INSTRUCTOR'S SOLUTION MANUAL
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Circuit Variables
1
Assessment Problems
AP 1.1 Use a product of ratios to convert two-thirds the speed of light from meters
per second to miles per second:
2 3 × 108 m 100 cm
1 in
1 ft
1 mile
124,274.24 miles
·
·
·
·
=
3
1s
1m
2.54 cm 12 in 5280 feet
1s
Now set up a proportion to determine how long it takes this signal to travel
1100 miles:
124,274.24 miles
1100 miles
=
1s
xs
Therefore,
x=
1100
= 0.00885 = 8.85 × 10−3 s = 8.85 ms
124,274.24
AP 1.2 To solve this problem we use a product of ratios to change units from
dollars/year to dollars/millisecond. We begin by expressing $10 billion in
scientific notation:
$100 billion = $100 × 109
Now we determine the number of milliseconds in one year, again using a
product of ratios:
1 year
1 day
1 hour 1 min
1 sec
1 year
·
·
·
·
=
365.25 days 24 hours 60 mins 60 secs 1000 ms 31.5576 × 109 ms
Now we can convert from dollars/year to dollars/millisecond, again with a
product of ratios:
$100 × 109
1 year
100
·
=
= $3.17/ms
9
1 year
31.5576 × 10 ms
31.5576
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1–2
CHAPTER 1. Circuit Variables
AP 1.3 Remember from Eq. (1.2), current is the time rate of change of charge, or
i = dq
In this problem, we are given the current and asked to find the total
dt
charge. To do this, we must integrate Eq. (1.2) to find an expression for
charge in terms of current:
q(t) =
t
Z
i(x) dx
0
We are given the expression for current, i, which can be substituted into the
above expression. To find the total charge, we let t → ∞ in the integral. Thus
we have
qtotal =
=
Z
∞
0
20e−5000x dx =
20 −5000x ∞
20
e
=
(e−∞ − e0)
−5000
−5000
0
20
20
(0 − 1) =
= 0.004 C = 4000 µC
−5000
5000
AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or
i = dq
. In this problem we are given an expression for the charge, and asked to
dt
find the maximum current. First we will find an expression for the current
using Eq. (1.2):
i=
dq
d 1
t
1
=
−
+ 2 e−αt
2
dt
dt α
α α
d 1
d t −αt
d 1 −αt
=
−
e
−
e
2
dt α
dt α
dt α2
1 −αt
t
1
= 0−
e
− α e−αt − −α 2 e−αt
α
α
α
= −
1
1 −αt
+t+
e
α
α
= te−αt
Now that we have an expression for the current, we can find the maximum
value of the current by setting the first derivative of the current to zero and
solving for t:
di
d
= (te−αt) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0
dt
dt
Since e−αt never equals 0 for a finite value of t, the expression equals 0 only
when (1 − αt) = 0. Thus, t = 1/α will cause the current to be maximum. For
this value of t, the current is
i=
1 −α/α
1
e
= e−1
α
α
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Problems
1–3
Remember in the problem statement, α = 0.03679. Using this value for α,
i=
1
e−1 ∼
= 10 A
0.03679
AP 1.5 Start by drawing a picture of the circuit described in the problem statement:
Also sketch the four figures from Fig. 1.6:
[a] Now we have to match the voltage and current shown in the first figure
with the polarities shown in Fig. 1.6. Remember that 4A of current
entering Terminal 2 is the same as 4A of current leaving Terminal 1. We
get
(a) v = −20 V,
(c) v = 20 V,
i = −4 A; (b) v = −20 V,
i = −4 A;
(d) v = 20 V,
i = 4A
i = 4A
[b] Using the reference system in Fig. 1.6(a) and the passive sign convention,
p = vi = (−20)(−4) = 80 W. Since the power is greater than 0, the box is
absorbing power.
[c] From the calculation in part (b), the box is absorbing 80 W.
AP 1.6 [a] Applying the passive sign convention to the power equation using the
voltage and current polarities shown in Fig. 1.5, p = vi. To find the time
at which the power is maximum, find the first derivative of the power
with respect to time, set the resulting expression equal to zero, and solve
for time:
p = (80,000te−500t)(15te−500t) = 120 × 104 t2 e−1000t
dp
= 240 × 104 te−1000t − 120 × 107 t2e−1000t = 0
dt
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1–4
CHAPTER 1. Circuit Variables
Therefore,
240 × 104 − 120 × 107 t = 0
Solving,
t=
240 × 104
= 2 × 10−3 = 2 ms
120 × 107
[b] The maximum power occurs at 2 ms, so find the value of the power at 2
ms:
p(0.002) = 120 × 104 (0.002)2 e−2 = 649.6 mW
[c] From Eq. (1.3), we know that power is the time rate of change of energy,
or p = dw/dt. If we know the power, we can find the energy by
integrating Eq. (1.3). To find the total energy, the upper limit of the
integral is infinity:
wtotal =
Z
∞
0
120 × 104 x2e−1000x dx
120 × 104 −1000x
=
e
[(−1000)2 x2 − 2(−1000)x + 2)
(−1000)3
=0−
∞
0
120 × 104 0
e (0 − 0 + 2) = 2.4 mJ
(−1000)3
AP 1.7 At the Oregon end of the line the current is leaving the upper terminal, and
thus entering the lower terminal where the polarity marking of the voltage is
negative. Thus, using the passive sign convention, p = −vi. Substituting the
values of voltage and current given in the figure,
p = −(800 × 103 )(1.8 × 103 ) = −1440 × 106 = −1440 MW
Thus, because the power associated with the Oregon end of the line is
negative, power is being generated at the Oregon end of the line and
transmitted by the line to be delivered to the California end of the line.
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Problems
1–5
Chapter Problems
P 1.1
[a] We can set up a ratio to determine how long it takes the bamboo to grow
10 µm First, recall that 1 mm = 103 µm. Let’s also express the rate of
growth of bamboo using the units mm/s instead of mm/day. Use a
product of ratios to perform this conversion:
250 mm
1 day
1 hour 1 min
250
10
·
·
·
=
=
mm/s
1 day 24 hours 60 min 60 sec
(24)(60)(60)
3456
Use a ratio to determine the time it takes for the bamboo to grow 10 µm:
10 × 10−6 m
10/3456 × 10−3 m
=
1s
xs
[b]
P 1.2
so
x=
10 × 10−6
= 3.456 s
10/3456 × 10−3
1 cell length 3600 s (24)(7) hr
·
·
= 175,000 cell lengths/week
3.456 s
1 hr
1 week
Volume = area × thickness
Convert values to millimeters, noting that 10 m2 = 106 mm2
106 = (10 × 106 )(thickness)
⇒ thickness =
106
= 0.10 mm
10 × 106
P 1.3
(260 × 106 )(540)
= 104.4 gigawatt-hours
109
P 1.4
[a]
20,000 photos
x photos
=
3
(11)(15)(1) mm
1 mm3
x=
[b]
16 × 230 bytes
x bytes
=
(11)(15)(1) mm3
(0.2)3 mm3
x=
P 1.5
(20,000)(1)
= 121 photos
(11)(15)(1)
(16 × 230 )(0.008)
= 832,963 bytes
(11)(15)(1)
(480)(320) pixels 2 bytes 30 frames
·
·
= 9.216 × 106 bytes/sec
1 frame
1 pixel
1 sec
(9.216 × 106 bytes/sec)(x secs) = 32 × 230 bytes
x=
32 × 230
= 3728 sec = 62 min ≈ 1 hour of video
9.216 × 106
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1–6
CHAPTER 1. Circuit Variables
5280 ft 2526 lb 1 kg
·
·
= 20.5 × 106 kg
1 mi 1000 ft 2.2 lb
P 1.6
(4 cond.) · (845 mi) ·
P 1.7
w = qV = (1.6022 × 10−19 )(6) = 9.61 × 10−19 = 0.961 aJ
P 1.8
n=
P 1.9
C/m3 =
35 × 10−6 C/s
= 2.18 × 1014 elec/s
1.6022 × 10−19 C/elec
1.6022 × 10−19 C 1029 electrons
×
= 1.6022 × 1010 C/m3
3
1 electron
1m
Cross-sectional area of wire = (0.4 × 10−2 m)(16 × 10−2 m) = 6.4 × 10−4 m2
C/m = (1.6022 × 1010 C/m3)(6.4 × 10−4 m2 ) = 10.254 × 106 C/m
C
C
m
Therefore, i
= (10.254 × 106 )
× avg vel
sec
m
s
Thus, average velocity =
P 1.10
i
1600
=
= 156.04 µm/s
6
10.254 × 10
10.254 × 106
First we use Eq. (1.2) to relate current and charge:
i=
dq
= 20 cos 5000t
dt
Therefore, dq = 20 cos 5000t dt
To find the charge, we can integrate both sides of the last equation. Note that
we substitute x for q on the left side of the integral, and y for t on the right
side of the integral:
Z
q(t)
q(0)
dx = 20
Z
t
0
cos 5000y dy
We solve the integral and make the substitutions for the limits of the integral,
remembering that sin 0 = 0:
q(t) − q(0) = 20
sin 5000y
5000
t
=
0
20
20
20
sin 5000t −
sin 5000(0) =
sin 5000t
5000
5000
5000
But q(0) = 0 by hypothesis, i.e., the current passes through its maximum
value at t = 0, so q(t) = 4 × 10−3 sin 5000t C = 4 sin 5000t mC
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Problems
P 1.11
1–7
[a] In Car A, the current i is in the direction of the voltage drop across the 12
V battery(the current i flows into the + terminal of the battery of Car
A). Therefore using the passive sign convention,
p = vi = (30)(12) = 360 W.
Since the power is positive, the battery in Car A is absorbing power, so
Car A must have the ”dead” battery.
[b] w(t) =
Z
t
0
w(60) =
p dx;
Z
1 min = 60 s
60
360 dx
0
w = 360(60 − 0) = 360(60) = 21,600 J = 21.6 kJ
P 1.12
p = (12)(100 × 10−3 ) = 1.2 W;
w(t) =
P 1.13
p = vi;
Z
0
t
p dt
w=
w(14,400) =
Z
0
t
4 hr ·
Z
3600 s
= 14,400 s
1 hr
14,400
0
1.2 dt = 1.2(14,400) = 17.28 kJ
p dx
Since the energy is the area under the power vs. time plot, let us plot p vs. t.
Note that in constructing the plot above, we used the fact that 40 hr
= 144,000 s = 144 ks
p(0) = (1.5)(9 × 10−3 ) = 13.5 × 10−3 W
p(144 ks) = (1)(9 × 10−3 ) = 9 × 10−3 W
1
w = (9 × 10−3 )(144 × 103 ) + (13.5 × 10−3 − 9 × 10−3 )(144 × 103 ) = 1620 J
2
P 1.14
Assume we are standing at box A looking toward box B. Then, using the
passive sign convention p = −vi, since the current i is flowing into the −
terminal of the voltage v. Now we just substitute the values for v and i into
the equation for power. Remember that if the power is positive, B is absorbing
power, so the power must be flowing from A to B. If the power is negative, B
is generating power so the power must be flowing from B to A.
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1–8
CHAPTER 1. Circuit Variables
[a] p = −(125)(10) = −1250 W
[b] p = −(−240)(5) = 1200 W
1250 W from B to A
1200 W from A to B
[c] p = −(480)(−12) = 5760 W
5760 W from A to B
[d] p = −(−660)(−25) = −16,500 W
P 1.15
16,500 W from B to A
[a]
p = vi = (40)(−10) = −400 W
Power is being delivered by the box.
[b] Entering
[c] Gaining
P 1.16
[a] p = vi = (−60)(−10) = 600 W, so power is being absorbed by the box.
[b] Entering
[c] Losing
P 1.17
[a] p = vi = (0.05e−1000t )(75 − 75e−1000t) = (3.75e−1000t − 3.75e−2000t) W
dp
= −3750e−1000t + 7500e−2000t = 0
dt
2 = e1000t
so
ln 2 = 1000t
so
thus
2e−2000t = e−1000t
p is maximum at t = 693.15 µs
pmax = p(693.15 µs) = 937.5 mW
[b] w =
Z
=
P 1.18
∞
0
[3.75e
−1000t
− 3.75e
−2000t
3.75 −1000t
3.75 −2000t ∞
] dt =
e
−
e
−1000
−2000
0
3.75
3.75
−
= 1.875 mJ
1000 2000
[a] p = vi = 0.25e−3200t − 0.5e−2000t + 0.25e−800t
p(625 µs) = 42.2 mW
[b]
w(t)
=
Z
0
t
(0.25e−3200t − 0.5e−2000t + 0.25e−800t )
= 140.625 − 78.125e−3200t + 250e−2000t − 312.5e−800t µJ
w(625 µs)
= 12.14 µJ
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Problems
1–9
[c] wtotal = 140.625 µJ
P 1.19
[a] 0 s ≤ t < 1 s:
v = 5 V;
i = 20t A;
p = 100t W
i = 20 A;
p=0W
1 s < t ≤ 3 s:
v = 0 V;
3 s ≤ t < 5 s:
v = −5 V; i = 80 − 20t A;
p = 100t − 400 W
5 s < t ≤ 7 s:
v = 5 V;
i = 20t − 120 A; p = 100t − 600 W
t > 7 s:
v = 0 V;
i = 20 A;
p=0W
[b] Calculate the area under the curve from zero up to the desired time:
P 1.20
w(1)
=
1
(1)(100)
2
= 50 J
w(6)
=
1
(1)(100)
2
− 12 (1)(100) + 12 (1)(100) − 12 (1)(100) = 0 J
w(10)
=
w(6) + 12 (1)(100) = 50 J
[a] v(10 ms) = 400e−1 sin 2 = 133.8 V
i(10 ms) = 5e−1 sin 2 = 1.67 A
p(10 ms) = vi = 223.80 W
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1–10
CHAPTER 1. Circuit Variables
[b]
= vi = 2000e−200t sin2 200t
1
−200t 1
= 2000e
− cos 400t
2 2
= 1000e−200t − 1000e−200t cos 400t
p
w
Z
=
∞
1000e
0
−200t
e−200t
= 1000
−200
(
∞
∞
0
1000e−200t cos 400t dt
0
e−200t
−1000
[−200 cos 400t + 400 sin 400t]
2 + (400)2
(200)
200
= 5 − 1000
= 5−1
4 × 104 + 16 × 104
= 4 J
w
P 1.21
dt −
Z
)
∞
0
[a]
p
= vi = [16,000t + 20)e−800t][(128t + 0.16)e−800t ]
= 2048 × 103 t2 e−1600t + 5120te−1600t + 3.2e−1600t
= 3.2e−1600t[640,000t2 + 1600t + 1]
dp
dt
= 3.2{e−1600t[1280 × 103 t + 1600] − 1600e−1600t [640,000t2 + 1600t + 1]}
= −3.2e−1600t[128 × 104 (800t2 + t)] = −409.6 × 104 e−1600tt(800t + 1)
dp
Therefore,
= 0 when t = 0
dt
so pmax occurs at t = 0.
[b] pmax
= 3.2e−0 [0 + 0 + 1]
= 3.2 W
[c]
w
w
3.2
=
=
Z
t
Z0 t
0
pdx
640,000x2 e−1600x dx +
Z
0
t
1600xe−1600x dx +
t
Z
0
t
e−1600x dx
640,000e−1600x
=
[256 × 104 x2 + 3200x + 2] +
−4096 × 106
0
t
t
1600e−1600x
e−1600x
(−1600x − 1) +
256 × 104
−1600 0
0
When t → ∞ all the upper limits evaluate to zero, hence
w
(640,000)(2)
1600
1
=
+
+
3.2
4096 × 106
256 × 104 1600
w = 10−3 + 2 × 10−3 + 2 × 10−3 = 5 mJ.
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Problems
P 1.22
[a]
p =
dp
dt
1–11
vi
=
400 × 103 t2 e−800t + 700te−800t + 0.25e−800t
=
e−800t[400,000t2 + 700t + 0.25]
=
{e−800t[800 × 103 t + 700] − 800e−800t [400,000t2 + 700t + 0.25]}
=
[−3,200,000t2 + 2400t + 5]100e−800t
dp
Therefore,
= 0 when 3,200,000t2 − 2400t − 5 = 0
dt
so pmax occurs at t = 1.68 ms.
[b] pmax
= [400,000(.00168)2 + 700(.00168) + 0.25]e−800(.00168)
= 666 mW
[c] w
=
w
=
Z
t
Z0 t
0
pdx
400,000x2 e−800x dx +
Z
0
t
700xe−800x dx +
t
Z
0
t
0.25e−800x dx
400,000e−800x
[64 × 104 x2 + 1600x + 2] +
−512 × 106
0
t
t
700e−800x
e−800x
(−800x − 1) + 0.25
64 × 104
−800 0
0
When t = ∞ all the upper limits evaluate to zero, hence
(400,000)(2)
700
0.25
w=
+
+
= 2.97 mJ.
512 × 106
64 × 104
800
=
P 1.23
[a] p = vi = 2000 cos(800πt) sin(800πt) = 1000 sin(1600πt) W
Therefore, pmax = 1000 W
[b] pmax (extracting) = 1000 W
[c] pavg
2.5×10−3
1
1000 sin(1600πt) dt
2.5 × 10−3 0
2.5×10−3
250
5 − cos 1600πt
4 × 10
=
[1 − cos 4π] = 0
1600π
π
0
Z
=
=
[d]
pavg
=
=
P 1.24
[a] q
Z 15.625×10−3
1
1000 sin(1600πt) dt
15.625 ×10−3 0
15.625×10−3
40
3 − cos 1600πt
64 × 10
= [1 − cos 25π] = 25.46 W
1600π
π
0
=
area under i vs. t plot
=
h
=
1
(5)(4)
2
i
+ (10)(4) + 12 (8)(4) + (8)(6) + 12 (3)(6) × 103
[10 + 40 + 16 + 48 + 9]103 = 123,000 C
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1–12
CHAPTER 1. Circuit Variables
[b] w
Z
=
p dt =
Z
vi dt
v = 0.2 × 10−3 t + 9
0 ≤ t ≤ 15 ks
0 ≤ t ≤ 4000s
i = 15 − 1.25 × 10−3 t
= 135 − 8.25 × 10−3 t − 0.25 × 10−6 t2
p
w1
=
4000
Z
0
(135 − 8.25 × 10−3 t − 0.25 × 10−6 t2) dt
= (540 − 66 − 5.3333)103 = 468.667 kJ
4000 ≤ t ≤ 12,000
i
= 12 − 0.5 × 10−3 t
p
= 108 − 2.1 × 10−3 t − 0.1 × 10−6 t2
w2
=
12,000
Z
4000
(108 − 2.1 × 10−3 t − 0.1 × 10−6 t2) dt
= (864 − 134.4 − 55.467)103 = 674.133 kJ
12,000 ≤ t ≤ 15,000
i =
p =
w3
=
wT
P 1.25
30 − 2 × 10−3 t
270 − 12 × 10−3 t − 0.4 × 10−6 t2
Z
15,000
12,000
(270 − 12 × 10−3 t − 0.4 × 10−6 t2) dt
=
(810 − 486 − 219.6)103 = 104.4 kJ
=
w1 + w2 + w3 = 468.667 + 674.133 + 104.4 = 1247.2 kJ
[a] We can find the time at which the power is a maximum by writing an
expression for p(t) = v(t)i(t), taking the first derivative of p(t)
and setting it to zero, then solving for t. The calculations are shown below:
p
p
dp
dt
dp
dt
t1
= 0 t < 0,
p = 0 t > 40 s
= vi = t(1 − 0.025t)(4 − 0.2t) = 4t − 0.3t2 + 0.005t3 W
0 ≤ t ≤ 40 s
= 4 − 0.6t + 0.015t2 = 0.015(t2 − 40t + 266.67)
= 0
when t2 − 40t + 266.67 = 0
= 8.453 s;
t2 = 31.547 s
(using the polynomial solver on your calculator)
p(t1 )
= 4(8.453) − 0.3(8.453)2 + 0.005(8.453)3 = 15.396 W
p(t2 ) = 4(31.547) − 0.3(31.547)2 + 0.005(31.547)3 = −15.396 W
Therefore, maximum power is being delivered at t = 8.453 s.
[b] The maximum power was calculated in part (a) to determine the time at
which the power is maximum: pmax = 15.396 W (delivered)
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Problems
1–13
[c] As we saw in part (a), the other “maximum” power is actually a
minimum, or the maximum negative power. As we calculated in part (a),
maximum power is being extracted at t = 31.547 s.
[d] This maximum extracted power was calculated in part (a) to determine
the time at which power is maximum: pmax = 15.396 W (extracted)
[e] w =
Z
0
t
pdx =
Z
0
t
(4x − 0.3x2 + 0.005x3 )dx = 2t2 − 0.1t3 + 0.00125t4
w(0)
=
0J
w(30)
= 112.5 J
w(10)
=
112.5 J
w(40)
= 0J
w(20) = 200 J
To give you a feel for the quantities of voltage, current, power, and energy
and their relationships among one another, they are plotted below:
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1–14
P 1.26
CHAPTER 1. Circuit Variables
We use the passive sign convention to determine whether the power equation
is p = vi or p = −vi and substitute into the power equation the values for v
and i, as shown below:
pa
=
vaia = (150 × 103 )(0.6 × 10−3 ) = 90 W
pb
=
vbib = (150 × 103 )(−1.4 × 10−3 ) = −210 W
pc
=
−vcic = −(100 × 103 )(−0.8 × 10−3 ) = 80 W
pd
=
vdid = (250 × 103 )(−0.8 × 10−3 ) = −200 W
pe
=
−veie = −(300 × 103 )(−2 × 10−3 ) = 600 W
pf = vf if = (−300 × 103 )(1.2 × 10−3 ) = −360 W
Remember that if the power is positive, the circuit element is absorbing
power, whereas is the power is negative, the circuit element is developing
power. We can add the positive powers together and the negative powers
together
— if the power balances, these power sums should be equal:
X
Pdev = 210 + 200 + 360 = 770 W;
X
Pabs = 90 + 80 + 600 = 770 W
Thus, the power balances and the total power developed in the circuit is 770
W.
P 1.27
pa
=
−vaia = −(990)(−0.0225) = 22.275 W
pb
=
−vbib = −(600)(−0.03) = 18 W
pc
=
vcic = (300)(0.06) = 18 W
pd
=
vdid = (105)(0.0525) = 5.5125 W
pe
=
−veie = −(−120)(0.03) = 3.6 W
pf
=
vf if = (165)(0.0825) = 13.6125 W
pg
=
−vgig = −(585)(0.0525) = −30.7125 W
ph = vhih = (−585)(0.0825) = −48.2625 W
Therefore,
X
Pabs = 22.275 + 18 + 18 + 5.5125 + 3.6 + 13.6125 = 81 W
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Problems
X
Pdel = 30.7125 + 48.2625 = 78.975 W
X
Pabs 6=
X
1–15
Pdel
Thus, the interconnection does not satisfy the power check.
P 1.28
[a] From the diagram and the table we have
pa
=
−vaia = −(46.16)(−6) = −276.96 W
pb
=
vbib = (14.16)(4.72) = 66.8352 W
pc
=
vcic = (−32)(−6.4) = 204.8 W
pd
=
−vdid = −(22)(1.28) = −28.16 W
pe
=
−veie = −(33.6)(1.68) = −56.448 W
pf
=
vf if = (66)(−0.4) = −26.4 W
pg
=
vg ig = (2.56)(1.28) = 3.2768 W
ph
=
−vhih = −(−0.4)(0.4) = 0.16 W
X
X
Pdel
= 276.96 + 28.16 + 56.448 + 26.4 = 387.968 W
Pabs
= 66.8352 + 204.8 + 3.2768 + 0.16 = 275.072 W
Therefore,
X
Pdel 6=
X
Pabs and the subordinate engineer is correct.
[b] The difference between the power delivered to the circuit and the power
absorbed by the circuit is
−387.986 + 275.072 = −112.896 W
One-half of this difference is −56.448 W, so it is likely that pe is in error.
Either the voltage or the current probably has the wrong sign. (In
Chapter 2, we will discover that using KCL at the node connecting
components b, c, and e, the current ie should be −1.68 A, not 1.68 A!) If
the sign of pe is changed from negative to positive, we can recalculate the
power delivered and the power absorbed as follows:
X
Pdel
= 276.96 + 28.16 + 26.4 = 331.52 W
X
Pabs = 66.8352 + 204.8 + 56.448 + 3.2768 + 0.16 = 331.52 W
Now the power delivered equals the power absorbed and the power
balances for the circuit.
P 1.29
[a] From an examination of reference polarities, elements a, e, f, and h use a
+ sign in the power equation, so would be expected to absorb power.
Elements b, c, d, and g use a − sign in the power equation, so would be
expected to supply power.
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1–16
CHAPTER 1. Circuit Variables
[b]
pa
=
va ia = (5)(2 × 10−3 ) = 10 mW
pb
=
−vbib = −(1)(3 × 10−3 ) = −3 mW
pc
=
−vcic = −(7)(−2 × 10−3 ) = 14 mW
pd
=
−vdid = −(−9)(1 × 10−3 ) = 9 mW
pe
=
ve ie = (−20)(5 × 10−3 ) = −100 mW
pf
=
vf if = (20)(2 × 10−3 ) = 40 mW
pg
=
−vg ig = −(−3)(−2 × 10−3 ) = −6 mW
ph
=
vhih = (−12)(−3 × 10−3 ) = 36 mW
X
Pabs = 10 + 14 + 9 + 40 + 36 = 109 mW
Pdel = 3 + 100 + 6 = 109 mW
Thus, 109 mW of power is delivered and 109 mW of power is absorbed,
and the power balances.
X
[c] Looking at the calculated power values, elements a, c, d, f, and h have
positive power, so are absorbing, while elements b, e, and g have negative
power so are supplying. These answers are different from those in part
(a) because the voltages and currents used in the power equation are not
all positive numbers.
P 1.30
pa
=
−vaia = −(1.6)(0.080) = −128 mW
pb
=
−vbib = −(2.6)(0.060) = −156 mW
pc
=
vcic = (−4.2)(−0.050) = 210 mW
pd
=
−vdid = −(1.2)(0.020) = −24 mW
pe
=
veie = (1.8)(0.030) = 54 mW
pf
=
−vf if = −(−1.8)(−0.040) = −72 mW
pg
=
vgig = (−3.6)(−0.030) = 108 mW
ph
=
vhih = (3.2)(−0.020) = −64 mW
pj
=
−vjij = −(−2.4)(0.030) = 72 mW
X
Pdel = 128 + 156 + 24 + 72 + 64 = 444 mW
Pabs = 210 + 54 + 108 + 72 = 444 mW
X
X
Therefore,
Pdel = Pabs = 444 mW
X
Thus, the interconnection satisfies the power check.
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Problems
P 1.31
pa
=
vaia = (120)(−10) = −1200 W
pb
=
−vbib = −(120)(9) = −1080 W
pc
=
vcic = (10)(10) = 100 W
pd
=
−vdid = −(10)(−1) = 10 W
pe
=
veie = (−10)(−9) = 90 W
pf
=
−vf if = −(−100)(5) = 500 W
pg
=
vgig = (120)(4) = 480 W
ph
=
vhih = (−220)(−5) = 1100 W
1–17
X
Pdel = 1200 + 1080 = 2280 W
Pabs = 100 + 10 + 90 + 500 + 480 + 1100 = 2280 W
X
X
Therefore,
Pdel = Pabs = 2280 W
X
Thus, the interconnection now satisfies the power check.
P 1.32
[a] The revised circuit model is shown below:
[b] The expression for the total power in this circuit is
va ia − vb ib − vf if + vg ig + vh ih
= (120)(−10) − (120)(10) − (−120)(3) + 120ig + (−240)(−7) = 0
Therefore,
120ig = 1200 + 1200 − 360 − 1680 = 360
so
360
=3A
120
Thus, if the power in the modified circuit is balanced the current in
component g is 3 A.
ig =
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Circuit Elements
2
Assessment
a
Problems
AP 2.1
[a] Note that the current ib is in the same circuit branch as the 8 A current
source; however, ib is defined in the opposite direction of the current
source. Therefore,
ib = −8 A
Next, note that the dependent voltage source and the independent
voltage source are in parallel with the same polarity. Therefore, their
voltages are equal, and
ib
−8
vg =
=
= −2 V
4
4
[b] To find the power associated with the 8 A source, we need to find the
voltage drop across the source, vi . Note that the two independent sources
are in parallel, and that the voltages vg and v1 have the same polarities,
so these voltages are equal:
vi = vg = −2 V
Using the passive sign convention,
ps = (8 A)(vi ) = (8 A)(−2 V) = −16 W
Thus the current source generated 16 W of power.
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2–1 system, or transmission in any form or by any means, electronic,
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2–2
CHAPTER 2. Circuit Elements
AP 2.2
[a] Note from the circuit that vx = −25 V. To find α note that the two
current sources are in the same branch of the circuit but their currents
flow in opposite directions. Therefore
αvx = −15 A
Solve the above equation for α and substitute for vx,
α=
−15 A
−15 A
=
= 0.6 A/V
vx
−25 V
[b] To find the power associated with the voltage source we need to know the
current, iv . Note that this current is in the same branch of the circuit as
the dependent current source and these two currents flow in the same
direction. Therefore, the current iv is the same as the current of the
dependent source:
iv = αvx = (0.6)(−25) = −15 A
Using the passive sign convention,
ps = −(iv )(25 V) = −(−15 A)(25 V) = 375 W.
Thus the voltage source dissipates 375 W.
AP 2.3
[a] The resistor and the voltage source are in parallel and the resistor voltage
and the voltage source have the same polarities. Therefore these two
voltages are the same:
vR = vg = 1 kV
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Problems
2–3
Note from the circuit that the current through the resistor is ig = 5 mA.
Use Ohm’s law to calculate the value of the resistor:
vR
1 kV
R=
=
= 200 kΩ
ig
5 mA
Using the passive sign convention to calculate the power in the resistor,
pR = (vR )(ig ) = (1 kV)(5 mA) = 5 W
The resistor is dissipating 5 W of power.
[b] Note from part (a) the vR = vg and iR = ig . The power delivered by the
source is thus
psource
−3 W
psource = −vg ig
so
vg = −
=−
= 40 V
ig
75 mA
Since we now have the value of both the voltage and the current for the
resistor, we can use Ohm’s law to calculate the resistor value:
R=
vg
40 V
=
= 533.33 Ω
ig
75 mA
The power absorbed by the resistor must equal the power generated by
the source. Thus,
pR = −psource = −(−3 W) = 3 W
[c] Again, note the iR = ig . The power dissipated by the resistor can be
determined from the resistor’s current:
pR = R(iR )2 = R(ig )2
Solving for ig ,
pr
480 mW
=
= 0.0016
R
300 Ω
Then, since vR = vg
i2g =
so
ig =
vR = RiR = Rig = (300 Ω)(40 mA) = 12 V
√
0.0016 = 0.04 A = 40 mA
so
vg = 12 V
AP 2.4
[a] Note from the circuit that the current through the conductance G is ig ,
flowing from top to bottom, because the current source and the
conductance are in the same branch of the circuit so must have the same
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2–4
CHAPTER 2. Circuit Elements
current. The voltage drop across the current source is vg , positive at the
top, because the current source and the conductance are also in parallel
so must have the same voltage. From a version of Ohm’s law,
ig
0.5 A
=
= 10 V
G
50 mS
Now that we know the voltage drop across the current source, we can
find the power delivered by this source:
vg =
psource = −vg ig = −(10)(0.5) = −5 W
Thus the current source delivers 5 W to the circuit.
[b] We can find the value of the conductance using the power, and the value
of the current using Ohm’s law and the conductance value:
pg = Gvg2
so
G=
pg
9
= 2 = 0.04 S = 40 mS
2
vg
15
ig = Gvg = (40 mS)(15 V) = 0.6 A
[c] We can find the voltage from the power and the conductance, and then
use the voltage value in Ohm’s law to find the current:
pg = Gvg2
Thus
so
vg =
q
vg2 =
pg
8W
=
= 40,000
G
200 µS
40,000 = 200 V
ig = Gvg = (200 µS)(200 V) = 0.04 A = 40 mA
AP 2.5 [a] Redraw the circuit with all of the voltages and currents labeled for every
circuit element.
Write a KVL equation clockwise around the circuit, starting below the
voltage source:
−24 V + v2 + v5 − v1 = 0
Next, use Ohm’s law to calculate the three unknown voltages from the
three currents:
v2 = 3i2 ;
v5 = 7i5;
v1 = 2i1
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Problems
2–5
A KCL equation at the upper right node gives i2 = i5 ; a KCL equation at
the bottom right node gives i5 = −i1; a KCL equation at the upper left
node gives is = −i2. Now replace the currents i1 and i2 in the Ohm’s law
equations with i5:
v2 = 3i2 = 3i5;
v5 = 7i5 ;
v1 = 2i1 = −2i5
Now substitute these expressions for the three voltages into the first
equation:
24 = v2 + v5 − v1 = 3i5 + 7i5 − (−2i5) = 12i5
Therefore i5 = 24/12 = 2 A
[b] v1 = −2i5 = −2(2) = −4 V
[c] v2 = 3i5 = 3(2) = 6 V
[d] v5 = 7i5 = 7(2) = 14 V
[e] A KCL equation at the lower left node gives is = i1. Since i1 = −i5,
is = −2 A. We can now compute the power associated with the voltage
source:
p24 = (24)is = (24)(−2) = −48 W
Therefore 24 V source is delivering 48 W.
AP 2.6 Redraw the circuit labeling all voltages and currents:
We can find the value of the unknown resistor if we can find the value of its
voltage and its current. To start, write a KVL equation clockwise around the
right loop, starting below the 24 Ω resistor:
−120 V + v3 = 0
Use Ohm’s law to calculate the voltage across the 8 Ω resistor in terms of its
current:
v3 = 8i3
Substitute the expression for v3 into the first equation:
−120 V + 8i3 = 0
so
i3 =
120
= 15 A
8
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2–6
CHAPTER 2. Circuit Elements
Also use Ohm’s law to calculate the value of the current through the 24 Ω
resistor:
i2 =
120 V
= 5A
24 Ω
Now write a KCL equation at the top middle node, summing the currents
leaving:
−i1 + i2 + i3 = 0
so
i1 = i2 + i3 = 5 + 15 = 20 A
Write a KVL equation clockwise around the left loop, starting below the
voltage source:
−200 V + v1 + 120 V = 0
so
v1 = 200 − 120 = 80 V
Now that we know the values of both the voltage and the current for the
unknown resistor, we can use Ohm’s law to calculate the resistance:
R =
v1
80
=
= 4Ω
i1
20
AP 2.7 [a] Plotting a graph of vt versus it gives
Note that when it = 0, vt = 25 V; therefore the voltage source must be 25
V. Since the plot is a straight line, its slope can be used to calculate the
value of resistance:
∆v
25 − 0
25
R=
=
=
= 100 Ω
∆i
0.25 − 0
0.25
A circuit model having the same v − i characteristic is a 25 V source in
series with a 100Ω resistor, as shown below:
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Problems
2–7
[b] Draw the circuit model from part (a) and attach a 25 Ω resistor:
To find the power delivered to the 25 Ω resistor we must calculate the
current through the 25 Ω resistor. Do this by first using KCL to recognize
that the current in each of the components is it , flowing in a clockwise
direction. Write a KVL equation in the clockwise direction, starting
below the voltage source, and using Ohm’s law to express the voltage
drop across the resistors in the direction of the current it flowing through
the resistors:
25
−25 V + 100it + 25it = 0
so
125it = 25
so
it =
= 0.2 A
125
Thus, the power delivered to the 25 Ω resistor is
p25 = (25)i2t = (25)(0.2)2 = 1 W.
AP 2.8 [a] From the graph in Assessment Problem 2.7(a), we see that when vt = 0,
it = 0.25 A. Therefore the current source must be 0.25 A. Since the plot
is a straight line, its slope can be used to calculate the value of resistance:
∆v
25 − 0
25
R=
=
=
= 100 Ω
∆i
0.25 − 0
0.25
A circuit model having the same v − i characteristic is a 0.25 A current
source in parallel with a 100Ω resistor, as shown below:
[b] Draw the circuit model from part (a) and attach a 25 Ω resistor:
Note that by writing a KVL equation around the right loop we see that
the voltage drop across both resistors is vt. Write a KCL equation at the
top center node, summing the currents leaving the node. Use Ohm’s law
to specify the currents through the resistors in terms of the voltage drop
across the resistors and the value of the resistors.
vt
vt
−0.25 +
+
= 0,
so
5vt = 25,
thus
vt = 5 V
100 25
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2–8
CHAPTER 2. Circuit Elements
p25 =
vt2
= 1 W.
25
AP 2.9 First note that we know the current through all elements in the circuit except
the 6 kΩ resistor (the current in the three elements to the left of the 6 kΩ
resistor is i1 ; the current in the three elements to the right of the 6 kΩ resistor
is 30i1 ). To find the current in the 6 kΩ resistor, write a KCL equation at the
top node:
i1 + 30i1 = i6k = 31i1
We can then use Ohm’s law to find the voltages across each resistor in terms
of i1 . The results are shown in the figure below:
[a] To find i1, write a KVL equation around the left-hand loop, summing
voltages in a clockwise direction starting below the 5V source:
−5 V + 54,000i1 − 1 V + 186,000i1 = 0
Solving for i1
54,000i1 + 186,000i1 = 6 V
so
240,000i1 = 6 V
Thus,
i1 =
6
= 25 µA
240,000
[b] Now that we have the value of i1 , we can calculate the voltage for each
component except the dependent source. Then we can write a KVL
equation for the right-hand loop to find the voltage v of the dependent
source. Sum the voltages in the clockwise direction, starting to the left of
the dependent source:
+v − 54,000i1 + 8 V − 186,000i1 = 0
Thus,
v = 240,000i1 − 8 V = 240,000(25 × 10−6 ) − 8 V = 6 V − 8 V = −2 V
We now know the values of voltage and current for every circuit element.
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Problems
2–9
Let’s construct a power table:
Element
Current
Voltage
Power
Power
(µA)
(V)
Equation
(µW)
5V
25
5
p = −vi
−125
54 kΩ
25
1.35
p = Ri2
33.75
1V
25
1
p = −vi
−25
6 kΩ
775
4.65
p = Ri2
3603.75
Dep. source
750
−2
p = −vi
1500
1.8 kΩ
750
1.35
p = Ri2
1012.5
8V
750
8
p = −vi
−6000
[c] The total power generated in the circuit is the sum of the negative power
values in the power table:
−125 µW + −25 µW + −6000 µW = −6150 µW
Thus, the total power generated in the circuit is 6150 µW.
[d] The total power absorbed in the circuit is the sum of the positive power
values in the power table:
33.75 µW + 3603.75 µW + 1500 µW + 1012.5 µW = 6150 µW
Thus, the total power absorbed in the circuit is 6150 µW.
AP 2.10 Given that iφ = 2 A, we know the current in the dependent source is
2iφ = 4 A. We can write a KCL equation at the left node to find the current in
the 10 Ω resistor. Summing the currents leaving the node,
−5 A + 2 A + 4 A + i10Ω = 0
so
i10Ω = 5 A − 2 A − 4 A = −1 A
Thus, the current in the 10 Ω resistor is 1 A, flowing right to left, as seen in
the circuit below.
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2–10
CHAPTER 2. Circuit Elements
[a] To find vs , write a KVL equation, summing the voltages counter-clockwise
around the lower right loop. Start below the voltage source.
−vs + (1 A)(10 Ω) + (2 A)(30 Ω) = 0
so
vs = 10 V + 60 V = 70 V
[b] The current in the voltage source can be found by writing a KCL equation
at the right-hand node. Sum the currents leaving the node
−4 A + 1 A + iv = 0
so
iv = 4 A − 1 A = 3 A
The current in the voltage source is 3 A, flowing top to bottom. The
power associated with this source is
p = vi = (70 V)(3 A) = 210 W
Thus, 210 W are absorbed by the voltage source.
[c] The voltage drop across the independent current source can be found by
writing a KVL equation around the left loop in a clockwise direction:
−v5A + (2 A)(30 Ω) = 0
so
v5A = 60 V
The power associated with this source is
p = −v5A i = −(60 V)(5 A) = −300 W
This source thus delivers 300 W of power to the circuit.
[d] The voltage across the controlled current source can be found by writing a
KVL equation around the upper right loop in a clockwise direction:
+v4A + (10 Ω)(1 A) = 0
so
v4A = −10 V
The power associated with this source is
p = v4A i = (−10 V)(4 A) = −40 W
This source thus delivers 40 W of power to the circuit.
[e] The total power dissipated by the resistors is given by
(i30Ω )2(30 Ω) + (i10Ω)2 (10 Ω) = (2)2 (30 Ω) + (1)2 (10 Ω) = 120 + 10 = 130 W
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Problems
2–11
Problems
P 2.1
The interconnect is valid since the voltage sources can all carry 5 A of current
supplied by the current source, and the current source can carry the voltage
drop required by the interconnection. Note that the branch containing the 10
V, 40 V, and 5 A sources must have the same voltage drop as the branch
containing the 50 V source, so the 5 A current source must have a voltage
drop of 20 V, positive at the right. The voltages and currents are summarize
in the circuit below:
P50V
=
(50)(5) = 250 W (abs)
P10V
=
(10)(5) = 50 W
P40V
=
−(40)(5) = −200 W (dev)
P5A
=
−(20)(5) = −100 W
X
(abs)
(dev)
Pdev = 300 W
P 2.2
The interconnection is not valid. Note that the 10 V and 20 V sources are
both connected between the same two nodes in the circuit. If the
interconnection was valid, these two voltage sources would supply the same
voltage drop between these two nodes, which they do not.
P 2.3
[a] Yes, independent voltage sources can carry the 5 A current required by the
connection; independent current source can support any voltage required
by the connection, in this case 5 V, positive at the bottom.
[b] 20 V source:
absorbing
15 V source:
developing (delivering)
5 A source:
developing (delivering)
[c] P20V
=
(20)(5) = 100 W (abs)
P15V
=
−(15)(5) = −75 W (dev/del)
P5A
=
X
Pabs =
−(5)(5) = −25 W
X
(dev/del)
Pdel = 100 W
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2–12
CHAPTER 2. Circuit Elements
[d] The interconnection is valid, but in this circuit the voltage drop across the
5 A current source is 35 V, positive at the top; 20 V source is developing
(delivering), the 15 V source is developing (delivering), and the 5 A
source is absorbing:
P20V
= −(20)(5) = −100 W (dev/del)
P15V
= −(15)(5) = −75 W (dev/del)
P5A
P 2.4
X
= (35)(5) = 175 W
Pabs =
X
(abs)
Pdel = 175 W
The interconnection is valid. The 10 A current source has a voltage drop of
100 V, positive at the top, because the 100 V source supplies its voltage drop
across a pair of terminals shared by the 10 A current source. The right hand
branch of the circuit must also have a voltage drop of 100 V from the left
terminal of the 40 V source to the bottom terminal of the 5 A current source,
because this branch shares the same terminals as the 100 V source. This
means that the voltage drop across the 5 A current source is 140 V, positive at
the top. Also, the two voltage sources can carry the current required of the
interconnection. This is summarized in the figure below:
From the values of voltage and current in the figure, the power supplied by the
current sources is calculated as follows:
P10A = −(100)(10) = −1000 W (1000 W supplied)
P5A
P 2.5
X
=
−(140)(5) = −700 W
(700 W supplied)
Pdev = 1700 W
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Problems
2–13
The interconnection is invalid. The voltage drop between the top terminal and
the bottom terminal on the left hand side is due to the 6 V and 8 V sources,
giving a total voltage drop between these terminals of 14 V. But the voltage
drop between the top terminal and the bottom terminal on the right hand side
is due to the 4 V and 12 V sources, giving a total voltage drop between these
two terminals of 16 V. The voltage drop between any two terminals in a valid
circuit must be the same, so the interconnection is invalid.
P 2.6
The interconnection is valid, since the voltage sources can carry the 20 mA
current supplied by the current source, and the current sources can carry
whatever voltage drop is required by the interconnection. In particular, note
the the voltage drop across the three sources in the right hand branch must be
the same as the voltage drop across the 15 mA current source in the middle
branch, since the middle and right hand branches are connected between the
same two terminals. In particular, this means that
v1(the voltage drop across the middle branch)
= −20V + 60V − v2
Hence any combination of v1 and v2 such that v1 + v2 = 40 V is a valid
solution.
P 2.7
The interconnection is invalid. In the middle branch, the value of the current
i∆ must be −25 A, since the 25 A current source supplies current in this
branch in the direction opposite the direction of the current i∆ . Therefore, the
voltage supplied by the dependent voltage source in the left hand branch is
6(−25) = −150 V. This gives a voltage drop from the top terminal to the
bottom terminal in the left hand branch of 50 − (−150) = 200 V. But the
voltage drop between these same terminals in the right hand branch is 250 V,
due to the voltage source in that branch. Therefore, the interconnection is
invalid.
P 2.8
First, 10va = 5 V, so va = 0.5 V. Then recognize that each of the three
branches is connected between the same two nodes, so each of these branches
must have the same voltage drop. The voltage drop across the middle branch
is 5 V, and since va = 0.5 V, vg = 0.5 − 5 = −4.5 V. Also, the voltage drop
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2–14
CHAPTER 2. Circuit Elements
across the left branch is 5 V, so 20 + v9A = 5 V, and v9A = −15 V, where v9A
is positive at the top. Note that the current through the 20 V source must be
9 A, flowing from top to bottom, and the current through the vg is 6 A flowing
from top to bottom. Let’s find the power associated with the left and middle
branches:
p9A = (9)(−15) = −135 W
p20V = (9)(20) = 180 W
pvg = −(6)(−4.5) = 27 W
p6A = (6)(0.5) = 3 W
Since there is only one component left, we can find the total power:
ptotal = −135 + 180 + 27 + 3 + pds = 75 + pds = 0
so pds must equal −75 W.
Therefore,
P 2.9
X
Pdev =
X
Pabs = 210 W
[a] Yes, each of the voltage sources can carry the current required by the
interconnection, and each of the current sources can carry the voltage
drop required by the interconnection. (Note that i∆ = −8 A.)
[b] No, because the voltage drop between the top terminal and the bottom
terminal cannot be determined. For example, define v1, v2 , and v3 as
shown:
The voltage drop across the left branch, the center branch, and the right
branch must be the same, since these branches are connected at the same
two terminals. This requires that
20 + v1 = v2 + 100 = v3
But this equation has three unknown voltages, so the individual voltages
cannot be determined, and thus the power of the sources cannot be
determined.
P 2.10
[a]
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Problems
[b]
P 2.11
Vbb
=
no-load voltage of battery
Rbb
=
internal resistance of battery
Rx
=
resistance of wire between battery and switch
Ry
=
resistance of wire between switch and lamp A
Ra
=
resistance of lamp A
Rb
=
resistance of lamp B
Rw
=
resistance of wire between lamp A and lamp B
Rg1
=
resistance of frame between battery and lamp A
Rg2
=
resistance of frame between lamp A and lamp B
S
=
switch
2–15
Since we know the device is a resistor, we can use Ohm’s law to calculate the
resistance. From Fig. P2.11(a),
v = Ri
so
R=
v
i
Using the values in the table of Fig. P2.11(b),
R=
−108
−54
54
108
162
=
=
=
=
= 27 kΩ
−0.004
−0.002
0.002
0.004
0.006
Note that this value is found in Appendix H.
P 2.12
The resistor value is the ratio of the power to the square of the current:
p
R = 2 . Using the values for power and current in Fig. P2.12(b),
i
5.5 × 10−3
22 × 10−3
49.5 × 10−3
88 × 10−3
=
=
=
(50 × 10−6 )2
(100 × 10−6 )2
(150 × 10−6 )2
(200 × 10−6 )2
=
137.5 × 10−3
198 × 10−3
=
= 2.2 MΩ
(250 × 10−6 )2
(300 × 10−6 )2
Note that this is a value from Appendix H.
P 2.13
Since we know the device is a resistor, we can use the power equation. From
Fig. P2.13(a),
p = vi =
v2
R
so
R=
v2
p
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2–16
CHAPTER 2. Circuit Elements
Using the values in the table of Fig. P2.13(b)
R=
(−10)2
(−5)2
(5)2
(10)2
=
=
=
17.86 × 10−3
4.46 × 10−3
4.46 × 10−3
17.86 × 10−3
=
(15)2
(20)2
=
≈ 5.6 kΩ
40.18 × 10−3
71.43 × 10−3
Note that this value is found in Appendix H.
P 2.14
[a] Plot the v—i characteristic:
From the plot:
R=
∆v
(180 − 100)
=
= 5Ω
∆i
(16 − 0)
When it = 0, vt = 100 V; therefore the ideal current source must have a
current of 100/5 = 20 A
[b] We attach a 20 Ω resistor to the device model developed in part (a):
Write a KCL equation at the top node:
20 + it = i1
Write a KVL equation for the right loop, in the direction of the two
currents, using Ohm’s law:
5i1 + 20it = 0
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Problems
2–17
Combining the two equations and solving,
5(20 + it ) + 20it = 0
so
25it = −100;
thus
it = −4 A
Now calculate the power dissipated by the resistor:
p20 Ω = 20i2t = 20(−4)2 = 320 W
P 2.15
[a] Plot the v − i characteristic
From the plot:
R=
∆v
(130 − 50)
=
= 8Ω
∆i
(10 − 0)
When it = 0, vt = 50 V; therefore the ideal voltage source has a voltage
of 50 V.
[b]
When vt = 0,
it =
−50
= −6.25A
8
Note that this result can also be obtained by extrapolating the v − i
characteristic to vt = 0.
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2–18
P 2.16
CHAPTER 2. Circuit Elements
[a]
[b] ∆v = 25V;
∆i = 2.5 mA;
[c] 10,000i1 = 2500is ,
R=
∆v
= 10 kΩ
∆i
i1 = 0.25is
0.02 = i1 + is = 1.25is ,
is = 16 mA
[d] vs (open circuit) = (20 × 10−3 )(10 × 103 ) = 200 V
[e] The open circuit voltage can be found in the table of values (or from the
plot) as the value of the voltage vs when the current is = 0. Thus,
vs (open circuit) = 140 V (from the table)
[f] Linear model cannot predict the nonlinear behavior of the practical
current source.
P 2.17
[a] Begin by constructing a plot of voltage versus current:
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Problems
2–19
[b] Since the plot is linear for 0 ≤ is ≤ 24 mA amd since R = ∆v/∆i, we can
calculate R from the plotted values as follows:
R=
∆v
24 − 18
6
=
=
= 250 Ω
∆i
0.024 − 0
0.024
We can determine the value of the ideal voltage source by considering the
value of vs when is = 0. When there is no current, there is no voltage
drop across the resistor, so all of the voltage drop at the output is due to
the voltage source. Thus the value of the voltage source must be 24 V.
The model, valid for 0 ≤ is ≤ 24 mA, is shown below:
[c] The circuit is shown below:
Write a KVL equation in the clockwise direction, starting below the
voltage source. Use Ohm’s law to express the voltage drop across the
resistors in terms of the current i:
−24 V + 250i + 1000i = 0
Thus,
i=
so
1250i = 24 V
24 V
= 19.2 mA
1250 Ω
[d] The circuit is shown below:
Write a KVL equation in the clockwise direction, starting below the
voltage source. Use Ohm’s law to express the voltage drop across the
resistors in terms of the current i:
−24 V + 250i = 0
Thus,
i=
so
250i = 24 V
24 V
= 96 mA
250 Ω
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2–20
CHAPTER 2. Circuit Elements
[e] The short circuit current can be found in the table of values (or from the
plot) as the value of the current is when the voltage vs = 0. Thus,
isc = 48 mA
(from table)
[f] The plot of voltage versus current constructed in part (a) is not linear (it
is piecewise linear, but not linear for all values of is ). Since the proposed
circuit model is a linear model, it cannot be used to predict the nonlinear
behavior exhibited by the plotted data.
P 2.18
[a] Write a KCL equation at the top node:
−1.5 + i1 + i2 = 0
so
i1 + i2 = 1.5
Write a KVL equation around the right loop:
−v1 + v2 + v3 = 0
From Ohm’s law,
v1 = 100i1 ,
v2 = 150i2 ,
v3 = 250i2
Substituting,
−100i1 + 150i2 + 250i2 = 0
so
− 100i1 + 400i2 = 0
Solving the two equations for i1 and i2 simultaneously,
i1 = 1.2 A
and
i2 = 0.3 A
[b] Write a KVL equation clockwise around the left loop:
−vo + v1 = 0
So
but
v1 = 100i1 = 100(1.2) = 120 V
vo = v1 = 120 V
[c] Calculate power using p = vi for the source and p = Ri2 for the resistors:
psource = −vo(1.5) = −(120)(1.5) = −180 W
p100Ω = 1.22 (100) = 144 W
p150Ω = 0.32 (150) = 13.5 W
p250Ω = 0.32 (250) = 22.5 W
X
Pdev = 180 W
X
Pabs = 144 + 13.5 + 22.5 = 180 W
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Problems
P 2.19
2–21
[a]
20ia
=
80ib
ia
=
4ib
ig = ia + ib = 5ib
50 =
4ig + 80ib = 20ib + 80ib = 100ib
ib
0.5 A, therefore, ia = 2 A
=
and
ig = 2.5 A
[b] ib = 0.5 A
[c] vo = 80ib = 40 V
[d]
p4Ω
= i2g (4) = 6.25(4) = 25 W
p20Ω
= i2a (20) = (4)(20) = 80 W
p80Ω
= i2b (80) = 0.25(80) = 20 W
[e] p50V (delivered) = 50ig = 125 W
Check:
X
P 2.20
X
Pdis = 25 + 80 + 20 = 125 W
Pdel = 125 W
[a] Use KVL for the right loop to calculate the voltage drop across the
right-hand branch vo . This is also the voltage drop across the middle
branch, so once vo is known, use Ohm’s law to calculate io :
vo
= 1000ia + 4000ia + 3000ia = 8000ia = 8000(0.002) = 16 V
16
= 2000io
16
io =
= 8 mA
2000
[b] KCL at the top node: ig = ia + io = 0.002 + 0.008 = 0.010 A = 10 mA.
