrms
Z A = Z B = Z C = 10 + j ( 2 × π × 60 ) (100 × 10−3 )
and
Z aA
= 10 + j 37.7 Ω
= Z bB = Z cC = 2 Ω
Mathcad Analysis (12p4_2.mcd):
3
Consequently, the line-to-line voltages at the source are:
Vab = Va × 3∠30° = 120∠0°× 3∠30° = 208∠30° Vrms,
Vbc = 208∠ − 120° Vrms and Vca = 208∠120° Vrms
The line-to-line voltages at the load are:
VAB = VA × 3∠30° = 118.3∠3°× 3∠30° = 205∠33° Vrms,
Vbc = 205∠ − 117° Vrms and Vca = 205∠123° Vrms
and the phase currents are
I a = I aA = 10∠ − 72° A rms, I b = I bB = 3∠168° A rms and I c = I cC = 3∠48° A rms
4
P 12.3-3 A Y-connected source and load are shown in Figure P 12.3-3. (a) Determine the rms
value of the current iA (t). (b) Determine the average power delivered to the load.
ia(t)
+
–
–
+
1H
1H
12 Ω
12 Ω
10 cos 16t V
10 cos (16t – 120°) V
Source
Load
12 Ω
–
+
10 cos (16t + 120°) V
1H
Figure P 12.3-3
Solution:
Balanced, three-wire, Y-Y circuit:
where
Va = 10∠0° V = 7.07∠0° V rms, Vb = 7.07∠ − 120° V rms and Vc = 7.07∠120° V rms
and
Z A = Z B = Z C = 12 + j (16 )(1) = 12 + j16 Ω
MathCAD analysis (12p4_3.mcd):
5
Vp :=
Describe the three-phase source:
10
2
π
j⋅
Va := Vp⋅ e
⋅0
j⋅
180
π
⋅ − 120
j⋅
180
Vb := Va⋅ e
Vc := Va⋅ e
ZA := 12 + j⋅ 16
Describe the balanced three-phase load:
π
⋅ 120
180
ZB := ZA
ZC := ZB
Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero:
VnN :=
ZA ⋅ ZC⋅ e
4
j⋅ ⋅ π
3
+ ZA ⋅ ZB⋅ e
IaA :=
IaA = 0.212 − 0.283i
IaA = 0.354
π
+ ZB⋅ ZC
ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC
Calculate the line currents:
180
2
j⋅ ⋅ π
3
⋅ arg( IaA ) = −53.13
Va − VnN
ZA
− 15
⋅ Vp
IbB :=
VnN = 1.675 × 10
Vb − VnN
IbB = −0.351 − 0.042i
IbB = 0.354
180
π
⋅ arg( IbB) = −173.13
Calculate the power delivered to the load:
⎯
⎯
SA := IaA ⋅ IaA ⋅ ZA
SB := IbB⋅ IbB⋅ ZB
SA = 1.5 + 2i
Total power delivered to the load:
SB = 1.5 + 2i
IcC :=
ZB
Vc − VnN
ZC
IcC = 0.139 + 0.325i
IcC = 0.354
180
π
⋅ arg( IcC) = 66.87
⎯
SC := IcC⋅ IcC⋅ ZC
SC = 1.5 + 2i
SA + SB + SC = 4.5 + 6i
Consequently
(a) The rms value of ia(t) is 0.354 A rms.
(b) The average power delivered to the load is P = Re {S} = Re {4.5 + j 6} = 4.5 W
6
P 12.3-4 An unbalanced Y-Y circuit is shown in Figure P 12.3-4. Find the average power
delivered to the load.
Hint:
VNn(ω) = 27.4 ∠–63.6 V
Answer: 436.4 W
a
10 Ω
5 mH
A
20 Ω
60 mH
b
10 Ω
5 mH
B
40 Ω
40 mH
C
60 Ω
+
–
100 cos
(377t)
–
N
+
n
100 cos
(377t +120°)
c
10 Ω
5 mH
20 mH
+
–
100 cos
(377t + 240°)
Source
Line
Load
Figure P 12.3-4
Solution:
Unbalanced, three-wire, Y-Y circuit:
where
Va = 100∠0° V = 70.7∠0° V rms, Vb = 70.7∠ − 120° V rms and Vc = 7.07∠120° V rms
Z A = 20 + j ( 377 ) ( 60 ×10−3 ) = 20 + j 22.6 Ω, Z B = 40 + j ( 377 ) ( 40 × 10−3 ) = 40 + j 15.1 Ω
Z C = 60 + j ( 377 ) ( 20 × 10−3 ) = 60 + j 7.54 Ω
and
Z aA = Z bB = Z cC = 10 + j ( 377 ) ( 5 × 10−3 ) = 10 + j 1.89 Ω
Mathcad Analysis (12p4_4.mcd):
7
Vp := 100
Describe the three-phase source:
π
j⋅
Va := Vp⋅ e
⋅0
j⋅
180
π
⋅ 120
180
Vb := Va⋅ e
π
j⋅
⋅ − 120
180
Vc := Va⋅ e
Enter the frequency of the 3-phase source: ω := 377
Describe the three-phase load:
ZA := 20 + j⋅ ω⋅ 0.06
ZB := 40 + j⋅ ω⋅ 0.04
Describe the three-phase line:
ZaA := 10 + j⋅ ω⋅ 0.005 ZbB := ZaA
ZC := 60 + j⋅ ω⋅ 0.02
ZcC := ZaA
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN :=
( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e
4
j⋅ ⋅ π
3
+ ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e
2
j⋅ ⋅ π
3
+ ( ZbB + ZB) ⋅ ( ZcC + ZC)
( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC)
VnN = 12.209 − 24.552i
Calculate the line currents:
IaA :=
IaA = 2.156 − 0.943i
IaA = 2.353
180
π
180
VnN = 27.42
Va − VnN
ZA + ZaA
IbB :=
π
IcC :=
ZB + ZbB
180
π
⋅ arg( IbB) = 100.492
SA = 55.382 + 62.637i
Total power delivered to the load:
(
ZC + ZcC
IcC = 1.244
Calculate the power delivered to the load:
⎯
⎯
IbB⋅ IbB
IaA ⋅ IaA
⋅ ZA
SB :=
⋅ ZB
SA :=
2
2
)
Vc − VnN
IcC = −0.99 − 0.753i
IbB = 2.412
⋅ arg( IaA ) = −23.619
(
⋅ arg( VnN) = −63.561
Vb − VnN
IbB = −0.439 + 2.372i
⋅ Vp
)
SB = 116.402 + 43.884i
180
π
⋅ arg( IcC) = −142.741
SC :=
⎯
(IcC
⋅ IcC)
2
⋅ ZC
SC = 46.425 + 5.834i
SA + SB + SC = 218.209 + 112.355i
The average power delivered to the load is P = Re {S} = Re {218.2 + j112.4} = 218.2 W
8
P 12.3-5 A balanced Y-Y circuit is shown in Figure P 12.3-5. Find the average power delivered
to the load.
a
10 Ω
5 mH
A
20 Ω
60 mH
b
10 Ω
5 mH
B
20 Ω
60 mH
C
20 Ω
+
–
100 cos
(377t)
–
N
+
n
100 cos
(377t +120°)
c
10 Ω
5 mH
60 mH
+
–
100 cos
(377t + 240°)
Source
Line
Load
Figure P 12.3-5
Solution:
Balanced, three-wire, Y-Y circuit:
where
Va = 100∠0° V = 70.7∠0° V rms, Vb = 70.7∠ − 120° V rms and Vc = 7.07∠120° V rms
Z A = Z B = Z C = 20 + j ( 377 ) ( 60 × 10−3 ) = 20 + j 22.6 Ω
and
Z aA = Z bB = Z cC = 10 + j ( 377 ) ( 5 × 10−3 ) = 10 + j 1.89 Ω
Mathcad Analysis (12p4_5.mcd):
9
Vp := 100
Describe the three-phase source:
π
j⋅
Va := Vp⋅ e
⋅0
j⋅
180
π
⋅ 120
180
Vb := Va⋅ e
π
j⋅
⋅ − 120
180
Vc := Va⋅ e
Enter the frequency of the 3-phase source: ω := 377
Describe the three-phase load:
ZA := 20 + j⋅ ω⋅ 0.06
ZB := ZA
ZC := ZA
Describe the three-phase line:
ZaA := 10 + j⋅ ω⋅ 0.005
ZbB := ZaA
ZcC := ZaA
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN :=
( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e
4
j⋅ ⋅ π
3
+ ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e
2
j⋅ ⋅ π
3
+ ( ZbB + ZB) ⋅ ( ZcC + ZC)
( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC)
− 15
VnN = −8.982 × 10
− 14
Calculate the line currents:
IaA :=
IaA = 1.999 − 1.633i
IaA = 2.582
180
π
− 14
+ 1.879i× 10
⋅ arg( IaA ) = −39.243
VnN = 2.083 × 10
Va − VnN
ZA + ZaA
IbB :=
180
π
⋅ arg( VnN) = 115.55
Vb − VnN
IcC :=
ZB + ZbB
IbB = 0.415 + 2.548i
π
⋅ arg( IbB) = 80.757
Vc − VnN
ZC + ZcC
IcC = −2.414 − 0.915i
IbB = 2.582
180
⋅ Vp
IcC = 2.582
180
π
⋅ arg( IcC) = −159.243
Calculate the power delivered to the load:
SA :=
⎯
(IaA
⋅ IaA )
⋅ ZA
2
SA = 66.645 + 75.375i
Total power delivered to the load:
SB :=
⎯
(IbB
⋅ IbB)
⋅ ZB
2
SB = 66.645 + 75.375i
SC :=
⎯
(IcC
⋅ IcC)
⋅ ZC
2
SC = 66.645 + 75.375i
SA + SB + SC = 199.934 + 226.125i
The average power delivered to the load is P = Re {S} = Re {200 + j 226} = 200 W
10
P 12.3-6 An unbalanced Y-Y circuit is shown in Figure P 12.3-6 Find the average power
delivered to the load.
Hint: VNn (ω) = 1.755∠–29.5 V
Answer: 436.4 W
a
A
4Ω
1H
b
B
2Ω
2H
c
C
4Ω
+
–
10 cos
(4t – 90°)
–
N
+
n
10 cos
(4t + 150°)
2H
+
–
10 cos
(4t + 30°)
Source
Line
Load
Figure P 12.3-6
Solution:
Unbalanced, three-wire, Y-Y circuit:
where
Va = 10∠ − 90° V = 7.07∠ − 90° V rms, Vb = 7.07∠150° V rms and Vc = 7.07∠30° V rms
and
Z A = 4 + j ( 4 )(1) = 4 + j 4 Ω, Z B = 2 + j ( 4 )( 2 ) = 2 + j 8 Ω and Z C = 4 + j ( 4 )( 2 ) = 4 + j 8 Ω
Mathcad Analysis (12p4_6.mcd):
11
Vp := 10
Describe the three-phase source:
π
j⋅
Va := Vp⋅ e
⋅ − 90
j⋅
180
π
⋅ 150
180
Vb := Vp⋅ e
π
j⋅
Vc := Vp⋅ e
⋅ 30
180
Enter the frequency of the 3-phase source: ω := 4
ZA := 4 + j⋅ ω⋅ 1
Describe the three-phase load:
ZB := 2 + j⋅ ω⋅ 2
ZC := 4 + j⋅ ω⋅ 2
Calculate the voltage at the neutral of the load with respect to the neutral of the source:
VnN :=
ZA ⋅ ZC⋅ Vb + ZA ⋅ ZB⋅ Vc + ZB⋅ ZC⋅ Va
ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC
VnN = 1.528 − 0.863i
Calculate the line currents:
IaA :=
IaA = −1.333 − 0.951i
IaA = 1.638
180
π
180
VnN = 1.755
Va − VnN
ZA
IbB :=
π
Vb − VnN
IbB = 0.39 + 1.371i
180
π
)
⋅ arg( IbB) = 74.116
SA = 5.363 + 5.363i
Total power delivered to the load:
(
Vc − VnN
ZC
IcC = 0.943 − 0.42i
IcC = 1.032
Calculate the power delivered to the load:
⎯
⎯
IaA ⋅ IaA
IbB⋅ IbB
SA :=
⋅ ZA
SB :=
⋅ ZB
2
2
(
IcC :=
ZB
IbB = 1.426
⋅ arg( IaA ) = −144.495
⋅ arg( VnN) = −29.466
)
SB = 2.032 + 8.128i
180
π
⋅ arg( IcC) = −24.011
SC :=
⎯
(IcC
⋅ IcC)
2
⋅ ZC
SC = 2.131 + 4.262i
SA + SB + SC = 9.527 + 17.754i
The average power delivered to the load is P = Re {S} = Re {9.527 + j17.754} = 9.527 W
12
P 12.3-7 A balanced Y-Y circuit is shown in Figure P 12.3-7. Find the average power delivered
to the load.
a
A
4Ω
2H
b
B
4Ω
2H
c
C
4Ω
+
–
10 cos
(4t – 90°)
–
N
+
n
10 cos
(4t + 150°)
2H
+
–
10 cos
(4t + 30°)
Source
Line
Load
Figure P 12.3-7
Solution:
Balanced, three-wire, Y-Y circuit:
where
Va = 10∠ − 90° V = 7.07∠ − 90° V rms, Vb = 7.07∠150° V rms and Vc = 7.07∠30° V rms
and
Z A = Z B = Z C = 4 + j ( 4 )( 2 ) = 4 + j 8 Ω
Mathcad Analysis (12p4_7.mcd):
13
Vp := 10
Describe the three-phase source:
j⋅
Va := Vp⋅ e
π
⋅ − 90
180
π
j⋅
⋅ 150
j⋅
180
Vb := Vp⋅ e
Vc := Vp⋅ e
π
⋅ 30
180
Enter the frequency of the 3-phase source: ω := 4
Describe the three-phase load:
ZA := 4 + j⋅ ω⋅ 2
ZB := ZA
ZC := ZA
The voltage at the neutral of the load with respect to the neutral of the source should be zero:
VnN :=
ZA ⋅ ZC⋅ Vb + ZA ⋅ ZB⋅ Vc + ZB⋅ ZC⋅ Va
Calculate the line currents:
IaA :=
IaA = −1 − 0.5i
π
Va − VnN
ZA
IbB :=
Vb − VnN
IbB = 1.118
⋅ arg( IaA ) = −153.435
180
π
)
⋅ arg( IbB) = 86.565
SA = 2.5 + 5i
Total power delivered to the load:
(
Vc − VnN
ZC
IcC = 0.933 − 0.616i
IcC = 1.118
Calculate the power delivered to the load:
⎯
⎯
IaA ⋅ IaA
IbB⋅ IbB
SA :=
⋅ ZA
SB :=
⋅ ZB
2
2
(
IcC :=
ZB
IbB = 0.067 + 1.116i
IaA = 1.118
180
− 15
VnN = 1.517 × 10
ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC
)
SB = 2.5 + 5i
180
π
⋅ arg( IcC) = −33.435
SC :=
⎯
(IcC
⋅ IcC)
2
⋅ ZC
SC = 2.5 + 5i
SA + SB + SC = 7.5 + 15i
The average power delivered to the load is P = Re {S} = Re {7.5 + j15} = 7.5 W
14
Section 12-4: The Δ- Connected Source and Load
A balanced three-phase Δ-connected load has one line current:
P 12.4-1
IB = 50 ∠–40° A rms
Find the phase currents IBC, IAB, and ICA. Draw the phasor diagram showing the line and phase
currents. The source uses the abc phase sequence.
Solution:
Given I B = 50∠ − 40° A rms and assuming the abc phase sequence
we have
I A = 50∠80° A rms and I C = 50∠200° A rms
From Eqn 12.6-4
I A = I AB × 3∠ − 30° ⇒ I AB =
so
IA
3∠ − 30°
50∠80°
= 28.9∠110° A rms
3∠−30°
= 28.9∠ − 10° A rms and ICA = 28.9∠ − 130° A rms
I AB =
I BC
P 12.4-2 A three-phase circuit has two parallel balanced Δ loads, one of 5-Ω resistors and one
of 20-Ω resistors. Find the magnitude of the total line current when the line-to-line voltage is 480
V rms.
Solution:
The two delta loads connected in parallel are equivalent to a single delta
load with
Z Δ = 5 || 20 = 4 Ω
The magnitude of phase current is
480
Ip =
= 120 A rms
4
The magnitude of line current is
I L = 3 I p = 208 A rms
1
Section 12-5: The Y- to Δ- Circuit
P 12.5-1
Consider a three-wire Y-to-Δ circuit. The voltages of the Y-connected source are
Va = (208/ 3 ) ∠–30°V rms, Vb = (208/ 3 ) ∠–150° V rms, and Vc = (208/ 3 ) ∠90°V rms.
The Δ-connected load is balanced. The impedance of each phase is Z = 12 ∠30° Ω. Determine
the line currents and calculate the power dissipated in the load.
Answer: P = 9360 W
Solution:
We have a delta load with Z = 12∠30° . One phase current is
I AB
⎛ 208
⎞ ⎛ 208
⎞
∠−30° ⎟ −⎜
∠−150° ⎟
⎜
V
V −V
3
⎠ ⎝ 3
⎠ = 208∠0° = 17.31∠ − 30° A rms
= AB = A B = ⎝
12∠30°
12∠30°
Z
Z
The other phase currents are
I BC = 17.31∠ − 150° A rms and I CA = 17.31∠90° A rms
One line currents is
I A = I AB × 3∠ − 30° = (17.31∠ − 30° ) ×
(
)
3∠ − 30° = 30∠ − 60° A rms
The other line currents are
I B = 30∠ − 180° A rms and I C = 30∠60° A rms
The power delivered to the load is
P = 3(
208
) (30) cos ( 60° − 30° ) = 9360 W
3
1
P 12.5-2 A balanced Δ-connected load is connected by three wires, each with a 4-Ω resistance,
to a Y source with
Va = (480/ 3 )∠–30° V rms, Vb = (480/ 3 ) ∠–150° V rms, and Vc = (480/ 3 )∠–90° V rms.
Find the line current IA when ZΔ = 39 ∠–40° Ω.
Answer: IA = 17 ∠0.9 A
Solution:
The balanced delta load with Z Δ = 39∠− 40° Ω is
equivalent to a balanced Y load with
ZY =
ZΔ
= 13∠ − 40° = 9.96 − j 8.36 Ω
3
Z T = Z Y + 4 = 13.96 − j 8.36 = 16.3∠ − 30.9 Ω
480
∠−30°
3
then I A =
= 17∠0.9° A rms
°
16.3 ∠−30.9
2
P 12.5-3 The balanced circuit shown in Figure P 12.5-3 has Vab = 380 ∠30°V rms. Determine
the phase currents in the load when Z = 3 + j4 Ω. Sketch a phasor diagram.
a
A
Z
n
+
–
–
+
Vab
Z
c
b
– +
Z
B
C
Figure P 12.5-3
Solution:
Vab = Va × 3∠30° ⇒ Va =
Vab
3∠30°
In our case, the given line-to-line voltage is
Vab = 380 ∠30° V rms
380 ∠30°
So one phase voltage is Va =
= 200∠0° V rms
3∠30°
So
VAB = 380∠30° V rms
VA = 220∠0° V rms
VBC = 380∠-90° V rms
VB = 220∠−120° V rms
VCA = 380∠150° V rms
VC = 220∠120° V rms
One phase current is
IA =
VA 220∠0°
= 44∠ − 53.1° A rms
3+ j4
Z
The other phase currents are
I B = 44∠−173.1° A rms amd I C = 44∠66.9° A rms
3
P 12.5-4 The balanced circuit shown in Figure P 12.5-3 has Vab = 380 ∠0° V rms. Determine
the line and phase currents in the load when Z = 9 + j12 Ω.
Solution:
Vab = Va × 3∠30° ⇒ Va =
Vab
3∠30°
In our case, the given line-to-line voltage is
Vab = 380 ∠0° V rms
Va =
So one phase voltage is
So
380 ∠0°
= 200∠ − 30° V rms
3∠30°
Vab = 380∠0° V rms
Va = 220∠ − 30° V rms
Vbc = 380∠-120° V rms
Vb = 220∠−150° V rms
Vca = 380∠120° V rms
Vc = 220∠90° V rms
One phase current is
IA =
Va 220∠−30°
=
= 14.67∠ − 83.1° A rms
Z
9+ j12
The other phase currents are
I B = 14.67∠ − 203.1° A rms and I C = 14.67∠36.9° A rms
4
Section 12-6: Balanced Three-Phase Circuits
P 12.6-1 The English Channel Tunnel rail link is supplied at 25 kV rms from the United
Kingdom and French grid systems. When there is a grid supply failure, each end is
capable of supplying the whole tunnel but in a reduced operational mode.
The tunnel traction system is a conventional catenary (overhead wire) system
similar to the surface main line electric railway system of the United Kingdom and
France. What makes the tunnel traction system different and unique is the high density of
traction load and the end-fed supply arrangement. The tunnel traction load is
considerable. For each half tunnel, the load is 180 MVA (Barnes and Wong, 1991).
Assume that each line-to-line voltage of the Y-connected source is 25 kV rms and
the three-phase system is connected to the traction motor of an electric locomotive. The
motor is a Y-connected load with Z = 150 ∠25° Ω. Find the line currents and the power
delivered to the traction motor.
Solution:
Va =
IA
25
×103 ∠0° Vrms
3
25
×103 ∠0°
Va
3
=
=
= 96∠ − 25° A rms
150 ∠25°
Z
⎛ 25
⎞
P = 3 Va I A cos (θ v -θ I ) = 3 ⎜
×103 ⎟ 96 cos(0 − 25°) = 3.77 MW
⎝ 3
⎠
P 12.6-2 A three-phase source with a line voltage of 45 kV rms is connected to two
balanced loads. The Y-connected load has Z = 10 + j20 Ω, and the Δ load has a branch
impedance of 50 Ω. The connecting lines have an impedance of 2 Ω. Determine the
power delivered to the loads and the power lost in the wires. What percentage of power is
lost in the wires?
Solution:
Convert the delta load to an equivalent Y connected
load:
ˆ = 50 Ω
ZΔ ⇒ Z
Y
3
To get the per-phase equivalent circuit shown to the
right:
The phase voltage of the source is
Z Δ = 50 Ω
Va =
45×103
∠0° = 26∠0° kV rms
3
The equivalent impedance of the load together with the line is
50
3 + 2 = 12 + j 5 = 13∠22.6° Ω
Z eq =
50
10 + j 20 +
3
(10 + j 20 )
The line current is
Ι aA
Va 26 × 103 ∠0°
=
=
= 2000∠ − 22.6° A rms
Z eq
13∠22.6°
The power delivered to the parallel loads (per phase) is
50 ⎫
⎧
⎪ (10 + j 20 ) 3 ⎪
2
6
PLoads = I aA × Re ⎨
⎬ = 4 ×10 × 10 = 40 MW
50
⎪10 + j 20 + ⎪
3 ⎭
⎩
The power lost in the line (per phase) is
PLine = I aA × Re {Z Line } = 4 × 106 × 2 = 8 MW
2
The percentage of the total power lost in the line is
PLine
8
× 100% =
× 100% = 16.7%
PLoad + PLine
40 +8
P 12.6-3 A balanced three-phase source has a Y-connected source with va = 5 cos (2t +
30°) connected to a three-phase Y load. Each phase of the Y-connected load consists of a
4-Ω resistor and a 4-H inductor. Each connecting line has a resistance of 2 Ω. Determine
the total average power delivered to the load.
Solution:
Ia =
Va 5∠30°
=
= 0.5∠ − 23° A ∴ I a = 0.5 A
Z T 6 + j8
2
PLoad
I
= 3 a Re {Z Load } = 3 × 0.125 × 4 = 1.5 W
2
also (but not required) :
PSource = 3
(5) (0.5)
cos(−30 − 23) = 2.25 W
2
2
Pline
I
= 3 a Re{Z Line } = 3×0.125× 2 = 0.75 W
2
Section 12.7
Load
P 12.7-1
Instantaneous and Average Power in a Balanced Three-Phase
Find the power absorbed by a balanced three-phase Y-connected load when
VCB = 208 ∠15° V rms
and
IB = 3 ∠110° A rms
The source uses the abc phase sequence.
Answer: P = 620 W
Solution:
Assuming the abc phase sequence:
VCB = 208∠15° V rms ⇒ VBC = 208∠195° V rms ⇒ VAB = 208∠315° V rms
Then
VA =
VAB
208∠315° 208
=
=
∠285° V rms
3∠30°
3∠30°
3
also
I B = 3∠110° A rms ⇒ I A = 3∠230° A rms
Finally
P = 3 VA I A cos (θ V − θ I ) = 3(
208
) (3) cos(285° − 230°) = 620 W
3
P 12.7-2 A three-phase motor delivers 20 hp operating from a 480-V rms line voltage. The
motor operates at 85 percent efficiency with a power factor equal to 0.8 lagging. Find the
magnitude and angle of the line current for phase A.
Hint: 1 hp = 745.7 W
Solution:
Assuming a lagging power factor:
cos θ = pf = 0.8 ⇒
θ = 36.9°
The power supplied by the three-phase source is given by
Pin =
Pout
η
=
Pin = 3 I A VA pf
20 ( 745.7 )
= 17.55 kW where 1 hp = 745.7 W
0.85
⇒
IA =
17.55 ×103
Pin
=
= 26.4 A rms
3 VA pf
⎛ 480 ⎞
3⎜
⎟ ( 0.8 )
⎝ 3⎠
480 °
I A = 26.4∠ − 36.9° A rms when VA =
∠0 V rms
3
1
P 12.7-3 A three-phase balanced load is fed by a balanced Y-connected source with a line-toline voltage of 220 V rms. It absorbs 1500 W at 0.8 power factor lagging. Calculate the phase
impedance if it is (a) Δ-connected and (b) Y-connected.
Solution:
(a) For a Δ-connected load, Eqn 12.7-5 gives
PT
1500
=
= 4.92 A rms
3 VP pf 3( 220 )(.8)
3
The phase current in the Δ-connected load is given by
PT = 3 VP I L pf
⇒ IL =
I
IL
4.92
⇒ IP = L =
= 2.84 A rms
3
3
3
The phase impedance is determined as:
IP =
Z=
V
VL VL
220
=
∠ (θ V − θ I ) = L ∠ cos −1 pf =
∠ cos −1 0.8 = 77.44∠36.9° Ω
IP
IP
IP
2.84
(b) For a Y-connected load, Eqn 12.7-4 gives
PT = 3 VP I L pf ⇒ I L =
PT
1500
=
= 4.92 A rms
220
3 VP I L pf 3(
)(.8)
3
The phase impedance is determined as:
220
V
V
V
Z = P = P ∠ (θ V − θ I ) = P ∠ cos −1 pf = 3 ∠ cos −1 0.8 = 25.8∠36.9° Ω
4.92
IP
IP
IP
2
P 12.7-4 A 600-V rms three-phase Y-connected source has two balanced Δ loads connected to
the lines. The load impedances are 40 ∠30° Ω and 50 ∠–60° Ω, respectively. Determine the line
current and the total average power.
Solution:
Parallel Δ loads
ZΔ =
Z1Z 2
(40∠30° ) (50∠−60° )
=
= 31.2 ∠−8.7° Ω
°
°
Ζ1 + Ζ 2
40∠30 + 50∠− 60
VL = VP , Ι P =
VP
600
=
= 19.2 A rms,
ZΔ
31.2
IL =
3 Ι P = 33.3 A rms
So P = 3 VL I L pf = 3 (600) (33.3) cos ( − 8.7° ) = 34.2 kW
P 12.7-5 A three-phase Y-connected source simultaneously supplies power to two separate
balanced three-phase loads. The first total load is Δ connected and requires 39 kVA at 0.7
lagging. The second total load is Y connected and requires 15 kW at 0.21 leading. Each line has
an impedance 0.038 + j0.072 Ω/phase. Calculate the line-to-line source voltage magnitude
required so that the loads are supplied with 208-V rms line-to-line.
Solution:
We will use
In our case:
S = S ∠θ = S cos θ + j S sin θ = S pf + j S sin ( cos −1 pf )
S 1 = 39 (0.7) + j 39 sin ( cos −1 ( 0.7 ) ) = 27.3 + j 27.85 kVA
15
sin ( cos −1 ( 0.21) ) = 15 − j 69.84 kVA
S 2 = 15 +
0.21
S
S 3φ = S 1 + S 2 = 42.3 − j 42.0 kVA ⇒ S φ = 3φ = 14.1− j 14.0 kVA
3
The line current is
*
⎛ S ⎞ (14100+ j 14000)
S = Vp I L ⇒ I L = ⎜ ⎟ =
= 117.5 + j 116.7 A rms = 167 ∠45° A rms
⎜V ⎟
208
⎝ p⎠
3
208
∠0° = 120∠0° V rms. The source must
The phase voltage at the load is required to be
3
provide this voltage plus the voltage dropped across the line, therefore
*
= 120∠0° + (0.038 + j 0.072)(117.5 + j 116.7) = 115.9 + j 12.9 = 116.6 ∠6.4° V rms
V
Sφ
Finally
V
= 116.6 V rms
Sφ
3
P 12.7-6 A building is supplied by a public utility at 4.16 kV rms. The building contains three
balanced loads connected to the three-phase lines:
(a)
(b)
(c)
Δ-connected, 500 kVA at 0.85 lagging
Y-connected, 75 kVA at 0.0 leading
Y connected; each phase with a 150-Ω resistor parallel to a 225-Ω inductive reactance
The utility feeder is five miles long with an impedance per phase of 1.69 + j0.78 Ω/mile. At what
voltage must the utility supply its feeder so that the building is operating at 4.16 kV rms?
Hint:
41.6 kV is the line-to-line voltage of the balanced Y-connected source.
Solution:
The required phase voltage at the load is VP =
4.16
∠0° = 2.402∠0° kVrms .
3
Let I1 be the line current required by the Δ-connected load. The apparent power per phase
500 kVA
required by the Δ-connected load is S1 =
= 167 kVA . Then
3
S1 = S1 ∠θ = S1 ∠ cos −1 ( pf ) = 167 ∠ cos−1 ( 0.85) = 167∠31.8° kVA
and
*
3
⎛ S1 ⎞ ⎛ (167 ×10 ) ∠31.8° ⎞
⎜
⎟ = 69.6∠ − 31.8° = 59 − j36.56 A rms
⇒ I1 = ⎜ ⎟ =
3
⎜
⎟
V
×
∠
°
2.402
10
0
(
)
⎝ P⎠ ⎝
⎠
*
S1 = VP I1
*
Let I2 be the line current required by the first Y-connected load. The apparent power per phase
75 kVA
required by this load is S 2 =
= 25 kVA . Then, noticing the leading power factor,
3
S 2 = S 2 ∠θ = S 2 ∠ cos −1 ( pf ) = 25 ∠ cos −1 ( 0 ) = 25∠ − 90° kVA
and
*
3
⎛ S 2 ⎞ ⎛ ( 25 ×10 ) ∠ − 90° ⎞
⎟ = 10.4∠90° = j10.4 A rms
⇒ I2 = ⎜ ⎟ = ⎜
3
⎝ VP ⎠ ⎜⎝ ( 2.402 ×10 ) ∠0° ⎟⎠
*
S 2 = VP I 2
*
Let I3 be the line current required by the other Y-connected load. Use Ohm’s law to determine I3
to be
2402∠0° 2402∠0°
I3 =
+
= 16 − j 10.7 A rms
150
j 225
The line current is
I L = I1 + I 2 + I 3 = 75− j 36.8 A rms
4
4.16
∠0° = 2.402∠0° kVrms .The source
3
must provide this voltage plus the voltage dropped across the line, therefore
The phase voltage at the load is required to be VP =
VSφ = 2402∠0° + (8.45 + j 3.9) (75 − j 36.8) = 3179 ∠−0.3° Vrms
Finally
VSL = 3 (3179) = 5506 Vrms
5
P 12.7-7 The diagram shown in Figure P 12.7-7 has two three-phase loads that form part of a
manufacturing plant. They are connected in parallel and require 4.16 kV rms. Load 1 is 1.5
MVA, 0.75 lag pf, Δ-connected. Load 2 is 2 MW, 0.8 lagging pf, Y-connected. The feeder from
the power utility’s substation transformer has an impedance of 0.4 + j0.8 Ω/phase. Determine the
following:
(a)
(b)
(c)
The required magnitude of the line voltage at the supply.
The real power drawn from the supply.
The percentage of the real power drawn from the supply that is consumed by the loads.
Three-phase
supply from
utility
0.4 Ω
j0.8 Ω
0.4 Ω
j0.8 Ω
0.4 Ω
j0.8 Ω
Load 2
Load 1
Figure P 12.7-7
Solution:
The required phase voltage at the load is VP =
4.16
∠0° = 2.402∠0° kVrms .
3
Let I1 be the line current required by the Δ-connected load. The apparent power per phase
1.5 MVA
required by the Δ-connected load is S1 =
= 0.5 MVA . Then
3
S1 = S1 ∠θ = S1 ∠ cos −1 ( pf ) = 0.5 ∠ cos −1 ( 0.75) = 0.5∠41.4° MVA
and
*
6
⎛ S1 ⎞ ⎛ ( 0.5 ×10 ) ∠41.4° ⎞
⎟ = 2081.6∠ − 41.4° = 1561.4 − j1376.6 A rms
⇒ I1 = ⎜ ⎟ = ⎜
3
⎝ VP ⎠ ⎜⎝ ( 2.402 ×10 ) ∠0° ⎟⎠
*
S1 = VP I1
*
Let I2 be the line current required by the first Y-connected load. The complex power, per phase,
is
0.67
S 2 = 0.67 +
sin ( cos −1 ( 0.8 ) ) = 0.67 + j 0.5 MVA
0.8
6
*
6
⎛ S 2 ⎞ ⎛ ( 0.67 + j 0.5 ) ×106 ⎞ ⎛ ( 0.833 ×10 ) ∠ − 36.9° ⎞
⎟ =⎜
⎟
I2 = ⎜ ⎟ = ⎜
3
3
⎝ VP ⎠ ⎜⎝ ( 2.402 ×10 ) ∠0° ⎟⎠ ⎜⎝ ( 2.402 ×10 ) ∠0° ⎟⎠
= 346.9∠ − 36.9° = 277.4 − j 208.3 A rms
The line current is
I L = I1 + I 2 = 433.7 − j 345.9 = 554.7∠ − 38.6 A rms
*
*
4.16
∠0° = 2.402∠0° kVrms .The source
3
must provide this voltage plus the voltage dropped across the line, therefore
The phase voltage at the load is required to be VP =
VSφ = 2402∠0° + (0.4 + j 0.8) (433.7 − j 345.9) = 2859.6 ∠ − 38.6° Vrms
Finally
VSL = 3 (2859.6) = 4953 Vrms
The power supplied by the source is
PS =
3 (4953) (554.7) cos (4.2° + 38.6° ) = 3.49 MW
The power lost in the line is
PLine = 3 × ( 554.7 2 ) × Re {0.4+ j 0.8} = 0.369 MW
The percentage of the power consumed by the loads is
3.49 − 0.369
× 100% = 89.4%
3.49
7
P 12.7-8 The balanced three-phase load of a large commercial building requires 480 kW at a
lagging power factor of 0.8. The load is supplied by a connecting line with an impedance of
5 + j25 mΩ for each phase. Each phase of the load has a line-to-line voltage of 600 V rms.
Determine the line current and the line voltage at the source. Also, determine the power factor at
the source. Use the line-to-neutral voltage as the reference with an angle of 0°.
Solution:
The required phase voltage at the load is VP =
600
∠0° = 346.4∠0° Vrms .
3
Let I be the line current required by the load. The complex power, per phase, is
S = 160 + j
160
sin ( cos −1 ( 0.8 ) ) = 160 + j 120 kVA
0.8
The line current is
*
*
⎛ S ⎞ ⎛ (160 + j 120 ) ×103 ⎞
I=⎜ ⎟ =⎜
⎟ = 461.9 − j 346.4 A rms
346.4∠0°
⎝ VP ⎠ ⎝
⎠
600
∠0° = 346.4∠0° Vrms .The source
3
must provide this voltage plus the voltage dropped across the line, therefore
The phase voltage at the load is required to be VP =
VSφ = 346.4∠0° + (0.005 + j 0.025) (461.9 − j 346.4) = 357.5 ∠1.6° Vrms
Finally
VSL = 3 (357.5) = 619.2 Vrms
The power factor of the source is
pf = cos (θ V − θ I ) = cos (1.6° − ( − 37°)) = 0.78
8
Section 12-8: Two-Wattmeter Power Measurement
P 12.8-1 The two-wattmeter method is used to determine the power drawn by a three-phase
440-V rms motor that is a Y-connected balanced load. The motor operates at 20 hp at 74.6
percent efficiency. The magnitude of the line current is 52.5 A rms. The wattmeters are
connected in the A and C lines. Find the reading of each wattmeter. The motor has a lagging
power factor.