[c] The voltage drop across the source is v0, seen by writing a KVL equation
for the left loop. Thus,
pg = −vo ig = −(16)(0.01) = −0.160 W = −160 mW.
Thus the source delivers 160 mW.
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2–22
P 2.21
CHAPTER 2. Circuit Elements
[a]
v2 = 150 − 50(1) = 100V
i2 =
v2
= 4A
25
i3 + 1 = i2 ,
i3 = 4 − 1 = 3A
v1 = 10i3 + 25i2 = 10(3) + 25(4) = 130V
i1 =
v1
130
=
= 2A
65
65
Note also that
i4 = i1 + i3 = 2 + 3 = 5 A
ig = i4 + io = 5 + 1 = 6 A
[b]
[c]
p4Ω
= 52 (4) = 100 W
p50Ω
= 12 (50) = 50 W
p65Ω
= 22 (65) = 260 W
p10Ω
= 32 (10) = 90 W
p25Ω
= 42 (25) = 400 W
X
Pdis = 100 + 50 + 260 + 90 + 400 = 900 W
Pdev = 150ig = 150(6) = 900 W
P 2.22
[a]
icd = 80/16 = 5 A
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Problems
vac = 125 − 80 = 45
iac + ibc = icd
so
so
2–23
iac = 45/15 = 3 A
ibc = 5 − 3 = 2 A
vab = 15iac − 5ibc = 15(3) − 5(2) = 35 V
so
iab = 35/7 = 5 A
ibd = iab − ibc = 5 − 2 = 3 A
Calculate the power dissipated by the resistors using the equation
pR = Ri2R :
p7Ω = (7)(5)2 = 175 W
p15Ω = (15)(3)2 = 135 W
p30Ω = (30)(3)2 = 270 W
p16Ω = (16)(5)2 = 400 W
p5Ω = (5)(2)2 = 20 W
[b] Calculate the current through the voltage source:
iad = −iab − iac = −5 − 3 = −8 A
Now that we have both the voltage and the current for the source, we can
calculate the power supplied by the source:
pg = 125(−8) = −1000 W
[c]
P 2.23
[a]
thus
pg (supplied) = 1000 W
X
Pdis = 175 + 270 + 135 + 400 + 20 = 1000 W
Therefore,
X
Psupp =
X
Pdis
va = (5 + 10)(4) = 60 V
−240 + va + vb = 0
so
vb = 240 − va = 240 − 60 = 180 V
ie = vb /(14 + 6) = 180/20 = 9 A
id = ie − 4 = 9 − 4 = 5 A
vc = 4id + vb = 4(5) + 180 = 200 V
ic = vc /10 = 200/10 = 20 A
vd = 240 − vc = 240 − 200 = 40 V
ia = id + ic = 5 + 20 = 25 A
R = vd /ia = 40/25 = 1.6 Ω
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2–24
CHAPTER 2. Circuit Elements
[b] ig = ia + 4 = 25 + 4 = 29 A
pg (supplied) = (240)(29) = 6960 W
P 2.24
id = 60/12 = 5 A; therefore, vcd = 60 + 18(5) = 150 V
−240 + vac + vcd = 0; therefore, vac = 240 − 150 = 90 V
ib = vac/45 = 90/45 = 2 A; therefore, ic = id − ib = 5 − 2 = 3 A
vbd = 10ic + vcd = 10(3) + 150 = 180 V;
therefore, ia = vbd/180 = 180/180 = 1 A
ie = ia + ic = 1 + 3 = 4 A
−240 + vab + vbd = 0 therefore, vab = 240 − 180 = 60 V
R = vab/ie = 60/4 = 15 Ω
CHECK: ig = ib + ie = 2 + 4 = 6 A
pdev = (240)(6) = 1440 W
X
Pdis = 12 (180) + 42 (15) + 32 (10) + 52 (12) + 52 (18) + 22 (45)
P 2.25
= 1440 W (CHECKS)
[a] Start by calculating the voltage drops due to the currents i1 and i2. Then
use KVL to calculate the voltage drop across and 35 Ω resistor, and
Ohm’s law to find the current in the 35 Ω resistor. Finally, KCL at each
of the middle three nodes yields the currents in the two sources and the
current in the middle 2 Ω resistor. These calculations are summarized in
the figure below:
p147(top)
= −(147)(28) = −4116 W
p147(bottom)
= −(147)(21) = −3087 W
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Problems
2–25
[b]
X
X
Pdis
= 784 + 98 + 441 + 2205 + 1960 + 1715 = 7203 W
Psup
Therefore,
P 2.26
[a]
= (28)2 (1) + (7)2 (2) + (21)2 (1) + (21)2 (5) + (14)2 (10) + (7)2 (35)
= 4116 + 3087 = 7203 W
X
Pdis =
X
Psup = 7203 W
v2 = 100 + 4(15) = 160 V;
i1 =
v1
100
=
= 5 A;
4 + 16
20
v1 = 160 − (9 + 11 + 10)(2) = 100 V
i3 = i1 − 2 = 5 − 2 = 3 A
vg = v1 + 30i3 = 100 + 30(3) = 190 V
i4 = 2 + 4 = 6 A
ig = −i4 − i3 = −6 − 3 = −9 A
[b] Calculate power using the formula p = Ri2 :
p9 Ω = (9)(2)2 = 36 W;
p10 Ω = (10)(2)2 = 40 W;
p11 Ω = (11)(2)2 = 44 W
p5 Ω = (5)(6)2 = 180 W
p30 Ω = (30)(3)2 = 270 W;
p4 Ω = (4)(5)2 = 100 W
p16 Ω = (16)(5)2 = 400 W;
p15 Ω = (15)(4)2 = 240 W
[c] vg = 190 V
[d] Sum the power dissipated by the resistors:
X
pdiss = 36 + 44 + 40 + 180 + 270 + 100 + 400 + 240 = 1310 W
The power associated with the sources is
pvolt−source = (100)(4) = 400 W
pcurr−source = vg ig = (190)(−9) = −1710 W
Thus the total power dissipated is 1310 + 400 = 1710 W and the total
power developed is 1710 W, so the power balances.
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2–26
P 2.27
CHAPTER 2. Circuit Elements
[a] io = 0 because no current can exist in a single conductor connecting two
parts of a circuit.
[b]
18 = (12 + 6)ig
ig = 1 A
v∆ = 6ig = 6V
v∆ /2 = 3 A
10i1 = 5i2 , so i1 + 2i1 = −3 A; therefore, i1 = −1 A
[c] i2 = 2i1 = −2 A.
P 2.28
First note that we know the current through all elements in the circuit except
the 200 Ω resistor (the current in the three elements to the left of the 200 Ω
resistor is iβ ; the current in the three elements to the right of the 200 Ω
resistor is 29iβ ). To find the current in the 200 Ω resistor, write a KCL
equation at the top node:
iβ + 29iβ = i200 Ω = 30iβ
We can then use Ohm’s law to find the voltages across each resistor in terms
of iβ . The results are shown in the figure below:
[a] To find iβ , write a KVL equation around the left-hand loop, summing
voltages in a clockwise direction starting below the 15.2V source:
−15.2 V + 10,000i1 − 0.8 V + 6000iβ = 0
Solving for iβ
10,000iβ + 6000iβ = 16 V
so
16,000iβ = 16 V
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Problems
2–27
Thus,
iβ =
16
= 1 mA
16,000
Now that we have the value of iβ , we can calculate the voltage for each
component except the dependent source. Then we can write a KVL
equation for the right-hand loop to find the voltage vy of the dependent
source. Sum the voltages in the clockwise direction, starting to the left of
the dependent source:
−vy − 14,500iβ + 25 V − 6000iβ = 0
Thus,
vy = 25 V − 20,500iβ = 25 V − 20,500(10−3 ) = 25 V − 20.5 V = 4.5 V
[b] We now know the values of voltage and current for every circuit element.
Let’s construct a power table:
Element
Current
Voltage
(mA)
(V)
Power
Power
Equation (mW)
15.2 V
1
15.2
p = −vi
−15.2
10 kΩ
1
10
p = Ri2
10
0.8 V
1
0.8
p = −vi
−0.8
200 Ω
30
6
p = Ri2
180
Dep. source
29
4.5
p = vi
130.5
500 Ω
29
14.5
p = Ri2
420.5
25 V
29
25
p = −vi
−725
The total power generated in the circuit is the sum of the negative power
values in the power table:
−15.2 mW + −0.8 mW + −725 mW = −741 mW
Thus, the total power generated in the circuit is 741 mW. The total
power absorbed in the circuit is the sum of the positive power values in
the power table:
10 mW + 180 mW + 130.5 mW + 420.5 mW = 741 mW
Thus, the total power absorbed in the circuit is 741 mW and the power in
the circuit balances.
P 2.29
40i2 +
5
5
+
= 0;
40 10
i2 = −15.625 mA
v1 = 80i2 = −1.25 V
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2–28
CHAPTER 2. Circuit Elements
25i1 +
(−1.25)
+ (−0.015625) = 0;
20
i1 = 3.125 mA
vg = 60i1 + 260i1 = 320i1
Therefore, vg = 1 V.
P 2.30
[a] −50 − 20iσ + 18i∆ = 0
−18i∆ + 5iσ + 40iσ = 0
Therefore,
so
18i∆ = 45iσ
− 50 − 20iσ + 45iσ = 0,
so
iσ = 2 A
18i∆ = 45iσ = 90; so i∆ = 5 A
vo = 40iσ = 80 V
[b] ig = current out of the positive terminal of the 50 V source
vd = voltage drop across the 8i∆ source
ig = i∆ + iσ + 8i∆ = 9i∆ + iσ = 47 A
vd = 80 − 20 = 60 V
X
X
Pgen
= 50ig + 20iσ ig = 50(47) + 20(2)(47) = 4230 W
Pdiss
= 18i2∆ + 5iσ (ig − i∆ ) + 40i2σ + 8i∆ vd + 8i∆ (20)
= (18)(25) + 10(47 − 5) + 4(40) + 40(60) + 40(20)
= 4230 W; Therefore,
P 2.31
X
Pgen
=
iE − iB − iC = 0
iC = βiB
X
Pdiss = 4230 W
therefore iE = (1 + β)iB
i2 = −iB + i1
Vo + iE RE − (i1 − iB )R2 = 0
−i1R1 + VCC − (i1 − iB )R2 = 0
Vo + iE RE + iB R2 −
or
i1 =
VCC + iB R2
R1 + R2
VCC + iB R2
R2 = 0
R1 + R2
Now replace iE by (1 + β)iB and solve for iB . Thus
iB =
[VCC R2 /(R1 + R2 )] − Vo
(1 + β)RE + R1 R2 /(R1 + R2 )
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Problems
P 2.32
2–29
Here is Equation 2.25:
iB =
(VCC R2 )/(R1 + R2 ) − V0
(R1 R2 )/(R1 + R2 ) + (1 + β)RE
VCC R2
(10)(60,000)
=
= 6V
R1 + R2
100,000
R1 R2
(40,000)(60,000)
=
= 24 kΩ
R1 + R2
100,000
iB =
6 − 0.6
5.4
=
= 0.18 mA
24,000 + 50(120)
30,000
iC = βiB = (49)(0.18) = 8.82 mA
iE = iC + iB = 8.82 + 0.18 = 9 mA
v3d = (0.009)(120) = 1.08V
vbd = Vo + v3d = 1.68V
i2 =
vbd
1.68
=
= 28 µA
R2
60,000
i1 = i2 + iB = 28 + 180 = 208 µA
vab = 40,000(208 × 10−6 ) = 8.32 V
iCC = iC + i1 = 8.82 + 0.208 = 9.028 mA
v13 + (8.82 × 10−3 )(750) + 1.08 = 10 V
v13 = 2.305 V
P 2.33
[a]
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2–30
CHAPTER 2. Circuit Elements
[b]
P 2.34
[a] From the simplified circuit model, using Ohm’s law and KVL:
400i + 50i + 200i − 250 = 0
so
i = 250/650 = 385 mA
This current is nearly enough to stop the heart, according to Table 2.1,
so a warning sign should be posted at the 250 V source.
[b] The closest value from Appendix H to 400 Ω is 390 Ω; the closest value
from Appendix H to 50 Ω is 47 Ω. There are two possibilites for replacing
the 200 Ω resistor with a value from Appendix H – 180 Ω and 220 Ω. We
calculate the resulting current for each of these possibilities, and
determine which current is closest to 385 mA:
390i + 47i + 180i − 250 = 0
so
i = 250/617 = 405.2 mA
390i + 47i + 220i − 250 = 0
so
i = 250/657 = 380.5 mA
Therefore, choose the 220 Ω resistor to replace the 200 Ω resistor in the
model.
P 2.35
P 2.36
[a] p = i2R
parm =
pleg =
ptrunk =
250
650
250
650
2
2
250
650
(400) = 59.17 W
(200) = 29.59 W
2
(50) = 7.40 W
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Problems
[b]
dT
dt
!
=
arm
!
tleg =
dT
dt
2.39 × 10−4 parm
= 35.36 × 10−4 ◦ C/s
4
5
× 104 = 1414.23 s or 23.57 min
35.36
tarm =
dT
dt
2–31
=
leg
2.39 × 10−4
Pleg = 7.07 × 10−4 ◦ C/s
10
5 × 104
= 7,071.13 s or 117.85 min
7.07
!
=
trunk
ttrunk =
2.39 × 10−4 (7.4)
= 0.707 × 10−4 ◦ C/s
25
5 × 104
= 70,711.30 s or 1,178.52 min
0.707
[c] They are all much greater than a few minutes.
P 2.37
[a] Rarms = 400 + 400 = 800 Ω
iletgo = 50 mA (minimum)
vmin = (800)(50) × 10−3 = 40 V
[b] No, 12/800 = 15 mA. Note this current is sufficient to give a perceptible
shock.
P 2.38
Rspace = 1 MΩ
ispace = 3 mA
v = ispaceRspace = 3000 V.
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3
Simple Resistive Circuits
Assessment Problems
AP 3.1
Start from the right hand side of the circuit and make series and parallel
combinations of the resistors until one equivalent resistor remains. Begin by
combining the 6 Ω resistor and the 10 Ω resistor in series:
6 Ω + 10 Ω = 16 Ω
Now combine this 16 Ω resistor in parallel with the 64 Ω resistor:
16 Ωk64 Ω =
(16)(64)
1024
=
= 12.8 Ω
16 + 64
80
This equivalent 12.8 Ω resistor is in series with the 7.2 Ω resistor:
12.8 Ω + 7.2 Ω = 20 Ω
Finally, this equivalent 20 Ω resistor is in parallel with the 30 Ω resistor:
20 Ωk30 Ω =
(20)(30)
600
=
= 12 Ω
20 + 30
50
Thus, the simplified circuit is as shown:
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3–1 system, or transmission in any form or by any means, electronic,
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3–2
CHAPTER 3. Simple Resistive Circuits
[a] With the simplified circuit we can use Ohm’s law to find the voltage across
both the current source and the 12 Ω equivalent resistor:
v = (12 Ω)(5 A) = 60 V
[b] Now that we know the value of the voltage drop across the current source,
we can use the formula p = −vi to find the power associated with the
source:
p = −(60 V)(5 A) = −300 W
Thus, the source delivers 300 W of power to the circuit.
[c] We now can return to the original circuit, shown in the first figure. In this
circuit, v = 60 V, as calculated in part (a). This is also the voltage drop
across the 30 Ω resistor, so we can use Ohm’s law to calculate the current
through this resistor:
60 V
iA =
=2A
30 Ω
Now write a KCL equation at the upper left node to find the current iB :
−5 A + iA + iB = 0
so
iB = 5 A − iA = 5 A − 2 A = 3 A
Next, write a KVL equation around the outer loop of the circuit, using
Ohm’s law to express the voltage drop across the resistors in terms of the
current through the resistors:
−v + 7.2iB + 6iC + 10iC = 0
So
16iC = v − 7.2iB = 60 V − (7.2)(3) = 38.4 V
38.4
= 2.4 A
16
Now that we have the current through the 10 Ω resistor we can use the
formula p = Ri2 to find the power:
Thus
iC =
p10 Ω = (10)(2.4)2 = 57.6 W
AP 3.2
[a] We can use voltage division to calculate the voltage vo across the 75 kΩ
resistor:
75,000
vo (no load) =
(200 V) = 150 V
75,000 + 25,000
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Problems
3–3
[b] When we have a load resistance of 150 kΩ then the voltage vo is across the
parallel combination of the 75 kΩ resistor and the 150 kΩ resistor. First,
calculate the equivalent resistance of the parallel combination:
(75,000)(150,000)
= 50,000 Ω = 50 kΩ
75,000 + 150,000
Now use voltage division to find vo across this equivalent resistance:
50,000
vo =
(200 V) = 133.3 V
50,000 + 25,000
75 kΩk150 kΩ =
[c] If the load terminals are short-circuited, the 75 kΩ resistor is effectively
removed from the circuit, leaving only the voltage source and the 25 kΩ
resistor. We can calculate the current in the resistor using Ohm’s law:
200 V
i=
= 8 mA
25 kΩ
Now we can use the formula p = Ri2 to find the power dissipated in the
25 kΩ resistor:
p25k = (25,000)(0.008)2 = 1.6 W
[d] The power dissipated in the 75 kΩ resistor will be maximum at no load
since vo is maximum. In part (a) we determined that the no-load voltage
is 150 V, so be can use the formula p = v 2/R to calculate the power:
p75k (max) =
(150)2
= 0.3 W
75,000
AP 3.3
[a] We will write a current division equation for the current throught the 80Ω
resistor and use this equation to solve for R:
R
i80Ω =
(20 A) = 4 A
so
20R = 4(R + 120)
R + 40 Ω + 80 Ω
480
Thus
16R = 480
and
R=
= 30 Ω
16
[b] With R = 30 Ω we can calculate the current through R using current
division, and then use this current to find the power dissipated by R,
using the formula p = Ri2 :
40 + 80
iR =
(20 A) = 16 A
so
pR = (30)(16)2 = 7680 W
40 + 80 + 30
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3–4
CHAPTER 3. Simple Resistive Circuits
[c] Write a KVL equation around the outer loop to solve for the voltage v,
and then use the formula p = −vi to calculate the power delivered by the
current source:
−v + (60 Ω)(20 A) + (30 Ω)(16 A) = 0
Thus,
so
v = 1200 + 480 = 1680 V
psource = −(1680 V)(20 A) = −33,600 W
Thus, the current source generates 33,600 W of power.
AP 3.4
[a] First we need to determine the equivalent resistance to the right of the
40 Ω and 70 Ω resistors:
1
1
1
1
1
Req = 20 Ωk30 Ωk(50 Ω + 10 Ω)
so
=
+
+
=
Req
20 Ω 30 Ω 60 Ω
10 Ω
Thus,
Req = 10 Ω
Now we can use voltage division to find the voltage vo :
vo =
40
(60 V) = 20 V
40 + 10 + 70
[b] The current through the 40 Ω resistor can be found using Ohm’s law:
vo
20 V
=
= 0.5 A
40
40 Ω
This current flows from left to right through the 40 Ω resistor. To use
current division, we need to find the equivalent resistance of the two
parallel branches containing the 20 Ω resistor and the 50 Ω and 10 Ω
resistors:
(20)(60)
20 Ωk(50 Ω + 10 Ω) =
= 15 Ω
20 + 60
Now we use current division to find the current in the 30 Ω branch:
15
i30Ω =
(0.5 A) = 0.16667 A = 166.67 mA
15 + 30
i40Ω =
[c] We can find the power dissipated by the 50 Ω resistor if we can find the
current in this resistor. We can use current division to find this current
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Problems
3–5
from the current in the 40 Ω resistor, but first we need to calculate the
equivalent resistance of the 20 Ω branch and the 30 Ω branch:
(20)(30)
= 12 Ω
20 + 30
Current division gives:
12
i50Ω =
(0.5 A) = 0.08333 A
12 + 50 + 10
20 Ωk30 Ω =
Thus,
p50Ω = (50)(0.08333)2 = 0.34722 W = 347.22 mW
AP 3.5 [a]
We can find the current i using Ohm’s law:
1V
i=
= 0.01 A = 10 mA
100 Ω
[b]
Rm = 50 Ωk5.555 Ω = 5 Ω
We can use the meter resistance to find the current using Ohm’s law:
imeas =
1V
= 0.009524 = 9.524 mA
100 Ω + 5 Ω
AP 3.6 [a]
Use voltage division to find the voltage v:
75,000
v=
(60 V) = 50 V
75,000 + 15,000
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3–6
CHAPTER 3. Simple Resistive Circuits
[b]
The meter resistance is a series combination of resistances:
Rm = 149,950 + 50 = 150,000 Ω
We can use voltage division to find v, but first we must calculate the
equivalent resistance of the parallel combination of the 75 kΩ resistor and
the voltmeter:
(75,000)(150,000)
75,000 Ωk150,000 Ω =
= 50 kΩ
75,000 + 150,000
Thus,
vmeas =
50,000
(60 V) = 46.15 V
50,000 + 15,000
AP 3.7 [a] Using the condition for a balanced bridge, the products of the opposite
resistors must be equal. Therefore,
(1000)(150)
= 1500 Ω = 1.5 kΩ
100
[b] When the bridge is balanced, there is no current flowing through the
meter, so the meter acts like an open circuit. This places the following
branches in parallel: The branch with the voltage source, the branch with
the series combination R1 and R3 and the branch with the series
combination of R2 and Rx . We can find the current in the latter two
branches using Ohm’s law:
100Rx = (1000)(150)
so
Rx =
5V
5V
= 20 mA;
iR2 ,Rx =
= 2 mA
100 Ω + 150 Ω
1000 + 1500
We can calculate the power dissipated by each resistor using the formula
p = Ri2 :
iR1 ,R3 =
p100Ω = (100 Ω)(0.02 A)2 = 40 mW
p150Ω = (150 Ω)(0.02 A)2 = 60 mW
p1000Ω = (1000 Ω)(0.002 A)2 = 4 mW
p1500Ω = (1500 Ω)(0.002 A)2 = 6 mW
Since none of the power dissipation values exceeds 250 mW, the bridge
can be left in the balanced state without exceeding the power-dissipating
capacity of the resistors.
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Problems
3–7
AP 3.8 Convert the three Y-connected resistors, 20 Ω, 10 Ω, and 5 Ω to three
∆-connected resistors Ra , Rb, and Rc . To assist you the figure below has both
the Y-connected resistors and the ∆-connected resistors
(5)(10) + (5)(20) + (10)(20)
= 17.5 Ω
20
(5)(10) + (5)(20) + (10)(20)
Rb =
= 35 Ω
10
(5)(10) + (5)(20) + (10)(20)
Rc =
= 70 Ω
5
Ra =
The circuit with these new ∆-connected resistors is shown below:
From this circuit we see that the 70 Ω resistor is parallel to the 28 Ω resistor:
70 Ωk28 Ω =
(70)(28)
= 20 Ω
70 + 28
Also, the 17.5 Ω resistor is parallel to the 105 Ω resistor:
17.5 Ωk105 Ω =
(17.5)(105)
= 15 Ω
17.5 + 105
Once the parallel combinations are made, we can see that the equivalent 20 Ω
resistor is in series with the equivalent 15 Ω resistor, giving an equivalent
resistance of 20 Ω + 15 Ω = 35 Ω. Finally, this equivalent 35 Ω resistor is in
parallel with the other 35 Ω resistor:
35 Ωk35 Ω =
(35)(35)
= 17.5 Ω
35 + 35
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3–8
CHAPTER 3. Simple Resistive Circuits
Thus, the resistance seen by the 2 A source is 17.5 Ω, and the voltage can be
calculated using Ohm’s law:
v = (17.5 Ω)(2 A) = 35 V
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Problems
3–9
Problems
P 3.1
[a] The 6 kΩ and 12 kΩ resistors are in series, as are the 9 kΩ and 7 kΩ
resistors. The simplified circuit is shown below:
[b] The 3 kΩ, 5 kΩ, and 7 kΩ resistors are in series. The simplified circuit is
shown below:
[c] The 300 Ω, 400 Ω, and 500 Ω resistors are in series. The simplified circuit is
shown below:
P 3.2
[a] The 10 Ω and 40 Ω resistors are in parallel, as are the 100 Ω and 25 Ω
resistors. The simplified circuit is shown below:
[b] The 9 kΩ, 18 kΩ, and 6 kΩ resistors are in parallel. The simplified
circuit is shown below:
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3–10
CHAPTER 3. Simple Resistive Circuits
[c] The 600 Ω, 200 Ω, and 300 Ω resistors are in parallel. The simplified circuit
is shown below:
P 3.3
Always work from the side of the circuit furthest from the source. Remember
that the current in all series-connected circuits is the same, and that the
voltage drop across all parallel-connected resistors is the same.
[a] Req = 6 + 12 + [4k(9 + 7)] = 6 + 12 + 4k16 = 6 + 12 + 3.2 = 21.2 Ω
[b] Req = 4 k + [10 kk(3 k + 5 k + 7 k)] = 4 k + 10 kk15 k = 4 k + 6 k = 10 kΩ
[c] Req = 300 + 400 + 500 + (600k1200) = 300 + 400 + 500 + 400 = 1600 Ω
P 3.4
Always work from the side of the circuit furthest from the source. Remember
that the current in all series-connected circuits is the same, and that the
voltage drop across all parallel-connected resistors is the same.
[a] Req = 18 + [100k25k(10k40 + 22)] = 18 + [100k25k(8 + 22)]
= 18 + [100k25k30] = 18 + 12 = 30 Ω
[b] Req = 10 kk[5 k + 2 k + (9 kk18 kk6 k)] = 10 kk[5 k + 2 k + 3 k]
= 10 kk10 k = 5 kΩ
[c] Req = 600k200k300k(250 + 150) = 600k200k300k400 = 80 Ω
P 3.5
[a] Rab = 10 + (5k20) + 6 = 10 + 4 + 6 = 20 Ω
[b] Rab = 30 kk60 kk[20 k + (200 kk50 k)] = 30 kk60 kk(20 k + 40 k)
= 30 kk60 kk60 k = 15 kΩ
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Problems
P 3.6
[a] 60k20 = 1200/80 = 15 Ω
15 + 8 + 7 = 30 Ω
3–11
12k24 = 288/36 = 8 Ω
30k120 = 3600/150 = 24 Ω
Rab = 15 + 24 + 25 = 64 Ω
[b] 35 + 40 = 75 Ω
75k50 = 3750/125 = 30 Ω
30 + 20 = 50 Ω
50k75 = 3750/125 = 30 Ω
30 + 10 = 40 Ω
40k60 + 9k18 = 24 + 6 = 30 Ω
30k30 = 15 Ω
Rab = 10 + 15 + 5 = 30 Ω
[c] 50 + 30 = 80 Ω
P 3.7
80k20 = 16 Ω
16 + 14 = 30 Ω
30 + 24 = 54 Ω
54k27 = 18 Ω
18 + 12 = 30 Ω
30k30 = 15 Ω
Rab = 3 + 15 + 2 = 20 Ω
[a] For circuit (a)
Rab = 4k(3 + 7 + 2) = 4k12 = 3 Ω
For circuit (b)
Rab = 6 + 2 + [8k(7 + 5k2.5k7.5k5k(9 + 6))] = 6 + 2 + 8k(7 + 1)
= 6 + 2 + 4 = 12 Ω
For circuit (c)
144k(4 + 12) = 14.4 Ω
14.4 + 5.6 = 20 Ω
20k12 = 7.5 Ω
7.5 + 2.5 = 10 Ω
10k15 = 6 Ω
14 + 6 + 10 = 30 Ω
Rab = 30k60 = 20 Ω
[b] Pa =
Pb =
152
= 75 W
3
482
= 192 W
12
Pc = 52 (20) = 500 W
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3–12
P 3.8
CHAPTER 3. Simple Resistive Circuits
[a] p4Ω
=
i2s 4 = (12)2 4 = 576 W
p18Ω = (4)2 18 = 288 W
p3Ω
=
(8)2 3 = 192 W
p6Ω = (8)2 6 = 384 W
[b] p120V (delivered) = 120is = 120(12) = 1440 W
[c] pdiss = 576 + 288 + 192 + 384 = 1440 W
P 3.9
[a] From Ex. 3-1: i1 = 4 A,
i2 = 8 A,
is = 12 A
at node b: −12 + 4 + 8 = 0,
at node d: 12 − 4 − 8 = 0
[b] v1
= 4is = 48 V
v3 = 3i2 = 24 V
v2 = 18i1 = 72 V
v4 = 6i2 = 48 V
loop abda: −120 + 48 + 72 = 0,
loop bcdb: −72 + 24 + 48 = 0,
loop abcda: −120 + 48 + 24 + 48 = 0
P 3.10
Req = 10k[6 + 5k(8 + 12)] = 10k(6 + 5k20) = 10k(6 + 4) = 5 Ω
v10A = v10Ω = (10 A)(5 Ω) = 50 V
Using voltage division:
v5Ω =
5k(8 + 12)
4
(50) =
(50) = 20 V
6 + 5k(8 + 12)
6+4
Thus, p5Ω =
P 3.11
2
v5Ω
202
=
= 80 W
5
5
[a]
Req = (10 + 20)k[12 + (90k10)] = 30k15 = 10 Ω
v2.4A = 10(2.4) = 24 V
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Problems
vo = v20Ω =
v90Ω =
io =
3–13
20
(24) = 16 V
10 + 20
90k10
9
(24) = (24) = 14.4 V
6 + (90k10)
15
14.4
= 0.16 A
90
(v2.4A − v90Ω)2
(24 − 14.4)2
=
= 15.36 W
6
6
[c] p2.4A = −(2.4)(24) = −57.6 W
Thus the power developed by the current source is 57.6 W.
[b] p6Ω =
P 3.12
[a] R + R = 2R
[b] R + R + R + · · · + R = nR
[c] R + R = 2R = 3000 so R = 1500 = 1.5 kΩ
This is a resistor from Appendix H.
[d] nR = 4000;
so if n = 4, R = 1 kΩ
This is a resistor from Appendix H.
P 3.13
[a] Req
R2
R
= RkR =
=
2R
2
[b] Req
=
=
=
RkRkRk · · · kR
(n R’s)
R
Rk
n−1
R2 /(n − 1)
R2
R
=
=
R + R/(n − 1)
nR
n
R
= 5000 so R = 10 kΩ
2
This is a resistor from Appendix H.
R
[d]
= 4000 so R = 4000n
n
If n = 3
r = 4000(3) = 12 kΩ
This is a resistor from Appendix H. So put three 12k resistors in parallel
to get 4kΩ.
[c]
P 3.14
4=
20R2
R2 + 40
so
R2 = 10 Ω
3=
20Re
40 + Re
so
Re =
Thus,
120
10RL
=
17
10 + RL
120
Ω
17
so
RL = 24 Ω
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3–14
P 3.15
CHAPTER 3. Simple Resistive Circuits
[a] vo =
160(3300)
= 66 V
(4700 + 3300)
[b] i = 160/8000 = 20 mA
PR1 = (400 × 10−6 )(4.7 × 103 ) = 1.88 W
PR2 = (400 × 10−6 )(3.3 × 103 ) = 1.32 W
[c] Since R1 and R2 carry the same current and R1 > R2 to satisfy the voltage
requirement, first pick R1 to meet the 0.5 W specification
iR1
160 − 66
=
,
R1
Thus, R1 ≥
Therefore,
942
0.5
or
94
R1
2
R1 ≤ 0.5
R1 ≥ 17,672 Ω
Now use the voltage specification:
R2
(160) = 66
R2 + 17,672
Thus, R2 = 12,408 Ω
P 3.16
[a] vo =
40R2
=8
R1 + R2
so
Let Re = R2 kRL =
vo =
R2 RL
R2 + RL
40Re
= 7.5
R1 + Re
so
Then, 4R2 = 4.33Re =
Thus, R2 = 300 Ω
R1 = 4R2
R1 = 4.33Re
4.33(3600R2 )
3600 + R2
and
R1 = 4(300) = 1200 Ω
[b] The resistor that must dissipate the most power is R1 , as it has the largest
resistance and carries the same current as the parallel combination of R2
and the load resistor. The power dissipated in R1 will be maximum when
the voltage across R1 is maximum. This will occur when the voltage
divider has a resistive load. Thus,
vR1 = 40 − 7.5 = 32.5 V
pR1 =
32.52
= 880.2 m W
1200
Thus the minimum power rating for all resistors should be 1 W.
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Problems
P 3.17
3–15
Refer to the solution to Problem 3.16. The voltage divider will reach the
maximum power it can safely dissipate when the power dissipated in R1 equals
1 W. Thus,
vR2 1
=1
1200
so
vR1 = 34.64 V
vo = 40 − 34.64 = 5.36 V
So,
40Re
= 5.36
1200 + Re
Thus,
and
(300)RL
= 185.68
300 + RL
Re = 185.68 Ω
and
RL = 487.26 Ω
The minimum value for RL from Appendix H is 560 Ω.
P 3.18
Begin by using the relationships among the branch currents to express all
branch currents in terms of i4:
i1 = 2i2 = 2(2i3 ) = 4(2i4 )
i2 = 2i3 = 2(2i4 )
i3 = 2i4
Now use KCL at the top node to relate the branch currents to the current
supplied by the source.
i1 + i2 + i3 + i4 = 1 mA
Express the branch currents in terms of i4 and solve for i4:
1 mA = 8i4 + 4i4 + 2i4 + i4 = 15i4
so
i4 =
0.001
A
15
Since the resistors are in parallel, the same voltage, 1 V appears across each of
them. We know the current and the voltage for R4 so we can use Ohm’s law
to calculate R4 :
R4 =
vg
1V
=
= 15 kΩ
i4
(1/15) mA
Calculate i3 from i4 and use Ohm’s law as above to find R3 :
i3 = 2i4 =
0.002
A
15
vg
1V
.·. R3 =
=
= 7.5 kΩ
i3
(2/15) mA
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3–16
CHAPTER 3. Simple Resistive Circuits
Calculate i2 from i4 and use Ohm’s law as above to find R2 :
i2 = 4i4 =
0.004
A
15
vg
1V
.·. R2 =
=
= 3750 Ω
i2
(4/15) mA
Calculate i1 from i4 and use Ohm’s law as above to find R1 :
i1 = 8i4 =
0.008
A
15
vg
1V
.·. R1 =
=
= 1875 Ω
i1
(8/15) mA
The resulting circuit is shown below:
P 3.19
[a ]
40 kΩ + 60 kΩ = 100 kΩ
25 kΩk100 kΩ = 20 kΩ
vo1 =
vo =
20,000
(380) = 80 V
(75,000 + 20,000)
60,000
(vo1 ) = 48 V
(100,000)
[b ]
i=
380
= 3.8 mA
100,000
25,000i = 95 V
vo =
60,000
(95) = 57 V
100,000
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Problems
3–17
[c] It removes loading effect of second voltage divider on the first voltage
divider. Observe that the open circuit voltage of the first divider is
0
vo1
=
25,000
(380) = 95 V
(100,000)
Now note this is the input voltage to the second voltage divider when the
current controlled voltage source is used.
P 3.20
(24)2
= 80,
R1 + R2 + R3
Therefore, R1 + R2 + R3 = 7.2 Ω
(R1 + R2 )24
= 12
(R1 + R2 + R3 )
Therefore, 2(R1 + R2) = R1 + R2 + R3
Thus, R1 + R2 = R3 ;
2R3 = 7.2;
R3 = 3.6 Ω
R2 (24)
=5
R1 + R2 + R3
4.8R2 = R1 + R2 + 3.6 = 7.2
Thus, R2 = 1.5 Ω;
P 3.21
R1 = 7.2 − R2 − R3 = 2.1 Ω
[a] Let vo be the voltage across the parallel branches, positive at the upper
terminal, then
ig = voG1 + voG2 + · · · + voGN = vo (G1 + G2 + · · · + GN )
It follows that
vo =
ig
(G1 + G2 + · · · + GN )
The current in the k th branch is
ik =
[b] i5 =
P 3.22
ik = vo Gk ;
Thus,
ig Gk
[G1 + G2 + · · · + GN ]
40(0.2)
= 3.2 A
2 + 0.2 + 0.125 + 0.1 + 0.05 + 0.025
[a] At no load:
At full load:
vo = kvs =
R2
vs .
R1 + R2
vo = αvs =
Re
vs ,
R1 + Re
where Re =
Ro R2
Ro + R2
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3–18
CHAPTER 3. Simple Resistive Circuits
Therefore k
=
α
=
Thus
1−α
α
R2
R1 + R2
Re
R1 + Re
R1 =
[b] R1
=
R2
=
and
(1 − k)
R2
k
(1 − α)
R1 =
Re
α
R1 =
R2 Ro
(1 − k)
=
R2
Ro + R2
k
Solving for R2 yields
Also,
and
R2 =
(1 − k)
R2
k
(k − α)
Ro
α(1 − k)
.·.
R1 =
(k − α)
Ro
αk
0.05
Ro = 2.5 kΩ
0.68
0.05
Ro = 14.167 kΩ
0.12
[c ]
Maximum dissipation in R2 occurs at no load, therefore,
PR2 (max)
[(60)(0.85)]2
=
= 183.6 mW
14,167
Maximum dissipation in R1 occurs at full load.
PR1 (max) =
[60 − 0.80(60)]2
= 57.60 mW
2500
[d ]
PR1
=
PR2
=
(60)2
= 1.44 W = 1440 mW
2500
(0)2
=0W
14,167
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Problems
P 3.23
3–19
[a] The equivalent resistance of the circuit to the right of the 18 Ω resistor is
100k25k[(40k10) + 22] = 100k25k30 = 12 Ω
Thus by voltage division,
v18 =
18
(60) = 36 V
18 + 12
[b] The current in the 18 Ω resistor can be found from its voltage using Ohm’s
law:
36
i18 =
=2A
18
[c] The current in the 18 Ω resistor divides among three branches – one
containing 100 Ω, one containing 25 Ω and one containing
(22 + 40k10) = 30 Ω. Using current division,
100k25k30
12
(i18) = (2) = 0.96 A
25
25
[d] The voltage drop across the 25 Ω resistor can be found using Ohm’s law:
i25 =
v25 = 25i25 = 25(0.96) = 24 V
[e] The voltage v25 divides across the 22 Ω resistor and the equivalent
resistance 40k10 = 8 Ω. Using voltage division,
v10 =
P 3.24
8
(24) = 6.4 V
8 + 22
[a] The equivalent resistance to the right of the 10 kΩ resistor is
5 k + 2 k + [9 kk18 kk6 k)] = 10 kΩ. Therefore,
10 kk10 k
(0.050) = 25 mA
10 k
[b] The voltage drop across the 10 kΩ resistor can be found using Ohm’s law:
i10k =
v10k = (10, 000)i10k = (10, 000)(0.025) = 250 V
[c] The voltage v10k drops across the 5 kΩ resistor, the 2 kΩ resistor and the
equivalent resistance of the 9 kΩ, 18 kΩ and 6 kΩ resistors in parallel.
Thus, using voltage division,
v6k =
2k
2
(250) = (250) = 50 V
5 k + 2 k + [9 kk18 kk6 k]
10
[d] The current through the 2 kΩ resistor can be found from its voltage using
Ohm’s law:
v2k
50
i2k =
=
= 25 mA
2000
2000
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3–20
CHAPTER 3. Simple Resistive Circuits
[e] The current through the 2 kΩ resistor divides among the 9 kΩ, 18 kΩ,
and 6 kΩ. Using current division,
i18k =
P 3.25
9 kk18 kk6 k
3
(0.025) = (0.025) = 4.167 mA
18 k
18
The equivalent resistance of the circuit to the right of the 90 Ω resistor is
Req = [(150k75) + 40]k(30 + 60) = 90k90 = 45 Ω
Use voltage division to find the voltage drop between the top and bottom
nodes:
vReq =
45
(3) = 1 V
45 + 90
Use voltage division again to find v1 from vReq:
v1 =
150k75
50
5
(1) = (1) = V
150k75 + 40
90
9
Use voltage division one more time to find v2 from vReq:
v2 =
P 3.26
30
1
(1) = V
30 + 60
3
i10k =
(18)(15 k)
= 6.75 mA
40 k
v15k = −(6.75 m)(15 k) = −101.25 V
i3k = 18 m − 6.75 m = 11.25 mA
v12k = −(12 k)(11.25 m) = −135 V
vo = −101.25 − (−135) = 33.75 V
P 3.27
[a] v6k =
6
(18) = 13.5 V
6+2
v3k =
3
(18) = 4.5 V
3+9
vx = v6k − v3k = 13.5 − 4.5 = 9 V
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Problems
3–21
6
[b] v6k = (Vs ) = 0.75Vs
8
3
v3k = (Vs ) = 0.25Vs
12
vx = (0.75Vs ) − (0.25Vs ) = 0.5Vs
P 3.28
5 Ωk20 Ω = 4 Ω;
Therefore, ig =
i6Ω =
P 3.29
4 Ω + 6 Ω = 10 Ω;
10k(15 + 12 + 13) = 8 Ω;
125
= 12.5 A
2+8
8
(12.5) = 10 A;
6+4
io =
5k20
(10) = 2 A
20
[a] The equivalent resistance seen by the voltage source is
60k[8 + 30k(4 + 80k20)] = 60k[8 + 30k20] = 60k20 = 15 Ω
Thus,
300
ig =
= 20 A
15
[b] Use current division to find the current in the 8 Ω division:
15
(20) = 15 A
20
Use current division again to find the current in the 30 Ω resistor:
12
i30 = (15) = 6 A
30
Thus,
p30 = (6)2 (30) = 1080 W
P 3.30
[a] The voltage across the 9 Ω resistor is 1(12 + 6) = 18 V.
The current in the 9 Ω resistor is 18/9 = 2 A. The current in the 2 Ω
resistor is 1 + 2 = 3 A. Therefore, the voltage across the 24 Ω resistor is
(2)(3) + 18 = 24 V.
The current in the 24 Ω resistor is 1 A. The current in the 3 Ω resistor is
1 + 2 + 1 = 4 A. Therefore, the voltage across the 72 Ω resistor is
24 + 3(4) = 36 V.
The current in the 72 Ω resistor is 36/72 = 0.5 A.
The 20 Ωk5 Ω resistors are equivalent to a 4 Ω resistor. The current in
this equivalent resistor is 0.5 + 1 + 3 = 4.5 A. Therefore the voltage
across the 108 Ω resistor is 36 + 4.5(4) = 54 V.
The current in the 108 Ω resistor is 54/108 = 0.5 A. The current in the
1.2 Ω resistor is 4.5 + 0.5 = 5 A. Therefore,
vg = (1.2)(5) + 54 = 60 V
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3–22
CHAPTER 3. Simple Resistive Circuits
[b] The current in the 20 Ω resistor is
(4.5)(4)
18
=
= 0.9 A
20
20
Thus, the power dissipated by the 20 Ω resistor is
i20 =
p20 = (0.9)2 (20) = 16.2 W
P 3.31
For all full-scale readings the total resistance is
RV + Rmovement =
full-scale reading
10−3
We can calculate the resistance of the movement as follows:
Rmovement =
Therefore,
20 mV
= 20 Ω
1 mA
RV = 1000 (full-scale reading) − 20
[a] RV = 1000(50) − 20 = 49, 980 Ω
[b] RV = 1000(5) − 20 = 4980 Ω
[c] RV = 1000(0.25) − 20 = 230 Ω
[d] RV = 1000(0.025) − 20 = 5 Ω
P 3.32
[a] vmeas = (50 × 10−3 )[15k45k(4980 + 20)] = 0.5612 V
[b] vtrue = (50 × 10−3 )(15k45) = 0.5625 V
% error =
P 3.33
0.5612
− 1 × 100 = −0.224%
0.5625
The measured value is
ig =
50
= 1.995526 A;
(15.05618 + 10)
The true value is
ig =
60k20.1 = 15.05618 Ω.
imeas =
60
(1.996) = 1.494768 A
80.1
60k20 = 15 Ω.
50
= 2 A;
(15 + 10)
itrue =
60
(2) = 1.5 A
80
1.494768
%error =
− 1 × 100 = −0.34878% ≈ −0.35%
1.5
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Problems
P 3.34
3–23
Begin by using current division to find the actual value of the current io :
itrue =
15
(50 mA) = 12.5 mA
15 + 45
imeas =
15
(50 mA) = 12.4792 mA
15 + 45 + 0.1
12.4792
% error =
− 1 100 = −0.166389% ≈ −0.17%
12.5
P 3.35
[a] The model of the ammeter is an ideal ammeter in parallel with a resistor
whose resistance is given by
100 mV
= 50 Ω.
2 mA
We can calculate the current through the real meter using current
division:
(25/12)
25
1
im =
(imeas ) =
(imeas ) = imeas
50 + (25/12)
625
25
Rm =
[b] At full scale, imeas = 5 A and im = 2 mA so 5 − 0.002 = 4998 mA flows
throught the resistor RA :
RA =
im =
100 mV
100
=
Ω
4998 mA 4998
(100/4998)
1
(imeas) =
(imeas)
50 + (100/4998)
2500
[c] Yes
P 3.36
Original meter:
Re =
50 × 10−3
= 0.01 Ω
5
Modified meter:
Re =
(0.02)(0.01)
= 0.00667 Ω
0.03
.·. (Ifs )(0.00667) = 50 × 10−3
.·. Ifs = 7.5 A
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3–24
P 3.37
CHAPTER 3. Simple Resistive Circuits
[a ]
20 × 103 i1 + 80 × 103 (i1 − iB ) = 7.5
80 × 103 (i1 − iB) = 0.6 + 40iB (0.2 × 103 )
. ·.
100i1 − 80iB = 7.5 × 10−3
80i1 − 88iB = 0.6 × 10−3
Calculator solution yields iB = 225 µA
[b] With the insertion of the ammeter the equations become
100i1 − 80iB = 7.5 × 10−3
(no change)
80 × 103 (i1 − iB) = 103 iB + 0.6 + 40iB (200)
80i1 − 89iB = 0.6 × 10−3
Calculator solution yields iB = 216 µA
216
[c] % error =
− 1 100 = −4%
225
P 3.38
The current in the shunt resistor at full-scale deflection is
iA = ifullscale = 2 × 10−3 A. The voltage across RA at full-scale deflection is
always 50 mV; therefore,
RA =
50 × 10−3
50
=
−3
ifullscale − 2 × 10
1000ifullscale − 2
50
= 5.001 mΩ
10,000 − 2
50
[b] RA =
= 50.1 mΩ
1000 − 2
50
[c] RA =
= 1.042 mΩ
50 − 2
[a] RA =
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Problems
[d] RA =
P 3.39
50
=∞
2−2
3–25
(open circuit)
At full scale the voltage across the shunt resistor will be 50 mV; therefore the
power dissipated will be
PA =
(50 × 10−3 )2
RA
Therefore RA ≥
(50 × 10−3 )2
= 5 mΩ
0.5
Otherwise the power dissipated in RA will exceed its power rating of 0.5 W
When RA = 5 mΩ, the shunt current will be
iA =
50 × 10−3
= 10 A
5 × 10−3
The measured current will be imeas = 10 + 0.001 = 10.001 A
.·. Full-scale reading for practical purposes is 10 A.
P 3.40
Rmeter = Rm + Rmovement =
750 V
= 500 kΩ
1.5 mA
vmeas = (25 kΩk125 kΩk50 kΩ)(30 mA) = (20 kΩ)(30 mA) = 600 V
vtrue = (25 kΩk125 kΩ)(30 mA) = (20.83 kΩ)(30 mA) = 625 V
600
% error =
− 1 100 = −4%
625
P 3.41
[a] Since the unknown voltage is greater than either voltmeter’s maximum
reading, the only possible way to use the voltmeters would be to connect
them in series.
[b ]
Rm1 = (300)(900) = 270 kΩ;
Rm2 = (150)(1200) = 180 kΩ
.·. Rm1 + Rm2 = 450 kΩ
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3–26
CHAPTER 3. Simple Resistive Circuits
i1
max
=
300
× 10−3 = 1.11 mA;
270
i2
max
=
150
× 10−3 = 0.833 mA
180
.·. imax = 0.833 mA since meters are in series
vmax = (0.833 × 10−3 )(270 + 180)103 = 375 V
Thus the meters can be used to measure the voltage.
320
[c] im =
= 0.711 mA
450 × 103
vm1 = (0.711)(270) = 192 V;
P 3.42
vm2 = (0.711)(180) = 128 V
The current in the series-connected voltmeters is
205.2
136.8
=
= 0.76 mA
270,000
180,000
im =
v50 kΩ = (0.76 × 10−3 )(50,000) = 38 V
Vpower
P 3.43
supply
= 205.2 + 136.8 + 38 = 380 V
[a] vmeter = 180 V
[b] Rmeter = (100)(200) = 20 kΩ
20k70 = 15.555556 kΩ
vmeter =
180
× 15.555556 = 78.75 V
35.555556
[c] 20k20 = 10 kΩ
vmeter =
180
(10) = 22.5 V
80
[d] vmeter a = 180 V
vmeter b + vmeter c = 101.26 V
No, because of the loading effect.
P 3.44
From the problem statement we have
Vs (10)
50 =
(1) Vs in mV; Rs in MΩ
10 + Rs
48.75 =
Vs (6)
6 + Rs
(2)
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Problems
[a] From Eq (1)
3–27
10 + Rs = 0.2Vs
.·. Rs = 0.2Vs − 10
Substituting into Eq (2) yields
48.75 =
6Vs
0.2Vs − 4
or
Vs = 52 mV
[b] From Eq (1)
50 =
520
10 + Rs
or
50Rs = 20
So Rs = 400 kΩ
P 3.45
[a] R1
=
(100/2)103 = 50 kΩ
R2
=
(10/2)103 = 5 kΩ
R3
=
(1/2)103 = 500 Ω
[b] Let ia
= actual current in the movement
id
= design current in the movement
ia
Then % error =
− 1 100
id
For the 100 V scale:
100
100
ia =
=
,
50,000 + 25
50,025
ia
50,000
=
= 0.9995
id
50,025
For the 10 V scale:
ia
5000
=
= 0.995
id
5025
For the 1 V scale:
ia
500
=
= 0.9524
id
525
P 3.46
id =
100
50,000
% error = (0.9995 − 1)100 = −0.05%
% error = (0.995 − 1.0)100 = −0.4975%
% error = (0.9524 − 1.0)100 = −4.76%
[a] Rmovement = 50 Ω
R1 + Rmovement =
30
= 30 kΩ
1 × 10−3
R2 + R1 + Rmovement =
.·. R1 = 29,950 Ω
150
= 150 kΩ
1 × 10−3
.·. R2 = 120 kΩ
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3–28
CHAPTER 3. Simple Resistive Circuits
R3 + R2 + R1 + Rmovement =
300
= 300 kΩ
1 × 10−3
.·. R3 = 150 kΩ
[b]
v1 = (0.96 m)(150 k) = 144 V
imove =
i1 =
144
= 0.96 mA
120 + 29.95 + 0.05
144
= 0.192 mA
750 k
i2 = imove + i1 = 0.96 m + 0.192 m = 1.152 mA
vmeas = vx = 144 + 150i2 = 316.8 V
[c] v1 = 150 V;
i2 = 1 m + 0.20 m = 1.20 mA
i1 = 150/750,000 = 0.20 mA
.·. vmeas = vx = 150 + (150 k)(1.20 m) = 330 V
P 3.47
[a] Rmeter = 300 kΩ + 600 kΩk200 kΩ = 450 kΩ
450k360 = 200 kΩ
Vmeter =
200
(600) = 500 V
240
[b] What is the percent error in the measured voltage?
True value =
% error =
360
(600) = 540 V
400
500
− 1 100 = −7.41%
540
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Problems
P 3.48
3–29
Since the bridge is balanced, we can remove the detector without disturbing
the voltages and currents in the circuit.
It follows that
i1 =
i2 =
ig (R2 + Rx )
ig (R2 + Rx )
X
=
R1 + R2 + R3 + Rx
R
ig (R1 + R3 )
ig (R1 + R3 )
X
=
R1 + R2 + R3 + Rx
R
v3 = R3 i1 = vx = i2Rx
R3 ig (R2 + Rx )
Rx ig (R1 + R3 )
X
X
.·.
=
R
R
.·. R3 (R2 + Rx ) = Rx (R1 + R3 )
From which Rx =
P 3.49
R2 R3
R1
Note the bridge structure is balanced, that is 15 × 5 = 3 × 25, hence there is
no current in the 5 kΩ resistor. It follows that the equivalent resistance of the
circuit is
Req = 750 + (15,000 + 3000)k(25,000 + 5000) = 750 + 11,250 = 12 kΩ
The source current is 192/12,000 = 16 mA.
The current down through the branch containing the 15 kΩ and 3 kΩ
resistors is
i3k =
11,250
(0.016) = 10 mA
18,000
.·. p3k = 3000(0.01)2 = 0.3 W
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3–30
P 3.50
CHAPTER 3. Simple Resistive Circuits
Redraw the circuit, replacing the detector branch with a short circuit.