Hint: 1 hp = 745.7 W
Solution:
W
= 14920 W
hp
P
14920
Pin = out =
= 20 kW
0.746
η
Pout = 20 hp × 746
Pin = 3 VL I L cos θ
⇒ cos θ =
Pin
20 × 103
=
= 0.50
3 VL I L
3 (440) (52.5)
⇒ θ cos -1 ( 0.5 ) = 60°
The powers read by the two wattmeters are
P1 = VL I L cos (θ + 30° ) = (440) (52.5)cos ( 60° + 30° ) = 0
and
P2 = VL I L cos (θ − 30° ) = (440) (52.5)cos ( 60° − 30° ) = 20 kW
1
P 12.8-2 A three-phase system has a line-to-line voltage of 4000 V rms and a balanced
Δ-connected load with Z = 40 + j30 Ω. The phase sequence is abc. Use the two wattmeters
connected to lines A and C, with line B as the common line for the voltage measurement.
Determine the total power measurement recorded by the wattmeters.
Answer: P = 768 kW
Solution:
VP = VL = 4000 V rms
IP =
VP
4000
=
= 80 A rms
ZΔ
50
Z
Δ
= 40 + j 30 = 50 ∠36.9°
Ι L = 3 I P = 138.6 A rms
pf = cos θ = cos (36.9° ) = 0.80
P1 = VL I L cos (θ + 30° ) = 4000 (138.6) cos 66.9° = 217.5 kW
P2 = VL I L cos (θ −30° ) = 4000 (138.6) cos 6.9° = 550.4 kW
PT = P1 + P2 = 767.9 kW
Check : PT = 3 Ι L VL cos θ =
3 (4000) (138.6) cos 36.9°
= 768 kW which checks
2
P 12.8-3 A three-phase system with a sequence abc and a line-to-line voltage of 200 V rms
feeds a Y-connected load with Z = 70.7 ∠45° Ω. Find the line currents. Find the total power by
using two wattmeters connected to lines B and C.
Answer: P = 400 W
Solution:
Vp = Vp =
200
= 115.47 Vrms
3
VA =115.47∠0° V rms, VB = 115.47∠−120° V rms
and VC = 115.47∠120° V rms
IA =
VA
115.47∠0°
=
= 1.633∠ − 45° A rms
Z
70.7∠45°
I B = 1.633 ∠ − 165° A rms and I C = 1.633 ∠75° A rms
PT =
3 VL I L cos θ = 3 (200) (1.633) cos 45° = 400 W
PB = VAC I A cos θ1 = 200 (1.633) cos (45° − 30° ) = 315.47 W
PC = VBC I B cos θ 2 = 200 (1.633) cos (45° + 30° ) = 84.53 W
3
P 12.8-4 A three-phase system with a line-to-line voltage of 208 V rms and phase sequence
abc is connected to a Y-balanced load with impedance 10 ∠–30° Ω and a balanced Δ load with
impedance 15 ∠30° Ω. Find the line currents and the total power using two wattmeters.
Solution:
ZY = 10∠ − 30° Ω and Z Δ = 15∠30° Ω
Convert Z Δ to Z Yˆ → Z Yˆ =
then Zeq =
ZΔ
= 5∠30° Ω
3
(10∠−30° ) ( 5∠30° ) =
10∠−30°+5∠30°
208
Vp = Vp =
= 120 V rms
3
VA = 120∠0° V rms ⇒ I A =
50∠0°
= 3.78∠10.9° Ω
13.228 ∠−10.9°
120∠0°
= 31.75 ∠−10.9°
3.78 ∠10.9°
I B = 31.75∠−130.9°
I C = 31.75∠109.1°
PT = 3VL I L cos θ = 3 ( 208 ) ( 31.75 ) cos (10.9 ) =11.23 kW
W1 = VL I L cos (θ −30°) = 6.24 kW
W2 = VL I L cos (θ + 30°) = 4.99 kW
P 12.8-5 The two-wattmeter method is used. The wattmeter in line A reads 920 W, and the
wattmeter in line C reads 460 W. Find the impedance of the balanced Δ-connected load. The
circuit is a three-phase 120-V rms system with an abc sequence.
Answer: ZΔ = 27.1 ∠–30° Ω
P12.8-5
PT = PA + PC = 920 + 460 = 1380 W
tan θ = 3
( −460 ) = −0.577 ⇒ θ = −30°
PC − PA
= 3
PC + PA
1380
PT = 3 VL I L cos θ so I L =
PT
1380
=
=7.67 A rms
3 VL cos θ
2 ×120×cos( −30 )
IL
= 4.43 A rms
3
120
= 27.1 Ω ο r Z Δ = 27.1 ∠−30°
4.43
IP =
∴ ZΔ =
4
P 12.8-6 Using the two-wattmeter method, determine the power reading of each wattmeter and
the total power for Problem P 12.5-1 when Z = 0.868 + j4.924 Ω. Place the current coils in the Ato-a and C-to-c lines.
Solution:
Z = 0.868 + j 4.924 = 5∠80°
VL = 380 V rms, VP =
I L = I P and I P =
⇒
θ = 80°
380
= 219.4 V rms
3
VP
= 43.9 A rms
Z
P1 = ( 380 ) ( 43.9 ) cos (θ −30° ) = 10,723 W
P2 = ( 380 ) ( 43.9 ) cos (θ + 30° ) = −5706 W
PT = P1 + P2 = 5017 W
5
Section 12.9 How Can We Check …?
P 12.9-1 A Y-connected source is connected to a Y-connected load (Figure 12.3-1) with
Z = 10 + j4 Ω. The line voltage is VL = 416 V rms. A student report states that the line current
IA = 38.63 A rms and that the power delivered to the load is 16.1 kW. Verify these results.
Solution:
416
= 240 V = VA
3
Z = 10 + j4 = 10.77 ∠21.8° Ω
VA =
VA
240
=
= 22.28 A rms ≠ 38.63 A rms
Z
10.77
38.63
= 22.3 . It appears that the line-to-line voltage was
The report is not correct. (Notice that
3
mistakenly used in place of the phase voltage.)
IA =
P 12.9-2 A Δ load with Z = 40 + j30 Ω has a three-phase source with VL = 240 V rms (Figure
12.3-2). A computer analysis program states that one phase current is 4.8 ∠–36.9° A. Verify this
result.
Solution:
VL = VP = 240∠0° Vrms
Z = 40 + j 30 = 50 ∠36.9° Ω
IP =
VP
240∠0°
=
= 4.8 ∠−36.9° A rms
°
Z
50∠36.9
The result is correct.
1
PSpice Problems
SP 12-1 Use PSpice to determine the power delivered to the load in the circuit shown in Figure
SP 12-1.
a
5Ω
5 mH
A
20 Ω
60 mH
b
5Ω
5 mH
B
20 Ω
60 mH
C
20 Ω
+
–
110 cos
(377t)
N
+
n
–
110 cos
(377t +120°)
c
5Ω
5 mH
60 mH
+
–
110 cos
(377t + 240°)
Source
Line
Load
Figure SP 12-1
Solution:
FREQ
6.000E+01
IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3)
3.142E+00 -1.644E+02 -3.027E+00 -8.436E-01
FREQ
IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1)
1
6.000E+01
3.142E+00
-4.443E+01
FREQ
6.000E+01
VM(N01496)
2.045E-14
FREQ
6.000E+01
IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2)
3.142E+00
7.557E+01
7.829E-01
3.043E+00
VP(N01496)
2.211E+01
2.244E+00
VR(N01496)
1.895E-14
-2.200E+00
VI(N01496)
7.698E-15
3.1422
20 = 98.7 W
2
3.1422
20 = 98.7 W
I B = 3.142∠75.57° A and RB = 20 Ω ⇒ PB =
2
3.1422
20 = 98.7 W
I C = 3.142∠ − 164.4° A and RC = 20 Ω ⇒ PC =
2
I A = 3.142∠ − 43.43° A and
RA = 20 Ω ⇒ PA =
P = 3 ( 98.7 ) = 696.1 W
2
SP 12-2
2.
Use PSpice to determine the power delivered to the load in the circuit shown in SP 12a
10 Ω
5 mH
A
20 Ω
60 mH
b
10 Ω
5 mH
B
30 Ω
25 mH
C
60 Ω
+
–
110 cos
(377t)
–
N
+
n
110 cos
(377t +120°)
c
10 Ω
5 mH
20 mH
+
–
110 cos
(377t + 240°)
Source
Line
Load
Figure SP 12-2
Solution:
FREQ
6.000E+01
IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3)
1.612E+00 -1.336E+02 -1.111E+00 -1.168E+00
FREQ
6.000E+01
IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1)
2.537E+00 -3.748E+01
2.013E+00 -1.544E+00
3
FREQ
6.000E+01
VM(N01496) VP(N01496)
1.215E+01 -1.439E+01
VR(N01496) VI(N01496)
1.177E+01 -3.018E+00
FREQ
6.000E+01
IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2)
2.858E+00
1.084E+02 -9.023E-01
2.712E+00
2.537 2
20 = 64.4 W
2
2.8582
30 = 122.5 W
I B = 2.858∠108.4° A and RB = 30 Ω ⇒ PB =
2
1.6122
60 = 78 W
I C = 1.612∠ − 133.6° A and RC = 600 Ω ⇒ PC =
2
I A = 2.537∠ − 37.48° A and
RA = 20 Ω ⇒ PA =
P = 64.4 + 122.5 + 78 = 264.7 V
4
Design Problems
DP 12-1 A balanced three-phase Y source has a line voltage of 208 V rms. The total power
delivered to the balanced Δ load is 1200 W with a power factor of 0.94 lagging. Determine the
required load impedance for each phase of the Δ load. Calculate the resulting line current. The
source is a 208-V rms ABC sequence.
Solution:
P = 400 W per phase,
0.94 = pf = cos θ
θ = cos -1 ( 0.94 ) =20°
⇒
208
I L 0.94 ⇒ I L = 3.5 A rms
3
I
I Δ = L = 2.04 A rms
3
V
208
Z = L =
= 101.8 Ω
IΔ
2.04
400 =
Z = 101.8 ∠20° Ω
DP 12-2 A three-phase 240-V rms circuit has a balanced Y-load impedance Z. Two wattmeters
are connected with current coils in lines A and C. The wattmeter in line A reads 1440 W, and the
wattmeter in line C reads zero. Determine the value of the impedance.
Solution:
VL = 240 V rms
PA = VL I L cos (30° + θ ) = 1440 W
PC = VL I L cos (30° − θ ) = 0 W ⇒
30−θ = 90° or θ = −60°
then 1440 = 240 I L cos (−30° )
I L = 6.93 A rms
IL
= IP =
VP
Z
⇒
⇒
240
V
Z = P = 3 = 20 Ω
IP
6.93
Finally, Z = 20 ∠ − 60° Ω
1
DP 12-3 A three-phase motor delivers 100 hp and operates at 80 percent efficiency with a 0.75
lagging power factor. Determine the required Δ-connected balanced set of three capacitors that
will improve the power factor to 0.90 lagging. The motor operates from 480-V rms lines.
Solution:
Pin =
Pout
η
100 hp × (746
=
W
)
hp
0.8
= 93.2 kW, P =
φ
Pin
= 31.07 kW
3
VL = 480 V rms, pfc = 0.9 and pf = 0.75. We need the impedance of the load so that we can use
Eqn 11.6-7 to calculate the value of capacitance needed to correct the power factor.
0.75 = pf = cos θ
⇒
θ = cos-1 ( 0.75) = 41.4°
480
I p ( 0.75 ) ⇒ I p = 149.5 Arms
3
480
VP
3 = 1.85 Ω
Z =
=
IP
149.5
31070 =
Z = 1.85 ∠41.4° Ω = 1.388 + j1.223 Ω
The capacitance required to correct the power factor is given by
⎡⎣ tan (cos −1 0.75) − tan (cos −1 0.9) ⎤⎦
1.365
C=
= 434 μ F
×
1.3652 +1.2042
377
(Checked using LNAPAC 6/12/03)
2
DP 12-4 A three-phase system has balanced conditions so that the per-phase circuit
representation can be utilized as shown in Figure DP 12-4. Select the turns ratio of the step-up
and step-down transformers so that the system operates with an efficiency greater than 99
percent. The load voltage is specified as 4 kV rms, and the load impedance is 4/3 Ω.
2.5 Ω
1 : n1
+
20 kV –
j40 Ω
IL
n2 : 1
+
+
+
V1
V2
VL
–
–
–
Load
Figure DP 12-4
Solution:
VL = 4∠0° kV rms
n2
25
VL = 4000∠0° = 100∠0° kVrms
n1
1
Try n2 = 25 then V2 =
VL 4×103∠0°
=
= 3∠0° kA rms
4
ZL
3
3000∠0°
= 120∠0° A rms
The line current in 2.5 Ω is I =
25
Thus V1 = ( R + j X ) I + V2
IL =
= (2.5 + j 40) (120∠0°) + 100×103 = 100.4 ∠2.7° kV
Step need : n1 =
Ploss = I
2
100.4 kV
= 5.02 ≅ 5
20 kV
R = 120
2
(2.5) = 36 kW, P = (4×103 ) (3× 103 ) = 12 MW
12 − .036
× 100% = 99.7 % of the power supplied by the source
12
is delivered to the load.
∴η =
3
Chapter 13: Frequency Response
Exercises
Exercise 13.2-1 The input to the circuit shown in Figure E 13.2-
R
1 is the source voltage, vs, and the response is the capacitor
voltage, vo. Suppose R = 10 kΩ and C = 1 μF. What are the values
+
vs
+
–
vo
C
–
of the gain and phase shift when the input frequency is ω = 100
rad/s?
Figure E 13.2-1
Answer: 0.707 and –45°
Solution:
H (ω ) =
gain =
1
Vo (ω )
=
1 + jω C R
Vs (ω )
1
1 + (ω C R) 2
phase shift = − tan −1 (ω C R )
When R = 104 Ω, ω = 100 rad/s, and C = 10−6 F, then gain =
1
= 0.707 and phase shift = − 45o
2
13Ex-1
Exercise 13.2-2 The input to the circuit shown in Figure E 13.2-2
L
is the source voltage, vs, and the response is the resistor voltage, vo.
+
R = 30 Ω and L = 2 H. Suppose the input frequency is adjusted until
vs +
vo
R
–
i
the gain is equal to 0.6. What is the value of the frequency?
–
Figure E 13.2-2
Answer: 20 rad/s
Solution:
H (ω ) =
gain =
Vo (ω )
R
=
Vs (ω )
R + jω L
R
R 2 + (ω L) 2
2
0.6 =
30
30 + (2ω ) 2
2
⎛ 30 ⎞
2
⎜ ⎟ − 30
⎝ .6 ⎠
⇒ ω=
= 20 rad s
2
Exercise 13.2-3 The input to the circuit shown in Figure E 13.2-2
L
is the source voltage, vs, and the response is the mesh current, i. R =
30 Ω and L = 2 H. What are the values of the gain and phase shift
when the input frequency is ω = 20 rad/s?
+
vs +
–
R
i
vo
–
Figure E 13.2-2
Answer: 0.02 A/V and –53.1°
Solution:
H (ω ) =
gain =
I (ω )
1
=
R + jω L
Vs (ω )
1
R + (ω L) 2
2
phase shift = − tan −1
ωL
R
When R = 30 Ω, L = 2 H, and ω = 20 rad/s, then
gain =
1
30 + 40
2
2
= 0.02
A
⎛ 40 ⎞
and phase shift = − tan −1 ⎜ ⎟ = − 53.1°
V
⎝ 30 ⎠
13Ex-2
Exercise 13.2-4 The input to the circuit shown in Figure E 13.2-1
R
is the source voltage, vs, and the response is the capacitor voltage,
+
vo. Suppose C = 1 μF. What value of R is required to cause a phase
+
–
vs
vo
C
–
shift equal to –45° when the input frequency is ω = 20 rad/s?
Answer: R = 50 kΩ
Figure E 13.2-1
Solution:
H (ω ) =
gain =
Vo (ω )
1
=
Vs (ω ) 1 + jω C R
1
1 + (ω C R) 2
phase shift = − tan −1 ω C R
−1
−6
−45° = − tan (20 ⋅10 ⋅ R )
tan (45° )
⇒ R=
= 50 ⋅103 = 50 kΩ
−6
20⋅10
Exercise 13.2-5 The input to the circuit shown in Figure E 13.2-1
R
is the source voltage, vs, and the response is the capacitor voltage,
vo. Suppose C = 1 μF. What value of R is required to cause a gain
+
+
–
vs
vo
C
–
equal to 1.5 when the input frequency is ω = 20 rad/s?
Figure E 13.2-1
Answer: No such value of R exists. The gain of this circuit will
never be greater than 1.
Solution:
H (ω ) =
gain =
Vo (ω )
1
=
Vs (ω )
1 + jω C R
1
1 + (ω C R) 2
ω , C , and R are all positive, or at least nonnegative, so gain ≤ 1. These specifications cannot be met.
13Ex-3
Exercise 13.3-1 (a) Convert the gain|Vo/Vs| = 2 to decibels. (b) Suppose |Vo/Vs| = –6.02 dB.
What is the value of this gain “not in dB”?
Answer: (a) + 6.02 dB (b) 0.5
Solution:
(a) dB = 20 log (2) = 6.02 dB
−6.02 20
(b) 10
= 0.5
Exercise 13.3-2 In a certain frequency range, the magnitude of the network function can be
approximated as H = 1/ω2. What is the slope of the Bode plot in this range, expressed in decibels
per decade?
Answer:–40 dB/decade
Solution:
⎛ 1 ⎞
20 log H = 20 log ⎜ 2 ⎟ = 20 log (ω )-2 = −40 log ω
⎝ω ⎠
⎛ω ⎞
= − 40 log ω2 + 40 log ω1 = − 40 log⎜ 2 ⎟
⎝ ω1 ⎠
let ω2 = 10 ω1 to consider 1 decade, then slope = − 40 log10 = − 40 dB
decade
slope = 20 log H (ω2 ) − 20 log H (ω1 )
13Ex-4
Exercise 13.3-3 Consider the network function
H (ω ) =
jω A
B + jωC
Find (a) the corner frequency, (b) the slope of the asymptotic magnitude Bode plot for ω above
the corner frequency in decibels per decade, (c) the slope of the magnitude Bode plot below the
corner frequency, and (d) the gain for ω above the corner frequency in decibels.
Answer: (a) ω0 = B/C (b) zero (c) 20 dB/decade (d) 20 log10 =
A
C
Solution:
When ω C >> B, H (ω ) −
(d )
(b)
jω A A
=
jω C C
⎛ A⎞
H (ω ) in dB = 20 log10 H (ω ) = 20 log10 ⎜ ⎟
⎝C ⎠
H (ω ) does not depend on ω so slope = 0
When ω C << B, H (ω ) −
jω A
⎛ A⎞
= jω ⎜ ⎟
B
⎝B⎠
⎛ A⎞
H (ω ) in dB = 20 log10 H (ω ) = 20 log10ω +20 log10 ⎜ ⎟
⎝B⎠
(c) The slope is the coefficient of log10 ω , that is, slope = 20 dB
decade
(a) The break frequency is the frequency at which ω0 C = B, that is, ω0 =
B
C
13Ex-5
Exercise 13.4-1 For the RLC parallel resonant circuit when R = 8 kΩ, L = 40 mH, and C = 0.25
μF, find (a) Q and (b) bandwidth.
Answer: (a) Q = 20 (b) BW = 500 rad/s
Solution:
a) Q = ω 0 RC = R
b) BW =
ωo
=
Q
C
2.5 × 10−7
= 8000
= 20
L
40 × 10−3
1
1
=
= 500 rad s
Q LC
20 (40×10−3 ) (2.5×10−7 )
Exercise 13.4-2 A high-frequency RLC parallel resonant circuit is required to operate at ω0 =
10 Mrad/s with a bandwidth of 200 krad/s. Determine the required Q and L when C = 10 pF.
Answer: Q = 50 and L = 1 mH
Solution:
ω0
107
=
= 50
Q=
BW
2 × 105
1
1
1
ω0 =
⇒ L =
=
= 1 mH
2
7 2
LC
ω0 C
(10 ) (10 × 10−12 )
Exercise 13.4-3 A series resonant circuit has L = 1 mH and C = 10 μF. Find the required Q and
R when it is desired that the bandwidth be 15.9 Hz.
Answer: Q = 100 and R = 0.1 Ω
Solution:
1
1
=
= 104 rad s
1
LC
⎡(10−3 )(10−5 ) ⎤ 2
⎣
⎦
ω0 =
Q =
R =
ω0
BW
ω 0L
Q
=
=
104
= 100
2π (15.9)
(104 )(10−3 )
= 0.1 Ω
100
13Ex-6
Exercise 13.4-4 A series resonant circuit has an inductor L = 10 mH. (a) Select C and R so that
ω0 = 106 rad/s and the bandwidth is BW = 103 rad/s. (b) Find the admittance Y of this circuit for a
signal at ω = 1.05×106 rad/s.
Answer: (a)
(b)
Solution:
a)
C = 100pF, R = 10 Ω
Y=
10
1 + j 97.6
ω0 =
Q=
b)
Q =
Y=
1
1
⇒ C =
= 100 pF
6 2
LC
(10 ) (0.01)
ω0
BW
ω0
BW
=
1
103
BW
⇒ R=
=
= 10 Ω
ω 0 RC
ω 02 C (106 )2 (10−10 )
=
106
= 1000
103
1
⎛ ω ω0 ⎞
−
1+ j Q⎜
⎟
⎜ω0 ω ⎟
⎝
⎠
=
1
⎡1.05×106
106 ⎤
−
1+ j 1000 ⎢
⎥
6
1.05×106 ⎦⎥
⎣⎢ 10
=
1
1+ j 97.6
13Ex-7
Section 13-2: Gain, Phase Shift, and the Network Function
P 13.2-1 The input to the circuit shown in Figure P 13.2-1 is the voltage of the voltage source, vi(t).
The output is the voltage, vo(t), across the parallel connection of the capacitor and 10-Ω resistor.
Determine the network function, H(ω) = Vo(ω)/Vi(ω), of this circuit.
Answer: H (ω ) =
0.2
1 + j 4ω
Figure P 13.2-1
Solution:
R 2 ||
H (ω ) =
1
jω C
=
Vo (ω )
=
Vi (ω )
R2
1 + jω C R 2
R2
1 + jω C R 2
R2
R1 +
1 + jω C R 2
R2
=
R1 + R 2
1 + jω C R p
where Rp = R1 || R2.
When R1 = 40 Ω, R2 = 10 Ω and C = 0.5 F
H (ω ) =
0.2
1 + j 4ω
(checked using ELab on 8/6/02)
13.2-1
P 13.2-2 The input to the circuit shown in Figure P 13.2-2 is the voltage of the voltage source, vi(t).
The output is the voltage, vo(t), across the series connection of the capacitor and 160-kΩ resistor.
Determine the network function, H(ω) = Vo(ω)/Vi(ω), of this circuit.
Answer: H (ω ) =
1 + j (0.004)ω
1 + j (0.005)ω
Figure P 13.2-2
Solution:
H (ω ) =
1
R2 +
Vo (ω )
jω C
=
Vi (ω ) R + R + 1
1
2
jω C
1 + jω C R 2
=
1 + jω C R1 + R 2
(
)
When R1 = 40 kΩ, R2 = 160 kΩ and C = 0.025 μF
H (ω ) =
1 + j ( 0.004 ) ω
1 + j ( 0.005 ) ω
(checked using ELab on 8/6/02)
13.2-2
P 13.2-3 The input to the circuit shown in Figure P 13.2-3 is the voltage of the voltage source, vi(t).
The output is the voltage, vo(t), across the 6-Ω resistor. Determine the network function, H(ω) =
Vo(ω)/Vi(ω), of this circuit.
Figure P 13.2-3
Solution:
H (ω ) =
R2
Vo (ω )
=
Vi (ω ) R1 + R 2 + jω L
R2
=
R1 + R 2
L
1 + jω
R1 + R 2
When R1 = 4 Ω, R2 = 6 Ω and L = 8 H
H (ω ) =
0.6
1 + j ( 0.8 ) ω
(checked using ELab on 8/6/02)
13.2-3
P 13.2-4 The input to the circuit shown in Figure P 13.2-4 is the voltage of the voltage source, vi(t).
The output is the voltage, vo(t), across the series connection of the inductor and 60-Ω resistor. The
network function that represents this circuit is
1+
V (ω )
H (ω ) = o
= (0.6)
Vi (ω )
1+
j
j
ω
12
ω
20
Determine the values of the inductance, L, and of the resistance, R.
Figure P 13.2-4
Solution:
H (ω ) =
R 2 + jω L
Vo (ω )
=
Vi (ω ) R + R 2 + jω L
L
⎛
1 + jω
⎜
⎛ R2 ⎞
R2
=⎜
⎟⎜
⎜ R + R2 ⎟⎜
L
⎝
⎠ ⎜ 1 + jω
R + R2
⎝
⎞
⎟
⎟
⎟
⎟
⎠
Comparing the given and derived network functions, we require
L
⎛
1 + jω
⎜
⎛ R2 ⎞
R2
⎜
⎟⎜
⎜
⎟
L
⎝ R + R 2 ⎠ ⎜⎜ 1 + jω
R + R2
⎝
Since R2 = 60 Ω, we have L =
⎞
⎟
1+
⎟ = ( 0.6 )
⎟
1+
⎟
⎠
j
j
ω
12
ω
20
⎧ R2
= 0.6
⎪
⎪ R + R2
⎪ R
⎪
2
⇒ ⎨
= 12
⎪ L
⎪ R + R2
= 20
⎪
L
⎪⎩
60
= 5 H , then R = ( 20 )( 5 ) − 60 = 40 Ω .
12
(checked using ELab on 8/6/02)
13.2-4
P 13.2-5 The input to the circuit shown in Figure P 13.2-5 is the voltage of the voltage source, vi(t).
The output is the voltage, vo(t), across the parallel connection of the capacitor and 2-Ω resistor. The
network function that represents this circuit is
H (ω ) =
Vo (ω )
0.2
=
Vi (ω ) 1 + j 4ω
Determine the values of the capacitance, C, and of the resistance, R.
Figure P 13.2-5
Solution:
R 2 ||
H (ω ) =
1
jω C
=
R2
1 + jω C R 2
Vo (ω )
=
Vi (ω )
R2
1 + jω C R 2
R2
R+
1 + jω C R 2
R2
=
R + R2
1 + jω C R p
where Rp = R || R2.
Comparing the given and derived network functions, we require
R2
R + R2
1 + jω C R p
Since R2 = 2 Ω, we have
C=
=
0.2
1+ j4 ω
⎧ R2
= 0.2
⎪
⇒ ⎨ R + R2
⎪ CR =4
p
⎩
( 2 )( 8) = 1.6 Ω . Finally,
2
= 0.2 ⇒ R = 8 Ω . Then R p =
R+2
2+8
4
= 2.5 F .
1.6
(checked using ELab on 8/6/02)
13.2-5
P 13.2-6 The input to the circuit shown in Figure P 13.2-6 is the voltage of the voltage source, vi(t).
The output is the voltage, vo(t), across the capacitor. Determine the network function, H(ω) =
Vo(ω)/Vi(ω), of this circuit.
Answer: H (ω ) =
0.6
( jω )(1 + j (0.2)ω )
Figure P 13.2-6
Solution:
Vi (ω )
⎫
⎪
R + jω L
⎪
⎬ ⇒
1
Vo (ω ) =
( A Ia (ω ) )⎪⎪
jω C
⎭
I a (ω ) =
A
Vo (ω )
CR
=
L
Vi (ω )
( j ω ) ⎛⎜1 + j ω ⎞⎟
R⎠
⎝
When R = 20 Ω, L = 4 H, A = 3 A/A and C = 0.25 F
H (ω ) =
0.6
( j ω ) (1 + j ( 0.2 ) ω )
(checked using LNAP on 12/29/02)
13.2-6
P 13.2-7 The input to the circuit shown in Figure P 13.2-7 is the voltage of the voltage source, vi(t).
The output is the voltage, vo(t), across the 30-kΩ resistor. The network function of this circuit is
H (ω ) =
Vo (ω )
4
=
Vi (ω ) 1 + j ω
100
Determine the value of the capacitance, C, and the value of the gain, A, of the VCVS.
Answer: C = 5 μF and A = 6 V/V
Figure P 13.2-7
Solution:
In the frequency domain, use voltage division on the left side of the circuit to get:
1
1
jω C
VC (ω ) =
Vi (ω ) =
Vi (ω )
1
1
j
C
R
+
ω
1
R1 +
jω C
Next, use voltage division on the right side of the circuit to get:
2
A
R3
2
3
A VC (ω ) = A VC (ω ) =
Vo (ω ) =
Vi (ω )
R 2 + R3
3
1 + jω C R1
Compare the specified network function to the calculated network function:
2
2
A
A
2
1
3
3
=
=
⇒ 4 = A and
= 2000 C
ω 1 + jω C R1 1 + jω C 2000
3
100
1+ j
100
4
Thus, C = 5 μF and A = 6 V/V.
(checked using ELab on 8/6/02)
13.2-7
P 13.2-8
The input to the circuit shown in Figure P 13.2-8 is the source voltage, vi(t), and the response
is the voltage across RL, vo(t). Find the network function.
Answer: H(ω) = –5/(1 + jω/10)
Figure P 13.2-8
Solution:
H (ω ) =
Vo (ω )
=−
Vi (ω )
R2
1
jω C
R1
⎛ R2 ⎞
−⎜
⎜ R1 ⎟⎟
⎝
⎠
=
1+ jω C R2
R
1
−5
When R = 10 k Ω, R = 50 k Ω, and C = 2 μ F , then 2 = 5 and R C =
so H (ω ) =
1
2
2
ω
R
10
1+ j
1
10
13.2-8
P 13.2-9
The input to the circuit shown in Figure P 13.2-9 is the source voltage, vi(t), and the response
is the voltage across RL, vo(t). Express the gain and phase shift as functions of the radian frequency, ω.
Figure P 13.2-9
Solution:
H (ω ) =
Vo (ω )
=−
Vi (ω )
R2
1
j ω C2
R1
1
j ω C1
R2
1 + jω C2 R2
= −
R1
1 + jω C1R1
⎛R ⎞
H (ω ) = − ⎜ 2 ⎟
⎝ R1 ⎠
⎛ 1 + jω C1R1 ⎞
⎜
⎟
⎝ 1 + jω C2 R2 ⎠
When R1 = 10 kΩ, R2 = 50 kΩ , C1 = 4 μ F and C2 = 2μ F,
then
R2
R1
= 5 , C1R1 =
1
1
and C2 R 2 =
so
25
10
ω ⎞
⎛
⎜ 1 + j 25 ⎟
H (ω ) = − 5 ⎜
ω ⎟⎟
⎜⎜ 1 + j
⎟
10 ⎠
⎝
13.2-9
gain = H(ω ) = ( 5 )
1+
1+
ω2
625
ω2
100
⎛ω⎞
⎛ω ⎞
phase shift = ∠Η (ω ) = 180 + tan −1 ⎜ ⎟ − tan −1 ⎜ ⎟
⎝ 25 ⎠
⎝ 10 ⎠
13.2-10
P 13.2-10
The input to the circuit shown in Figure P 13.2-10 is the source voltage, vi(t), and the
response is the voltage across RL, vo(t). The resistance, R1, is 10 kΩ. Design this circuit to satisfy the
following two specifications:
(a)
The gain at low frequencies is 5.
(b)
The gain at high frequencies is 2.
Answer: R2 = 20 kΩ and R3 = 30 kΩ
Figure P 13.2-10
Solution
R3
1
=
jωC
H (ω ) = −
R2 +
1
R3
jωC
=
1
1 + jωC R 3
R3 +
jωC
R3
R3
1+ jω C R3
R + R + jω R2 R3C
=− 2 3
R1
R1 + jω R1 R3C
5 = lim H (ω ) =
R2 + R3
R1
2 = lim H (ω ) =
R2
⇒ R2 = 2 R1 = 20 kΩ
R1
ω →0
ω →∞
then
R3 = 5 R1 − R2 = 30 kΩ
13.2-11
P 13.2-11
The input to the circuit shown in Figure P 13.2-11 is the source voltage, vi(t), and the
response is the voltage across RL, vo(t). Design this circuit to satisfy the following two specifications:
(a)
The phase shift at ω = 1000 rad/s is 135°.
(b)
The gain at high frequencies is 10.
Answer: R1 = 1 kΩ and R2 = 10 kΩ
Figure P 13.2-11
Solution:
H (ω ) = −
R2 +
1
jω C
R1
=−
1 + jω C R 2
jω C R1
∠H (ω ) = 180° + tan −1 (ω C R 2 ) − 90°
∠H (ω ) = 135° ⇒ tan −1 (ωCR2 ) = 45° ⇒ ω C R 2 = 1
⇒ R2 =
10 = lim H (ω ) =
ω →∞
1
= 10 kΩ
10 ×10−7
3
R2
R
⇒ R1 = 2 = 1 kΩ
R1
10
13.2-12
P 13.2-12
The input to the circuit shown in Figure P 13.2-12 is the source voltage, vi(t), and the
response is the voltage across RL, vo(t). Design this circuit to satisfy the following two specifications:
(a)
The phase shift at ω = 1000 rad/s is 225°.
(b)
The gain at high frequencies is 10.
Answer: R1 = 10 kΩ and R2 = 100 kΩ
Figure P 13.2-12
Solution:
−R2
jω C R 2
Vo (ω )
= H (ω ) =
= −
1
1+ jω C R1
Vs (ω )
R1 +
jω C
R
10 = lim H (ω ) = 2 ⇒ R 2 = 10 R1
R1
ω →∞
∠ H (1000) = 180° + 90° − tan −1 1000 C R1 ⇒ R1 =
tan(270° − 225°)
= 10 kΩ
1000 C
R2 = 100 kΩ
13.2-13
P 13.2-13 The input to the circuit of Figure P 13.2-13 is
vs = 50 + 30 cos (500t + 115°) − 20 cos(2500t + 30°) mV
Find the steady-state output voltage, vo, for (a) C = 0.1μF and (b) C = 0.01μF. Assume an ideal op amp.
Figure P 13.2-13
Solution:
1
jω C2
V (ω )
=−
H (ω ) = o
1
Vs (ω )
R1 +
jω C1
R2
=
( −C1R2 ) jω
(1 + jω R1C1 ) (1 + jω R2C2 )
When R1 = 5 kΩ, C1 = 1 μ F,
R2 = 10 kΩ and C2 = 0.1 μ F,
then
H (ω ) =
( −0.01) jω
ω ⎞
⎛
⎜1 + j
⎟
200 ⎠
⎝
ω ⎞
⎛
⎜1 + j
⎟
1000 ⎠
⎝
so
ω
0
500
2500
H (ω )
0
1.66
0.74
∠H (ω )
−90°
175°
116°
Then
13.2-14
v ( t ) = (0) 50 + (1.66) ( 30 ) cos(500t + 115° + 175°) − (0.74) ( 20 ) cos(2500t + 30° + 116°)
o
=49.8cos(500t − 70°) −14.8cos (2500t +146°) mV
When R1 =5 kΩ, C1 =1 μ F, R 2 =10 kΩ and C2 =0.01 μ F, then
H (ω )= − 0.01
jω
ω ⎞⎛
ω ⎞
⎛
⎜1+ j
⎟ ⎜1+ j
200 ⎠ ⎝
10,000 ⎟⎠
⎝
So
ω
0
500
2500
H (ω )
0
1.855
1.934
∠H (ω )
−90°
−161°
170°
Then
v (t ) = (0) ( 50 ) + (1.855) ( 30 ) cos(500t + 115° − 161°) − (1.934) ( 20 ) cos(2500t + 30° + 170°)
o
= 55.65 cos(500t − 46°) − 38.68cos(2500t + 190°) mV
13.2-15
P 13.2-14
The source voltage, vs, shown in the circuit of Figure P 13.2-14a is a sinusoid having a
frequency of 500 Hz and an amplitude of 8 V. The circuit is in steady state. The oscilloscope traces
show the input and output waveforms as shown in Figure P 13.2-14b.
Figure P 13.2-14
(a)
Determine the gain and phase shift of the circuit at 500 Hz.
(b)
Determine the value of the capacitor.