6 kΩk30 kΩ = 5 kΩ
12 kΩk20 kΩ = 7.5 kΩ
is =
75
= 6 mA
12,500
v1 = 0.006(5000) = 30 V
v2 = 0.006(7500) = 45 V
i1 =
30
= 5 mA
6000
i2 =
45
= 3.75 mA
12,000
id = i1 − i2 = 1.25 mA
P 3.51
[a]
The condition for a balanced bridge is that the product of the opposite
resistors must be equal:
(500)(Rx ) = (1000)(750)
so
Rx =
(1000)(750)
= 1500 Ω
500
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Problems
3–31
[b] The source current is the sum of the two branch currents. Each branch
current can be determined using Ohm’s law, since the resistors in each
branch are in series and the voltage drop across each branch is 24 V:
is =
24 V
24 V
+
= 28.8 mA
500 Ω + 750 Ω 1000 Ω + 1500 Ω
[c] We can use Ohm’s law to find the current in each branch:
ileft =
24
= 19.2 mA
500 + 750
iright =
24
= 9.6 mA
1000 + 1500
Now we can use the formula p = Ri2 to find the power dissipated by each
resistor:
p500 = (500)(0.0192)2 = 184.32 mW
p1000 = (1000)(0.0096)2 = 92.16 mW
p750 = (750)(0.0192)2 = 276.18 mW
p1500 = (1500)(0.0096)2 = 138.24 mW
Thus, the 750 Ω resistor absorbs the most power; it absorbs 276.48 mW
of power.
[d] From the analysis in part (c), the 1000 Ω resistor absorbs the least power;
it absorbs 92.16 mW of power.
P 3.52
In order that all four decades (1, 10, 100, 1000) that are used to set R3
contribute to the balance of the bridge, the ratio R2 /R1 should be set to 0.001.
P 3.53
Begin by transforming the ∆-connected resistors (10 Ω, 40 Ω, 50 Ω) to
Y-connected resistors. Both the Y-connected and ∆-connected resistors are
shown below to assist in using Eqs. 3.44 – 3.46:
Now use Eqs. 3.44 – 3.46 to calculate the values of the Y-connected resistors:
R1 =
(40)(10)
= 4 Ω;
10 + 40 + 50
R2 =
(10)(50)
= 5 Ω;
10 + 40 + 50
R3 =
(40)(50)
= 20 Ω
10 + 40 + 50
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3–32
CHAPTER 3. Simple Resistive Circuits
The transformed circuit is shown below:
The equivalent resistance seen by the 24 V source can be calculated by making
series and parallel combinations of the resistors to the right of the 24 V source:
Req = (15 + 5)k(1 + 4) + 20 = 20k5 + 20 = 4 + 20 = 24 Ω
Therefore, the current i in the 24 V source is given by
i=
24 V
=1A
24 Ω
Use current division to calculate the currents i1 and i2 . Note that the current
i1 flows in the branch containing the 15 Ω and 5 Ω series connected resistors,
while the current i2 flows in the parallel branch that contains the series
connection of the 1 Ω and 4 Ω resistors:
i1 =
4
4
(i) = (1 A) = 0.2 A,
15 + 5
20
and
i2 = 1 A − 0.2 A = 0.8 A
Now use KVL and Ohm’s law to calculate v1. Note that v1 is the sum of the
voltage drop across the 4 Ω resistor, 4i2 , and the voltage drop across the 20 Ω
resistor, 20i:
v1 = 4i2 + 20i = 4(0.8 A) + 20(1 A) = 3.2 + 20 = 23.2 V
Finally, use KVL and Ohm’s law to calculate v2 . Note that v2 is the sum of
the voltage drop across the 5 Ω resistor, 5i1 , and the voltage drop across the
20 Ω resistor, 20i:
v2 = 5i1 + 20i = 5(0.2 A) + 20(1 A) = 1 + 20 = 21 V
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Problems
P 3.54
3–33
[a] After the 20 Ω—100 Ω—50 Ω wye is replaced by its equivalent delta, the
circuit reduces to
Now the circuit can be reduced to
96
(1000) = 240 mA
400
400
io =
(240) = 96 mA
1000
80
[b] i1 =
(240) = 48 mA
400
[c] Now that io and i1 are known return to the original circuit
i=
v2 = (50)(0.048) + (600)(0.096) = 60 V
i2 =
v2
60
=
= 600 mA
100
100
[d] vg = v2 + 20(0.6 + 0.048) = 60 + 12.96 = 72.96 V
pg = −(vg )(1) = −72.96 W
Thus the current source delivers 72.96 W.
P 3.55
The top of the pyramid can be replaced by a resistor equal to
R1 =
(18)(9)
= 6 kΩ
27
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3–34
CHAPTER 3. Simple Resistive Circuits
The lower left and right deltas can be replaced by wyes. Each resistance in the
wye equals 3 kΩ. Thus our circuit can be reduced to
Now the 12 kΩ in parallel with 6 kΩ reduces to 4 kΩ.
.·. Rab = 3 k + 4 k + 3 k = 10 kΩ
P 3.56
[a] Calculate the values of the Y-connected resistors that are equivalent to the
10 Ω, 40 Ω, and 50Ω ∆-connected resistors:
RX =
(10)(50)
= 5 Ω;
10 + 40 + 50
RZ =
(10)(40)
= 4Ω
10 + 40 + 50
RY =
(50)(40)
= 20 Ω;
10 + 40 + 50
Replacing the R2 —R3—R4 delta with its equivalent Y gives
Now calculate the equivalent resistance Rab by making series and parallel
combinations of the resistors:
Rab = 13 + 5 + [(8 + 4)k(20 + 4)] + 7 = 33 Ω
[b] Calculate the values of the ∆-connected resistors that are equivalent to
the 10 Ω, 8 Ω, and 40 Ω Y-connected resistors:
(10)(8) + (8)(40) + (10)(40)
800
RX =
=
= 100 Ω
8
8
(10)(8) + (8)(40) + (10)(40)
800
RY =
=
= 80 Ω
10
10
(10)(8) + (8)(40) + (10)(40)
800
RZ =
=
= 20 Ω
40
40
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Problems
3–35
Replacing the R2 , R4 , R5 wye with its equivalent ∆ gives
Make series and parallel combinations of the resistors to find the
equivalent resistance Rab:
100 Ωk50 Ω = 33.33 Ω;
80 Ωk4 Ω = 3.81 Ω
.·. 20k(33.33 + 3.81) = 13 Ω
.·. Rab = 13 + 13 + 7 = 33 Ω
[c] Convert the delta connection R4 —R5—R6 to its equivalent wye.
Convert the wye connection R3 —R4 —R6 to its equivalent delta.
P 3.57
[a] Convert the upper delta to a wye.
R1 =
(50)(50)
= 12.5 Ω
200
R2 =
(50)(100)
= 25 Ω
200
R3 =
(100)(50)
= 25 Ω
200
Convert the lower delta to a wye.
R4 =
(60)(80)
= 24 Ω
200
R5 =
(60)(60)
= 18 Ω
200
R6 =
(80)(60)
= 24 Ω
200
Now redraw the circuit using the wye equivalents.
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3–36
CHAPTER 3. Simple Resistive Circuits
Rab = 1.5 + 12.5 +
(120)(80)
+ 18 = 14 + 48 + 18 = 80 Ω
200
[b] When vab = 400 V
400
ig =
=5A
80
48
i31 = (5) = 3 A
80
p31Ω = (31)(3)2 = 279 W
P 3.58
Replace the upper and lower deltas with the equivalent wyes:
R1U =
(10)(50)
(50)(40)
(10)(40)
= 5 Ω; R2U =
= 20 Ω; R3U =
= 4Ω
100
100
100
R1L =
(10)(60)
(60)(30)
(10)(30)
= 6 Ω; R2L =
= 18 Ω; R3L =
= 3Ω
100
100
100
The resulting circuit is shown below:
Now make series and parallel combinations of the resistors:
(4 + 6)k(20 + 32 + 20 + 18) = 10k90 = 9 Ω
Rab = 33 + 5 + 9 + 3 + 40 = 90 Ω
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Problems
P 3.59
3–37
8 + 12 = 20 Ω
20k60 = 15 Ω
15 + 20 = 35 Ω
35k140 = 28 Ω
28 + 22 = 50 Ω
50k75 = 30 Ω
30 + 10 = 40 Ω
ig = 240/40 = 6 A
io = (6)(50)/125 = 2.4 A
i140Ω = (6 − 2.4)(35)/175 = 0.72 A
p140Ω = (0.72)2 (140) = 72.576 W
P 3.60
[a] Replace the 60—120—20 Ω delta with a wye equivalent to get
is =
750
750
=
= 10 A
5 + (24 + 36)k(14 + 6) + 12 + 43
75
i1 =
(24 + 36)k(14 + 6)
15
(10) = (10) = 2.5 A
24 + 36
60
[b] io = 10 − 2.5 = 7.5 A
v = 36i1 − 6io = 36(2.5) − 6(7.5) = 45 V
v
45
= 7.5 +
= 8.25 A
60
60
[d] Psupplied = (750)(10) = 7500 W
[c] i2 = io +
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3–38
CHAPTER 3. Simple Resistive Circuits
P 3.61
25k6.25 = 5 Ω
i1 =
60k30 = 20 Ω
(6)(15)
= 2.25 A;
(40)
vx = 20i1 = 45 V
vg = 25i1 = 56.25 V
v6.25 = vg − vx = 11.25 V
11.252 452 56.252
Pdevice =
+
+
= 298.6875 W
6.25
30
15
P 3.62
[a] Subtracting Eq. 3.42 from Eq. 3.43 gives
R1 − R2 = (Rc Rb − Rc Ra )/(Ra + Rb + Rc ).
Adding this expression to Eq. 3.41 and solving for R1 gives
R1 = Rc Rb/(Ra + Rb + Rc).
To find R2, subtract Eq. 3.43 from Eq. 3.41 and add this result to
Eq. 3.42. To find R3 , subtract Eq. 3.41 from Eq. 3.42 and add this result
to Eq. 3.43.
[b] Using the hint, Eq. 3.43 becomes
R1 + R3 =
Rb [(R2/R3 )Rb + (R2 /R1 )Rb ]
Rb(R1 + R3 )R2
=
(R2/R1 )Rb + Rb + (R2 /R3 )Rb
(R1 R2 + R2 R3 + R3 R1 )
Solving for Rb gives Rb = (R1 R2 + R2R3 + R3 R1 )/R2 . To find Ra : First
use Eqs. 3.44–3.46 to obtain the ratios (R1/R3 ) = (Rc /Ra ) or
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Problems
3–39
Rc = (R1 /R3 )Ra and (R1/R2 ) = (Rb /Ra ) or Rb = (R1/R2 )Ra. Now use
these relationships to eliminate Rb and Rc from Eq. 3.42. To find Rc , use
Eqs. 3.44–3.46 to obtain the ratios Rb = (R3 /R2 )Rc and
Ra = (R3 /R1 )Rc . Now use the relationships to eliminate Rb and Ra from
Eq. 3.41.
P 3.63
P 3.64
1
R1
=
Ra
R1 R2 + R2 R3 + R3R1
1/G1
=
(1/G1 )(1/G2 ) + (1/G2 )(1/G3 ) + (1/G3 )(1/G1 )
(1/G1 )(G1 G2 G3 )
G2 G3
=
=
G1 + G2 + G3
G1 + G2 + G3
Similar manipulations generate the expressions for Gb and Gc .
Ga
=
[a] Rab = 2R1 +
Therefore
Thus
R2 (2R1 + RL )
= RL
2R1 + R2 + RL
2R1 − RL +
R2 (2R1 + RL )
=0
2R1 + R2 + RL
R2L = 4R21 + 4R1 R2 = 4R1 (R1 + R2 )
When Rab = RL , the current into terminal a of the attenuator will be
vi /RL
Using current division, the current in the RL branch will be
vi
R2
·
RL 2R1 + R2 + RL
Therefore
and
vo =
vi
R2
·
RL
RL 2R1 + R2 + RL
vo
R2
=
vi
2R1 + R2 + RL
[b] (600)2 = 4(R1 + R2 )R1
9 × 104 = R21 + R1 R2
vo
R2
= 0.6 =
vi
2R1 + R2 + 600
.·. 1.2R1 + 0.6R2 + 360 = R2
0.4R2 = 1.2R1 + 360
R2 = 3R1 + 900
.·. 9 × 104 = R21 + R1 (3R1 + 900) = 4R21 + 900R1
.·. R21 + 225R1 − 22,500 = 0
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3–40
CHAPTER 3. Simple Resistive Circuits
R1 = −112.5 ±
.·. R1 = 75 Ω
q
(112.5)2 + 22,500 = −112.5 ± 187.5
.·. R2 = 3(75) + 900 = 1125 Ω
[c] From Appendix H, choose R1 = 68 Ω and R2 = 1.2 kΩ. For these values,
Rab = RL =
q
(4)(68)(68 + 1200) = 587.3 Ω
587.3
% error =
− 1 100 = −2.1%
600
vo
1200
=
= 0.624
vi
2(68) + 1200 + 587.3
0.624
% error =
− 1 100 = 4%
0.6
P 3.65
[a] After making the Y-to-∆ transformation, the circuit reduces to
Combining the parallel resistors reduces the circuit to
Now note:
Therefore
0.75R +
Rab
2.25R2 + 3.75RRL
3R
3R + RL
3R(3R + 5RL )
! =
=
2
2.25R + 3.75RRL
15R + 9RL
3R +
3R + RL
!
If R = RL , we have
Therefore
3RRL
2.25R2 + 3.75RRL
=
3R + RL
3R + RL
Rab =
3RL (8RL )
= RL
24RL
Rab = RL
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Problems
3–41
[b] When R = RL , the circuit reduces to
io =
ii(3RL )
1
1 vi
=
ii =
,
4.5RL
1.5
1.5 RL
Therefore
P 3.66
1
vo = 0.75RL io = vi ,
2
vo
= 0.5
vi
[a] 3.5(3R − RL ) = 3R + RL
10.5R − 1050 = 3R + 300
7.5R = 1350,
R = 180 Ω
2(180)(300)2
R2 =
= 4500 Ω
3(180)2 − (300)2
[b ]
vo =
vi
42
=
= 12 V
3.5 3.5
io =
12
= 40 mA
300
i1 =
42 − 12
30
=
= 6.67 mA
4500
4500
ig =
42
= 140 mA
300
i2 = 140 − 6.67 = 133.33 mA
i3 = 40 − 6.67 = 33.33 mA
i4 = 133.33 − 33.33 = 100 mA
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3–42
CHAPTER 3. Simple Resistive Circuits
p4500
top
= (6.67 × 10−3 )2(4500) = 0.2 W
= (133.33 × 10−3 )2 (180) = 3.2 W
p180
left
p180
right
p180
vertical
p300
load
= (33.33 × 10−3 )2 (180) = 0.2 W
= (100 × 10−3 )2 (180) = 0.48 W
= (40 × 10−3 )2 (300) = 0.48 W
The 180 Ω resistor carrying i2
[c] p180
left
= 3.2 W
[d] Two resistors dissipate minimum power – the 4500 Ω resistor and the 180
Ω resistor carrying i3.
[e] They both dissipate 0.2 W.
P 3.67
[a ]
va =
vinR4
Ro + R4 + ∆R
vb =
R3
vin
R2 + R3
vo = va − vb =
R4 vin
R3
−
vin
Ro + R4 + ∆R R2 + R3
When the bridge is balanced,
R4
R3
vin =
vin
Ro + R4
R2 + R3
. ·.
R4
R3
=
Ro + R4
R2 + R3
Thus,
vo
=
=
=
≈
R4 vin
R4 vin
−
Ro + R4 + ∆R Ro + R4
1
1
R4 vin
−
Ro + R4 + ∆R Ro + R4
R4 vin(−∆R)
(Ro + R4 + ∆R)(Ro + R4)
−(∆R)R4 vin
,
since ∆R << R4
(Ro + R4 )2
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Problems
3–43
[b] ∆R = 0.03Ro
Ro =
R2 R4
(1000)(5000)
=
= 10,000 Ω
R3
500
∆R = (0.03)(104 ) = 300 Ω
. ·. v o ≈
[c] vo
=
=
=
P 3.68
−300(5000)(6)
= −40 mV
(15,000)2
−(∆R)R4 vin
(Ro + R4 + ∆R)(Ro + R4 )
−300(5000)(6)
(15,300)(15,000)
−39.2157 mV
−(∆R)R4 vin
(Ro + R4 )2
[a] approx value =
true value =
. ·.
−(∆R)R4vin
(Ro + R4 + ∆R)(Ro + R4 )
approx value
(Ro + R4 + ∆R)
=
true value
(Ro + R4 )
.·. % error =
Ro + R4
−∆R
− 1 × 100 =
× 100
Ro + R4 + ∆R
Ro + R4
Note that in the above expression, we take the ratio of the true value to
the approximate value because both values are negative.
But Ro =
R2 R4
R3
.·. % error =
[b] % error =
P 3.69
−R3∆R
R4 (R2 + R3 )
−(500)(300)
× 100 = −2%
(5000)(1500)
∆R(R3 )(100)
= 0.5
(R2 + R3)R4
∆R(500)(100)
= 0.5
(1500)(5000)
.·. ∆R = 75 Ω
% change =
75
× 100 = 0.75%
10,000
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3–44
P 3.70
CHAPTER 3. Simple Resistive Circuits
[a] From Eq 3.64 we have
i1
i2
2
=
R22
R21 (1 + 2σ)2
Substituting into Eq 3.63 yields
R2 =
R22
R1
R21 (1 + 2σ)2
Solving for R2 yields
R2 = (1 + 2σ)2 R1
[b] From Eq 3.67 we have
i1
R2
=
ib
R1 + R2 + 2Ra
But R2 = (1 + 2σ)2 R1 and Ra = σR1 therefore
i1
(1 + 2σ)2 R1
(1 + 2σ)2
=
=
ib
R1 + (1 + 2σ)2 R1 + 2σR1
(1 + 2σ) + (1 + 2σ)2
=
1 + 2σ
2(1 + σ)
It follows that
i1
ib
2
=
(1 + 2σ)2
4(1 + σ)2
Substituting into Eq 3.66 gives
Rb =
P 3.71
(1 + 2σ)2 Ra
(1 + 2σ)2 σR1
=
4(1 + σ)2
4(1 + σ)2
From Eq 3.69
i1
R2 R3
=
i3
D
But D = (R1 + 2Ra )(R2 + 2Rb ) + 2Rb R2
where Ra = σR1 ; R2 = (1 + 2σ)2 R1 and Rb =
(1 + 2σ)2σR1
4(1 + σ)2
Therefore D can be written as
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Problems
D
=
=
=
=
3–45
2(1 + 2σ)2 σR1
(R1 + 2σR1 ) (1 + 2σ) R1 +
+
4(1 + σ)2
"
#
(1 + 2σ)2 σR1
2
2(1 + 2σ) R1
4(1 + σ)2
"
#
σ
(1 + 2σ)σ
3 2
(1 + 2σ) R1 1 +
+
2(1 + σ)2
2(1 + σ)2
(1 + 2σ)3R21
{2(1 + σ)2 + σ + (1 + 2σ)σ}
2(1 + σ)2
(1 + 2σ)3R21
{1 + 3σ + 2σ 2}
(1 + σ)2
"
#
2
(1 + 2σ)4R21
D=
(1 + σ)
i1
.·.
i3
R2 R3 (1 + σ)
(1 + 2σ)4R21
(1 + 2σ)2R1 R3 (1 + σ)
=
(1 + 2σ)4R21
(1 + σ)R3
=
(1 + 2σ)2R1
When this result is substituted into Eq 3.69 we get
R3 =
=
(1 + σ)2R23 R1
(1 + 2σ)4 R21
Solving for R3 gives
R3 =
P 3.72
(1 + 2σ)4R1
(1 + σ)2
From the dimensional specifications, calculate σ and R3 :
y
0.025
σ= =
= 0.025;
x
1
Vdc2
122
R3 =
=
= 1.2 Ω
p
120
Calculate R1 from R3 and σ:
R1 =
(1 + σ)2
R3 = 1.0372 Ω
(1 + 2σ)4
Calculate Ra , Rb , and R2 :
Ra = σR1 = 0.0259 Ω
Rb =
(1 + 2σ)2σR1
= 0.0068 Ω
4(1 + σ)2
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3–46
CHAPTER 3. Simple Resistive Circuits
R2 = (1 + 2σ)2R1 = 1.1435 Ω
Using symmetry,
R4 = R2 = 1.1435 Ω
R5 = R1 = 1.0372 Ω
Rc = Rb = 0.0068 Ω
Rd = Ra = 0.0259 Ω
Test the calculations by checking the power dissipated, which should be 120
W/m. Calculate D, then use Eqs. (3.58)-(3.60) to calculate ib , i1, and i2:
D = (R1 + 2Ra )(R2 + 2Rb ) + 2R2 Rb = 1.2758
ib =
Vdc (R1 + R2 + 2Ra )
= 21 A
D
i1 =
Vdc R2
= 10.7561 A
D
i2 =
Vdc (R1 + 2Ra )
= 10.2439 A
D
It follows that i2b Rb = 3 W and the power dissipation per meter is
3/0.025 = 120 W/m. The value of i21R1 = 120 W/m. The value of i22 R2 = 120
W/m. Finally, i21Ra = 3 W/m.
P 3.73
From the solution to Problem 3.72 we have ib = 21 A and i3 = 10 A. By
symmetry ic = 21 A thus the total current supplied by the 12 V source is
21 + 21 + 10 or 52 A. Therefore the total power delivered by the source is p12 V
(del) = (12)(52) = 624 W. We also have from the solution that
pa = pb = pc = pd = 3 W. Therefore the total power delivered to the vertical
resistors is pV = (8)(3) = 24 W. The total power delivered to the five
horizontal resistors is pH = 5(120) = 600 W.
.·.
P 3.74
X
pdiss = pH + pV = 624 W =
X
pdel
[a] σ = 0.03/1.5 = 0.02
Since the power dissipation is 200 W/m the power dissipated in R3 must
be 200(1.5) or 300 W. Therefore
122
= 0.48 Ω
300
From Table 3.1 we have
(1 + σ)2R3
R1 =
= 0.4269 Ω
(1 + 2σ)4
R3 =
Ra = σR1 = 0.0085 Ω
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Problems
3–47
R2 = (1 + 2σ)2 R1 = 0.4617 Ω
Rb =
(1 + 2σ)2σR1
= 0.0022 Ω
4(1 + σ)2
Therefore
R4 = R2 = 0.4617 Ω
R5 = R1 = 0.4269 Ω
Rc = Rb = 0.0022 Ω
Rd = Ra = 0.0085 Ω
[b] D = [0.4269 + 2(0.0085)][0.4617 + 2(0.0022)] + 2(0.4617)(0.0022) = 0.2090
i1 =
Vdc R2
= 26.51 A
D
i21 R1 = 300 W or 200 W/m
i2 =
R1 + 2Ra
Vdc = 25.49 A
D
i22 R2 = 300 W or 200 W/m
i21 Ra = 6 W or 200 W/m
ib =
R1 + R2 + 2Ra
Vdc = 52 A
D
i2b Rb = 6 W or 200 W/m
isource = 52 + 52 +
12
= 129 A
0.48
pdel = 12(129) = 1548 W
pH = 5(300) = 1500 W
pV = 8(6) = 48 W
X
pdel =
X
pdiss = 1548 W
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4
Techniques of Circuit Analysis
Assessment Problems
AP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages:
The two node voltage equations are
v1
v1 v1 − v2
−15 +
+
+
= 0
60 15
5
v2 v2 − v1
5+
+
= 0
2
5
Place
in standard
these equations
form:
1
1
1
1
v1
+
+
+ v2 −
= 15
60 15 5
5
1
1 1
v1 −
+ v2
+
= −5
5
2 5
Solving, v1 = 60 V and v2 = 10 V;
Therefore, i1 = (v1 − v2)/5 = 10 A
[b] p15A = −(15 A)v1 = −(15 A)(60 V) = −900 W = 900 W(delivered)
[c] p5A = (5 A)v2 = (5 A)(10 V) = 50 W= −50 W(delivered)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
4–1 system, or transmission in any form or by any means, electronic,
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4–2
CHAPTER 4. Techniques of Circuit Analysis
AP 4.2 Redraw the circuit, choosing the node voltages and reference node as shown:
The two node voltage equations are:
v1 v1 − v2
−4.5 +
+
= 0
1
6+2
v2 v2 − v1 v2 − 30
+
+
= 0
12
6+2
4
Place
form:
theseequationsin standard
1
1
v1 1 +
+ v2 −
= 4.5
8
8
1
1
1 1
v1 −
+ v2
+ +
= 7.5
8
12 8 4
Solving, v1 = 6 V
v2 = 18 V
To find the voltage v, first find the current i through the series-connected 6 Ω
and 2 Ω resistors:
i=
v1 − v2
6 − 18
=
= −1.5 A
6+2
8
Using a KVL equation, calculate v:
v = 2i + v2 = 2(−1.5) + 18 = 15 V
AP 4.3 [a] Redraw the circuit, choosing the node voltages and reference node as
shown:
The node voltage equations are:
v1 − 50 v1 v1 − v2
+
+
− 3i1 =
6
8
2
v2 v2 − v1
−5 +
+
+ 3i1 =
4
2
0
0
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Problems
4–3
The dependent source requires the following constraint equation:
50 − v1
i1 =
6
Place these equations in standard form:
1 1 1
1
50
v1
+ +
+ v2 −
+ i1(−3) =
6 8 2
2
6
1
1 1
v1 −
+ v2
+
+ i1(3)
= 5
2
4 2
1
50
v1
+ v2(0)
+ i1(1)
=
6
6
Solving, v1 = 32 V;
v2 = 16 V;
i1 = 3 A
Using these values to calculate the power associated with each source:
p50V = −50i1
=
−150 W
p5A = −5(v2)
=
−80 W
p3i1 = 3i1 (v2 − v1)
=
−144 W
[b] All three sources are delivering power to the circuit because the power
computed in (a) for each of the sources is negative.
AP 4.4 Redraw the circuit and label the reference node and the node at which the
node voltage equation will be written:
The node voltage equation is
vo
vo − 10 vo + 20i∆
+
+
=0
40
10
20
The constraint equation required by the dependent source is
i∆ = i10 Ω + i30 Ω =
10 − vo 10 + 20i∆
+
10
30
Place these equations in standard form:
1
1
1
vo
+
+
+ i∆ (1)
=
40 10 20
1
20
vo
+ i∆ 1 −
=
10
30
Solving,
i∆ = −3.2 A
and
1
1+
10
30
vo = 24 V
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4–4
CHAPTER 4. Techniques of Circuit Analysis
AP 4.5 Redraw the circuit identifying the three node voltages and the reference node:
Note that the dependent voltage source and the node voltages v and v2 form a
supernode. The v1 node voltage equation is
v1
v1 − v
+
− 4.8 = 0
7.5
2.5
The supernode equation is
v − v1
v
v2
v2 − 12
+
+
+
=0
2.5
10 2.5
1
The constraint equation due to the dependent source is
ix =
v1
7.5
The constraint equation due to the supernode is
v + ix = v2
Place this set of equations in standard form:
1
1
1
v1
+
+ v −
+ v2 (0)
+
7.5 2.5
2.5
1
1
1
1
v1 −
+ v
+
+ v2
+1
+
2.5
2.5 10
2.5
1
v1 −
+ v(0)
+ v2 (0)
+
7.5
v1(0)
+
v(1)
+
v2 (−1)
+
ix (0)
=
4.8
ix (0)
=
12
ix (1)
=
0
ix (1)
=
0
Solving this set of equations gives v1 = 15 V, v2 = 10 V, ix = 2 A, and
v = 8 V.
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Problems
4–5
AP 4.6 Redraw the circuit identifying the reference node and the two unknown node
voltages. Note that the right-most node voltage is the sum of the 60 V source
and the dependent source voltage.
The node voltage equation at v1 is
v1 − 60 v1 v1 − (60 + 6iφ )
+
+
=0
2
24
3
The constraint equation due to the dependent source is
iφ =
60 + 6iφ − v1
3
Place these two equations in standard form:
1
1
1
v1
+
+
+ iφ(−2)
= 30 + 20
2 24 3
1
v1
+ iφ(1 − 2) = 20
3
Solving,
iφ = −4 A
and
v1 = 48 V
AP 4.7 [a] Redraw the circuit identifying the three mesh currents:
The mesh current equations are:
−80 + 5(i1 − i2 ) + 26(i1 − i3)
=
0
30i2 + 90(i2 − i3) + 5(i2 − i1)
=
0
8i3 + 26(i3 − i1 ) + 90(i3 − i2)
=
0
Place these equations in standard form:
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4–6
CHAPTER 4. Techniques of Circuit Analysis
31i1 − 5i2 − 26i3
=
80
−5i1 + 125i2 − 90i3
=
0
−26i1 − 90i2 + 124i3
Solving,
=
0
i1 = 5 A;
i2 = 2 A;
i3 = 2.5 A
p80V = −(80)i1 = −(80)(5) = −400 W
Therefore the 80 V source is delivering 400 W to the circuit.
[b] p8Ω = (8)i23 = 8(2.5)2 = 50 W, so the 8 Ω resistor dissipates 50 W.
AP 4.8 [a] b = 8,
n = 6,
b−n+1 = 3
[b] Redraw the circuit identifying the three mesh currents:
The three mesh-current equations are
−25 + 2(i1 − i2) + 5(i1 − i3) + 10
= 0
−(−3vφ ) + 14i2 + 3(i2 − i3 ) + 2(i2 − i1 )
= 0
1i3 − 10 + 5(i3 − i1 ) + 3(i3 − i2 )
= 0
The dependent source constraint equation is
vφ = 3(i3 − i2)
Place these four equations in standard form:
7i1 − 2i2 − 5i3 + 0vφ
=
15
−2i1 + 19i2 − 3i3 + 3vφ
=
0
−5i1 − 3i2 + 9i3 + 0vφ
=
10
0i1 + 3i2 − 3i3 + 1vφ
=
0
Solving
i1 = 4 A;
i2 = −1 A;
i3 = 3 A;
vφ = 12 V
pds = −(−3vφ)i2 = 3(12)(−1) = −36 W
Thus, the dependent source is delivering 36 W, or absorbing −36 W.
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Problems
4–7
AP 4.9 Redraw the circuit identifying the three mesh currents:
The mesh current equations are:
−25 + 6(ia − ib) + 8(ia − ic)
=
0
2ib + 8(ib − ic) + 6(ib − ia)
=
0
5iφ + 8(ic − ia) + 8(ic − ib)
=
0
The dependent source constraint equation is iφ = ia. We can substitute this
simple expression for iφ into the third mesh equation and place the equations
in standard form:
14ia − 6ib − 8ic
=
25
−6ia + 16ib − 8ic
=
0
−3ia − 8ib + 16ic
=
0
Solving,
ia = 4 A;
ib = 2.5 A;
ic = 2 A
Thus,
vo = 8(ia − ic) = 8(4 − 2) = 16 V
AP 4.10 Redraw the circuit identifying the mesh currents:
Since there is a current source on the perimeter of the i3 mesh, we know that
i3 = −16 A. The remaining two mesh equations are
−30 + 3i1 + 2(i1 − i2) + 6i1
=
0
8i2 + 5(i2 + 16) + 4i2 + 2(i2 − i1)
=
0
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4–8
CHAPTER 4. Techniques of Circuit Analysis
Place these equations in standard form:
11i1 − 2i2
=
30
−2i1 + 19i2
=
−80
Solving: i1 = 2 A, i2 = −4 A, i3 = −16 A
The current in the 2 Ω resistor is i1 − i2 = 6 A .·.
Thus, the 2 Ω resistors dissipates 72 W.
p2 Ω = (6)2 (2) = 72 W
AP 4.11 Redraw the circuit and identify the mesh currents:
There are current sources on the perimeters of both the ib mesh and the ic
mesh, so we know that
ib = −10 A;
ic =
2vφ
5
The remaining mesh current equation is
−75 + 2(ia + 10) + 5(ia − 0.4vφ ) = 0
The dependent source requires the following constraint equation:
vφ = 5(ia − ic) = 5(ia − 0.4vφ )
Place the mesh current equation and the dependent source equation is
standard form:
7ia − 2vφ
=
55
5ia − 3vφ
=
0
Solving: ia = 15 A;
Thus, ia = 15 A.
ib = −10 A;
ic = 10 A;
vφ = 25 V
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Problems
4–9
AP 4.12 Redraw the circuit and identify the mesh currents:
The 2 A current source is shared by the meshes ia and ib. Thus we combine
these meshes to form a supermesh and write the following equation:
−10 + 2ib + 2(ib − ic) + 2(ia − ic) = 0
The other mesh current equation is
−6 + 1ic + 2(ic − ia) + 2(ic − ib) = 0
The supermesh constraint equation is
ia − ib = 2
Place these three equations in standard form:
2ia + 4ib − 4ic
= 10
−2ia − 2ib + 5ic
= 6
ia − ib + 0ic
= 2
Solving,
Thus,
ia = 7 A;
ib = 5 A;
ic = 6 A
p1 Ω = i2c(1) = (6)2 (1) = 36 W
AP 4.13 Redraw the circuit and identify the reference node and the node voltage v1:
The node voltage equation is
v1 − 20
v1 − 25
−2+
=0
15
10
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4–10
CHAPTER 4. Techniques of Circuit Analysis
Rearranging and solving,
v1
1
1
20 25
+
= 2+
+
15 10
15 10
.·. v1 = 35 V
p2A = −35(2) = −70 W
Thus the 2 A current source delivers 70 W.
AP 4.14 Redraw the circuit and identify the mesh currents:
There is a current source on the perimeter of the i3 mesh, so i3 = 4 A. The
other two mesh current equations are
−128 + 4(i1 − 4) + 6(i1 − i2 ) + 2i1
=
0
30ix + 5i2 + 6(i2 − i1 ) + 3(i2 − 4)
=
0
The constraint equation due to the dependent source is
ix = i1 − i3 = i1 − 4
Substitute the constraint equation into the second mesh equation and place
the resulting two mesh equations in standard form:
12i1 − 6i2
=
144
24i1 + 14i2
=
132
Solving,
i1 = 9 A;
i2 = −6 A;
i3 = 4 A;
ix = 9 − 4 = 5 A
.·. v4A = 3(i3 − i2 ) − 4ix = 10 V
p4A = −v4A (4) = −(10)(4) = −40 W
Thus, the 2 A current source delivers 40 W.
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Problems
4–11
AP 4.15 [a] Redraw the circuit with a helpful voltage and current labeled:
Transform the 120 V source in series with the 20 Ω resistor into a 6 A
source in parallel with the 20 Ω resistor. Also transform the −60 V source
in series with the 5 Ω resistor into a −12 A source in parallel with the 5 Ω
resistor. The result is the following circuit:
Combine the three current sources into a single current source, using
KCL, and combine the 20 Ω, 5 Ω, and 6 Ω resistors in parallel. The
resulting circuit is shown on the left. To simplify the circuit further,
transform the resulting 30 A source in parallel with the 2.4 Ω resistor into
a 72 V source in series with the 2.4 Ω resistor. Combine the 2.4 Ω resistor
in series with the 1.6 Ω resisor to get a very simple circuit that still
maintains the voltage v. The resulting circuit is on the right.
Use voltage division in the circuit on the right to calculate v as follows:
8
(72) = 48 V
12
[b] Calculate i in the circuit on the right using Ohm’s law:
v=
v
48
=
=6A
8
8
Now use i to calculate va in the circuit on the left:
i=
va = 6(1.6 + 8) = 57.6 V
Returning back to the original circuit, note that the voltage va is also the
voltage drop across the series combination of the 120 V source and 20 Ω
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4–12
CHAPTER 4. Techniques of Circuit Analysis
resistor. Use this fact to calculate the current in the 120 V source, ia:
ia =
120 − va
120 − 57.6
=
= 3.12 A
20
20
p120V = −(120)ia = −(120)(3.12) = −374.40 W
Thus, the 120 V source delivers 374.4 W.
AP 4.16 To find RTh , replace the 72 V source with a short circuit:
Note that the 5 Ω and 20 Ω resistors are in parallel, with an equivalent
resistance of 5k20 = 4 Ω. The equivalent 4 Ω resistance is in series with the 8 Ω
resistor for an equivalent resistance of 4 + 8 = 12 Ω. Finally, the 12 Ω
equivalent resistance is in parallel with the 12 Ω resistor, so
RTh = 12k12 = 6 Ω.
Use node voltage analysis to find vTh . Begin by redrawing the circuit and
labeling the node voltages:
The node voltage equations are
v1 − 72 v1 v1 − vTh
+
+
= 0
5
20
8
vTh − v1 vTh − 72
+
= 0
8
12
Place these equations in standard form:
1
1
1
1
72
v1
+
+
+ vTh −
=
5 20 8
8
5
1
1
1
v1 −
+ vTh
+
= 6
8
8 12
Solving, v1 = 60 V and vTh = 64.8 V. Therefore, the Thévenin equivalent
circuit is a 64.8 V source in series with a 6 Ω resistor.
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Problems
4–13
AP 4.17 We begin by performing a source transformation, turning the parallel
combination of the 15 A source and 8 Ω resistor into a series combination of a
120 V source and an 8 Ω resistor, as shown in the figure on the left. Next,
combine the 2 Ω, 8 Ω and 10 Ω resistors in series to give an equivalent 20 Ω
resistance. Then transform the series combination of the 120 V source and the
20 Ω equivalent resistance into a parallel combination of a 6 A source and a
20 Ω resistor, as shown in the figure on the right.
Finally, combine the 20 Ω and 12 Ω parallel resistors to give
RN = 20k12 = 7.5 Ω. Thus, the Norton equivalent circuit is the parallel
combination of a 6 A source and a 7.5 Ω resistor.
AP 4.18 Find the Thévenin equivalent with respect to A, B using source
transformations. To begin, convert the series combination of the −36 V source
and 12 kΩ resistor into a parallel combination of a −3 mA source and 12 kΩ
resistor. The resulting circuit is shown below:
Now combine the two parallel current sources and the two parallel resistors to
give a −3 + 18 = 15 mA source in parallel with a 12 kk60 k= 10 kΩ resistor.
Then transform the 15 mA source in parallel with the 10 kΩ resistor into a
150 V source in series with a 10 kΩ resistor, and combine this 10 kΩ resistor
in series with the 15 kΩ resistor. The Thévenin equivalent is thus a 150 V
source in series with a 25 kΩ resistor, as seen to the left of the terminals A,B
in the circuit below.
Now attach the voltmeter, modeled as a 100 kΩ resistor, to the Thévenin
equivalent and use voltage division to calculate the meter reading vAB :
vAB =
100,000
(150) = 120 V
125,000
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4–14
CHAPTER 4. Techniques of Circuit Analysis
AP 4.19 Begin by calculating the open circuit voltage, which is also vTh , from the
circuit below:
Summing the currents away from the node labeled vTh We have
vTh
vTh − 24
+ 4 + 3ix +
=0
8
2
Also, using Ohm’s law for the 8 Ω resistor,
ix =
vTh
8
Substituting the second equation into the first and solving for vTh yields
vTh = 8 V.
Now calculate RTh . To do this, we use the test source method. Replace the
voltage source with a short circuit, the current source with an open circuit,
and apply the test voltage vT , as shown in the circuit below:
Write a KCL equation at the middle node:
iT = ix + 3ix + vT /2 = 4ix + vT /2
Use Ohm’s law to determine ix as a function of vT :
ix = vT /8
Substitute the second equation into the first equation:
iT = 4(vT /8) + vT /2 = vT
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Problems
4–15
Thus,
RTh = vT /iT = 1 Ω
The Thévenin equivalent is an 8 V source in series with a 1 Ω resistor.
AP 4.20 Begin by calculating the open circuit voltage, which is also vTh , using the
node voltage method in the circuit below:
The node voltage equations are
v
v − (vTh + 160i∆ )
+
− 4 = 0,
60
20
vTh vTh vTh + 160i∆ − v
+
+
= 0
40
80
20
The dependent source constraint equation is
i∆ =
vTh
40
Substitute the constraint equation into the node voltage equations and put the
two equations in standard form:
1
1
5
v
+
+ vTh −
= 4
60 20
20
1
1
1
5
v −
+ vTh
+
+
= 0
20
40 80 20
Solving, v = 172.5 V and vTh = 30 V.
Now use the test source method to calculate the test current and thus RTh .
Replace the current source with a short circuit and apply the test source to
get the following circuit:
Write a KCL equation at the rightmost node:
iT =
vT vT vT + 160i∆
+
+
80 40
80
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4–16
CHAPTER 4. Techniques of Circuit Analysis
The dependent source constraint equation is
i∆ =
vT
40
Substitute the constraint equation into the KCL equation and simplify the
right-hand side:
iT =
vT
10
Therefore,
RTh =
vT
= 10 Ω
iT
Thus, the Thévenin equivalent is a 30 V source in series with a 10 Ω resistor.
AP 4.21 First find the Thévenin equivalent circuit. To find vTh , create an open circuit
between nodes a and b and use the node voltage method with the circuit
below:
The node voltage equations are:
vTh − (100 + vφ) vTh − v1
+
= 0
4
4
v1 − 100 v1 − 20 v1 − vTh
+
+
= 0
4
4
4
The dependent source constraint equation is
vφ = v1 − 20
Place these three equations in standard form:
1 1
1
1
vTh
+
+ v1 −
+ vφ −
4 4
4
4
1
1 1 1
vTh −
+ v1
+ +
+ vφ (0)
4
4 4 4
vTh (0)
+ v1 (1)
+
vφ (−1)
=
25
=
30
=
20
Solving, vTh = 120 V, v1 = 80 V, and vφ = 60 V.
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Problems
4–17
Now create a short circuit between nodes a and b and use the mesh current
method with the circuit below:
The mesh current equations are
−100 + 4(i1 − i2) + vφ + 20
=
0
−vφ + 4i2 + 4(i2 − isc) + 4(i2 − i1)
=
0
−20 − vφ + 4(isc − i2)
=
0
The dependent source constraint equation is
vφ = 4(i1 − isc)
Place these four equations in standard form:
4i1 − 4i2 + 0isc + vφ
=
80
−4i1 + 12i2 − 4isc − vφ
=
0
0i1 − 4i2 + 4isc − vφ
=
20
4i1 + 0i2 − 4isc − vφ
=
0
Solving, i1 = 45 A, i2 = 30 A, isc = 40 A, and vφ = 20 V. Thus,
RTh =
vTh
120
=
= 3Ω
isc
40
[a] For maximum power transfer, R = RTh = 3 Ω
[b] The Thévenin voltage, vTh = 120 V, splits equally between the Thévenin
resistance and the load resistance, so
120
= 60 V
2
Therefore,
vload =
pmax =
2
vload
602
=
= 1200 W
Rload
3
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4–18
CHAPTER 4. Techniques of Circuit Analysis
AP 4.22 Sustituting the value R = 3 Ω into the circuit and identifying three mesh
currents we have the circuit below:
The mesh current equations are:
−100 + 4(i1 − i2 ) + vφ + 20
=
0
−vφ + 4i2 + 4(i2 − i3) + 4(i2 − i1)
=
0
−20 − vφ + 4(i3 − i2) + 3i3
=
0
The dependent source constraint equation is
vφ = 4(i1 − i3)
Place these four equations in standard form:
4i1 − 4i2 + 0i3 + vφ
=
80
−4i1 + 12i2 − 4i3 − vφ
=
0
0i1 − 4i2 + 7i3 − vφ
=
20
4i1 + 0i2 − 4i3 − vφ
=
0
Solving, i1 = 30 A, i2 = 20 A, i3 = 20 A, and vφ = 40 V.
[a] p100V = −(100)i1 = −(100)(30) = −3000 W. Thus, the 100 V source is
delivering 3000 W.
[b] pdepsource = −vφi2 = −(40)(20) = −800 W. Thus, the dependent source is
delivering 800 W.
[c] From Assessment Problem 4.21(b), the power delivered to the load resistor
is 1200 W, so the load power is (1200/3800)100 = 31.58% of the
combined power generated by the 100 V source and the dependent source.
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Problems
4–19
Problems
P 4.1
[a] 11 branches, 8 branches with resistors, 2 branches with independent
sources, 1 branch with a dependent source
[b] The current is unknown in every branch except the one containing the 8 A
current source, so the current is unknown in 10 branches.
[c] 9 essential branches – R4 − R5 forms an essential branch as does R8 − 10
V. The remaining seven branches are essential branches that contain a
single element.
[d] The current is known only in the essential branch containing the current
source, and is unknown in the remaining 8 essential branches
[e] From the figure there are 6 nodes – three identified by rectangular boxes,
two identified with single black dots, and one identified by a triangle.
[f] There are 4 essential nodes, three identified with rectangular boxes and
one identified with a triangle
[g] A mesh is like a window pane, and as can be seen from the figure there are
6 window panes or meshes.
P 4.2 [a] From Problem 4.1(d) there are 8 essential branches where the current is
unknown, so we need 8 simultaneous equations to describe the circuit.
[b] From Problem 4.1(f), there are 4 essential nodes, so we can apply KCL at
(4 − 1) = 3 of these essential nodes. There would also be a dependent
source constraint equation.
[c] The remaining 4 equations needed to describe the circuit will be derived
from KVL equations.
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4–20
CHAPTER 4. Techniques of Circuit Analysis
[d] We must avoid using the topmost mesh and the leftmost mesh. Each of
these meshes contains a current source, and we have no way of
determining the voltage drop across a current source.
P 4.3
[a] As can be seen from the figure, the circuit has 2 separate parts.
[b] There are 5 nodes – the four black dots and the node betweem the voltage
source and the resistor R1.
[c] There are 7 branches, each containing one of the seven circuit components.
[d] When a conductor joins the lower nodes of the two separate parts, there is
now only a single part in the circuit. There would now be 4 nodes,
because the two lower nodes are now joined as a single node. The
number of branches remains at 7, where each branch contains one of the
seven individual circuit components.
P 4.4 [a] There are six circuit components, five resistors and the current source.
Since the current is known only in the current source, it is unknown in
the five resistors. Therefore there are five unknown currents.
[b] There are four essential nodes in this circuit, identified by the dark black
dots in Fig. P4.4. At three of these nodes you can write KCL equations
that will be independent of one another. A KCL equation at the fourth
node would be dependent on the first three. Therefore there are three
independent KCL equations.
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Problems
4–21
[c]
Sum the currents at any three of the four
essential nodes a, b, c, and d. Using nodes a, b, and c we get
−ig + i1 + i2 = 0
−i1 + i4 + i3 = 0
i5 − i2 − i3 = 0
[d] There are three meshes in this circuit: one on the left with the
components ig , R1 , and R4 ; one on the top right with components R1 ,
R2 , and R3 ; and one on the bottom right with components R3 , R4, and
R5 . We cannot write a KVL equation for the left mesh because we don’t
know the voltage drop across the current source. Therefore, we can write
KVL equations for the two meshes on the right, giving a total of two
independent KVL equations.
[e] Sum the voltages around two independent closed paths, avoiding a path
that contains the independent current source since the voltage across the
current source is not known. Using the upper and lower meshes formed
by the five resistors gives
R1 i1 + R3i3 − R2i2 = 0
R3 i3 + R5i5 − R4i4 = 0
P 4.5
[a] At node a:
− ig + i1 + i2 = 0
At node b:
− i1 + i3 + i4 = 0
At node c:
− i2 − i3 + i5 = 0
At node d:
ig − i4 − i5 = 0
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4–22
CHAPTER 4. Techniques of Circuit Analysis
[b] There are many possible solutions. For example, solve the equations at
nodes a and d for ig :
ig = i4 + i5
ig = i1 + i2
so
i1 + i2 = i4 + i5
Solve this expression for i1 :
i1 = i4 + i5 − i2
Substitute this expression for i1 into the equation for node b:
−(i4 + i5 − i2) + i3 + i4 = 0
so
− i2 − i3 + i5 = 0
The result above is the equation at node c.
P 4.6
v1 − 144 v1 v1 − v2
+
+
=0
4
10
80
v2 − v1 v2
−3 +
+
=0
80
5
Solving, v1 = 100 V;
so
29v1 − v2 = 2880
so
−v1 + 17v2 = 240
v2 = 20 V
P 4.7
−2 +
vo
vo − 45
+
=0
50
1+4
vo = 50 V
p2A = −(50)(2) = −100 W (delivering)
The 2 A source extracts −100 W from the circuit, because it delivers 100 W
to the circuit.
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Problems
P 4.8 −6 +
4–23
v1 v1 − v2
+
=0
40
8
v2 − v1 v2
v2
+
+
+1 =0
8
80 120
Solving, v1 = 120 V;
CHECK:
(120)2
= 360 W
40
p40Ω =
p8Ω =
v2 = 96 V
(120 − 96)2
= 72 W
8
p80Ω =
(96)2
= 115.2 W
80
p120Ω =
(96)2
= 76.8 W
120
p6A = −(6)(120) = −720 W
p1A = (1)(96) = 96 W
X
pabs = 360 + 72 + 115.2 + 76.8 + 96 = 720 W
X
pdev = 720 W (CHECKS)
P 4.9 Use the lower terminal of the 25 Ω resistor as the reference node.
vo − 24
vo
+
+ 0.04 = 0
20 + 80 25
Solving,
vo = 4 V
P 4.10 [a] From the solution to Problem 4.9 we know vo = 4 V, therefore
p40mA = 0.04vo = 0.16 W
.·. p40mA (developed) = −160 mW
[b] The current into the negative terminal of the 24 V source is
ig =
24 − 4
= 0.2 A
20 + 80
p24V = −24(0.2) = −4.8 W
.·. p24V (developed) = 4800 mW
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4–24
CHAPTER 4. Techniques of Circuit Analysis
[c] p20Ω = (0.2)2 (20) = 800 mW
p80Ω = (0.2)2 (80) = 3200 mW
p25Ω = (4)2 /25 = 640 mW
X
X
pdev = 4800 mW
pdis = 160 + 800 + 3200 + 640 = 4800 mW
v0 − 24
vo
+
+ 0.04 = 0; vo = 4 V
20 + 80 25
[b] Let vx = voltage drop across 40 mA source
P 4.11 [a]
vx = vo − (50)(0.04) = 2 V
p40mA = (2)(0.04) = 80 mW so p40mA (developed) = −80 mW
[c] Let ig = be the current into the positive terminal of the 24 V source
ig = (4 − 24)/100 = −0.2 A
p24V = (−0.2)(24) = −4800 mW so p24V (developed) = 4800 mW
[d]
X
pdis = (0.2)2 (20) + (0.2)2 (80) + (4)2 /25 + (0.04)2 (50) + 0.08
= 4800 mW
[e] vo is independent of any finite resistance connected in series with the 40
mA current source
P 4.12 [a]
v1 − 125 v1 − v2 v1 − v3
+
+
1
6
24
v2 − v1 v2 v2 − v3
+
+
6
2
12
v3 + 125 v3 − v2 v3 − v1
+
+
1
12
24
=
0
=
0
=
0
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Problems
4–25
In standard form:
1 1
1
1
1
+ +
+ v2 −
+ v3 −
=
1 6 24
6
24
1
1 1
1
1
v1 −
+ v2
+ +
+ v3 −
=
6
6 2 12
12
1
1
1
1
1
v1 −
+ v2 −
+ v3
+
+
=
24
12
1 12 24
v1
Solving, v1 = 101.24 V;
v2 = 10.66 V;
125 − v1
= 23.76 A
1
v2
i2 =
= 5.33 A
2
v3 + 125
i3 =
= 18.43 A
1
Thus, i1 =
[b]
P 4.13 [a]
X
125
0
−125
v3 = −106.57 V
v1 − v2
= 15.10 A
6
v2 − v3
i5 =
= 9.77 A
12
v1 − v3
i6 =
= 8.66 A
24
i4 =
Pdev = 125i1 + 125i3 = 5273.09 W
X
Pdis = i21(1) + i22 (2) + i23 (1) + i24(6) + i25(12) + i26(24) = 5273.09 W
v1 − 128 v1 v1 − v2
+
+
= 0
5
60
4
v2 − v1 v2 v2 − 320
+
+
= 0
4
80
10
In standard
form,
1
1
1
1
v1
+
+
+ v2 −
=
5 60 4
4
1
1
1
1
v1 −
+ v2
+
+
=
4
4 80 10
Solving, v1 = 162 V;
128
5
320
10
v2 = 200 V
ia =
128 − 162
= −6.8 A
5
ib =
162
= 2.7 A
60
ic =
162 − 200
= −9.5 A
4
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4–26
CHAPTER 4. Techniques of Circuit Analysis
id =
200
= 2.5 A
80
ie =
200 − 320
= −12 A
10
[b] p128V = −(128)(−6.8) = 870.4 W (abs)
p320V = (320)(−12) = −3840 W (dev)
Therefore, the total power developed is 3840 W.