(c)
If the frequency of the input is changed, then the gain and phase shift of the circuit will change.
What are the values of the gain and phase shift at the frequency 200 Hz? At 2000 Hz? At what
frequency will the phase shift be –45°? At what frequency will the phase shift be –135°?
(d)
What value of capacitance would be required to make the phase shift at 500 Hz be –60°? What
value of capacitance would be required to make the phase shift at 500 Hz be –300°?
(e)
Suppose the phase shift had been –120°at 500 Hz. What would be the value of the capacitor?
Answer: (b) C = 0.26 μF (e) this circuit can’t be designed to produce a phase shift = –120°
Solution:
⎛2V⎞
(8 div) ⎜
⎟
⎝ div ⎠ = 8 V and
(a) Vs =
2
⎛2V⎞
(6.2 div) ⎜
⎟
⎝ div ⎠ = 6.2 V so gain = Vo = 6.2 = 0.775
Vo =
2
Vs
8
13.2-16
1
(b) H (ω ) =
Vo (ω )
1
jω C
=
=
.
1
Vs (ω )
1 + jω C R
R+
jω C
1
1
Let g = H (ω ) =
then C =
2 2 2
ωR
1+ ω C R
In this case ω = 2π ⋅ 500 = 3142 rad s , H (ω )
(c) ∠ H (ω )= − tan −1 (ω R C ) so ω =
2
⎛1⎞
⎜ ⎟ −1
⎝g⎠
= 0.775 and R = 1000 Ω so C = 0.26 μ F.
tan( −∠ H (ω ))
RC
Recalling that R = 1000 Ω and C=0.26µF, we calculate
ω
2π (200)
2π (2000)
H (ω )
0.95
0.26
∠H (ω )
−18°
−73°
( )
tan ⎛⎜ − −45° ⎞⎟
⎝
⎠ = 3846 rad s
∠ H (ω )= − 45° requires ω =
−
1000 .26×10 6
(
∠ Η (ω ) = −135° requires ω =
)
(
)
tan ( − (−135°))
= − 3846 rad s
(1000)(0.26×10−6 )
A negative frequency is not acceptable. We conclude that this circuit cannot produce a phase
shift equal to −135°.
(d)
tan (−60°)
⎧
C=
= 0.55μ F
⎪
(2π ⋅ 500) (1000)
tan (−∠H (ω ))
⎪
⇒ ⎨
C=
ωR
⎪C = tan (−(−300°)) = −0.55μ F
⎪⎩
(2π ⋅ 500 ) (1000)
A negative value of capacitance is not acceptable and indicates that this
circuit cannot be designed to produce a phase shift at −300° at a frequency of
500 Hz.
(e)
C =
tan( − (−120° ))
= −0.55 μ F
(2π ⋅ 500)(1000)
This circuit cannot be designed to produce a phase shift of −120° at 500 Hz.
13.2-17
P 13.2-15 The input to the circuit in Figure P 13.215 is the voltage of the voltage source, vi(t). The
output is the voltage vo(t). The network function of
this circuit is
H (ω ) =
Vo (ω )
(−0.1) jω
=
Vi (ω ) ⎛
ω ⎞⎛
ω ⎞
⎜1 + j p ⎟ ⎜ 1 + j 125 ⎟
⎠
⎝
⎠⎝
Determine the values of the capacitance, C, and the
pole, p.
Figure P 13.2-15
Solution:
R2
1
jω C2
1+ j ω C 2 R2
H (ω ) = −
=−
1
j ω C 1 R1 + 1
R1 +
j ω C1
j ω C1
R 2 ||
=
( −C
1
1
1
R2 ) j ω
(1 + j ω C R )(1 + j ω C
2
R2 )
⎧
⎪ −C R = −0.1
1
2
⎪
C
R
j
−
ω
( 1 2)
⎪
( −0.1) jω
1
1
or
=
⇒ ⎨ C 1 R1 =
p
125
(1 + j ω C1 R1 )(1 + j ω C 2 R 2 ) ⎛⎜1 + j ω ⎞⎟ ⎛⎜1 + j ω ⎞⎟
⎪
⎪
p⎠⎝
125 ⎠
1
1
⎝
or
⎪C 2 R 2 =
125
p
⎩
Since C 1 = 5 μF, R1 = 8 kΩ and R 2 = 20 kΩ
C 1 R1 = ( 5 ×10−6 )( 8 × 103 ) =
1
= C 2 R2
125
⇒ C2 =
40
1
1
=
≠
⇒
1000 25 125
p = 25 rad/s
1
1
=
= 0.4 × 10−6 = 0.4 μF
125 R 2 125 ( 20 ×103 )
13.2-18
P 13.2-16 The input to the circuit in Figure P
13.2-16 is the voltage of the voltage source, vs(t).
The output is the voltage vo(t). The network
function of this circuit is
ω
1+ j
V (ω )
z
=k
H (ω ) = o
ω
Vs (ω )
1+ j
p
Figure P 13.2-16
Determine expressions that relate the network function parameters k, z, and p to the circuit parameters
R1, R2, L, N1, and N2.
I 1 (ω ) =
Solution:
V s (ω )
⎛ N1 ⎞
R1 + ⎜
⎟
⎜ N2 ⎟
⎝
⎠
2
( R2 + j ω L)
⎛ N1 ⎞
⎜
⎟ R 2 + j ω L V s (ω )
⎛ N1
⎞ ⎜⎝ N 2 ⎟⎠
V o (ω ) = − R 2 + j ω L I 2 (ω ) = − R 2 + j ω L ⎜ −
I (ω ) ⎟ =
2
⎜ N2 1
⎟
⎛ N1 ⎞
⎝
⎠
R1 + ⎜
⎟ R2 + jω L
⎜ N2 ⎟
⎝
⎠
⎛ N1 ⎞
⎛ N2 ⎞
L
⎜
⎟ R2 + jω L
⎜
⎟ R2
1+ jω
⎜
⎟
⎜
⎟
R2
V (ω )
⎝ N2 ⎠
⎝ N1 ⎠
H (ω ) = o
=
=
2
2
L
Vs (ω )
⎛ N1 ⎞
⎛ N2 ⎞
1+ jω
2
R1 + ⎜
⎟ R2 + jω L
⎜
⎟ R + R2
⎛ N2 ⎞
⎜ N2 ⎟
⎜ N1 ⎟ 1
⎝
⎠
⎝
⎠
⎜
⎟ R + R2
⎜ N1 ⎟ 1
⎝
⎠
Comparing to the given network function:
(
)
(
(
⎛ N2 ⎞
⎜
⎟R
⎜ N1 ⎟ 2
⎝
⎠
2
)
)
(
)
)
(
k=
(
⎛ N2 ⎞
⎜
⎟ R + R2
⎜ N1 ⎟ 1
⎝
⎠
)
2
⎛ N2 ⎞
⎜
⎟ R + R2
⎜ N1 ⎟ 1
R2
⎠
, z=
and p = ⎝
.
L
L
13.2-19
P 13.2-17 The input to the circuit in Figure P 13.2-17 is the voltage of the voltage source, vs(t). The
output is the voltage vo(t). The network function of this circuit is
Vo (ω )
jω
=k
ω
Vs (ω )
1+ j
p
Determine expressions that relate the network function parameters k and p to the circuit parameters R1,
R2, M, L1, and L2.
H (ω ) =
Figure P 13.2-17
Solution: Mesh equations:
(
)
0 = ( R 2 + j ω L 2 ) I 2 (ω ) + j ω M I 1 (ω )
V s (ω ) = R1 + j ω L1 I 1 (ω ) + j ω M I 2 (ω )
Solving the mesh equations
R2 + jω L2
I 1 (ω ) = −
⎡
V s (ω ) = ⎢ − R1 + j ω L1
⎣
(
I 2 (ω ) =
− jω M
=
V o (ω )
V s (ω )
=
jω M R2
( R1 + j ω L1 )( R 2 + j ω L 2 ) + ω 2 M 2
(
jω M R2
)
M R2
R1 R 2 + ω
2
(M
2
M R2
(
V s (ω )
(
R1 R 2 + ω 2 M 2 − L1 L 2 + j ω R1 L 2 + L1 R 2
)
− L1 L 2 1 + j ω
Comparing to the given network function:
k=
V s (ω )
( R1 + j ω L1 )( R 2 + j ω L 2 ) + ω 2 M 2
V o (ω ) = − R 2 I 2 (ω ) =
H (ω ) =
)
I 2 (ω )
jω M
R2 + jω L2
⎤
+ j ω M ⎥ I 2 (ω )
jω M
⎦
R1 R 2 + ω 2 M 2 − L1 L 2
)
and p =
)
V s (ω )
jω
R1 L 2 + L1 R 2
(
R1 R 2 + ω 2 M 2 − L1 L 2
(
R1 R 2 + ω 2 M 2 − L1 L 2
)
)
R1 L 2 + L1 R 2
13.2-20
P 13.2-18 The input to the circuit in Figure P 13.2-18 is the voltage of the voltage source, vi(t). The
output is the voltage vo(t). The network function of this circuit is
H (ω ) =
Vo (ω )
jω
=k
Vi (ω )
⎛
ω ⎞⎛
ω⎞
⎜1 + j ⎟ ⎜1 + j ⎟
p1 ⎠ ⎝
p2 ⎠
⎝
Determine expressions that relate the network function parameters k, p1, and p2 to the circuit parameters
R1, R2, R3, R4, A, C, and L.
Figure P 13.2-18
P13.2-18
Using voltage division twice gives:
A V2 (ω )
=
Vi (ω )
Vo (ω )
=
A V2 (ω )
and
R2
A R2
A R2
R1 + R 2
1 + jω C R 2
A=
=
R2
C R1 R 2
R1 + R 2 + jω C R1 R 2
1 + jω
R1 +
1 + jω C R 2
R1 + R 2
jω L R 4
R 4 + jω L
jω L R 4
L
jω
=
=
jω L R 4
R 3 R 4 + jω L ( R 3 + R 4 ) R 3
L ( R3 + R 4 )
R3 +
ω
+
1
j
R 4 + jω L
RR
3
4
Combining these equations gives
ALR 2
Vo (ω )
jω
=
Vi (ω ) R 3 ( R1 + R 2 ) ⎛
L ( R3 + R 4 ) ⎞ ⎛
C R1 R 2 ⎞
⎜1 + jω
⎟ ⎜1 + jω
⎟
⎜
⎟ ⎜⎝
R3 R 4
R1 + R 2 ⎟⎠
⎝
⎠
ALR 2
R 3R 4
and either p 1 =
and
Comparing to the given network function gives k =
R 3 R1 + R 2
L R3 + R4
H (ω ) =
(
p2 =
R1 + R 2
CR1R 2
or p 1 =
R1 + R 2
CR1R 2
and p 2 =
(
R3R 4
L R3 + R 4
)
)
(
)
.
13.2-21
P 13.2-19 The input to the circuit shown in Figure P 13.2-19 is the voltage of the voltage source, vs.
The output of the circuit is the capacitor voltage, vo. Determine the values of the resistances R1, R2, R3,
and R4 required to cause the network function of the circuit to be
H (ω ) =
Vo (ω )
21
=
ω ⎞⎛
ω ⎞
Vs (ω ) ⎛
⎜1 + j ⎟ ⎜1 + j
⎟
5 ⎠⎝
200 ⎠
⎝
Figure P 13.2-19
Solution: Represent the circuit in the frequency domain.
Apply KCL at the top node of the left capacitor, C1, to get
V a − Vs
R1
+ j ω C 1 Va = 0 ⇒ Va =
1
Vs
1 + j ω C 1 R1
The op amp, together with resistors R2 and R3, comprise a noninverting amplifier so
⎛ R3 ⎞
Vb = ⎜1 +
V
⎜ R 2 ⎟⎟ a
⎝
⎠
13.2-22
(Alternately, this equation can be obtained by applying KCL at the inverting input node of the op amp.)
Apply KCL at the top node of the right capacitor, C2, to get
Vo − V b
R4
+ j ω C 2 Vo = 0 ⇒ Vo =
1
Vb
1+ j ω C 2 R4
Combining these equations gives
H (ω ) =
V o (ω )
V s (ω )
1+
=
R3
R2
(1 + j ω C R ) (1 + jω C
1
1
2
R4 )
Comparing to the specified network function gives
1+
R3
R2
(1 + j ω C R ) (1 + jω C
1
1
2
R4 )
=
21
ω⎞⎛
ω ⎞
⎛
⎜1 + j ⎟ ⎜1 + j
⎟
5 ⎠⎝
200 ⎠
⎝
The solution is not unique. For example, we can require
1+
R3
R2
= 21 , C 1 R1 =
1
1
= 0.2 , C 2 R 4 =
= 0.005
5
200
With the given values of capacitance, and choosing R2 = 10 kΩ, we have
R1 = 200 kΩ, R2 = 10 kΩ, R3 = 200 kΩ and R4 = 5 kΩ
(checked using LNAP 9/14/04)
13.2-23
P 13.2-20
The input to the circuit shown in Figure
P 13.2-20 is the voltage of the voltage source, vs. The
output of the circuit is the voltage vo. Determine the
network function
H (ω ) =
Vo (ω )
Vs (ω )
of the circuit.
Figure P 13.2-20
Solution:
Represent the circuit in the frequency domain. Apply KCL at the
inverting input node of the op amp to get
V o − Vs
R1
or
(R
1
+ j ω C 1 ( V o − Vs ) +
Vs
R2
=0
+ R 2 + j ω C 1 R1 R 2 ) V s = ( R 2 + j ω C 1 R1 R 2 ) V o
so
H=
Vo
Vs
=
R1 + R 2 + j ω C 1 R1 R 2
R 2 + j ω C 1 R1 R 2
=
R1 + R 2
R2
1 + j ω C1
×
R1 R 2
R1 + R 2
1 + j ω C 1 R1
With the given values
H=
Vo
Vs
=
1+ j
ω
25 + j ω
25
=5
ω
5 + jω
1+ j
5
(checked using LNAP 7/24/05)
13.2-24
P 13.2-21
The input to the circuit shown in
Figure P 13.2-21 is the voltage of the voltage
source, vs. The output of the circuit is the
capacitor voltage, vo. Determine the network
function
H (ω ) =
Vo (ω )
Vs (ω )
Figure P 13.2-21
of the circuit.
Solution:
Represent the circuit in the frequency domain. After
determining some equivalent impedances, the network
function can be determined using voltage division.
⎛
1 ⎞
⎜⎜ R 2 +
⎟
j ω C 1 ⎟⎠
⎛
1 + j ω C1 R 2
1
1 ⎞
⎝
=
=
|| ⎜ R 2 +
⎟
j ω C 2 ⎜⎝
j ω C 1 ⎟⎠ R + 1 + 1
j ω ( C1 + C 2 ) − ω 2 R 2 C1 C 2
2
j ω C1 j ω C 2
1
jω C2
Next, using voltage division gives
1 + j ω C1 R 2
j ω ( C1 + C 2 ) − ω 2 R 2 C1 C 2
1 + j ω C1 R 2
H=
=
=
1 + j ω C1 R 2
Vs
R1 ⎡⎣ j ω ( C 1 + C 2 ) − ω 2 R 2 C 1 C 2 ⎤⎦ + 1 + j ω C 1 R 2
R1 +
With the
j ω ( C1 + C 2 ) − ω 2 R 2 C1 C 2
Vo
=
1 + j ω C1 R 2
1 − ω C 1 C 2 R1 R 2 + j ω ( C 1 R1 + C 2 R1 + R 2 C 1 )
2
given values
H=
Vo
Vs
1+ j
=
1−
ω
2
100
ω
10
+j
3ω
10
=
100 + j 10 ω
100 − ω 2 + j 30 ω
13.2-25
P 13.2-22
The input to the circuit shown in Figure
P 13.2-22 is the voltage of the voltage source, vs.
The output of the circuit is the capacitor voltage, vo.
The network function of the circuit is
H (ω ) =
Vo (ω )
Ho
=
Vs (ω ) 1+j ω
p
Figure P 13.2-22
Determine the values of Ho and p.
Solution:
Represent the circuit in the frequency domain. Apply
KVL to the left mesh to get
Vs = 8 I a + 4 I a
⇒ Ia =
Vs
12
Voltage division gives
40
4
4
jω
Vo =
Ia =
4I a =
jω
jω
40
8+
1+
1+
5
5
jω
1
V
⎛ s⎞
3 V
⎜ ⎟=
s
12
⎝ ⎠ 1+ jω
5
The network function of the circuit is
1
H=
= 3
Vs 1 + j ω
5
Vo
Comparing this network function to the specified network function gives
Ho =
1
and p = 5
3
(checked using LNAP 9/19/04)
13.2-26
P 13.2-23
The input to the circuit shown in Figure P 13.2-23 is the current of the current source, is.
The output of the circuit is the resistor current, io. The network function of the circuit is
H (ω ) =
I o (ω )
0.8
=
I s (ω ) 1+j ω
40
Determine the values of the resistances R1 and R2.
Figure P 13.2-23
Solution:
Represent the circuit in the frequency domain. Apply
KCL at the top node of R1 to get
Is =
Va
R1
+ K Va
⇒ Va =
R1
1 + K R1
Is
Current division gives
1
Io =
⎛ R1
⎞
K
K
jω C
K Va =
Va =
Is ⎟
⎜⎜
1
1 + j ω C R2
1 + j ω C R 2 ⎝ 1 + K R1 ⎟⎠
+ R2
jω C
The network function of the circuit is
K R1
H=
Io
Is
=
1 + K R1
1+ j ω C R2
Comparing this network function to the specified network function gives
K R1
1 + K R1
= 0.8 and C R 2 =
1
40
With the given values
0.2 R1
1 + 0.2 R1
= 0.8 ⇒ R1 = 20 Ω and 0.001 R 2 =
1
⇒ R 2 = 25 Ω
40
(checked using LNAP 9/19/04)
13.2-27
P 13.2-24
The input to the circuit shown in Figure P
13.2-24 is the voltage of the voltage source, vs. The output
of the circuit is the resistor voltage, vo. Specify values for
L1, L2, R, and K that cause the network function of the
circuit to be
H (ω ) =
Vo (ω )
1
=
ω ⎞⎛
ω⎞
Vs (ω ) ⎛
⎜1 + j ⎟ ⎜1 + j ⎟
20 ⎠ ⎝
50 ⎠
⎝
Figure P 13.2-24
Solution:
Represent the circuit in the frequency domain. Apply
KVL to the left mesh to get
V s = j ω L1 I a + K I a
⇒ Ia =
Vs
K + j ω L1
Voltage division gives
Vo =
⎛
⎞
Vs
R
R
RK
=
K Ia =
K⎜
Vs
⎟
R + j ω L2
R + j ω L 2 ⎝⎜ K + j ω L1 ⎠⎟ ( R + j ω L 2 )( K + j ω L1 )
The network function of the circuit is
H=
Vo
Vs
=
1
L2 ⎞ ⎛
L1 ⎞
⎛
⎜1 + j ω ⎟ ⎜1 + j ω ⎟
R ⎠⎝
K⎠
⎝
Comparing this network function to the specified network function gives
L2
R
=
L1 1
L1 1
L
1
1
and
and
=
or 2 =
=
K 50
K 20
20
R 50
These equations do not have a unique solution. One solution is
L1 = 0.1 H, L2 = 0.1 H, R = 5 Ω and K = 2 V/A
(checked using LNAP 9/19/04)
13.2-28
P 13.2-25
The input to the circuit shown in Figure P
13.2-25 is the voltage of the voltage source, vs. The
output of the circuit is the resistor voltage, vo. Specify
values for R and C that cause the network function of
the circuit to be
H (ω ) =
Vo (ω )
−8
=
Vs (ω ) 1 + j ω
250
Figure P 13.2-24
Solution: Represent the circuit in the frequency domain.
The node equations are
V a − Vs
R1
+
and
Va
Va
R2
+
= 0 ⇒ Va =
Vs
1
R2
R1 + R 2 + j ω C R1 R 2
jω C
Va Vo
R3
+
= 0 ⇒ Vo = −
Va
R 2 R3
R2
The network function is
H=
Vo
Vs
−
=
R3
R2
−
R2
R1 + R 2 + j ω C R1 R 2
=
R3
R1 + R 2
R R
1+ jω C 1 2
R1 + R 2
Using the given values for R1 and R2 and letting R3 = R gives
R
−
Vo
4 × 104
H=
=
V s 1 + j ω C (104 )
Comparing this network function to the specified network function gives
C (104 ) =
1
250
⇒ C = 0.4 μF and
R
= 8 ⇒ R = 320 kΩ
4 ×104
(checked using LNAP 9/19/04)
13.2-29
P13.2-26
The network function of a circuit is Η ( ω ) =
Vo ( ω )
Vs ( ω )
=
j 40 ω
. When the input to this circuit is
120 + j 20 ω
v s ( t ) = 5cos ( 5 t + 15° ) V , the output is v o ( t ) = A cos ( 5 t + 65.194° ) V . On the other hand, when the
input to this circuit is v s ( t ) = 5cos ( 8 t + 15° ) V , the output is v o ( t ) = 8cos ( 8 t + θ ) V . Determine the
values of A and θ.
Answers: A = 6.4018 V and θ = 51.87 °
Solution:
When the input to this circuit is v s ( t ) = 5cos ( 5 t + 15° ) V :
Η ( 5) =
j 40 ( 5 )
= 1.2804∠50.194° ⇒ v o ( t ) = 5 (1.2804 ) cos ( 5 t + 15° + 50.194° )
120 + j 20 ( 5 )
= 6.4018cos ( 5 t + 65.194° ) V
When the input to this circuit is v s ( t ) = 8cos ( 8 t − 15° ) V
Η (8) =
j 40 ( 8 )
= 1.6∠36.87° ⇒ v o ( t ) = 5 (1.6 ) cos ( 8 t + 15° + 36.87° )
120 + j 20 ( 8 )
= 8cos ( 8 t + 51.87° ) V
13.2-30
P13.2-27
The network function of a circuit is Η ( ω ) =
Vo ( ω )
Vs ( ω )
=
k
ω
1+ j
p
where k > 0 and p > 0. When the input to
v s ( t ) = 12 cos (120 t + 30° ) V
this circuit is
v o ( t ) = 42.36 cos (120 t − 48.69° ) V .
the output is
Determine the values of k and p.
Answers: k = 18 and p = 24 rad/s.
Solution:
Vo 42.36∠ − 48.69°
⎛ 120 ⎞
k
k
=
∠ − tan −1 ⎜
and
=
= 3.53∠ − 78.69°
⎟
2
120
12
30
V
∠
°
p
⎝
⎠
s
⎛ 120 ⎞
1+ j
1+ ⎜
p
⎟
⎝ p ⎠
so
⎛ 120 ⎞
120
− tan −1 ⎜
= tan ( 78.69° ) = 5 ⇒
⎟ = −78.69° ⇒
p
⎝ p ⎠
p=
120
= 24 rad/s
5
and
k
⎛ 120 ⎞
1+ ⎜
⎟
⎝ p ⎠
2
=
k
1 + ( 5)
2
= 3.53 ⇒ k = 3.53 26 = 18
13.2-31
P13.2-28
The network function of a circuit is H (ω ) =
20
. When the input to this circuit is sinusoidal, the
8+ jω
output is also sinusoidal. Let ω 1 be the frequency at which the output sinusoid is twice as large as the
input sinusoid and let ω 2 be the frequency at which output sinusoid is delayed by one tenth period with
respect to the input sinusoid. Determine the values of ω 1 and ω 2.
Solution:
The gain is 2 at the frequency ω 1 so 2 =
When the frequency is ω 2, the period is
20
8 2 + ω12
2π
ω2
2
⎛ 20 ⎞
and ω1 = ⎜ ⎟ − 8 2 = 6 rad/s .
⎝ 2 ⎠
. Also a delay t o corresponds to a phase shift −ω 2 t o. In this
⎛ 2π ⎞
⎛ ω2 ⎞
case, t 0 = 0.1⎜
so the phase shift is -0.2 π. Then −0.2 π = − tan −1 ⎜
⎟
⎟ so
⎜ ω2 ⎟
⎝ 8 ⎠
⎝
⎠
ω 2 = 8 tan ( 0.2 π ) = 5.8123 rad/s .
13.2-32
P13.2-29
The input to the circuit in Figure P13.3-29 is the voltage source voltage, v s ( t ) . The output is the voltage
v o ( t ) . When the input is v s ( t ) = 8cos ( 40 t ) V , the output is v o ( t ) = 2.5cos ( 40 t + 14° ) V . Determine
the values of the resistances R1 and R 2 .
Figure P13.2-29
Solution:
jω L R 2
L
R 2 + jω L
R1
Vo (ω )
=
=
jω L R 2
L
Vi (ω )
1 + jω
R1 +
Rp
R 2 + jω L
where R p =
Vo (ω )
Vi (ω )
In this case the angle of
the magnitude of
jω
R1 R 2
R1 + R 2
ω
=
L
R1
⎛
j ⎜ 90 − tan −1 ω
⎛ L ⎞
1+ ⎜ω
⎜ R p ⎟⎟
⎝
⎠
2
e
⎜
⎝
L ⎞⎟
R p ⎟⎠
Vo (ω )
L L ( R1 + R 2 ) tan ( 90° − 14° )
=
=
= 0.1 and
is specified to be 14° so
40
Rp
R1 R 2
Vi (ω )
Vo (ω )
2.5
is specified to be
so
8
Vi (ω )
40
L
R1
=
2.5
⇒
8
1 + 16
that satisfies these two equations is L = 1 H, R1 = 31 Ω, R 2 = 14.76 Ω .
L
= 0.0322 . One set of values
R1
13.2-33
P13.2-30
The input to the circuit shown in Figure P13.2-30 is the voltage source voltage, v s ( t ) . The output is the
voltage v o ( t ) . The input v s ( t ) = 2.5cos (1000 t ) V causes the output to be
v o ( t ) = 8cos (1000 t + 104° ) V . Determine the values of the resistances R1 and R 2 .
Answers: R1 = 1515 Ω and R 2 = 20 kΩ.
Figure P13.2-30
Solution:
R2
1
jω C
=
R2
R2
1 + jω CR 2
R2
1 + jω CR 2
R1
Vo (ω )
=−
=−
Vi (ω )
R1
1 + jω CR 2
R2
R1
Vo (ω )
j (180− tan −1 ω CR 2 )
e
=
2
Vi (ω )
1 + (ω CR 2 )
In this case the angle of
magnitude of
Vo (ω )
tan (180° − 104° )
is specified to be 104° so CR 2 =
= 0.004 and the
1000
Vi (ω )
Vo (ω )
8
so
is specified to be
2.5
Vi (ω )
R2
R1
1 + 16
=
8
⇒
2.5
R2
R1
= 13.2 . One set of values that
satisfies these two equations is C = 0.2 μ F, R1 = 1515 Ω, R 2 = 20 kΩ .
13.2-34
Section 13-3: Bode Plots
P 13.3-1
Sketch the magnitude Bode plot of H (ω ) =
4(5 + jω )
ω⎞
⎛
⎜1 + j ⎟
50 ⎠
⎝
Solution:
ω⎞
⎛
20 ⎜ 1 + j ⎟
5⎠
H (ω )= ⎝
ω⎞
⎛
⎜1 + j ⎟
50 ⎠
⎝
⎧
⎪
⎪
20
⎪
⎪
⎪
⎪ 20 ⎛ j ω ⎞
⎪
⎜
⎟
≈⎨
⎝ 5⎠
⎪
⎪
⎪ ⎛ ω⎞
⎪ 20 ⎜ j 5 ⎟
⎠ = 200
⎪ ⎝
⎪ ⎛jω⎞
⎪⎩ ⎜⎝ 50 ⎟⎠
ω <5
5 < ω < 50
50 < ω
13.3-1
P 13.3-2 Compare the magnitude Bode plots of H1 (ω ) =
Solution:
ω
10 ( 5 + j ω ) 1 + j 5
H1 (ω ) =
=
ω
50 + j ω
1+ j
50
10(5 + jω )
100(5 + jω )
and H 2 (ω ) =
50 + jω
50 + jω
ω
and
1+ j
100 ( 5 + j ω )
5
H 2 (ω ) =
= 10
ω
50 + j ω
1+ j
50
Both H1(ω) and H2(ω) have a pole at ω = 50 rad/s and a zero at ω = 5 rad/s. The slopes of both
magnitude Bode plots increase by 20 dB/decade at ω = 5 rad/s and decrease by 20 dB/decade at ω =
50rad/s. The difference is that for ω < 5rad/s
H1 (ω )
1 = 0 dB and
H 2 (ω )
10 = 20 dB
13.3-2
P 13.3-3 The input to the circuit shown in Figure P 13.3-3 is the source voltage, vin (t), and the
response is the voltage across R3, vout(t). The component values are R1 = 5 kΩ, R2 = 10 kΩ, C1 = 0.1 μF,
and C2 = 0.1 μF. Sketch the asymptotic magnitude Bode plot for the network function.
C2
R1
R2
C1
–
+
+
vin(t)
+
–
vout(t)
R3
–
Figure P 13.3-3
Solution:
R2
jω
1+ jω C2 R2
H (ω ) = −
= −C1 R2
1
(1+ jω R1C1 )(1+ jω R2C2 )
R1 +
jω C1
This network function has poles at
1
1
p1 =
= 2000 rad s and p2 =
= 1000 rad s
R1C1
R2C2
so
H (ω )
⎧
⎪−(C R ) jω
ω < p2
1 2
⎪
⎪
jω
R
=− 2 =−2
p2 < ω < p1
⎨−(C1 R2 )
jωC1 R1
R1
⎪
⎪
1
jω
=−
ω > p1
⎪−(C1 R2 )
( jωC1 R1 )( jωC2 R2 )
jωC2 R1
⎩
Here’s the Bode plot:
13.3-3
C1
C2
R1
R2
P 13.3-4 The input to the circuit shown in Figure P 13.3-4
is the source voltage, vs(t), and the response is the voltage
across R3, vo(t). Determine H(ω) and sketch the Bode
diagram.
–
vs +
+
+
–
R3
vo
–
Figure P 13.3-4
Solution:
R2
R (1+ jω C1 R1 )
R
1
1
1+ jω C2 R2
H (ω ) = −
=− 2
so K = − 2 , z =
and p =
R1
R1 (1+ jω C2 R2 )
R1
C1 R1
C2 R2
1+ jω C1 R1
When z < p
When z > p
13.3-4
P 13.3-5 The input to the circuit shown in Figure P 13.3-5a is the voltage, vi(t), of the independent
voltage source. The output is the voltage, vo(t), across the capacitor. Design this circuit to have the Bode
plot shown in Figure P 13.3-5b.
Hint: First show that the network function of the circuit is
⎛ AL R4 ⎞
jω ⎜
⎟
R1 ( R3 + R4 ⎠
Vo (ω )
⎝
=
H (ω ) =
Vi (ω ) ⎛
C R3 R4 ⎞
L( R1 + R2 ) ⎞ ⎛
⎟
⎜1 + jω
⎟ ⎜1 + jω
R1 R2 ⎠ ⎝
R3 + R4 ⎠
⎝
R3
R1
+
–
v1(t)
L
+
v2(t)
–
+
–
R2
Av2(t)
R4
+
vo (t)
–
C
20 log10 |H(ω)|, (dB)
(a)
20
0
200
20
20k
200k
ω (rad/s, log scale)
(b)
Figure P 13.3-5
Solution: Using voltage division twice gives:
V2 (ω )
=
Vi (ω )
and
jω L R 2
R 2 + jω L
jω L R 2
L
jω
=
=
jω L R 2
R1 R 2 + jω L ( R1 + R 2 ) R1
L ( R1 + R 2 )
R1 +
+
ω
j
1
R 2 + jω L
R1 R 2
Vo (ω )
=
V2 (ω )
R4
A R4
A R4
R3 + R 4
1 + jω C R 4
A=
=
R4
C R3 R 4
R3 + R 4 + jω C R3 R 4
R3 +
1 + jω
R3 + R 4
1 + jω C R 4
13.3-5
Combining these equations gives
H (ω ) =
ALR 4
Vo (ω )
jω
=
Vi (ω ) R1 ( R 3 + R 4 ) ⎛
L ( R1 + R 2 ) ⎞ ⎛
CR 3 R 4 ⎞
⎜1 + jω
⎟ ⎜ 1 + jω
⎟
⎜
⎟ ⎜⎝
R1 R 2
R 3 + R 4 ⎟⎠
⎝
⎠
The Bode plot corresponds to the network function:
H (ω ) =
k jω
k jω
=
ω ⎞⎛
ω ⎞
⎛
ω ⎞⎛
ω ⎞ ⎛
⎜1 + j ⎟ ⎜1 + j ⎟ ⎜⎝1 + j 200 ⎟⎠ ⎜⎝1 + j 20000 ⎟⎠
p1 ⎠ ⎝
p2 ⎠
⎝
⎧
⎪
⎪
⎪ k jω
= k jω
⎪
1 ⋅1
⎪
⎪ k jω
H (ω ) ≈ ⎨
= k p1
jω
⎪
⋅1
p1
⎪
⎪ k jω
k p1 p2
⎪
=
jω
⎪ jω ⋅ jω
⎪ p p
⎩ 1
2
ω ≤ p1
p1 ≤ ω ≤ p2
ω ≥ p2
This equation indicates that |H(ω)|=k p1 when p1 ≤ ω ≤ p2. The Bode plot indicates that |H(ω)|=20 dB =
10 when p1 ≤ ω ≤ p2. Consequently
10 10
k=
=
= 0.05
p1 200
Finally,
H (ω ) =
0.05 jω
ω ⎞⎛
ω ⎞
⎛
⎜1 + j
⎟ ⎜1 + j
⎟
200 ⎠ ⎝
20000 ⎠
⎝
Comparing the equation for H(ω) obtained from the circuit to the equation for H(ω)obtained from the
Bode plot gives:
R3 + R 4
ALR 4
R1 R 2
, 200 =
0.05 =
and 20000 =
C R3 R 4
R1 ( R 3 + R 4 )
L ( R1 + R 2 )
Pick L = 1 H, and R1 = R2 , then R1 = R2 = 400 Ω. Let C = 0.1 μF and R3 = R4 , then R3 = R4 = 1000 Ω.
Finally, A=40.
(Checked using ELab 3/5/01)
13.3-6
P 13.3-6
The input to the circuit shown in Figure P 13.3-6b is the voltage of the voltage source, vi(t).
The output is the voltage vo(t). The network function of this circuit is H(ω) = Vo(ω)/Vi(ω). Determine
the values of R2, C1, and C2 that are required to make this circuit have the magnitude Bode plot shown in
Figure P 13.3-6a.
20 log10|H(ω)|, dB
Answer: R2 = 400 kΩ, C1 = 25 nF, and C2 = 6.25 nF
32
v1(t)
R1 = 10 kΩ
R2
C1
C2
–
+
–
+
12
10 kΩ
40
400
4k
+
vo(t)
–
40 k
ω (rad/s, log scale)
(a)
(b)
Figure P 13.3-6
Solution: From Table 13.3-2:
R2
R1
= k = 32 dB = 40 R 2 = 40 (10 × 103 ) = 400 kΩ
1
1
= p = 400 rad/s ⇒ C 2 =
= 6.25 nF
C 2 R2
( 400 ) ( 400 ×103 )
1
1
= z = 4000 rad/s ⇒ C 1 =
= 25 nF
C 1 R1
( 4000 ) (10 ×103 )
13.3-7
The input to the circuit shown in Figure P 13.3-7b is the voltage of the voltage source, vi(t).
P 13.3-7
The output is the voltage vo(t). The network function of this circuit is H(ω) = Vo(ω)/Vi(ω). The
magnitude Bode plot is shown in Figure p 13.3-7a. Determine values of the corner frequencies, z and p.
Determine value of the low-frequency gain, k.
20 log10|H( ω)|, dB
8Ω
+
0
2Ω
vi(t)
20 log10(k)
+
–
vo(t)
0.4 H
z
p
–
ω (rad/s, log scale)
(a)
(b)
Figure P 13.3-7
Solution:
H (ω ) =
R 2 + jω L
Vo (ω )
=
Vi (ω ) R + R 2 + jω L
L
⎛
1 + jω
⎜
⎛ R2 ⎞
R2
=⎜
⎟⎜
⎜ R + R2 ⎟⎜
L
⎝
⎠ ⎜ 1 + jω
R + R2
⎝
H (ω ) =
( 0.2 ) (1 + j ( 0.2 ) ω )
1 + j ( 0.04 ) ω
⎞
⎟
⎟
⎟
⎟
⎠
⎧
⎪ k = 0.2
⎪
1
⎪
⇒ ⎨ z=
=5
0.2
⎪
1
⎪
⎪⎩ p = 0.04 = 25
13.3-8
P 13.3-8
Determine H( jω) from the asymptotic Bode diagram in Figure P 13.3-8.