P 4.14
v1 + 40 v1 v1 − v2
+
+
+5 =0
12
25
20
v2 − v1
v2 − v1
−5+
+ −7.5 = 0
20
40
v3 v3 − v2
+
+ 7.5 = 0
40
40
Solving, v1 = −10 V;
v2 = 132 V;
v3 = −84 V;
i40V =
−10 + 40
= 2.5 A
12
p5A = 5(v1 − v2) = 5(−10 − 132) = −710 W (del)
p7.5A = (−84 − 132)(7.5) = −1620 W (del)
p40V = −(40)(2.5) = −100 W (del)
p12Ω = (2.5)2 (12) = 75 W
p25Ω =
v12
102
=
=4W
25
25
p20Ω =
(v1 − v2)2
1422
=
= 1008.2 W
20
20
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Problems
p40Ω (lower) =
(v3)2
842
=
= 176.4 W
40
40
p40Ω (right) =
(v2 − v3)2
2162
=
= 1166.4 W
40
40
X
pdiss = 75 + 4 + 1008.2 + 176.4 + 1166.4 = 2430 W
X
pdev = 710 + 1620 + 100 = 2430 W
4–27
(CHECKS)
The total power dissipated in the circuit is 2430 W.
P 4.15 [a]
v1 v1 − 40 v1 − v2
+
+
=0
so 31v1 − 20v2 + 0v3 = 400
40
4
2
v2 − v1 v2 − v3
+
− 28 = 0
so −2v1 + 3v2 − v3 = 112
2
4
v3 v3 − v2
+
+ 28 = 0
so 0v1 − v2 + 3v3 = −112
2
4
Solving, v1 = 60 V; v2 = 73 V; v3 = −13 V,
[b] ig =
40 − 60
= −5 A
4
pg = (40)(−5) = −200 W
Thus the 40 V source delivers 200 W of power.
P 4.16 [a]
vo − v1 vo − v2 vo − v3
vo − vn
+
+
+ ··· +
=0
R
R
R
R
.·. nvo = v1 + v2 + v3 + · · · + vn
. ·. v o =
1
1 Xn
[v1 + v2 + v3 + · · · + vn ] =
vk
n
n k=1
1
[b] vo = (100 + 80 − 60) = 40 V
3
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4–28
CHAPTER 4. Techniques of Circuit Analysis
P 4.17 [a] −25 +
v1
v1
v1 − v2
+
+
= 0 so 21v1 − 16v2 + 0i∆ = 4000
40 160
10
v2 − v1 v2
v2 − 84i∆
+
+
= 0 so − 16v1 + 44v2 − 1680i∆ = 0
10
20
8
v1
i∆ =
so v1 + (0)v2 − 160i∆ = 0
160
Solving, v1 = 352 V;
idepsource =
v2 = 212 V;
i∆ = 2.2 A;
212 − 84(2.2)
= 3.4 A
8
p84i∆ = 84(2.2)(3.4) = 628.32 W(abs)
p25A = −25(352) = −8800 W(del)
.·. pdev = 8800 W
[b]
X
pabs =
. ·.
P 4.18 −3 +
(352)2 (352)2 (352 − 212)2 (212)2
+
+
+
40
160
10
20
+(3.4)2 (8) + 628.32 = 8800 W
X
pdev =
X
pabs = 8800 W
vo
vo + 5i∆ vo − 80
+
+
= 0;
200
10
20
i∆ =
vo − 80
20
[a] Solving, vo = 50 V
vo + 5i∆
[b] ids =
10
i∆ = (50 − 80)/20 = −1.5 A
.·. ids = 4.25 A;
5i∆ = −7.5 V :
pds = (−5i∆ )(ids) = 31.875 W
[c] p3A = −3vo = −3(50) = −150 W (del)
p80V = 80i∆ = 80(−1.5) = −120 W (del)
X
pdel = 150 + 120 = 270 W
CHECK:
p200Ω = 2500/200 = 12.5 W
p20Ω = (80 − 50)2 /20 = 900/20 = 45 W
p10Ω = (4.25)2 (10) = 180.625 W
X
pdiss = 31.875 + 180.625 + 12.5 + 45 = 270 W
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Problems
4–29
P 4.19
vo − 160
vo
vo − 150iσ
+
+
= 0;
10
100
50
Solving, vo = 100 V;
io =
iσ = −
vo
100
iσ = −1 A
100 − (150)(−1)
=5A
50
p150iσ = 150iσ io = −750 W
.·. The dependent voltage source delivers 750 W to the circuit.
P 4.20 [a]
io =
v2
40
v1 v1 − v2
+
=0
20
5
v2 − v1 v2 v2 − v3
+
+
5
40
10
v3 − v2 v3 − 11.5io v3 − 96
+
+
=0
10
5
4
−5io +
Solving, v1 = 156 V;
[b] io =
v2 = 120 V;
so
10v1 − 13v2 + 0v3 = 0
so
−8v1 + 13v2 − 4v3 = 0
so
0v1 − 63v2 + 220v3 = 9600
v3 = 78 V
v2
120
=
=3A
40
40
i3 =
v3 − 11.5io
78 − 11.5(3)
=
= 8.7 A
5
5
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4–30
CHAPTER 4. Techniques of Circuit Analysis
ig =
78 − 96
= −4.5 A
4
p5io = −5io v1 = −5(3)(156) = −2340 W(dev)
p11.5io = 11.5io i3 = 11.5(3)(8.7) = 300.15 W(abs)
p96V = 96(−4.5) = −432 W(dev)
X
pdev = 2340 + 432 = 2772 W
CHECK
X
. ·.
P 4.21
pdis
=
1562 (156 − 120)2 1202 (120 − 78)2
+
+
+
20
5
40
50
+(8.7)2(5) + (4.5)2 (4) + 300.15 = 2772 W
X
pdev =
X
pdis = 2772 W
v1
v1 − v2 v1 − 20
+
+
=0
30,000
5000
2000
v2
v2 − v1 v2 − 20
+
+
=0
1000
5000
5000
Solving, v1 = 15 V;
Thus, io =
so
22v1 − 6v2 = 300
so
−v1 + 7v2 = 20
v2 = 5 V
v1 − v2
= 2 mA
5000
P 4.22 [a]
There is only one node voltage equation:
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Problems
va + 30
va
va − 80
+
+
+ 0.01 = 0
5000
500
1000
Solving,
va + 30 + 10va + 5va − 400 + 50 = 0
. ·.
va = 20 V
Calculate the currents:
i1 = (−30 − 20)/5000 = −10 mA
i2
=
20/500 = 40 mA
i4
=
80/4000 = 20 mA
i5
=
(80 − 20)/1000 = 60 mA
so
4–31
16va = 320
i3 + i4 + i5 − 10 mA = 0 so i3 = 0.01 − 0.02 − 0.06 = −0.07 = −70 mA
[b] p30V
=
(30)(−0.01) = −0.3 W
p10mA
=
(20 − 80)(0.01) = −0.6 W
p80V
=
(80)(−0.07) = −5.6 W
p5k
=
(−0.01)2 (5000) = 0.5 W
p500Ω
=
(0.04)2 (500) = 0.8 W
p1k
=
(80 − 20)2 /(1000) = 3.6 W
p4k
=
(80)2 /(4000) = 1.6 W
X
P 4.23 [a]
X
pabs = 0.5 + 0.8 + 3.6 + 1.6 = 6.5 W
pdel = 0.3 + 0.6 + 5.6 = 6.5 W (checks!)
v2 − 230 v2 − v4 v2 − v3
+
+
=0
1
1
1
v3 − v2 v3 v3 − v5
+
+
=0
1
1
1
so
3v2 − 1v3 − 1v4 + 0v5 = 230
so
−1v2 + 3v3 + 0v4 − 1v5 = 0
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4–32
CHAPTER 4. Techniques of Circuit Analysis
v4 − v2 v4 − 230 v4 − v5
+
+
=0
1
6
2
v5 − v3 v5 v5 − v4
+
+
=0
1
6
2
Solving, v2 = 150 V;
i2Ω =
v3 = 80 V;
so
−12v2 + 0v3 + 20v4 − 6v5 = 460
so
0v2 − 12v3 − 6v4 + 20v5 = 0
v4 = 140 V;
v5 = 90 V
v4 − v5
140 − 90
=
= 25 A
2
2
p2Ω = (25)2 (2) = 1250 W
[b]
i230V
v1 − v2 v1 − v4
+
1
6
230 − 150 230 − 140
+
= 80 + 15 = 95 A
1
6
=
=
p230V = (230)(95) = 21,850 W
Check:
X
Pdis
=
(80)2 (1) + (70)2 (1) + (80)2 (1) + (15)2 (6) + (10)2 (1)
+(10)2 (1) + (25)2 (2) + (15)2 (6) = 21,850 W
P 4.24
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Problems
The two node voltage equations are:
v1 − 50 v1 v1 − v2
+
+
= 0
80
50
40
v2 − v1
v2
v2 − 50
− 0.75 +
+
= 0
40
200
800
Place these equations in standard form:
1
1
1
1
v1
+
+
+ v2 −
=
80 50 40
40
1
1
1
1
v1 −
+ v2
+
+
=
40
40 200 800
4–33
50
80
0.75 +
50
800
Solving, v1 = 34 V;
v2 = 53.2 V.
Thus, vo = v2 − 50 = 53.2 − 50 = 3.2 V.
POWER CHECK:
ig
= (50 − 34)/80 + (50 − 53.2)/800 = 196 m A
p50V
=
−(50)(0.196) = −9.8 W
p80Ω
=
(50 − 34)2 /80 = 3.2 W
p800Ω
=
(50 − 53.2)2 /800 = 12.8 m W
p40Ω
=
(53.2 − 34)2 /40 = 9.216 W
p50Ω
=
342 /50 = 23.12 W
p200Ω
=
53.22 /200 = 14.1512 W
p0.75A
=
−(53.2)(0.75) = −39.9 W
X
pabs = 3.2 + .0128 + 9.216 + 23.12 + 14.1512 = 49.7 W =
9.8 + 39.9 = 49.7
X
pdel =
P 4.25
The two node voltage equations are:
vb vb − vc
7+
+
= 0
3
1
vc − vb vc − 4
−2vx +
+
= 0
1
2
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4–34
CHAPTER 4. Techniques of Circuit Analysis
The constraint equation for the dependent source is:
vx = vc − 4
Place these equations in standard form:
1
vb
+1
+ vc(−1)
+ vx (0)
=
3
1
vb(−1)
+ vc 1 +
+ vx (−2) =
2
vb(0)
+
vc(1)
+
vx (−1) =
−7
4
2
4
Solving, vc = 9 V, vx = 5 V, and vo = vb = 1.5 V
P 4.26 [a]
This circuit has a supernode includes the nodes v1, v2 and the 25 V
source. The supernode equation is
v1
v2
v2
2+
+
+
=0
50 150 75
The supernode constraint equation is
v1 − v2 = 25
Place these two equations in standard form:
1
1
1
v1
+ v2
+
= −2
50
150 75
v1(1)
+
v2(−1)
= 25
Solving, v1 = −37.5 V and v2 = −62.5 V, so vo = v1 = −37.5 V.
p2A = (2)vo = (2)(−37.5) = −75 W
The 2 A source delivers 75 W.
[b]
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Problems
4–35
This circuit now has only one non-reference essential node where the
voltage is not known – note that it is not a supernode. The KCL
equation at v1 is
v1 v1 + 25 v1 + 25
+
+
=0
50
150
75
Solving, v1 = 37.5 V so v0 = −v1 = −37.5 V.
−2 +
p2A = (2)vo = (2)(−37.5) = −75 W
The 2 A source delivers 75 W.
[c] The choice of a reference node in part (b) resulted in one simple KCL
equation, while the choice of a reference node in part (a) resulted in a
supernode KCL equation and a second supernode constraint equation.
Both methods give the same result but the choice of reference node in
part (b) yielded fewer equations to solve, so is the preferred method.
P 4.27 Place 5v∆ inside a supernode and use the lower node as a reference. Then
v∆ − 15 v∆ v∆ − 5v∆ v∆ − 5v∆
+
+
+
=0
10
2
20
40
12v∆ = 60;
v∆ = 5 V
vo = v∆ − 5v∆ = −4(5) = −20 V
P 4.28
Node equations:
v1 v1 − 20 v3 − v2 v3
+
+
+
+ 3.125v∆ = 0
20
2
4
80
v2 v2 − v3 v2 − 20
+
+
=0
40
4
1
Constraint equations:
v∆ = 20 − v2
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4–36
CHAPTER 4. Techniques of Circuit Analysis
v1 − 35iφ = v3
iφ = v2 /40
Solving, v1 = −20.25 V;
v2 = 10 V;
v3 = −29 V
Let ig be the current delivered by the 20 V source, then
ig =
20 − (20.25) 20 − 10
+
= 30.125 A
2
1
pg (delivered) = 20(30.125) = 602.5 W
P 4.29 For the given values of v3 and v4:
v∆ = 120 − v3 = 120 − 108 = 12 V
iφ =
v4 − v3
81.6 − 108
=
= −3.3 A
8
8
40
iφ = −44 V
3
v1 = v4 +
40
iφ = 81.6 − 44 = 37.6 V
3
Let ia be the current from right to left through the dependent voltage source:
ia =
v1 v1 − v2
+
= 1.88 − 20.6 = −18.72 A
20
4
Let ib be the current supplied by the 120 V source:
ib =
120 − 37.6 120 − 108
+
= 20.6 + 6 = 26.6 A
4
2
Then
p120V = −(120)(26.6) = −3192 W
pCCVS = [(40/3)(−3.3)](−18.72) = −823.68 W
pVCVS = (81.6)[1.75(12)] = 1713.6 W
.·.
X
pdev = 3192 + 823.68 = 4015.68 W
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Problems
4–37
The total power dissipated by the resistors is
(37.6)2 (82.4)2 (12)2 (108)2
+
+
+
2
4
2
40
pR =
= +(3.3)2 (8) +
.·.
X
Thus,
(81.6)2
= 2302.08 W
80
pdiss = 2302.08 + 1713.6 = 4015.68 W
X
pdev =
X
pdiss;
Agree with analyst
P 4.30 From Eq. 4.16, iB = vc /(1 + β)RE
From Eq. 4.17, iB = (vb − Vo )/(1 + β)RE
From Eq. 4.19,
"
#
1
VCC (1 + β)RE R2 + Vo R1 R2
iB =
− Vo
(1 + β)RE R1 R2 + (1 + β)RE (R1 + R2 )
=
VCC R2 − Vo (R1 + R2 )
[VCC R2 /(R1 + R2 )] − Vo
=
R1 R2 + (1 + β)RE (R1 + R2)
[R1R2 /(R1 + R2)] + (1 + β)RE
P 4.31 [a]
The three mesh current equations are:
−125 + 1i1 + 6(i1 − i6) + 2(i1 − i3)
=
0
24i6 + 12(i6 − i3) + 6(i6 − i1)
=
0
−125 + 2(i3 − i1) + 12(i3 − i6) + 1i3
=
0
Place these equations in standard form:
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4–38
CHAPTER 4. Techniques of Circuit Analysis
i1 (1 + 6 + 2) + i3(−2) + i6 (−6)
= 125
i1 (−6) + i3(−12) + i6(24 + 12 + 6)
= 0
i1 (−2) + i3(2 + 12 + 1) + i6(−12)
= 125
Solving, i1 = 23.76 A;
i3 = 18.43 A;
i6 = 8.66 A
Now calculate the remaining branch currents:
i2
=
i1 − i3 = 5.33 A
i4
=
i1 − i6 = 15.10 A
i5
=
i3 − i6 = 9.77 A
[b] psources = ptop + pbottom = −(125)(23.76) − (125)(18.43)
= −2970 − 2304 = −5274 W
Thus, the power developed in the circuit is 5274 W.
Now calculate the power absorbed by the resistors:
p1top = (23.76)2 (1) = 564.54 W
p2 = (5.33)2 (2) = 56.82 W
p1bot = (18.43)2 (1) = 339.66 W
p6 = (15.10)2 (6) = 1368.06 W
p12 = (9.77)2 (12) = 1145.43 W
p24 = (8.66)2 (24) = 1799.89 W
The power absorbed by the resistors is
564.54 + 56.82 + 339.66 + 1368.06 + 1145.43 + 1799.89 = 5274 W so the
power balances.
P 4.32 [a]
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Problems
4–39
The three mesh current equations are:
−128 + 5ia + 60(ia − ic )
= 0
4ic + 80(ic − ie) + 60(ic − ia )
= 0
320 + 80(ie − ic ) + 10ie
= 0
Place these equations in standard form:
ia (5 + 60) + ic(−60) + ie(0)
=
128
ia (−60) + ic(4 + 80 + 60) + ie(−80)
=
0
ia (0) + ic(−80) + ie(80 + 10)
=
−320
Solving, ia = −6.8 A;
ic = −9.5 A;
ie = −12 A
Now calculate the remaining branch currents:
ib
=
ia − ic = 2.7 A
id
=
ic − ie = 2.5 A
[b] p128V = −(128)ia = −(128)(−6.8) = 870.4 W (abs)
p320V = (320)ie = (320)(−12) = −3840 W (dev)
Thus, the power developed in the circuit is 3840 W. Note that the
resistors cannot develop power!
P 4.33 [a]
60 = 15i1 − 10i2
−20 = −10i1 + 15i2
Solving, i1 = 5.6 A;
ia = i1 = 5.6 A;
i2 = 2.4 A
ib = i1 − i2 = 3.2 A;
ic = −i2 = −2.4 A
[b] If the polarity of the 60 V source is reversed, we have
−60 = 15i1 − 10i2
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4–40
CHAPTER 4. Techniques of Circuit Analysis
−20 = −10i1 + 15i2
i1 = −8.8 A and i2 = −7.2 A
ia = i1 = −8.8 A;
ib = i1 − i2 = −1.6 A;
ic = −i2 = 7.2 A
P 4.34 [a]
230 − 115 = 7i1 − 1i2 − 2i3
0 = −1i1 + 10i2 − 3i3
115 − 460 = −2i1 − 3i2 + 10i3
Solving, i1 = 4.4 A;
i2 = −10.6 A;
i3 = −36.8 A
p230 = −230i1 = −1012 W(del)
p115 = 115(i1 − i3) = 4738 W(abs)
p460 = 460i3 = −16,928 W(del)
. ·.
X
pdev = 17,940 W
[b] p6Ω = (10.6)2 (6) = 674.16 W
p1Ω = (15)2 (1) = 225 W
p3Ω = (26.2)2 (3) = 2059.32 W
p2Ω = (41.2)2 (2) = 3394.88 W
p4Ω = (4.4)2 (4) = 77.44 W
p5Ω = (36.8)2 (5) = 6771.2 W
. ·.
X
pabs = 4738 + 674.16 + 225 + 2059.32 + 3394.88
+77.44 + 6771.2 = 17,940 W
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Problems
4–41
P 4.35
The three mesh current equations are:
−20 + 2000(i1 − i2) + 30,000(i1 − i3)
=
0
5000i2 + 5000(i2 − i3) + 2000(i2 − i1)
=
0
1000i3 + 30,000(i3 − i1) + 5000(i3 − i2)
=
0
Place these equations in standard form:
i1(32,000) + i2(−2000) + i3 (−30,000)
= 20
i1(−2000) + i2(12,000) + i3 (−5000)
= 0
i1(−30,000) + i2 (−5000) + i3(36,000)
= 0
Solving, i1 = 5.5 mA;
i2 = 3 mA;
Thus, io = i3 − i2 = 2 mA.
i3 = 5 mA
P 4.36 [a]
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4–42
CHAPTER 4. Techniques of Circuit Analysis
The four mesh current equations are:
−230 + 1(i1 − i2) + 1(i1 − i3) + 1(i1 − i4)
=
0
6i2 + 1(i2 − i3) + 1(i2 − i1)
=
0
2i3 + 1(i3 − i4) + 1(i3 − i1) + 1(i3 − i2)
=
0
6i4 + 1(i4 − i1) + 1(i4 − i3)
=
0
Place these equations in standard form:
i1 (3) + i2(−1) + i3 (−1) + i4(−1)
=
230
i1 (−1) + i2(8) + i3 (−1) + i4(0)
=
0
i1 (−1) + i2(−1) + i3 (5) + i4(−1)
=
0
i1 (−1) + i2(0) + i3 (−1) + i4(8)
=
0
Solving, i1 = 95 A;
i2 = 15 A;
i3 = 25 A;
The power absorbed by the 5 Ω resistor is
i4 = 15 A
p5 = i23(2) = (25)2 (2) = 1250 W
[b] p230 = −(230)i1 = −(230)(95) = −21,850 W
P 4.37 [a]
25 = 30i1 − 20i2 + 0i∆
0 = −20i1 + 44i2 + 6i∆
21 = 0i1 + 0i2 + 14i∆
Solving, i1 = 1 A;
i2 = 0.25 A;
i∆ = 1.5 A
vo = 20(i1 − i2) = 20(0.75) = 15 V
[b] p6i∆ = 6i∆ i2 = (6)(1.5)(0.25) = 2.25 W (abs)
.·. p6i∆ (deliver) = −2.25 W
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
4–43
P 4.38
−135 + 25i1 − 3i2 − 20i3 + 0iσ = 0
−3i1 + 12i2 − 4i3 + 0iσ = 0
−20i1 − 4i2 + 25i3 + 10iσ = 0
1i1 − 1i2 + 0i3 + 1iσ = 0
Solving, i1 = 64.8 A
i2 = 39 A
i3 = 68.4 A
iσ = −25.8 A
p20Ω = (68.4 − 64.8)2 (20) = 259.2 W
P 4.39
660 = 30i1 − 10i2 − 15i3
20iφ = −10i1 + 60i2 − 50i3
0 = −15i1 − 50i2 + 90i3
iφ = i2 − i3
Solving, i1 = 42 A;
i2 = 27 A;
i3 = 22 A;
iφ = 5 A
20iφ = 100 V
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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4–44
CHAPTER 4. Techniques of Circuit Analysis
p20iφ = −100i2 = −100(27) = −2700 W
.·. p20iφ (developed) = 2700 W
CHECK:
p660V = −660(42) = −27,720 W (dev)
.·.
X
Pdev
X
Pdis
=
27,720 + 2700 = 30,420 W
=
(42)2 (5) + (22)2 (25) + (20)2 (15) + (5)2 (50)+
(15)2 (10)
=
30,420 W
P 4.40
Mesh equations:
53i∆ + 8i1 − 3i2 − 5i3 = 0
0i∆ − 3i1 + 30i2 − 20i3 = 30
0i∆ − 5i1 − 20i2 + 27i3 = 30
Constraint equations:
i∆ = i2 − i3
Solving, i1 = 110 A;
i2 = 52 A;
i3 = 60 A;
i∆ = −8 A
pdepsource = 53i∆ i1 = (53)(−8)(110) = −46,640 W
Therefore, the dependent source is developing 46,640 W.
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
4–45
CHECK:
p30V = −30i2 = −1560 W (left source)
p30V = −30i3 = −1800 W (right source)
X
pdev = 46,640 + 1560 + 1800 = 50 k W
p3Ω = (110 − 52)2 (3) = 10,092 W
p5Ω = (110 − 60)2 (5) = 12,500 W
p20Ω = (−8)2 (20) = 1280 W
p7Ω = (52)2 (7) = 18,928 W
p2Ω = (60)2 (2) = 7200 W
X
pdiss = 10,092 + 12,500 + 1280 + 18,928 + 7200 = 50 kW
P 4.41
Mesh equations:
128i1 − 80i2 = 240
−80i1 + 200i2 = 120
Solving,
i1 = 3 A;
i2 = 1.8 A
Therefore,
v1 = 40(6 − 3) = 120 V;
v2 = 120(1.8 − 1) = 96 V
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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4–46
CHAPTER 4. Techniques of Circuit Analysis
P 4.42
120 = 32i1 − 20i2 − 7i3
−80 = −20i1 + 25i2 − 1i3
−4 = 0i1 + 0i2 + 1i3
Solving, i1 = 1.55 A;
i2 = −2.12 A;
i3 = −4 A
[a] v4A = 7(−4 − 1.55) + 1(−4 + 2.12)
= −40.73 V
p4A = 4v4A = 4(−40.73) = −162.92 W
Therefore, the 4 A source delivers 162.92 W.
[b] p120V = −120(1.55) = −186 W
p80V = −80(−2.12) = 169.6 W
Therefore, the total power delivered is 162.92 + 186 + 169.6 = 518.52 W
[c]
P 4.43 [a]
X
presistors = (1.55)2 (5) + (2.12)2 (4) + (3.67)2 (20) + (5.55)2 (7) + (1.88)2 (1)
X
= 518.52 W
pabs = 518.52 W =
X
pdel (CHECKS)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
4–47
Mesh equations:
65i1 − 40i2 + 0i3 − 100io = 0
−40i1 + 55i2 − 5i3 + 11.5io = 0
0i1 − 5i2 + 9i3 − 11.5io = 0
−1i1 + 1i2 + 0i3 + 1io = 0
Solving,
i1 = 7.2 A;
i2 = 4.2 A;
i3 = −4.5 A;
io = 3 A
Therefore,
v1 = 20[5(3) − 7.2] = 156 V;
v2 = 40(7.2 − 4.2) = 120 V
v3 = 5(4.2 + 4.5) + 11.5(3) = 78 V
[b] p5io = −5io v1 = −5(3)(156) = −2340 W
p11.5io = 11.5io (i2 − i3 ) = 11.5(3)(4.2 + 4.5) = 300.15 W
p96V = 96i3 = 96(−4.5) = −432 W
Thus, the total power dissipated in the circuit, which equals the total
power developed in the circuit is 2340 + 432 = 2772 W.
P 4.44 [a]
The mesh current equation for the right mesh is:
5400(i1 − 0.005) + 3700i1 − 150(0.005 − i1) = 0
Solving,
9250i1 = 27.75
.·. i1 = 3 mA
Then,
i∆ = 5 − 3 = 2 mA
[b] vo = (0.005)(10,000) + (5400)(0.002) = 60.8 V
p5mA = −(60.8)(0.005) = −304 mW
Thus, the 5 mA source delivers 304 mW
[c] pdep source = −150i∆ i1 = (−150)(0.002)(0.003) = −0.9 mW
The dependent source delivers 0.9 mW.
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4–48
CHAPTER 4. Techniques of Circuit Analysis
P 4.45
Mesh equations:
10i∆ − 4i1 = 0
−4i∆ + 24i1 + 6.5i∆ = 400
Solving, i1 = 15 A;
i∆ = 16 A
v20A = 1i∆ + 6.5i∆ = 7.5(16) = 120 V
p20A = −20v20A = −(20)(120) = −2400 W (del)
p6.5i∆ = 6.5i∆ i1 = (6.5)(16)(15) = 1560 W (abs)
Therefore, the independent source is developing 2400 W, all other elements are
absorbing power, and the total power developed is thus 2400 W.
CHECK:
p1Ω = (16)2 (1) = 256 W
p5Ω = (20 − 16)2 (5) = 80 W
p4Ω = (1)2 (4) = 4 W
p20Ω = (20 − 15)2 (20) = 500 W
X
pabs = 1560 + 256 + 80 + 4 + 500 = 2400 W (CHECKS)
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Problems
4–49
P 4.46 [a]
Mesh equations:
−50 + 6i1 − 4i2 + 9i∆ = 0
−9i∆ − 4i1 + 29i2 − 20i3 = 0
Constraint equations:
i∆ = i2 ;
i3 = −1.7v∆ ;
Solving, i1 = −5 A;
v∆ = 2i1
i2 = 16 A;
i3 = 17 A;
v∆ = −10 V
9i∆ = 9(16) = 144 V
ia = i2 − i1 = 21 A
ib = i2 − i3 = −1 A
vb = 20ib = −20 V
p50V = −50i1 = 250 W (absorbing)
p9i∆ = −ia(9i∆ ) = −(21)(144) = −3024 W (delivering)
p1.7V = −1.7v∆ vb = i3 vb = (17)(−20) = −340 W (delivering)
[b]
X
Pdev = 3024 + 340 = 3364 W
X
Pdis = 250 + (−5)2 (2) + (21)2 (4) + (16)2 (5) + (−1)2 (20)
= 3364 W
P 4.47 [a]
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4–50
CHAPTER 4. Techniques of Circuit Analysis
Supermesh equations:
1000ib + 4000(ic − id ) + 500(ic − ia) = 0
ic − ib = 0.01
Two remaining mesh equations:
5500ia − 500ic = −30
4000id − 4000ic = −80
In standard form,
−500ia + 1000ib + 4500ic − 4000id = 0
0ia − 1ib + 1ic + 0id = 0.01
5500ia + 0ib − 500ic + 0id = −30
0ia + 0ib − 4000ic + 4000id = −80
Solving:
ia = −10 mA;
ib = −60 mA;
ic = −50 mA;
id = −70 mA
Then,
i1 = ia = −10 mA;
i2 = ia − ic = 40 mA;
i3 = id = −70 mA
[b] psources = 30(−0.01) + [1000(−0.06)](0.01) + 80(−0.07) = −6.5 W
presistors = 1000(0.06)2 + 5000(0.01)2 + 500(0.04)2
+4000(−0.05 + 0.07)2 = 6.5 W
P 4.48
−20 + 4i1 + 9i2 − 90 + 6i2 + 1i1 = 0;
Solving, i1 = 10 A;
i1 − i2 = 6
i2 = 4 A
p20V = −20i1 = −200 W (diss)
p4 Ω = (10)2 (4) = 400 W
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Problems
4–51
p1 Ω = (10)2 (1) = 100 W
p9 Ω = (4)2 (9) = 144 W
p6 Ω = (4)2 (6) = 96 W
vo = 9(4) − 90 + 6(4) = −30 V
p6A = 6vo = −180 W
p90V = −90i2 = −360 W
X
pdev = 200 + 180 + 360 = 740 W
X
pdiss = 400 + 100 + 144 + 96 = 740 W
Thus the total power dissipated is 740 W.
P 4.49 [a] Summing around the supermesh used in the solution to Problem 4.48 gives
−60 + 4i1 + 9i2 − 90 + 6i2 + 1i1 = 0;
.·. i1 = 12 A;
i1 − i2 = 6
i2 = 6 A
p60V = −60(12) = −720 W (del)
vo = 9(6) − 90 + 6(6) = 0 V
p6A = 6vo = 0 W
p90V = −90i2 = −540 W (del)
X
X
pdiss = (12)2 (4 + 1) + (6)2 (9 + 6) = 1260 W
pdev = 720 + 0 + 540 = 1260 W =
X
pdiss
[b] With 6 A current source replaced with a short circuit
5i1 = 60;
15i2 = 90
Solving,
i1 = 12 A,
. ·.
X
i2 = 6 A
Psources = −(60)(12) − (90)(6) = −1260 W
[c] A 6 A source with zero terminal voltage is equivalent to a short circuit
carrying 6 A.
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4–52
CHAPTER 4. Techniques of Circuit Analysis
P 4.50 [a]
−4id + 10(ie − id ) + 5(ie − ix ) = 0
5(ix − ie ) + 10(id − ie ) − 240 + 40(ix − 19) = 0
id − ix = 2ib = 2(ie − ix )
Solving, id = 10 A;
ie = 18 A;
ia = 19 − ix = −7 A;
ix = 26 A
ib = ie − ix = −8 A;
[b] va = 40ia = −280 V;
ic = ie − id = 8 A;
vb = 5ib + 40ia = −320 V
p19A = −19va = 5320 W
p4id = −4id ie = −720 W
p2ia = −2ib vb = −5120 W
p240V = −240id = −2400 W
p40Ω = (7)2 (40) = 1960 W =
p5Ω = (8)2 (5) = 320 W
p10Ω = (8)2 (10) = 640 W
X
X
Pgen = 720 + 5120 + 2400 = 8240 W
Pdiss = 5320 + 1960 + 320 + 640 = 8240 W
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Problems
4–53
P 4.51 [a]
200 = 85i1 − 25i2 − 50i3
0 = −75i1 + 35i2 + 150i3
(supermesh)
i3 − i2 = 4.3(i1 − i2)
Solving, i1 = 4.6 A;
ia = i2 = 5.7 A;
i2 = 5.7 A;
i3 = 0.97 A
ib = i1 = 4.6 A
ic = i3 = 0.97 A;
id = i1 − i2 = −1.1 A
ie = i1 − i3 = 3.63 A
[b] 10i2 + vo + 25(i2 − i1) = 0
.·. vo = −57 − 27.5 = −84.5 V
p4.3id = −vo (4.3id) = −(−84.5)(4.3)(−1.1) = −399.685 W(dev)
p200V = −200(4.6) = −920 W(dev)
X
Pdev
X
Pdis
=
1319.685 W
=
(5.7)2 10 + (1.1)2 (25) + (0.97)2 100 + (4.6)2 (10)+
(3.63)2 (50)
=
. ·.
X
Pdev =
1319.685 W
X
Pdis = 1319.685 W
P 4.52 [a] The node voltage method requires summing the currents at two
supernodes in terms of four node voltages and using two constraint
equations to reduce the system of equations to two unknowns. If the
connection at the bottom of the circuit is used as the reference node,
then the voltages controlling the dependent sources are node voltages.
This makes it easy to formulate the constraint equations. The current in
the 20 V source is obtained by summing the currents at either terminal of
the source.
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4–54
CHAPTER 4. Techniques of Circuit Analysis
The mesh current method requires summing the voltages around the two
meshes not containing current sources in terms of four mesh currents. In
addition the voltages controlling the dependent sources must be
expressed in terms of the mesh currents. Thus the constraint equations
are more complicated, and the reduction to two equations and two
unknowns involves more algebraic manipulation. The current in the 20 V
source is found by subtracting two mesh currents.
Because the constraint equations are easier to formulate in the node
voltage method, it is the preferred approach.
[b]
Node voltage equations:
v1
v2
+ 0.003v∆ +
− 0.2 = 0
100
250
v3
v4
0.2 +
+
− 0.003v∆ = 0
100 200
Constraints:
v2 = va ;
v3 = v∆ ;
v4 − v3 = 0.4va ;
Solving, v1 = 24 V; v2 = 44 V;
v2
io = 0.2 −
= 24 m A
250
v2 − v1 = 20
v3 = −72 V;
v4 = −54 V.
p20V = 20(0.024) = 480 m W
Thus, the 20 V source absorbs 480 mW.
P 4.53 [a] There are three unknown node voltages and only two unknown mesh
currents. Use the mesh current method to minimize the number of
simultaneous equations.
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Problems
4–55
[b]
The mesh current equations:
2500(i1 − 0.01) + 2000i1 + 1000(i1 − i2 )
= 0
5000(i2 − 0.01) + 1000(i2 − i1 ) + 1000i2
= 0
Place the equations in standard form:
i1 (2500 + 2000 + 1000) + i2 (−1000)
= 25
i1 (−1000) + i2(5000 + 1000 + 1000)
= 50
Solving, i1 = 6 mA;
i2 = 8 mA
Find the power in the 1 kΩ resistor:
i1k = i1 − i2 = −2 m A
p1k = (−0.002)2 (1000) = 4 mW
[c] No, the voltage across the 10 A current source is readily available from the
mesh currents, and solving two simultaneous mesh-current equations is
less work than solving three node voltage equations.
[d] vg = 2000i1 + 1000i2 = 12 + 8 = 20 V
p10mA = −(20)(0.01) = −200 m W
Thus the 10 mA source develops 200 mW.
P 4.54 [a] There are three unknown node voltages and three unknown mesh currents,
so the number of simultaneous equations required is the same for both
methods. The node voltage method has the advantage of having to solve
the three simultaneous equations for one unknown voltage provided the
connection at either the top or bottom of the circuit is used as the
reference node. Therefore recommend the node voltage method.
[b]
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4–56
CHAPTER 4. Techniques of Circuit Analysis
The node voltage equations are:
v1
v1 − v2 v1 − v3
+
+
= 0
5000
2500
1000
v2
v2 − v1 v2 − v3
−0.01 +
+
+
= 0
4000
2500
2000
v3 − v1 v3 − v2
v3
+
+
= 0
1000
2000
1000
Putthe equations in standard
form:
1
1
1
1
1
v1
+
+
+ v2 −
+ v3 −
=
5000 2500 1000
2500
1000
1
1
1
1
1
v1 −
+ v2
+
+
+ v3 −
=
2500
4000 2500 2000
2000
1
1
1
1
1
v1 −
+ v2 −
+ v3
+
+
=
1000
2000
2000 1000 1000
Solving,
v1 = 6.67 V; v2 = 13.33 V; v3 = 5.33 V
p10m = −(13.33)(0.01) = −133.33 m W
Therefore, the 10 mA source is developing 133.33 mW
0
0.01
0
P 4.55 [a] Both the mesh-current method and the node-voltage method require three
equations. The mesh-current method is a bit more intuitive due to the
presence of the voltage sources. We choose the mesh-current method,
although technically it is a toss-up.
[b]
125
=
10i1 − 0.5i2 − 9.2i3
125
=
−0.5i1 + 20i2 − 19.2i3
0
=
−9.2i1 − 19.2i2 + 40i3
Solving, i1 = 32.25 A;
i2 = 26.29 A;
v1
=
9.2(i1 − i3) = 112.35 V
v2
=
19.2(i2 − i3) = 120.09 V
v3
=
11.6i3 = 232.44 V
i3 = 20.04 A
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Problems
[c] pR1
=
(i1 − i3)2 (9.2) = 1371.93 W
pR2
=
(i2 − i3)2 (19.2) = 751.13 W
pR3
=
i23(11.6) = 4657.52 W
[d]
X
4–57
pdev = 125(i1 + i2) = 7317.72 W
X
pload = 6780.58 W
% delivered =
6780.58
× 100 = 92.66%
7317.72
[e]
250 = 29i1 − 28.4i2
0 = −28.4i1 + 40i2
Solving, i1 = 28.29 A;
i2 = 20.09 A
i1 − i2
= 8.2 A
v1
= (8.2)(9.2) = 75.44 V
v2
= (8.2)(19.2) = 157.44 V
Note v1 is low and v2 is high. Therefore, loads designed for 125 V would
not function properly, and could be damaged.
P 4.56
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4–58
CHAPTER 4. Techniques of Circuit Analysis
The mesh current equations:
125
=
(R + 0.8)ia − 0.5ib − Ric
125
=
−0.5ia + (R + 0.8)ib − Ric
.·. (R + 0.8)ia − 0.5ib − Ric = −0.5ia + (R + 0.8)ib − Ric
.·. (R + 0.8)ia − 0.5ib = −0.5ia + (R + 0.8)ib
.·. (R + 1.3)ia = (R + 1.3)ib
Thus
ia = ib
so
io = ib − ia = 0
P 4.57 [a]
Write the mesh current equations. Note that if io = 0, then i1 = 0:
−23 + 5(−i2) + 10(−i3 ) + 46
=
0
30i2 + 15(i2 − i3 ) + 5i2
=
0
Vdc + 25i3 − 46 + 10i3 + 15(i3 − i2)
=
0
i2 (−5) + i3(−10) + Vdc(0)
=
−23
i2 (30 + 15 + 5) + i3(−15) + Vdc (0)
=
0
i2 (−15) + i3(25 + 10 + 15) + Vdc (1)
=
46
Place the equations in standard form:
Solving,
i2 = 0.6 A;
i3 = 2 A;
Vdc = −45 V
Thus, the value of Vdc required to make io = 0 is −45 V.
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Problems
4–59
[b] Calculate the power:
p23V
= −(23)(0) = 0 W
p46V
= −(46)(2) = −92 W
pVdc
= (−45)(2) = −90 W
p30Ω
= (30)(0.6)2 = 10.8 W
p5Ω
= (5)(0.6)2 = 1.8 W
p15Ω
= (15)(2 − 0.6)2 = 29.4 W
p10Ω
= (10)(2)2 = 40 W
p20Ω
= (20)(0)2 = 0 W
p25Ω
= (25)(2)2 = 100 W
X
X
pdev = 92 + 90 = 182 W
pdis = 10.8 + 1.8 + 29.4 + 40 + 0 + 100 = 182 W(checks)
P 4.58 Choose the reference node so that a node voltage is identical to the voltage
across the 4 A source; thus:
Since the 4 A source is developing 0 W, v1 must be 0 V.
Since v1 is known, we can sum the currents away from node 1 to find v2; thus:
0 − (240 + v2 ) 0 − v2
0
+
+
−4=0
12
20
15
.·. v2 = −180 V
Now that we know v2 we sum the currents away from node 2 to find v3; thus:
v2 + 240 − 0 v2 − 0 v2 − v3
+
+
=0
12
20
40
.·. v3 = −340 V
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4–60
CHAPTER 4. Techniques of Circuit Analysis
Now that we know v3 we sum the currents away from node 3 to find idc; thus:
v3 v3 − v2
+
= idc
50
40
.·. idc = −10.8 A
P 4.59 [a] Apply source transformations to both current sources to get
io =
−(5.4 + 0.6)
= −1 mA
2700 + 2300 + 1000
[b]
The node voltage equations:
v1
v1 − v2
−2 × 10−3 +
+
= 0
2700
2300
v2
v2 − v1
+
+ 0.6 × 10−3 = 0
1000
2300
Place
standard form:
these equations
in 1
1
1
v1
+
+ v2 −
= 2 × 10−3
2700 2300
2300
1
1
1
v1 −
+ v2
+
= −0.6 × 10−3
2300
1000 2300
Solving, v1 = 2.7 V;
v2 = 0.4 V
v2 − v1
. ·. i o =
= −1 mA
2300
P 4.60 [a] Applying a source transformation to each current source yields
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Problems
4–61
Now combine the 12 V and 5 V sources into a single voltage source and
the 6 Ω, 6 Ω and 5 Ω resistors into a single resistor to get
Now use a source transformation on each voltage source, thus
which can be reduced to
. ·. i o = −
8.5
(1) = −0.85 A
10
[b]
34ia − 17ib = 12 + 5 + 34 = 51
−17ia + 18.5ib = −34
Solving, ib = −0.85 A = io
P 4.61 [a]
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4–62
CHAPTER 4. Techniques of Circuit Analysis
io =
120
= 4 mA
30,000
[b]
va
=
ia
=
ib
=
vb
=
ig
=
p100V
=
(15,000)(0.004) = 60 V
va
= 1 mA
60,000
12 − 1 − 4 = 7 mA
60 − (0.007)(4000) = 32 V
32
0.007 −
= 6.6 mA
80,000
−(100)(6.6 × 10−3 ) = −660 mW
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
Check:
p12mA
X
X
4–63
=
−(60)(12 × 10−3 ) = −720 mW
Pdev
=
660 + 720 = 1380 mW
Pdis
=
(20,000)(6.6 × 10−3 )2 + (80,000)(0.4 × 10−3 )2 + (4000)(7 × 10−3 )2
+(60,000)(1 × 10−3 )2 + (15,000)(4 × 10−3 )2
=
1380 mW
P 4.62 [a] First remove the 16 Ω and 260 Ω resistors:
Next use a source transformation to convert the 1 A current source and
40 Ω resistor:
which simplifies to
250
(480) = 400 V
300
[b] Return to the original circuit with vo = 400 V:
. ·. v o =
ig =
520
+ 1.6 = 3.6 A
260
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4–64
CHAPTER 4. Techniques of Circuit Analysis
p520V = −(520)(3.6) = −1872 W
Therefore, the 520 V source is developing 1872 W.
[c] v1 = −520 + 1.6(4 + 250 + 6) = −104 V
vg = v1 − 1(16) = −104 − 16 = −120 V
p1A = (1)(−120) = −120 W
Therefore the 1 A source is developing 120 W.
[d]
X
pdev = 1872 + 120 = 1992 W
X
. ·.
P 4.63 vTh =
pdiss = (1)2 (16) +
X
pdiss =
X
(104)2 (520)2
+
+ (1.6)2 (260) = 1992 W
40
260
pdev
30
(80) = 60 V
40
RTh = 2.5 +
(30)(10)
= 10 Ω
40
P 4.64 First we make the observation that the 10 mA current source and the 10 kΩ
resistor will have no influence on the behavior of the circuit with respect to
the terminals a,b. This follows because they are in parallel with an ideal
voltage source. Hence our circuit can be simplified to
or
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Problems
4–65
Therefore the Norton equivalent is
P 4.65 [a] Open circuit:
v2 − 9 v2
+
− 1.8 = 0
20
70
v2 = 35 V
vTh =
60
v2 = 30 V
70
Short circuit:
v2 − 9 v2
+
− 1.8 = 0
20
10
.·. v2 = 15 V
ia =
9 − 15
= −0.3 A
20
isc = 1.8 − 0.3 = 1.5 A
RTh =
30
= 20 Ω
1.5
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4–66
CHAPTER 4. Techniques of Circuit Analysis
[b]
RTh = (20 + 10k60 = 20 Ω (CHECKS)
P 4.66 After making a source transformation the circuit becomes
500 = 20i1 − 8i2
300 = −8i1 + 43.2i2
.·. i1 = 30 A and i2 = 12.5 A
vTh = 12i1 + 5.2i2 = 425 V
RTh = (8k12 + 5.2)k30 = 7.5 Ω
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Problems
4–67
P 4.67
50i1 − 40isc = 60 + 40
−40i1 + 48iscs = 32
Solving,
RTh = 8 +
isc = 7 A
(10)(40)
= 16 Ω
50
P 4.68 First, find the Thévenin equivalent with respect to Ro .
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4–68
CHAPTER 4. Techniques of Circuit Analysis
Ro (Ω)
io (A)
vo (V)
10
1.2
12
15
1.067
16
22
0.923
20.31
33
0.762
25.14
47
0.623
29.30
68
0.490
33.31
P 4.69
12.5 = vTh − 2RTh
11.7 = vTh − 18RTh
Solving the above equations for VTh and RTh yields
vTh = 12.6 V,
.·. IN = 252 A,
RTh = 50 mΩ
RN = 50 mΩ
P 4.70
i1 = 100/20 = 5 A
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Problems
100 = vTh − 5RTh ,
4–69
vTh = 100 + 5RTh
i2 = 200/50 = 4 A
200 = vTh − 4RTh ,
vTh = 200 + 4RTh
.·. 100 + 5RTh = 200 + 4RTh
so
RTh = 100Ω
vTh = 100 + 500 = 600 V
P 4.71 [a] First, find the Thévenin equivalent with respect to a,b using a succession
of source transformations.
.·. vTh = 54 V
RTh = 4.5 kΩ
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4–70
CHAPTER 4. Techniques of Circuit Analysis
54
(85.5) = 51.3 V
90
51.3 − 54
[b] %error =
× 100 = −5%
54
vmeas =
P 4.72
v1 =
200
(18) = 5.143 V
700
v2 =
1200
(18) = 5.139 V
4203
vTh = v1 − v2 = 5.143 − 5.139 = 3.67 mV
RTh =
igal =
(500)(200) (3003)(1200)
+
= 1000.24 Ω
700
4203
3.67 × 10−3
= 3.5 µA
1050.24
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4–71
Problems
P 4.73
The node voltage equations are:
v1 − 40
v1
v1 − v2
+
+
2000
20,000
5000
v2 − v1
v2
v2 − v3
v1
+
+
+ 30
5000
50,000
10,000
20,000
v3 − v2
v3
v1
+
− 30
10,000
40,000
20,000
=
0
=
0
=
0
In standard form:
!
1
1
1
1
40
v1
+
+
+ v2 −
+ v3(0) =
2000 20,000 5000
5000
2000
!
!
v1
1
30
1
1
1
1
−
+
+ v2
+
+
+ v3 −
5000 20,000
5000 50,000 10,000
10,000
v1
30
−
20,000
!
+ v2
!
1
1
1
−
+ v3
+
10,000
10,000 40,000
Solving,
v1 = 24 V;
VTh = v3 = 280 V
v2 = −10 V;
!
!
=0
=0
v3 = 280 V
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4–72
CHAPTER 4. Techniques of Circuit Analysis
The mesh current equations are:
−40 + 2000i1 + 20,000(i1 − i2)
= 0
5000i2 + 50,000(i2 − isc) + 20,000(i2 − i1)
= 0
50,000(isc − i2 ) + 10,000(isc − 30i∆ )
= 0
The constraint equation is:
i∆ = i1 − i2
Put these equations in standard form:
i1(22,000) + i2(−20,000) + isc (0) + i∆ (0)
=
40
i1(−20,000) + i2 (75,000) + isc (−50,000) + i∆ (0)
=
0
i1(0) + i2(−50,000) + isc (60,000) + i∆ (−300,000)
=
0
i1(−1) + i2 (1) + isc(0) + i∆ (1)
=
0
Solving,
RTh
i1 = 13.6 mA;
280
=
= 20 kΩ
0.014
i2 = 12.96 mA;
isc = 14 mA;
i∆ = 640 µA
P 4.74
OPEN CIRCUIT
v2 = −80ib (50 × 103 ) = −40 × 105 ib
4 × 10−5 v2 = −160ib
1310ib + 4 × 10−5 v2 = 1310ib − 160ib = 1150ib
So 1150ib is the voltage across the 100 Ω resistor.
From KCL at the top left node,
500 µA =
1150ib
+ ib = 12.5ib
100
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Problems
4–73
500 × 10−6
.·. ib =
= 40 µA
12.5
vTh = −40 × 105 (40 × 10−6 ) = −160 V
SHORT CIRCUIT
v2 = 0;
ib =
isc = −80ib
100
(500 × 10−6 ) = 35.46 µA
100 + 1310
isc = −80(35.46) = −2837 µA
RTh =
−160
= 56.4 kΩ
−2837 × 10−6
P 4.75 [a] Use source transformations to simplify the left side of the circuit.
ib =
Let
. ·.
7.7 − 5.5
= 0.1 mA
22,000
Ro = Rmeter k1.3 kΩ = 5.5/4.4 = 1.25 kΩ
(Rmeter)(1.3)
= 1.25;
Rmetere + 1.3
Rmeter =
(1.25)(1.3)
= 32.5 kΩ
0.05
[b] Actual value of ve :
ib =
7.7
= 0.0972 mA
22 + (44)(1.3)
ve = 44ib (1.3) = 5.56 V
% error =
5.5 − 5.56
× 100 = −1.1%
5.56
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4–74
CHAPTER 4. Techniques of Circuit Analysis
P 4.76 [a] Find the Thévenin equivalent with respect to the terminals of the
ammeter. This is most easily done by first finding the Thévenin with
respect to the terminals of the 4.8 Ω resistor.
Thévenin voltage: note iφ is zero.
vTh vTh vTh vTh − 16
+
+
+
=0
100
25
20
2
Solving, vTh = 20 V.
Short-circuit current:
isc = 12 + 2isc ,
RTh =
.·. isc = −12 A
20
= −(5/3) Ω
−12
Rtotal =
20
= 3.33 Ω
6
Rmeter = 3.33 − 3.13 = 0.2 Ω
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Problems
4–75
[b] Actual current:
iactual =
20
= 6.38 A
3.13
% error =
6 − 6.38
× 100 = −6%
6.38
P 4.77 VTh = 0, since circuit contains no independent sources.
v1 − 10i∆
v1
v1 − vT
+
+
=0
10
2.5
12
vT − v1 vT − 10i∆
+
−1=0
12
6
i∆ =
vT − v1
12
In standard form:
v1
1
1
1
1
+
+
+ vT −
+ i∆ (−1) = 0
10 2.5 12
12
v1 −
1
1
1
10
+ vT
+
+ i∆ −
=1
12
12 6
6
v1(1) + vT (−1) + i∆ (12) = 0
Solving,
v1 = 2 V;
vT = 8 V;
i∆ = 0.5 A
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4–76
CHAPTER 4. Techniques of Circuit Analysis
vT
.·. RTh =
= 8Ω
1 A
P 4.78 VTh = 0 since there are no independent sources in the circuit. Thus we need
only find RTh .
iT =
vT
+ ia
10
ia = i∆ − 21i∆ = −20i∆
i∆ =
vT − 300i∆
,
700
1000i∆ = vT
vT
vT
.·. iT =
− 20
= 0.08vT
10
1000
vT
= 1/0.08 = 12.5 Ω
iT
.·. RTh = 12.5 Ω
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Problems
4–77
P 4.79 [a]
v − 12 v − 10
v
+
+
=0
12,000 20,000 12,500
Solving,
v = 7.03125 V
10,000
v10k =
(7.03125) = 5.625 V
12,500
.·. VTh = v − 10 = −4.375 V
RTh = [(12,000k20,000) + 2500] = 5 kΩ
Ro = RTh = 5 kΩ
[b]
pmax = (−437.5 × 10−6 )2(5000) = 957 µW
[c] The resistor closest to 5 kΩ from Appendix H has a value of 4.7 kΩ. Use
voltage division to find the voltage drop across this load resistor, and use
the voltage to find the power delivered to it:
4700
v4.7k =
(−4.375) = −2.12 V
4700 + 5000
(−2.12)2
p4.7k =
= 956.12 µW
4700
The percent error between the maximum power and the power delivered
to the best resistor from Appendix H is
956
% error =
− 1 (100) = −0.1%
957
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4–78
CHAPTER 4. Techniques of Circuit Analysis
P 4.80 Write KCL equations at each of the labeled nodes, place them in standard
form, and solve:
v1
v1 − v2
+
=0
4000
8000
At v1 :
− 3 × 10−3 +
At v2 :
v2 − v1 v2 − 10 v2 − v3
+
+
=0
8000
20,000
2500
At v3 :
v3 − v2
v3
v3 − 10
+
+
=0
2500
10,000
5000
Standard form:
v1
v1
1
1
1
+
+ v2 −
+ v3(0) = 0.003
4000 8000
8000
!
1
1
1
1
1
−
+ v2
+
+
+ v3 −
8000
8000 20,000 2500
2500
v1(0) + v2
1
1
1
1
−
+ v3
+
+
2500
2500 10,000 5000
!
=
=
10
20,000
10
5000
Calculator solution:
v1 = 10.890625 V
v2 = 8.671875 V
v3 = 7.8125 V
Calculate currents:
i2 =
10 − v2
= 66.40625 µ A
20,000
i3 =
10 − v3
= 437.5 µ A
5000
Calculate power delivered by the sources:
p3mA = (3 × 10−3 )v1 = (3 × 10−3 )(10.890625) = 32.671875 mW
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Problems
4–79
p10Vmiddle = i2(10) = (66.40625 × 10−6 )(10) = 0.6640625 mW
p10Vtop = i3(10) = (437.5 × 10−6 )(10) = 4.375 mW
pdeliveredtotal = 32.671875 + 0.6640625 + 4.375 = 37.7109375 mW
Calculate power absorbed by the 5 kΩ resistor and the percentage power:
p5k = i23 (5000) = (437.5 × 10−6 )2(5000) = 0.95703125 mW
0.95793125
(100) = 2.54%
37.7109375
% delivered to Ro :
P 4.81 [a] Since 0 ≤ Ro ≤ ∞ maximum power will be delivered to the 6 Ω resistor
when Ro = 0.
302
[b] P =
= 150 W
6
P 4.82 [a] From the solution to Problem 4.68 we have
Ro (Ω)
po (W)
10
14.4
15
17.07
22
18.75
33
19.16
47
18.26
68 16.31
The 33 Ω resistor dissipates the most power, because its value is closest to
the Thévenin equivalent resistance of the circuit.
[b]
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4–80
CHAPTER 4. Techniques of Circuit Analysis
[c] Ro = 33 Ω,
po = 19.16 W. Compare this to Ro = RTh = 30 Ω, which
then gives the maximum power delivered to the load, po (max) = 19.2 W.
P 4.83 We begin by finding the Thévenin equivalent with respect to Ro . After making
a couple of source transformations the circuit simplifies to
i∆ =
160 − 30i∆
;
50
i∆ = 2 A
VTh = 20i∆ + 30i∆ = 50i∆ = 100 V
Using the test-source method to find the Thévenin resistance gives
iT =
vT vT − 30(−vT /30)
+
30
20
iT
1
1
4
2
=
+
=
=
vT
30 10
30
15
RTh =
vT
15
=
= 7.5 Ω
iT
2
Thus our problem is reduced to analyzing the circuit shown below.
p=
100
7.5 + Ro
2
Ro = 250
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Problems
4–81
104
Ro = 250
R2o + 15Ro + 56.25
104 Ro
= R2o + 15Ro + 56.25
250
40Ro = R2o + 15Ro + 56.25
R2o − 25Ro + 56.25 = 0
Ro = 12.5 ±
√
156.25 − 56.25 = 12.5 ± 10
Ro = 22.5 Ω
Ro = 2.5 Ω
P 4.84 [a] From the solution of Problem 4.73 we have RTh = 20 kΩ and VTh = 280 V.
Therefore
Ro = RTh = 20 kΩ
[b] p =
(140)2
= 980 mW
20,000
[c]
The node voltage equations are:
v1 − 40
v1
v1 − v2
+
+
= 0
2000
20,000
5000
v2 − v1
v2
v2 − v3
+
+
+ 30i∆ = 0
5000
50,000
10,000
v3 − v2
v3
v3
+
− 30i∆ +
= 0
10,000
40,000
20,000
The dependent source constraint equation is:
v1
i∆ =
20,000
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4–82
CHAPTER 4. Techniques of Circuit Analysis
Place these equations in standard
form:
!
1
1
1
1
40
v1
+
+
+ v2 −
+ v3 (0) + i∆ (0) =
2000 20,000 5000
5000
2000
v1
v1 (0) + v2
v1
!
!
1
1
1
1
1
−
+ v2
+
+
+ v3 −
+ i∆ (30) = 0
4000
4000 50,000 10,000
10,000
1
−
10,000
!
+ v3
!
1
1
1
+
+
+ i∆ (−30) = 0
10,000 40,000 20,000
!
−1
+ v2(0) + v3(0) + i∆ (1) = 0
20,000
Solving, v1 = 18.4 V; v2 = −31 V; v3 = 140 V;
Calculate the power:
40 − 18.4
ig =
= 10.8 mA
2000
p40V = −(40)(10.8 × 10−3 ) = −432 mW
pX
dep source = (v2 − v3 )(30i∆ ) = −4719.6 mW
pdev = 432 + 4719.6 = 5151.6 mW
i∆ = 920 µA
980 × 10−3
× 100 = 19.02%
5151.6 × 10−3
[d] There are two resistor values in Appendix H that fit the criterion – 18 kΩ
and 22 kΩ. Let’s use the Thévenin equivalent circuit to calculate the
power delivered to each in turn, first by calculating the current through
the load resistor and then using the current to calculate to power
delivered to the load:
280
i18k =
= 7.368 m A
20,000 + 18,000
% delivered =
p18k = (7.368)2 (18,000) = 977.17 m W
i22k =
280
= 6.667 m A
20,000 + 22,000
p22k = (6.667)2 (22,000) = 977.88 m W
We select the 22 kΩ resistor, as the power delivered to it is closer to the
maximum power of 980 mW.
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Problems
4–83
[e] Now substitute the 22 kΩ resistor into the original circuit and calculate
the power developed by the sources in this circuit:
The node voltage equations are:
v1 − 40
v1
v1 − v2
+
+
= 0
2000
20,000
5000
v2 − v1
v2
v2 − v3
+
+
+ 30i∆ = 0
5000
50,000
10,000
v3 − v2
v3
v3
+
− 30i∆ +
= 0
10,000
40,000
22,000
The dependent source constraint equation is:
v1
i∆ =
20,000
Place these equations in standard
form:
!
1
1
1
1
40
v1
+
+
+ v2 −
+ v3 (0) + i∆ (0) =
2000 20,000 5000
5000
2000
v1
v1 (0) + v2
v1
!
!
1
1
1
1
1
−
+ v2
+
+
+ v3 −
+ i∆ (30) = 0
5000
5000 50,000 10,000
10,000
1
−
10,000
!
+ v3
!
1
1
1
+
+
+ i∆ (−30) = 0
10,000 40,000 22,000
!
−1
+ v2(0) + v3(0) + i∆ (1) = 0
20,000
Solving, v1 = 18.67 V; v2 = −30 V; v3 = 146.67 V;
Calculate the power:
40 − 18.67
ig =
= 10.67 mA
2000
p40V = −(40)(10.67 × 10−3 ) = −426.67 mW
pX
dep source = (v2 − v3 )(30i∆ ) = −4946.67 mW
pdev = 426.67 + 4946.67 = 5373.33 mW
i∆ = 933.3 µA
pL = (146.67)2 /22,000 = 977.78 mW
977.78 × 10−3
% delivered =
× 100 = 18.20%
5373.33 × 10−3
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4–84
CHAPTER 4. Techniques of Circuit Analysis
P 4.85 [a] Open circuit voltage
Node voltage equations:
v1 − 24 v1 − 6i∆ v1 − v2
+
+
=0
2
4
5
v2 − v1
− 0.85v∆ = 0
5
Constraint equations:
v1 − v2
i∆ =
;
v∆ = 24 − v1
5
Solving, v2 = 84 V = vTh
Thévenin resistance using a test source:
v1 v1 − 6i∆ v1 − vT
+
+
=0
2
4
5
vT − v1
− 0.85v∆ − 1 = 0
5
i∆ =
v1 − vT
;
5
v∆ = −v1
Solving, vT = 10
vT
RTh =
= 10 Ω
1
.·. Ro = RTh = 10 Ω
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Problems
4–85
[b]
pmax =
(42)2
= 176.4 W
10
[c]
v1 − 24 v1 − 6i∆ v1 − 42
+
+
=0
2
4
5
i∆ =
v1 − 42
5
Solving, v1 = 12 V;
i24V =
i∆ = −6 A;
v∆ = 24 − v1 = 24 − 12 = 12 V
24 − v1
=6A
2
p24V = −24i24V = −24(6) = −144 W
iCCVS =
v1 − 6i∆
= 12 A
4
pCCVS = [6(−6)](12) = −432 W
pVCCS = −[0.85(12)](42) = −428.4 W
X
pdev = 144 + 432 + 428.4 = 1004.4 W
% delivered =
176.4
× 100 = 17.56%
1004.4
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4–86
CHAPTER 4. Techniques of Circuit Analysis
P 4.86 Find the Thévenin equivalent with respect to the terminals of Ro .
Open circuit voltage:
(440 − 220) = 5ia − 2ib − 3ic
0 = −2ia + 10ib − 1ic
ic = 0.5v∆ ;
v∆ = 2(ia − ib )
Solving, ib = 26.4 A
.·. vTh = 7ib = 184.8 V
Short circuit current:
440 − 220 = 5ia − 2isc − 3ic
0 = −2ia + 3isc − 1ic
ic = 0.5v∆ ;
v∆ = 2(ia − isc)
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Problems
Solving, isc = 60 A;
ia = 80 A;
4–87
ic = 20 A
RTh = vTh /isc = 184.8/60 = 3.08 Ω
Ro = 3.08 Ω
Therefore, the Thévenin equivalent circuit configured for maximum power to
the load is
From this circuit,
pmax =
(92.4)2
= 2772 W
3.08
With Ro equal to 3.08 Ω the original circuit becomes
440 − 220 = 5ia − 2ib − 3ic
ic = 0.5v∆ ;
v∆ = 2(ia − ib )
−92.4 = −2ia + 3ib − 1ic
Solving, ia = 88.4 A;
ib = 43.2 A;
ic = 45.2 A
v∆ = 2(88.4 − 43.2) = 90.4 V
p440V = −(440)(88.4) = −38,896 W
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4–88
CHAPTER 4. Techniques of Circuit Analysis
p220V = (220)(88.4 − 45.2) = 9504 W
pdep.source = (440 − 92.4)[0.5(90.4)] = 15,711.52 W
Therefore, only the 440 V source supplies power to the circuit, and the power
supplied is 38,896 W.
% delivered =
2772
= 7.13%
38,896
P 4.87 [a] Find the Thévenin equivalent with respect to the terminals of RL .
Open circuit voltage:
The mesh current equations are:
−240 + 3(i1 − i2) + 20(i1 − i3) + 2i1
=
0
2i2 + 4(i2 − i3) + 3(i2 − i1)
=
0
10iβ + 1i3 + 20(i3 − i1) + 4(i3 − i2)
=
0
The dependent source constraint equation is:
iβ = i2 − i1
Place these equations in standard form:
i1 (3 + 20 + 2) + i2 (−3) + i3(−20) + iβ (0)
=
240
i1 (−3) + i2(2 + 4 + 3) + i3 (−4) + iβ (0)
=
0
i1 (−20) + i2(−4) + i3(1 + 20 + 4) + iβ (10)
=
0
i1 (−1) + i2(1) + i3 (0) + iβ (−1)
=
0
Solving, i1 = 99.6 A; i2 = 78 A;
VTh = 20(i1 − i3 ) = −24 V
i3 = 100.8 A;
iβ = 21.6 A
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Problems
4–89
Short-circuit current:
The mesh current equations are:
−240 + 3(i1 − i2) + 2i1
= 0
2i2 + 4(i2 − i3) + 3(i2 − i1)
= 0
10iβ + 1i3 + 4(i3 − i2)
= 0
The dependent source constraint equation is:
iβ = i2 − i1
Place these equations in standard form:
i1 (3 + 2) + i2 (−3) + i3(0) + iβ (0)
=
240
i1 (−3) + i2(2 + 4 + 3) + i3 (−4) + iβ (0)
=
0
i1 (0) + i2(−4) + i3 (4 + 1) + iβ (10)
=
0
i1 (−1) + i2(1) + i3 (0) + iβ (−1)
=
0
Solving,
i1 = 92 A;
isc = i1 − i3 = −4 A;
i2 = 73.33 A; i3 = 96 A; iβ = 18.67 A
VTh
−24
RTh =
=
= 6Ω
isc
−4
RL = RTh = 6 Ω
[b] pmax =
122
= 24 W
6
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4–90
CHAPTER 4. Techniques of Circuit Analysis
P 4.88 [a] First find the Thévenin equivalent with respect to Ro .
Open circuit voltage: iφ = 0; 50iφ = 0
v1
v1 − 280 v1 − 280
v1
+
+
+
+ 0.5125v∆ = 0
100
10
25
400
(280 − v1)
5 = 56 − 0.2v1
25
v1 = 210 V;
v∆ = 14 V
v∆ =
VTh = 280 − v∆ = 280 − 14 = 266 V
Short circuit current
v1
v1 − 280 v2
v2
+
+
+
+ 0.5125(280) = 0
100
10
20 400
v∆ = 280 V
v2 + 50iφ = v1
280 v2
+
= 56 + 0.05v2
5
20
v2 = −968 V;
v1 = −588 V
iφ =
iφ = isc = 56 + 0.05(−968) = 7.6 A
RTh = VTh /isc = 266/7.6 = 35 Ω
.·. Ro = 35 Ω
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Problems
4–91
[b]
pmax = (133)2 /35 = 505.4 W
[c]
v1
v1 − 280 v2 − 133
v2
+
+
+
+ 0.5125(280 − 133) = 0
100
10
20
400
v2 + 50iφ = v1;
iφ = 133/35 = 3.8 A
Therefore, v1 = −189 V and v2 = −379 V; thus,
ig =
280 − 133 280 + 189
+
= 76.30 A
5
10
p280V (dev) = (280)(76.3) = 21,364 W
P 4.89 [a] We begin by finding the Thévenin equivalent with respect to the terminals
of Ro .
Open circuit voltage
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4–92
CHAPTER 4. Techniques of Circuit Analysis
The mesh current equations are:
−100 + 4(i1 − i2) + 80(i1 − i3) + 16i1
=
0
124i∆ + 8(i2 − i3) + 4(i2 − i1)
=
0
50 + 12i3 + 80(i3 − i1 ) + 8(i3 − i2 )
=
0
The constraint equation is:
i∆ = i3 − i1
Place these equations in standard form:
i1 (4 + 80 + 16) + i2(−4) + i3(−80) + i∆ (0)
=
100
i1 (−4) + i2(8 + 4) + i3(−8) + i∆ (124)
=
0
i1 (−80) + i2(−8) + i3(12 + 80 + 8) + i∆ (0)
=
−50
i1 (1) + i2(0) + i3(−1) + i∆ (1)
=
0
Solving, i1 = 4.7 A;
i2 = 10.5 A;
Also,
VTh = vab = −80i∆ = 48 V
Now find the short-circuit current.
i3 = 4.1 A;
i∆ = −0.6 A
Note with the short circuit from a to b that i∆ is zero, hence 124i∆ is
also zero.
The mesh currents are:
−100 + 4(i1 − i2) + 16i1 = 0
8(i2 − i3) + 4(i2 − i1)
=
0
50 + 12i3 + 8(i3 − i2)
=
0
Place these equations in standard form:
i1 (4 + 16) + i2(−4) + i3 (0)
=
100
i1 (−4) + i2(8 + 4) + i3(−8)
=
0
i1 (0) + i2(−8) + i3 (12 + 8)
=
−50
Solving,
i1 = 5 A; i2 = 0 A;
Then,
isc = i1 − i3 = 7.5 A
RTh = 48/7.5 = 6.4 Ω
i3 = −2.5 A
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Problems
4–93
For maximum power transfer Ro = RTh = 6.4 Ω
242
[b] pmax =
= 90 W
6.4
[c] The resistor from Appendix H that is closest to the Thévenin resistance is
10 Ω. To calculate the power delivered to a 10 Ω load resistor, calculate
the current using the Thévenin circuit and use it to find the power
delivered to the load resistor:
48
i10 =
= 2.927 A
6.4 + 10
p10 = 10(2.927)2 = 85.7 W
Thus, using a 10 Ω resistor selected from Appendix H will cause 85.7 W
of power to be delivered to the load, compared to the maximum power of
90 W that will be delivered if a 6.4 Ω resistor is used.
P 4.90 From the solution of Problem 4.89 we know that when Ro is 6.4 Ω, the voltage
across Ro is 24 V, positive at the upper terminal. Therefore our problem
reduces to the analysis of the following circuit. In constructing the circuit we
have used the fact that i∆ is −0.3 A, and hence 124i∆ is −37.2 V.
Using the node voltage method to find v1 and v2 yields
4.05 +
24 − v1 24 − v2
+
=0
4
8
2v1 + v2 = 104.4;
v1 + 37.2 = v2
Solving, v1 = 22.4 V;
v2 = 59.6 V.
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4–94
CHAPTER 4. Techniques of Circuit Analysis
It follows that
22.4 − 100
i g1
=
= −4.85 A
16
59.6 − 50
i g2
=
= 0.8 A
12
59.6 − 24
i2
=
= 4.45 A
8
ids
=
−4.45 − 0.8 = −5.25 A
p100V
=
100ig1 = −485 W
p50V
=
50ig2 = 40 W
pds
=
37.2ids = −195.3 W
.·.
X
pdev = 485 + 195.3 = 680.3 W
.·. % delivered =
90
(100) = 13.23%
680.3
.·. 13.23% of developed power is delivered to load
P 4.91 [a] 110 V source acting alone:
10(14)
35
=
Ω
24
6
110
132
i0 =
=
A
5 + 35/6
13
Re =
35
132
770
=
V = 59.231 V
6
13
13
4 A source acting alone:
v0 =
5 Ωk10 Ω = 50/15 = 10/3 Ω
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Problems
4–95
10/3 + 2 = 16/3 Ω
16/3k12 = 48/13 Ω
Hence our circuit reduces to:
It follows that
va00 = 4(48/13) = (192/13) V
and
−va00
5
v =
(10/3) = − va00 = −(120/13) V = −9.231 V
(16/3)
8
00
. ·.
[b] p =
v = v 0 + v 00 =
770 120
−
= 50 V
13
13
v2
= 250 W
10
P 4.92 Voltage source acting alone:
io1 =
10
10
=
= 0.2 A
45 + (5 + 5)k10
45 + 5
vo1 =
20
(−10) = −2.5 V
20 + 60
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4–96
CHAPTER 4. Techniques of Circuit Analysis
Current source acting alone:
v2
v2 − v3
+2+
=0
5
5
v3 v3 − v2 v3
+
+
=0
10
5
45
Solving, v2 = −7.25 V = vo2;
io2 = −
i20 =
v3 = −4.5 V
v3
= −0.1 A
45
60k20
(2) = 1.5 A
20
vo2 = −20i20 = −20(1.5) = −30 V
.·. vo = vo1 + vo2 = −2.5 − 30 = −32.5 V
io = io1 + io2 = 0.2 + 0.1 = 0.3 A
P 4.93 240 V source acting alone:
vo1 =
20k5
(240) = 60 V
5 + 7 + 20k5
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Problems
4–97
84 V source acting alone:
vo2 =
20k12
(−84) = −50.4 V
1 + 4 + 20k12
16 A current source acting alone:
v1 − v2 v1
+
− 16 = 0
5
7
v2 − v1 v2 v2 − v3
+
+
=0
5
20
4
v3 − v2 v3
+
+ 16 = 0
4
1
Solving, v2 = 18.4 V = vo3. Therefore,
vo = vo1 + vo2 + vo3 = 60 − 50.4 + 18.4 = 28 V
P 4.94 6 A source:
30 Ωk5 Ωk60 Ω = 4 Ω
.·. io1 =
20
(6) = 4.8 A
20 + 5
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4–98
CHAPTER 4. Techniques of Circuit Analysis
10 A source:
io2 =
4
(10) = 1.6 A
25
75 V source:
io3 = −
4
(15) = −2.4 A
25
io = io1 + io2 + io3 = 4.8 + 1.6 − 2.4 = 4 A
P 4.95 [a] By hypothesis i0o + i00o = 3 mA.
i000
o = −5
(2)
= −1.25 mA;
(8)
.·. io = 3.5 − 1.25 = 2.25 mA
[b] With all three sources in the circuit write a single node voltage equation.
vb vb − 8
+
+ 5 − 10 = 0
6
2
.·. vb = 13.5 V
io =
vb
= 2.25 mA
6
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Problems
4–99
P 4.96 70-V source acting alone:
v 0 = 70 − 4i0b
i0s =
vb0
v0
+
= i0a + i0b
2
10
70 = 20i0a + vb0
i0a =
.·.
70 − vb0
20
i0b =
vb0
v0
70 − vb0
11
v0
+
−
= vb0 +
− 3.5
2
10
20
20
10
v 0 = vb0 + 2i0b
.·. vb0 = v 0 − 2i0b
11
v0
.·. i0b = (v 0 − 2i0b ) +
− 3.5
20
10
13 0 70
.·. v 0 = 70 − 4
v −
42
42
or
or
i0b =
v0 =
13 0 70
v −
42
42
3220
1610
=
V = 34.255 V
94
47
50-V source acting alone:
v 00 = −4i00b
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4–100
CHAPTER 4. Techniques of Circuit Analysis
v 00 = vb00 + 2i00b
v 00 = −50 + 10i00d
v 00 + 50
.·. i00d =
10
i00s =
vb00 v 00 + 50
+
2
10
i00b =
vb00
v 00 v 00 v 00 + 50
11
v 00 + 50
+ i00s = b + b +
= vb00 +
20
20
2
10
20
10
vb00 = v 00 − 2i00b
11
v 00 + 50
.·. i00b = (v 00 − 2i00b ) +
20
10
Thus,
Hence,
13 00 100
v = −4
v +
42
42
00
v = v 0 + v 00 =
13 00 100
v +
42
42
or
i00b =
or
v 00 = −
200
V = −4.255 V
47
1610 200
1410
−
=
= 30 V
47
47
47
P 4.97 Voltage source acting alone:
vo1 − 25
vo1
vo1 − 25
+
− 2.2
=0
4000
20,000
4000
Simplifying
5vo1 − 125 + vo1 − 11vo1 + 275 = 0
.·. vo1 = 30 V
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Problems
4–101
Current source acting alone:
vo2
vo2
vo2
+
+ 0.005 − 2.2
4000 20,000
4000
Simplifying
=0
5vo2 + vo2 + 100 − 11vo2 = 0
.·. vo2 = 20 V
vo = vo1 + vo2 = 30 + 20 = 50 V
P 4.98
100 = 6ia − 1ib + 0ic − 2id − 2ie + 0if − 1ig
0 = −1ia + 4ib − 2ic + 0id + 0ie + 0if + 0ig
0 = 0ia − 2ib + 13ic − 3id + 0ie + 0if + 0ig
0 = −2ia + 0ib − 3ic + 9id − 4ie + 0if + 0ig
0 = −2ia + 0ib + 0ic − 4id + 9ie − 3if + 0ig
0 = 0ia + 0ib + 0ic + 0id − 3ie + 13if − 2ig
0 = −1ia + 0ib + 0ic + 0id + 0ie − 2if + 4ig
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4–102
CHAPTER 4. Techniques of Circuit Analysis
A computer solution yields
ia = 30 A;
ie = 15 A;
ib = 10 A;
if = 5 A;
ic = 5 A;
ig = 10 A;
id = 15 A
.·. i = id − ie = 0 A
CHECK:
p1T = p1B = (ib )2 = (ig )2 = 100 W
p1L = (ia − ib )2 = (ia − ig )2 = 400 W
p2C = 2(ib − ic )2 = (ig − if )2 = 50 W
p3 = 3(ic − id )2 = 3(ie − if )2 = 300 W
p4 = 4(id − ie )2 = 0 W
p8 = 8(ic )2 = 8(if )2 = 200 W
p2L = 2(ia − id )2 = 2(ia − ie )2 = 450 W
X
P 4.99
X
pabs
pgen
=
100 + 400 + 50 + 200 + 300 + 450 + 0 + 450 + 300+
200 + 50 + 400 + 100 = 3000 W
=
100ia = 100(30) = 3000 W (CHECKS)
The mesh equations are:
−125 + 0.15ia + 18.4(ia − ic) + 0.25(ia − ib )
= 0
−125 + 0.25(ib − ia) + 38.4(ib − id) + 0.15ib
= 0
0.15ic + 18.4(ic − ie) + 0.25(ic − id ) + 18.4(ic − ia)
= 0
0.15id + 38.4(id − ib) + 0.25(id − ic) + 38.4(id − ie )
= 0
11.6ie + 38.4(ie − id) + 18.4(ie − ic )
= 0
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Problems
4–103
Place these equations in standard form:
ia(18.8) + ib(−0.25) + ic(−18.4) + id(0) + ie (0)
=
125
ia(−0.25) + ib(38.8) + ic(0) + id(−38.4) + ie (0)
=
125
ia(−18.4) + ib(0) + ic (37.2) + id(−0.25) + ie (−18.4)
=
0
ia(0) + ib(−38.4) + ic (−0.25) + id (77.2) + ie (−38.4)
=
0
ia(0) + ib(0) + ic (−18.4) + id(−38.4) + ie(68.4)
=
0
Solving,
ia = 32.77 A; ib = 26.46 A;
Find the requested voltages:
v1 = 18.4(ic − ie) = 113.90 V
v2 = 38.4(id − ie) = 120.19 V
v3 = 11.6ie = 233.62 V
ic = 26.33 A;
id = 23.27 A;
ie = 20.14 A
P 4.100
KCL equations at nodes B, D, and E:
vB − vA vB − vE
+
− 0.1 = 0
4
7
0.1 +
vD vD + 13v∆
+
−5=0
2
3
vE − vB vE − vA vE
+
+
+5 =0
7
5
6
Multiply the first equation by 28, the second by 6, and the third by 42 to get
−7vA + 11vB − 4vE = 2.8
5vD + 26v∆ = 29.4
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4–104
CHAPTER 4. Techniques of Circuit Analysis
−8.4vA − 6vB + 21.4vE = −210
Constraint equations:
vA = 3vx ;
vσ =
vx = vE − vC − 0.9;
vA − vB
= 0.25vA − 0.25vB ;
4
v∆ = vB − vE
5iσ = vB = vC
Use the constraint equations to solve for vA , vB and v∆ in terms of vC and vE :
vA = 3vE − 3vC − 2.7
vB =
15
11
vE − vC − 1.5
9
9
6
11
v∆ = vE − vC − 1.5
9
9
Substitute these three expressions into the previous three equations to yield:
68vC + 0vD − 60vE = 3.6
−286vC + 45vD + 156vE = 615.6
292.8vC + 0vD − 124.2vE = −2175.12
Solving,
vC = −14.3552 V;
vD = −20.9474 V;
vC = 16.3293 V
From the circuit diagram,
p5A = 5v5A = 5(vE − vD) = 23.09 W
Therefore the 5 A source is absorbing 23.09 W of power.
P 4.101 [a] In studying the circuit in Fig. P4.101 we note it contains six meshes and
six essential nodes. Further study shows that by replacing the parallel
resistors with their equivalent values the circuit reduces to four meshes
and four essential nodes as shown in the following diagram.
The node Voltage approach will require solving three node Voltage
equations along with equations involving v∆ and iβ .
The mesh-current approach will require writing one supermesh equation
plus three constraint equations involving the three current sources. Thus
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Problems
4–105
at the outset we know the supermesh equation can be reduced to a single
unknown current. Since we are interested in the power developed by the
1 V source, we will retain the mesh current ib and eliminate the mesh
currents ia, ic And id.
The supermesh is denoted by the dashed line in the following figure.
[b] Summing the voltages around the supermesh yields
4
−9iβ + ia + 0.75ib + 1 + 5ib + 7(ic − id ) + 8ic = 0
3
Note that iβ = ib; make that substitution and multiply the equation by
12:
−108ib + 16ia + 9ib + 12 + 60ib + 84(ic − id) + 96ic = 0
or
16ia − 39ib + 180ic − 84id = −12
Use the following constraints:
ia − ic = −2;
. ·.
ib − ic = 3ib
ia = −2 + ic = −2 − 2ib
Therefore,
16(−2 − 2ib ) − 39ib + 180(−2ib ) − 84id = −12
so
−431ib − 84id = 20
Finally use the following constraint:
4
id = −6v∆ = −6 − ia = 8ia = −16 − 16ib
3
Thus,
−431ib − 84(−16 − 16ib ) = 20
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4–106
CHAPTER 4. Techniques of Circuit Analysis
so
913ib = −1324
and
ib = −1.45 A
Finally,
p1V = 1ib = −1.45 W
The 1 V source delivers 1.45 W of power.
P 4.102 [a]
v − v1
v
v − v2
+ +
=0
2xr
R 2r(L − x)
"
#
1
1
1
v1
v2
v
+ +
=
+
2xr R 2r(L − x)
2xr 2r(L − x)
v=
v1RL + xR(v2 − v1)
RL + 2rLx − 2rx2
[b] Let D = RL + 2rLx − 2rx2
dv
(RL + 2rLx − 2rx2 )R(v2 − v1) − [v1RL + xR(v2 − v1)]2r(L − 2x)
=
dx
D2
dv
= 0 when numerator is zero.
dx
The numerator simplifies to
x2 +
2Lv1
RL(v2 − v1) − 2rv1 L2
x+
=0
(v2 − v1)
2r(v2 − v1)
Solving for the roots of the quadratic yields

L 
x=
−v1 ±
v2 − v1 

L 
[c] x =
−v1 ±
v2 − v1 
v2 = 1200 V,
s
s


R
v1 v2 −
(v2 − v1)2

2rL


R
v1 v2 −
(v1 − v2 )2

2rL
v1 = 1000 V,
r = 5 × 10−5 Ω/m;
L = 16 km
R = 3.9 Ω
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Problems
L
16,000
=
= 80;
v2 − v1
1200 − 1000
4–107
v1v2 = 1.2 × 106
R
3.9(−200)2
2
(v1 − v2 ) =
= 0.975 × 105
−5
3
2rL
(10 × 10 )(16 × 10 )
√
x = 80{−1000 ± 1.2 × 106 − 0.0975 × 106 }
= 80{−1000 ± 1050} = 80(50) = 4000 m
[d]
vmin
=
v1RL + R(v2 − v1)x
RL + 2rLx − 2rx2
=
(1000)(3.9)(16 × 103 ) + 3.9(200)(4000)
(3.9)(16,000) + 10 × 10−5 (16,000)(4000) − 10 × 10−5 (16 × 106 )
=
975 V
P 4.103 [a]
voc = VTh = 75 V;
Therefore
[b] iL =
60
= 3 A;
20
iL =
75 − 60
15
=
RTh
RTh
15
= 5Ω
3
vo
VTh − vo
=
RL
RTh
Therefore
P 4.104
RTh =
iL =
RTh
VTh − vo
VTh
=
=
− 1 RL
vo /RL
vo
dv1
−R1 [R2(R3 + R4 ) + R3 R4 ]
=
dIg1
(R1 + R2 )(R3 + R4) + R3R4
dv1
R1 R3 R4
=
dIg2
(R1 + R2 )(R3 + R4) + R3R4
dv2
−R1 R3 R4
+
dIg1 (R1 + R2 )(R3 + R4 ) + R3 R4
dv2
R3 R4 (R1 + R2)
=
dIg2
(R1 + R2 )(R3 + R4) + R3R4
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4–108
CHAPTER 4. Techniques of Circuit Analysis
P 4.105 From the solution to Problem 4.104 we have
dv1
−25[5(125) + 3750]
175
=
=−
V/A = −14.5833 V/A
dIg1
30(125) + 3750
12
and
dv2
−(25)(50)(75)
=
= −12.5 V/A
dIg1
30(125) + 3750
By hypothesis, ∆Ig1 = 11 − 12 = −1 A
175
175
.·. ∆v1 = (−
)(−1) =
= 14.583 V
12
12
Thus, v1 = 25 + 14.583 = 39.583 V
Also,
∆v2 = (−12.5)(−1) = 12.5 V
Thus, v2 = 90 + 12.5 = 102.5 V
The PSpice solution is
v1 = 39.583 V
and
v2 = 102.5 V
These values are in agreement with our predicted values.
P 4.106 From the solution to Problem 4.104 we have
dv1
(25)(50)(75)
=
= 12.5 V/A
dIg2
30(125) + 3750
and
dv2
(50)(75)(30)
=
= 15 V/A
dIg2
30(125) + 3750
By hypothesis, ∆Ig2 = 17 − 16 = 1 A
.·. ∆v1 = (12.5)(1) = 12.5 V
Thus, v1 = 25 + 12.5 = 37.5 V
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Problems
4–109
Also,
∆v2 = (15)(1) = 15 V
Thus, v2 = 90 + 15 = 105 V
The PSpice solution is
v1 = 37.5 V
and
v2 = 105 V
These values are in agreement with our predicted values.
P 4.107 From the solutions to Problems 4.104 — 4.106 we have
dv1
175
=−
V/A;
dIg1
12
dv1
= 12.5 V/A
dIg2
dv2
= −12.5 V/A;
dIg1
dv2
= 15 V/A
dIg2
By hypothesis,
∆Ig1 = 11 − 12 = −1 A
∆Ig2 = 17 − 16 = 1 A
Therefore,
∆v1 =
175
+ 12.5 = 27.0833 V
12
∆v2 = 12.5 + 15 = 27.5 V
Hence
v1 = 25 + 27.0833 = 52.0833 V
v2 = 90 + 27.5 = 117.5 V
The PSpice solution is
v1 = 52.0830 V
and
v2 = 117.5 V
These values are in agreement with our predicted values.
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4–110
CHAPTER 4. Techniques of Circuit Analysis
P 4.108 By hypothesis,
∆R1 = 27.5 − 25 = 2.5 Ω
∆R2 = 4.5 − 5 = −0.5 Ω
∆R3 = 55 − 50 = 5 Ω
∆R4 = 67.5 − 75 = −7.5 Ω
So
∆v1 = 0.5833(2.5) − 5.417(−0.5) + 0.45(5) + 0.2(−7.5) = 4.9168 V
.·. v1 = 25 + 4.9168 = 29.9168 V
∆v2 = 0.5(2.5) + 6.5(−0.5) + 0.54(5) + 0.24(−7.5) = −1.1 V
.·. v2 = 90 − 1.1 = 88.9 V
The PSpice solution is
v1 = 29.6710 V
and
v2 = 88.5260 V
Note our predicted values are within a fraction of a volt of the actual values.
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5
The Operational Amplifier
Assessment Problems
AP 5.1 [a] This is an inverting amplifier, so
vo = (−Rf /Ri )vs = (−80/16)vs ,
vs ( V)
0.4
2.0
vo ( V) −2.0 −10.0
so
vo = −5vs
3.5 −0.6 −1.6 −2.4
−15.0
3.0
8.0
10.0
Two of the values, 3.5 V and −2.4 V, cause the op amp to saturate.
[b] Use the negative power supply value to determine the largest input
voltage:
−15 = −5vs ,
vs = 3 V
Use the positive power supply value to determine the smallest input
voltage:
10 = −5vs ,
Therefore
vs = −2 V
− 2 ≤ vs ≤ 3 V
AP 5.2 From Assessment Problem 5.1
vo = (−Rf /Ri )vs = (−Rx /16,000)vs = (−Rx /16,000)(−0.640)
= 0.64Rx /16,000 = 4 × 10−5 Rx
Use the negative power supply value to determine one limit on the value of Rx :
4 × 10−5 Rx = −15
so
Rx = −15/4 × 10−5 = −375 kΩ
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
5–1 system, or transmission in any form or by any means, electronic,
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5–2
CHAPTER 5. The Operational Amplifier
Since we cannot have negative resistor values, the lower limit for Rx is 0. Now
use the positive power supply value to determine the upper limit on the value
of Rx :
4 × 10−5 Rx = 10
so
Rx = 10/4 × 10−5 = 250 kΩ
Therefore,
0 ≤ Rx ≤ 250 kΩ
AP 5.3 [a] This is an inverting summing amplifier so
vo = (−Rf /Ra )va + (−Rf /Rb )vb = −(250/5)va − (250/25)vb = −50va − 10vb
Substituting the values for va and vb:
vo = −50(0.1) − 10(0.25) = −5 − 2.5 = −7.5 V
[b] Substitute the value for vb into the equation for vo from part (a) and use
the negative power supply value:
vo = −50va − 10(0.25) = −50va − 2.5 = −10 V
Therefore 50va = 7.5,
so va = 0.15 V
[c] Substitute the value for va into the equation for vo from part (a) and use
the negative power supply value:
vo = −50(0.10) − 10vb = −5 − 10vb = −10 V;
Therefore 10vb = 5,
so vb = 0.5 V
[d] The effect of reversing polarity is to change the sign on the vb term in
each equation from negative to positive.
Repeat part (a):
vo = −50va + 10vb = −5 + 2.5 = −2.5 V
Repeat part (b):
vo = −50va + 2.5 = −10 V;
50va = 12.5,
va = 0.25 V
Repeat part (c), using the value of the positive power supply:
vo = −5 + 10vb = 15 V;
10vb = 20;
vb = 2.0 V
AP 5.4 [a] Write a node voltage equation at vn ; remember that for an ideal op amp,
the current into the op amp at the inputs is zero:
vn
vn − vo
+
=0
4500
63,000
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
5–3
Solve for vo in terms of vn by multiplying both sides by 63,000 and
collecting terms:
14vn + vn − vo = 0
so
vo = 15vn
Now use voltage division to calculate vp. We can use voltage division
because the op amp is ideal, so no current flows into the non-inverting
input terminal and the 400 mV divides between the 15 kΩ resistor and
the Rx resistor:
vp =
Rx
(0.400)
15,000 + Rx
Now substitute the value Rx = 60 kΩ:
vp =
60,000
(0.400) = 0.32 V
15,000 + 60,000
Finally, remember that for an ideal op amp, vn = vp , so substitute the
value of vp into the equation for v0
vo = 15vn = 15vp = 15(0.32) = 4.8 V
[b] Substitute the expression for vp into the equation for vo and set the
resulting equation equal to the positive power supply value:
0.4Rx
vo = 15
15,000 + Rx
!
=5
15(0.4Rx ) = 5(15,000 + Rx ) so Rx = 75 kΩ
AP 5.5 [a] Since this is a difference amplifier, we can use the expression for the
output voltage in terms of the input voltages and the resistor values
given in Eq. 5.22:
vo =
20(60)
50
vb − va
10(24)
10
Simplify this expression and subsitute in the value for vb:
vo = 5(vb − va) = 20 − 5va
Set this expression for vo to the positive power supply value:
20 − 5va = 10 V so va = 2 V
Now set the expression for vo to the negative power supply value:
20 − 5va = −10 V so va = 6 V
Therefore 2 ≤ va ≤ 6 V
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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5–4
CHAPTER 5. The Operational Amplifier
[b] Begin as before by substituting the appropriate values into Eq. 5.22:
vo =
8(60)
vb − 5va = 4vb − 5va
10(12)
Now substitute the value for vb:
vo = 4(4) − 5va = 16 − 5va
Set this expression for vo to the positive power supply value:
16 − 5va = 10 V so va = 1.2 V
Now set the expression for vo to the negative power supply value:
16 − 5va = −10 V so va = 5.2 V
Therefore 1.2 ≤ va ≤ 5.2 V
AP 5.6 [a] Replace the op amp with the more realistic model of the op amp from Fig.
5.15:
Write the node voltage equation at the left hand node:
vn
vn − vg
vn − vo
+
+
=0
500,000
5000
100,000
Multiply both sides by 500,000 and simplify:
vn + 100vn − 100vg + 5vn − 5v0 = 0 so 21.2vn − vo = 20vg
Write the node voltage equation at the right hand node:
vo − 300,000(−vn )
vo − vn
+
=0
5000
100,000
Multiply through by 100,000 and simplify:
20vo + 6 × 106 vn + vo − vn = 0 so 6 × 106 vn + 21vo = 0
Use Cramer’s method to solve for vo :
∆=
21.2 −1
6 × 106 21
= 6,000,445.2
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
No =
vo =
21.2 20vg
6
6 × 10
0
5–5
= −120 × 106 vg
No
= −19.9985vg ;
∆
so
vo
= −19.9985
vg
[b] Use Cramer’s method again to solve for vn :
N1 =
vn =
20vg −1
0
21
= 420vg
N1
= 6.9995 × 10−5 vg
∆
vg = 1 V,
vn = 69.995 µ V
[c] The resistance seen at the input to the op amp is the ratio of the input
voltage to the input current, so calculate the input current as a function
of the input voltage:
vg − vn
vg − 6.9995 × 10−5 vg
ig =
=
5000
5000
Solve for the ratio of vg to ig to get the input resistance:
Rg =
vg
5000
=
= 5000.35 Ω
ig
1 − 6.9995 × 10−5
[d] This is a simple inverting amplifier configuration, so the voltage gain is
the ratio of the feedback resistance to the input resistance:
vo
100,000
=−
= −20
vg
5000
Since this is now an ideal op amp, the voltage difference between the two
input terminals is zero; since vp = 0, vn = 0
Since there is no current into the inputs of an ideal op amp, the
resistance seen by the input voltage source is the input resistance:
Rg = 5000 Ω
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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5–6
CHAPTER 5. The Operational Amplifier
Problems
P 5.1
[a] The five terminals of the op amp are identified as follows:
[b] The input resistance of an ideal op amp is infinite, which constrains the
value of the input currents to 0. Thus, in = 0 A.
[c] The open-loop voltage gain of an ideal op amp is infinite, which constrains
the difference between the voltage at the two input terminals to 0. Thus,
(vp − vn ) = 0.
[d] Write a node voltage equation at vn :
vn + 3 vn − vo
+
=0
5000
15,000
But vp = 0 and vn = vp = 0. Thus,
3
vo
−
= 0 so vo = 9 V
5000 15,000
P 5.2
vo = −(0.5 × 10−3 )(10,000) = −5 V
vo
−5
.·. io =
=
= −1 mA
5000
5000
P 5.3
vb − va
vb − vo
+
= 0,
20,000
100,000
therefore vo = 6vb − 5va
[a] va = 4 V,
vb = 0 V,
[b] va = 2 V,
vb = 0 V,
[c] va = 2 V,
vb = 1 V,
[d] va = 1 V,
vb = 2 V,
[e] va = 1.5 V,
[f] If vb = 1.6 V,
vb = 4 V,
vo = −15 V (sat)
vo = −10 V
vo = −4 V
vo = 7 V
vo = 15 V (sat)
vo = 9.6 − 5va = ±15
.·. −1.08 V ≤ va ≤ 4.92 V
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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Problems
P 5.4
vp =
5–7
3000
(3) = 1 V = vn
3000 + 6000
vn − 5 vn − vo
+
=0
10,000
5000
(1 − 5) + 2(1 − vo ) = 0
vo = −1.0 V
iL =
vo
1
=−
= −250 × 10−6
4000
4000
iL = −250 µA
P 5.5
Since the current into the inverting input terminal of an ideal op-amp is zero,
the voltage across the 2.2 MΩ resistor is (2.2 × 106 )(3.5 × 10−6 ) or 7.7 V.
Therefore the voltmeter reads 7.7 V.
P 5.6
[a] i2 =
150 × 10−3
= 75 µA
2000
v1 = −40 × 103 i2 = −3 V
v1
v1
v1 − vo
[b]
+
+
=0
20,000 40,000
50,000
.·. vo = 4.75v1 = −14.25 V
[c] i2 = 75 µA, (from part [a])
−vo
v1 − vo
[d] io =
+
= 795 µ A
25,000
50,000
P 5.7
[a] First, note that vn = vp = 2.5 V
Let vo1 equal the voltage output of the op-amp. Then
2.5 − vg 2.5 − vo1
+
= 0,
5000
10,000
.·. vo1 = 7.5 − 2vg
Also note that vo1 − 2.5 = vo ,
.·. vo = 5 − 2vg
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5–8
CHAPTER 5. The Operational Amplifier
[b] Yes, the circuit designer is correct!
P 5.8
[a] The gain of an inverting amplifier is the ratio of the feedback resistor to
the input resistor. If the gain of the inverting amplifier is to be 3, the
feedback resistor must be 3 times as large as the input resistor. There are
many possible designs that use a resistor value chosen from Appendix H.
We present two here that use 3.3 kΩ resistors. Use a single 3.3 kΩ
resistor as the input resistor, and use three 3.3 kΩ resistors in series as
the feedback resistor to give a total of 9.9 kΩ.
Alternately, use a single 3.3 kΩ resistor as the feedback resistor and use
three 3.3 kΩ resistors in parallel as the input resistor to give a total of
1.1 kΩ.
[b] To amplify a 5 V signal without saturating the op amp, the power supply
voltages must be greater than or equal to the product of the input
voltage and the amplifier gain. Thus, the power supplies should have a
magnitude of (5)(3) = 15 V.
P 5.9
[a] Replace the combination of vg , 1.6 kΩ, and the 6.4 kΩ resistors with its
Thévenin equivalent.
−[12 + σ50]
(0.20)
1.28
At saturation vo = −5 V;
Then vo =
−
12 + σ50
(0.2) = −5,
1.28
or
therefore
σ = 0.4
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Problems
5–9
Thus for 0 ≤ σ ≤ 0.40 the operational amplifier will not saturate.
−(12 + 13.6)
[b] When σ = 0.272,
vo =
(0.20) = −4 V
1.28
vo
vo
Also
+
+ io = 0
10 25.6
. ·. i o = −
P 5.10
vo
4
4
vo
−
=
+
mA = 556.25 µA
10 25.6
10 25.6
[a] Let v∆ be the voltage from the potentiometer contact to ground. Then
0 − vg 0 − v∆
+
=0
2000
50,000
.·. v∆ = −25(40 × 10−3 ) = −1 V
−25vg − v∆ = 0,
v∆
v∆ − 0
v∆ − vo
+
+
=0
αR∆ 50,000 (1 − α)R∆
v∆
v∆ − vo
+ 2v∆ +
=0
α
1−α
v∆
1
1
+2+
α
1−α
=
vo
1−α
"
(1 − α)
.·. vo = −1 1 + 2(1 − α) +
α
When α = 0.2,
When α = 1,
#
vo = −1(1 + 1.6 + 4) = −6.6 V
vo = −1(1 + 0 + 0) = −1 V
.·. −6.6 V ≤ vo ≤ −1 V
"
#
(1 − α)
[b] −1 1 + 2(1 − α) +
= −7
α
α + 2α(1 − α) + (1 − α) = 7α
α + 2α − 2α2 + 1 − α = 7α
.·. 2α2 + 5α − 1 = 0
P 5.11
α∼
= 0.186
"
Rf
Rf
Rf
vo = −
(0.2) +
(0.15) +
(0.4)
4000
5000
20,000
−6 = −0.1 × 10−3 Rf ;
P 5.12
so
Rf = 60 kΩ;
#
.·. 0 ≤ Rf ≤ 60 kΩ
[a] This circuit is an example of an inverting summing amplifier.
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5–10
CHAPTER 5. The Operational Amplifier
220
220
220
va −
vb −
vc = −5 − 12 + 11 = −6 V
44
27.5
80
[c] vo = −6 − 8vb = ±10
[b] vo = −
.·. vb = −0.5 V when vo = 10 V;
vb = 2 V when vo = −10 V
.·. −0.5 V ≤ vb ≤ 2 V
P 5.13
We want the following expression for the output voltage:
vo = −(3va + 5vb + 4vc + 2vd )
This is an inverting summing amplifier, so each input voltage is amplified by a
gain that is the ratio of the feedback resistance to the resistance in the
forward path for the input voltage. Pick a feedback resistor with divisors of 3,
5, 4, and 2 – say 60 kΩ:
"
60k
60k
60k
60k
vo = −
va +
vb +
vc +
vd
Ra
Rb
Rc
Rd
#
Solve for each input resistance value to yield the desired gain:
.·.
Ra = 60,000/3 = 20 kΩ Rc = 60,000/4 = 15 kΩ
Rb = 60,000/5 = 12 kΩ
Rd = 60,000/2 = 30 kΩ
Now create the 5 resistor values needed from the realistic resistor values in
Appendix H. Note that Rb = 12 kΩ and Rc = 15 kΩ are already values from
Appendix H. Create Rf = 60 kΩ by combining 27 kΩ and 33 kΩ in series.
Create Ra = 20 kΩ by combining two 10 kΩ resistors in series. Create
Rd = 30 kΩ by combining 18 kΩ and 12 kΩ in series. Of course there are many
other acceptable possibilities. The final circuit is shown here:
P 5.14
[a] Write a KCL equation at the inverting input to the op amp:
vd − va vd − vb
vd − vc
vd
vd − vo
+
+
+
+
=0
40,000
22,000
100,000 352,000 220,000
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Problems
5–11
Multiply through by 220,000, plug in the values of input voltages, and
rearrange to solve for vo:
vo = 220,000
4
−1
−5
+
+
40,000 22,000 100,000
8
8
+
+
352,000 220,000
!
= 14 V
[b] Write a KCL equation at the inverting input to the op amp. Use the given
values of input voltages in the equation:
8 − va
8−9
8 − 13
8
8 − vo
+
+
+
+
=0
40,000 22,000 100,000 352,000 220,000
Simplify and solve for vo :
44 − 5.5va − 10 − 11 + 5 + 8 − vo = 0 so vo = 36 − 5.5va
Set vo to the positive power supply voltage and solve for va :
36 − 5.5va = 15
.·.
va = 3.818 V
Set vo to the negative power supply voltage and solve for va :
36 − 5.5va = −15
. ·.
va = 9.273 V
Therefore,
3.818 V ≤ va ≤ 9.273 V
P 5.15
[a]
8−4
8−9
8 − 13
8
8 − v0
+
+
+
+
=0
40,000 22,000 100,000 352,000
Rf
8 − vo
= −2.7272 × 10−5
Rf
For vo = 15 V,
For vo = −15 V,
so Rf =
8 − vo
−2.727 × 10−5
Rf = 256.7 kΩ
Rf < 0
so this solution is not possible.
"
#
15 − 8
15
[b] io = −(if + i10k) = −
+
= −1527 µA
3
256.7 × 10
10,000
P 5.16
[a] The circuit shown is a non-inverting amplifier.
[b] We assume the op amp to be ideal, so vn = vp = 3 V. Write a KCL
equation at vn :
3
3 − vo
+
=0
40,000 80,000
Solving,
vo = 9 V.
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5–12
P 5.17
CHAPTER 5. The Operational Amplifier
[a] This circuit is an example of the non-inverting amplifier.
[b] Use voltage division to calculate vp:
vp =
10,000
vs
vs =
10,000 + 30,000
4
Write a KCL equation at vn = vp = vs /4:
vs /4 vs /4 − vo
+
=0
4000
28,000
Solving,
vo = 7vs /4 + vs /4 = 2vs
[c] 2vs = 8
so
2vs = −12
vs = 4 V
so
vs = −6 V
Thus, −6 V ≤ vs ≤ 4 V.
P 5.18
[a] vp = vn =
. ·.
68
vg = 0.85vg
80
0.85vg
0.85vg − vo
+
= 0;
30,000
63,000
.·. vo = 2.635vg = 2.635(4),
vo = 10.54 V
[b] vo = 2.635vg = ±12
vg = ±4.55 V,
[c]
−4.55 ≤ vg ≤ 4.55 V
0.85vg
0.85vg − vo
+
=0
30,000
Rf
!
0.85Rf
+ 0.85 vg = vo = ±12
30,000
.·. 1.7Rf + 51 = ±360;
P 5.19
1.7Rf = 360 − 51;
Rf = 181.76 kΩ
[a] From the equation for the non-inverting amplifier,
Rs + Rf
=4
Rs
so
Rs + Rf = 4Rs
and therefore
Rf = 3Rs
Choose Rf = 30 kΩ and implement this choice from components in
Appendix H by combining two 15 kΩ resistors in series. Choose
Rs = Rg = 10 kΩ, which is a component in Appendix H. The resulting
non-inverting amplifier circuit is shown here:
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Problems
[b] vo = 4vg = 12
so
vo = 4vg = −12
5–13
vg = 3 V
so
vg = −3 V
Therefore,
−3 V ≤ vg ≤ 3 V
P 5.20
[a] This circuit is an example of a non-inverting summing amplifier.
[b] Write a KCL equation at vp and solve for vp in terms of vs :
vp − vs vp − 6
+
=0
15,000
30,000
2vp − 2vs + vp − 6 = 0
so
vp = 2vs /3 + 2
Now write a KCL equation at vn and solve for vo :
vn
vn − vo
+
=0
so
vo = 4vn
20,000
60,000
Since we assume the op amp is ideal, vn = vp. Thus,
vo = 4(2vs /3 + 2) = 8vs /3 + 8
[c] 8vs /3 + 8 = 16
so
8vs /3 + 8 = −12
vs = 3 V
so
vs = −7.5 V
Thus, −7.5 V ≤ vs ≤ 3 V.
P 5.21
[a] This is a non-inverting summing amplifier.
vp − va
vp − vb
[b]
+
=0
3
13 × 10
27 × 103
.·. 40vp = 27va + 13vb
so
vp = 0.675va + 0.325vb
vn
vn − vo
+
=0
11,000 110,000
.·. vo = 11vn = 11vp = 11(0.675va + 0.325vb )
= 11[0.675(0.8) + 0.325(0.4)] = 7.37 V
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5–14
CHAPTER 5. The Operational Amplifier
[c] vp = vn =
ia =
ib =
vo
= 0.667 V
11
va − vp
= 10 µA
13 × 103
vb − vp
= −10 µA
27 × 103
[d] 7.425 for va ;
P 5.22
[a]
3.575 for vb
vp − va vp − vb vp − vc
+
+
=0
Ra
Rb
Rc
. ·. v p =
RbRc
Ra Rc
Ra Rb
va +
vb +
vc
D
D
D
where D = Rb Rc + Ra Rc + Ra Rb
vn
vn − vo
+
=0
20,000 100,000
!