40
20
dB
0
–20
–40
0.1
1
10
100
1000
10,000
ω (rad/s)
Figure P 13.3-8
Solution
• The slope is 40dB/decade for low frequencies, so the numerator will include the factor (jω)2 .
• The slope decreases by 40 dB/decade at ω = 0.7rad/sec. So there is a second order pole at ω 0 = 0.7
rad/s. The damping factor of this pole cannot be determined from the asymptotic Bode plot; call it δ
1. The denominator of the network function will contain the factor
ω
⎛ ω ⎞
−⎜
1 + 2 δ1 j
⎟
0.7 ⎝ 0.7 ⎠
•
•
•
2
The slope increases by 20 dB/decade at ω = 10 rad/s, indicating a zero at 10 rad/s.
The slope decreases by 20 dB/decade at ω = 100 rad/s, indicating a pole at 100 rad/s.
The slope decreases by 40 dB/decade at ω = 600 rad/s, indicating a second order pole at ω 0 =
600rad/s. The damping factor of this pole cannot be determined from an asymptotic Bode plot; call it
δ 2. The denominator of the network function will contain the factor
ω
⎛ ω ⎞
−⎜
1 + 2δ 2 j
⎟
600 ⎝ 600 ⎠
H (ω ) =
K (1+ j
ω
2
)( jω ) 2
10
2
2
⎛
⎞⎛
ω ⎛ω ⎞
ω ⎛ ω ⎞ ⎞⎛
ω ⎞
−⎜
1+ 2δ 2 j
−⎜
⎜⎜ 1+ 2δ1 j
⎟ ⎟⎜
⎟ ⎟⎟⎜1+ j
⎟
⎟⎜
0.7 ⎝ 0.7 ⎠ ⎠⎝
600 ⎝ 600 ⎠ ⎠⎝
100 ⎠
⎝
To determine K , notice that H (ω ) = 1 dB=0 when 0.7 < ω < 10. That is
1=
K (1) ω 2
2
⎛ ω ⎞
−⎜
⎟ (1)(1)
⎝ 0.7 ⎠
= K (0.7) 2 ⇒ K = 2
13.3-9
P 13.3-9 A circuit has a voltage ratio
H (ω ) =
k (1 + jω / z )
jω
(a)
Find the high- and low-frequency asymptotes of the magnitude Bode plot.
(b)
The high- and low-frequency asymptotes comprise the magnitude Bode plot. Over what ranges
of frequencies is the asymptotic magnitude Bode plot of H(ω) within 1 percent of the actual
value of H(ω) in decibels?
Solution:
(a)
ω⎞
⎛
K ⎜1+ j ⎟
z⎠
⎝
H (ω ) =
jω
H (ω ) =
K
ω
⎛ω ⎞
1+ ⎜ ⎟
⎝z⎠
H (ω ) dB = 20 log10
K
ω
2
⎛ω ⎞
1+⎜ ⎟
⎝z⎠
2
= 20 log10 K − 20 log10 ω + 20 log10
Let
⎛ω ⎞
1+⎜ ⎟
⎝z⎠
2
H L (ω ) dB = 20 log10 K − 20 log10 ω
K
z
⎪⎧ H L (ω ) dB
Then H (ω ) dB ~_ ⎨
⎪⎩ H H (ω ) dB
and H H (ω ) dB = 20 log10
ω << z
ω >> z
So H L (ω ) dB and H H (ω ) dB are the required low and high-frequency asymptotes.
13.3-10
The Bode plot will be within 1% of |H(ω)| dB both for ω << z and for ω >> z. The range when ω << z is
characterized by
H L (ω ) = 0.99 H (ω )
(gains not in dB)
or equivalently
20 log10 ( 0.99 ) = H L (ω ) dB − H (ω ) dB
(gains in dB)
= 20 log10 K − 20 log10 ω − 20 log10
⎛ω ⎞
1+⎜ ⎟
ω
⎝z⎠
K
2
⎛ω ⎞
= − 20 log10 1+⎜ ⎟ = 20 log10
⎝z⎠
2
1
⎛ω ⎞
1+ ⎜ ⎟
⎝z⎠
2
Therefore
0.99 =
2
1
⎛ω ⎞
1+⎜ ⎟
⎝z⎠
2
z
⎛ 1 ⎞
⇒ ω= z ⎜
⎟ −1 = 0.14 z −
7
⎝ .99 ⎠
The range when ω >> z is characterized by
H H (ω ) = .99 H (ω )
(gains not in dB)
or equivalently
20 log10 0.99 = H H (ω ) dB − H (ω ) dB
(gains in dB)
⎛ω ⎞
= 20 log10 K − 20 log10 z − 20 log10
1+ ⎜ ⎟
ω
⎝z⎠
K
2
2
1
⎛ω ⎞
= − 20 log10
1+ ⎜ ⎟ = 20 log10
2
ω
⎝z⎠
⎛z⎞
⎜ ⎟ +1
⎝ω ⎠
z
Therefore
2
⎛ 1 ⎞
= ⎜
⎟ −1 ⇒
ω
⎝ .99 ⎠
z
The error is less than 1% when ω <
ω=
z
2
⎛ 1 ⎞
⎜
⎟ −1
⎝ .99 ⎠
=
z
− 7z
0.14
z
and when ω > 7 z.
7
13.3-11
P 13.3-10
Physicians use tissue electrodes to form the
R1
interface that conducts current to the target tissue of the
human body. The electrode in tissue can be modeled by the
RC circuit shown in Figure P 13.3-10. The value of each
+
vs
+
–
C
Rt
vo
–
element depends on the electrode material and physical
Figure P 13.3-10
construction as well as the character of the tissue being
probed. Find the Bode diagram for Vo/Vs = H(jω) when R1 =
1 kΩ, C = 1 μF, and the tissue resistance is Rt = 5 kΩ.
Solution:
H (ω ) =
=
Vo (ω )
=
Vs (ω )
Rt
R t + R1
1
jω C
=
Rt +
⎛ Rt
=⎜
R1 + R t + jω C R1 R t ⎜⎝ R1 + R t
R t (1+ jω C R1 )
Rt
R1
1+ jω C R1
⎞ 1+ jω C R1
⎟⎟
⎠ 1+ jω ⎛ C R1 R t
⎜⎜
⎝ R1 + R t
⎞
⎟⎟
⎠
When R1 = 1 kΩ, C = 1 μ F and R t = 5 kΩ
ω ⎞
⎛
1+ j
5⎜
1000 ⎟ ⇒
H (ω ) = ⎜
⎟
6 ⎜ 1+ j ω ⎟
1200 ⎠
⎝
⎧5
ω <1000
⎪6
⎪
⎪⎛ 5 ⎞ ω
1000<ω <1200
H (ω ) ≅ ⎨⎜ ⎟ j
⎪⎝ 6 ⎠ 1000
⎪1
ω <1200
⎪
⎩
13.3-12
R1
P 13.3-11
L1
+
Figure P 13.3-11 shows a circuit and
M
corresponding asymptotic magnitude Bode plot.
+
–
The input to this circuit shown is the source
L2
vo(t)
vin(t)
voltage vin(t), and the response is the voltage vo(t).
R2
The component values are R1 = 80 Ω, R2 = 20 Ω,
–
L1 = 0.03 H, L2 = 0.07 H, and M = 0.01 H.
Answer: K1 = 0.75, K2 = 0.2, z = 333 rad/s, and p =
1250 rad/s
|H (ω )|, dB
Determine the values of K1, K2, p, and z.
20 log10 (K1)
20 log10 (K2)
z
p
ω , rad/sec
Figure P 13.3-11
Solution:
Mesh equations:
Vin (ω ) = I (ω ) [ R1 + ( jω L1 − jω M ) + (− jω M + jω L2 ) + R2 ]
Vo (ω ) = I (ω ) [(− jω M + jω L2 ) + R2 ]
Solving yields:
Vo (ω )
R2 + jω ( L2 − M )
=
Vin (ω ) R1 + R2 + jω ( L1 + L2 − 2M )
Comparing to the given Bode plot yields:
H (ω ) =
K1 = ωlim
→∞ |H (ω )| =
z =
L2 − M
R2
= 0.75 and K 2 = lim | H (ω ) | =
= 0.2
ω →0
L1 + L2 − 2 M
R1 + R2
R2
R1 + R2
= 333 rad s and p =
=1250 rad s
L2 − M
L1 + L2 − 2 M
13.3-13
P 13.3-12 The input to the circuit shown in Figure P 13.3-12 is the source voltage vin(t), and the
response is the voltage across R3, vout(t). The component values are R1 = 10 kΩ, C1 = 0.025 μF, and C2 =
0.05 μF. Sketch the asymptotic magnitude Bode plot for the network function.
vin(t)
C1
C2
R1
–
+
+
+
–
vout(t)
R3
–
Figure P 13.3-12
Solution:
1
1+ jω R1 C1
1 (1+ jω R1 C1 )
jω C2
=−
=−
H (ω ) = −
jω R1 C2
R1 C2
jω
1
R1
jω C1
⎧ 1 ⎛ 1 ⎞
⎪−
⎜
⎟
⎪ R C jω ⎠
H (ω ) − ⎨ 1 2 ⎝
⎪− 1 ( R C ) = − C1
⎪⎩ R1 C2 1 1
C2
ω<
ω>
1
R1 C1
1
R1 C1
With the given values:
C1 1
= = −6 dB,
C2 2
1
= 4000 rad / s
R1 C1
Here’s the Bode plot:
13.3-14
magnitude Bode plot shown in Figure P 13.3-13.
20 log10|H( ω )|, dB
P 13.3-13 Design a circuit that has the asymptotic
14
20
dB
decade
200
500
ω (rad/s logarithmic scale)
Figure P 13.3-13
Solution: Pick the appropriate circuit from Table 13.3-2.
We require
200 = z =
1
1
p C
, 500 = p =
and 14 dB = 5 = k = 1
C 1 R1
C2 R 2
z C2
Pick C1 = 1 μ F, then C2 = 0.2 μ F, R1 = 5 kΩ and R 2 = 10 kΩ.
13.3-15
magnitude Bode plot shown in Figure P 13.3-14.
20 log10|H( ω )|, dB
P 13.3-14 Design a circuit that has the asymptotic
34
20
dB
decade
500
ω (rad/s logarithmic scale)
Figure P 13.3-14
Solution: Pick the appropriate circuit from Table 13.3-2.
We require
500 = p =
R2
1
and 34 dB = 50 =
C R2
R1
Pick C = 0.1 μ F, then R 2 = 20 kΩ and R1 = 400 Ω.
13.3-16
magnitude Bode plot shown in Figure P 13.3-15.
20 log10|H( ω )|, dB
P 13.3-15 Design a circuit that has the asymptotic
14
20
dB
decade
200
500
ω (rad/s logarithmic scale)
Figure P 13.3-15
Solution: Pick the appropriate circuit from Table 13.3-2.
We require
500 = z =
1
1
p C
, 200 = p =
and 14 dB = 5 = k = 1
C 1 R1
C2 R 2
z C2
Pick C1 = 0.1 μ F, then C2 = 0.02 μ F, R1 = 20 kΩ and R 2 = 250 kΩ.
13.3-17
magnitude Bode plot shown in Figure P 13.3-16.
20 log10|H( ω )|, dB
P 13.3-16 Design a circuit that has the asymptotic
34
20
40
200
dB
decade
dB
decade
500
ω (rad/s logarithmic scale)
Figure P 13.3-16
Solution: Let’s try designing the circuit as a cascade circuit using circuits from Table 13.3-2.
A Cascade Circuit
From Table 13.3-2
Η1 ( ω ) =
and
Η 2 (ω ) =
Consequently
Η (ω ) =
k
R2
V2 (ω )
1
1
=−
where k 1 =
and p1 =
ω
R1
C1R 2
V1 (ω )
1+ j
p1
R4
V3 (ω )
k2
1
=−
where k2 =
and p2 =
ω
V2 (ω )
R3
C 2R4
1+ j
p2
V3 (ω ) V3 (ω ) V2 (ω )
k1k2
=
= Η 1 ( ω )i Η 2 ( ω ) =
i
V1 (ω ) V2 (ω ) V1 (ω )
⎛
ω ⎞⎛
ω⎞
⎜⎜ 1 + j ⎟⎟ ⎜ 1 + j ⎟
p1 ⎠ ⎝
p2 ⎠
⎝
(Perhaps we should verify that this transfer function is correct before proceeding.. Analysis of the
cascade circuit, e.g. using node equations, shows that the transfer function is indeed correct.)
Next, from the Bode plot we see that
13.3-18
Η (ω ) =
k
⎛
ω ⎞⎛
ω⎞
⎜⎜1 + j ⎟⎟ ⎜ 1 + j ⎟
p1 ⎠ ⎝
p2 ⎠
⎝
where
200 rad/s = p1 , 500 rad/s = p 2 and 34 dB = 50 = k
Pick C2 = C1 = 0.1 μ F . Then
1
200 =
−7
(10 ) R 2
⇒ R 2 = 50 kΩ and 500 =
Next
50 = k = k 1k2 =
R2 R4
R1 R 3
=
1
(10 ) R
⇒ R 4 = 20 kΩ
−7
4
( 20 kΩ )( 50 kΩ )
R1 R 3
The solution is not unique. For example, we can choose R1 = 4 kΩ and R 3 = 5 kΩ .
13.3-19
P 13.3-17
The cochlear implant is intended for patients with deafness due to malfunction of the
sensory cells of the cochlea in the inner ear (Loeb, 1985). These devices use a microphone for picking
up the sound and a processor for converting to electrical signals, and they transmit these signals to the
nervous system. A cochlear implant relies on the fact that many of the auditory nerve fibers remain
intact in patients with this form of hearing loss. The overall transmission from microphone to nerve cells
is represented by the gain function
H ( jω ) =
10( jω / 50 + 1)
( jω / 2 + 1)( jω / 20 + 1)( jω / 80 + 1)
Plot the Bode diagram for H(jω) for 1 ≤ ω ≤ 100.
Solution:
H (ω ) =
10(1+ jω 50)
(1+ jω 2)(1+ jω 20)(1+ jω 80)
ϕ = ∠H (ω ) = tan −1 (ω 50 ) − ( tan −1 (ω 2 ) + tan −1 (ω 20 ) + tan −1 (ω 80 ) )
13.3-20
R2
P 13.3-18
An operational amplifier circuit is shown in
Figure P 13.3-18, where R2 = 5 kΩ and C = 0.02 μF.
(a)
Find the expression for the network function H =
C
500 Ω
–
Vo/Vs and sketch the asymptotic Bode diagram.
+
(b)
What is the gain of the circuit, |Vo/Vs|, for ω = 0?
(c)
At what frequency does |Vo/Vs| fall to 1 2 of its low-
vs
+
–
+
vo
–
frequency value?
Figure P 13.3-18
Answer: (b) 20 dB and (c) 10,000 rad/s
Solution:
(a)
H (ω ) =
Vo (ω )
R2 R1
=−
Vs (ω )
1+ jω R2C
10
=−
1+ j
(b)
10 = 20 dB
(c)
10,000 rad/s
ω
10,000
13.3-21
1 μF
P 13.3-19
Determine the network
function H(ω) for this op amp circuit
and plot the Bode diagram.
1.25 MΩ
–
0.8 MΩ
–
+
+
+
vs
+
–
1 μF
vo
–
Solution:
⎫
⎪
⎪
⎪
⎬ ⇒ Vo (1 + jω C 1 R1 )(1 + jω C 2 R 2 ) = jω C 1 R1Vo + Vs
⎪
⎪
Va (ω ) − Vs (ω )
+ jω C 1 (Va (ω ) − Vo (ω )) ⎪
0=
R1
⎭
1
jω C 2
Vo (ω ) =
Va (ω )
1
R+
jω C 2
T(ω ) =
Vo (ω )
1
1
=
=
2
2
Vs (ω ) 1+ C 2 R 2 jω −ω C 1C 2 R1 R 2
−ω + 0.8 jω +1
This is a second order transfer function with ωo = 1 and δ = 0.4 .
13.3-22
P 13.3-20
The network function of a circuit is H (ω ) =
−3(5 + jω )
. Sketch the asymptotic magnitude
jω (2 + jω )
Bode plot corresponding to H.
Solution:
−3 ( 5 + j ω )
=
H (ω ) =
jω (2 + jω )
15 ⎛
ω⎞
⎜1 + j ⎟
2⎝
5⎠
ω⎞
⎛
j ω ⎜1 + j ⎟
2⎠
⎝
−
There is a zero at 5 rad/s and poles at 0 and 2 rad/s. To obtain the asymptotic magnitude Bode plot, use
⎧1
⎪
1 + j = ⎨ω
p ⎪
⎩p
ω
Then
for ω < p
for ω > p
⎧
⎪
⎪ 15 (1)
7.5
⎪ 2
⎪ ω (1) = ω for ω < 2
⎪
⎪ 15
⎪⎪ (1) 15
H = H =⎨ 2
= 2 for 2 < ω < 5
⎪ ω ⎛⎜ ω ⎞⎟ ω
⎪ ⎝2⎠
⎪
⎪ 15 ⎛⎜ ω ⎞⎟
⎪ 2 ⎝5⎠ 3
for ω > 5
=
⎪
⎛ω ⎞ ω
⎪ ω⎜ ⎟
⎪⎩
⎝2⎠
⎧ 20 log10 ( 7.5 ) − 20 log10 (ω ) for ω < 2
⎪
20 log10 H = ⎨20 log10 (15 ) − 2 ⎡⎣ 20 log10 (ω ) ⎤⎦ for 2 < ω < 5
⎪
20 log10 ( 3) − 20 log10 (ω ) for ω > 5
⎩
The slope of the asymptotic magnitude Bode plot is −20 db/decade for ω < 2 and ω > 5 rad/s and is −40
db/decade for 2 < ω < 5 rad/s. Also, at ω = 1 rad/s
⎧ 7.5
⎪ 1 = 7.5 at ω = 1 rad/s
⎧ 20 log10 ( 7.5 ) = 17.5 dB at ω = 1 rad/s
⎪
⎪ 7.5
⎪
= 3.75 at ω = 2 rad/s ⇒ 20 log10 H = ⎨ 20 log10 ( 3.75 ) = 11.5 dB at ω = 2 rad/s
H =⎨
⎪ 2
⎪20 log ( 0.6 ) = −4.44 dB at ω = 5 rad/s
10
⎩
⎪ 3
0.6
at
5
rad/s
=
=
ω
⎪ 5
⎩
The asymptotic magnitude Bode plot for H is
13.3-23
13.3-24
P 13.3-21
The network function of a circuit is H (ω ) =
( jω )3
. Sketch the asymptotic magnitude
(4 + j 2ω )
Bode plot corresponding to H.
Solution:
1
3
3
ω
j
(
)
ω
j
( ) =4
H (ω ) =
( 4 + j 2ω ) 1 + j ω
2
There is a pole at 2 rad/s and three zeros at 0 rad/s. To obtain the asymptotic magnitude Bode plot, use
⎧1
⎪
1 + j = ⎨ω
p ⎪
⎩p
ω
Then
⎧1 3
⎪ 4ω
1
= ω3
⎪
4
⎪⎪ (1)
H = H = ⎨1 3
⎪ 4ω
1
= ω2
⎪
⎪ ⎛⎜ ω ⎞⎟ 2
⎪⎩ ⎝ 2 ⎠
for ω < p
for ω > p
for ω < 2
for ω > 2
⎧⎪ 20 log10 ( 0.25 ) + 3 ⎡⎣ 20 log10 (ω ) ⎤⎦ for ω < 2
20 log10 H = ⎨
⎩⎪20 log10 ( 0.50 ) + 2 ⎡⎣ 20 log10 (ω ) ⎤⎦ for ω > 2
The slope of the asymptotic magnitude Bode plot is 60 db/decade for ω < 2 rad/s and is
40 db/decade for ω > 2 rad/s. Also,
20 log10 H = 20 log10 ( 0.25 ) + 3 ⎡⎣ 20 log10 (1) ⎤⎦ = −12 dB at ω = 1 rad/s
20 log10 H = 20 log10 ( 0.25 ) + 3 ⎡⎣ 20 log10 ( 2 ) ⎤⎦ = 6 dB at ω = 2 rad/s
The asymptotic magnitude Bode plot for H is
13.3-25
P 13.3-22
The network function of a circuit is H (ω ) =
2( j 2ω + 5)
. Sketch the asymptotic
(4 + j 3ω )( jω + 2)
magnitude Bode plot corresponding to H.
Solution:
5 ⎛⎜
ω ⎞⎟
1+ j
5 ⎟
4⎜
2 ( j 2ω + 5)
2⎠
⎝
H (ω ) =
=
⎞
( 4 + j 3ω )( j ω + 2 ) ⎛
⎜1 + j ω ⎟ ⎛⎜ 1 + j ω ⎞⎟
4 ⎟⎝
2⎠
⎜
3⎠
⎝
There is a zero at 2.5 rad/s and poles at 1.33 and 2 rad/s. To obtain the asymptotic magnitude Bode plot,
use
⎧ 1 for ω < p
ω ⎪
1 + j = ⎨ω
for ω > p
p ⎪
⎩p
Then
5
⎧
(1)
⎪
4
for ω < 4 rad/s
=5
⎪
4
3
1
1
(
)(
)
⎪
⎪ 5
5
(1)
⎪
⎪ 4
= 3 for 4 <ω < 2 rad/s
3
ω
⎪ ⎛ω ⎞
⎜
⎟
⎪
(1)
⎪ ⎜⎝ 4 3 ⎟⎠
⎪
⎪⎪ 5
10
H = H = ⎨ 4 (1)
3 for 2<ω < 5 rad/s
=
2
⎪⎛
2
ω
⎞
⎪⎜ ω ⎟ ⎛ ω ⎞
⎪ ⎜ 4 ⎟ ⎝⎜ 2 ⎠⎟
⎪⎝ 3 ⎠
⎪
⎪
5 ⎛⎜ ω ⎞⎟
⎪
4
4⎜ 5 ⎟
⎪
⎝ 2 ⎠ = 3 for ω > 2.5 rad/s
⎪ ⎛
ω
⎞
⎪ ⎜ ω ⎟⎛ ω ⎞
⎪ ⎜ 4 ⎟ ⎜⎝ 2 ⎟⎠
⎩⎪ ⎝ 3 ⎠
13.3-26
( )
⎧
20 log10 5
for ω < 4
4
3
⎪
⎪
5
4
⎪ 20 log10 3 − 20 log10 (ω ) for 3 < ω < 2
20 log10 H = ⎨
⎪20 log10 10 3 − 40 log10 (ω ) for 2 < ω < 5 2
⎪
⎪ 20 log 4 − 20 log (ω ) for ω > 5
10
10
⎪⎩
3
2
( )
( )
( )
The slope of the asymptotic magnitude Bode plot is −20 db/decade for 4/3 < ω < 2 rad/s and ω > 5/2
rad/s and is −40 db/decade for 2 < ω < 5/2 rad/s. Also,
( 4 ) = 1.9 dB
20 log10 H = 20 log10 5
( 3 ) − 20 log
20 log10 H = 20 log10 5
10
for ω ≤ 4
( 2 ) = −1.6 dB
( 3 ) − 40 log ( 5 2 ) = −5.4 dB
20 log10 H = 20 log10 10
10
3
rad/s
at ω = 2 rad/s
at ω = 5
2
rad/s
The asymptotic magnitude Bode plot for H is
13.3-27
P 13.3-23 The network function of a circuit is H (ω ) =
4(20 + jω )(20, 000 + jω )
Sketch the asymptotic
(200 + jω )(2000 + jω )
magnitude Bode plot corresponding to H.
Solution:
ω ⎞⎛
ω ⎞
⎛
4 ⎜1 + j ⎟ ⎜1 + j
4 ( 20 + j ω ) ( 20, 000 + j ω )
20 ⎠ ⎝
20, 000 ⎟⎠
⎝
H (ω ) =
=
ω ⎞⎛
ω ⎞
⎛
( 200 + j ω ) ( 2000 + j ω )
⎜1 + j
⎟ ⎜1 + j
⎟
200 ⎠ ⎝
2000 ⎠
⎝
There are zeros at 20 and 20,000 rad/s and poles at 200 and 2000 rad/s. To obtain the asymptotic
magnitude Bode plot, use
⎧ 1 for ω < p
ω ⎪
1 + j = ⎨ω
for ω > p
p ⎪
⎩p
Then
4 (1)(1)
⎧
=4
for ω < 20 rad/s
⎪
1
1
(
)(
)
⎪
⎪
⎛ω ⎞
4 ⎜ ⎟ (1)
⎪
ω
⎝ 20 ⎠
⎪
for 20 < ω < 200 rad/s
=
⎪
5
(1)(1)
⎪
⎪
⎛ω ⎞
4 ⎜ ⎟ (1)
⎪
⎝ 20 ⎠
for 200 < ω < 2000 rad/s
= 40
⎪
⎛ ω ⎞
⎪⎪
⎜
⎟ (1)
H = H =⎨
⎝ 200 ⎠
⎪
⎪ 4 ⎛ ω ⎞ (1)
⎜ ⎟
⎪
80000
⎝ 20 ⎠
for 2000 < ω < 20, 000 rad/s
=
⎪ ω
ω
ω
⎛
⎞
⎛
⎞
⎪⎜
⎟⎜
⎟
⎪ ⎝ 200 ⎠ ⎝ 2000 ⎠
⎪
⎪ 4 ⎛⎜ ω ⎞⎟ ⎛ ω ⎞
⎜
⎟
⎪
⎝ 20 ⎠ ⎝ 20, 000 ⎠
⎪ ⎛ ω ⎞ ⎛ ω ⎞ = 4 for ω > 2000 rad/s
⎪ ⎜
⎟
⎟⎜
⎩⎪ ⎝ 200 ⎠ ⎝ 2000 ⎠
⎧
20 log10 ( 4 ) for ω < 20
⎪
20 log10 (ω ) − 20 log10 ( 5 ) for 20 < ω < 200
⎪⎪
20 log10 H = ⎨
20 log10 ( 40 ) for 200 < ω < 2000
⎪20 log ( 80000 ) − 20 log (ω ) for 2000 < ω < 20, 000
10
10
⎪
20 log10 ( 4 ) for ω > 20, 000
⎪⎩
13.3-28
The slope of the asymptotic magnitude Bode plot is 20 db/decade for 20 < ω < 200 rad/s and is −20
db/decade for 2000 < ω < 20,000 rad/s and is 0 db/decade for ω < 20 and 200 < ω < 2000 rad/s, and ω >
20,000 rad/s. Also,
20 log10 H = 20 log10 ( 4 ) = 12 dB for ω ≤ 20 and ω ≥ 20, 000 rad/s
20 log10 H = 20 log10 ( 40 ) = 32 dB for 200 ≤ ω ≤ 2000 rad/s
The asymptotic magnitude Bode plot for H is
13.3-29
R3
–
+
+
–
vs
R1
+
+
0.5 μF
va
vb
–
–
R4
+
0.5 μF
vo
20 log10|H(ω)| (dB)
R2
−20
32
dB
decade
0
−40
dB
decade
–
320
8
ω (rad/s logarithmic scale)
(a)
(b)
Figure P 13.3-24
P 13.3-24
The input to the circuit shown in Figure P 13.3-24a is the voltage of the voltage source, vs.
The output of the circuit is the capacitor voltage, vo. The network function of the circuit is
H (ω ) =
Vo (ω )
Vs (ω )
Determine the values of the resistances R1, R2, R3, and R4 required to cause the network function of the
circuit to correspond to the asymptotic Bode plot shown in Figure P 13.3-24b.
Solution:
From Figure P13.3-24b, H (ω ) has poles at 8 and 320 rad/s and has a low frequency gain equal to 32 dB
= 40. Consequently, the network function corresponding to the Bode plot is
H (ω ) =
±40
ω ⎞⎛
ω ⎞
⎛
⎜1 + j ⎟ ⎜1 + j
⎟
8 ⎠⎝
320 ⎠
⎝
Next, we find the network function corresponding to the circuit. Represent the circuit in the frequency
domain.
Apply KCL at the top node of the left capacitor, C1, to get
13.3-30
V a − Vs
R1
+ j ω C 1 Va = 0 ⇒ Va =
1
Vs
1 + j ω C 1 R1
The op amp, together with resistors R2 and R3, comprise a noninverting amplifier so
⎛ R3 ⎞
Vb = ⎜1 +
V
⎜ R 2 ⎟⎟ a
⎝
⎠
(Alternately, this equation can be obtained by applying KCL at the inverting input node of the op amp.)
Apply KCL at the top node of the right capacitor, C2, to get
Vo − V b
R4
+ j ω C 2 Vo = 0 ⇒ Vo =
1
Vb
1+ j ω C 2 R4
Combining these equations gives
H (ω ) =
V o (ω )
V s (ω )
1+
=
R3
R2
(1 + j ω C R ) (1 + jω C
1
1
2
R4 )
Comparing to the specified network function gives
1+
R3
R2
(1 + j ω C R ) (1 + jω C
1
1
2
R4 )
=
±40
ω⎞⎛
ω ⎞
⎛
⎜1 + j ⎟ ⎜1 + j
⎟
8 ⎠⎝
320 ⎠
⎝
The solution is not unique. For example, we can require
1+
R3
R2
= 40 , C 1 R1 =
1
1
= 0.125 , C 2 R 4 =
= 0.00758
8
320
With the given values of capacitance, and choosing R2 = 10 kΩ, we have
R1 = 250 kΩ, R2 = 10 kΩ, R3 = 390 kΩ and R4 = 6.25 kΩ
(checked using LNAP 10/1/04)
13.3-31
P 13.3-25
R1
The input to the circuit shown in Figure
R2
R3
–
P 13.3-25a is the voltage of the voltage source, vs.
+
–
The output of the circuit is the voltage, vo. The
0.2 μF
vs
+
+
20 kΩ
network function of the circuit is
vo
–
V (ω )
H (ω ) = o
Vs (ω )
(a)
20 log10|H(ω)| (dB)
Determine the values of the resistances R1, R2, and
R3 required to cause the network function of the
circuit to correspond to the asymptotic Bode plot
shown in Figure P 13.3-25b.
−20
18
dB
decade
500
ω (rad/s logarithmic scale)
(b)
Figure P 13.3-25
Solution:
From Figure P13.3-25b, H (ω ) has a pole at 500 rad/s and a low frequency gain of 18 dB = 8.
Consequently, the network function corresponding to the Bode plot is
H (ω ) =
±8
ω ⎞
⎛
⎜1 + j
⎟
500 ⎠
⎝
Next, we find the network function corresponding to the circuit. Represent the circuit in the frequency
domain.
The node equations are
V a − Vs
R1
+
Va
Va
R2
+
= 0 ⇒ Va =
Vs
1
R2
R1 + R 2 + j ω C R1 R 2
jω C
13.3-32
Va
and
R2
+
Vo
R3
= 0 ⇒ Vo = −
R3
R2
Va
The network function is
H=
Vo
Vs
−
=
R3
R2
−
R2
R1 + R 2 + j ω C R1 R 2
=
R3
R1 + R 2
R R
1+ jω C 1 2
R1 + R 2
Comparing to the specified network function gives
−
R3
R1 + R 2
±8
=
R R
ω ⎞
⎛
1+ jω C 1 2
⎜1 + j
⎟
500 ⎠
R1 + R 2 ⎝
We require
R3
R1 + R 2
= 8 and C
R1 R 2
R1 + R 2
=
1
= 0.002
500
The solution is not unique. With the given values of capacitance, and choosing R1 = R2, we have
R1 = R2 = 20 kΩ and R3 = 320 kΩ
(checked using LNAP 10/1/04)
13.3-33
+
–
The input to the circuit shown in Figure P 13.3-26a is
P 13.3-26
R1
the voltage of the voltage source, vs. The output of the circuit is the
vs
voltage, vo. The network function of the circuit is
+
–
V (ω )
H (ω ) = o
Vs (ω )
(a)
1 μF
–
Determine the values of the resistances, R1 and R2, required
to cause the network function of the circuit to correspond to
Determine the values of the gains K1 and K2 in Figure P
13.3-26b.
(a)
20 log10|H( ω )| (dB)
(b)
vo
R2
the asymptotic Bode plot shown in Figure P 13.3-26b.
+
K1
−20
dB
decade
K2
500
20
ω (rad/s logarithmic scale)
(b)
Figure P 13.3-26
Solution:
From Figure P13.3-26b, H (ω ) has a pole at 20 rad/s and a zero at 500 rad/s. Consequently, the network
function corresponding to the Bode plot is
ω ⎞
⎛
⎜1 + j
⎟
500 ⎠
H (ω ) = ± K ⎝
.
ω⎞
⎛
⎜1 + j ⎟
20 ⎠
⎝
Next, we find the network function corresponding to the circuit.
Represent the circuit in the frequency domain. Apply KCL at the
inverting input node of the op amp to get
V o − Vs
or
(R
R1
1
+ j ω C 1 ( V o − Vs ) +
Vs
R2
=0
+ R 2 + j ω C 1 R1 R 2 ) Vs = ( R 2 + j ω C 1 R1 R 2 ) V o
so
H=
Vo
Vs
=
R1 + R 2 + j ω C 1 R1 R 2
R 2 + j ω C 1 R1 R 2
=
R1 + R 2
R2
1 + j ω C1
×
R1 R 2
R1 + R 2
1 + j ω C 1 R1
13.3-34
a. Comparing to the specified network function gives
R1 + R 2
R2
C 1 R1 =
We require
×
R1 R 2
R1 + R 2
1 + j ω C 1 R1
=K
1+ j
ω
500
1+ j
ω
20
R1 R 2
1
1
=
= 0.002
= .05 and C
R1 + R 2 500
20
K=
Notice that
1 + j ω C1
R1 + R 2
R2
1
C 1 R1
×
= 20 = 25
R1 R 2
1
C1
R1 + R 2 500
The solution is not unique. For example, choosing C = 1 μF
b. The network function is
R1 = 50 kΩ and R2 = 2.083 kΩ
ω ⎞
⎛
⎜1 + j
⎟
500 ⎠
H (ω ) = 25 ⎝
ω⎞
⎛
⎜1 + j ⎟
20 ⎠
⎝
so
20 ⎞
⎛
K 1 = 20 log10 ( 25 ) = 28 dB and K 2 = 20 log10 ⎜ 25 ×
⎟ = 0 dB
500 ⎠
⎝
(checked using LNAP 10/1/04)
13.3-35
R2
R1
P 13.3-27
+
The input to the circuit shown in Figure P
+
–
vs
13.3-27a is the voltage of the voltage source, vs. The
Ria
C
function of the circuit is
vo
–
ia
output of the circuit is the voltage, vo. The network
(a)
Vo (ω )
Vs (ω )
20 log10|H( ω )| (dB)
H (ω ) =
+
–
Determine the values of R, C, R1, and R2 required to cause
the network function of the circuit to correspond to the
asymptotic Bode plot shown in Figure P 13.3-27b.
−12
−20
dB
decade
250
ω (rad/s logarithmic scale)
(b)
Figure P 13.3-27
Solution:
From Figure P13.3-27b, H (ω ) has a pole at 250 rad/s and a low frequency gain equal to −12 dB = 0.25.
Consequently, the network function corresponding to the Bode plot is
H (ω ) =
±0.25
1+ j
ω
.
250
Next, we find the network function corresponding to the
circuit. Represent the circuit in the frequency domain.
Apply KVL to the left mesh to get
Vs = R1 I a + R I a
⇒ Ia =
Vs
R1 + R
Voltage division gives
R
R1 + R
1
Vo =
R
jω C
RIa =
Ia =
Vs
1
1+ j ω C R2
1+ j ω C R2
R2 +
jω C
The network function of the circuit is
H=
Vo
Vs
=
R
R1 + R
1 + j ω C R2
Comparing to the specified network function gives
13.3-36
R
R1 + R
1+ j ω C R2
=
±0.25
1+ j
ω
250
The solution is not unique. We require
R
1
1
= and C R 2 =
= 0.004
R1 + R 4
250
Choosing R = 100 Ω and C = 10 μF we have R1 = 300 Ω and R2 = 400 Ω
(checked using LNAP 10/2/04)
13.3-37
P 13.3-28
The input to the circuit shown in Figure P
Gva
13.3-28a is the current of the current source, is. The
io
+
output of the circuit is the current io. The network
va
is
function of the circuit is
R1
R2
C
–
I o (ω )
I s (ω )
(a)
20 log10|H( ω )| (dB)
H (ω ) =
Determine the values of G, C, R1, and R2 required to
cause the network function of the circuit to correspond
to the asymptotic Bode plot shown in Figure P 13.328b.