100,000
+ 1 vn = 6vn = vo
20,000
. ·. v o =
6Rb Rc
6Ra Rc
6Ra Rb
va +
vb +
vc
D
D
D
By hypothesis,
6Rb Rc
= 1;
D
Then
6Ra Rb /D
3
=
6Ra Rc /D
2
6Ra Rc
= 2;
D
so
6Ra Rb
=3
D
Rb = 1.5Rc
But from the circuit
Rb = 15 kΩ
so
Rc = 10 kΩ
Similarly,
6Rb Rc /D
1
=
6Ra Rb /D
3
so
3Rc = Ra
Thus,
Ra = 30 kΩ
[b] vo = 1(0.7) + 2(0.4) + 3(1.1) = 4.8 V
vn = vo /6 = 0.8 V = vp
ia =
va − vp
0.7 − 0.8
=
= −3.33 µA
30,000
30,000
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Problems
ib =
vb − vp
0.4 − 0.8
=
= −26.67 µA
15,000
15,000
ic =
vc − vp
1.1 − 0.8
=
= 30 µA
10,000
10,000
5–15
Check:
ia + ib + ic = 0?
P 5.23
[a]
− 3.33 − 26.67 + 30 = 0 (checks)
vp − va vp − vb vp − vc
vp
+
+
+
=0
Ra
Rb
Rc
Rg
. ·. v p =
RbRc Rg
Ra Rc Rg
Ra Rb Rg
va +
vb +
vc
D
D
D
where D = Rb Rc Rg + Ra Rc Rg + Ra Rb Rg + Ra Rb Rc
vn
vn − vo
+
=0
Rs
Rf
vn
1
1
+
Rs Rf
. ·. v o = 1 +
=
vo
Rf
Rf
vn = kvn
Rs
Rf
where k = 1 +
Rs
vp = vn
.·. vo = kvp
or
kRg Rb Rc
kRg Ra Rc
kRg Ra Rb
va +
vb +
vc
D
D
D
vo =
kRg Rb Rc
=6
D
. ·.
kRg Ra Rc
=3
D
Rb
6
= =2
Ra
3
Since
Rc
3
= = 0.75
Rb
4
Ra = 1 kΩ
Rb = 2 kΩ
kRg Ra Rb
=4
D
Rc
6
= = 1.5
Ra
4
Rc = 1.5 kΩ
.·. D = [(2)(1.5)(3) + (1)(1.5)(3) + (1)(2)(3) + (1)(2)(1.5)] × 109 = 22.5 × 109
k(3)(2)(1.5) × 109
=6
22.5 × 109
k=
135 × 109
= 15
9 × 109
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5–16
CHAPTER 5. The Operational Amplifier
.·. 15 = 1 +
Rf
Rs
Rf
= 14
Rs
Rf = (14)(15,000) = 210 kΩ
[b] vo = 6(0.5) + 3(2.5) + 4(1) = 14 V
vn = vp =
P 5.24
14.5
= 0.967 V
15
ia =
0.5 − 0.967
= −466.67 µA
1000
ib =
2.5 − 0.967
= 766.67 µA
2000
ic =
1 − 0.967
= 22.22 µA
1500
ig =
0.967
= 322.22 µA
3000
is =
vn
0.967
=
= 64.44 µA
15,000
15,000
[a] Assume va is acting alone. Replacing vb with a short circuit yields vp = 0,
therefore vn = 0 and we have
0 − va 0 − vo0
+
+ in = 0,
Ra
Rb
Therefore
vo0
va
=− ,
Rb
Ra
vo0 = −
in = 0
Rb
va
Ra
Assume vb is acting alone. Replace va with a short circuit. Now
vp = vn =
vbRd
Rc + Rd
vn
vn − vo00
+
+ in = 0,
Ra
Rb
1
1
+
Ra Rb
vo00
Rd
v 00
vb − o = 0
Rc + Rd
Rb
Rb
=
+1
Ra
in = 0
vo = vo0 + vo00 =
Rd
Rd
vb =
Rc + Rd
Ra
Rd
Ra
Ra + Rb
vb
Rc + Rd
Ra + Rb
Rb
vb −
va
Rc + Rd
Ra
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
Rd
[b]
Ra
Ra + Rb
Rc + Rd
=
Rb
,
Ra
therefore Rd(Ra + Rb ) = Rb(Rc + Rd )
Rd Ra = Rb Rc ,
therefore
Rd
When
Ra
Ra + Rb
Rc + Rd
=
[a] vo =
Ra
Rc
=
Rb
Rd
Rb
Ra
Eq. (5.22) reduces to vo =
P 5.25
5–17
Rb
Rb
Rb
vb −
va =
(vb − va ).
Ra
Ra
Ra
Rd (Ra + Rb)
Rb
120(24 + 75)
75
vb −
va =
(5) − (8)
Ra (Rc + Rd )
Ra
24(130 + 120)
24
vo = 9.9 − 25 = −15.1 V
[b]
v1 − 8 v1 − 15.1
+
=0
24,000
75,000
ia =
so
v1 = 2.4 V
8 − 2.4
= 233 µ A
24,000
Rina =
va
8
=
= 34.3 kΩ
ia
233 × 10−6
[c] Rin b = Rc + Rd = 250 kΩ
P 5.26
Use voltage division to find vp :
vp =
2000
(5) = 1 V
2000 + 8000
Write a KCL equation at vn and solve it for vo :
vn − va vn − vo
+
=0
5000
Rf
so
Rf
Rf
+ 1 vn −
va = vo
5000
5000
Since the op amp is ideal, vn = vp = 1V, so
vo =
Rf
Rf
+1 −
va
5000
5000
To satisfy the equation,
Rf
+1 =5
5000
and
Rf
=4
5000
Thus, Rf = 20 kΩ.
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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5–18
P 5.27
CHAPTER 5. The Operational Amplifier
vbRb
= vn
Ra + Rb
vp =
vn − va vn − vo
+
=0
4700
Rf
vn
vaRf
Rf
+1 −
= vo
4700
4700
Rf
Rb
Rf
+1
vb −
va = vo
4700
Ra + Rb
4700
.·.
.·.
Rf
= 10;
4700
Rf = 47 kΩ
(a value from Appendix H)
Ra + Rb = 220 kΩ
Thus,
47
1+
4700
.·.
Rb
220,000
Rb = 200 kΩ
!
= 10
and
Ra = 220 − 200 = 20 kΩ
Use two 100 kΩ resistors in series for Rb and use two 10 kΩ resistors in series
for Ra .
P 5.28
[a]
vp
vp − vc vp − vd
+
+
=0
20,000
30,000
20,000
.·. 8vp = 2vc + 3vd = 8vn
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Problems
5–19
vn − va vn − vb
vn − vo
+
+
=0
20,000
18,000
180,000
. ·. v o
=
20vn − 9va − 10vb
=
20[(1/4)vc + (3/8)vd ] − 9va − 10vb
=
20(0.75 + 1.5) − 9(1) − 10(2) = 16 V
[b] vo = 5vc + 30 − 9 − 20 = 5vc + 1
±20 = 5vc + 1
.·. vb = −4.2 V
and
vb = 3.8 V
.·. −4.2 V ≤ vb ≤ 3.8 V
P 5.29
vp = 1000ib
1000ib 1000ib − vo
+
− ia = 0
Ra
Rf
.·.
.·.
1000ib
1000ib
1
1
+
Ra Rf
!
Rf
1+
Ra
− Rf ia = vo
− ia =
vo
Rf
By hypopthesis, vo = 5000(ib − ia). Therefore,
Rf = 5 kΩ
1000 1 +
(use two 10 kΩ resistors in parallel)
Rf
Ra
= 5000
so
Ra = 1250 Ω
To construct the 1250 Ω resistor, combine a 1.2 kΩ resistor in series with a
parallel combination of two 100 Ω resistors.
P 5.30
vo =
Rd (Ra + Rb )
Rb
vb −
va
Ra (Rc + Rd )
Ra
By hypothesis: Rb /Ra = 4;
Rc + Rd = 470 kΩ;
Rd (Ra + 4Ra )
.·.
=3
Ra 470,000
Rd = 282 kΩ;
so
Rd (Ra + Rb)
=3
Ra (Rc + Rd)
Rc = 188 kΩ
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5–20
CHAPTER 5. The Operational Amplifier
Create Rd = 282 kΩ by combining a 270 kΩ resistor and a 12 kΩ resistor in
series. Create Rc = 188 kΩ by combining a 120 kΩ resistor and a 68 kΩ resistor
in series. Also, when vo = 0 we have
vn − va
vn
+
=0
Ra
Rb
Ra
.·. vn 1 +
Rb
ia =
= va ;
vn = 0.8va
va − 0.8va
va
= 0.2 ;
Ra
Ra
.·. Ra = 4.4 kΩ;
Rin =
va
= 5Ra = 22 kΩ
ia
Rb = 17.6 kΩ
Create Ra = 4.4 kΩ by combining two 2.2 kΩ resistors in series. Create
Rb = 17.6 kΩ by combining a 12 kΩ resistor and a 5.6 kΩ resistor in series.
P 5.31
vp =
1500
(−18) = −3 V = vn
9000
−3 + 18 −3 − vo
+
=0
1600
Rf
.·. vo = 0.009375Rf − 3
vo = 9 V;
Rf = 1280 Ω
vo = −9 V;
But
P 5.32
Rf = −640 Ω
Rf ≥ 0,
.·. Rf = 1.28 kΩ
αRg
[a] vp =
vg
αRg + (Rg − αRg )
vn = vp = αvg
vn − vg vn − vo
+
=0
R1
Rf
(vn − vg )
Rf
+ vn − vo = 0
R1
vo
=
!
Rf
Rf
1+
αvg −
vg
Rg
R1
=
(αvg − vg )4 + αvg
=
[(α − 1)4 + α]vg
=
(5α − 4)vg
=
(5α − 4)(2) = 10α − 8
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Problems
α
vo
α
vo
α
5–21
vo
0.0 −8 V
0.4 −4 V
0.8 0 V
0.1 −7 V
0.5 −3 V
0.9 1 V
0.2 −6 V
0.6 −2 V
1.0 2 V
0.3 −5 V
0.7 −1 V
[b] Rearranging the equation for vo from (a) gives
vo =
Rf
Rf
+ 1 vg α + −
vg
R1
R1
Therefore,
Rf
slope =
+ 1 vg ;
R1
Rf
intercept = −
vg
R1
[c] Using the equations from (b),
−6 =
Rf
+ 1 vg ;
R1
4=−
Rf
vg
R1
Solving,
Rf
=2
R1
vg = −2 V;
P 5.33
Acm =
(20)(50) − (50)Rx
20(50 + Rx )
Adm =
50(20 + 50) + 50(50 + Rx )
2(20)(50 + Rx )
Adm
Rx + 120
=
Acm
2(20 − Rx )
.·.
Rx + 120
= ±1000 for the limits on the value of Rx
2(20 − Rx )
If we use +1000 Rx = 19.93 kΩ
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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5–22
CHAPTER 5. The Operational Amplifier
If we use −1000 Rx = 20.07 kΩ
19.93 kΩ ≤ Rx ≤ 20.07 kΩ
P 5.34
[a] Adm =
(24)(26) + (25)(25)
= 24.98
(2)(1)(25)
(1)(24) − 25(1)
= −0.04
1(25)
24.98
[c] CMRR =
= 624.50
0.04
[b] Acm =
P 5.35
[a] vp = vs ,
vn =
R1 vo
,
R1 + R2
Therefore vo =
vn = vp
R1 + R2
R2
vs = 1 +
vs
R1
R1
[b] vo = vs
[c] Because vo = vs , thus the output voltage follows the signal voltage.
P 5.36
It follows directly from the circuit that vo = −(120/7.5)vg = −16vg
From the plot of vg we have vg = 0, t < 0
vg
=
t
0 ≤ t ≤ 0.5
vg
=
1−t
0.5 ≤ t ≤ 1.5
vg
=
t−2
1.5 ≤ t ≤ 2.5
vg
=
3−t
2.5 ≤ t ≤ 3.5
vg = t − 4
Therefore
vo = −16t
3.5 ≤ t ≤ 4.5,
etc.
0 ≤ t ≤ 0.5
vo
=
16t − 16
0.5 ≤ t ≤ 1.5
vo
=
32 − 16t
1.5 ≤ t ≤ 2.5
vo
=
16t − 48
2.5 ≤ t ≤ 3.5
vo = 64 − 16t
3.5 ≤ t ≤ 4.5, etc.
These expressions for vo are valid as long as the op amp is not saturated.
Since the peak values of vo are ±5, the output is clipped at ±5. The plot is
shown below.
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Problems
P 5.37
vp =
5–23
5.6
vg = 0.7vg = 7 sin(π/3)t V
8.0
vn
vn − vo
+
=0
15,000
75,000
6vn = vo ;
vn = vp
.·. vo = 42 sin(π/3)t V
vo = 0
0≤t≤∞
t≤0
At saturation
42 sin
π
t = ±21;
3
π
π
.·.
t= ,
3
6
t = 0.50 s,
5π
,
6
2.50 s,
π
sin t = ±0.5
3
7π
,
6
3.50 s,
11π
,
6
etc.
5.50 s,
etc.
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5–24
P 5.38
CHAPTER 5. The Operational Amplifier
[a]
vn − va vn − vo
+
=0
R
R
2vn − va = vo
va
va − vn va − vo
+
+
=0
Ra
R
R
va
1
2
vn
vo
+
−
=
Ra R
R
R
va 2 +
R
− vn = vo
Ra
vn = vp = va + vg
.·. 2vn − va = 2va + 2vg − va = va + 2vg
.·. va − vo = −2vg
2va + va
(1)
R
− va − vg = vo
Ra
R
. ·. v a 1 +
− vo = vg
Ra
(2)
Now combining equations (1) and (2) yields
−va
R
= −3vg
Ra
or va = 3vg
Ra
R
Hence ia =
va
3vg
=
Ra
R
Q.E.D.
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
5–25
[b] At saturation vo = ± Vcc
.·. va = ± Vcc − 2vg
(3)
and
R
. ·. v a 1 +
Ra
= ± Vcc + vg
(4)
Dividing Eq (4) by Eq (3) gives
1+
R
± Vcc + vg
=
Ra
± Vcc − 2vg
. ·.
R
± Vcc + vg
3vg
=
−1=
Ra
± Vcc − 2vg
± Vcc − 2vg
or Ra =
P 5.39
(± Vcc − 2vg )
R
3vg
Q.E.D.
(320 × 10−3 )2
= 6.4 µW
(16 × 103 )
16
[b] v16 kΩ =
(320) = 80 mV
64
[a] p16 kΩ =
p16 kΩ =
(80 × 10−3 )2
= 0.4 µW
(16 × 103 )
pa
6.4
=
= 16
pb
0.4
[d] Yes, the operational amplifier serves several useful purposes:
[c]
• First, it enables the source to control 16 times as much power
delivered to the load resistor. When a small amount of power controls
a larger amount of power, we refer to it as power amplification.
• Second, it allows the full source voltage to appear across the load
resistor, no matter what the source resistance. This is the voltage
follower function of the operational amplifier.
• Third, it allows the load resistor voltage (and thus its current) to be
set without drawing any current from the input voltage source. This
is the current amplification function of the circuit.
P 5.40
[a] Assume the op-amp is operating within its linear range, then
iL =
8
= 2 mA
4000
For RL = 4 kΩ
vo = (4 + 4)(2) = 16 V
Now since vo < 20 V our assumption of linear operation is correct,
therefore
iL = 2 mA
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5–26
CHAPTER 5. The Operational Amplifier
[b] 20 = 2(4 + RL );
RL = 6 kΩ
[c] As long as the op-amp is operating in its linear region iL is independent of
RL . From (b) we found the op-amp is operating in its linear region as
long as RL ≤ 6 kΩ. Therefore when RL = 6 kΩ the op-amp is saturated.
We can estimate the value of iL by assuming ip = in iL . Then
iL = 20/(4000 + 16,000) = 1 mA. To justify neglecting the current into
the op-amp assume the drop across the 50 kΩ resistor is negligible, since
the input resistance to the op-amp is at least 500 kΩ. Then
ip = in = (8 − 4)/(500 × 103 ) = 8 µA. But 8 µA 1 mA, hence our
assumption is reasonable.
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Problems
5–27
[d]
P 5.41
i1 =
15 − 10
= 1 mA
5000
i2 + i1 + 0 = 10 mA;
i2 = 9 mA
vo2 = 10 + (400)(9) × 10−3 = 13.6 V
i3 =
15 − 13.6
= 0.7 mA
2000
i4 = i3 + i1 = 1.7 mA
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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5–28
CHAPTER 5. The Operational Amplifier
vo1 = 15 + 1.7(0.5) = 15.85 V
P 5.42
[a] Let vo1 = output voltage of the amplifier on the left. Let vo2 = output
voltage of the amplifier on the right. Then
vo1 =
ia =
−47
(1) = −4.7 V;
10
−220
(−0.15) = 1.0 V
33
vo2 − vo1
= 5.7 mA
1000
[b] ia = 0 when vo1 = vo2
Thus
−47
(vL) = 1
10
vL = −
P 5.43
vo2 =
so from (a) vo2 = 1 V
10
= −212.77 mV
47
[a] Replace the op amp with the model from Fig. 5.15:
Write two node voltage equations, one at the left node, the other at the
right node:
vn − vg
vn − vo
vn
+
+
=0
5000
100,000 500,000
vo + 3 × 105 vn
vo − vn
vo
+
+
=0
5000
100,000 500
Simplify and place in standard form:
106vn − 5vo = 100vg
(6 × 106 − 1)vn + 221vo = 0
Let vg = 1 V and solve the two simultaneous equations:
vo = −19.9844 V;
vn = 736.1 µV
Thus the voltage gain is vo/vg = −19.9844.
[b] From the solution in part (a), vn = 736.1 µV.
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Problems
[c] ig =
5–29
vg − vn
vg − 736.1 × 10−6 vg
=
5000
5000
Rg =
vg
5000
=
= 5003.68 Ω
ig
1 − 736.1 × 10−6
[d] For an ideal op amp, the voltage gain is the ratio between the feedback
resistor and the input resistor:
100,000
vo
=−
= −20
vg
5000
For an ideal op amp, the difference between the voltages at the input
terminals is zero, and the input resistance of the op amp is infinite.
Therefore,
vn = vp = 0 V;
P 5.44
Rg = 5000 Ω
[a]
vn − vg vn − vo
+
=0
2000
10,000
.·. vo = 6vn − 5vg
Also
vo = A(vp − vn ) = −Avn
. ·. v n =
−vo
A
6
. ·. v o 1 +
= −5vg
A
vo =
−5A
vg
(6 + A)
−5(194)(1)
= −4.85 V
200
−5
[c] vo =
(1) = −5 V
1 + (6/A)
[b] vo =
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5–30
CHAPTER 5. The Operational Amplifier
[d]
P 5.45
[a]
−5A
(1) = −0.99(5)
A+6
.·. −0.05A = −29.7
so
− 5A = −4.95(A + 6)
so
A = 594
vn
vn − vg
vn − vo
+
+
=0
16,000 800,000 200,000
or
55vn − 4vo = vg
Eq (1)
vo
vo − vn
vo − 50,000(vp − vn )
+
+
=0
20,000 200,000
8000
36vo − vn − 125 × 104 (vp − vn ) = 0
vp = vg +
(vn − vg )(240)
= (0.7)vg + (0.3)vn
800
36vo − vn − 125 × 104 [(0.7)vg − (0.7)vn ] = 0
36vo + 874,999vn = 875,000vg
Eq (2)
Let vg = 1 V and solve Eqs. (1) and (2) simultaneously:
vn = 999.446 mV
and
vo
. ·.
= 13.49
vg
vo = 13.49 V
[b] From part (a), vn = 999.446 mV.
vp = (0.7)(1000) + (0.3)(999.446) = 999.834 mV
[c] vp − vn = 387.78 µV
(1000 − 999.83)10−3
[d] ig =
= 692.47 pA
24 × 103
vg
vg − vo
[e]
+
= 0,
since vn = vp = vg
16,000 200,000
vo
.·. vo = 13.5vg ,
= 13.5
vg
vn = vp = 1 V;
P 5.46
vp − vn = 0 V;
ig = 0 A
[a]
vn − 0.88
vn
vn − vTh
+
+
=0
1600
500,000
24,000
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Problems
5–31
vTh + 105 vn vTh − vn
+
=0
2000
24,000
Solving, vTh = −13.198 V
Short-circuit current calculation:
vn
vn − 0.88 vn − 0
+
+
=0
500,000
1600
24,000
.·. vn = 0.8225 V
isc =
vn
105
−
vn = −41.13 A
24,000 2000
RTh =
vTh
= 320.9 mΩ
isc
[b] The output resistance of the inverting amplifier is the same as the
Thévenin resistance, i.e.,
Ro = RTh = 320.9 mΩ
[c]
330
vo =
(−13.2) = −13.18 V
330.3209
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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5–32
CHAPTER 5. The Operational Amplifier
vn − 0.88
vn
vn + 13.18
+
+
=0
1600
500,000
24,000
.·. vn = 942 µV
0.88 − 942 × 10−6
ig =
= 549.41 µA
1600
Rg =
P 5.47
0.88
= 1601.71 Ω
ig
[a] vTh = −
24,000
(0.88) = −13.2 V
1600
RTh = 0, since op-amp is ideal
[b] Ro = RTh = 0 Ω
[c] Rg = 1.6 kΩ
P 5.48
since
vn = 0
From Eq. 5.57,
vref
1
1
1
= vn
+
+
R + ∆R
R + ∆R R − ∆R Rf
!
−
vo
Rf
Substituting Eq. 5.59 for vp = vn :
1
1
vref R+∆R
+ R−∆R
+ R1f
vref
=
1
1
R + ∆R
(R − ∆R) R+∆R
+ R−∆R
+
1
Rf
−
vo
Rf
Rearranging,
vo
1
1
= vref
−
Rf
R − ∆R R + ∆R
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Problems
5–33
Thus,
vo = vref
P 5.49
2∆R
Rf
R2 − ∆R2
[a] Use Eq. 5.61 to solve for Rf ; note that since we are using 1% strain gages,
∆ = 0.01:
(5)(120)
vo R
Rf =
=
= 2 kΩ
2∆vref
(2)(0.01)(15)
[b] Now solve for ∆ given vo = 50 mV:
∆=
vo R
(0.05)(120)
=
= 100 × 10−6
2Rf vref
2(2000)(15)
The change in strain gage resistance that corresponds to a 50 mV change
in output voltage is thus
∆R = ∆R = (100 × 10−6 )(120) = 12 mΩ
P 5.50
[a]
Let R1 = R + ∆R
vp
vp vp − vin
+
+
=0
Rf
R
R1
. ·. v p
"
#
1
1
1
vin
+ +
=
Rf
R R1
R1
. ·. v p =
RRf vin
= vn
RR1 + Rf R1 + Rf R
vn vn − vin vn − vo
+
+
=0
R
R
Rf
vn
"
#
1
1
1
vo
vin
+ +
−
=
R R Rf
Rf
R
. ·. v n
"
#
R + 2Rf
vin
vo
−
=
RRf
R
Rf
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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5–34
CHAPTER 5. The Operational Amplifier
. ·.
"
vo
R + 2Rf
=
Rf
RRf
#"
#
RRf vin
vin
−
RR1 + Rf R1 + Rf R
R
"
#
vo
R + 2Rf
1
. ·.
=
−
vin
Rf
RR1 + Rf R1 + Rf R R
[R2 + 2RRf − R1 (R + Rf ) − RRf ]Rf
·
. . vo =
vin
R[R1(R + Rf ) + RRf ]
Now substitute R1 = R + ∆R and get
vo =
−∆R(R + Rf )Rf vin
R[(R + ∆R)(R + Rf ) + RRf ]
If ∆R R
(R + Rf )Rf (−∆R)vin
vo ≈
R2 (R + 2Rf )
[b] vo ≈
P 5.51
47 × 104 (48 × 104 )(−95)15
≈ −3.384 V
108 (95 × 104 )
[c] vo =
−95(48 × 104 )(47 × 104 )15
= −3.368 V
104 [(1.0095)104 (48 × 104 ) + 47 × 108 ]
[a] vo ≈
(R + Rf )Rf (−∆R)vin
R2 (R + 2Rf )
vo =
. ·.
(R + Rf )(−∆R)Rf vin
R[(R + ∆R)(R + Rf ) + RRf ]
approx value
R[(R + ∆R)(R + Rf ) + RRf ]
=
true value
R2 (R + 2Rf )
.·. Error =
R[(R + ∆R)(R + Rf ) + RRf ] − R2 (R + 2Rf )
R2 (R + 2Rf )
=
.·. % error =
[b] % error =
P 5.52
1=
∆R (R + Rf )
R (R + 2Rf )
∆R(R + Rf )
× 100
R(R + 2Rf )
95(48 × 104 ) × 100
= 0.48%
104 (95 × 104 )
∆R(48 × 104 )
× 100
104 (95 × 104 )
9500
.·. ∆R =
= 197.91667 Ω
48
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Problems
5–35
197.19667
.·. % change in R =
× 100 ≈ 1.98%
104
P 5.53
[a] It follows directly from the solution to Problem 5.50 that
vo =
[R2 + 2RRf − R1 (R + Rf ) − RRf ]Rf vin
R[R1 (R + Rf ) + RRf ]
Now R1 = R − ∆R. Substituting into the expression gives
vo =
(R + Rf )Rf (∆R)vin
R[(R − ∆R)(R + Rf ) + RRf ]
Now let ∆R R and get
vo ≈
(R + Rf )Rf ∆Rvin
R2 (R + 2Rf )
[b] It follows directly from the solution to Problem 5.50 that
. ·.
approx value
R[(R − ∆R)(R + Rf ) + RRf ]
=
true value
R2 (R + 2Rf )
.·. Error =
=
(R − ∆R)(R + Rf ) + RRf − R(R + 2Rf )
R(R + 2Rf )
−∆R(R + Rf )
R(R + 2Rf )
.·. % error =
−∆R(R + Rf )
× 100
R(R + 2Rf )
[c] R − ∆R = 9810 Ω
.·. ∆R = 10,000 − 9810 = 190 Ω
(48 × 104 )(47 × 104 )(190)(15)
·
. . vo ≈
≈ 6.768 V
108 (95 × 104 )
−190(48 × 104 )(100)
[d] % error =
= −0.96%
104 (95 × 104 )
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6
Inductance, Capacitance, and
Mutual Inductance
Assessment Problems
AP 6.1 [a] ig = 8e−300t − 8e−1200tA
v=L
dig
= −9.6e−300t + 38.4e−1200tV,
dt
t > 0+
v(0+ ) = −9.6 + 38.4 = 28.8 V
[b] v = 0 when 38.4e−1200t = 9.6e−300t
or t = (ln 4)/900 = 1.54 ms
[c] p = vi = 384e−1500t − 76.8e−600t − 307.2e−2400t W
dp
[d]
= 0 when e1800t − 12.5e900t + 16 = 0
dt
Let x = e900t and solve the quadratic x2 − 12.5x + 16 = 0
x = 1.44766,
t=
ln 1.45
= 411.05 µs
900
x = 11.0523,
t=
ln 11.05
= 2.67 ms
900
p is maximum at t = 411.05 µs
[e] pmax = 384e−1.5(0.41105) − 76.8e−0.6(0.41105) − 307.2e−2.4(0.41105) = 32.72 W
[f] W is max when i is max, i is max when di/dt is zero.
When di/dt = 0, v = 0, therefore t = 1.54 ms.
[g] imax = 8[e−0.3(1.54) − e−1.2(1.54)] = 3.78 A
wmax = (1/2)(4 × 10−3 )(3.78)2 = 28.6 mJ
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6–1 system, or transmission in any form or by any means, electronic,
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6–2
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
AP 6.2 [a] i = C
dv
d
= 24 × 10−6 [e−15,000t sin 30,000t]
dt
dt
= [0.72 cos 30,000t − 0.36 sin 30,000t]e−15,000t A,
π
[b] i
ms = −31.66 mA,
80
i(0+ ) = 0.72 A
π
v
ms = 20.505 V,
80
p = vi = −649.23 mW
[c] w =
1
Cv 2 = 126.13 µJ
2
1
AP 6.3 [a] v =
C
=
t
Z
i dx + v(0− )
0−
1
0.6 × 10−6
Z
t
0−
3 cos 50,000x dx = 100 sin 50,000t V
[b] p(t) = vi = [300 cos 50,000t] sin 50,000t
= 150 sin 100,000t W,
[c] w(max) =
p(max) = 150 W
1
2
Cvmax
= 0.30(100)2 = 3000 µJ = 3 mJ
2
60(240)
= 48 mH
300
[b] i(0+ ) = 3 + −5 = −2 A
Z
125 t
[c] i =
(−0.03e−5x ) dx − 2 = 0.125e−5t − 2.125 A
6 0+
Z
50 t
[d] i1 =
(−0.03e−5x ) dx + 3 = 0.1e−5t + 2.9 A
+
3 0
AP 6.4 [a] Leq =
25
i2 =
6
Z
t
0+
(−0.03e−5x ) dx − 5 = 0.025e−5t − 5.025 A
i1 + i2 = i
AP 6.5 v1 = 0.5 × 106
Z
240 × 10−6 e−10x dx − 10 = −12e−10t + 2 V
0+
v2 = 0.125 × 106
v1(∞) = 2 V,
t
Z
t
0+
240 × 10−6 e−10x dx − 5 = −3e−10t − 2 V
v2 (∞) = −2 V
1
1
W = (2)(4) + (8)(4) × 10−6 = 20 µJ
2
2
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Problems
6–3
AP 6.6 [a] Summing the voltages around mesh 1 yields
di1
d(i2 + ig )
+8
+ 20(i1 − i2 ) + 5(i1 + ig ) = 0
dt
dt
or
4
di1
di2
dig
4
+ 25i1 + 8
− 20i2 = − 5ig + 8
dt
dt
dt
!
Summing the voltages around mesh 2 yields
16
d(i2 + ig )
di1
+8
+ 20(i2 − i1) + 780i2 = 0
dt
dt
or
di1
di2
dig
8
− 20i1 + 16
+ 800i2 = −16
dt
dt
dt
[b] From the solutions given in part (b)
i1 (0) = −0.4 − 11.6 + 12 = 0;
i2 (0) = −0.01 − 0.99 + 1 = 0
These values agree with zero initial energy in the circuit. At infinity,
i1 (∞) = −0.4A;
i2(∞) = −0.01A
When t = ∞ the circuit reduces to
7.8
7.8
7.8
+
= −0.4A; i2(∞) = −
= −0.01A
20
780
780
From the solutions for i1 and i2 we have
.·. i1(∞) = −
di1
= 46.40e−4t − 60e−5t
dt
di2
= 3.96e−4t − 5e−5t
dt
Also,
dig
= 7.84e−4t
dt
Thus
di1
4
= 185.60e−4t − 240e−5t
dt
25i1 = −10 − 290e−4t + 300e−5t
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6–4
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
8
di2
= 31.68e−4t − 40e−5t
dt
20i2 = −0.20 − 19.80e−4t + 20e−5t
5ig = 9.8 − 9.8e−4t
8
dig
= 62.72e−4t
dt
Test:
185.60e−4t − 240e−5t − 10 − 290e−4t + 300e−5t + 31.68e−4t − 40e−5t
?
+0.20 + 19.80e−4t − 20e−5t = −[9.8 − 9.8e−4t + 62.72e−4t ]
−9.8 + (300 − 240 − 40 − 20)e−5t
?
+(185.60 − 290 + 31.68 + 19.80)e−4t = −(9.8 + 52.92e−4t )
?
−9.8 + 0e−5t + (237.08 − 290)e−4t = −9.8 − 52.92e−4t
−9.8 − 52.92e−4t = −9.8 − 52.92e−4t
(OK)
Also,
8
di1
= 371.20e−4t − 480e−5t
dt
20i1 = −8 − 232e−4t + 240e−5t
16
di2
= 63.36e−4t − 80e−5t
dt
800i2 = −8 − 792e−4t + 800e−5t
16
dig
= 125.44e−4t
dt
Test:
371.20e−4t − 480e−5t + 8 + 232e−4t − 240e−5t + 63.36e−4t − 80e−5t
?
−8 − 792e−4t + 800e−5t = −125.44e−4t
(8 − 8) + (800 − 480 − 240 − 80)e−5t
?
+(371.20 + 232 + 63.36 − 792)e−4t = −125.44e−4t
?
(800 − 800)e−5t + (666.56 − 792)e−4t = −125.44e−4t
−125.44e−4t = −125.44e−4t
(OK)
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Problems
6–5
Problems
P 6.1
0 ≤ t ≤ 2s :
103
iL =
2.5
Z
t
3 × 10−3 e−4x dx + 1 = 1.2
0
= −0.3e−4t + 1.3 A,
e−4x
−4
t
+1
0
0 ≤ t ≤ 2s
iL (2) = −0.3e−8 + 1.3 = 1.3 A
t ≥ 2s :
103
iL =
2.5
Z
t
2
−3 × 10−3 e−4(x−2) dx + 1.3 = −1.2
= 0.3e−4(t−2) + 1 A,
P 6.2
[a] v = L
e−4(x−2)
−4
t
+ 1.3
2
t ≥ 2s
di
dt
= (50 × 10−6 )(18)[e−10t − 10te−10t ] = 900e−10t (1 − 10t) µV
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6–6
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[b] i(200 ms) = 18(0.2)(e−2 ) = 487.21 mA
v(200 ms) = 900(e−2 )(1 − 2) = −121.8 µV
p(200 ms) = vi = (487.21 × 10−3 )(−121.8 × 10−6 ) = −59.34 µW
[c] delivering 59.34 µW
[d] i(200 ms) = 487.21 mA
(from part [b])
1
1
w = Li2 = (50 × 10−6 )(0.48721)2 = 5.93 µJ
2
2
[e] The energy is a maximum where the current is a maximum:
diL
= 0 when 1 − 10t = 0
dt
or
t = 0.1 s
imax = 18(0.1)e−1 = 662.18 mA
1
wmax = (50 × 10−6 )(0.66218)2 = 10.96 µJ
2
P 6.3
[a] 0 ≤ t ≤ 2 ms :
1
i=
L
=
Z
0
t
106
vs dx + i(0) =
200
Z
0
t
5 × 10−3 dx + 0
5000 t
x = 25t A
200 0
2 ms ≤ t < ∞ :
[b] i = 25t mA,
106
i=
200
0 ≤ t ≤ 2 ms;
Z
t
2×10−3
(0) dx + 2 × 10−3 = 50 mA
i = 50 mA,
t ≥ 2 ms
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Problems
P 6.4
[a] i =
0
t<0
i =
50t A
0 ≤ t ≤ 5 ms
i =
0.5 − 50t A
5 ≤ t ≤ 10 ms
i =
0
10 ms < t
[b] v = L
di
= 20 × 10−3 (50) = 1 V
dt
6–7
0 ≤ t ≤ 5 ms
v = 20 × 10−3 (−50) = −1 V
5 ≤ t ≤ 10 ms
v
=
0
t<0
v
=
1V
0 < t < 5 ms
v
=
−1 V
5 < t < 10 ms
v
=
0
10 ms < t
p = vi
p =
0
t<0
p =
(50t)(1) = 50t W
0 < t < 5 ms
p =
(0.5 − 50t)(−1) = 50t − 0.5 W
5 < t < 10 ms
p =
0
10 ms < t
w
w
w
w
P 6.5
=
0
t
=
Z
t<0
t
=
Z
=
25x2 − 0.5x
2
(50x) dx = 50
0
0.005
x
2
t
= 25t2 J
0 < t < 5 ms
0
(50x − 0.5) dx + 0.625 × 10−3
t
0.005
+0.625 × 10−3
=
25t2 − 0.5t + 2.5 × 10−3 J
5 < t < 10 ms
=
0
10 ms < t
[a] 0 ≤ t ≤ 1 s :
v = −100t
i=
1
5
Z
0
t
−100x dx + 0 = −20
x2
2
t
0
2
i = −10t A
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6–8
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
1s ≤ t ≤ 3s :
v = 100t − 200
i(1) = −10 A
. ·. i =
=
1
5
t
Z
1
Z
20
(100x − 200) dx − 10
t
1
(x − 2) dx − 10
= 10t2 − 40t + 20 A
3s ≤ t ≤ 5s :
v = 100 V
i(3) = 90 − 120 + 20 = −10 A
i =
=
1
5
Z
t
3
(100) dx − 10
20(t − 3) − 10
= 20t − 70 A
5s ≤ t ≤ 6s :
v = 600t − 100
i(5) = 100 − 70 = 30 A
i =
1
5
Z
t
5
(600x − 100) dx + 30
t
Z
=
20
=
120t − 600 − 10t2 + 250 + 30
=
t ≥ 6s :
5
(6 − x) dx + 30
−10t2 + 120t − 320 A
v=0
i(6) = 720 − 360 − 320 = 40 A
i =
=
1
5
Z
t
6
0 dx + 40
40 A
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Problems
6–9
[b] v = 0 at t = 2 s and t = 6 s
i(2) = 10(4) − 40(2) + 20 = −20 A
i(6) = 40 A
[c]
P 6.6
[a] i(0) = A1 + A2 = 0.04
di
= −10,000A1 e−10,000t − 40,000A2 e−40,000t
dt
v = −200A1 e−10,000t − 800A2 e−40,000t V
v(0) = −200A1 − 800A2 = 28
Solving, A1 = 0.1
and A2 = −0.06
Thus,
i1 = (100e−10,000t − 60e−40,000t) mA
v = −20e−10,000t + 48e−40,000t V,
t≥0
t≥0
[b] i = 0 when 100e−10,000t = 60e−40,000t
Therefore
e30,000t = 0.6 so t = −17.03 µs
which is not possible!
v = 0 when 20e−10,000t = 48e−40,000t
Therefore
e30,000t = 2.4 so t = 29.18 µs
Thus the power is zero at t = 29.18 µs.
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6–10
P 6.7
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[a] From Problem 6.6 we have
i = A1 e−10,000t + A2e−40,000t A
v = −20A1 e−10,000t + 48A2 e−40,000t V
i(0) = A1 + A2 = 0.04
v(0) = −200A1 − 800A2 = −68
Solving,
Thus,
A1 = −0.06;
A2 = 0.1
i = −60e−10,000t + 100e−40,000t mA t ≥ 0
v = 12e−10,000t − 80e−40,000t V t ≥ 0
[b] i = 0 when 60e−10,000t = 100e−40,000t
.·. e30,000t = 5/3 so t = 17.03 µs
Thus,
i > 0 for 0 ≤ t ≤ 17.03 µs
and
i < 0 for 17.03 µs ≤ t < ∞
v = 0 when 12e−10,000t = 80e−40,000t
.·. e30,000t = 20/3 so t = 63.24 µs
Thus,
v < 0 for 0 ≤ t ≤ 63.24 µs
and
v > 0 for 63.24 µs ≤ t < ∞
and
63.24 µs ≤ t < ∞
Therefore,
p < 0 for
0 ≤ t ≤ 17.03 µs
(inductor delivers energy)
p > 0 for
17.03 µs ≤ t ≤ 63.24 µs
(inductor stores energy)
[c] The energy stored at t = 0 is
1
1
w(0) = L[i(0)]2 = (0.02)(0.04)2 = 16 µJ
2
2
p = vi = 6e−50,000t − 8e−80,000t − 0.72e−20,000t W
For t > 0:
w=
Z
∞
0
6e−50,000t dt −
6e−50,000t
=
−50,000
∞
0
Z
∞
0
8e−80,000t dt −
8e−80,000t
−
−80,000
∞
0
Z
∞
0
0.72e−20,000t dt
0.72e−20,000t
−
−20,000
∞
0
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Problems
6–11
= (1.2 − 1 − 0.36) × 10−4
= −16 µJ
Thus, the energy stored equals the energy extracted.
P 6.8
[a] v = L
di
dt
v = −25 × 10−3
d
[10 cos 400t + 5 sin 400t]e−200t
dt
= −25 × 10−3 (−200e−200t [10 cos 400t + 5 sin 400t]
+e−200t[−4000 sin 400t + 2000 cos 400t])
v = −25 × 10−3 e−200t(−1000 sin 400t − 4000 sin 400t)
= −25 × 10−3 e−200t(−5000 sin 400t)
= 125e−200t sin 400t V
dv
= 125(e−200t(400) cos 400t − 200e−200t sin 400t)
dt
= 25,000e−200t (2 cos 400t − sin 400t) V/s
dv
=0
when
dt
.·. tan 400t = 2,
2 cos 400t = sin 400t
400t = 1.11;
t = 2.77 ms
[b] v(2.77 ms) = 125e−0.55 sin 1.11 = 64.27 V
P 6.9
[a] i =
=
=
[b]
p
1000
20
Z
−2500
t
0
− 50 sin 250x dx + 10
− cos 250x
250
t
+ 10
0
10 cos 250t A
= vi = (−50 sin 250t)(10 cos 250t)
= −500 sin 250t cos 250t
p
w
= −250 sin 500t W
1 2
=
Li
2
=
1
(20 × 10−3 )(10 cos 250t)2
2
= 1000 cos 2 250t mJ
w
= (500 + 500 cos 500t) mJ
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6–12
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[c] Absorbing power:
P 6.10
Delivering power:
2π ≤ t ≤ 4π ms
0 ≤ t ≤ 2π ms
6π ≤ t ≤ 8π ms
4π ≤ t ≤ 6π ms
i = (B1 cos 4t + B2 sin 4t)e−t/2
i(0) = B1 = 10 A
di
= (B1 cos 4t + B2 sin 4t)(−0.5e−t/2 ) + e−t/2(−4B1 sin 4t + 4B2 cos 4t)
dt
= [(4B2 − 0.5B1 ) cos 4t − (4B1 + 0.5B2 ) sin 4t]e−t/2
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Problems
v=4
6–13
di
= [(16B2 − 2B1 ) cos 4t − (16B1 + 2B2 ) sin 4t]e−t/2
dt
v(0) = 60 = 16B2 − 2B1 = 16B2 − 20
.·. B2 = 5 A
Thus,
i = (10 cos 4t + 5 sin 4t)e−t/2, A,
t≥0
v = (60 cos 4t − 170 sin 4t)e−t/2 V,
i(1) = −26, A;
t≥0
v(1) = 54.25 V
p(1) = (−26)(54.25) = −339.57 W delivering
P 6.11
p = vi = 40t[e−10t − 10te−20t − e−20t]
W=
Z
0
∞
p dx =
Z
0
∞
40x[e−10x − 10xe−20x − e−20x ] dx = 0.2 J
This is energy stored in the inductor at t = ∞.
P 6.12
[a] v(20 µs) =
v(20 µs) =
=
v(40 µs) =
=
12.5 × 109 (20 × 10−6 )2 = 5 V (end of first interval)
106 (20 × 10−6 ) − (12.5)(400) × 10−3 − 10
5 V (start of second interval)
106 (40 × 10−6 ) − (12.5)(1600) × 10−3 − 10
10 V (end of second interval)
[b] p(10µs) = 62.5 × 1012 (10−5 )3 = 62.5 mW,
i(10µs) = 50 mA,
v(10 µs) = 1.25 V,
p(10 µs) = vi = (1.25)(50 m) = 62.5 mW (checks)
p(30 µs) = 437.50 mW,
v(30 µs) = 8.75 V,
i(30 µs) = 0.05 A
p(30 µs) = vi = (8.75)(0.05) = 62.5 mW (checks)
[c] w(10 µs) = 15.625 × 1012 (10 × 10−6 )4 = 0.15625 µJ
w = 0.5Cv 2 = 0.5(0.2 × 10−6 )(1.25)2 = 0.15625 µJ
w(30 µs) = 7.65625 µJ
w(30 µs) = 0.5(0.2 × 10−6 )(8.75)2 = 7.65625 µJ
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6–14
P 6.13
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
For 0 ≤ t ≤ 1.6 s:
iL =
1
5
t
Z
0
3 × 10−3 dx + 0 = 0.6 × 10−3 t
iL (1.6 s) = (0.6 × 10−3 )(1.6) = 0.96 mA
Rm = (20)(1000) = 20 kΩ
vm (1.6 s) = (0.96 × 10−3 )(20 × 103 ) = 19.2 V
P 6.14
[a] i =
400 × 10−3
t = 80 × 103 t
−6
5 × 10
i = 400 × 10−3
i=
q
0 ≤ t ≤ 5 µs
5 ≤ t ≤ 20 µs
300 × 10−3
t − 0.5 = 104 t − 0.5
30 × 10−6
5×10−6
=
Z
=
8 × 104
0
4
8 × 10 t dt +
t2
2
Z
15×10−6
5×10−6
20 µs ≤ t ≤ 50 µs
0.4 dt
5×10−6
+0.4(10 × 10−6 )
0
=
4 × 104 (25 × 10−12 ) + 4 × 10−6
=
5 µC
[b] v = 4 × 106
Z
5×10−6
0
6
+ 4 × 10
= 4 × 106
Z
8 × 104 x dx + 4 × 106
30×10−6
20×10−6
20×10−6
5×10−6
0.4 dx
(104 x − 0.5) dx
x2
8 × 104
2
"
Z
5×10−6
20×10−6
+0.4x
0
5×10−6
x2
+104
2
30×10−6
20×10−6
30×10−6
−0.5x
20×10−6
#
= 4 × 106 [4 × 104 (25 × 10−12 ) + 0.4(15 × 10−6 )
+ 5000(900 × 10−12 400 × 10−12 ) − 0.5(10 × 10−6 )]
= 18 V
v(30 µs) = 18 V
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Problems
6–15
[c] v(50 µs) = 4 × 106 [10−6 + 6 × 10−6 + 5000(2500 × 10−12 − 400 × 10−12 )
− 0.5(30 × 10−6 )]
= 10 V
1
1
w = Cv 2 = (0.25 × 10−6 )(10)2 = 12.5 µJ
2
2
P 6.15
[a]
v
=
1
0.5 × 10−6
Z
3e
= 100 × 10
500×10−6
0
−2000t
−2000
50 × 10−3 e−2000t dt − 20
500×10−6
−20
0
= 50(1 − e−1 ) − 20 = 11.61 V
w
=
1
Cv 2
2
= 12 (0.5)(10−6 )(11.61)2 = 33.7 µJ
[b] v(∞) = 50 − 20 = 30V
1
w(∞) = (0.5 × 10−6 )(30)2 = 225 µJ
2
P 6.16
[a] 0 ≤ t ≤ 10 µs
1
= 10 × 106
C
C = 0.1 µF
v = 10 × 106
t
Z
0
− 0.05 dx + 15
v = −50 × 104 t + 15 V
0 ≤ t ≤ 10 µs
v(10 µs) = −5 + 15 = 10 V
[b] 10 µs ≤ t ≤ 20 µs
6
v = 10 × 10
Z
t
10×10−6
v = 106 t V
0.1 dx + 10 = 106 t − 10 + 10
10 ≤ t ≤ 20 µs
v(20 µs) = 106 (20 × 10−6 ) = 20 V
[c] 20 µs ≤ t ≤ 40 µs
6
v = 10 × 10
Z
t
20×10−6
1.6 dx + 20 = 1.6 × 106 t − 32 + 20
v = 1.6 × 106 t − 12 V,
20 µs ≤ t ≤ 40 µs
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6–16
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[d] 40 µs ≤ t < ∞
v(40 µs) = 64 − 12 = 52 V
P 6.17
P 6.18
40 µs ≤ t < ∞
iC = C(dv/dt)
0 < t < 0.5 :
iC = 20 × 10−6 (60)t = 1.2t mA
0.5 < t < 1 :
iC = 20 × 10−6 (60)(t − 1) = 1.2(t − 1) mA
1
1
[a] w(0) = C[v(0)]2 = (0.20) × 10−6 (150)2 = 2.25 mJ
2
2
−5000t
[b] v = (A1t + A2)e
v(0) = A2 = 150 V
dv
dt
= −5000e−5000t (A1t + A2) + e−5000t(A1)
= (−5000A1 t − 5000A2 + A1 )e−5000t
dv
(0) = A1 − 5000A2
dt
i=C
dv
,
dt
i(0) = C
dv(0)
dt
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Problems
6–17
dv(0)
i(0)
250 × 10−3
=
=
= 1250 × 103
dt
C
0.2 × 10−6
. ·.
.·. 1.25 × 106 = A1 − 5000(150)
Thus, A1 = 1.25 × 106 + 75 × 104 = 2 × 106
V
s
[c] v = (2 × 106 t + 150)e−5000t
dv
d
= 0.2 × 10−6 (2 × 106 t + 150)e−5000t
dt
dt
d
=
[(0.4t + 10 × 30−6 )e−5000t]
dt
i=C
i
= (0.4t + 30 × 10−6 )(−5000)e−5000t + e−5000t(0.4)
= (−2000t − 150 × 10−3 + 0.4)e−5000t
= (0.25 − 2000t)e−5000t A,
P 6.19
t≥0
dv
= 0, t < 0
dt
dv
d
[b] i = C
= 4 × 10−6 [100 − 40e−2000t (3 cos 1000t + sin 1000t)]
dt
dt
[a] i = C
= 4 × 10−6 [−40(−2000)e−2000t (3 cos 1000t + sin 1000t)
−40(1000)e−2000t (−3 sin 1000t + cos 1000t)]
= 0.32e−2000t(3 cos 1000t + sin 1000t) − 0.16(−3 sin 1000t + cos 1000t)
= 0.8e−2000t[cos 1000t + sin 1000t] A,
[c] no,
[d] yes,
t≥0
v(0− ) = −20 V
v(0+ ) = 100 − 40(1)(3) = −20 V
i(0− ) = 0 A
i(0+ ) = 0.8 A
[e] v(∞) = 100 V
1
1
w = Cv 2 = (4 × 10−6 )(100)2 = 20 mJ
2
2
P 6.20
30k20 = 12 H
80k(8 + 12) = 16 H
60k(14 + 16) = 20 H
15k(20 + 10) = 20 H
Lab = 5 + 10 = 15 H
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6–18
P 6.21
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
5k(12 + 8) = 4 H
4k4 = 2 H
15k(8 + 2) = 6 H
3k6 = 2 H
6 + 2 = 8H
P 6.22
[a] Combine three 1 mH inductors in series to get a 3 mH equivalent inductor.
[b] Combine two 100 µH inductors in parallel to get a 50 µH inductor. Then
combine this parallel pair in series with two more 100 µH inductors:
100 µk100 µ + 100 µ + 100 µ = 50 µ + 100 µ + 100 µ = 250 µH
[c] Combine two 100 µH inductors in parallel to get a 50 µH inductor. Then
combine this parallel pair with a 10 µH inductor in series:
100 µk100 µ + 10 µ = 50 µ + 10 µ = 60 µH
P 6.23
[a] io(0) = −i1 (0) − i2(0) = 6 − 1 = 5 A
[b]
io
= −
1
4
Z
t
0
2000e−100x dx + 5 = −500
= 5(e−100t − 1) + 5 = 5e−100t A,
e−100x
−100
t
+5
0
t≥0
[c]
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Problems
va
=
3.2(−500e−100t ) = −1600e−100t V
vc
=
va + vb = −1600e−100t + 2000e−100t
=
=
400e−100t V
Z
1 t
400e−100x dx − 6
1 0
=
−4e−100t + 4 − 6
=
−4e−100t − 2 A
i1
i1
[d] i2
1
4
=
Z
6–19
t≥0
t
0
400e−100x dx + 1
= −e−100t + 2 A,
t≥0
1
1
1
[e] w(0) = (1)(6)2 + (4)(1)2 + (3.2)(5)2 = 60 J
2
2
2
1
[f] wdel = (4)(5)2 = 50 J
2
[g] wtrapped = 60 − 50 = 10 J
or
P 6.24
1
1
wtrapped = (1)(2)2 + (4)(2)2 + 10 J (check)
2
2
vb = 2000e−100t V
io = 5e−100t A
p = 10,000e−200t W
w=
Z
0
t
104 e−200x dx = 10,000
e−200x
−200
t
0
= 50(1 − e−200t) W
wtotal = 50 J
80%wtotal = 40 J
Thus,
50 − 50e−200t = 40;
e200t = 5;
.·. t = 8.05 ms
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6–20
P 6.25
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[a]
3.2
di
= 64e−4t
dt
Z
i(t) =
20
=
20
i(t) =
[b] 4
di
= 20e−4t
dt
so
t
0
e−4x dx − 5
t
e−4x
−4
−5
0
−5e−4t A
di1
= 64e−4t
dt
Z t
i1 (t) = 16 e−4x dx − 10
0
=
i1 (t) =
[c] 16
16
0
−10
−4e−4t − 6 A
di2
= 64e−4t
dt
Z
i2 (t) =
4
=
4
i2 (t) =
t
e−4x
−4
di2
= 4e−4t
dt
so
t
0
e−4x dx + 5
t
e−4x
−4
+5
0
−e−4t + 6 A
[d] p = −vi = (−64e−4t )(−5e−4t ) = 320e−8t W
w
∞
=
Z
=
320
=
0
p dt =
e−8t
−8
Z
∞
0
320e−8t dt
∞
0
40 J
1
1
[e] w = (4)(−10)2 + (16)(5)2 = 400 J
2
2
[f] wtrapped = winitial − wdelivered = 400 − 40 = 360 J
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
1
1
[g] wtrapped = (4)(−6)2 + (16)(6)2 = 360 J
2
2
P 6.26
1
1
1
1
=
+
= ;
C1
48 16
12
6–21
checks
C1 = 12 µF
C2 = 3 + 12 = 15 µF
1
1
1
1
=
+
= ;
C3
30 15
10
C3 = 10 µF
C4 = 10 + 10 = 20 µF
1
1
1
1
1
= +
+ = ;
C5
5 20 4
2
C5 = 2 µF
Equivalent capacitance is 2 µF with an initial voltage drop of +25 V.