−20
−6
dB
decade
200
ω (rad/s logarithmic scale)
(b)
Figure P 13.3-28
Solution:
From Figure P13.3-28b, H (ω ) has a pole at 200 rad/s and a low frequency gain equal to −6 dB = 0. 5.
Consequently, the network function corresponding to the Bode plot is
H (ω ) =
±0.5
1+ j
ω
.
200
Next, we find the network function corresponding to the
circuit. Represent the circuit in the frequency domain.
Apply KCL at the top node of R1 to get
Is =
Va
R1
+ G Va
⇒ Va =
R1
1 + G R1
Is
Current division gives
1
Io =
⎛ R1
⎞
G
G
jω C
Va =
Is ⎟
G Va =
⎜⎜
1
1+ j ω C R2
1 + j ω C R 2 ⎝ 1 + G R1 ⎟⎠
+ R2
jω C
The network function of the circuit is
G R1
H=
Io
Is
=
1 + G R1
1+ j ω C R2
Comparing this network function to the specified network function gives
13.3-38
G R1
1 + G R1
= 0.5 and C R 2 =
1
200
The solution is not unique. Choosing G = 0.01 A/V and C = 10 μF gives R1 = 100 Ω and R2 = 500 Ω
(checked using LNAP 10/2/04)
13.3-39
P 13.3-29
A first-order circuit is shown in Figure P
R2
R1
13.3-29. Determine the ratio Vo/Vs and sketch the Bode
diagram when RC = 0.1 and R1/R2 = 3.
⎛
R ⎞
1
Answer: H = ⎜ 1 + 1 ⎟
⎝ R2 ⎠ 1 + jω RC
–
R
+
vs
+
–
C
+
vo
–
Figure P 13.3-29
Solution:
⎛ R ⎞
Vo (ω ) = ⎜ 1+ 1 ⎟ Vc (ω )
⎝ R2 ⎠
⎛ R ⎞⎛
⎞
1
= ⎜ 1+ 1 ⎟ ⎜
⎟ Vs (ω )
⎝ R2 ⎠ ⎝ 1 + jω C R ⎠
H (ω ) =
⎞
Vo (ω ) ⎛ R1 ⎞⎛
1
=⎜1+ ⎟⎜
⎟
Vs (ω ) ⎝ R2 ⎠⎝ 1+ jω C R ⎠
When R C = 0.1 and
then H (ω ) =
4
1+j
R1
= 3,
R2
ω
10
13.3-40
4Ω
P 13.3-30 (a) Draw the Bode diagram of the network function Vo/Vs
+
for the circuit of Figure P 13.3-30.
2Ω
v s +–
vo
30 mF
(b) Determine vo(t) when
–
Figure P 13.3-30
vs = 10 cos 20t V.
Answer: (b) vo = 4.18 cos (20t – 24.3°) V
Solution:
a)
Zo = R2 +
1
jω C
ω
Vo
Zo
ω1
jω C
=
=
=
ω
1
Vs
R1 + Z o
R1 + R2 +
1+j
ω2
jω C
R2 +
where ω1 =
and ω 2 =
1
R2 C
1
1+ j
= 16.7 rad/s
1
= 5.56 rad/s
( R1 + R2 )C
vs ( t ) = 10 cos 20 t or Vs = 10∠0°
(
(
)
)
1+ j 20
Vo
16.7
∴
=
Vs
1+ j 20
5.56
1+ j 1.20
=
= 0.418 ∠− 24.3°
1+ j 3.60
b) So
Vo = 4.18 ∠ − 24.3°
vo (t ) = 4.18cos(20t − 24.3°) V
13.3-41
P 13.3-31
Draw the asymptotic magnitude Bode diagram for
H (ω ) =
10(1 + jω )
jω (1 + j 0.5ω )(1 + j 0.6(ω / 50) + ( jω / 50) 2 )
Hint: At ω = 0.1 rad/s the value of the gain is 40 dB and the slope of the asymptotic Bode plot is –20
dB/decade. There is a zero at 1 rad/s, a pole at 2 rad/s, and a second-order pole at 50 rad/s. The slope of
the asymptotic magnitude Bode diagram increases by 20 dB/decade as the frequency increases past the
zero, decreases by 20 dB/decade as the frequency increases past the pole, and, finally, decreases by 40
dB/decade as the frequency increases past the second-order pole.
P13.3-31
13.3-42
Section 13-4: Resonant Circuits
P 13.4-1
For a parallel RLC circuit with R = 10 kΩ, L = 1/120 H, and C = 1/30 μF, find ω0, Q,
ω1, ω2, and the bandwidth BW.
Answer: ω0 = 60 krad/s, Q = 20, ω1 = 58.519 krad/s, ω2 = 61.519 krad/s, and BW = 3 krad/s
Solution:
For the parallel resonant RLC circuit with R = 10 kΩ, L = 1/120 H, and C = 1/30 µF we have
1
1
=
= 60 k rad sec
LC
⎛ 1 ⎞⎛ 1
−6 ⎞
x 10 ⎟
⎜
⎟⎜
⎝ 120 ⎠ ⎝ 30
⎠
ω0 =
1
× 10−6
C
30
Q= R
= 10, 000
= 20
1
L
120
ω0
2
2
⎛ω ⎞
⎛ω ⎞
ω
2
2
ω1 = −
+ ⎜ 0 ⎟ +ω 0 = 58.52 k rad s and ω 2 = 0 + ⎜ 0 ⎟ +ω 0 = 61.52 k rad s
2Q
2Q
⎝ 2Q ⎠
⎝ 2Q ⎠
1
1
=
= 3 krad s
BW =
RC
⎛ 1
−6 ⎞
(10000 )⎜ ×10 ⎟
⎝ 30
⎠
Notice that BW = ω 2 − ω1 =
ω0
Q
.
13.4-1
P 13.4-2
A parallel resonant RLC circuit is driven by a current source is = 20 cos ωt mA and
shows a maximum response of 8 V at ω = 1000 rad/s and 4 V at 897.6 rad/s. Find R, L, and C.
Answer: R = 400 Ω, L = 50 mH, and C = 20 μF
Solution: For the parallel resonant RLC circuit we have
H (ω ) =
k
⎛ω ω ⎞
1+ Q ⎜ − 0 ⎟
⎝ ω0 ω ⎠
2
2
so
R = k = H(ω 0 ) =
8
= 400 Ω and ω 0 = 1000 rad s
20⋅10−3
4
= 200, so
20.10−3
400
200 =
2
2 ⎛ 897.6 1000 ⎞
1+ Q ⎜
−
⎟
⎝ 1000 897.6 ⎠
At ω = 897.6 rad s , H (ω ) =
⇒ Q =8
Then
1
⎫
= ω 0 = 1000 ⎪
LC
⎪
C = 20 μ F
⎬ ⇒ L = 50 mH
C
=Q=8 ⎪
400
⎪⎭
L
13.4-2
P 13.4-3
A series resonant RLC circuit has L = 10 mH, C = 0.01 μF, and R = 100 Ω. Determine
ω0, Q, and BW.
Answer: ω0 = 105, Q = 10, and BW = 104
Solution:
For the series resonant RLC circuit with R = 100 Ω, L = 10 mH, and C = 0.01 µF we have
ω0 =
P 13.4-4
1
1 L
R
= 105 rad s , Q =
= 10, BW = = 104 rad s
R C
L
LC
A quartz crystal exhibits the property that when mechanical stress is applied across its
faces, a potential difference develops across opposite faces. When an alternating voltage is
applied, mechanical vibrations occur and electromechanical resonance is exhibited. A crystal can
be represented by a series RLC circuit. A specific crystal has a model with L = 1 mH, C = 10 μF,
and R = 1 Ω. Find ω0, Q, and the bandwidth.
Answer: ω0 = 104 rad/s, Q = 10, and BW = 103 rad/s
Solution:
For the series resonant RLC circuit with R = 1 Ω, L = 1 mH, and C = 10 µF we have
ω0 =
1
1 L
R
= 104 rad s , Q =
= 10, BW = = 103 rad s
R C
L
LC
13.4-3
P 13.4-5 Design a parallel resonant circuit to have ω0 = 2500 rad/s, Z(ω0) = 100 Ω, and BW =
500 rad/s.
Answer: R = 100 Ω, L = 8 mH, and C = 20 μF
Solution:
For the parallel resonant RLC circuit we have
R = Z (ω 0 ) = 100 Ω
1
= BW = 500 rad/s ⇒ C = 20 μ F
100 C
1
= ω0 = 2500 rad/s ⇒ L = 8 mH
−6
( 20⋅10 ) L
P 13.4-6
Design a series resonant circuit to have ω0 = 2500 rad/s, Y(ω0) = 1/100 Ω, and BW =
500 rad/s.
Answer: R = 100 Ω, L = 0.2 H, and C = 0.8 μF
P13.4-6
For the series resonant RLC circuit we have
R=
1
Y (ω 0 )
= 100 Ω
100
= BW = 500 rad/s ⇒ L = 0.2 H
L
1
= ω 0 = 2500 rad/s ⇒ C = 0.8 μ F
( 0.2 )C
13.4-4
a
P 13.4-7 The circuit shown in Figure P 13.4-7
10 μ H
represents a capacitor, coil, and resistor in parallel.
22 kΩ
600 pF
1.8 Ω
Coil
resistance
Calculate the resonant frequency, bandwidth, and Q
for the circuit.
b
Figure P 13.4-7
Solution:
C = 600 pF
L = 10 µH
R1 = 1.8 Ω
R 2 = 22 kΩ
1
1
+
R1 + jω L R 2
Y (ω ) = jω C +
(R +R
=
1
=
2
−ω 2C L R 2 ) + jω ( L +C R1 R 2 ) R1 − jω L
×
R1 − jω L
R 2 ( R1 + jω L )
R1 ( R1 + R 2 −ω 2C L R 2 ) +ω 2 L( L + C R1 R 2 )+ jω R1 ( L + C R1 R 2 ) − jω L( R1 + R 2 −ω 2C L R 2 )
R 2 ( R1 −ω 2 L2 )
ω = ω 0 is the frequency at which the imaginary part of Y (ω ) is zero :
R1 ( L +C R1 R 2 ) − L ( R1 + R 2 −ω02 C L R 2 ) = 0
⇒ ω0 =
L R 2 −C R12 R 2
C L2 R 2
= 12.9 M rad sec
13.4-5
P 13.4-8 Consider the simple model of an electric
L
power system as shown in Figure P 13.4-8. The
+
Power line
vs
inductance, L = 0.25 H, represents the power line
+
–
C
RL
vo
–
and transformer. The customer’s load is RL = 100
Power
plant
Ω, and the customer adds C = 25 μF to increase the
Customer
load
Figure P 13.4-8
magnitude of Vo.
The source is vs = 1000 cos 400t V, and it is desired that |Vo| also be 1000 V.
(a)
Find |V0| for RL = 100 Ω.
(b)
When the customer leaves for the night, he turns off much of his load, making RL = 1 kΩ,
at which point sparks and smoke begin to appear in the equipment still connected to the
power line. The customer calls you in as a consultant. Why did the sparks appear when
RL = 1 kΩ?
Solution: In the frequency domain we have:
(a) Using voltage division yields
(
Vo = 1000∠0°
)
(100 )( − j100 )
100
105
∠− 45°
∠− 45°
100 − j100
°
2
2
= 1000∠0
=
= 1000∠ − 90° V
100
(100 )( − j100 ) + j100
∠− 45°+ j100 50 2∠45°
2
100 − j100
∴|Vo | = 1000 V
(
)
(b) Do a source transformation to obtain
13.4-6
This is a resonant circuit with ω 0 = 1
LC = 400 rad/s. Since this also happens to be the
frequency of the input, so this circuit is being operated at resonance. At resonance the
admittances of the capacitor and inductor cancel each other, leaving the impedance of the
resistor. Increasing the resistance by a factor of 10 will increase the voltage Vo by a factor of 10.
This increased voltage will cause increased currents in both the inductance and the capacitance,
causing the sparks and smoke.
P 13.4-9
a
L
b
R1
Consider the circuit in Figure P 13.4-9. R1 = R2 = 1 Ω.
Select C and L to obtain a resonant frequency of ω0 = 100 rad/s.
R2
C
Figure P 13.4-9
Solution:
Let G 2 =
1
. Then
R2
Z = R1 + jω L +
(R G
=
1
2
1
G 2 + jω C
+ 1 − ω 2 L C ) + j (ω LG 2 + ω C R1 )
G 2 + jω C
At resonance, ∠Z = 0° so
tan −1
ω L G2 +ω C R1
ωC
= tan −1
2
G2
( R1G 2 +1−ω L C )
so
C − L G 22
ω L G 2 +ω C R1
ωC
2
=
⇒
ω
=
LC2
( R1 G 2 +1−ω 2 L C ) G 2
and C > G 22 L
C−L
. Then choose C and calculate L:
LC2
C = 10 mF ⇒ L = 5 mH
2
With R1 = R 2 = 1 Ω and ω 0 = 100 rad s , ω0 = 104 =
Since C > G 22 L , we are done.
13.4-7
R
P 13.4-10
For the circuit shown in Figure P 13.4-10,
(a) derive an expression for the magnitude response |Zin|
C
L
versus ω,
Zin
(b) sketch |Zin| versus ω, and
(c) find |Zin| at ω = 1/ LC .
Figure P 13.4-10
Solution:
(a)
R
R −ω 2 R L C ) + jω L
(
jω C
Z in = jω L +
=
1
1+ jω R C
R+
jω C
Consequently,
( R −ω R L C ) +(ω L )
2
| Zin | =
2
1+ (ω R C )
2
2
(b)
(c)
ω=
1
LC
⇒ | Zin | =
1
C
L
⎛ R2 C ⎞
⎜1 +
⎟
L ⎠
⎝
13.4-8
P 13.4-11 The circuit shown in Figure P
13.4-11 shows an experimental setup that
could be used to measure the parameters k, Q,
Oscilloscope
and ω0 of this series resonant circuit. These
parameters can be determined from a
magnitude frequency response plot for Y =
I/V. It is more convenient to measure node
voltages than currents, so the node voltages V
and V2 have been measured. Express |Y| as a
v
+
–
R
i
function of V and V2.
Hint: Let V = A and V2 = B∠θ
Then I =
+
v2
–
L
C
Figure P 13.4-11
( A − B cos θ ) − jB sin θ
R
( A − B cos θ ) 2 + ( B sin θ ) 2
Answer: | Y | =
AR
Solution: Let V (ω ) = A∠0 and V2 (ω ) = B∠θ . Then
I (ω )
=
Y (ω ) =
V (ω )
V (ω ) − V2 (ω )
A − B∠θ A − B cos θ − j B sin θ
R
=
=
AR
AR
V (ω )
( A − B cosθ ) + ( B sin θ )
2
| Y (ω ) | =
2
AR
13.4-9
Section 13-6: Plotting Bode Plots Using MATLAB
P 13.6-1
The input to the circuit shown in Figure P 13.6-1 is the voltage of the voltage source,
vs. The output of the circuit is the voltage, vo. Use MATLAB to plot the gain and phase shift of
this circuit as a function of frequency for frequencies in the range 1 < ω < 1000 rad/s.
10 Ω
20 Ω
+
–
vs
+
vo
–
1 mF
0.5 H
Figure P 13.6-1
Solution:
Using voltage division twice gives
1
jω L
jω C
V o (ω ) =
V s (ω ) −
V s (ω )
1
R
j
L
+
ω
1
R2 +
jω C
so
V o (ω )
1
jω L
H (ω ) =
=
−
Vs (ω ) 1 + j ω C R 2 R1 + j ω L
Modify the MATLAB script given in Section 13.7 of the text:
% P13_7_1.m - plot the gain and phase shift of a circuit
%--------------------------------------------------------------%
Create a list of logarithmically spaced frequencies.
%--------------------------------------------------------------wmin=1;
wmax=1000;
% starting frequency, rad/s
% ending frequency, rad/s
w = logspace(log10(wmin),log10(wmax));
%--------------------------------------------------------------%
Enter values of the parameters that describe the circuit.
%--------------------------------------------------------------R1 = 10;
% Ohms
R2 = 20;
% Ohms
C = 0.001; % Farads
13.6-1
L = 0.5;
% Henries
%--------------------------------------------------------------% Calculate the value of the network function at each frequency.
%
Calculate the magnitude and angle of the network function.
%--------------------------------------------------------------for k=1:length(w)
H(k) = 1/(1+j*R2*C*w(k)) - j*L*w(k)/(R1+j*L*w(k));
gain(k) = abs(H(k));
phase(k) = angle(H(k))*180/pi;
end
%--------------------------------------------------------------%
Plot the frequency response.
%--------------------------------------------------------------subplot(2,1,1), semilogx(w, gain)
xlabel('Frequency, rad/s'), ylabel('Gain, V/V')
title('Frequency Response Plots')
subplot(2,1,2), semilogx(w, phase)
xlabel('Frequency, rad/s'), ylabel('Phase, deg')
Here are the plots produced by MATLAB:
13.6-2
P 13.6-2 The input to the circuit shown in
Figure P 13.6-2 is the voltage of the voltage
vs
20 Ω
1 mF
0.5 H
10 Ω
+
–
source, vs. The output of the circuit is the
+
vo
–
voltage, vo. Use MATLAB to plot the gain and
Figure P 13.6-2
phase shift of this circuit as a function of
frequency for frequencies in the range 1 < ω < 1000 rad/s.
Solution:
Let
Z s = R2 +
1
jω C
and Z p =
R1Z s
R1 + Z s
Using voltage division twice gives
1
V a (ω ) =
Zp
jω L + Z p
so
H (ω ) =
Vs (ω ) and Vo (ω ) =
V o (ω )
V s (ω )
=
jω C
V a (ω )
1
R2 +
jω C
Zp
( j ω L + Z ) (1 + j ω C R )
p
2
Modify the MATLAB script given in Section 13.7 of the text:
% P13_7_2.m - plot the gain and phase shift of a circuit
pi = 3.14159;
%--------------------------------------------------------------%
Create a list of logarithmically spaced frequencies.
%--------------------------------------------------------------wmin=1;
wmax=1000;
% starting frequency, rad/s
% ending frequency, rad/s
w = logspace(log10(wmin),log10(wmax));
%--------------------------------------------------------------%
Enter values of the parameters that describe the circuit.
%--------------------------------------------------------------R1 = 10;
R2 = 20;
C = 0.001;
L = 0.5;
%
%
%
%
Ohms
Ohms
Farads
Henries
13.6-3
%--------------------------------------------------------------% Calculate the value of the network function at each frequency.
%
Calculate the magnitude and angle of the network function.
%--------------------------------------------------------------for k=1:length(w)
Zs(k) = R2+1/(j*w(k)*C);
Zp(k) = R1*Zs(k)/(R1+Zs(k));
H(k) = Zp(k)/((j*w(k)*L+Zp(k))*(1+j*w(k)*C*R2));
gain(k) = abs(H(k));
phase(k) = angle(H(k))*180/pi;
end
%--------------------------------------------------------------%
Plot the frequency response.
%--------------------------------------------------------------subplot(2,1,1), semilogx(w, gain)
xlabel('Frequency, rad/s'), ylabel('Gain, V/V')
title('Frequency Response Plots')
subplot(2,1,2), semilogx(w, phase)
xlabel('Frequency, rad/s'), ylabel('Phase, deg')
Here are the plots produced by MATLAB:
13.6-4
P 13.6-3
40 Ω
The input to the circuit shown in Figure
P 13.6-3 is the voltage of the voltage source, vs. The
output of the circuit is the voltage, vo. Use
20 Ω
25 mF
MATLAB to plot the gain and phase shift of this
+
–
+
25 Ω
vo
vs
circuit as a function of frequency for frequencies in
0.2 H
–
the range 1 < ω < 1000 rad/s.
Figure P 13.6-3
Solution:
Let
Z1 = R2 +
Using voltage division gives
V a (ω ) =
Z2
Z1 + Z 2
V s (ω ) ⇒ H ( ω ) =
R1
j ω C R1
V o (ω )
V s (ω )
=
and Z 2 = R 3 + j ω L
Z2
Z1 + Z 2
Modify the MATLAB script given in Section 13.7 of the text:
% P13_7_3.m - plot the gain and phase shift of a circuit
pi = 3.14159;
%--------------------------------------------------------------%
Create a list of logarithmically spaced frequencies.
%--------------------------------------------------------------wmin=1;
wmax=1000;
% starting frequency, rad/s
% ending frequency, rad/s
w = logspace(log10(wmin),log10(wmax));
%--------------------------------------------------------------%
Enter values of the parameters that describe the circuit.
%--------------------------------------------------------------R1 = 40;
R2 = 20;
R3 = 25;
C = 0.025;
L = 0.2;
%
%
%
%
%
Ohms
Ohms
Ohms
Farads
Henries
%---------------------------------------------------------------
13.6-5
% Calculate the value of the network function at each frequency.
%
Calculate the magnitude and angle of the network function.
%--------------------------------------------------------------for k=1:length(w)
Z1(k) = R2+R1/(j*w(k)*C*R1);
Z2(k) = R3+j*w(k)*L;
H(k) = Z2(k)/(Z1(k)+Z2(k));
gain(k) = abs(H(k));
phase(k) = angle(H(k))*180/pi;
end
%--------------------------------------------------------------%
Plot the frequency response.
%--------------------------------------------------------------subplot(2,1,1), semilogx(w, gain)
xlabel('Frequency, rad/s'), ylabel('Gain, V/V')
title('Frequency Response Plots')
subplot(2,1,2), semilogx(w, phase)
xlabel('Frequency, rad/s'), ylabel('Phase, deg')
Here are the plots produced by MATLAB:
13.6-6
Section 13.8 How Can We Check…?
P 13.8-1 Circuit analysis contained in a lab report indicates that the network function of a
circuit is
H (ω ) =
1+ j
ω
630
ω ⎞
⎛
10 ⎜1 + j
⎟
6300 ⎠
⎝
This lab report contains the following frequency response data from measurements made on the
circuit. Do these data seem reasonable?
ω, rad/s
200
400
795
1585
3162
|H(ω)|
0.105
0.12
0.16
0.26
0.460
ω, rad/s
6310
12,600
25,100
50,000
100,000
|H(ω)|
0.71
1.0
1.0
1.0
1.0
P13.8-1
When ω < 630 rad/s, H(ω) ≅ 0.1, which agrees with the tabulated values of | H(ω)|
corresponding to ω = 200 and 400 rad/s.
When ω > 6300 rad/s, H(ω) ≅ 1.0, which agrees with the tabulated values of | H(ω)|
corresponding to ω = 12600, 25000, 50000 and 100000 rad/s.
At ω = 6300 rad/s, we expect | H(ω)| = −3 dB = 0.707. This agrees with the tabulated value of |
H(ω)| corresponding to ω = 6310 rad/s.
At ω = 630 rad/s, we expect | H(ω)| = −20 dB = 0.14. This agrees with the tabulated values of |
H(ω)| corresponding to ω = 400 and 795 rad/s.
This data does seem reasonable.
13.8-1
P 13.8-2
A parallel resonant circuit (see Figure 13.4-2) has Q = 70 and a resonant frequency ω0
= 10,000 rad/s. A report states that the bandwidth of this circuit is 71.43 rad/s. Verify this result.
Solution:
BW =
P 13.8-3
ω0
Q
=
10,000
= 143 ≠ 71.4 rad s . Consequently, this report is not correct.
70
A series resonant circuit (see Figure 13.4-4) has L = 1 mH, C = 10 μF, and R = 0.5 Ω.
A software program report states that the resonant frequency is f0 = 1.59 kHz and the bandwidth
is BW = 79.6 Hz. Are these results correct?
Solution:
1
1 L
R
= 10 k rad s = 1.59 kHz, Q =
= 20 and BW = = 500 rad s = 79.6 Hz
R C
L
LC
The reported results are correct.
ω0 =
An old lab report contains the approximate
Bode plot shown in Figure P 13.8-4 and concludes that the
network function is
ω ⎞
⎛
40 ⎜ 1 + j
⎟
200 ⎠
H (ω ) = ⎝
ω ⎞
⎛
⎜1 + j
⎟
800 ⎠
⎝
Do you agree?
32
|H ( ω )|, dB
P 13.8-4
20
200
800
ω , rad/sec
log scale
Figure P 13.8-4
Solution:
The network function indicates a zero at 200 rad/s and a pole at 800 rad/s. In contrast, the Bode
plot indicates a pole at 200 rad/s and a zero at 800 rad/s. Consequently, the Bode plot and
network function don’t correspond to each other.
13.8-2
PSpice Problems
4 kΩ
SP 13-1 The input to the circuit shown in
Figure SP 13-1 is the voltage of the voltage
+
source, vi(t). The output is the voltage, vo(t),
+
vo(t)
vi(t) –
5 μF
1 kΩ
across the parallel connection of the capacitor and
1-kΩ resistor. The network function that
–
represents this circuit is
Figure SP 13-1
V (ω )
k
H (ω ) = o
=
Vi (ω ) 1 + j ω
p
Use PSpice to plot the frequency response of this circuit. Determine the values of the pole, p, and
of the dc gain, k.
Solution:
Here are the magnitude and phase frequency response plots:
From the magnitude plot, the low frequency gain is k = 200m = 0.2.
From the phase plot, the angle is -45° at p = 2π ( 39.891) = 251 rad/s .
13SP-1
SP 13-2 The input to the circuit shown in Figure SP 13-2 is the
voltage of the voltage source, vi(t). The output is the voltage, vo(t),
across the series connection of the inductor and 60-Ω resistor.
The network function that represents this circuit is
ω
1+ j
V (ω )
z
H (ω ) = o
=k =
ω
Vi (ω )
1+ j
p
Use PSpice to plot the frequency response of this circuit.
Determine the values of the pole, p, of the zero, z, and of the dc
gain, k.
Answer: p = 20 rad/s, z = 12 rad/s, and k = 0.6 V/V
40 Ω
+
60 Ω
vi(t)
+
–
vo(t)
5H
–
Figure SP 13-2
Solution: Here is the magnitude frequency response plot:
The low frequency gain is 0.6 = lim H (ω ) = k ⇒ k = 0.6 .
ω →0
The high frequency gain is 1 = lim H (ω ) = k
ω →∞
At ω = 2π ( 2.8157 ) = 17.69 rad/s ,
p
z
⇒ z = ( 0.6 ) p
2
⎛ 17.69 ⎞
1+ ⎜
0.6 p ⎟⎠
⎝
0.8 = 0.6
2
⎛ 17.69 ⎞
1+ ⎜
⎟
⎝ p ⎠
⇒
16 p 2 + 869
16 2
= 2
⇒
p + 313 = p 2 + 869 ⇒
9
9
p + 313
⇒
p = 20 rad/s and z = 12 rad/s
(
)
( 0.77778) p 2 = 312.56
13SP-1
SP 13-3 The input to the circuit shown in Figure SP 13-3 is the voltage of the voltage source,
vi(t). The output is the voltage, vo(t), across 30-kΩ resistor. The network function that represents
H (ω ) =
this circuit is
Vo (ω )
k
=
Vi (ω ) 1 + j ω
p
Use PSpice to plot the frequency response of this circuit. Determine the values of the pole, p, and
of the dc gain, k.
Answer: p = 100 rad/s and k = 4 V/V
2 kΩ
15 kΩ
+
+
vi(t)
+
–
vC(t)
5 μF 6 vC(t)
+
–
30 kΩ
vo(t)
–
–
Figure SP 13-3
Solution:
From the magnitude plot, the low frequency gain is k = 4.0. Also, the gain is 4/sqrt(2) = 2.828 at
15.914 hertz.
From the phase plot, the angle is -45° at p = 2π (15.998 ) = 100.5 rad/s .
13SP-1
SP 13-4 The input to the circuit shown in
Figure SP 13-4 is the voltage of the voltage
source, vi(t). The output is the voltage, vo(t),
across 20-kΩ resistor. The network function that
represents this circuit is
H (ω ) =
Vo (ω )
k
=
Vi (ω ) 1 + j ω
p
50 kΩ
10 kΩ
2 μF
vi(t)
+
–
–
+
20 kΩ
Use PSpice to plot the frequency response of this
circuit. Determine the values of the pole, p, and
of the dc gain, k.
Answer: p = 10 rad/s and k = 5 V/V
+
vo(t)
–
Figure SP 13-4
Solution:
Here’s the circuit drawn in the PSpice workspace:
Here are the frequency response plots:
13SP-1
From the magnitude plot, the low frequency gain is k = 5.0.
From the phase plot, the angle is 180°-45°=135° at p = 2π (1.5849 ) = 9.958 rad/s .
13SP-1
SP 13-5 Figure SP 13-5 shows a circuit and a frequency response. The frequency response
plots were made using PSpice and Probe. V(R3:2) and Vp(R3:2) denote the magnitude and angle
of the phasor corresponding to vo(t). V(V1:+) and Vp(V1:+) denote the magnitude and angle of
the phasor corresponding to vi(t). Hence V(R3:2)/V(V1:+) is the gain of the circuit and Vp(R3:2)
– Vp(V1:+) is the phase shift of the circuit.
Determine values for R and C required to cause the circuit correspond to the frequency
response.
Hint: PSpice and Probe use m for milli or 10–3. Hence, the label (159.513, 892.827m) indicates
that the gain of the circuit is 892.827*10–3 = 0.892827 at a frequency of 159.513 Hz ≈
1000rad/sec.
Answer: R = 5kΩ and C = 0.2 μF
2.0
(159.513, 892.827 m)
(31.878, 1.8565)
(318.784, 484.412 m)
1.0
V(R3:2)/V(V1:+)
0
10 kΩ
R
175 d
C
(159.513, 116.515)
150 d
(31.878, 158.169)
vi(t)
+
–
(318.784, 104.017)
–
125 d
+
+
20 kΩ
vo(t)
–
100 d
Vp(R3:2)– Vp(V1:+)
10 Hz
30 Hz
100 Hz
300 Hz
1.0 KHz
Frequency
(a)
FIGURE
(b)
SP 13-5 (a) A circuit and (b) the corresponding frequency response.
Solution: From the circuit
104
104
R
R
H (ω ) = −
=
∠ − tan −1 (ω C 104 )
1 + jω C 104
4 2
1 + (ω C 10 )
From the plot, at ω = 200 rad/sec = 31.83 Hertz H(ω) is
1.8565∠158° =
104
R
1 + (ω C 10
)
4 2
∠ − tan −1 (ω C 104 )
13SP-1
Equating phase shifts gives
C R 104
ω C 10 = 10
= tan(22°) = 0.404 ⇒ C = 0.2 μ F
R + 104
4
3
Equating gains gives
1.8565 =
104
R
1 + (ω C 104 )
2
=
104
R
1 + ( 0.404 )
2
⇒ R = 5 kΩ
13SP-1
SP 13-6 Figure SP 13-6 shows a circuit and a frequency response. The frequency response
plots were made using PSpice and Probe. V(R2:2) and Vp(R2:2) denote the magnitude and angle
of the phasor corresponding to vo(t). V(V1:+) and Vp(V1:+) denote the magnitude and angle of
the phasor corresponding to vi(t). Hence V(R2:2)/V(V1:+) is the gain of the circuit, and
Vp(R2:2) – Vp(V1:+) is the phase shift of the circuit.
Determine values for R and C required to make the circuit correspond to the frequency
response.
Hint: PSpice and Probe use m for milli or 10–3. Hence, the label (159.268, 171.408m) indicates
that the gain of the circuit is 171.408*10–3 = 0.171408 at a frequency of 159.268 Hz ≈ 1000
rad/sec.
Answer: R = 20 kΩ and C = 0.25 μF
400 m
(79.239, 256.524 m)
(159.268, 171.406 m)
200 m
(316.228, 96.361 m)
V(R2:2)/V(V1:+)
0
0d
(79.239, –39.685)
(159.268, –59.055)
(316.228, –73.197)
–50 d
R
+
vi(t)
+
–
10 kΩ
C
Vp(R2:2)– Vp(V1:+)
vo(t) –100 d
10 Hz
–
100 Hz
1.0 KHz
10 KHz
Frequency
(a)
FIGURE
(b)
SP 13-6 (a) A circuit and (b) the corresponding frequency response.
Solution: From the circuit
104
1 + jω C R 2
104
104
C R 104 ⎞
R + 104
R + 104
−1 ⎛
H (ω ) =
=
=
∠
−
tan
ω
⎜
4 ⎟
104
C R 104
4 2
⎝ R + 10 ⎠
⎛
⎞
10
C
R
+
R
1
+
j
ω
1 + ⎜ω
1 + jω C 104
R + 104
4 ⎟
⎝ R + 10 ⎠
From the plot, at ω = 1000 rad/sec = 159.1 Hertz H(ω) is
13SP-1
104
R + 104
⎛ C R 104 ⎞
∠ − tan ⎜ ω
0.171408∠ − 59° =
4 ⎟
2
⎝ R + 10 ⎠
⎛ C R 104 ⎞
1+ ⎜ω
4 ⎟
⎝ R + 10 ⎠
−1
Equating phase shifts gives
ω
4
C R 104
3 C R 10
10
=
= tan(59°) = 1.665
R + 104
R + 104
Equating gains gives
104
R + 104
0.171408 =
⎛ C R 10 ⎞
1+ ⎜ω
4 ⎟
⎝ R + 10 ⎠
4
2
=
104
R + 104
1 + (1.665 )
2
⇒ R = 20 kΩ
Substitute this value of R into the equation for phase shift to get:
C ( 20 ×103 ) 104
C R 104
3
1.665 = 10
= 10
R + 104
( 20 ×103 ) + 104
3
⇒ C = 0.25 μ F
13SP-1
Design Problems
DP 13-1 Design a circuit that has a low-frequency gain of 2, a high-frequency gain of 5, and
makes the transition of H = 2 to H = 5 between the frequencies of 1 kHz and 10 kHz.
Solution:
Pick the appropriate circuit from Table 13.4-2.
We require
2π × 1000 < z =
R2
1
1
p C
, 2π × 10000 > p =
, 2=k =
and 5 = k = 1
C 1 R1
C2 R 2
z C2
R1
Try z = 2π × 2000. Pick C1 = 0.05 μ F. Then
R1 =
Check: p =
1
C
C
= 1.592 kΩ, R 2 = 2 R1 = 3.183 kΩ and C2 = 1 = 1 = 0.01 μ F
p
C1 z
2
k
z
1
= 31.42 k rad s < 2π ⋅10, 000 rad s.
C 2 R2
13DP-1
DP 13-2 Determine L and C for the circuit of Figure DP 13-2 in order to obtain a low-pass
filter with a gain of –3 dB at 100 kHz.
L
+
vs
+
–
C
1 kΩ
vo
–
Figure DP 13-2
Solution:
1
1
V (ω )
jω C
1+ jω C R
LC
H (ω ) = o
=
=
=
1
1
R
Vs (ω )
⎛ 1
⎞
−ω 2 + jω
+
jω L + ⎜
|| R ⎟ jω L +
1+ jω C R
RC LC
⎝ jω C ⎠
Pick
|| R
R
1
= ω 0 = 2π (100 ⋅103 ) rad s . When ω = ω 0
LC
1
LC
H 0 (ω ) =
1
1
1
1
−
+j
+
LC
LC RC LC
So H (ω 0 ) = R
C
. We require
L
−3 dB = 0.707 = H (ω 0 ) = R
C
C
= 1000
L
L
Finally
1
⎫
= 2π (100⋅103 ) ⎪
LC
C =1.13 nF
⎪
⎬⇒
C
⎪ L = 2.26 mH
0.707 =1000
⎪⎭
L
13DP-2
0.47 μ F
DP 13-3 British Rail has
constructed an instrumented
8.06 kΩ
railcar that can be pulled over
2.37 MΩ
its tracks at speeds up to 180
10 kΩ
1 MΩ
km/hr and will measure the
–
Circuit
–
B
track-grade geometry. Using
+ v
+
s
–
Circuit
A
+
+
such a railcar, British Rail can
vo
0.1 μ F
monitor and track gradual
–
degradation of the rail grade,
866 kΩ
499 kΩ
especially the banking of
–
curves, and permit preventive
+
maintenance to be scheduled as
Circuit C
needed well in advance of
track-grade failure.
Figure DP 13-3
The instrumented railcar has
numerous sensors, such as
angular-rate sensors (devices that output a signal proportional to rate of rotation) and
accelerometers (devices that output a signal proportional to acceleration), whose signals are
filtered and combined in a fashion to create a composite sensor called a compensated
accelerometer (Lewis, 1988).