P 6.27
1 1
5
+ =
4 6
12
.·. Ceq = 2.4 µF
1
1
4
+
=
4 12
12
.·. Ceq = 3 µF
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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6–22
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
1
1
4
+ =
24 8
24
P 6.28
.·. Ceq = 6 µF
[a] Combine two 220 µF capacitors in series to get a 110 µF capacitor. Then
combine the series pair in parallel with another 220 µF capacitor to get
330 µF:
(220 µ + 220 µ)k220 µ = 110 µk220 µ = 330 µF
[b] Create a 1500 nF capacitor as follows:
(1 µ + 1 µ)k1 µ = 500 nk1000 n = 1500 nF
Create a second 1500 nF capacitor using the same three resistors. Place
these two 1500 nF in series:
1500 n + 1500 n = 750 nF
[c] Combine two 100 pF capacitors in series to get a 50 pF capacitor. Then
combine the series pair in parallel with another 100 pF capacitor to get
150 pF:
(100 p + 100 p)k100 p = 50 pk100 p = 150 pF
P 6.29
1
1 1
1
10
= + +
=
=2
Ce
1 5 1.25
5
.·. C2 = 0.5 µF
vb = 20 − 250 + 30 = −200 V
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
6–23
[a]
vb
Z
−
=
10,000
=
[b] va
106
0.5
=
t
0
− 5 × 10−3 e−50x dx − 200
e−50x
−50
t
0
−200
−200e−50t V
= −
106
0.5
Z
t
0
− 5 × 10−3 e−50x dx − 20
= 20(e−50t − 1) − 20
= 20e−50t − 40 V
[c] vc
[d] vd
=
106 Z t
− 5 × 10−3 e−50x dx − 30
1.25 0
=
80(e−50t − 1) − 30
=
80e−50t − 110 V
6
= 10
Z
t
0
− 5 × 10−3 e−50x dx + 250
= 100(e−50t − 1) + 250
= 100e−50t + 150 V
CHECK: vb
[e] i1
[f] i2
=
−vc − vd − va
=
−200e−50t V (checks)
d
[100e−50t + 150]
dt
=
0.2 × 10−6
=
0.2 × 10−6 (−5000e−50t )
=
−e−50t mA
=
0.8 × 10−6
=
−4e−50t mA
d
[100e−50t + 150]
dt
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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6–24
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
CHECK: ib = i1 + i2 = −5e−50t mA
P 6.30
[a] w(0)
1
(0.2
2
=
=
× 10−6 )(250)2 + 12 (0.8 × 10−6 )(250)2 + 12 (5 × 10−6 )(20)2
1
(1.25
2
+
(OK)
× 10−6 )(30)2
32,812.5 µJ
[b] w(∞) = 12 (5 × 10−6 )(40)2 + 12 (1.25 × 10−6 )(110)2 + 12 (0.2 × 10−6 )(150)2
+ 12 (0.8 × 10−6 )(150)2
= 22,812.5 µJ
1
[c] w = (0.5 × 10−6 )(200)2 = 10,000 µJ
2
CHECK: 32,812.5 − 22,812.5 = 10,000 µJ
10,000
[d] % delivered =
× 100 = 30.48%
32,812.5
[e] w
t
=
Z
=
10(1 − e−100t) mJ
0
(−0.005e−50x )(−200e−50x ) dx =
.·. 10−2 (1 − e−100t) = 7.5 × 10−3 ;
Thus, t =
P 6.31
Z
t
0
e−100x dx
e−100t = 0.25
ln 4
= 13.86 ms.
100
[a]
vo
=
106
1.6
Z
t
0
800 × 10−6 e−25x dx − 20
e−25x
= 500
−25
t
0
−20
= −20e−25t V,
[b] v1
=
t≥0
106
e−25x
(800 × 10−6 )
2
−25
= −16e−25t + 21 V,
t
+5
0
t≥0
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
[c] v2
106
e−25x
(800 × 10−6 )
8
−25
=
[d] p
t
0
−25
=
−4e−25t − 21 V,
=
−vi = −(−20e−25t)(800 × 10−6 )e−25t
=
16 × 10−3 e−50t
w
[e] w
t≥0
∞
=
Z
=
16 × 10−3
=
6–25
0
16 × 10−3 e−50t dt
e−50t
−50
∞
0
−0.32 × 10−3 (0 − 1) = 320 µJ
=
1
(2
2
× 10−6 )(5)2 + 12 (8 × 10−6 )(25)2
=
2525 µJ
[f] wtrapped = winitial − wdelivered = 2525 − 320 = 2205 µJ
[g] wtrapped
=
1
(2
2
× 10−6 )(21)2 + 12 (8 × 10−6 )(−21)2
= 2205 µJ
P 6.32
From Figure 6.17(a) we have
v=
1
C1
v=
Z
0
i dx + v1(0) +
1
1
+
+ ···
C1 C2
Therefore
P 6.33
t
Z
1
C2
Z
0
t
i dx + v2(0) + · · ·
t
0
i dx + v1(0) + v2(0) + · · ·
1
1
1
=
+
+ ··· ,
Ceq
C1 C2
veq(0) = v1(0) + v2(0) + · · ·
From Fig. 6.18(a)
i = C1
dv
dv
dv
+ C2 + · · · = [C1 + C2 + · · ·]
dt
dt
dt
Therefore Ceq = C1 + C2 + · · ·. Because the capacitors are in parallel, the
initial voltage on every capacitor must be the same. This initial voltage would
appear on Ceq.
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6–26
P 6.34
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
dio
dt
= (5){e−2000t[−8000 sin 4000t + 4000 cos 4000t]
+(−2000e−2000t )[2 cos 4000t + sin 4000t]}
= e−2000t{− 50,000 sin 4000t} V
dio +
(0 ) = (1)[sin(0)] = 0
dt
.·.
10 × 10−3
dio +
(0 ) = 0
dt
so
v2 (0+ ) = 0
v1(0+ ) = 40io (0+ ) + v2(0+ ) = 40(10) = 0 = 400 V
P 6.35
vc
vL
vo
P 6.36
1
0.625 × 10−6
Z
t
t
−
=
150(e−16,000t − 1) − 200(e−4000t − 1) − 50
=
150e−16,000t − 200e−4000t V
dio
25 × 10−3
dt
=
0
1.5e−16,000x dx −
Z
=
0
0.5e−4000x dx − 50
=
25 × 10−3 (−24,000e−16,000t + 2000e−4000t )
=
−600e−16,000t + 50e−4000t V
=
vc − vL
=
(150e−16,000t − 200e−4000t ) − (−600e−16,000t + 50e−4000t)
=
750e−16,000t − 250e−4000t V, t > 0
[a] Rearrange by organizing the equations by di1/dt, i1, di2/dt, i2 and transfer
the ig terms to the right hand side of the equations. We get
4
di1
di2
dig
+ 25i1 − 8
− 20i2 = 5ig − 8
dt
dt
dt
−8
di1
di2
dig
− 20i1 + 16
+ 80i2 = 16
dt
dt
dt
[b] From the given solutions we have
di1
= −320e−5t + 272e−4t
dt
di2
= 260e−5t − 204e−4t
dt
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Problems
6–27
Thus,
4
di1
= −1280e−5t + 1088e−4t
dt
25i1 = 100 + 1600e−5t − 1700e−4t
8
di2
= 2080e−5t − 1632e−4t
dt
20i2 = 20 − 1040e−5t + 1020e−4t
5ig = 80 − 80e−5t
8
dig
= 640e−5t
dt
Thus,
−1280e−5t + 1088e−4t + 100 + 1600e−5t − 1700e−4t − 2080e−5t
?
+1632e−4t − 20 + 1040e−5t − 1020e−4t = 80 − 80e−5t − 640e−5t
80 + (1088 − 1700 + 1632 − 1020)e−4t
?
+(1600 − 1280 − 2080 + 1040)e−5t = 80 − 720e−5t
80 + (2720 − 2720)e−4t + (2640 − 3360)e−5t = 80 − 720e−5t
8
(OK)
di1
= −2560e−5t + 2176e−4t
dt
20i1 = 80 + 1280e−5t − 1360e−4t
16
di2
= 4160e−5t − 3264e−4t
dt
80i2 = 80 − 4160e−5t + 4080e−4t
16
dig
= 1280e−5t
dt
2560e−5t − 2176e−4t − 80 − 1280e−5t + 1360e−4t + 4160e−5t − 3264e−4t
?
+80 − 4160e−5t + 4080e−4t = 1280e−5t
(−80 + 80) + (2560 − 1280 + 4160 − 4160)e−5t
?
+(1360 − 2176 − 3264 + 4080)e−4t = 1280e−5t
0 + 1280e−5t + 0e−4t = 1280e−5t
P 6.37
(OK)
[a] Yes, using KVL around the lower right loop
vo = v20Ω + v60Ω = 20(i2 − i1) + 60i2
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6–28
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[b] vo
=
20(1 − 52e−5t + 51e−4t − 4 − 64e−5t + 68e−4t )+
60(1 − 52e−5t + 51e−4t )
=
20(−3 − 116e−5t + 119e−4t ) + 60 − 3120e−5t + 3060e−4t
vo
=
−5440e−5t + 5440e−4t V
[c] vo
=
L2
d
d
(15 + 36e−5t − 51e−4t ) + 8 (4 + 64e−5t − 68e−4t )
dt
dt
−5t
−4t
−5t
−2880e + 3264e − 2560e + 2176e−4t
=
16
=
P 6.38
d
di1
(ig − i2) + M
dt
dt
vo
=
−5440e−5t + 5440e−4t V
[a] vg
=
5(ig − i1) + 20(i2 − i1) + 60i2
=
5(16 − 16e−5t − 4 − 64e−5t + 68e−4t )+
20(1 − 52e−5t + 51e−4t − 4 − 64e−5t + 68e−4t )+
60(1 − 52e−5t + 51e−4t )
=
60 + 5780e−4t − 5840e−5t V
[b] vg (0) = 60 + 5780 − 5840 = 0 V
[c] pdev
=
vg ig
=
960 + 92,480e−4t − 94,400e−5t − 92,480e−9t +
93,440e−10t W
[d] pdev (∞) = 960 W
[e] i1(∞) = 4 A;
i2(∞) = 1 A;
ig (∞) = 16 A;
p5Ω = (16 − 4)2 (5) = 720 W
p20Ω = 32 (20) = 180 W
p60Ω = 12 (60) = 60 W
X
. ·.
P 6.39
[a] −2
pabs = 720 + 180 + 60 = 960 W
X
pdev =
X
pabs = 960 W
dig
di2
+ 16
+ 32i2 = 0
dt
dt
16
di2
dig
+ 32i2 = 2
dt
dt
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Problems
6–29
[b] i2 = e−t − e−2t A
di2
= −e−t + 2e−2t A/s
dt
ig = 8 − 8e−t A
dig
= 8e−t A/s
dt
.·. −16e−t + 32e−2t + 32e−t − 32e−2t = 16e−t
[c] v1
dig
di2
−2
dt
dt
=
4
=
4(8e−t ) − 2(−e−t + 2e−2t )
=
34e−t − 4e−2t V,
t>0
[d] v1(0) = 34 − 4 = 30 V;
Also
dig
di2
v1(0) = 4 (0) − 2 (0)
dt
dt
= 4(8) − 2(−1 + 2) = 32 − 2 = 30 V
Yes, the initial value of v1 is consistent with known circuit behavior.
P 6.40
[a] vab = L1
di
di
di
di
di
+ L2 + M + M = (L1 + L2 + 2M)
dt
dt
dt
dt
dt
It follows that Lab = (L1 + L2 + 2M)
[b] vab = L1
di
di
di
di
di
− M + L2 − M = (L1 + L2 − 2M)
dt
dt
dt
dt
dt
Therefore Lab = (L1 + L2 − 2M)
P 6.41
[a] vab = L1
0 = L1
d(i1 − i2)
di2
+M
dt
dt
d(i2 − i1)
di2
d(i1 − i2)
di2
−M
+M
+ L2
dt
dt
dt
dt
Collecting coefficients of [di1/dt] and [di2/dt], the two mesh-current
equations become
vab = L1
di1
di2
+ (M − L1 )
dt
dt
and
di1
di2
+ (L1 + L2 − 2M)
dt
dt
Solving for [di1/dt] gives
0 = (M − L1 )
di1
L1 + L2 − 2M
=
vab
dt
L1 L2 − M 2
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6–30
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
from which we have
vab =
L1 L2 − M 2
L1 + L2 − 2M
.·. Lab =
!
di1
dt
!
L1 L2 − M 2
L1 + L2 − 2M
[b] If the magnetic polarity of coil 2 is reversed, the sign of M reverses,
therefore
Lab =
P 6.42
L1 L2 − M 2
L1 + L2 + 2M
When the switch is opened the induced voltage is negative at the dotted
terminal. Since the voltmeter kicks upscale, the induced voltage across the
voltmeter must be positive at its positive terminal. Therefore, the voltage is
negative at the negative terminal of the voltmeter.
Thus, the lower terminal of the unmarked coil has the same instantaneous
polarity as the dotted terminal. Therefore, place a dot on the lower terminal
of the unmarked coil.
P 6.43
[a] Dot terminal 1; the flux is up in coil 1-2, and down in coil 3-4. Assign the
current into terminal 4; the flux is down in coil 3-4. Therefore, dot
terminal 4. Hence, 1 and 4 or 2 and 3.
[b] Dot terminal 2; the flux is up in coil 1-2, and right-to-left in coil 3-4.
Assign the current into terminal 4; the flux is right-to-left in coil 3-4.
Therefore, dot terminal 4. Hence, 2 and 4 or 1 and 3.
[c] Dot terminal 2; the flux is up in coil 1-2, and right-to-left in coil 3-4.
Assign the current into terminal 4; the flux is right-to-left in coil 3-4.
Therefore, dot terminal 4. Hence, 2 and 4 or 1 and 3.
[d] Dot terminal 1; the flux is down in coil 1-2, and down in coil 3-4. Assign
the current into terminal 4; the flux is down in coil 3-4. Therefore, dot
terminal 4. Hence, 1 and 4 or 2 and 3.
P 6.44
[a] W = (0.5)L1 i21 + (0.5)L2 i22 + Mi1 i2
q
M = 0.85 (18)(32) = 20.4 mH
W = [9(36) + 16(81) + 20.4(54)] = 2721.6 mJ
[b] W = [324 + 1296 + 1101.6] = 2721.6 mJ
[c] W = [324 + 1296 − 1101.6] = 518.4 mJ
[d] W = [324 + 1296 − 1101.6] = 518.4 mJ
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Problems
P 6.45
q
[a] M = 1.0 (18)(32) = 24 mH,
i1 = 6 A
Therefore 16i22 + 144i2 + 324 = 0,
9
Therefore i2 = −
±
2
6–31
s
2
9
2
i22 + 9i2 + 20.25 = 0
− 20.25 = −4.5 ±
√
0
Therefore i2 = −4.5 A
[b] No, setting W equal to a negative value will make the quantity under the
square root sign negative.
P 6.46
M
22.8
=√
= 0.95
L1 L2
576
√
= 576 = 24 mH
[a] k = √
[b] Mmax
[c]
L1
N 2 P1
N1
= 12
=
L2
N2 P2
N2
2
N1 2
60
=
= 6.25
N2
9.6
N1 √
= 6.25 = 2.5
N2
. ·.
P 6.47
[a] L1 = N12 P1 ;
P1 =
dφ11
P11
=
= 0.2;
dφ21
P21
72 × 10−3
= 1152 nWb/A
6.25 × 104
P21 = 2P11
.·. 1152 × 10−9 = P11 + P21 = 3P11
P11 = 192 nWb/A;
q
P21 = 960 nWb/A
q
M = k L1 L2 = (2/3) (0.072)(0.0405) = 36 mH
N2 =
M
36 × 10−3
=
= 150 turns
N1P21
(250)(960 × 10−9 )
L2
40.5 × 10−3
=
= 1800 nWb/A
N22
(150)2
[c] P11 = 192 nWb/A [see part (a)]
φ22
P22
P2 − P12
P2
[d]
=
=
=
−1
φ12
P12
P12
P12
[b] P2 =
P21 = P21 = 960 nWb/A;
P2 = 1800 nWb/A
φ22
1800
=
− 1 = 0.875
φ12
960
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6–32
P 6.48
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
M2
k 2 L1
[a] L2 =
s
N1
=
N2
[b] P1 =
P1 =
=
(0.09)2
= 50 mH
(0.75)2 (0.288)
s
L1
=
L2
288
= 2.4
50
L1
0.288
=
= 0.2 × 10−6 Wb/A
2
N1
(1200)2
P2 =
P 6.49
!
L2
0.05
=
= 0.2 × 10−6 Wb/A
2
N2
(500)2
L1
= 2 nWb/A;
N12
P12 = P21 =
P2 =
L2
= 2 nWb/A;
N22
q
M = k L1 L2 = 180 µH
M
= 1.2 nWb/A
N1 N2
P11 = P1 − P21 = 0.8 nWb/A
P 6.50
1
P11
P22
P11
[a] 2 = 1 +
1+
= 1+
k
P12
P12
P21
Therefore
P12P21
k2 =
(P21 + P11)(P12 + P22)
P22
1+
P12
Now note that
φ1 = φ11 + φ21 = P11N1i1 + P21N1 i1 = N1i1 (P11 + P21)
and similarly
φ2 = N2 i2 (P22 + P12)
It follows that
(P11 + P21) =
φ1
N1 i1
and
(P22 + P12) =
φ2
N2i2
!
Therefore
(φ12/N2 i2 )(φ21/N1 i1 )
φ12φ21
k2 =
=
(φ1 /N1 i1 )(φ2/N2 i2)
φ1 φ2
or
v
!
u
u φ21
k=t
φ1
φ12
φ2
!
[b] The fractions (φ21/φ1 ) and (φ12/φ2 ) are by definition less than 1.0,
therefore k < 1.
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Problems
P 6.51
6–33
When the button is not pressed we have
C2
dv
d
= C1 (vs − v)
dt
dt
or
(C1 + C2)
dv
dvs
= C1
dt
dt
dv
C1
dvs
=
dt
(C1 + C2) dt
Assuming C1 = C2 = C
dv
dvs
= 0.5
dt
dt
or
v = 0.5vs (t) + v(0)
When the button is pressed we have
C1
dv
dv
d(v − vs )
+ C3 + C2
=0
dt
dt
dt
dv
C2
dvs
.·.
=
dt
C1 + C2 + C3 dt
Assuming C1 = C2 = C3 = C
dv
1 dvs
=
dt
3 dt
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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6–34
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
1
v = vs (t) + v(0)
3
Therefore interchanging the fixed capacitor and the button has no effect on
the change in v(t).
P 6.52
With no finger touching and equal 10 pF capacitors
v(t) =
10
(vs (t)) + 0 = 0.5vs (t)
20
With a finger touching
Let Ce = equivalent capacitance of person touching lamp
Ce =
Then
(10)(100)
= 9.091 pF
110
C + Ce = 10 + 9.091 = 19.091 pF
.·. v(t) =
10
vs = 0.344vs
29.091
.·. ∆v(t) = (0.5 − 0.344)vs = 0.156vs
P 6.53
With no finger on the button the circuit is
C1
d
d
(v − vs ) + C2 (v + vs ) = 0
dt
dt
when
C1 = C2 = C
(2C)
dv
=0
dt
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
6–35
With a finger on the button
C1
d(v − vs )
d(v + vs )
dv
+ C2
+ C3
=0
dt
dt
dt
(C1 + C2 + C3 )
when
dv
dvs
dvs
+ C2
− C1
=0
dt
dt
dt
C1 = C2 = C3 = C
(3C)
dv
=0
dt
.·. there is no change in the output voltage of this circuit.
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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7
Response of First-Order RL and
RC Circuits
Assessment Problems
AP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a
short circuit, effectively eliminating the 2 Ω resistor from the circuit.
First combine the 30 Ω and 6 Ω resistors in parallel:
30k6 = 5 Ω
Use voltage division to find the voltage drop across the parallel resistors:
5
v=
(120) = 75 V
5+3
Now find the current using Ohm’s law:
v
75
i(0− ) = − = − = −12.5 A
6
6
1
1
[b] w(0) = Li2 (0) = (8 × 10−3 )(12.5)2 = 625 mJ
2
2
[c] To find the time constant, we need to find the equivalent resistance seen
by the inductor for t > 0. When the switch opens, only the 2 Ω resistor
remains connected to the inductor. Thus,
L
8 × 10−3
τ=
=
= 4 ms
R
2
[d] i(t) = i(0− )et/τ = −12.5e−t/0.004 = −12.5e−250t A,
t≥0
[e] i(5 ms) = −12.5e−250(0.005) = −12.5e−1.25 = −3.58 A
So w (5 ms) = 12 Li2 (5 ms) = 12 (8) × 10−3 (3.58)2 = 51.3 mJ
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
7–1 system, or transmission in any form or by any means, electronic,
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7–2
CHAPTER 7. Response of First-Order RL and RC Circuits
w (dis) = 625 − 51.3 = 573.7 mJ
573.7
% dissipated =
100 = 91.8%
625
AP 7.2 [a] First, use the circuit for t < 0 to find the initial current in the inductor:
Using current division,
10
i(0− ) =
(6.4) = 4 A
10 + 6
Now use the circuit for t > 0 to find the equivalent resistance seen by the
inductor, and use this value to find the time constant:
L
0.32
=
= 0.1 s
Req
3.2
Use the initial inductor current and the time constant to find the current
in the inductor:
i(t) = i(0− )e−t/τ = 4e−t/0.1 = 4e−10t A, t ≥ 0
Use current division to find the current in the 10 Ω resistor:
Req = 4k(6 + 10) = 3.2 Ω,
io (t) =
.·.
τ=
4
4
(−i) = (−4e−10t ) = −0.8e−10t A,
4 + 10 + 6
20
t ≥ 0+
Finally, use Ohm’s law to find the voltage drop across the 10 Ω resistor:
vo (t) = 10io = 10(−0.8e−10t ) = −8e−10t V, t ≥ 0+
[b] The initial energy stored in the inductor is
1
1
w(0) = Li2(0− ) = (0.32)(4)2 = 2.56 J
2
2
Find the energy dissipated in the 4 Ω resistor by integrating the power
over all time:
di
v4Ω (t) = L = 0.32(−10)(4e−10t ) = −12.8e−10t V,
t ≥ 0+
dt
p4Ω (t) =
2
v4Ω
= 40.96e−20t W,
4
t ≥ 0+
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Problems
w4Ω (t) =
Z
7–3
∞
0
40.96e−20t dt = 2.048 J
Find the percentage of the initial energy in the inductor dissipated in the
4 Ω resistor:
2.048
% dissipated =
100 = 80%
2.56
AP 7.3 [a] The circuit for t < 0 is shown below. Note that the capacitor behaves like
an open circuit.
Find the voltage drop across the open circuit by finding the voltage drop
across the 50 kΩ resistor. First use current division to find the current
through the 50 kΩ resistor:
80 × 103
i50k =
(7.5 × 10−3 ) = 4 mA
3
3
3
80 × 10 + 20 × 10 + 50 × 10
Use Ohm’s law to find the voltage drop:
v(0− ) = (50 × 103 )i50k = (50 × 103 )(0.004) = 200 V
[b] To find the time constant, we need to find the equivalent resistance seen
by the capacitor for t > 0. When the switch opens, only the 50 kΩ
resistor remains connected to the capacitor. Thus,
τ = RC = (50 × 103 )(0.4 × 10−6 ) = 20 ms
[c] v(t) = v(0− )e−t/τ = 200e−t/0.02 = 200e−50t V, t ≥ 0
1
1
[d] w(0) = Cv 2 = (0.4 × 10−6 )(200)2 = 8 mJ
2
2
1 2
1
[e] w(t) = Cv (t) = (0.4 × 10−6 )(200e−50t )2 = 8e−100t mJ
2
2
The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains:
8 × 10−3 e−100t = 2 × 10−3 ,
e100t = 4,
t = (ln 4)/100 = 13.86 ms
AP 7.4 [a] This circuit is actually two RC circuits in series, and the requested
voltage, vo , is the sum of the voltage drops for the two RC circuits. The
circuit for t < 0 is shown below:
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7–4
CHAPTER 7. Response of First-Order RL and RC Circuits
Find the current in the loop and use it to find the initial voltage drops
across the two RC circuits:
15
i=
= 0.2 mA,
v5(0− ) = 4 V,
v1(0− ) = 8 V
75,000
There are two time constants in the circuit, one for each RC subcircuit.
τ5 is the time constant for the 5 µF – 20 kΩ subcircuit, and τ1 is the time
constant for the 1 µF – 40 kΩ subcircuit:
τ5 = (20 × 103 )(5 × 10−6 ) = 100 ms;
τ1 = (40 × 103 )(1 × 10−6 ) = 40 ms
Therefore,
v5 (t) = v5(0− )e−t/τ5 = 4e−t/0.1 = 4e−10t V, t ≥ 0
v1 (t) = v1(0− )e−t/τ1 = 8e−t/0.04 = 8e−25t V, t ≥ 0
Finally,
vo (t) = v1(t) + v5(t) = [8e−25t + 4e−10t] V,
t≥0
[b] Find the value of the voltage at 60 ms for each subcircuit and use the
voltage to find the energy at 60 ms:
v1 (60 ms) = 8e−25(0.06) ∼
v5 (60 ms) = 4e−10(0.06) ∼
= 1.79 V,
= 2.20 V
w1 (60 ms) = 21 Cv12(60 ms) = 12 (1 × 10−6 )(1.79)2 ∼
= 1.59 µJ
w5 (60 ms) = 12 Cv52(60 ms) = 12 (5 × 10−6 )(2.20)2 ∼
= 12.05 µJ
w(60 ms) = 1.59 + 12.05 = 13.64 µJ
Find the initial energy from the initial voltage:
w(0) = w1 (0) + w2 (0) = 21 (1 × 10−6 )(8)2 + 12 (5 × 10−6 )(4)2 = 72 µJ
Now calculate the energy dissipated at 60 ms and compare it to the
initial energy:
wdiss = w(0) − w(60 ms) = 72 − 13.64 = 58.36 µJ
% dissipated = (58.36 × 10−6 /72 × 10−6 )(100) = 81.05 %
AP 7.5 [a] Use the circuit at t < 0, shown below, to calculate the initial current in
the inductor:
i(0− ) = 24/2 = 12 A = i(0+ )
Note that i(0− ) = i(0+ ) because the current in an inductor is continuous.
[b] Use the circuit at t = 0+ , shown below, to calculate the voltage drop
across the inductor at 0+ . Note that this is the same as the voltage drop
across the 10 Ω resistor, which has current from two sources — 8 A from
the current source and 12 A from the initial current through the inductor.
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Problems
7–5
v(0+ ) = −10(8 + 12) = −200 V
[c] To calculate the time constant we need the equivalent resistance seen by
the inductor for t > 0. Only the 10 Ω resistor is connected to the inductor
for t > 0. Thus,
τ = L/R = (200 × 10−3 /10) = 20 ms
[d] To find i(t), we need to find the final value of the current in the inductor.
When the switch has been in position a for a long time, the circuit
reduces to the one below:
Note that the inductor behaves as a short circuit and all of the current
from the 8 A source flows through the short circuit. Thus,
if = −8 A
Now,
i(t) = if + [i(0+ ) − if ]e−t/τ = −8 + [12 − (−8)]e−t/0.02
= −8 + 20e−50t A, t ≥ 0
[e] To find v(t), use the relationship between voltage and current for an
inductor:
di(t)
v(t) = L
= (200 × 10−3 )(−50)(20e−50t ) = −200e−50t V,
t ≥ 0+
dt
AP 7.6 [a]
From Example 7.6,
vo (t) = −60 + 90e−100t V
Write a KCL equation at the top node and use it to find the relationship
between vo and vA :
vA − vo
vA
vA + 75
+
+
=0
8000
160,000
40,000
20vA − 20vo + vA + 4vA + 300 = 0
25vA = 20vo − 300
vA = 0.8vo − 12
Use the above equation for vA in terms of vo to find the expression for vA :
vA (t) = 0.8(−60 + 90e−100t ) − 12 = −60 + 72e−100t V,
t ≥ 0+
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7–6
CHAPTER 7. Response of First-Order RL and RC Circuits
[b] t ≥ 0+ , since there is no requirement that the voltage be continuous in a
resistor.
AP 7.7 [a] Use the circuit shown below, for t < 0, to calculate the initial voltage drop
across the capacitor:
i=
40 × 103
(10 × 10−3 ) = 3.2 mA
125 × 103
!
vc (0− ) = (3.2 × 10−3 )(25 × 103 ) = 80 V
so vc (0+ ) = 80 V
Now use the next circuit, valid for 0 ≤ t ≤ 10 ms, to calculate vc (t) for
that interval:
For 0 ≤ t ≤ 100 ms:
τ = RC = (25 × 103 )(1 × 10−6 ) = 25 ms
vc (t) = vc (0− )et/τ = 80e−40t V 0 ≤ t ≤ 10 ms
[b] Calculate the starting capacitor voltage in the interval t ≥ 10 ms, using
the capacitor voltage from the previous interval:
vc (0.01) = 80e−40(0.01) = 53.63 V
Now use the next circuit, valid for t ≥ 10 ms, to calculate vc (t) for that
interval:
For t ≥ 10 ms :
Req = 25 kΩk100 kΩ = 20 kΩ
τ = ReqC = (20 × 103 )(1 × 10−6 ) = 0.02 s
Therefore vc (t) = vc (0.01+ )e−(t−0.01)/τ = 53.63e−50(t−0.01) V,
t ≥ 0.01 s
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Problems
7–7
[c] To calculate the energy dissipated in the 25 kΩ resistor, integrate the
power absorbed by the resistor over all time. Use the expression
p = v 2/R to calculate the power absorbed by the resistor.
w25 k =
Z
0.01
0
[80e−40t ]2
dt +
25,000
[53.63e−50(t−0.01)]2
dt = 2.91 mJ
25,000
0.01
Z
∞
[d] Repeat the process in part (c), but recognize that the voltage across this
resistor is non-zero only for the second interval:
w100 kΩ =
[53.63e−50(t−0.01)]2
dt = 0.29 mJ
100,000
0.01
Z
∞
We can check our answers by calculating the initial energy stored in the
capacitor. All of this energy must eventually be dissipated by the 25 kΩ
resistor and the 100 kΩ resistor.
Check: wstored = (1/2)(1 × 10−6 )(80)2 = 3.2 mJ
wdiss = 2.91 + 0.29 = 3.2 mJ
AP 7.8 [a] Prior to switch a closing at t = 0, there are no sources connected to the
inductor; thus, i(0− ) = 0.
At the instant A is closed, i(0+ ) = 0.
For 0 ≤ t ≤ 1 s,
The equivalent resistance seen by the 10 V source is 2 + (3k0.8). The
current leaving the 10 V source is
10
= 3.8 A
2 + (3k0.8)
The final current in the inductor, which is equal to the current in the
0.8 Ω resistor is
3
IF =
(3.8) = 3 A
3 + 0.8
The resistance seen by the inductor is calculated to find the time
constant:
L
2
[(2k3) + 0.8]k3k6 = 1 Ω
τ=
= = 2s
R
1
Therefore,
i = iF + [i(0+ ) − iF ]e−t/τ = 3 − 3e−0.5t A,
0 ≤ t ≤ 1s
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7–8
CHAPTER 7. Response of First-Order RL and RC Circuits
For part (b) we need the value of i(t) at t = 1 s:
i(1) = 3 − 3e−0.5 = 1.18 A
.
[b] For t > 1 s
Use current division to find the final value of the current:
9
i=
(−8) = −4.8 A
9+6
The equivalent resistance seen by the inductor is used to calculate the
time constant:
L
2
3k(9 + 6) = 2.5 Ω
τ=
=
= 0.8 s
R
2.5
Therefore,
i = iF + [i(1+ ) − iF ]e−(t−1)/τ
= −4.8 + 5.98e−1.25(t−1) A,
t ≥ 1s
AP 7.9 0 ≤ t ≤ 32 ms:
1
vo = −
RCf
Z
0
32×10−3
1
−10 dt + 0 = −
(−10t)
RCf
RCf = (200 × 103 )(0.2 × 10−6 ) = 40 × 10−3
32×10−3
=−
0
so
1
(−320 × 10−3 )
RCf
1
= 25
RCf
vo = −25(−320 × 10−3 ) = 8 V
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Problems
7–9
t ≥ 32 ms:
1 Zt
1
vo = −
5 dy + 8 = −
(5y)
−3
RCf 32×10
RCf
t
+8 = −
32×10−3
RCf = (250 × 103 )(0.2 × 10−6 ) = 50 × 10−3
so
1
5(t − 32 × 10−3 ) + 8
RCf
1
= 20
RCf
vo = −20(5)(t − 32 × 10−3 ) + 8 = −100t + 11.2
The output will saturate at the negative power supply value:
−15 = −100t + 11.2
.·.
t = 262 ms
AP 7.10 [a] Use RC circuit analysis to determine the expression for the voltage at the
non-inverting input:
vp = Vf + [Vo − Vf ]e−t/τ = −2 + (0 + 2)e−t/τ
τ = (160 × 103 )(10 × 10−9 ) = 10−3 ;
vp = −2 + 2e−625t V;
1/τ = 625
vn = vp
Write a KVL equation at the inverting input, and use it to determine vo:
vn
vn − vo
+
=0
10,000
40,000
.·. vo = 5vn = 5vp = −10 + 10e−625t V
The output will saturate at the negative power supply value:
−10 + 10e−625t = −5;
e−625t = 1/2;
t = ln 2/625 = 1.11 ms
[b] Use RC circuit analysis to determine the expression for the voltage at the
non-inverting input:
vp = Vf + [Vo − Vf ]e−t/τ = −2 + (1 + 2)e−625t = −2 + 3e−625t V
The analysis for vo is the same as in part (a):
vo = 5vp = −10 + 15e−625t V
The output will saturate at the negative power supply value:
−10 + 15e−625t = −5;
e−625t = 1/3;
t = ln 3/625 = 1.76 ms
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7–10
CHAPTER 7. Response of First-Order RL and RC Circuits
Problems
P 7.1
v
= 25 Ω
i
1
[b] τ =
= 100 ms
10
L
= 0.1
[c] τ =
R
L = (0.1)(25) = 2.5 H
[a] R =
1
1
[d] w(0) = L[i(0)]2 = (2.5)(6.4)2 = 51.2 J
2
2
Z t
e−20x t
[e] wdiss =
1024e−20x dx = 1024
= 51.2(1 − e−20t) J
−20 0
0
% dissipated =
51.2(1 − e−20t)
(100) = 100(1 − e−20t)
51.2
.·. 100(1 − e−20t) = 60
Therefore t =
P 7.2
so
e−20t = 0.4
1
ln 2.5 = 45.81 ms
20
[a] Note that there are several different possible solutions to this problem,
and the answer to part (c) depends on the value of inductance chosen.
L
τ
Choose a 10 mH inductor from Appendix H. Then,
0.01
R=
= 10 Ω which is a resistor value from Appendix H.
0.001
R=
[b] i(t) = Io e−t/τ = 10e−1000t mA,
t≥0
1
1
[c] w(0) = LIo2 = (0.01)(0.01)2 = 0.5 µJ
2
2
1
w(t) = (0.01)(0.01e−1000t )2 = 0.5 × 10−6 e−2000t
2
1
So 0.5 × 10−6 e−2000t = w(0) = 0.25 × 10−6
2
e−2000t = 0.5
then
e2000t = 2
ln 2
. ·. t =
= 346.57 µs
(for a 10 mH inductor)
2000
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Problems
P 7.3
[a] iL(0) =
7–11
125
= 2.5 A
50
io (0+ ) =
125
− 2.5 = 5 − 2.5 = 2.5 A
25
io (∞) =
125
= 5A
25
[b] iL = 2.5e−t/τ ;
τ=
L
50 × 10−3
=
= 2 ms
R
25
iL = 2.5e−500t A
t ≥ 0+
io = 5 − iL = 5 − 2.5e−500t A,
[c] 5 − 2.5e−500t = 3
2 = 2.5e−500t
e500t = 1.25
P 7.4
.·. t = 446.29 µs
[a] t < 0
2 kΩk6 kΩ = 1.5 kΩ
Find the current from the voltage source by combining the resistors in
series and parallel and using Ohm’s law:
ig (0− ) =
40
= 20 mA
(1500 + 500)
Find the branch currents using current division:
i1 (0− ) =
2000
(0.02) = 5 mA
8000
i2 (0− ) =
6000
(0.02) = 15 mA
8000
[b] The current in an inductor is continuous. Therefore,
i1 (0+ ) = i1(0− ) = 5 mA
i2 (0+ ) = −i1(0+ ) = −5 mA
(when switch is open)
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7–12
CHAPTER 7. Response of First-Order RL and RC Circuits
[c] τ =
L
0.4 × 10−3
=
= 5 × 10−5 s;
R
8 × 103
i1 (t) = i1(0+ )e−t/τ = 5e−20,000t mA,
1
= 20,000
τ
t≥0
when t ≥ 0+
[d] i2(t) = −i1(t)
.·. i2(t) = −5e−20,000t mA,
t ≥ 0+
[e] The current in a resistor can change instantaneously. The switching
operation forces i2 (0− ) to equal 15 mA and i2(0+ ) = −5 mA.
P 7.5
[a] io(0− ) = 0
since the switch is open for t < 0.
[b] For t = 0− the circuit is:
120 Ωk60 Ω = 40 Ω
. ·. i g =
12
= 0.24 A = 240 mA
10 + 40
iL (0− ) =
120
ig = 160 mA
180
[c] For t = 0+ the circuit is:
120 Ωk40 Ω = 30 Ω
. ·. i g =
ia =
12
= 0.30 A = 300 mA
10 + 30
120
300 = 225 mA
160
.·. io(0+ ) = 225 − 160 = 65 mA
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Problems
7–13
[d] iL(0+ ) = iL(0− ) = 160 mA
[e] io (∞) = ia = 225 mA
[f] iL(∞) = 0,
since the switch short circuits the branch containing the 20
Ω resistor and the 100 mH inductor.
L
100 × 10−3
1
[g] τ =
=
= 5 ms;
= 200
R
20
τ
.·. iL = 0 + (160 − 0)e−200t = 160e−200t mA,
[h] vL(0− ) = 0
t≥0
since for t < 0 the current in the inductor is constant
[i] Refer to the circuit at t = 0+ and note:
20(0.16) + vL(0+ ) = 0;
[j] vL(∞) = 0,
.·. vL (0+ ) = −3.2 V
since the current in the inductor is a constant at t = ∞.
[k] vL (t) = 0 + (−3.2 − 0)e−200t = −3.2e−200t V,
[l] io (t) = ia − iL = 225 − 160e−200t mA,
P 7.6
t ≥ 0+
t ≥ 0+
For t < 0
ig =
−48
= −6.5 A
6 + (18k1.5)
iL (0− ) =
18
(−6.5) = −6 A = iL (0+ )
18 + 1.5
For t > 0
iL (t) = iL (0+ )e−t/τ A,
τ=
t≥0
L
0.5
=
= 0.0125 s;
R
10 + 12.45 + (54k26)
1
= 80
τ
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7–14
CHAPTER 7. Response of First-Order RL and RC Circuits
iL (t) = −6e−80t A,
io (t) =
P 7.7
t≥0
54
54
(−iL(t)) = (6e−80t) = 4.05e−80t V,
80
80
t ≥ 0+
24
= 2A
12
L
1.6
[b] τ =
=
= 20 ms
R
80
[c] i = 2e−50t A,
t≥0
[a] i(0) =
v1 = L
d
(2e−50t ) = −160e−50t V
dt
v2 = −72i = −144e−50t V
t ≥ 0+
t≥0
1
[d] w(0) = (1.6)(2)2 = 3.2 J
2
w72Ω =
Z
0
t
72(4e−100x ) dx = 288
e−100x
−100
t
= 2.88(1 − e−100t) J
0
w72Ω (15 ms) = 2.88(1 − e−1.5 ) = 2.24 J
% dissipated =
P 7.8
2.24
(100) = 69.92%
3.2
1
w(0) = (10 × 10−3 )(5)2 = 125 mJ
2
0.9w(0) = 112.5 mJ
1
w(t) = (10 × 10−3 )i(t)2,
2
i(t) = 5e−t/τ A
.·. w(t) = 0.005(25e−2t/τ ) = 125e−2t/τ ) mJ
w(10 µs) = 125e−20×10
−6 /τ
mJ
−6
.·. 125e−20×10 /τ = 112.5
so
τ=
20 × 10−6
L
=
ln(10/9)
R
R=
10 × 10−3 ln(10/9)
= 52.68 Ω
20 × 10−6
e20×10
−6/τ
=
10
9
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Problems
P 7.9
7–15
1
[a] w(0) = LIg2
2
e−2t/τ to
(−2/τ ) 0
0
1
1
= Ig2Rτ (1 − e−2to /τ ) = Ig2L(1 − e−2to /τ )
2
2
to
Z
wdiss =
Ig2Re−2t/τ dt = Ig2R
wdiss = σw(0)
1 2
1 2
. ·.
LIg (1 − e−2to/τ ) = σ
LI
2
2 g
1 − e−2to/τ = σ;
"
e2to/τ =
#
2to
1
= ln
;
τ
(1 − σ)
R=
[b] R =
1
(1 − σ)
R(2to )
= ln[1/(1 − σ)]
L
L ln[1/(1 − σ)]
2to
(10 × 10−3 ) ln[1/0.9]
20 × 10−6
R = 52.68 Ω
P 7.10
[a] vo(t) = vo(0+ )e−t/τ
.·. vo(0+ )e−10
.·. e10
−3 /τ
. ·. τ =
−3 /τ
= 0.5vo (0+ )
=2
L
10−3
=
R
ln 2
10 × 10−3
·
.. L=
= 14.43 mH
ln 2
[b] vo(0+ ) = −10iL (0+ ) = −10(1/10)(30 × 10−3 ) = −30 mV
vo (t) = −0.03e−t/τ V
p10Ω =
vo2
= 9 × 10−5 e−2t/τ
10
w10Ω =
Z
τ=
10−3
0
1
1000 ln 2
9 × 10−5 e−2t/τ dt = 4.5τ × 10−5 (1 − e−2×10
. ·.
−3 /τ
)
w10Ω = 48.69 nJ
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7–16
CHAPTER 7. Response of First-Order RL and RC Circuits
1
1
wL (0) = Li2L (0) = (14.43 × 10−3 )(3 × 10−3 )2 = 64.92 nJ
2
2
48.69
× 100 = 75%
64.92
% diss in 1 ms =
P 7.11
[a] t < 0
iL (0− ) =
150
(12) = 10 A
180
t≥0
τ=
1.6 × 10−3
= 200 × 10−6 ;
8
1/τ = 5000
io = −10e−5000t A t ≥ 0
1
[b] wdel = (1.6 × 10−3 )(10)2 = 80 mJ
2
[c] 0.95wdel = 76 mJ
.·. 76 × 10−3 =
Z
to
0
8(100e−10,000t) dt
.·. 76 × 10−3 = −80 × 10−3 e−10,000t
to
= 80 × 10−3 (1 − e−10,000to )
0
.·. e−10,000to = 0.05 so to = 299.57 µs
. ·.
to
299.57 × 10−6
=
= 1.498
τ
200 × 10−6
so
to ≈ 1.498τ
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Problems
P 7.12
7–17
t < 0:
240
= 10 A;
16 + 8
iL (0+ ) =
iL(0− ) = 10
40
= 8A
50
t > 0:
Re =
τ=
(10)(40)
+ 10 = 18 Ω
50
L
72 × 10−3
=
= 4 ms;
Re
18
1
= 250
τ
.·. iL = 8e−250t A
vo = 8io = 64e−250t V,
P 7.13
t ≥ 0+
p40Ω =
vo2
(64)2 −500t
=
e
= 102.4e−500t W
40
40
w40Ω =
Z
0
∞
102.4e−500t dt = 102.4
e−500t
−500
∞
= 204.8 mJ
0
1
w(0) = (72 × 10−3 )(8)2 = 2304 mJ
2
% diss =
204.8
(100) = 8.89%
2304
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7–18
P 7.14
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] t < 0 :
iL (0) = −
72
= −2.4 A
24 + 6
t > 0:
i∆ = −
100
5
iT = − iT
160
8
vT = 20i∆ + iT
(100)(60)
= −12.5iT + 37.5iT
160
vT
= RTh = −12.5 + 37.5 = 25 Ω
iT
L
250 × 10−3
=
R
25
τ=
iL = −2.4e−100t A,
1
= 100
τ
t≥0
[b] vL = 250 × 10−3 (240e−100t ) = 60e−100t V,
[c] i∆ = 0.625iL = −1.5e−100t A
P 7.15
t ≥ 0+
t ≥ 0+
1
w(0) = (250 × 10−3 )(−2.4)2 = 720 mJ
2
p60Ω = 60(−1.5e−100t)2 = 135e−200t W
w60Ω =
Z
0
∞
135e−200t dt = 135
% dissipated =
e−200t
−200
∞
= 675 mJ
0
675
(100) = 93.75%
720
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Problems
P 7.16
7–19
t<0
iL (0− ) = iL (0+ ) = 4 A
t>0
Find Thévenin resistance seen by inductor:
iT = 4vT ;
vT
1
= RTh = = 0.25 Ω
iT
4
L
5 × 10−3
τ=
=
= 20 ms;
R
0.25
io = 4e−50t A,
vo = L
1/τ = 50
t≥0
dio
= (5 × 10−3 )(−200e−50t ) = −e−50t V,
dt
t ≥ 0+
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7–20
P 7.17
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] t < 0 :
t = 0+ :
120 = iab + 18 + 12,
iab = 90 A,
t = 0+
[b] At t = ∞:
iab = 240/2 = 120 A,
[c] i1 (0) = 18,
i2 (0) = 12,
τ1 =
t=∞
2 × 10−3
= 0.2 ms
10
τ2 =
6 × 10−3
= 0.4 ms
15
i1 (t) = 18e−5000t A,
t≥0
i2 (t) = 12e−2500t A,
t≥0
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Problems
iab = 120 − 18e−5000t − 12e−2500t A,
7–21
t≥0
120 − 18e−5000t − 12e−2500t = 114
6 = 18e−5000t + 12e−2500t
x = e−2500t
Let
P 7.18
so
Solving x =
1
= e−2500t
3
.·. e2500t = 3
and
6 = 18x2 + 12x
t=
ln 3
= 439.44 µs
2500
[a] t < 0
1 kΩk4 kΩ = 0.8 kΩ
20 kΩk80 kΩ = 16 kΩ
(105 × 10−3 )(0.8 × 103 ) = 84 V
iL (0− ) =
84
= 5 mA
16,800
t>0
τ=
L
6
=
× 10−3 = 250 µs;
R
24
1
= 4000
τ
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7–22
CHAPTER 7. Response of First-Order RL and RC Circuits
iL (t) = 5e−4000t mA,
t≥0
p4k = 25 × 10−6 e−8000t(4000) = 0.10e−8000t W
wdiss =
t
Z
0.10e−8000x dx = 12.5 × 10−6 [1 − e−8000t] J
0
1
w(0) = (6)(25 × 10−6 ) = 75 µJ
2
0.10w(0) = 7.5 µJ
.·. e8000t = 2.5
12.5(1 − e−8000t) = 7.5;
t=
ln 2.5
= 114.54 µs
8000
[b] wdiss(total) = 75(1 − e−8000t) µJ
wdiss(114.54 µs) = 45 µJ
% = (45/75)(100) = 60%
P 7.19
[a] t > 0:
Leq = 1.25 +
60
= 5H
16
iL (t) = iL (0)e−t/τ mA;
iL (t) = 2e−1500t A,
iL (0) = 2 A;
1
R
7500
=
=
= 1500
τ
L
5
t≥0
vR (t) = RiL (t) = (7500)(2e−1500t ) = 15,000e−1500t V,
vo = −3.75
[b] io =
P 7.20
−1
6
Z
0
diL
= 11,250e−1500t V,
dt
t ≥ 0+
t ≥ 0+
t
11,250e−1500x dx + 0 = 1.25e−1500t − 1.25 A
[a] From the solution to Problem 7.19,
1
1
w(0) = Leq[iL (0)]2 = (5)(2)2 = 10 J
2
2
1
1
[b] wtrapped = (10)(1.25)2 + (6)(1.25)2 = 12.5 J
2
2
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Problems
P 7.21
v
= 8 kΩ
i
1
1
1
=
= 500;
C=
= 0.25 µF
τ
RC
(500)(8000)
1
τ=
= 2 ms
500
1
w(0) = (0.25 × 10−6 )(72)2 = 648 µJ
2
Z to
(72)2 e−1000t
wdiss =
dt
(800)
0
[a] R =
[b]
[c]
[d]
[e]
= 0.648
e−1000t
−1000
%diss = 100(1 − e
. ·. t =
P 7.22
to
= 648(1 − e−1000to ) µJ
0
−1000to
) = 68
e1000to = 3.125
so
ln 3.125
= 1139 µs
1000
[a] Note that there are many different possible correct solutions to this
problem.
τ
R=
C
Choose a 100 µF capacitor from Appendix H. Then,
0.05
R=
= 500 Ω
100 × 10−6
Construct a 500 Ω resistor by combining two 1 kΩ resistors in parallel:
[b] v(t) = Vo e−t/τ = 50e−20t V,
[c] 50e−20t = 10
. ·. t =
P 7.23
7–23
so
t≥0
e20t = 5
ln 5
= 80.47 ms
20
[a] v1(0− ) = v1 (0+ ) = 40 V
v2(0+ ) = 0
Ceq = (1)(4)/5 = 0.8 µF
τ = (25 × 103 )(0.8 × 10−6 ) = 20ms;
1
= 50
τ
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7–24
CHAPTER 7. Response of First-Order RL and RC Circuits
i=
40 −50t
e
= 1.6e−50t mA,
25,000
t ≥ 0+
−1 Z t
1.6 × 10−3 e−50x dx + 40 = 32e−50t + 8 V,
t≥0
10−6 0
Z t
1
v2 =
1.6 × 10−3 e−50x dx + 0 = −8e−50t + 8 V,
t≥0
−6
4 × 10
0
v1 =
1
[b] w(0) = (10−6 )(40)2 = 800 µJ
2
1
1
[c] wtrapped = (10−6 )(8)2 + (4 × 10−6 )(8)2 = 160 µJ.
2
2
The energy dissipated by the 25 kΩ resistor is equal to the energy
dissipated by the two capacitors; it is easier to calculate the energy
dissipated by the capacitors:
1
wdiss = (0.8 × 10−6 )(40)2 = 640 µJ.
2
Check: wtrapped + wdiss = 160 + 640 = 800 µJ;
w(0) = 800 µJ.
P 7.24
[a] t < 0:
i1 (0− ) = i2(0− ) =
3
= 100 mA
30
[b] t > 0:
i1 (0+ ) =
0.2
= 100 mA
2
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Problems
i2 (0+ ) =
7–25
−0.2
= −25 mA
8
[c] Capacitor voltage cannot change instantaneously, therefore,
i1 (0− ) = i1(0+ ) = 100 mA
[d] Switching can cause an instantaneous change in the current in a resistive
branch. In this circuit
i2 (0− ) = 100 mA and i2(0+ ) = 25 mA
[e] vc = 0.2e−t/τ V,
t≥0
τ = Re C = 1.6(2 × 10−6 ) = 3.2 µs;
vc = 0.2e−312,000t V,
[f] i2 =
P 7.25
t≥0
vc
= 0.1e−312,000t A,
2
i1 =
1
= 312,500
τ
t≥0
−vc
= −25e−312,000t mA,
8
t ≥ 0+
[a] t < 0:
Req = 12 kk8 k = 10.2 kΩ
vo (0) =
10,200
(−120) = −102 V
10,200 + 1800
t > 0:
τ = [(10/3) × 10−6 )(12,000) = 40 ms;
vo = −102e−25t V,
p=
1
= 25
τ
t≥0
vo2
= 867 × 10−3 e−50t W
12,000
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7–26
CHAPTER 7. Response of First-Order RL and RC Circuits
wdiss =
12×10−3
Z
0
867 × 10−3 e−50t dt
= 17.34 × 10−3 (1 − e−50(12×10
1
[b] w(0) =
2
−3)
) = 7824 µJ
10
(102)2 × 10−6 = 17.34 mJ
3
0.75w(0) = 13 mJ
Z
0
to
867 × 10−3 e−50x dx = 13 × 10−3
.·. 1 − e−50to = 0.75;
P 7.26
e50to = 4;
so to = 27.73 ms
[a] t < 0:
io (0− ) =
6000
(40 m) = 24 mA
6000 + 4000
vo (0− ) = (3000)(24 m) = 72 V
i2 (0− ) = 40 − 24 = 16 mA
v2 (0− ) = (6000)(16 m) = 96 V
t>0
τ = RC = (1000)(0.2 × 10−6 ) = 200 µs;
io (t) =
24
e−t/τ = 24e−5000t mA,
1 × 103
1
= 5000
τ
t ≥ 0+
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Problems
7–27
[b]
Z t
1
24 × 10−3 e−5000x dx + 72
0.6 × 10−6 0
e−5000x t
= (40,000)
+72
−5000 0
= −8e−5000t + 8 + 72
vo = [−8e−5000t + 80] V,
t≥0
vo =
[c] wtrapped = (1/2)(0.3 × 10−6 )(80)2 + (1/2)(0.6 × 10−6 )(80)2
wtrapped = 2880 µJ.