A component of this composite sensor signal is obtained by integrating and high-pass
filtering an accelerometer signal. A first-order low-pass filter will approximate an integrator at
frequencies well above the break frequency. This can be seen by computing the phase shift of the
filter-transfer function at various frequencies. At sufficiently high frequencies, the phase shift
will approach 90°, the phase characteristic of an integrator.
A circuit has been proposed to filter the accelerometer signal, as shown in Figure DP 133. The circuit is comprised of three sections, labeled A, B, and C. For each section, find an
expression for and name the function performed by that section. Then find an expression for the
gain function of the entire circuit, Vo/Vs. For the component values, evaluate the magnitude and
phase of the circuit response at 0.01, 0.02, 0.05, 0.1, 0.2, 0.5, 1.0, 2.0, 5.0, and 10.0 Hz. Draw a
Bode diagram. At what frequency is the phase response approximately equal to 0°? What is the
significance of this frequency?
13DP-3
Solution:
R1 = 10 kΩ
R 2 = 866 kΩ
R 3 = 8.06 kΩ
R 4 = 1 MΩ
R 5 = 2.37 MΩ
R 6 = 499 kΩ
C 1 = 0.47 μ F
C 2 = 0.1 μ F
Va = −
Circuit A
R3
R2
Vc −
R3
R1
Vs = −H 1 Vc − H 2 Vs
R5
Circuit B
Vo = −
Circuit C
Vc = −
R4
1 + jω C 1 R 5
Va = − H 3 Va
1
Vo = −H 4 Vo
jω C 2 R 6
Then
Vc = H 3 H 4 Va
Va = −H 2 Vs − H1 H 3 H 4 Va
Vo = −H 3 Va =
⇒ Va =
−H 2
Vs
1 + H1 H 3 H 4
H 2 H3
Vs
1 + H1 H 3 H 4
After some algebra
jω
Vo =
R3
R1 R 4 C 1
R3
R 2 R 4 R 6 C1 C 2
−ω + j
2
ω
Vs
R5 C1
This MATLAB program plots the Bode plot:
R1=10;
R2=866;
R3=8.060;
R4=1000;
% units: kOhms and mF so RC has units of sec
13DP-4
R5=2370;
R6=449;
C1=0.00047;
C2=0.0001;
pi=3.14159;
fmin=5*10^5;
fmax=2*10^6;
f=logspace(log10(fmin),log10(fmax),200);
w=2*pi*f;
b1=R3/R1/R4/C1;
a0=R3/R2/R4/R6/C1/C2;
a1=R5/C1;
for k=1:length(w)
H(k)=(j*w(k)*b1)/(a0-w(k)*w(k)+j+w(k)*a1);
gain(k)=abs(H(k));
phase(k)=angle(H(k));
end
subplot(2,1,1), semilogx(f, 20*log10(gain))
xlabel('Frequency, Hz'), ylabel('Gain, dB')
title('Bode Plot')
subplot(2,1,2), semilogx(f, phase*180/pi)
xlabel('Frequency, Hz'), ylabel('Phase, deg')
13DP-5
DP 13-4 Design a circuit that has the network function
H (ω ) =
jω
ω ⎞⎛
ω ⎞
⎛
⎜1 + j
⎟ ⎜1 + j
⎟
200 ⎠ ⎝
500 ⎠
⎝
Hint: Use two circuits from Table 13.4-1. Connect the circuits in cascade. That means that the
output of one circuit is used as the input to the next circuit. H(ω) will be the product of the
network functions of the two circuits from Table 13.3-2.
Solution:
Pick the appropriate circuits from Table 13.4-2.
We require
10 = − k1k2 = R 2C1
R4
R3
, 200 = p1 =
1
1
and 500 = p2 =
R1 C 1
C 2 R4
Pick C 1 = 1 μ F. Then R1 =
1
1
= 5 kΩ. Pick C 2 = 0.1 μF. Then R 4 =
= 20 kΩ.
p1 C1
p2C2
Next
10 =
R2
R3
(10−6 )(20 ⋅103 ) ⇒
R2
R3
= 500
Let R 2 = 500 kΩ and R 3 = 1 kΩ.
13DP-6
DP 13-5 Strain-sensing instruments can be used to measure orientation and magnitude of
strains running in more than one direction. The search for a way to predict earthquakes focuses
on identifying precursors, or changes, in the ground that reliably warn of an impending event.
Because so few earthquakes have occurred precisely at instrumented locations, it has been a slow
and frustrating quest. Laboratory studies show that before rock actually ruptures—precipitating
an earthquake—its rate of internal strain increases. The material starts to fail before it actually
breaks. This prelude to outright fracture is called “tertiary creep” (Brown, 1989).
The frequency of strain signals varies from 0.1 to 100 rad/s. A circuit called a band-pass
filter is used to pass these frequencies. The network function of the band-pass filter is
Kjω
H (ω ) =
⎛
ω ⎞⎛
ω⎞
⎜ 1 + j ⎟⎜ 1 + j ⎟
ω1 ⎠⎝
ω2 ⎠
⎝
Specify ω1, ω2, and K so that the following are the case:
1.
The gain is at least 17 dB over the range 0.1 to 100 rad/s.
2.
The gain is less than 17 dB outside the range 0.1 to 100 rad/s.
3.
The maximum gain is 20 dB.
Solution:
Pick the appropriate circuits from Table 13.4-2.
13DP-7
We require
20 dB = 10 = − k1k2 = R 2C1
Pick C 1 = 20 μ F. Then R1 =
R4
R3
, 0.1 = p1 =
1
1
and 100 = p2 =
R1 C 1
C 2 R4
1
1
= 500 kΩ. Pick C 2 = 1 μF. Then R 4 =
= 10 kΩ.
p1 C 1
p2C2
Next
10 =
R2
R3
(20 ⋅10−6 )(10 ⋅103 ) ⇒
R2
R3
= 50
Let R 2 = 200 kΩ and R 3 = 4 kΩ.
13DP-8
DP 13-6 Is it possible to design the circuit shown in Figure DP 13-6 to have a phase shift of –
45°and a gain of 2 V/V both at a frequency of 1000 radians/second using a 0.1 microfarad
capacitor and resistors from the range 1 k ohm to 200 k ohm?
R3
R2
–
R1
+
A cos (ω t + θ) +–
+
100 kΩ
C
vo(t)
–
Figure DP 13-6
Solution:
1+
The network function of this circuit is H (ω ) =
R2
R3
1+ jω R1C
The phase shift of this network function is θ = − tan −1 ω R1C
1+
The gain of this network function is G = H (ω ) =
R3
R2
1+ (ω R1C ) 2
1+
=
R3
R2
1+ ( tan θ )
2
Design of this circuit proceeds as follows. Since the frequency and capacitance are known, R1 is
tan(−θ )
. Next pick R2 = 10kΩ (a convenient value) and calculated R3 using
calculated from R1 =
ωC
R 3 = (G ⋅ 1+ (tan θ ) 2 − 1) ⋅ R 2 . Finally
θ = −45 deg, G = 2, ω = 1000 rad s ⇒ R1 = 10 kΩ, R 2 = 10 kΩ, R 3 = 18.284 kΩ, C = 0.1 μF
13DP-9
DP 13-7 Design the circuit shown in Figure DP 13-7a to have the asymptotic Bode plot shown
in Figure DP 13-7b.
R2
0.5 μ F
C
vin(t)
–
32
| H(ω ) |, dB
+
–
R1
+
100 kΩ
vout(t)
+
–
20
200
800
ω , rad/sec
log scale
(a)
(b)
Figure DP 13-7
Solution: From Table 13.4-2 and the Bode plot:
800 = z =
1
⇒ R1 = 2.5 kΩ
R1 (0.5×10−6 )
32 dB = 40 =
200 = p =
R2
R1
⇒ R 2 = 100 kΩ
1
1
⇒ C =
= 0.05μ F
R2C
(200)(100×103 )
(Check: 20 dB = 10 = k
p 0.5×10−6
0.5×10−6
=
=
)
z
C
0.05×10−6
13DP-10
DP 13-8 For the circuit of Figure DP 13.-8, select R1 and R2 so that the gain at high frequencies
is 10 V/V and the phase shift is 195° at ω = 1000 rad/s. Determine the gain at ω = 10 rad/s.
R1
R2
0.1 μ F
vs
+
–
–
+
+
vo
–
Figure DP 13.-8
Solution:
−R2
jω C R 2
= −
1
1+ jω C R1
1+
jω C
tan(270°−195°)
195° = 180 + 90 − tan −1 ω C R1 ⇒ R1 =
= 37.3 kΩ
(1000)(0.1×10−6 )
R
10 = lim H (ω ) = 2 ⇒ R 2 = 10 R1 = 373 kΩ
ω →∞
R1
H (ω ) =
13DP-11
Chapter 14: The Laplace Transform
Exercises
Exercise 14.7-1 Determine the voltage vC(t) and the current iC(t) for t ≥ 0 for the circuit of
Figure 14.7-1.
Hint: vC(0) = 4 V
Answer:
vC(t) = (6–2e–0.67t) u(t)
V and
iC(t) =
2 –0.67t
e
u(t) A
3
Figure 14.7-1
Solution:
KCL at top node:
VC ( s )
3
s
2
+ VC ( s ) = + 2
2
s
VC ( s ) =
(
6
2
−
s s+ 2
3
v C (t ) = 6 − 2 e
2
VC ( s )
I C (s) =
−2= 3
2
2
s+
3
s
⇒ iC (t ) =
− ( 2 / 3) t
) u(t ) V
2 − ( 2 / 3) t
e
u (t ) A
3
14-1
Exercise 14.8-1 The transfer function of a circuit is H(s) =
–5s
. Determine the
s + 15s + 50
2
impulse response and step response of this circuit.
Answer:
5
10 ⎤
(a) impulse response = ℒ –1 ⎡
= (5e–5t–10e–10t)u(t)
⎢⎣ s + 5 – s + 10 ⎥⎦
1 ⎤
⎡ 1
(b) step response = ℒ –1 ⎢⎣ s + 10 – s + 5 ⎥⎦ = (e–10t – e–5t)u(t)
Solution:
10 ⎤
⎡ 5
(a) impulse response = L−1 ⎢
−
= ( 5 e − 5 t − 10 e −10 t ) u (t )
⎥
5
10
s
s
+
+
⎣
⎦
1 ⎤
⎡ 1
(b) step response = L−1 ⎢
−
= ( e−10 t − e − 5 t ) u (t )
⎥
⎣ s + 10 s + 5 ⎦
Exercise 14.8-2 The impulse response of a circuit is h(t) = 5e–2t sin (4t)u(t). Determine the step
response of this circuit.
Hint:
H(s) = ℒ [5e–2t sin (4t)u(t)] =
Answer:
5(4)
20
= 2
2
2
( s + 2) + 4
s + 4s + 20
s+4 ⎤
⎡1
⎡ H (s) ⎤
step response = ℒ –1 ⎢⎣ s ⎥⎦ = ℒ –1 ⎢⎣ s – s 2 + 4s + 20 ⎥⎦
1
⎛
⎞⎞
–2 t ⎛
= ⎜1 – e ⎜⎝ cos 4t + 2 sin 4t ⎟⎠ ⎟ u (t )
⎝
⎠
Solution:
H ( s ) = L ⎡⎣5 e − 2 t sin ( 4 t ) u ( t ) ⎤⎦ =
5 ( 4)
( s + 2)
2
+4
2
=
20
s + 4 s + 20
2
⎡ H (s) ⎤
⎤
s+4
1
-1 ⎡ 1
step response = L-1 ⎢
= (1 − e − 2 t (cos 4t − sin 4t )) u (t )
⎥=L ⎢ − 2
⎥
2
⎣ s s + 4 s + 20 ⎦
⎣ s ⎦
14-2
Exercise 14.10-1 The input to a circuit is the voltage vi(t). The output is the voltage vo(t). The
transfer function of this circuit is
H ( s) =
Vo ( s )
ks
= 2
Vi ( s ) s + (3 − k ) s + 2
Determine the following:
(a)
The steady-state response when vi(t) = 5 cos 2t V and the gain of the VCVS is k = 2 V/V
(b)
The impulse response when k = 3– 2 2 = 0.17 V/V
(c)
The impulse response when k = 3 + 2 2 = 5.83 V/V
Answer: (a) vo(t) = 7.07 cos (2t–45°) V
(b) h(t) = 0.17e −
(c) h(t) = 5.83e
(1 − 2t ) u(t )
(1 + 2t ) u(t )
2t
2t
Solution:
The poles of the transfer function are p1,2 =
a.) When k = 2 V/V, the poles are p1,2 =
− (3 − k ) ±
(3 − k )
2
2
−8
.
−1 ± −7
so the circuit is stable. The transfer function is
2
H (s) =
Vo ( s )
Vi ( s )
=
2s
s +s+2
2
The circuit is stable when k =2 V/V so we can determine the network function from the transfer
function by letting s = jω.
V o (ω )
V i (ω )
= H (ω ) = H ( s ) s = j ω =
2s
2 jω
=
s + s + 2 s = j ω ( 2 − ω 2 ) + jω
2
The input is v i ( t ) = 5cos 2 t V . The phasor of the steady state response is determine by
multiplying the phasor of the input by the network function evaluated at ω = 2 rad/s.
⎛
2 jω
V o (ω ) = H (ω ) ω = 2 × V i (ω ) = ⎜
⎜ ( 2 − ω 2 ) + jω
⎝
⎞
⎟ ( 5∠0° ) = ⎛⎜ j 4 ⎞⎟ ( 5∠0° ) = 7.07∠ − 45°
⎟
⎝ −2 + j 2 ⎠
ω =2 ⎠
14-3
The steady state response is vo ( t ) = 7.07 cos ( 2 t − 45° ) V .
b. When k = 3 − 2 2 , the poles are p1,2 =
−2 2 ± 0
= − 2, − 2 so the circuit is stable. The
2
transfer function is
H (s) =
0.17 s
=
0.17
−
0.17 2
(s + 2) (s + 2) (s + 2)
2
The impulse response is
h ( t ) = L -1⎡⎣ H ( s ) ⎤⎦ = 0.17 e −
2t
(1 −
2
)
2 t u (t )
We see that when k = 3 − 2 2 the circuit is stable and lim h(t ) = 0 .
t →∞
c. When, k = 3 + 2 2 the poles are p1,2 =
2 2± 0
= 2, 2 so the circuit is not stable. The
2
transfer function is
H (s) =
5.83s
=
5.83
+
5.83 2
(s − 2) (s − 2) (s − 2)
2
The impulse response is
h ( t ) = L -1⎡⎣ H ( s ) ⎤⎦ = 5.83 e
2t
(1 +
2
)
2 t u (t )
We see that when k = 3 + 2 2 the circuit is unstable and lim h(t ) = ∞ .
t →∞
14-4
Exercise 14.12-1
A circuit is specified to have a transfer function of
H(s) =
25
s + 10s + 125
2
and a unit step response of
vo(t) = 0.1(2 – e–5t(2 cos 10t + 3 sin 10t))u(t)
Verify that these specifications are consistent.
Solution:
For the poles to be in the left half of the s-plane, the s-term needs to be positive.
⎡2 ⎛
⎞⎤
10
2s + 20
s +5
⎡2
⎤
V0 ( s ) = 0.1 ⎢ −⎜ 2
+
⎟ ⎥ = 0.1 ⎢ − 2
2
2
2
2 ⎟
⎜
⎣ s s +10 s + 125 ⎥⎦
⎢⎣ s ⎝ ( s + 5 ) +10
( s +5) +10 ⎠ ⎥⎦
= 0.1
= 0.1
2( s 2 +10s +125 )−( 2s + 20 ) s
s ( s 2 +10s +125 )
250
25
=
2
s ( s +10s +125 )
s ( s +10s +125 )
2
Then
H (s) =
V0 ( s )
L ⎡⎣u ( t ) ⎤⎦
=
(
25
2
s s +10 s +125
1
s
)=
25
s 2 +10 s +125
These specifications are consistent.
14-5
Section 14.2 Laplace Transforms
P14.2‐1 Determine the Laplace Transform of v ( t ) = (17 e − 4 t − 147 e− 5 t ) u ( t ) V
Answer: V ( s ) =
3 s + 29
s + 9 s + 20
2
Solution:
L ⎡⎣17 e− 4 t − 14 e − 5 t ⎤⎦ =
17
14 17 ( s + 5 ) − 14 ( s + 4 )
3 s + 29
−
=
= 2
s+4 s+5
s + 9 s + 20
( s + 4 )( s + 5)
P14.2‐2 Determine the Laplace Transform of v ( t ) = 13cos ( 6 t − 22.62° ) V.
Answer: V ( s ) =
12 s + 30
s 2 + 36
Solution:
v ( t ) = 13cos ( 6 t − 22.62° ) = 13cos ( −22.62° ) cos ( 6 t ) − 13sin ( −22.62° ) sin ( 6 t )
= 12 cos ( 6 t ) + 5sin ( 6 t ) V
V ( s ) = L ⎡⎣12 cos ( 6 t ) + 5sin ( 6 t ) ⎤⎦ = L ⎡⎣12 cos ( 6 t ) ⎤⎦ + L ⎡⎣ +5sin ( 6 t ) ⎤⎦
= (12 )
s
6
12 s + 30
+ ( 5) 2
= 2
s + 36
s + 36 s + 36
2
P14.2‐3 Determine the Laplace Transform of v ( t ) = 10 e −5 t cos ( 4 t + 36.86° ) u ( t ) V.
Answer: V ( s ) =
8 s + 16
s + 25 s + 41
2
Solution:
10 cos ( 4 t + 36.86° ) = 10 cos ( 36.86° ) cos ( 4 t ) − 10sin ( 36.86° ) sin ( 4 t )
= 8cos ( 4 t ) − 6sin ( 4 t ) V
L ⎡⎣10 cos ( 4 t + 36.86° ) ⎤⎦ = ( 8 )
s
4
8s − 4
− ( 6) 2
= 2
s + 16
s + 16 s + 16
2
V ( s ) = L ⎡⎣10 e −5 t cos ( 4 t + 36.86° ) ⎤⎦ = L ⎡⎣10 cos ( 4 t + 36.86° ) ⎤⎦
=
s ← s +5
8 ( s + 5) − 4
8s − 4
8 s + 16
=
= 2
2
2
s + 16 s ← s +5 ( s + 5 ) + 16 s + 25 s + 41
P14.2‐4 Determine the Laplace Transform of v ( t ) = 3 t e − 2 t u ( t ) V
Answer: V ( s ) =
3
s + 4s + 4
2
Solution:
L ⎡⎣3 t e − 2 t ⎤⎦ = L [3 t ] s ←s + 2 =
3
s2
=
s←s + 2
3
( s + 2)
2
=
3
s + 4s + 4
2
P14.2‐5 Determine the Laplace Transform of v ( t ) = 16 (1 − 2 t ) e − 4 t u ( t ) V.
Answer: V ( s ) =
16 ( s + 2 )
s 2 + 8 s + 16
Solution:
16
32
⎛ 16 32 ⎞
L ⎡⎣16 (1 − 2 t ) e − 4 t ⎤⎦ = L [16 − 32 t ] s ←s + 4 = ⎜ − 2 ⎟
=
−
2
⎝ s s ⎠ s ←s + 4 s + 4 ( s + 4 )
=
16 ( s + 4 ) − 32
16 ( s + 2 )
= 2
2
s + 8 s + 16
s + 8 s + 16
Section 14-3: Pulse Inputs
P 14.3-1 Determine the Laplace transform of f(t) shown
in Figure P 14.3-1.
5 ⎞
5 ⎛ 21 ⎞ ⎛ 21 ⎞
⎛
Hint: f ( t ) = ⎜ 5 − t ⎟ u ( t ) + ⎜ t − ⎟ u ⎜ t − ⎟
3 ⎠
3⎝
5⎠ ⎝
5⎠
⎝
5
f (t)
0
–2
5e−4.2 s + 15s − 5
Answer: F ( s ) =
3s 2
3
t
Figure P 14.3-1
Solution:
⎛ 5
⎞
⎛ 5
⎞
f ( t ) = ⎜ − t + 5 ⎟ u ( t ) − ⎜ − ( t − 4.2 ) ⎟ u ( t − 4.2 )
⎝ 3
⎠
⎝ 3
⎠
⎛ 5 5⎞
⎛ 5 ⎞ 15 s + 5 ( e
F ( s ) = ⎜ − 2 + ⎟ − e −4.2 s ⎜ − 2 ⎟ =
s⎠
3 s2
⎝ 3s
⎝ 3s ⎠
−4.2 s
P 14.3-2 Use the Laplace transform to obtain the
transform of the signal f(t) shown in Figure P14.3-2.
Answer: F ( s ) =
− 1)
3
f (t)
3 (1 − e −2 s )
s
0
2
t
Figure P14.3-2
Solution:
∞
F (s) = ∫0
3 e − st
f (t ) e dt = ∫ 0 3 e dt =
−s
− st
2
2
− st
=
0
3(1−e −2 s )
s
1
P 14.3-3 Determine the Laplace transform of f(t) shown
in Figure P 14.3-3.
5
Answer: F ( s ) = 2 (1 − e −2 s − 2 se −2 s )
2s
5
f (t)
0
1
2
t
Figure P 14.3-3
Solution:
⎧5 2 t 0<t < 2
f (t ) = ⎨
otherwise
⎩0
5
5
5
5
t ⎡⎣u ( t ) −u ( t − 2 ) ⎤⎦ = t u ( t ) − t u ( t − 2 ) = ⎡⎣t u ( t ) −( t − 2 )u ( t − 2 )− 2u (t − 2) ⎤⎦
2
2
2
2
−2 s
−2 s
5⎡ 1 e
2e ⎤ 5 1
⎡1− e −2 s − 2 se −2 s ⎤⎦
∴F ( s ) = L ⎡⎣ f ( t ) ⎤⎦ = ⎢ 2 − 2 −
=
s
s ⎥⎦ 2 s 2 ⎣
2⎣s
f (t ) =
f(t)
P 14.3-4
Consider the pulse shown in Figure P 14.3-4,
where the time function follows eat for 0 < t < T. Find
eat
F(s) for the pulse.
Answer: F ( s ) =
1− e (
s−a
− s − a )T
1
0
T
t
Figure P 14.3-4
Solution:
f ( t ) = ⎡⎣u ( t ) −u ( t −T ) ⎤⎦ e at
⇒ F ( s ) = L ⎡⎣ eat ⎡⎣u ( t ) −u ( t −T ) ⎤⎦ ⎤⎦
1− e − sT
s
at
L ⎡⎣e g ( t ) ⎤⎦ =G ( s − a )
L ⎡⎣u ( t ) −u ( t −T ) ⎤⎦ =
⎫
1− e( s − a )T
⎪
⎬ ⇒ F (s) =
( s −a )
⎪
⎭
2
P 14.3-5
Find the Laplace transform for g(t) = e–tu(t–0.5).
Solution:
g ( t ) = e − t u ( t − 0.5 ) = e − ( t + (0.5−0.5))u ( t − 0.5 ) = e−0.5 e − (t −0.5)u ( t − 0.5 )
L ⎡⎣e −0.5 e − (t −0.5)u ( t − 0.5 ) ⎤⎦ = e−0.5 L ⎡⎣e− (t − 0.5)u ( t − 0.5 ) ⎤⎦ = e−0.5 e−0.5 s L ⎡⎣ e− t u ( t ) ⎤⎦ =
e0.5 e −0.5 s e0.5 − 0.5 s
=
s +1
s +1
Find the Laplace transform for
− (t − T )
f (t ) =
u (t − T )
T
−1e − sT
Answer: F ( s ) =
Ts 2
Solution:
P 14.3-6
− sT
e − sT
⎡ t −T
⎤
⎡ t
⎤ e
L ⎢−
L ⎡⎣ − t u ( t ) ⎤⎦ = − 2
u ( t − T ) ⎥ = e− sT L ⎢ − u ( t ) ⎥ =
T
Ts
⎣ T
⎦
⎣ T
⎦
3
Section 14-4: Inverse Laplace Transform
P 14.4-1
Find f(t) when
F (s) =
s+3
s + 3s 2 + 6s + 4
3
2
2
1 −t
Answer: f ( s ) = e − t − e −t cos 3t +
e × sin 3t , t ≥ 0
3
3
3
Solution
F (s) =
Where
s +3
s +3
A
Bs + C
=
=
+ 2
2
2
s + 3s + 6s + 4 ( s +1) ⎡( s +1) + 3⎤ s +1 s + 2 s + 4
⎣
⎦
s +3
2
A=
=
2
( s +1) +3 s =−1 3
3
Then
2
Bs +C
2
8
⎛4
⎞
= 3 + 2
⇒ ( s +3) = ( + B) s 2 + ⎜ + B + C ⎟ s + + C
2
3
3
⎝3
⎠
( s +1) ( s + 2s + 4 ) s +1 s + 2s + 4
( s + 3)
Equating coefficient yields
2
2
+ B ⇒ B= −
3
3
4 2
1
s : 1= − + C ⇒ C =
3 3
3
s2 : 0 =
Then
1
2
2 1
2
2
3
− s+
− ( s +1)
3
3
3
3
3
3
F (s) =
+
=
+
+
s +1 ( s +1)2 + 3 s +1 ( s +1)2 + 3 ( s +1)2 + 3
Taking the inverse Laplace transform yields
f (t ) =
2 −t 2 −t
1 −t
e − e cos 3t +
e sin 3 t , t ≥ 0
3
3
3
1
P 14.4-2
Find f(t) when
s 2 − 2s + 1
s 3 + 3s 2 + 4s + 2
F (s) =
Solution:
s 2 − 2s + 1
s 2 − 2s + 1
a
a*
b
F (s) = 3
=
=
+
+
2
s + 3s + 4s + 2 ( s +1) ( s +1− j )( s +1+ j ) s +1− j s +1+ j s +1
where
s 2 − 2 s +1
b=
( s +1)
a=
2
+1 s =−1
=4
3− j 4 3
s 2 − 2s +1
=
=− + j 2
( s +1) ( s +1+ j ) s =−1+ j −2 2
3
a* = − − j 2
2
Then
3
− + j2
F (s) = 2
+
s +1− j
3
− − j2
4
2
+
s +1+ j s +1
Next
m=
From Equation 14.5-8
( −3 2 )
2
+ ( 2)
2
⎛
⎞
⎜
⎟
5
2
=
= 126.9°
and θ = tan −1 ⎜
3⎟
2
⎜− ⎟
⎝ 2⎠
f ( t ) = ⎡⎣5 e − t cos ( t + 127° ) + 4 e −t ⎤⎦ u ( t )
2
P 14.4-3
Find f(t) when
F (s) =
5s − 1
s − 3s − 2
3
Answer: f(t) = – e– t + 2te– t + e2t, t ≥ 0
Solution:
F ( s) =
B=
where
A=
Then
Finally
F (s) =
5 s −1
( s +1) ( s − 2 )
5 s −1
s −2
2
=
A
B
C
+
+
2
s +1 ( s +1)
s −2
= 2 and C =
s =−1
d ⎡
2
( s +1) F ( s )⎤⎦
⎣
ds
−1
2
1
+
+
⇒
2
s +1 ( s +1)
s−2
s =−1
=
5 s −1
( s +1)
2
=1
s=2
−9
( s−2)
= −1
2
s =−1
f ( t ) = ⎡⎣ −e − t + 2 t e −t + e2t ⎤⎦ u ( t )
3
P 14.4-4
Find the inverse transform of
Y(s) =
1
s + 3s + 4s + 2
3
2
Answer: y(t) = e–t(1 – cos t), t ≥ 0
Solution:
Y (s) =
1
1
A
Bs +C
=
=
+
2
( s +1) ( s + 2s + 2 ) ( s +1) ⎡⎣( s +1) + 1⎤⎦ s +1 ( s +1)2 +1
A=
where
Next
2
1
s + 2s + 2
2
=1
s =−1
1
1
Bs +C
=
+ 2
⇒ 1 = s 2 + 2 s + 2 + ( Bs + C ) ( s + 1)
( s +1) ( s + 2s + 2 ) s +1 s + 2s + 2
2
⇒ 1 = ( B +1) s 2 + ( B + C + 2 ) s + C + 2
Equating coefficients:
s 2 : 0 = B + 1 ⇒ B = −1
s : 0 = B + C + 2 ⇒ C =−1
Finally
P 14.4-5
Y (s) =
1
s +1
−
⇒
s +1 ( s +1)2 +1
y ( t ) = ⎡⎣ e− t − e − t cos t ⎤⎦ u ( t )
Find the inverse transform of
F (s) =
Solution:
F (s) =
2s + 6
( s + 1) ( s 2 + 2s + 5)
2( s + 3)
( s +1) ( s 2 + 2 s +5)
=
−( s +1)
1
2
+
+
2
s +1 ( s +1) + 4 ( s +1)2 + 4
f ( t ) = ⎡⎣e −t − e− t cos ( 2t ) + e− t sin ( 2t ) ⎤⎦ u ( t )
4
P 14.4-6
Find the inverse transform of
F (s) =
2s + 6
s ( s 2 + 3s + 2 )
Answer: f(t) = [3 – 4e–t+e–2t] u(t)
Solution:
F (s) =
where
A = sF ( s ) s =0 =
2( s+3)
A
B
C
= +
+
s( s+1) ( s + 2 ) s s +1 s + 2
2( s + 3 )
( s +1) ( s + 2 )
s =0
= 3, B = ( s +1) F ( s ) s =−1 =
and
( s + 2 ) F ( s ) s = −2 =
Finally
3 −4
1
+
⇒
F (s) = +
s s +1 s + 2
2( s + 3)
s ( s +1)
s =−2
2( s + 3 )
s( s + 2 )
s =−1
=−4
= C =1
f ( t ) = ( 3 − 4e− t + e −2t ) u ( t )
5
P 14.4-7
Find the inverse transform of F(s) expressing f(t) in cosine and angle forms.
(a) F ( s ) =
8s − 3
2
s + 4s + 13
(b) F ( s ) =
3e− s
s 2 + 2s + 17
Answer: (a) f(t) = 10.2e–2t cos (3t + 38.4°), t ≥ 0
(b) f(t) = 34 e
Solution:
(a)
F (s) =
− ( t −1)
sin [4(t–1)], t ≥ 1
8s −3
1 2( 8s −3)
= ×
s + 4s +13 2 ( s + 2 )2 + 9
2
∴ a = 2, c =8, ω =3 & ca −ω d =−3 ⇒ d =
−3−( 8 )( 2 )
= 7.33
−3
2
2
⎛ 6.33 ⎞
°
∴ θ = tan −1 ⎜
⎟ =38.4 , m = ( 8 ) +( 6.33) =10.85
⎝ 8 ⎠
⇒ f ( t ) =10.85 e −2 t cos( 3 t + 42.5 ) u ( t )
(b)
Given F ( s ) =
( 2( 3) ) .
3e − s
3
1
= ×
, first consider F1 ( s ) = 2
2
s + 2s +17
s + 2s +17 2 ( s +1)2 +16
⎛ −3 4 ⎞
Identify a =1, c = 0, ω = 4 and −ω d =3 ⇒ d =−3 4. Then m =|d |=3 4, θ = tan −1 ⎜
⎟=−90°
⎝ 0 ⎠
So f1 (t ) = (3 4)e −t sin 4t u ( t ) . Next, F ( s ) = e− s F1 ( s ) ⇒ f ( t ) = f1 ( t −1) . Finally
∴ f ( t ) =(3 4)e − (t −1) sin ⎡⎣ 4( t −1) ⎤⎦ u ( t −1)
6
Find the inverse transform of F(s).
P 14.4-8
(a) f ( s ) =
s2 − 5
s ( s + 1)
4s 2
(b) f ( s ) =
2
( s + 3)
3
Answer: (a) f(t) = –5 + 6e–t + 4te–t, t ≥ 0
(b) f(t) = 4e–3t – 24te–3t + 18t2e–3t, t ≥ 0
Solution:
(a)
F (s) =
where
A = sF ( s ) |s = 0 =
Multiply both sides by s ( s + 1)
s 2 −5
s ( s +1)
=
2
A B
C
+
+
s s +1 ( s +1)2
−5
1−5
2
= −5 and C = ( s +1) F ( s ) |s =−1 =
=4
1
−1
2
s 2 − 5 = −5 ( s +1) + Bs ( s +1) + 4 s ⇒
2
Then
F (s) =
Finally
−5 6
4
+
+
s s +1 ( s +1)2
f ( t ) = ( −5 + 6 e − t + 4 t e − t ) ,
(b)
F (s) =
4s 2
( s + 3)
3
B=6
=
t≥0
A
B
C
+
+
2
( s + 3 ) ( s + 3 ) ( s + 3 )3
where
A=
1 d2 ⎡
d ⎡
3
3
s + 3) F ( s ) ⎤
= 4, B =
( s +3) F ( s )⎤⎦ s =−3 = −24
2 ⎣(
⎦
⎣
3
s
=−
2 ds
ds
and
C = ( s +3) F ( s ) s =−3 = 36
3
Then
F (s) =
Finally
4
−24
36
+
+
2
( s + 3 ) ( s + 3 ) ( s + 3 )3
f ( t ) = ( 4 − 24 t + 18t 2 ) e−3t ,
t≥0
7
Section 14-5: Initial and Final Value Theorems
P 14.5-1
A function of time is represented by
2 s 2 − 3s + 4
F (s) = 3
s + 3s 2 + 2s
(a) Find the initial value of f(t) at t = 0.
(b) Find the value of f(t) as t approaches infinity.
Solution:
(a)
2 s 2 −3s + 4 2s 2
f ( 0 ) = lim sF ( s ) = lim
= 2 =2
2
s
s →∞
s →∞ s +3s + 2
(b)
4
f ( ∞ ) = lim sF ( s ) = = 2
2
s →0
P 14.5-2
Find the initial and final values of v(t) when
V (s) =
( s + 16 )
s + 4s + 12
2
Answer: v(0) = 1, v(∞) = 0 V
Solution:
Initial value:
v ( 0 ) = lim sV ( s ) = lim
s →∞
Final value:
s →∞
s( s +16 )
s 2 + 16 s
=
lim
=1
s 2 + 4 s + 12 s →∞ s 2 + 4s + 12
⎛
⎞
s +16
s 2 + 16 s
=
=0
v ( ∞ ) = lim s ⎜ 2
lim
⎟
2
s→ 0
⎝ s + 4 s + 12 ⎠ s → 0 s + 4 s +12
(Check: V(s) is stable because Re { pi } < 0 since pi = − 2 ± 2.828 j . We
expect the final value to exist.)
14-1
P 14.5-3
Find the initial and final values of v(t) when
V (s) =
( s + 10 )
(3s + 2s 2 + 1s )
2
Answer: v(0) = 0, v(∞) = 10 V
Solution:
Initial value:
v(0) = lim sV ( s ) = lim
s →∞
Final value:
s →∞
v ( ∞ ) = lim sV ( s ) = lim
s →0
s →0
s 2 +10 s
= 0
3 s3 + 2 s 2 + s
s ( s +10 )
s ( 3s 2 + 2s +1)
= 10
(Check: V(s) is stable because p = − 0.333 ± 0.471 j . We expect the
i
final value to exist.)
Find the initial and final values of f(t) when
P 14.5-4
F (s) =
−2 ( s + 7 )
s 2 − 2 s + 10
Answer: initial value = –2; final value does not exist
Solution:
Initial value:
f ( 0 ) = lim s F ( s ) = lim
s →∞
Final value:
s →∞
−2 s 2 −14 s
= −2
s 2 − 2 s + 10
F(s) is not stable because Re { pi } > 0 since pi = 1 ± 3 j . No final value of
f ( t ) exists.
14-2
P14.5-5
as+b
where v ( t ) is the voltage shown in Figure P14.5-5, determine the
Given that L ⎡⎣v ( t ) ⎤⎦ = 2
s +8s
values of a and b.
Figure P14.5-5
Solution:
From the plot, v(0) = 4 V and lim v ( t ) = 12 V . From the final value theorem,
t →∞
lim v ( t ) = lim sV ( s ) = lim s
t →∞
Consequently, 12 =
s →0
s →0
as+b
as+b b
= lim
= .
2
s + 8 s s →0 s + 8 8
b
⇒ b = 96 . From the initial value theorem
8
as+b
as+b
lim v ( t ) = lim sV ( s ) = lim s 2
= lim
=a
t →0
s →∞
s →∞ s + 8 s
s →∞ s + 8
Consequently a = 4.
14-3
P14.5-6
as+b
Given that L ⎡⎣v ( t ) ⎤⎦ = 2
where v ( t ) is the voltage shown in Figure P14.5-6, determine
2 s + 40 s
the values of a and b.
Figure P14.5-6
Solution:
From the plot, v(0+) = 10 V and lim v ( t ) = 2 V . From the final value theorem,
t →∞
lim v ( t ) = lim sV ( s ) = lim s
t →∞
s →0
s →0
as+b
as+b
b
= lim
=
.
2
2 s + 40 s s→0 2 s + 40 40
b
⇒ b = 80 . From the initial value theorem
40
as+b
as+b a
lim v ( t ) = lim sV ( s ) = lim s 2
= lim
=
t →0 +
s →∞
s →∞ 2 s + 40 s
s →∞ 2 s + 40
2
a
⇒ a = 20 .
Consequently 10 =
2
Consequently, 2 =
14-4
Section 14.6 Solution of Differential Equations Describing a Circuit
Figure P14.6-1
P14.6-1 The circuit shown in Figure P14.6-1 is at steady state before the switch closes at time t = 0.
Determine the inductor current, i(t), after the switch closes.
Solution:
The initial inductor current is
i ( 0) =
12
=2A
6
d
i (t ) + 4 i (t )
dt
Take the Laplace Transform of both sides of this equation:
Apply KVL after the switch closes
12 = 2
12
= 2 ⎡⎣ s I ( s ) − 2 ⎤⎦ + 4 I ( s )
s
Solve for I(s):
Taking the Inverse Laplace Transform:
I (s) =
2s +6 3
1
= −
s ( s + 2) s s + 2
i ( t ) = 3 − e −2t A
Figure P14.6-2
P14.6-2 The circuit shown in Figure P14.6-1 is represented by the differential equation
d 2v ( t )
d v (t )
+7
+ 10 v ( t ) = 120
2
dt
dt
after time t = 0. The initial conditions are
i(0) = 0 and v(0) = 4 V
Determine the capacitor, v(t), after time t = 0.
Solution:
Let’s take the Laplace Transform of both sides of the differential equation.
⎡ d v (t ) ⎤
£⎢
⎥ = sV ( s ) − 4
⎣ dt ⎦
First, using Table 14.2-2
Next, notice that
Using Table 14.2-2 again
Now we have:
or:
Solve for V(s):
d v (t )
d v ( 0) i ( 0)
⇒
=
=0
dt
dt
0.1
⎡ d 2v ( t ) ⎤
⎡ d v (t ) ⎤ d v ( 0)
£⎢
= s 2 V (s) − 4 s
⎥−
2 ⎥ = s£ ⎢
dt
⎣ dt ⎦
⎣ dt ⎦
i ( t ) = 0.1
s 2 V ( s ) − 4 s + 7 ( sV ( s ) − 4 ) + 10V ( s ) =
120
s
120
+ 28 + 4 s
s
40 16
−
2
4 s + 28 s + 120 3
1
12
V (s) =
= −
= + 3 + 3
s ( s + 2 )( s + 5 ) s s + 2 s s + 2 s + 5
(s
2
+ 7 s + 10 ) V ( s ) =
Taking the inverse Laplace transform: v ( t ) = 12 −
40 − 2 t 16 −5t
e + e A
3
3
Figure P14.6-3
P14.6-3 The circuit shown in Figure P14.6-3 is at steady state before time t = 0. The input to the circuit is
v s ( t ) = 2.4 u ( t ) V
Consequently, the initial conditions are i1(0)=0 and i2(0)=0. Determine the inductor current, i2(t), after
time t = 0.
Solution:
First, let’s simplify the circuit by replacing the 3 12-Ω resistors at the right of the circuit by an equivalent
resistor:
vs (t ) = 2
Write the mesh equations
and
0=2
d i 2 (t )
dt
d i1 ( t )
dt
+ 12 ( i1 ( t ) − i 2 ( t ) )
+ 8 i 2 ( t ) − 12 ( i1 ( t ) − i 2 ( t ) ) = 2
d i2 (t )
dt
+ 20 i 2 ( t ) − 12 i1 ( t )
Taking the Laplace Transforms of these equations:
2.4
= 2 s I 1 ( s ) + 12 ( I 1 ( s ) − I 2 ( s ) )
s
0 = 2 s I 2 ( s ) + 20 I 2 ( s ) − 12 I 1 ( s )
and
1
5
5⎞
⎛1
s I 2 (s) + I 2 (s) = ⎜ s + ⎟ I 2 (s)
6
3
3⎠
⎝6
2.4
5⎞
⎛1
= ( 2 s + 12 ) I 1 ( s ) − 12 I 2 ( s ) = ( 2 s + 12 ) ⎜ s + ⎟ I 2 ( s ) − 12 I 2 ( s )
s
3⎠
⎝6
7.2
0.3 0.03794 0.33974
=
+
−
I 2 (s) =
2
s
s + 14.3 s + 1.68
2 s + 16 s + 24
Next, some algebra:
Solving for I2(s):
I1 (s) =
(
)
Taking the Inverse Laplace Transform gives i2(t) for t ≥ 0:
i 2 ( t ) = 300 + 39.74 e −14.3t − 339.74 e −1.68t A
Figure P14.6-4
P14.6-4 The circuit shown in Figure P14.6-4 is at steady state before the switch opens at time t = 0.
Determine the capacitor voltage, v(t), after the switch opens.
Solution:
v (0)
3
4
=− A
(12 ) = 4 V and i ( 0 ) = −
3
3
3+9
2
1 d v (t ) 3 d v (t )
+
+ v (t )
Apply KVL after the switch opens 0 =
2 dt 2
2 dt
The initial conditions are
v (0) =
d 2 v (t )
d v (t )
That is
0=
+3
+ 2 v (t )
2
dt
dt
Let’s take the Laplace Transform of both sides of the differential equation.
⎡ d v (t ) ⎤
£⎢
⎥ = sV ( s ) − 4
⎣ dt ⎦
First, using Table 14.2-2
d v (t )
d v ( 0)
i ( 0) 8
⇒
=−
= A
dt
dt
0.5 3
⎡ d 2v ( t ) ⎤
⎡ d v (t ) ⎤ d v ( 0)
2
Using Table 14.2-2 again £ ⎢
= s 2 V (s) − 4 s −
⎥−
2 ⎥ = s£ ⎢
dt
3
⎣ dt ⎦
⎣ dt ⎦
Next, notice that
Now we have:
Solve for V(s):
i ( t ) = −0.5
8⎞
⎛ 2
⎜ s V ( s ) − 4 s − ⎟ + 3 ( sV ( s ) − 4 ) + 2V ( s ) = 0
3⎠
⎝
44
32
20
4s+
3
= 3 − 3
V (s) =
( s + 1)( s + 2 ) s + 1 s + 2
Taking the Inverse Laplace Transform: v ( t ) =
32 − t 20 − 2 t
e − e
V
3
3
Figure P14.6-5
P14.6-5 The circuit shown in Figure P14.6-5 is at steady state before the switch closes at time t = 0.
Determine the capacitor voltage, v(t), after the switch closes.
Solution:
v (0) =
The initial capacitor voltage is
40
(12 ) = 0.98 V
10 + 40
Write a node equation after the switch closes:
vs (t ) − v (t )
10 × 10
3
= ( 2 × 10− 6 )
1200 − 100 v ( t ) = 2
600 =
v (t )
d
v (t ) +
dt
10 × 10 3
d
v ( t ) + 100 v ( t )
dt
d
v ( t ) + 100 v ( t )
dt
Take the Laplace Transform of both sides of this equation:
600
= ⎡⎣ sV ( s ) − 0.98⎤⎦ + 100V ( s )
s
Solve for V(s):
V (s) =
Taking the Inverse Laplace Transform:
600 s + 0.98 6
5.02
= −
s ( s + 100 )
s s + 100
v ( t ) = 6 − 5.02e −100 t V
Section 14.7 Circuit Analysis Using Impedance and Initial Conditions
P14.7-1
Figure P14.7-1a shows a circuit represented in the time domain. Figure P14.7-1b shows the same circuit,
now represented in the complex frequency domain. Figure P14.7-1c shows a plot of the inductor current.
(b)
(a)
(c)
Figure P14-7-1
Determine the values of D and E used to represent the circuit in the complex frequency domain. Determine
the values of the resistance R 2 and the inductance L.
Solution:
First, we find the values of D and E used to represent the circuit in
the complex frequency domain:
L ⎡⎣12 − 6 u ( t ) ⎤⎦ = L ⎡⎣6 u ( t ) ⎤⎦ =
6
⇒ D = 6 A.
s
E is the initial inductor current = 8 A from the plot.
Next, we find the values of the resistance R 2 and the inductance L:
The circuit is at steady state before t = 0, so the inductor acts like a short circuit. Using current division,
⎛ 30 ⎞
8=⎜
12 ⇒ R 2 = 15 Ω . Similarly, the circuit will at steady state for t → ∞. Again, the inductor
⎜ 30 + R 2 ⎟⎟
⎝
⎠
acts like a short circuit. Using current division,
14.7-1
⎛ 30 ⎞
4=⎜
6 ⇒ R 2 = 15 Ω .
⎜ 30 + R 2 ⎟⎟
⎝
⎠
The inductor current can be represented as v ( t ) = 4 + 4 e − at for t ≥ 0. From the plot,
then
⎛ 4.54 − 4 ⎞
ln ⎜
⎟
4 ⎠
4.54 = 4 + 4 e − a ( 0.5) so a = ⎝
= 4.005 ≅ 4 1/s .
−0.5
1
45
L
=τ =
⇒ L=
= 11.25 H .
4
15 + 30
4
As a check, apply KVL to the center mesh in the complex frequency domain to get
R D
E
D
L s + R1
Es+ 1
E
D
⎛
⎞
⎛
⎞
s
s =
L
R 2 I ( s ) + L s ⎜ I ( s ) − ⎟ + R1 ⎜ I ( s ) − ⎟ = 0 ⇒ I ( s ) =
R
+
R2 ⎞
s⎠
s⎠
L s + R1 + R 2
⎛
⎝
⎝
s⎜s + 1
⎟
L ⎠
⎝
8 s + 16 4
4
= +
s ( s + 4) s s + 4
Taking the inverse Laplace transform gives
Substituting values gives
I (s) =
4 ⎤
⎡4
i ( t ) = L−1 ⎡⎣ I ( s ) ⎤⎦ = L−1 ⎢ +
= 4 + 4 e−4 t A for t ≥ 0
⎥
⎣ s s + 4⎦
14.7-2
P14.7-2
Figure P14.7-2a shows a circuit represented in the time domain. Figure P14.7-2b shows the same circuit,
now represented in the complex frequency domain. Figure P14.7-2c shows a plot of the inductor current.
(a)
(b)
(c)
Figure P14-7-2
Determine the values of D and E used to represent the circuit in the complex frequency domain. Determine
the values of the resistance R 1 and the capacitance C.
Solution:
First, we find the values of D and E used to represent the circuit
in the complex frequency domain:
L ⎡⎣6 + 12 u ( t ) ⎤⎦ = L ⎡⎣18 u ( t ) ⎤⎦ =
18
⇒ D = 18 V.
s
E is the initial capacitor voltage = 4 V from the plot.
Next, we determine the values of the resistance R 1 and the
capacitance C:
The circuit is at steady state before t = 0, so the capacitor acts like an open circuit. Using voltage
⎛ 30 ⎞
6 ⇒ R1 = 15 Ω . Similarly, the circuit will at steady state for t → ∞. Again, the
division, 4 = ⎜
⎜ R1 + 30 ⎟⎟
⎝
⎠
capacitor acts like an open circuit. Using voltage division,
14.7-3
⎛ 30 ⎞
12 = ⎜
18 ⇒ R1 = 15 Ω .
⎜ R1 + 30 ⎟⎟
⎝
⎠
The capacitor voltage can be represented as v ( t ) = 12 − 8 e − at for t ≥ 0. From the plot,
⎛ 11.6 − 12 ⎞
ln ⎜
⎟
−8 ⎠
= 7.9886 ≅ 8 1/s .
11.6 = 12 − 8 e − a ( 0.375) so a = ⎝
−0.375
1
1
then
= τ = (15 || 30 ) C = 10 C ⇒ C =
= 12.5 mF .
8
80
As a check, apply KCL at the top node of the 30 Ω resistor, R 2 to get
D
D
Es+
R1 C
s + V ( s ) + C s ⎛V ( s ) − E ⎞ = 0 ⇒ V ( s ) =
⎜
⎟
R1
R2
s⎠
⎛
R + R2 ⎞
⎝
s⎜s + 1
⎟
⎜
R1 R 2 C ⎟⎠
⎝
Substituting values and performing a partial fraction expansion gives
V (s) −
V (s) =
4 s + 96 12
8
= −
s ( s + 8) s s + 8
Taking the inverse Laplace transform gives
8 ⎤
⎡12
= 12 − 8 e− 8t V for t ≥ 0
v ( t ) = L−1 ⎡⎣V ( s ) ⎤⎦ = L−1 ⎢ −
⎥
⎣ s s + 8⎦
14.7-4
P14.7-3
Figure P14.7-3a shows a circuit represented in the time domain. Figure P14.7-3b shows the same circuit,
now represented in the complex frequency domain. Determine the values of a, b and d used to represent the
circuit in the complex frequency domain.
(a)
(b)
Figure P14.7-3
Solution:
24 − 36 u ( t ) = −12 for t > 0
L [ −12] =
−12
⇒ a = −12 V.
s
The circuit is at steady state before t = 0, and the input
is constant, so the capacitor acts like an open circuit
and the inductor acts like a short circuit.
24
⎛ 8 ⎞
v (0) = ⎜
=2A
⎟ 24 = 16 V and i ( 0 ) =
4+8
⎝ 4+8⎠
Consequently, b = v ( 0 ) = 16 V and d = L i ( 0 ) = ( 6 )( 2 ) = 12 .
14.7-5
P 14.7-4
The input to the circuit shown in
t=0
Figure P 14.7-4 is the voltage of the voltage
source, 12 V. The output of this circuit is the
voltage, vo(t), across the capacitor. Determine
12 V
vo(t) for t > 0.
–
+
6Ω
6Ω
+
0.5 F
6Ω
–t/2
Answer: vo (t) = – (4 + 2e
vo(t)
–
)V for t > 0
Figure P 14.7-4
Solution:
t < 0
time domain
frequency domain
Mesh equations in the frequency domain:
6 I 1 ( s ) + 6 ( I 1 ( s ) − I 2 ( s )) + 6 I 1 ( s ) +
12
2
2
= 0 ⇒ I1 (s) = I 2 (s) −
s
3
3s
2
6
2⎞
6
⎛
I 2 ( s ) − − 6 ( I 1 ( s ) − I 2 ( s )) = 0 ⇒ ⎜ 6 + ⎟ I 2 ( s ) − 6 I 1 ( s ) =
s
s
s⎠
s
⎝
1
⎛2
2⎞
2 ⎞ 6
⎛
⇒ I 2 (s) = 2
Solving for I2(s):
⎜ 6 + ⎟ I 2 (s) − 6⎜ I 2 (s) − ⎟ =
1
s⎠
3s ⎠ s
⎝
⎝3
s+
2
⎛ 1 ⎞
−2
1
6 1⎜ 2 ⎟ 6
4
Calculate for Vo(s):
Vo ( s ) = I 2 ( s ) − = ⎜
− =
−
⎟
s 2⎜ s+ 1 ⎟ s s+ 1 s
2
⎝
2⎠
2
Take the Inverse Laplace transform:
(
vo ( t ) = − 4 + 2 e −t /2
)
V for t > 0
(Checked using LNAP, 12/29/02)
14.7-6
P 14.7-5
The input to the circuit shown in Figure
t=0
P 14.7-5 is the voltage of the voltage source, 12 V.
2Ω
The output of this circuit is the current, i(t), in the
12 V
inductor. Determine i(t) for t > 0.
–
+
2Ω
5H
Answer: i(t) = –3(1 + e–0.8t) A for t > 0
i(t)
Figure P 14.7-5
Solution:
t <0
frequency domain
time domain
Writing a mesh equation:
2⎞
⎛
⎛
⎞
−6 ⎜ s + ⎟
⎜
⎟
12
3
3
( 4 + 5 s ) I ( s ) + 30 + = 0 ⇒ I ( s ) = ⎝ 45 ⎠ = − ⎜ + 4 ⎟
s
s s+ ⎟
⎛
⎞
s⎜s + ⎟
⎜⎜
⎟
5⎠
⎝
5⎠
⎝
Take the Inverse Laplace transform:
(
i ( t ) = −3 1 + e −0.8t
)
A for t > 0
(Checked using LNAP, 12/29/02)
14.7-7
P 14.7-6 The input to the circuit shown in Figure P 14.7-6 is the voltage of the voltage source, 18 V. The
output of this circuit, the voltage across the capacitor, is given by
vo (t) = 6 + 12e–2t V when t > 0
Determine the value of the capacitance, C, and the value of the resistance, R.
Figure P 14.7-6
Solution:
Steady-state for t < 0:
Steady-state for t > 0:
From the equation for vo(t):
vo ( ∞ ) = 6 + 12 e
From the circuit:
Therefore:
6=
− 2 (∞)
vo ( ∞ ) =
=6 V
3
(18)
R+3
3
(18) ⇒ R = 6 Ω
R+3
⎛
1 ⎞ 18 6
−6
I (s)⎜ 2 +
⎟ + − = 0 ⇒ I (s) =
1
Cs⎠ s s
⎝
s+
2C
14.7-8
⎛
1
18 1 ⎜ −6
Vo ( s ) =
I (s) + =
⎜
Cs
s Cs ⎜ s + 1
⎜
2C
⎝
Taking the inverse Laplace transform:
⎞
⎟ 18 −12
12
18
12
6
+
+ =
+
⎟+ =
1
s
s s+ 1
s
⎟⎟ s
s+
2C
2C
⎠
vo ( t ) = 6 + 12 e − t / 2C V for t > 0
Comparing this to the given equation for vo(t), we see that 2 =
1
2C
⇒ C = 0.25 F .
(Checked using LNAP, 12/29/02)
14.7-9
P 14.7-7
The input to the circuit shown in Figure P 14.7-7 is the voltage source voltage
vs(t) = 3 – u(t)
V
The output is the voltage
vo(t) = 10 + 5e–100t V for t ≥ 0
Determine the values of R1 and R2.
Figure P 14.7-7
Solution:
We will determine Vo ( s ) , the Laplace transform of the output, twice, once from the given
equation and once from the circuit. From the given equation for the output, we have
Vo ( s ) =
10
5
+
s s + 100
Next, we determine Vo ( s ) from the circuit. For t ≥ 0 , we represent the circuit in the frequency
domain using the Laplace transform. To do so we need to determine the initial condition for the
capacitor.
When t < 0 and the circuit is at steady state,
the capacitor acts like an open circuit. Apply
KCL at the noninverting input of the op amp to
get
3 − v (0 −)
= 0 ⇒ v (0 −) = 3 V
R1
The initial condition is
v (0 +) = v (0 −) = 3 V
Now we can represent the circuit in the frequency domain, using Laplace transforms.
Apply KCL at the noninverting input of the op
am to get
2
3
−V ( s) V ( s) −
s
s
=
6
R1
10
s
Solving gives
106
3s + 2
R1
2
1
V (s) =
= +
⎛ 106 ⎞ s ⎛ 106 ⎞
s ⎜s+
⎟
⎜s+
⎟
⎜
⎜
R1 ⎟⎠
R1 ⎟⎠
⎝
⎝
Apply KCL at the inverting input of the op amp to get
⎛
⎞
⎜
⎟
Vo ( s) −V ( s) V ( s)
R2 ⎞
R2 ⎞⎜ 2
⎛
⎛
⎟
1
=
⇒ V o ( s ) = ⎜1 +
⎟V ( s ) = ⎜1 +
⎟⎜ +
⎟
R2
1000
⎝ 1000 ⎠
⎝ 1000 ⎠ ⎜ s ⎛ 106 ⎞ ⎟
⎜s+
⎟⎟
⎜
⎜
⎟
R
1
⎝
⎠⎠
⎝
The expressions for Vo(s) must be equal, so
⎛
⎞
⎜
⎟
R
⎛
⎞⎜
⎟
10
5
2
1
2
+
= ⎜1 +
⎟⎜ +
⎟
s s + 100 ⎝ 1000 ⎠ ⎜ s ⎛ 106 ⎞ ⎟
⎜s+
⎟
⎜
⎜
R1 ⎟⎠ ⎟
⎝
⎝
⎠
Equating coefficients gives
106
1+
= 5 ⇒ R 2 = 4 kΩ and
= 100 ⇒ R1 = 10 kΩ
1000
R1
R2
(checked using LNAPTR 7/31/04)
P 14.7-8 Determine the inductor current, iL(t), in the circuit shown in Figure P 14.7-8 for each
of the following cases:
(a)
R = 2 Ω, L = 4.5 H, C = 1/9 F, A = 5 mA, B = – 2 mA
(b)
R = 1 Ω, L = 0.4 H, C = 0.1 F, A = 1 mA, B = – 2 mA
(c)
R = 1 Ω, L = 0.08 H, C = 0.1 F, A = 0.2 mA, B = – 2 mA
Figure P 14.7-8
Solution:
For t < 0, The input is constant. At steady state,
the capacitor acts like an open circuit and the
inductor acts like a short circuit.
The circuit is at steady state at time t = 0 − so
vC ( 0 − ) = 0 and i L ( 0 − ) = B
The capacitor voltage and inductor current are continuous so vC ( 0 + ) = vC ( 0 − ) and
iL (0 +) = iL (0 −) .
For t < 0, represent the circuit in the
frequency domain using the Laplace
transform as shown. V C ( s ) is the node
voltage at the top node of the circuit.
Writing a node equation gives
A + B VC ( s ) B VC ( s )
=
+ +
+ C sV C ( s )
s
R
s
Ls
so
A L s + R + R L C s2
=
VC ( s )
s
RLs
Then
A
AR L
C
=
VC ( s ) =
R L C s2 + L s + R s2 + 1 s + 1
RC
LC
A
B
B
LC
and
I L (s) =
+ =
+
Ls
s
⎛
1
1 ⎞ s
s ⎜ s2 +
s+
⎟
RC
LC ⎠
⎝
1
a.) When R = 2 Ω, L = 4.5 H, C = F, A = 5 mA and B = −2 mA , then
9
5
40
−2 3
10
+
= + 7 − 7
I L (s) =
2
s s+4 s+ 1
s ( s + 4.5 s + 2 ) s
2
Taking the inverse Laplace transform gives
VC ( s )
5
5
i L ( t ) = 3 + e− 4 t − e − 0.5 t mA for t ≥ 0
7
7
b.) When R = 1 Ω, L = 0.4 H, C = 0.1 F, A = 1 mA and B = −2 mA , then
I L (s) =
⎛1
25
25
5
1 ⎞
−2
−2
+
=
+
=
−
+
+
⎜
⎟
2
⎜ s ( s + 5)2 s + 5 ⎟
s
s ( s 2 + 10 s + 25 ) s
s ( s + 5)
⎝
⎠
Taking the inverse Laplace transform gives
i L ( t ) = − (1 + 5 t e − 5 t − e− 5 t ) mA for t ≥ 0
c.) When R = 1 Ω, L = 0.08 H, C = 0.1 F, A = 0.2 mA and B = −2 mA , then
I L (s) =
25
10
s+5
−2 −1.8
−0.2 s − 2
−1.8
0.2
0.1
+
=
+
=
−
−
2
2
2
s
s ( s + 10 s + 125 ) s
( s + 5) + 102 s
( s + 5) + 102
( s + 5) + 102
2
Taking the inverse Laplace transform gives
i L ( t ) = −1.8 − e − 5 t ( 0.2 cos (10 t ) + 0.1sin (10 t ) ) mA for t ≥ 0
P 14.7-9 Determine the capacitor current, ic(t), in the circuit shown in Figure P 14.7-9 for each
of the following cases:
(a)
R = 3 Ω, L = 2 H, C = 1/24 F, A = 12 V
(b)
R = 2 Ω, L = 2 H, C = 1/8 F, A = 12 V
(c)
R = 10 Ω, L = 2 H, C = 1/40 F, A = 12 V
Figure P 14.7-9
Solution:
For t < 0, the switch is open and the circuit is at
steady state. and the circuit is at steady state. At
steady state, the capacitor acts like an open
circuit.
A
A
and vC ( t ) =
i (t ) =
2R
2
Consequently,
A
A
and vC ( 0 − ) =
i (0 −) =
2R
2
Also
iC (0 −) = 0
The capacitor voltage and inductor current are continuous so vC ( 0 + ) = vC ( 0 − ) and
iL (0 +) = iL (0 −) .
For t > 0, the voltage source voltage is
12 V. Represent the circuit in the
frequency domain using the Laplace
transform as shown.
I L ( s ) and I C ( s ) are mesh currents.
Writing a mesh equations gives
AL
A
+ R ( I L ( s ) − I C ( s )) − = 0
2R
s
1
A
+ R ( I L ( s ) − I C ( s )) = 0
I C (s) −
Cs
2s
⎛ AL A⎞
−R ⎞
⎛Ls+ R
⎜ 2R + s ⎟
I
s
⎛
⎞
(
)
L
⎜
⎟
⎟
1 ⎟ ⎜⎜
⎟⎟ = ⎜
⎜ −R
R
+
I
s
⎜
A
⎟
(
)
⎜
⎠
C s ⎟⎠ ⎝ C
⎝
⎜ − 2s ⎟
⎝
⎠
L s I L (s) −
Or, in matrix form
⎛
I C (s) =
⎛ AL A⎞
A⎞
A
+ ⎟
⎟ + R⎜
2L
⎝ 2s ⎠
⎝ 2R s ⎠ =
1
1
⎛
1 ⎞
s2 +
s+
( L s + R ) ⎜ R + ⎟ − R2
RC
LC
Cs⎠
⎝
( L s + R)⎜ −
a.) When R = 3 Ω, L = 2 H, C =
1
F and A = 12 V ,
24
3
3
3
2
I C (s) = 2
=
= 4 − 4 .
s + 8 s + 12 ( s + 2 )( s + 6 ) s + 2 s + 8
Taking the inverse Laplace transform gives
3
⎛3
⎞
i C ( t ) = ⎜ e− 2t − e− 6t ⎟ u ( t ) A
4
⎝4
⎠
b.) R = 2 Ω, L = 2 H, C =
1
F and A = 12 V ,
8
3
3
I C (s) = 2
=
s + 4 s + 4 ( s + 2 )2
Taking the inverse Laplace transform gives
i C ( t ) = 3 t e− 2t u ( t ) A
1
F and A = 12 V
40
3
3
3
4
I C (s) = 2
=
= ×
2
s + 4 s + 20 ( s + 2 ) + 16 4 ( s + 2 )2 + 16
c.) R = 10 Ω, L = 2 H, C =
Taking the inverse Laplace transform gives
i C (t ) =
3 − 2t
e sin ( 4 t ) u ( t ) A
4
(checked using LNAP 4/11/01)
P 14.7-10
The voltage source voltage in the circuit shown in Figure P 14.7-10 is
Determine v(t) for t ≥ 0.
vs(t) = 12 – 6u(t) V
Figure P 14.7-10
Solution:
For t < 0, The input is 12 V. At steady state, the
capacitor acts like an open circuit.
Notice that v(t) is a node voltage. Express the
controlling voltage of the dependent source as a
function of the node voltage:
va = −v(t)
Writing a node equation:
⎛ 12 − v ( t ) ⎞ v ( t ) ⎛ 3
⎞
−⎜
+ ⎜ − v (t ) ⎟ = 0
⎟+
8
4
⎝ 4
⎠
⎝
⎠
−12 + v ( t ) + 2 v ( t ) − 6 v ( t ) = 0 ⇒ v ( t ) = −4 V
v ( 0 + ) = v ( 0 − ) = −4 V
For t < 0, represent the circuit in the frequency
domain using the Laplace transform as shown.
V ( s ) is a node voltage. Express the controlling
voltage of the dependent source in terms of the
node voltages
V a ( s ) = −V ( s )
Writing a node equation gives
6
s + V ( s ) + 3 s ⎛ V ( s ) + 4 ⎞ = 0.75 V ( s )
⎜
⎟
8
4
40 ⎝
s⎠
V (s) −
Solving gives
−2
10
10
4
2
4
1 ⎞
⎛1
− 4 ⇒ V (s) =
−
=
+
−
= −2 ⎜ +
⎟
s
s ( s − 5) s − 5 s s − 5 s − 5
⎝ s s −5⎠
Taking the inverse Laplace transform gives
( s − 5)V ( s ) =
v ( t ) = −2 (1 + e5 t ) V for t ≥ 0
This voltage becomes very large as time goes on.
P 14.7-11
Determine the output voltage, vo(t), in the circuit shown in Figure P 14.7-11.
Figure P 14.7-11
Solution:
For t < 0, the voltage source voltage is 2 V
and the circuit is at steady state. At steady
state, the capacitor acts like an open circuit.
i (0 −) =
2−0
= 0.04 mA
10 × 103 + 40 × 103
and
vC ( 0 − ) = ( 40 × 103 )( 0.04 × 10−3 ) = 1.6 V
The capacitor voltage is continuous so vC ( 0 + ) = vC ( 0 − ) .
For t > 0, the voltage source voltage is
12 V. Represent the circuit in the
frequency domain using the Laplace
transform as shown.
V C ( s ) and V o ( s ) are node voltages.
Writing a node equation gives
12
1.6
VC ( s ) −
s +
s + V C ( s ) = 0 ⇒ 4 ⎛ V ( s ) − 12 ⎞ + 0.08 s ⎛ V ( s ) − 1.6 ⎞ + V ( s ) = 0
⎜ C
⎟
⎜ C
⎟ C
6
3
0.5 × 10
10 × 10
40 × 103
s ⎠
s ⎠
⎝
⎝
s
48
80 s + 48
1.6 s + 600 9.6
−8
V C ( s )( 0.08 s + 5 ) =
+ 0.128 ⇒ V C ( s ) =
=
=
+
s
s ( 0.08 s + 5 ) s ( s + 62.5 )
s s + 62.5
VC ( s ) −
Taking the inverse Laplace transform gives
v C ( t ) = 9.6 − 8 e − 62.5 t V for t ≥ 0
The 40 kΩ resistor, 50 kΩ resistor and op amp comprise an inverting amplifier so
v o (t ) = −
50
50
v C ( t ) = − ( 9.6 − 8 e − 62.5 t ) = −12 + 10 e − 62.5 t V for t ≥ 0
40
40
so
⎧ −2 V for t ≤ 0
v o (t ) = ⎨
− 62.5 t
V for t ≥ 0
⎩ −12 + 10 e
(checked using LNAP 10/11/04)
P 14.7-12
Determine the capacitor voltage, v(t), in the circuit shown in Figure P 14.7-12.
Figure P 14.7-12
Solution:
For t < 0, the voltage source voltage is 5 V and the
circuit is at steady state. At steady state, the capacitor
acts like an open circuit. Using voltage division twice
v (0 −) =
and
32
30
5−
5 = 0.25 V
32 + 96
120 + 30
v ( 0 + ) = v ( 0 − ) = 0.25 V
For t > 0, the voltage source voltage is 20 V.
Represent the circuit in the frequency domain using
the Laplace transform as shown.
We could write mesh or node equations, but finding a
Thevenin equivalent of the part of the circuit to the left
of terminals a-b seems promising.
Using voltage division twice
⎛ 32 ⎞ 20 ⎛ 30 ⎞ 20 5 − 4 1
V oc ( s ) = ⎜
=
= V
⎟ −⎜
⎟
s
s
⎝ 32 + 96 ⎠ s ⎝ 120 + 30 ⎠ s
Z t = ( 96 || 32 ) + (120 || 30 ) = 24 + 24 = 48 Ω
After replacing the part of the circuit to the left of
terminals a-b by its Thevenin equivalent circuit as shown
1 0.25
−
s
s = 0.75
I (s) =
80 48 s + 80
48 +
s
V (s) =
80
0.25 ⎛ 80 ⎞ 0.75
0.25
I (s) +
=⎜ ⎟
+
s
s
s
⎝ s ⎠ 48 s + 80
V (s) =
60
0.25
1.25
0.25 0.75 −0.75 0.25 1 −0.75
+
=
+
=
+
+
= +
s ( 48 s + 80 )
s
s ( s + 1.67 )
s
s
s + 1.67
s
s s + 1.67
Taking the inverse Laplace transform gives
v ( t ) = 1 − 0.75e −1.67 t V for t ≥ 0
Then
⎧0.25 V for t ≤ 0
v (t ) = ⎨
−1.67 t
V for t ≥ 0
⎩ 1 − 0.75e
(checked using LNAP 7/1/04)
P 14.7-13
Determine the voltage vo(t) for t ≥ 0 for the circuit of Figure P 14.7-13.
Hint: vC(0) = 4 V
Answer: vo(t) = 24e0.75t u(t) V (This circuit is unstable.)
Figure P 14.7-13
Solution:
Mesh Equations:
4 1
4 ⎛
1 ⎞
− − I C (s) − 6 I (s) + I C (s) = 0 ⇒ − = ⎜ 6 + ⎟ I C (s) + 6 I (s)
2s ⎠
s 2s
s ⎝
10
6 ( I ( s ) − I C ( s )) + 3 I ( s ) + 4 I C ( s ) = 0 ⇒ I ( s ) = − I C ( s )
9
Solving for I C(s):
4 ⎛ 2 1 ⎞
6
− = ⎜ − + ⎟ I C (s) ⇒ I C (s) =
3
s ⎝ 3 2s ⎠
s−
4
(
So Vo(s) is
Back in the time domain:
)
Vo ( s ) = 4 I C ( s ) =
24
3
s−
4
v o ( t ) = 24 e0.75t u (t ) V for t ≥ 0
P 14.7-14
Determine the current iL(t) for t ≥ 0 for the circuit of Figure P 14.7-14.
Hint: vC(0) = 8 V and iL(0) = 1 A
1
⎛
⎞
Answer: iL ( t ) = ⎜ e − t cos 2t + e− t sin 2t ⎟ u ( t ) A
2
⎝
⎠
Figure P 14.7-14
Solution:
KVL:
8
⎛ 20
⎞
+ 4 = ⎜ + 8 + 4s ⎟ I L ( s )
s
⎝ s
⎠
so
I L ( s) =
( s + 1) + 1
2+ s
=
s + 2 s + 5 ( s + 1) 2 + 4
2
Taking the inverse Laplace transform:
1
⎛
⎞
i L ( t ) = ⎜ e − t cos 2 t + e − t sin 2 t ⎟ u ( t ) A
2
⎝
⎠
P14.7-15
The circuit shown in Figure P14.7-23 is at steady
state before the switch opens at time
t = 0. Determine the inductor voltage v(t)
for t > 0.
Figure P14.7-15
Solution:
The circuit shown in Figure P14.7-23 is at steady
state before the switch opens at time
t = 0. Determine the inductor voltage v(t)
for t > 0.
Determine the initial conditions, i.e. the inductor
current and capacitor voltage at t = 0, as shown in
the circuit on the left below. Use those initial
conditions to represent the circuit in the s-domain
as s shown in the circuit on the left below.
Figure P14.7-15
Analysis of the s-domain circuit shows that
⎛
⎞
⎜
⎟⎛ 4
16 ( s + 2 )
4
⎞ 16 ( s + 2 )
V (s) = ⎜
+ 2⎟ = 2
=
⎟
⎜
2
8
⎠ s + 8 s + 16 ( s + 4 )
⎜⎜ 0.5 s + 4 + ⎟⎟ ⎝ s
s⎠
⎝
Performing the partial fraction expansion, we get
16 ( s + 2 )
( s + 4)
Finally
2
=
k
−32
+
s + 4 ( s + 4 )2
⇒ 16 ( s + 2 ) = k ( s + 2 ) − 32 ⇒ k = 16
V (s) =
16
−32
+
s + 4 ( s + 4 )2
Now use linearity, e − at f ( t ) ↔ F ( s + a ) and t ↔
1
to find the inverse Laplace transform
s2
1 2
v ( t ) = 16 e −4 t ℒ -1[ − 2 ] = 16 (1 − 2 t ) e − 4 t for t > 0
s s
P14.7-16
The circuit shown in Figure P14.7-16 is at
steady state before time t = 0. Determine the
resistor voltage v(t) for t > 0.
Figure P14.7-16
Solution:
The circuit shown in Figure 14.7-16 is at
steady state before time t = 0. Determine the
resistor voltage v(t) for t > 0.
Determine the initial conditions, i.e. the
inductor current and capacitor voltage at t =
0, as shown in the circuit on the left below.
Use those initial conditions to represent the
circuit in the s-domain as s shown in the
circuit on the left below.
Figure P14.7-16
Analysis of the s-domain circuit shows that
V (s) V (s) 2 ⎛ s ⎞
+
+ + ⎜ ⎟V ( s ) = 0 ⇒
5
6s
s ⎝ 30 ⎠
( 6 s + 5 + s )V ( s ) = −60
2
Performing the partial fraction expansion, we get
V (s) =
−60
−15 15
=
+
s + 6 s + 5 s +1 s + 5
2
The inverse Laplace transform is
v ( t ) = 15 ( e −5t − e − t ) V for t > 0
⇒ V (s) =
−60
s +6s +5
2
P14.7.17
The input to the circuit shown in Figure P14.7-17
is the voltage source voltage
⎧10 V when t < 0
v i ( t ) = 10 + 5 u ( t ) V = ⎨
⎩15 V when t > 0
Determine the response, vo(t). Assume that the
circuit is at steady state when t < 0. Sketch vo(t) as
a function of t.
Figure P14.7-17
Solution:
⎧10 V when t < 0
v i ( t ) = 10 + 5 u ( t ) V = ⎨
⎩15 V when t > 0
The circuit is at steady state when t < 0 so the capacitors
act like open circuits. Since the current in the resistor is
0 A, vo(0−) = 0 V. Then KVL gives v1(0−) = 10 V.
Use the initial conditions to represent the circuit for t>0
in the s-domain:
Calculate
5 ||
500
500
=
s
s + 100
and
500
0.8 s
s
=
125
500 s + 20
+ 5 ||
s
s
5 ||
Use voltage division to write
500
4
⎛ 15 10 ⎞ 0.8 s ⎛ 5 ⎞
s
Vo ( s ) =
− ⎟=
=
⎜
⎜
⎟
125
500 ⎝ s
s ⎠ s + 20 ⎝ s ⎠ s + 20
+ 5 ||
s
s
Taking the inverse Laplace transforms gives v o ( t ) = 4 e − 20 t V when t > 0 .
5 ||
In summary
when t < 0
⎧ 0V
v o ( t ) = ⎨ − 20 t
V when t > 0
⎩4 e
Let t = 0.05 s and calculate
v o ( 0.05 ) = 4 e − 20 ( 0.05) =1.47 V
P14.7-18
The input to the circuit shown in Figure P14.7-18
is the current source current
⎧25 mA when t < 0
i ( t ) = 25 − 15 u ( t ) mA = ⎨
⎩10 mA when t > 0
Determine the response, i2(t). Assume that the
circuit is at steady state when t < 0. Sketch i2(t) as
a function of t.
Figure P14.7-18
Solution:
⎧25 mA when t < 0
i ( t ) = 25 − 15 u ( t ) mA = ⎨
⎩10 mA when t > 0
The circuit is at steady state when t < 0 so the
inductors act like short circuits. Since the voltage
across the resistor is 0 V, i2(0−) = 0 V. Then KCL
gives i1(0−) = 25 mA.
Use the initial conditions to represent the circuit for
t>0 in the s-domain:
Use current division to write
I2 ( s ) =
1.25 s
⎛ 0.01 0.025 ⎞ 0.2 s ⎛ 0.015 ⎞ −0.003
−
⎜
⎟=
⎜−
⎟=
1.25 s + ( 25 + 5 s ) ⎝ s
s ⎠ s+4⎝
s ⎠ s + 20
Taking the inverse Laplace transforms gives i 2 ( t ) = 3 e − 4 t mA when t > 0 .
In summary
⎧ 0 mA when t < 0
i 2 ( t ) = ⎨ − 4t
mA when t > 0
⎩3 e
Let t = 0.1 s and calculate
i 2 ( 0.1) = 3 e − 4 ( 0.1) = 2.01 mA
P 14.7-19 All new homes are required to install a device called a ground fault circuit
interrupter (GFCI) that will provide protection from shock. By monitoring the current going to
and returning from a receptacle, a GFCI senses when normal flow is interrupted and switches off
the power in 1/40 second. This is particularly important if you are holding an appliance shorted
through your body to ground. A circuit model of the GFCI acting to interrupt a short is shown in
Figure P 14.7-19. Find the current flowing through the person and the appliance, i(t), for t ≥ 0
when the short is initiated at t = 0. Assume v = 160 cos 400t and the capacitor is intially
uncharged.
FIGURE
P 14.7-19
Circuit model of person and appliance shorted to ground
Solution:
We are given v ( t ) = 160 cos 400 t .
The capacitor is initially uncharged, so
v C ( 0 ) = 0 V . Then
i ( 0) =
10−3
KCL yields
dvC
dt
+
vC
100
160 cos ( 400 × 0 ) − 0
= 160 A
1
=i
Apply Ohm’s law to the 1 Ω resistor to get
v −v C
i=
⇒ vC = v− i
1
di
+ 1010 i = 1600 cos 400t − ( 6.4 × 104 ) sin 400t
Solving yields
dt
Taking the Laplace transform yields
s I ( s ) − i (0) + (1010 ) I ( s ) =
so
Next
I (s) =
s 2 + ( 400 )
( 6.4×10 ) ( 400 )
−
2
1600s
2
s 2 + ( 400 )
160
1600s − 2.5×107
+
s + 1010 ( s + 1010 ) ⎡⎣ s 2 + (400) 2 ⎤⎦
2
1600s − 2.5×107
A
B
B*
=
+
+
( s + 1010 ) ⎡⎣ s 2 + (400)2 ⎤⎦ s + 1010 s + j 400 s − j 400
where
A =
B =
1600 s − 2.5 x 107
( s +1010 ) ( s − j 400 )
1600 s − 2.5×107
s 2 + ( 400 )
=
s = − j 400
Then
I (s) =
= − 23.1 ,
2
s = −1010
2.56 × 107 ∠1.4°
8.69 × 10 ∠68.4
5
°
= 11.5 − j 27.2 and B* = 11.5 + j 27.2
136.9
11.5− j 27.2 11.5 + j 27.2
+
+
s + 1010
s + j 400
s − j 400
Finally
i ( t ) = 136.9e−1010t + 2 (11.5 ) cos 400t − 2 ( 27.2 ) sin 400t for t > 0
= 136.9e −1010t + 23.0 cos 400t − 54.4sin 400t for t > 0
P 14.7-20 Using the Laplace transform, find vc(t) for t > 0 for the circuit shown in Figure P 14.720. The initial conditions are zero.
Hint:Use a source transformation to obtain a single mesh circuit.
Answer: vc =–5e–2t + 5 (cos 2t + sin 2t) V
Figure P 14.7-20.
Solution:
vC (0) = 0
vc +15×103 i = 10 cos 2t ⎫
⎪
⎬ ⇒
⎛ 1
−3 ⎞ d vc
i = ⎜ ×10 ⎟
⎪
⎝ 30
⎠ dt ⎭
d vc
+ 2 vc = 20 cos 2t
dt
Taking the Laplace Transform yields:
sVC ( s ) − vC ( 0 ) + 2VC ( s ) = 20
20s
s
A
B
B*
⇒
V
s
=
=
+
+
(
)
C
s2+ 4
( s + 2 )( s 2 + 4 ) s + 2 s + j 2 s − j 2
where
A=
20 s
s2 +4
=
s = −2
−40
20s
= −5, B =
8
( s + 2 )( s − j 2 )
=
s = − j2
5
5
5
5
5
= + j and B* = − j
1− j 2
2
2
2
Then
5 5 5 5
+j
−j
−5 2 2 2 2
VC ( s ) =
+
+
s+2 s+ j2 s− j2
⇒ vC ( t ) = −5e−2t + 5 ( cos 2t + sin 2t ) V
P 14.7-21
Determine the inductor current, i(t), in the circuit shown in Figure P 14.7-21.
Figure P 14.7-21
Solution:
After the switch opens, apply KCL and KVL to
get
d
⎛
⎞
R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) = Vs
dt
⎝
⎠
Apply KVL to get
d
i (t ) + R2 i (t )
dt
Substituting v ( t ) into the first equation gives
v (t ) = L
d⎛ d
d
⎛
⎞⎞
R1 ⎜ i ( t ) + C ⎜ L i ( t ) + R 2 i ( t ) ⎟ ⎟ + L i ( t ) + R 2 i ( t ) = Vs
dt ⎝ dt
dt
⎠⎠
⎝
2
d
d
then
R 1 C L 2 i ( t ) + R1 C R 2 + L
i ( t ) + R1 + R 2 i ( t ) = Vs
dt
dt
Dividing by R1 C L :
(
)
(
)
⎛ R1 C R 2 + L ⎞ d
⎛ R1 + R 2 ⎞
Vs
i
t
i
t
+
+
⎜
⎟
⎜
⎟ i (t ) =
(
)
(
)
2
⎜
⎟
⎜
⎟
R1 C L
dt
⎝ R1 C L ⎠ dt
⎝ R1 C L ⎠
d2
With the given values:
d2
dt
Taking the Laplace transform:
2
i ( t ) + 25
d
i ( t ) + 156.25 i ( t ) = 125
dt
125
⎡ 2
⎛d
⎞⎤
⎢ s I ( s ) − ⎜ dt i ( 0 + ) + s i ( 0 + ) ⎟ ⎥ + 25 ⎡⎣ s I ( s ) − i ( 0 + ) ⎤⎦ + 156.25 I ( s ) = s
⎝
⎠⎦
⎣
We need the initial conditions. For t < 0, the switch is
closed and the circuit is at steady state. At steady state,
the capacitor acts like an open circuit and the inductor
acts like a short circuit. Using voltage division
v (0 −) =
9
20 = 14.754 V
9 + (16 || 4 )
Then, using current division
⎛ 4 ⎞ v (0 −)
i (0 −) = ⎜
= 0.328 A
⎟
⎝ 16 + 4 ⎠ 9
The capacitor voltage and inductor current are continuous so v ( 0 + ) = v ( 0 − ) and i ( 0 + ) = i ( 0 − ) .
After the switch opens
v (t ) = L
d
i (t ) + R2 i (t ) ⇒
dt
v ( 0 + ) 9 i ( 0 + ) 14.754 9 ( 0.328 )
d
i (0 +) =
+
=
+
= 29.508
dt
0.4
0.4
0.4
0.4
Substituting these initial conditions into the Laplace transformed differential equation gives
⎡ s 2 I ( s ) − ( 29.508 + 0.328 s ) ⎤ + 25 ⎡ s I ( s ) − 0.328⎤ + 156.25 I ( s ) = 125
⎣
⎦
⎣
⎦
s
( s2 + 25 s + 156.25) I ( s ) = 125s + ( 29.508 + 0.328 s ) + 25 ( 0.328)
so
I (s) =
=
0.328 s 2 + ( 29.508 + 25 ( 0.328 ) ) + 125
(
s s 2 + 25 s + 156.25
)
0.328 s 2 + ( 29.508 + 25 ( 0.328 ) ) + 125
s ( s + 12.5 )
2
=
−0.471
23.6
0.8
+
+
2
s + 12.5 ( s + 12.5 )
s
Taking the inverse Laplace transform
i ( t ) = 0.8 + e −12.5 t ( 23.6 t − 0.471) A for t ≥ 0
So
⎧0.328 A for t ≤ 0
i (t ) = ⎨
−12.5 t
( 23.6 t − 0.471) A for t ≥ 0
⎩ 0.8 + e
(checked using LNAP 10/11/04)
P 14.7-22
Find v2(t) for the circuit of Figure P 14.7-22 for t ≥ 0.
Hint: Write the node equations at a and b in terms of v1 and v2. The initial conditions are v1(0) =
10 V and v2(0) = 25 V. The source is vs = 50 cos 2t u(t) V.
Answer: v2 ( t ) =
23 − t 16 −4t
e + e + 12 cos 2t + 12 sin 2t V t ≥ 0
3
3
Figure P 14.7-22
Solution:
Apply KCL at node a to get
1 d v1 v 2 − v1
=
48 dt
24
⇒ 2 v1 +
d v1
dt
= 2 v2
Apply KCL at node b to get
v 2 − 50 cos 2 t
20
+
v 2 − v1
24
+
v2
30
+
d v2
1 d v2
= 0 ⇒ − v1 + 3 v 2 +
= 60 cos 2 t
24 dt
dt
Take the Laplace transforms of these equations, using v1 (0) = 10 V and v2 (0) = 25 V , to get
( 2+ s ) V1 ( s) − 2V2 ( s) = 10 and − V1 ( s) + ( 3+ s ) V2 ( s) =
25s 2 + 60s +100
s2 + 4
Solve these equations using Cramer’s rule to get
⎛ 25s 2 + 60s +100 ⎞
( 2+ s ) ⎜
⎟ +10 ( 2+ s ) ( 25s 2 + 60 s +100 )+10 ( s 2 + 4 )
s2 +4
⎝
⎠
V2 ( s ) =
=
( 2+ s ) (3+ s)− 2
( s 2 + 4 ) ( s +1)( s + 4 )
=
Next, partial fraction expansion gives
25s 3 +120s 2 + 220s + 240
( s 2 + 4 ) ( s +1)( s + 4 )
V2 ( s ) =
A
A*
B
C
+
+
+
s + j 2 s − j 2 s +1 s + 4
where
A =
25s 3 +120 s 2 + 220 s + 240
( s +1) ( s + 4 ) ( s − j 2 )
s =− j 2
=
−240− j 240
= 6 + j6
−40
A* = 6 − j 6
25s 3 +120 s 2 + 220 s + 240
B =
( s2 +4) ( s+4)
25s 3 +120 s 2 + 220 s + 240
C =
( s 2 + 4 ) ( s +1)
Then
V2 ( s ) =
s =−1
s =−4
=
115 23
=
15
3
=
−320 16
=
−60
3
6+ j 6 6− j 6 23 3 16 3
+
+
+
s + j 2 s − j 2 s +1 s + 4
Finally
v2 (t ) = 12 cos 2 t + 12sin 2 t +
23 − t 16 −4t
e + e V t≥0
3
3
P 14.7-23 The motor circuit for driving the snorkel shown in Figure P 14.7-23a is shown in
Figure P 14.7-23b. Find the motor current I2(s) when the initial conditions are i1(0 –) = 2 A and
i2(0 –) = 3 A. Determine i2(t) and sketch it for 10 s. Does the motor current smoothly drive the
snorkel?
Solution:
Here are the equations describing the coupled coils:
di1
di
+M 2
dt
dt
di
di
v2 (t ) = L2 2 + M 1
dt
dt
v1 (t ) = L1
⇒ V1 ( s) = 3 ( s I1 ( s) − 2 ) + ( sI 2 ( s) − 3) = 3s I1 ( s) + sI 2 ( s ) − 9
⇒ V2 ( s) = s( I1 ( s ) − 2 ) + 2( sI 2 ( s) −3) = sI1 ( s) + 2sI 2 ( s ) − 8
Writing mesh equations:
5
5
= 2 ( I1 ( s ) + I 2 ( s ) ) + V1 = 2 ( I1 ( s ) + I 2 ( s ) ) + 3s I1 ( s ) + sI 2 ( s ) − 9 ⇒ ( 3s + 2 ) I1 + ( s + 2 ) I 2 = 9 +
s
s
V1 ( s ) = V2 ( s ) + 1I 2 ( s ) ⇒ 3s I1 ( s ) + sI 2 ( s ) − 9 = sI1 ( s ) + 2 sI 2 ( s ) − 8+ I 2 ( s ) ⇒ 2s I1 − ( s +1) I 2 =1
Solving the mesh equations for I2(s):
I2 ( s ) =
15s + 8
3s + 1.6
0.64
2.36
=
+
=
2
s + 0.26
s + 1.54
5s + 9s + 2 ( s + 0.26 )( s +1.54 )
Taking the inverse Laplace transform:
i2 (t ) = 0.64e −0.26t + 2.36e −1.54t A for t > 0
P14.7-24
Using Laplace transforms, find vo(t) for t > 0 for the circuit shown in Figure 14.7-24.
Figure 14.7-24
Solution:
Find vo(t) for t > 0 for this circuit:
For t < 0, with the circuit at steady state, we have
v ( 0 ) = 2 V and i ( 0 ) =
For t > 0 we have
Apply KVL to the left mesh to get
⎛ 1 d
⎞
10 ⎜
v C (t ) ⎟ + v C (t ) − 8 = 0 ⇒
⎝ 20 dt
⎠
Apply KVL to the right mesh to get
1 d
vC (t ) + vC (t ) = 8
2 dt
1
A
2
4
d
d
i L ( t ) + 12 i L ( t ) − 3 v C ( t ) = 0 ⇒ 4 i L ( t ) + 12 i L ( t ) = 3 v C ( t )
dt
dt
Take the Laplace transform of these differential equations to get
1
8
⎡⎣ sVC ( s ) − v C ( 0 ) ⎤⎦ + VC ( s ) =
⇒
2
s
1
8
2 s + 16
⎡⎣ sVC ( s ) − 2 ⎤⎦ + VC ( s ) =
⇒ VC ( s ) =
2
s
s ( s + 2)
and
1⎤
⎡
4 ⎡⎣ s I L ( s ) − i L ( 0 ) ⎤⎦ + 12 I L ( s ) = 3VC ( s ) ⇒ 4 ⎢ s I L ( s ) − ⎥ + 12 I L ( s ) = 3VC ( s )
2⎦
⎣
3
1
⇒ ( s + 3) I L ( s ) = VC ( s ) +
4
2
3 ⎛ 2 s + 16 ⎞ 1
⇒ ( s + 3) I L ( s ) = ⎜⎜
⎟+
4 ⎝ s ( s + 2 ) ⎟⎠ 2
1 2 5
s + s + 12
2
⇒ IL ( s ) = 2
s ( s + 2 )( s + 3)
Taking the inverse Laplace transform gives
Finally
9
−
2
3
⇒ IL ( s ) = + 2 +
s s+2 s+3
9
⎛
⎞
i L ( t ) = ⎜ 2 − e − 2 t + 3 e − 3t ⎟ u ( t ) A
2
⎝
⎠
−2t
v o ( t ) = 12 i L ( t ) = 24 − 54 e + 36 e − 3t u ( t ) V
(
)
(checked with LNAP 2/28/05)
P14.7-25
The circuit shown in Figure P14.7-25 is at steady
state before the switch opens at time t = 0.
Determine the inductor voltage v(t) for t > 0.
Figure P14.7-25
Solution:
Determine the initial conditions, i.e. the inductor current and capacitor voltage at t = 0, as shown
in the circuit on the left below. Use those initial conditions to represent the circuit in the sdomain as shown in the circuit on the left below.
Analysis of the s-domain circuit shows that
⎛
1000 ⎞
1000 ⎞
⎛
⎟ (1.5 ) ⎜ 40 s +
⎟
3.846 s ⎠
12 s + 78
3.846 ⎠
⎝
⎝
V (s) = −
=
= 2
1000
1000
s + 8 s + 52
5 s + 40 +
5 s 2 + 40 s +
3.846 s
3.846
The denominator does not factor any further in the real numbers. Let’s complete the square in the
denominator
12 ( s + 4 ) + 30
12 ( s + 4 )
5 ( 6)
12 s + 78
12 s + 78
12 s + 78
V (s) = 2
= 2
=
=
=
+
2
2
2
2
s + 8 s + 52 ( s + 8 s + 16 ) + 36 ( s + 4 ) + 36 ( s + 4 ) + 36 ( s + 4 ) + 6 ( s + 4 )2 + 62
( −1.5) ⎜ 40 +
Now use the property e − at f ( t ) ↔ F ( s + a ) and the Laplace transform pairs
sin ωt for t ≥ 0 ↔
ω
s + ω2
2
and cos ωt for t ≥ 0 ↔
s
s + ω2
2
to find the inverse Laplace transform
v ( t ) = e − 4 t ℒ-1[
5 ( 6)
12 s
+ 2
] = e − 4t [ 12 cos(6t) + 5 sin (6t) ] for t > 0
2
2
s +6 s +6
2
P14.7-26
The circuit shown in Figure P14.7-26 is at
steady state before the switch opens at time
t = 0. Determine the inductor voltage v(t)
for t > 0.
Figure P14.7-26
Solution
The circuit shown in Figure P14.7-22 is at
steady state before the switch opens at time
t = 0. Determine the inductor voltage v(t)
for t > 0.
Determine the initial conditions, i.e. the
inductor current and capacitor voltage at t = 0,
as shown in the circuit on the left below. Use
those initial conditions to represent the circuit
in the s-domain as s shown in the circuit on the
left below.
Figure P14.7-26
Analysis of the s-domain circuit shows that
⎛
⎞
⎜
⎟ ⎛ 12 ⎞
−12
−144
−60
⎟⎜ ⎟ =
V (s) = ⎜
= 2
⎜ 2.4 s + 12 + 1000 ⎟ ⎝ s ⎠ 2.4 s 2 + 12 s + 1000 s + 5 s + 48.5
⎜
8.59 s ⎟⎠
8.59
⎝
The denominator does not factor any further in the real numbers. Let’s complete the square in the
denominator
V (s) =
−9.23 ( 6.5 )
−60
−60
−60
=
=
=
2
2
2
2
s + 5 s + 48.5 ( s + 2.5 ) + 42.25 ( s + 2.5 ) + 6.5
( s + 2.5) + 6.5 2
2
Now use e − at f ( t ) ↔ F ( s + a ) and sin ωt for t > 0 ↔
transform
v ( t ) = e − 2.5t ℒ-1[
−9.23 ( 6.5 )
( s + 2.5)
2
+ 6.5
2
ω
to find the inverse Laplace
s + ω2
2
] = −9.23 e − 2.5t sin (6.5 t) for t > 0
Section 14.8 Transfer Functions
P14.8‐1 The input to the circuit shown in
Figure P14.8‐1 is the voltage v i ( t ) and
the output is the voltage v o ( t ) .
Determine the values of L, C, k, R1 and R2
that cause the step response of this
circuit to be:
(
)
v o ( t ) = 5 + 20 e−5000 t − 25 e−4000t u ( t ) V
Figure 14.8‐1
Answer: One solution is R1 = 400 Ω, L = 0.1 H, k = 5 V/V, C = 0.1 μF, R1 = 2 kΩ.
Solution: First, determine the transfer function from the circuit. To do so, represent the circuit in the s‐
domain as shown below.
Applying voltage division twice
Va ( s ) =
R1
R1 + L s
R1
Vi ( s ) =
L V (s)
R i
s+ 1
L
k
R 2C
and V o ( s ) =
k Va ( s ) =
V (s)
1
1 a
R2 +
s+
Cs
R 2C
1
Cs
Substituting Va(s) from the first equation into the second, the transfer function of this circuit is
H (s) =
Vo ( s )
Vi ( s )
k R1
=
R 2C L
R1 ⎞ ⎛
⎛
1 ⎞
⎟
⎜ s + ⎟ ⎜⎜ s +
L ⎠⎝
R 2C ⎟⎠
⎝
Next, determine the transfer function from the step response.
H (s)
5
20
25
10 8
.
= L ⎡⎣5 + 20 e −5000 t − 25 e−4000 t ⎤⎦ = +
−
=
s
s s + 5000 s + 4000 s ( s + 5000 )( s + 4000 )
Performing partial fraction expansion:
k R1
R 2C L
10 8
= H (s) =
( s + 5000 )( s + 4000 )
R1 ⎞ ⎛
⎛
1 ⎞
⎟
⎜ s + ⎟ ⎜⎜ s +
L ⎠⎝
R 2C ⎟⎠
⎝
Equality requires
R1
L
and
= 4000 and
1
= 5000 or
R 2C
R1
L
= 5000 and
1
= 5000
R 2C
108 = k ( 4000 )( 5000 ) ⇒ k = 5 V/V
The solution is not unique. One solution is R1 = 400 Ω, L = 0.1 H, k = 5 V/V, C = 0.1 μF, R1 = 2 kΩ.
P14.8‐2 The input to the circuit shown in Figure P14.8‐2 is
the voltage v i ( t ) and the output is the voltage v o ( t ) .
Determine the step response of this circuit.
Figure 14.8‐2
Solution:
The transfer function of this circuit is
1
C s ⎛ R2 ⎞
H (s) =
⎜1 +
⎟
1 ⎜
R1 ⎟⎠
⎝
R+ Ls+
Cs
1
10 s 2 +
L C ⎛ R2 ⎞
=
⎜1 +
⎟⎟
R
1 ⎜
R
2
1
⎝
⎠
s + s+
L
LC
Ls +
s 2 + 500, 000
= 2
s + 200 s + 50, 000
⎡ 10 s 2 + 500, 000
⎡ H (s) ⎤
−1
The step response is v o ( t ) = L ⎢
⎥=L ⎢
s
⎢⎣ s s 2 + 200 s + 50, 000
⎣
⎦
−1
(
)
⎤
⎥.
⎥⎦
Let’s do some algebra:
−2000
10 s 2 + 500, 000
10
= + 2
2
s s + 200 s + 50, 000
s s + 200 s + 50, 000
(
)
=
−2000
10
+
s ( s + 100 )2 + 2002
=
10
200
− 10
2
s
( s + 100 ) + 2002
The step response is
⎡10
⎤
200
v o ( t ) = L −1 ⎢ − 10
⎥ = 10 − e−100 t ⎡⎣10sin ( 200 t ) ⎤⎦ = ⎣⎡10 + 10 e−100 t cos ( 200 t + 90° ) ⎦⎤ u ( t ) V
2
2
( s + 100 ) + 200 ⎦⎥
⎣⎢ s
P14.8‐3 The input to the circuit shown in Figure P14.8‐3
is the voltage v i ( t ) and the output is the voltage v o ( t ) .
Determine the impulse response of this circuit.
Figure 14.8‐3
Solution:
Represent the circuit in the s‐domain:
The transfer function of this circuit is
Vo ( s )
5s
s2
H (s) =
=
=
V i ( s ) 25 + 20 + 5 s s 2 + 5 s + 4
s
⎡
⎤
s2
The impulse response is h ( t ) = L −1 ⎣⎡ H ( s ) ⎦⎤ = L −1 ⎢ 2
⎥.
⎣s + 5s + 4⎦
Let’s do some algebra:
−1 ⎞
⎛ −16
⎜ −3
⎟
s2
5s + 4
5s + 4
= 1− 2
= 1−
= 1− ⎜
+ 3 ⎟
2
s +5s + 4
s +5s + 4
( s + 4 )( s + 1)
⎜ s + 4 s +1⎟
⎝
⎠
The impulse response is
16
1 ⎤
⎡
⎢
⎥
1 ⎞
⎛ 16
h ( t ) = L −1 ⎢1 − 3 + 3 ⎥ = δ ( t ) + ⎜ − e −4 t + et ⎟ u ( t ) V
3 ⎠
⎝ 3
⎢ s + 4 s + 1⎥
⎣
⎦
P14.8‐4 The input to the circuit shown in Figure P14.8‐4
is the voltage v i ( t ) and the output is the voltage v o ( t ) .
Determine the step response of this circuit.
Answer: step response = 5 − ( 5 + 20 t ) e −4 t u ( t )
(
)
Figure 14.8‐4
Solution:
Represent the circuit in the s‐domain as
shown.
Using voltage division
10
16
s
Va ( s ) =
Vi ( s ) = 2
Vi ( s )
20 5
+
8
s
+
s
16
+ s+5
s 8
Recognizing the noninverting amplifier:
80
⎛ 20 ⎞
V o ( s ) = ⎜1 + ⎟ V a ( s ) = 2
Vi ( s )
s + 8 s + 16
5 ⎠
⎝
The transfer function of this circuit is
H (s) =
Vo ( s )
Vi ( s )
=
80
80
=
s + 8 s + 16 ( s + 4 ) 2
2
⎡ 80 ⎤
⎡ H (s) ⎤
−1
step response = L −1 ⎢
⎥=L ⎢
2⎥.
⎣ s ⎦
⎣⎢ s ( s + 4 ) ⎦⎥
Performing partial fraction expansion:
80
80
80
A
16 +
+ −4 2
2 =
s s + 4 ( s + 4)
s ( s + 4)
The step response is
Multiplying both sides by s ( s + 4 ) and equating coefficients of like powers of s:
2
80 = 5 ( s + 4 ) + A s ( s + 4 ) + ( −20 ) s ⇒
2
A = −5
The step response is
⎡ 5 −5
−20 ⎤
−4 t
+
u (t ) V
step response = L −1 ⎢ +
2 ⎥ = 5 − ( 5 + 20 t ) e
⎣⎢ s s + 4 ( s + 4 ) ⎦⎥
(
)
P14.8-5 The input to the circuit shown in Figure P14.8-5 is the voltage, vi(t), of the independent voltage
source. The output is the voltage, vo(t), across the 5-kΩ resistor. Specify values of the resistance, R, the
capacitance, C, and the inductance, L, such that the transfer function of this circuit is given by
H (s) =
Vo ( s )
Vi ( s )
=
15 × 106
( s + 2000 )( s + 5000 )
Answer: R = 5k Ω, C = 0.5 μF, and L = 1 H (one possible solution)
Figure P14.8‐5
Solution:
The transfer function can also be calculated from the circuit itself. The circuit can be represented in the
frequency domain as
We can save ourselves some work be noticing that the 10000 ohm resistor, the resistor labeled R and
the op amp comprise a non‐inverting amplifier. Thus
⎛
R ⎞⎟
Va ( s ) = ⎜⎜1 +
V s
⎜⎝ 10000 ⎠⎟⎟ c ( )
Now, writing node equations,
Vc ( s ) −Vi ( s )
Vo ( s ) −Va ( s ) Vo ( s )
+ CsVc ( s ) = 0 and
+
=0
1000
Ls
5000
Solving these node equations gives
1 ⎛⎜
R ⎞⎟ 5000
⎟
⎜⎜⎝1 +
1000 C
10000 ⎠⎟ L
H (s) =
⎛
⎞⎛
⎞
⎜⎜ s + 1 ⎟⎟ ⎜⎜ s + 5000 ⎟⎟
⎟⎜
⎜⎝ 1000C ⎠⎝
L ⎠⎟
Comparing these two equations for the transfer function gives
⎛
⎞
⎛
⎞
⎜⎜ s + 1 ⎟⎟ = ( s + 2000) or ⎜⎜ s + 1 ⎟⎟ = ( s + 5000)
⎝⎜ 1000C ⎠⎟
⎝⎜ 1000C ⎠⎟
⎛
⎞
⎛
⎞
⎜⎜ s + 5000 ⎟⎟ = ( s + 2000) or ⎜⎜ s + 5000 ⎟⎟ = ( s + 5000)
⎜⎝
⎝⎜
L ⎟⎠
L ⎠⎟
R ⎞ 5000
1 ⎛
= 15 × 10 6
⎜1 +
⎟
1000C ⎝ 10000 ⎠ L
The solution isn’t unique, but there are only two possibilities. One of these possibilities is
⎛
⎞
⎜⎜ s + 1 ⎟⎟ = ( s + 2000) ⇒ C = 0.5 μ F
⎜⎝ 1000C ⎠⎟
5000 ⎞
⎛
⎜s +
⎟ = ( s + 5000) ⇒ L = 1 H
⎝
L ⎠
⎛
⎞
⎜⎜1 + R ⎟⎟ 5000 = 15×106 ⇒ R = 5 kΩ
⎟
⎜
1000 0.5×10 ⎝ 10000 ⎠ 1
(
1
6
)
(Checked using LNAP, 12/29/02)
P14.8-6 The input to the circuit shown in Figure P14.8-6
is the voltage, vi(t), of the independent voltage source.
The output is the voltage, vo(t), across the 10-kΩ resistor.
Specify values of the resistances, R1 and R2, such that the
step response of this circuit is given by
vo(t) = –4(1 – e–250t)u(t) V
Answer: R1 = 10 kΩ and R2 = 40 kΩ
Figure P 14.8‐6
Solution:
The transfer function of the circuit is
R2
1
1+ R2 C s
R1 C
H (s) = −
=−
1
R1
s+
R2 C
The give step response is vo ( t ) = −4 (1 − e −250 t ) u ( t ) V . The correspond transfer function is calculated as
H (s)
4 ⎞
−1000
−1000
⎛4
= L −4 (1 − e − 250 t ) u ( t ) = − ⎜ −
⇒ H (s) =
⎟=
s
s + 250
⎝ s s + 250 ⎠ s ( s + 250 )
{
}
Comparing these results gives
1
1
1
= 250 ⇒ R 2 =
=
= 40 kΩ
R2 C
250 C 250 ( 0.1× 10 − 6 )
1
1
1
= 1000 ⇒ R1 =
=
= 10 kΩ
R1 C
1000 C 1000 ( 0.1×10 − 6 )
(Checked using LNAP, 12/29/02)
P14.8‐7 The input to the circuit shown in Figure P14.8‐7 is
the voltage v i ( t ) and the output is the voltage v o ( t ) .
Determine the step response of this circuit.
Answer: v o ( t ) = ( 4 × 10 3 ) t u ( t ) V
Figure 14.8‐7
Solution: First, determine the transfer function from the
circuit. To do so, represent the circuit in the s‐domain as
shown.
Notice that Va(s) is the node voltage at both node a and
node b. Apply KCL at node a to get
Va ( s )
10, 000
+
Va ( s ) − Vo ( s )
30, 000
=0 ⇒
⎛ 30, 000 ⎞
V o ( s ) = ⎜1 +
⎟V a ( s ) = 4V a ( s )
⎝ 10, 000 ⎠
Apply KCL at node b to get
Vi ( s ) − Va ( s )
10, 000
=
Vi ( s )
Va ( s ) − Vo ( s ) Va ( s )
V a ( s ) V a ( s ) − 4V a ( s ) sV a ( s )
+
⇒
=
+
+
7
10
30, 000
10, 000 10, 000
30, 000
10 7
s
V i ( s ) sV a ( s ) sV o ( s )
⇒
=
=
10, 000
10 7
4 ×10 7
⇒ Vo ( s ) =
H (s) =
The transfer function is
The step response is
Vo ( s )
Vi ( s )
=
4 ×10 7
4 × 10 3
Vi ( s ) =
Vi ( s )
10, 000 s
s
4 × 10 3
s
3
⎡ H (s) ⎤
−1 ⎡ 4 × 10 ⎤
3
L ⎢
⎥=L ⎢
⎥ = ( 4 ×10 ) t u ( t ) V .
2
⎣ s
⎦
⎣ s ⎦
−1
P14.8‐8 The input to the circuit shown in Figure P14.8‐8
is the voltage v i ( t ) and the output is the voltage
v o ( t ) . Determine the step response of this circuit.
⎡ ⎛4
2
⎞⎤
Answer: v o ( t ) = ⎢ 2 − ⎜ e −1000 t + e −4000 t ⎟ ⎥ u ( t ) V
3
⎠⎦
⎣ ⎝3
Figure 14.8‐8
Solution: First, determine the transfer function from
the circuit. To do so, represent the circuit in the s‐
domain as shown.
Recognizing the noninverting amplifier we
⎛ 10 7
⎛
10 7 ⎞
4 ⎞
||10
⎜
⎟
⎜
3 |⎟
s
s
+
10
V a ( s ) = ⎜1 +
⎟ V ( s ) = ⎜1 +
⎟V ( s )
10 4 ⎟ i
10 4 ⎟ i
⎜
⎜
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎛ s + 2000 ⎞
=⎜
⎟Vi ( s )
⎝ s + 1000 ⎠
(Alternately, this equation can obtained by applying
KCL at the inverting input node of the op amp.)
Use voltage division to write
Vo ( s ) =
The transfer function is
2000
4000
Va ( s ) =
Va ( s )
2000 + 0.5 s
s + 2000
H (s) =
Vo ( s )
Vi ( s )
=
4000 ( s + 2000 )
( s + 1000 )( s + 4000 )
The step response is
⎡ 4000 ( s + 2000 ) ⎤
⎡ H (s) ⎤
−4 3
−2 3 ⎤
−1
−1 ⎡ 2
.
+
v o ( t ) = L −1 ⎢
⎥=L ⎢ +
⎥=L ⎢
⎣ s s + 1000 s + 4000 ⎥⎦
⎣ s ⎦
⎣ s ( s + 1000 )( s + 4000 ) ⎦
⎡ ⎛4
2
⎞⎤
v o ( t ) = ⎢ 2 − ⎜ e−1000t + e−4000 t ⎟ ⎥ u ( t ) V
3
⎠⎦
⎣ ⎝3
P14.8‐9 The input to the circuit shown in Figure P14.8‐9 is the voltage, vi(t), of the independent voltage
source. The output is the voltage, vo(t). The step response of this circuit is
vo(t) = 0.5(1 + e–4t)u(t) V
Determine the values of the inductance, L, and the resistance, R.
Answer: L = 6 H and R = 12 Ω
Figure P14.8‐9
Solution:
R
s+
Vo ( s )
R+ L