Check:
1
wdiss = (0.2 × 10−6 )(24)2 = 57.6 µJ
2
1
1
w(0) = (0.3 × 10−6 )(96)2 + (0.6 × 10−6 )(72)2 = 2937.6 µJ.
2
2
wtrapped + wdiss = w(0)
2880 + 57.6 = 2937.6
P 7.27
OK.
[a] At t = 0− the voltage on each capacitor will be 150 V(5 × 30), positive at
the upper terminal. Hence at t ≥ 0+ we have
.·. isd (0+ ) = 5 +
150 150
+
= 1055 A
0.2
0.5
At t = ∞, both capacitors will have completely discharged.
.·. isd (∞) = 5 A
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7–28
CHAPTER 7. Response of First-Order RL and RC Circuits
[b] isd (t) = 5 + i1 (t) + i2 (t)
τ1 = 0.2(10−6 ) = 0.2 µs
τ2 = 0.5(100 × 10−6 ) = 50 µs
.·. i1(t) = 750e−5×10 t A,
6
i2(t) = 300e−20,000t A,
t ≥ 0+
t≥0
.·. isd = 5 + 750e−5×10 t + 300e−20,000t A,
6
P 7.28
t ≥ 0+
[a]
vT = 20 × 103 (iT + αv∆ ) + 5 × 103 iT
v∆ = 5 × 103 iT
vT = 25 × 103 iT + 20 × 103 α(5 × 103 iT )
RTh = 25,000 + 100 × 106 α
τ = RTh C = 40 × 10−3 = RTh (0.8 × 10−6 )
RTh = 50 kΩ = 25,000 + 100 × 106 α
α=
25,000
= 2.5 × 10−4 A/V
6
100 × 10
[b] vo(0) = (−5 × 10−3 )(3600) = −18 V
t > 0:
vo = −18e−25t V,
t<0
t≥0
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Problems
7–29
v∆
v∆ − vo
+
+ 2.5 × 10−4 v∆ = 0
5000
20,000
4v∆ + v∆ − vo + 5v∆ = 0
vo
= −1.8e−25t V,
10
. ·. v ∆ =
P 7.29
t ≥ 0+
[a]
pds = (16.2e−25t )(−450 × 10−6 e−25t ) = −7290 × 10−6 e−50t W
wds =
Z
∞
pds dt = −145.8 µJ.
0
.·. dependent source is delivering 145.8 µJ.
[b] w5k =
Z
∞
(5000)(0.36 × 10−3 e−25t)2 dt = 648 × 10−6
0
w20k =
Z
0
∞
(16.2e−25t )2
dt = 13,122 × 10−6
20,000
Z
0
Z
0
∞
e−50t dt = 12.96 µJ
∞
e−50t dt = 262.44 µJ
1
wc (0) = (0.8 × 10−6 )(18)2 = 129.6 µJ
2
X
X
wdiss = 12.96 + 262.44 = 275.4 µJ
wdev = 145.8 + 129.6 = 275.4 µJ.
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7–30
P 7.30
CHAPTER 7. Response of First-Order RL and RC Circuits
t<0
t>0
.·.
vT = −5io − 15io = −20io = 20iT
τ = RC = 40 µs;
vo = 15e−25,000t V,
io = −
P 7.31
RTh =
vT
= 20 Ω
iT
1
= 25,000
τ
t≥0
vo
= −0.75e−25,000t A,
20
t ≥ 0+
[a] The equivalent circuit for t > 0:
τ = 2 ms;
1/τ = 500
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Problems
vo = 10e−500t V,
t≥0
io = e−500t mA,
t ≥ 0+
i24kΩ = e−500t
16
= 0.4e−500t mA,
40
7–31
t ≥ 0+
p24kΩ = (0.16 × 10−6 e−1000t)(24,000) = 3.84e−1000t mW
w24kΩ =
∞
Z
3.84 × 10−3 e−1000t dt = −3.84 × 10−6 (0 − 1) = 3.84 µJ
0
1
1
w(0) = (0.25 × 10−6 )(40)2 + (1 × 10−6 )(50)2 = 1.45 mJ
2
2
% diss (24 kΩ) =
3.84 × 10−6
× 100 = 0.26%
1.45 × 10−3
[b] p400Ω = 400(1 × 10−3 e−500t)2 = 0.4 × 10−3 e−1000t
w400Ω =
∞
Z
p400 dt = 0.40 µJ
0
% diss (400 Ω) =
i16kΩ = e
−500t
0.4 × 10−6
× 100 = 0.03%
1.45 × 10−3
24
= 0.6e−500t mA,
40
t ≥ 0+
p16kΩ = (0.6 × 10−3 e−500t)2 (16,000) = 5.76 × 10−3 e−1000t W
w16kΩ =
Z
0
∞
5.76 × 10−3 e−1000t dt = 5.76 µJ
% diss (16 kΩ) = 0.4%
[c]
X
wdiss = 3.84 + 5.76 + 0.4 = 10 µJ
wtrapped = w(0) −
% trapped =
X
wdiss = 1.45 × 10−3 − 10 × 10−6 = 1.44 mJ
1.44
× 100 = 99.31%
1.45
Check: 0.26 + 0.03 + 0.4 + 99.31 = 100%
P 7.32
[a] Ce =
(2 + 1)6
= 2 µF
2+1+6
vo (0) = −5 + 30 = 25 V
τ = (2 × 10−6 )(250 × 103 ) = 0.5 s;
1
=2
τ
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7–32
CHAPTER 7. Response of First-Order RL and RC Circuits
t ≥ 0+
vo = 25e−2t V,
1
1
[b] wo = (3 × 10−6 )(30)2 + (6 × 10−6 )(5)2 = 1425 µJ
2
2
1
wdiss = (2 × 10−6 )(25)2 = 625 µJ
2
625
% diss =
× 100 = 43.86%
1425
vo
[c] io =
= 100e−2t µA
250 × 10−3
1
6 × 10−6
e−2x
= −16.67
−2
Z
v1 = −
t
0
t
Z
100 × 10−6 e−2x dx − 5 = −16.67
−5 = 8.33e−2t − 13.33 V
t
0
e−2x dx − 5
t≥0
0
[d] v1 + v2 = vo
v2 = vo − v1 = 25e−2t − 8.33e−2t + 13.33 = 16.67e−2t + 13.33 V t ≥ 0
1
1
[e] wtrapped = (6 × 10−6 )(13.33)2 + (3 × 10−6 )(13.33)2 = 800 µJ
2
2
wdiss + wtrapped = 625 + 800 = 1425 µJ
(check)
P 7.33
[a] From Eqs. (7.35) and (7.42)
Vs
Vs −(R/L)t
i=
+ Io −
e
R
R
v = (Vs − Io R)e−(R/L)t
. ·.
Vs
= 4;
R
Io −
Vs − Io R = −80;
Vs
=4
R
R
= 40
L
Vs
= 8A
R
Now since Vs = 4R we have
. ·. I o = 4 +
4R − 8R = −80;
Vs = 80 V;
R = 20 Ω
L=
R
= 0.5 H
40
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Problems
7–33
i2 = 16 + 32e−40t + 16e−80t
[b] i = 4 + 4e−40t ;
1
1
w = Li2 = (0.5)[16 + 32e−40t + 16e−80t] = 4 + 8e−40t + 4e−80t
2
2
.·. 4 + 8e−40t + 4e−80t = 9 or e−80t + 2e−40t − 1.25 = 0
Let x = e−40t:
x2 + 2x − 1.25 = 0;
Solving, x = 0.5;
x = −2.5
But x ≥ 0 for all t. Thus,
e40t = 2;
e−40t = 0.5;
P 7.34
t = 25 ln 2 = 17.33 ms
[a] Note that there are many different possible solutions to this problem.
L
τ
Choose a 1 mH inductor from Appendix H. Then,
R=
0.001
= 125 Ω
8 × 10−6
Construct the resistance needed by combining 100 Ω, 10 Ω, and 15 Ω
resistors in series:
R=
[b] i(t) = If + (Io − If )e−t/τ
Io = 0 A;
If =
Vf
25
=
= 200 mA
R
125
.·. i(t) = 200 + (0 − 200)e−125,000t mA = 200 − 200e−125,000t mA,
t≥0
[c] i(t) = 0.2 − 0.2e−125,000t = (0.75)(0.2) = 0.15
e−125,000t = 0.25
. ·. t =
so
e125,000t = 4
ln 4
= 11.09 µs
125,000
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7–34
P 7.35
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] t < 0
iL (0− ) = −5 A
t>0
iL (∞) =
τ=
40 − 80
= −2 A
4 + 16
L
4 × 10−3
=
= 200 µs;
R
4 + 16
1
= 5000
τ
iL = iL(∞) + [iL(0+ ) − iL (∞)]e−t/τ
= −2 + (−5 + 2)e−5000t = −2 − 3e−5000t A,
t≥0
vo = 16iL + 80 = 16(−2 − 3e−5000t) + 80 = 48 − 48e−5000t V,
[b] vL = L
diL
= 4 × 10−3 (−5000)[−3e−5000t] = 60e−5000t V,
dt
t≥0
t ≥ 0+
vL (0+ ) = 60 V
From part (a)
vo (0+ ) = 0 V
Check: at t = 0+ the circuit is:
vL (0+ ) = 40 + (5 A)(4 Ω) = 60 V,
vo (0+ ) = 80 − (16 Ω)(5 A) = 0 V
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Problems
P 7.36
7–35
[a] For t < 0, calculate the Thévenin equivalent for the circuit to the left and
right of the 75 mH inductor. We get
i(0− ) =
5 − 120
= −5 mA
15 k + 8 k
i(0− ) = i(0+ ) = −5 mA
[b] For t > 0, the circuit reduces to
Therefore i(∞) = 5/15,000 = 0.333 mA
[c] τ =
L
75 × 10−3
=
= 5 µs
R
15,000
[d] i(t) = i(∞) + [i(0+ ) − i(∞)]e−t/τ
= 0.333 + [−5 − 0.333]e−200,000t = 0.333 − 5.333e−200,000t mA,
P 7.37
t≥0
[a] t < 0
KVL equation at the top node:
vo vo
vo
50 =
+
+
8
40 10
Multiply by 40 and solve:
2000 = (5 + 1 + 4)vo ;
.·. io(0− ) =
vo = 200 V
vo
= 200/10 = 20 A
10
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7–36
CHAPTER 7. Response of First-Order RL and RC Circuits
t>0
Use voltage division to find the Thévenin voltage:
40
VTh = vo =
(800) = 200 V
40 + 120
Remove the voltage source and make series and parallel combinations of
resistors to find the equivalent resistance:
RTh = 10 + 120k40 = 10 + 30 = 40 Ω
The simplified circuit is:
L
40 × 10−3
1
=
= 1 ms;
= 1000
R
40
τ
200
io (∞) =
= 5A
40
.·. io = io (∞) + [io(0+ ) − io(∞)]e−t/τ
τ=
= 5 + (20 − 5)e−1000t = 5 + 15e−1000t A,
[b] vo
vo
P 7.38
t≥0
=
dio
dt
10(5 + 15e−1000t) + 0.04(−1000)(15e−1000t )
=
50 + 150e−1000t − 600e−1000t
=
50 − 450e−1000t V,
=
10io + L
t ≥ 0+
[a]
−
Vs
v
1
+ +
R R L
Z
t
0
v dt + Io = 0
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Problems
7–37
Differentiating both sides,
1 dv
1
+ v=0
R dt L
dv R
+ v=0
dt
L
. ·.
[b]
dv
R
=− v
dt
L
dv
R
dt = − v dt
dt
L
R
dv = − v dt
L
so
dv
R
= − dt
v
L
v(t) dx
Z
Vo
ln
. ·.
P 7.39
x
=−
RZ t
dy
L 0
v(t)
R
=− t
Vo
L
v(t) = Vo e−(R/L)t = (Vs − RIo )e−(R/L)t
[a] vo(0+ ) = −Ig R2 ;
τ=
L
R1 + R2
vo (∞) = 0
vo (t) = −Ig R2 e−[(R1 +R2 )/L]t V,
t ≥ 0+
[b] vo(0+ ) → ∞, and the duration of vo(t) → zero
L
[c] vsw = R2 io ;
τ=
R1 + R2
io (0+ ) = Ig ;
Therefore
Therefore
[d] |vsw (0+ )| → ∞;
P 7.40
io (∞) = Ig
R1
R1 + R2
io (t) =
Ig R1
R1 +R2
io (t) =
R1 Ig
(R1 +R2 )
vsw =
h
+ Ig −
+
R1Ig
(1+R1 /R2)
Ig R1
R1 +R2
i
e−[(R1+R2 )/L]t
R2Ig
e−[(R1 +R2 )/L]t
(R1+R2 )
+
R2 Ig
e−[(R1+R2 )/L]t,
(1+R1 /R2)
t ≥ 0+
duration → 0
Opening the inductive circuit causes a very large voltage to be induced across
the inductor L. This voltage also appears across the switch (part [d] of
Problem 7.39), causing the switch to arc over. At the same time, the large
voltage across L damages the meter movement.
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7–38
P 7.41
CHAPTER 7. Response of First-Order RL and RC Circuits
t > 0;
calculate vo (0+ )
va va − vo (0+ )
+
= 20 × 10−3
15
5
.·. va = 0.75vo (0+ ) + 75 × 10−3
15 × 10−3 +
vo(0+ ) − va vo (0+ )
+
− 9i∆ + 50 × 10−3 = 0
5
8
13vo (0+ ) − 8va − 360i∆ = −2600 × 10−3
i∆ =
vo(0+ )
− 9i∆ + 50 × 10−3
8
vo (0+ )
·
. . i∆ =
+ 5 × 10−3
80
.·. 360i∆ = 4.5vo (0+ ) + 1800 × 10−3
8va = 6vo (0+ ) + 600 × 10−3
.·. 13vo (0+ ) − 6vo (0+ ) − 600 × 10−3 − 4.5vo (0+ )−
1800 × 10−3 = −2600 × 10−3
2.5vo (0+ ) = −200 × 10−3 ;
vo(0+ ) = −80 mV
vo (∞) = 0
Find the Thévenin resistance seen by the 4 mH inductor:
iT =
vT
vT
+
− 9i∆
20
8
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Problems
i∆ =
vT
− 9i∆
8
vT
.·. 10i∆ =
;
8
iT =
vT
10vT
9vT
+
−
20
80
80
i∆ =
7–39
vT
80
iT
1
1
5
1
=
+
=
=
S
vT
20 80
80
16
.·. RTh = 16Ω
τ=
4 × 10−3
= 0.25 ms;
16
1/τ = 4000
.·. vo = 0 + (−80 − 0)e−4000t = −80e−4000t mV,
P 7.42
t ≥ 0+
For t < 0
vx
vx − 480
− 0.8vφ +
=0
15
21
vφ =
vx − 480
21
vx
vx − 480
vx − 480
− 0.8
+
15
21
21
=
.·.
vx
vs − 480
+ 0.2
15
21
= 21vx + 3(vx − 480) = 0
24vx = 1440 so vx = 60 V
io (0− ) =
vx
= 4A
15
t>0
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7–40
CHAPTER 7. Response of First-Order RL and RC Circuits
Find Thévenin equivalent with respect to a, b
VTh − 320
VTh − 320
− 0.8
5
5
=0
vT = (iT + 0.8vφ )(5) = iT + 0.8
vT = 5iT + 0.8vT
vT
5
VTh = 320 V
(5)
.·. 0.2vT = 5iT
vT
= RTh = 25 Ω
iT
io (∞) = 320/40 = 8 A
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Problems
τ=
80 × 10−3
= 2 ms;
40
1/τ = 500
io = 8 + (4 − 8)e−500t = 8 − 4e−500t A,
P 7.43
7–41
t≥0
For t < 0,
i80mH (0) = 50 V/10 Ω = 5 A
For t > 0, after making a Thévenin equivalent we have
Vs
Vs −t/τ
i=
+ Io −
e
R
R
1
R
8
=
=
= 80
τ
L
100 × 10−3
Io = 5 A;
If =
Vs
−80
=
= −10 A
R
8
i = −10 + (5 + 10)e−80t = −10 + 15e−80t A,
vo = 0.08
P 7.44
di
= 0.08(−1200e−80t ) = −96e−80t V,
dt
t≥0
t ≥ 0+
[a] Let v be the voltage drop across the parallel branches, positive at the top
node, then
−Ig +
v
1
+
Rg L1
Z
v
1
1
+
+
Rg
L1 L2
v
1
+
Rg Le
Z
0
0
t
v dx +
Z
0
1
L2
Z
0
t
v dx = 0
t
v dx = Ig
t
v dx = Ig
1 dv
v
+
=0
Rg dt Le
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7–42
CHAPTER 7. Response of First-Order RL and RC Circuits
dv Rg
+
v=0
dt
Le
Therefore v = Ig Rg e−t/τ ;
τ = Le /Rg
Thus
Z
1 t
Ig Rg e−x/τ t Ig Le
i1 =
Ig Rg e−x/τ dx =
=
(1 − e−t/τ )
L1 0
L1 (−1/τ ) 0
L1
i1 =
Ig L2
Ig L1
(1 − e−t/τ ) and i2 =
(1 − e−t/τ )
L1 + L2
L1 + L2
[b] i1(∞) =
P 7.45
L2
Ig ;
L1 + L2
i2(∞) =
L1
Ig
L1 + L2
[a] t < 0
t>0
iL (0− ) = iL (0+ ) = 25 mA;
τ=
24 × 10−3
= 0.2 ms;
120
1
= 5000
τ
iL (∞) = −50 mA
iL = −50 + (25 + 50)e−5000t = −50 + 75e−5000t mA,
vo = −120[75 × 10−3 e−5000t] = −9e−5000t V,
t≥0
t ≥ 0+
t
1
[b] i1 =
−9e−5000x dx + 10 × 10−3 = (30e−5000t − 20) mA,
60 × 10−3 0
Z t
1
[c] i2 =
−9e−5000x dx + 15 × 10−3 = (45e−5000t − 30) mA,
40 × 10−3 0
Z
P 7.46
t≥0
t≥0
t>0
τ=
1
40
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Problems
io = 5e−40t A,
7–43
t≥0
t > 0+
vo = 40io = 200e−40t V,
200e−40t = 100;
e40t = 2
1
.·. t =
ln 2 = 17.33 ms
40
P 7.47
1
1
[a] wdiss = Le i2 (0) = (1)(5)2 = 12.5 J
2
2
Z
1 t
[b] i3H =
(200)e−40x dx − 5
3 0
= 1.67(1 − e−40t) − 5 = −1.67e−40t − 3.33 A
i1.5H =
1 Zt
(200)e−40x dx + 0
1.5 0
= −3.33e−40t + 3.33 A
1
wtrapped = (4.5)(3.33)2 = 25 J
2
1
[c] w(0) = (3)(5)2 = 37.5 J
2
P 7.48
[a] v = Is R + (Vo − Is R)e−t/RC
.·. Is R = 40,
i = Is −
Vo −t/RC
e
R
Vo − Is R = −24
.·. Vo = 16 V
Is −
Vo
= 3 × 10−3 ;
R
Is −
.·. Is − 0.4Is = 3 × 10−3 ;
R=
16
= 3 × 10−3 ;
R
R=
40
Is
Is = 5 mA
40
× 103 = 8 kΩ
5
1
= 2500;
RC
C=
1
10−3
=
= 50 nF;
2500R
20 × 103
τ = RC =
1
= 400 µs
2500
[b] v(∞) = 40 V
1
w(∞) = (50 × 10−9 )(1600) = 40 µJ
2
0.81w(∞) = 32.4 µJ
v 2(to ) =
32.4 × 10−6
= 1296;
25 × 10−9
40 − 24e−2500to = 36;
v(to) = 36 V
e2500to = 6;
.·. to = 716.70 µs
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7–44
P 7.49
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] Note that there are many different possible solutions to this problem.
τ
R=
C
Choose a 10 µH capacitor from Appendix H. Then,
0.25
= 25 kΩ
10 × 10−6
Construct the resistance needed by combining 10 kΩ and 15 kΩ resistors
in series:
R=
[b] v(t) = Vf + (Vo − Vf )e−t/τ
Vo = 100 V;
Vf = (If )(R) = (1 × 10−3 )(25 × 103 ) = 25 V
.·. v(t) = 25 + (100 − 25)e−4t V = 25 + 75e−4t V,
[c] v(t) = 25 + 75e−4t = 50
. ·. t =
P 7.50
so
e−4t =
t≥0
1
3
ln 3
= 274.65 ms
4
[a]
io (0+ ) =
−36
= −7.2 mA
5000
[b] io(∞) = 0
[c] τ = RC = (5000)(0.8 × 10−6 ) = 4 ms
[d] io = 0 + (−7.2)e−250t = −7.2e−250t mA,
t ≥ 0+
[e] vo = −[36 + 1800(−7.2 × 10−3 e−250t)] = −36 + 12.96e−250t V,
t ≥ 0+
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Problems
P 7.51
7–45
[a] Simplify the circuit for t > 0 using source transformation:
Since there is no source connected to the capacitor for t < 0
vo (0− ) = vo (0+ ) = 0 V
From the simplified circuit,
vo (∞) = 60 V
τ = RC = (20 × 103 )(0.5 × 10−6 ) = 10 ms
1/τ = 100
vo = vo(∞) + [vo(0+ ) − vo (∞)]e−t/τ = (60 − 60e−100t ) V,
t≥0
dvo
dt
ic = 0.5 × 10−6 (−100)(−60e−100t ) = 3e−100t mA
[b] ic = C
v1 = 8000ic + vo = (8000)(3 × 10−3 )e−100t + (60 − 60e−100t ) = 60 − 36e−100t V
v1
io =
= 1 − 0.6e−100t mA, t ≥ 0+
3
60 × 10
[c] i1 (t) = io + ic = 1 + 2.4e−100t mA, t ≥ 0+
v1
[d] i2(t) =
= 4 − 2.4e−100t mA, t ≥ 0+
15 × 103
[e] i1(0+ ) = 1 + 2.4 = 3.4 mA
At t = 0+ :
Re = 15 kk60 kk8 k = 4800 Ω
v1 (0+ ) = (5 × 10−3 )(4800) = 24 V
i1 (0+ ) =
P 7.52
v1 (0+ ) v1(0+ )
+
= 0.4 m + 3 m = 3.4 mA (checks)
60,000
8000
[a] vo(0− ) = vo (0+ ) = 120 V
vo (∞) = −150 V;
τ = 2 ms;
1
= 500
τ
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7–46
CHAPTER 7. Response of First-Order RL and RC Circuits
vo = −150 + (120 − (−150))e−500t
vo = −150 + 270e−500t V,
t≥0
[b] io = −0.04 × 10−6 (−500)[270e−500t ] = 5.4e−500t mA,
t ≥ 0+
[c] vg = vo − 12.5 × 103 io = −150 + 202.5e−500t V
[d] vg (0+ ) = −150 + 202.5 = 52.5 V
Checks:
vg (0+ ) = io (0+ )[37.5 × 103 ] − 150 = 202.5 − 150 = 52.5 V
vg
= −3 + 4.05e−500t mA
50k
vg
=
= −1 + 1.35e−500t mA
150k
i50k =
i150k
-io + i50k + i150k + 4 = 0
P 7.53
(ok)
For t < 0
Simplify the circuit:
80/10,000 = 8 mA,
10 kΩk40 kΩk24 kΩ = 6 kΩ
8 mA − 3 mA = 5 mA
5 mA × 6 kΩ = 30 V
Thus, for t < 0
.·. vo (0− ) = vo(0+ ) = 30 V
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
7–47
t>0
Simplify the circuit:
8 mA + 2 mA = 10 mA
10 kk40 kk24 k = 6 kΩ
(10 mA)(6 kΩ) = 60 V
Thus, for t > 0
vo (∞) = −10 × 10−3 (6 × 103 ) = −60 V
τ = RC = (10 k)(0.05 µ) = 0.5 ms;
1
= 2000
τ
vo = vo (∞) + [vo(0+ ) − vo(∞)]e−t/τ = −60 + [30 − (−60)]e−2000t
= −60 + 90e−2000t V
P 7.54
t≥0
t < 0:
io (0− ) =
20
(10 × 10−3 ) = 2 mA;
100
vo(0− ) = (2 × 10−3 )(50,000) = 100 V
t = ∞:
io (∞) = −5 × 10
−3
20
= −1 mA;
100
vo (∞) = io (∞)(50,000) = −50 V
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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7–48
CHAPTER 7. Response of First-Order RL and RC Circuits
RTh = 50 kΩk50 kΩ = 25 kΩ;
C = 16 nF
1
= 2500
τ
τ = (25,000)(16 × 10−9 ) = 0.4 ms;
.·. vo (t) = −50 + 150e−2500t V,
t≥0
ic = C
dvo
= −6e−2500t mA,
dt
i50k =
vo
= −1 + 3e−2500t mA,
50,000
t ≥ 0+
io = ic + i50k = −(1 + 3e−2500t) mA,
P 7.55
t ≥ 0+
t ≥ 0+
[a] vc (0+ ) = 50 V
[b] Use voltage division to find the final value of voltage:
vc (∞) =
20
(−30) = −24 V
20 + 5
[c] Find the Thévenin equivalent with respect to the terminals of the
capacitor:
VTh = −24 V,
RTh = 20k5 = 4 Ω,
Therefore τ = ReqC = 4(25 × 10−9 ) = 0.1 µs
The simplified circuit for t > 0 is:
[d] i(0+ ) =
−24 − 50
= −18.5 A
4
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
7–49
[e] vc = vc (∞) + [vc (0+ ) − vc(∞)]e−t/τ
7
= −24 + [50 − (−24)]e−t/τ = −24 + 74e−10 t V,
[f] i = C
P 7.56
t≥0
dvc
7
7
= (25 × 10−9 )(−107 )(74e−10 t ) = −18.5e−10 t A,
dt
t ≥ 0+
[a] Use voltage division to find the initial value of the voltage:
9k
(120) = 90 V
9k + 3k
vc (0+ ) = v9k =
[b] Use Ohm’s law to find the final value of voltage:
vc (∞) = v40k = −(1.5 × 10−3 )(40 × 103 ) = −60 V
[c] Find the Thévenin equivalent with respect to the terminals of the
capacitor:
VTh = −60 V,
RTh = 10 k + 40 k = 50 kΩ
τ = RTh C = 1 ms = 1000 µs
[d] vc = vc (∞) + [vc(0+ ) − vc (∞)]e−t/τ
= −60 + (90 + 60)e−1000t = −60 + 150e−1000t V,
t≥0
We want vc = −60 + 150e−1000t = 0:
Therefore t =
P 7.57
ln(150/60)
= 916.3 µs
1000
Use voltage division to find the initial voltage:
vo (0) =
60
(50) = 30 V
40 + 60
Use Ohm’s law to find the final value of voltage:
vo (∞) = (−5 mA)(20 kΩ) = −100 V
τ = RC = (20 × 103 )(250 × 10−9 ) = 5 ms;
1
= 200
τ
vo = vo (∞) + [vo(0+ ) − vo(∞)]e−t/τ
= −100 + (30 + 100)e−200t = −100 + 130e−200t V,
t≥0
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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7–50
P 7.58
CHAPTER 7. Response of First-Order RL and RC Circuits
For t < 0,
t > 0:
vo (0) = 80 V
3
3
vTh = 30 × 10 i∆ + 0.8(100) = 30 × 10
−100
+ 80 = 50 V
100 × 103
vT = 30 × 103 i∆ + 16 × 103 iT = 30 × 103 (0.8)iT + 16 × 103 iT = 40 × 103 iT
RTh =
vT
= 40 kΩ
iT
t>0
vo = 50 + (80 − 50)e−t/τ
τ = RC = (40 × 103 )(5 × 10−9 ) = 200 × 10−6 ;
vo = 50 + 30e−5000t V,
P 7.59
vo (0) = 50 V;
1
= 5000
τ
t≥0
vo(∞) = 80 V
RTh = 16 kΩ
τ = (16)(5 × 10−6 ) = 80 × 10−6 ;
1
= 12,500
τ
v = 80 + (50 − 80)e−12,500t = 80 − 30e−12,500t V,
t≥0
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
P 7.60
7–51
For t > 0
VTh = (−25)(16,000)ib = −400 × 103 ib
ib =
33,000
(120 × 10−6 ) = 49.5 µA
80,000
VTh = −400 × 103 (49.5 × 10−6 ) = −19.8 V
RTh = 16 kΩ
vo(0+ ) = 0
vo (∞) = −19.8 V;
τ = (16, 000)(0.25 × 10−6 ) = 4 ms;
vo = −19.8 + 19.8e−250t V,
1/τ = 250
t≥0
1
w(t) = (0.25 × 10−6 )vo2 = w(∞)(1 − e−250t)2 J
2
(1 − e−250t)2 =
0.36w(∞)
= 0.36
w(∞)
1 − e−250t = 0.6
e−250t = 0.4
P 7.61
. ·.
t = 3.67 ms
[a]
1 t
Is R = Ri +
i dx + Vo
C 0+
di
i
0=R + +0
dt C
di
i
. ·.
+
=0
dt RC
Z
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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7–52
CHAPTER 7. Response of First-Order RL and RC Circuits
[b]
di
i
=−
;
dt
RC
Z
i(t)
i(0+ )
ln
di
dt
=−
i
RC
dy
1
=−
y
RC
t
Z
0+
dx
i(t)
−t
=
+
i(0 )
RC
i(t) = i(0+ )e−t/RC ;
i(0+ ) =
Is R − Vo
Vo
= Is −
R
R
Vo −t/RC
.·. i(t) = Is −
e
R
P 7.62
[a] Let i be the current Zin the clockwise
direction around the circuit. Then
Z
1 t
1 t
Vg = iRg +
i dx +
i dx
C1 0
C2 0
1
1
iRg +
+
C1 C2
=
Z
1
i dx = iRg +
Ce
0
t
Z
t
0
i dx
Now differentiate the equation
0 = Rg
di
i
+
dt Ce
Therefore i =
1
v1 (t) =
C1
0
t
di
1
+
i=0
dt Rg Ce
Vg −t/Rg Ce
Vg −t/τ
e
=
e
;
Rg
Rg
Vg −x/τ
Vg e−x/τ
e
dx =
Rg
Rg C1 −1/τ
τ = Rg Ce
t
=−
0
v1 (t) =
Vg C2
(1 − e−t/τ );
C1 + C2
τ = Rg Ce
v2 (t) =
Vg C1
(1 − e−t/τ );
C1 + C2
τ = Rg Ce
[b] v1(∞) =
P 7.63
Z
or
C2
Vg ;
C1 + C2
v2 (∞) =
Vg Ce −t/τ
(e
− 1)
C1
C1
Vg
C1 + C2
[a] For t > 0:
τ = RC = 250 × 103 × 8 × 10−9 = 2 ms;
1
= 500
τ
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
7–53
t ≥ 0+
vo = 50e−500t V,
vo
50e−500t
[b] io =
=
= 200e−500t µA
250,000
250,000
−1
× 200 × 10−6
v1 =
40 × 10−9
P 7.64
Z
t
0
e−500x dx + 50 = 10e−500t + 40 V,
t≥0
[a] t < 0
t>0
vo (0− ) = vo (0+ ) = 40 V
vo (∞) = 80 V
τ = (0.16 × 10−6 )(6.25 × 103 ) = 1 ms;
vo = 80 − 40e−1000t V,
[b] io = −C
t≥0
dvo
= −0.16 × 10−6 [40,000e−1000t ]
dt
= −6.4e−1000t mA;
−1
[c] v1 =
0.2 × 10−6
Z
0
t ≥ 0+
t
−6.4 × 10−3 e−1000x dx + 32
= 64 − 32e−1000t V,
[d] v2 =
1/τ = 1000
t≥0
Z t
−1
−6.4 × 10−3 e−1000x dx + 8
0.8 × 10−6 0
= 16 − 8e−1000t V,
t≥0
1
1
[e] wtrapped = (0.2 × 10−6 )(64)2 + (0.8 × 10−6 )(16)2 = 512 µJ.
2
2
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7–54
P 7.65
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] Leq =
τ=
(3)(15)
= 2.5 H
3 + 15
Leq
2.5
1
=
= s
R
7.5
3
io (0) = 0;
120
= 16 A
7.5
io (∞) =
.·. io = 16 − 16e−3t A,
t≥0
t ≥ 0+
vo = 120 − 7.5io = 120e−3t V,
i1 =
1
3
Z
0
40 40 −3t
− e A,
3
3
t
120e−3x dx =
i2 = io − i1 =
8 8 −3t
− e A,
3 3
t≥0
t≥0
[b] io(0) = i1 (0) = i2(0) = 0, consistent with initial conditions.
vo (0+ ) = 120 V, consistent with io (0) = 0.
vo = 3
di1
= 120e−3t V,
dt
t ≥ 0+
or
di2
= 120e−3t V,
t ≥ 0+
dt
The voltage solution is consistent with the current solutions.
vo = 15
λ1 = 3i1 = 40 − 40e−3t Wb-turns
λ2 = 15i2 = 40 − 40e−3t Wb-turns
.·. λ1 = λ2 as it must, since
vo =
dλ1
dλ2
=
dt
dt
λ1 (∞) = λ2 (∞) = 40 Wb-turns
λ1 (∞) = 3i1 (∞) = 3(40/3) = 40 Wb-turns
λ2 (∞) = 15i2 (∞) = 15(8/3) = 40 Wb-turns
.·. i1(∞) and i2(∞) are consistent with λ1 (∞) and λ2 (∞).
P 7.66
[a] Leq = 5 + 10 − 2.5(2) = 10 H
τ=
L
10
1
=
= ;
R
40
4
i = 2 − 2e−4t A,
1
=4
τ
t≥0
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
7–55
di1
di
di
− 2.5 = 2.5 = 2.5(8e−4t ) = 20e−4t V, t ≥ 0+
dt
dt
dt
di1
di
di
[c] v2 (t) = 10
− 2.5 = 7.5 = 7.5(8e−4t ) = 60e−4t V, t ≥ 0+
dt
dt
dt
[d] i(0) = 2 − 2 = 0, which agrees with initial conditions.
[b] v1(t) = 5
80 = 40i1 + v1 + v2 = 40(2 − 2e−4t ) + 20e−4t + 60e−4t = 80 V
Therefore, Kirchhoff’s voltage law is satisfied for all values of t ≥ 0.
Thus, the answers make sense in terms of known circuit behavior.
P 7.67
[a] Leq = 5 + 10 + 2.5(2) = 20 H
τ=
L
20
1
=
= ;
R
40
2
i = 2 − 2e−2t A,
1
=2
τ
t≥0
di1
di
di
+ 2.5 = 7.5 = 7.5(4e−2t ) = 30e−2t V, t ≥ 0+
dt
dt
dt
di1
di
di
[c] v2 (t) = 10
+ 2.5 = 12.5 = 12.5(4e−2t ) = 50e−2t V, t ≥ 0+
dt
dt
dt
[d] i(0) = 0, which agrees with initial conditions.
[b] v1(t) = 5
80 = 40i1 + v1 + v2 = 40(2 − 2e−2t ) + 30e−2t + 50e−2t = 80 V
Therefore, Kirchhoff’s voltage law is satisfied for all values of t ≥ 0.
Thus, the answers make sense in terms of known circuit behavior.
P 7.68
[a] From Example 7.10,
Leq =
τ=
L1 L2 − M 2
50 − 25
=
= 1H
L1 + L2 + 2M
15 + 10
L
1
= ;
R
20
1
= 20
τ
.·. io(t) = 4 − 4e−20t A,
t≥0
[b] vo = 80 − 20io = 80 − 80 + 80e−20t = 80e−20t V,
di1
di2
[c] vo = 5
−5
= 80e−20t V
dt
dt
t ≥ 0+
io = i1 + i2
dio
di1 di2
=
+
= 80e−20t A/s
dt
dt
dt
. ·.
di2
di1
= 80e−20t −
dt
dt
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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7–56
CHAPTER 7. Response of First-Order RL and RC Circuits
.·. 80e−20t = 5
.·. 10
Z
di1
= 480e−20t ;
dt
t1
dx =
0
i1 =
di1
di1
− 400e−20t + 5
dt
dt
Z
di1 = 48e−20t dt
t
0
48e−20y dy
48 −20y t
e
= 2.4 − 2.4e−20t A,
−20
0
t≥0
[d] i2 = io − i1 = 4 − 4e−20t − 2.4 + 2.4e−20t
= 1.6 − 1.6e−20t A,
t≥0
[e] io (0) = i1(0) = i2(0) = 0, consistent with zero initial stored energy.
vo = Leq
dio
= 1(80)e−20t = 80e−20t V,
dt
t ≥ 0+ (checks)
Also,
vo = 5
di1
di2
−5
= 80e−20t V,
dt
dt
vo = 10
di2
di1
−5
= 80e−20t V,
dt
dt
t ≥ 0+ (checks)
t ≥ 0+ (checks)
vo (0+ ) = 80 V, which agrees with io (0+ ) = 0 A
io (∞) = 4 A;
io (∞)Leq = (4)(1) = 4 Wb-turns
i1 (∞)L1 + i2 (∞)M = (2.4)(5) + (1.6)(−5) = 4 Wb-turns (ok)
i2 (∞)L2 + i1 (∞)M = (1.6)(10) + (2.4)(−5) = 4 Wb-turns (ok)
Therefore, the final values of io , i1 , and i2 are consistent with
conservation of flux linkage. Hence, the answers make sense in terms of
known circuit behavior.
P 7.69
[a] From Example 7.10,
Leq =
τ=
L1 L2 − M 2
0.125 − 0.0625
=
= 50 mH
L1 + L2 + 2M
0.75 + 0.5
L
1
=
;
R
5000
1
= 5000
τ
.·. io(t) = 40 − 40e−5000t mA,
t≥0
[b] vo = 10 − 250io = 10 − 250(0.04 + 0.04e−5000t = 10e−5000t V,
t ≥ 0+
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
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Problems
[c] vo = 0.5
7–57
di1
di2
− 0.25
= 10e−5000t V
dt
dt
io = i1 + i2
dio
di1 di2
=
+
= 200e−5000t A/s
dt
dt
dt
di2
di1
= 200e−5000t −
dt
dt
. ·.
.·. 10e−5000t = 0.5
.·. 0.75
Z
di1
= 60e−5000t;
dt
t1
dx =
0
i1 =
di1
di1
− 50e−5000t + 0.25
dt
dt
Z
0
di1 = 80e−5000t dt
t
80e−5000y dy
80 −5000y t
e
= 16 − 16e−5000t mA,
−5000
0
t≥0
[d] i2 = io − i1 = 40 − 40e−5000t − 16 + 16e−5000t
= 24 − 24e−5000t mA,
t≥0
[e] io (0) = i1(0) = i2(0) = 0, consistent with zero initial stored energy.
vo = Leq
dio
= (0.05)(200)e−5000t = 10e−5000t V,
dt
t ≥ 0+ (checks)
Also,
vo = 0.5
di1
di2
− 0.25
= 10e−5000t V,
dt
dt
vo = 0.25
di2
di1
− 0.25
= 10e−5000t V,
dt
dt
t ≥ 0+ (checks)
t ≥ 0+ (checks)
vo (0+ ) = 10 V, which agrees with io (0+ ) = 0 A
io (∞) = 40 mA;
io (∞)Leq = (0.04)(0.05) = 2 mWb-turns
i1 (∞)L1 + i2 (∞)M = (16 m)(500) + (24 m)(−250) = 2 mWb-turns (ok)
i2 (∞)L2 + i1 (∞)M = (24 m)(250) + (16 m)(−250) = 2 mWb-turns (ok)
Therefore, the final values of io , i1 , and i2 are consistent with
conservation of flux linkage. Hence, the answers make sense in terms of
known circuit behavior.
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
www.EngineeringEBooksPdf.com
7–58
P 7.70
CHAPTER 7. Response of First-Order RL and RC Circuits
t < 0:
iL (0− ) = 10 V/5 Ω = 2 A = iL(0+ )
0 ≤ t ≤ 5:
τ = 5/0 = ∞
iL (t) = 2e−t/∞ = 2e−0 = 2
iL (t) = 2 A 0 ≤ t ≤ 5 s
5 ≤ t < ∞:
τ=
5
= 5 s;
1
1/τ = 0.2
iL (t) = 2e−0.2(t −5) A,
P 7.71
t ≥ 5s
For t < 0:
i(0) =
10
(15) = 10 A
15
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Problems
7–59
0 ≤ t ≤ 10 ms:
i = 10e−100t A
i(10 ms) = 10e−1 = 3.68 A
10 ms ≤ t ≤ 20 ms:
Req =
(5)(20)
= 4Ω
25
1
R
4
=
=
= 80
τ
L
50 × 10−3
i = 3.68e−80(t−0.01) A
20 ms ≤ t < ∞:
i(20 ms) = 3.68e−80(0.02−0.01) = 1.65 A
i = 1.65e−100(t−0.02) A
vo = L
di
;
dt
L = 50 mH
di
= 1.65(−100)e−100(t−0.02) = −165e−100(t−0.02)
dt
vo = (50 × 10−3 )(−165)e−100(t−0.02)
= −8.26e−100(t−0.02) V,
t > 20+ ms
vo (25 ms) = −8.26e−100(0.025−0.02) = −5.013 V
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7–60
P 7.72
CHAPTER 7. Response of First-Order RL and RC Circuits
From the solution to Problem 7.71, the initial energy is
1
w(0) = (50 mH)(10 A)2 = 2.5 J
2
0.04w(0) = 0.1 J
1
.·.
(50 × 10−3 )i2L = 0.1 so iL = 2 A
2
Again, from the solution to Problem 7.73, t must be between 10 ms and 20 ms
since
i(10 ms) = 3.68 A and
i(20 ms) = 1.65 A
For 10 ms ≤ t ≤ 20 ms:
i = 3.68e−80(t−0.01) = 2
e80(t−0.01) =
P 7.73
3.68
2
so t − 0.01 = 0.0076
.·.
t = 17.6 ms
[a] t < 0:
Using Ohm’s law,
800
ig =
= 12.5 A
40 + 60k40
Using current division,
60
i(0− ) =
(12.5) = 7.5 A = i(0+ )
60 + 40
[b] 0 ≤ t ≤ 1 ms:
i = i(0+ )e−t/τ = 7.5e−t/τ
1
R
40 + 120k60
=
=
= 1000
τ
L
80 × 10−3
i = 7.5e−1000t
i(200µs) = 7.5e−10
3 (200×10−6 )
= 7.5e−0.2 = 6.14 A
[c] i(1 ms) = 7.5e−1 = 2.7591 A
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Problems
7–61
1 ms ≤ t < ∞:
1
R
40
=
=
= 500
τ
L
80 × 10−3
i = i(1 ms)e−(t−1 ms)/τ = 2.7591e−500(t−0.001) A
i(6ms) = 2.7591e−500(0.005) = 2.7591e−2.5 = 226.48 mA
[d] 0 ≤ t ≤ 1 ms:
i = 7.5e−1000t
v=L
di
= (80 × 10−3 )(−1000)(7.5e−1000t ) = −600e−1000t V
dt
v(1− ms) = −600e−1 = −220.73 V
[e] 1 ms ≤ t ≤ ∞:
i = 2.759e−500(t−0.001)
v=L
di
= (80 × 10−3 )(−500)(2.759e−500(t−0.001))
dt
= −110.4e−500(t−0.001) V
v(1+ ms) = −110.4 V
P 7.74
0 ≤ t ≤ 10 µs:
τ = RC = (4 × 103 )(20 × 10−9 ) = 80 µs;
vo (0) = 0 V;
1/τ = 12,500
vo (∞) = −20 V
vo = −20 + 20e−12,500t V
0 ≤ t ≤ 10 µs
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7–62
CHAPTER 7. Response of First-Order RL and RC Circuits
10 µs ≤ t < ∞:
t → ∞:
i=
−50 V
= −2.5 mA
20 kΩ
vo (∞) = (−2.5 × 10−3 )(16,000) + 30 = −10 V
vo (10 µs) = −20 + 20−0.125 = −2.35 V
vo = −10 + (−2.35 + 10)e−(t − 10×10
−6 )/τ
RTh = 4 kΩk16 kΩ = 3.2 kΩ
τ = (3200)(20 × 10−9 ) = 64 µs;
vo = −10 + 7.65e−15,625(t − 10×10
P 7.75
1/τ = 15,625
−6 )
10 µs ≤ t < ∞
0 ≤ t ≤ 200 µs;
Re = 150k100 = 60 kΩ;
10
τ=
× 10−9 (60,000) = 200 µs
3
vc = 300e−5000t V
vc (200 µs) = 300e−1 = 110.36 V
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Problems
7–63
200 µs ≤ t < ∞:
Re = 30k60 + 120k40 = 20 + 30 = 50 kΩ
10
τ=
× 10−9 (50,000) = 166.67 µs;
3
1
= 6000
τ
vc = 110.36e−6000(t − 200 µs) V
vc (300 µs) = 110.36e−6000(100 µs) = 60.57 V
io (300 µs) =
i1 =
60.57
= 1.21 mA
50,000
60
2
io = io ;
90
3
i2 =
40
1
io = io
160
4
2
1
5
5
isw = i1 − i2 = io − io = io = (1.21 × 10−3 ) = 0.50 mA
3
4
12
12
P 7.76
Note that for t > 0, vo = (4/6)vc , where vc is the voltage across the 0.5 µF
capacitor. Thus we will find vc first.
t<0
vc(0) =
3
(−75) = −15 V
15
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7–64
CHAPTER 7. Response of First-Order RL and RC Circuits
0 ≤ t ≤ 800 µs:
τ = Re C,
Re =
(6000)(3000)
= 2 kΩ
9000
1
= 1000
τ
τ = (2 × 103 )(0.5 × 10−6 ) = 1 ms,
vc = −15e−1000t V,
t≥0
vc(800 µs) = −15e−0.8 = −6.74 V
800 µs ≤ t ≤ 1.1 ms:
1
= 333.33
τ
τ = (6 × 103 )(0.5 × 10−6 ) = 3 ms,
vc = −6.74e−333.33(t−800×10
−6)
V
1.1 ms ≤ t < ∞:
τ = 1 ms,
1
= 1000
τ
−6
vc(1.1ms) = −6.74e−333.33(1100−800)10
vc = −6.1e−1000(t−1.1×10
−3)
= −6.74e−0.1 = −6.1 V
V
vc(1.5ms) = −6.1e−1000(1.5−1.1)10
−3
= −6.1e−0.4 = −4.09 V
vo = (4/6)(−4.09) = −2.73 V
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Problems
P 7.77
7–65
1
w(0) = (0.5 × 10−6 )(−15)2 = 56.25 µJ
2
0 ≤ t ≤ 800 µs:
vc2 = 225e−2000t
vc = −15e−1000t;
p3k = 75e−2000t mW
w3k =
Z
800×10−6
75 × 10−3 e−2000t dt
0
−3 e
800×10−6
−2000t
= 75 × 10
−2000 0
= −37.5 × 10−6 (e−1.6 − 1) = 29.93 µJ
1.1 ms ≤ t ≤ ∞:
vc = −6.1e−1000(t−1.1×10
−3)
V;
−3)
mW
p3k = 12.4e−2000(t−1.1×10
w3k =
Z
vc2 = 37.19e−2000(t−1.1×10
−3 )
∞
1.1×10−3
12.4 × 10−3 e−2000(t−1.1×10
e−2000(t−1.1×10 )
−2000
−6
= −6.2 × 10 (0 − 1) = 6.2 µJ
−3
−3)
dt
∞
= 12.4 × 10−3
1.1×10−3
w3k = 29.93 + 6.2 = 36.13 µJ
%=
P 7.78
36.13
(100) = 64.23%
56.25
t < 0:
vc (0− ) = −(5)(1000) × 10−3 = −5 V = vc (0+ )
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7–66
CHAPTER 7. Response of First-Order RL and RC Circuits
0 ≤ t ≤ 5 s:
τ = ∞;
1/τ = 0;
vo = −5e−0 = −5 V
5 s ≤ t < ∞:
τ = (100)(0.1) = 10 s;
vo = −5e−0.1(t − 5) V
1/τ = 0.1;
Summary:
vo = −5 V,
0 ≤ t ≤ 5s
vo = −5e−0.1(t − 5) V,
P 7.79
5s ≤ t < ∞
[a] 0 ≤ t ≤ 2.5 ms
vo (0+ ) = 80 V;
vo (∞) = 0
L
= 2 ms;
R
1/τ = 500
τ=
vo (t) = 80e−500t V,
0+ ≤ t ≤ 2.5− ms
vo (2.5− ms) = 80e−1.25 = 22.92 V
io (2.5− ms) =
(80 − 22.92)
= 2.85 A
20
vo (2.5+ ms) = −20(2.85) = −57.08 V
vo (∞) = 0;
τ = 2 ms;
vo = −57.08e−500(t − 0.0025) V
1/τ = 500
t ≥ 2.5+ ms
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Problems
7–67
[b]
[c] vo (5 ms) = −16.35 V
io =
P 7.80
+16.35
= 817.68 mA
20
[a] io(0) = 0;
io (∞) = 25 mA
1
R
2000
=
=
× 103 = 8000
τ
L
250
io = (25 − 25e−8000t) mA,
vo = 0.25
0 ≤ t ≤ 75 µs
dio
= 50e−8000t V,
dt
0 ≤ t ≤ 75 µs
75 µs ≤ t < ∞:
io (75µs) = 25 − 25e−0.6 = 11.28 mA;
io = 11.28e−8000(t−75×10
vo = (0.25)
−6 )
io (∞) = 0
mA
dio
= −22.56e−8000(t−75µs)
dt
. ·. t < 0 :
vo
= 0
0 ≤ t ≤ 75 µs :
vo
= 50e−8000t V
75 µs ≤ t < ∞ :
vo
= −22.56e−8000(t−75µs)
[b] vo(75− µs) = 50e−0.6 = 27.44 V
vo (75+ µs) = −22.56 V
[c] io (75− µs) = io(75+ µs) = 11.28 mA
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7–68
P 7.81
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] 0 ≤ t ≤ 1 ms:
vc (0+ ) = 0;
vc (∞) = 50 V;
RC = 400 × 103 (0.01 × 10−6 ) = 4 ms;
1/RC = 250
vc = 50 − 50e−250t
vo = 50 − 50 + 50e−250t = 50e−250t V,
0 ≤ t ≤ 1 ms
1 ms ≤ t < ∞:
vc (1 ms) = 50 − 50e−0.25 = 11.06 V
vc (∞) = 0 V
τ = 4 ms;
1/τ = 250
vc = 11.06e−250(t − 0.001) V
vo = −vc = −11.06e−250(t − 0.001) V,
t ≥ 1 ms
[b]
P 7.82
[a] t < 0;
vo = 0
0 ≤ t ≤ 4 ms:
τ = (200 × 103 )(0.025 × 10−6 ) = 5 ms;
vo = 100 − 100e−200t V,
1/τ = 200
0 ≤ t ≤ 4 ms
vo (4 ms) = 100(1 − e−0.8 ) = 55.07 V
4 ms ≤ t ≤ 8 ms:
vo = −100 + 155.07e−200(t−0.004) V,
4 ms ≤ t ≤ 8 ms
vo (8 ms) = −100 + 155.07e−0.8 = −30.32 V
t ≥ 8 ms:
vo = −30.32e−200(t−0.008) V,
t ≥ 8 ms
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Problems
7–69
[b]
[c] t ≤ 0 :
vo = 0
0 ≤ t ≤ 4 ms:
τ = (50 × 103 )(0.025 × 10−6 ) = 1.25 ms 1/τ = 800
vo = 100 − 100e−800t V,
0 ≤ t ≤ 4 ms
vo (4 ms) = 100 − 100e−3.2 = 95.92 V
4 ms ≤ t ≤ 8 ms:
vo = −100 + 195.92e−800(t−0.004) V,
4 ms ≤ t ≤ 8 ms
vo (8 ms) = −100 + 195.92e−3.2 = −92.01 V
t ≥ 8 ms:
vo = −92.01e−800(t−0.008) V,
P 7.83
t ≥ 8 ms
[a] τ = RC = (20,000)(0.2 × 10−6 ) = 4 ms;
io = vo = 0
t<0
16
io (0 ) = 20
= 16 mA,
20
+
1/τ = 250
.·. io = 16e−250t mA
io(∞) = 0
0+ ≤ t ≤ 2− ms
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7–70
CHAPTER 7. Response of First-Order RL and RC Circuits
i16kΩ = 20 − 16e−250t mA
.·. vo = 320 − 256e−250t V
0+ ≤ t ≤ 2− ms
vc = vo − 4 × 103 io = 320 − 320e−250t V
0 ≤ t ≤ 2 ms
vc (2 ms) = 320 − 320e−0.5 = 125.91 V
.·. io(2+ ms) = 16e−0.5 = 9.7 mA
io (∞) = 0
vc = 125.91e−250(t−0.002),
io = C
t ≥ 2 ms
dvc
= (0.2 × 10−6 )(−250)(125.91)e−250(t−0.002)
dt
= −6.3e−250(t−0.002) mA,
t ≥ 2+ ms
vo = 4000io + vc = 100.73e−250(t −0.002) V
t ≥ 2+ ms
Summary part (a)
io = 0
t<0
(0+ ≤ t ≤ 2− ms)
io = 16e−250t mA
t ≥ 2+ ms
io = −6.3e−250(t −0.002) mA
vo = 0
t<0
vo = 320 − 256e−250t V,
0+ ≤ t ≤ 2− ms
vo = 100.73e−250(t −0.002) V,
t ≥ 2+ ms
[b] io(0− ) = 0
io (0+ ) = 16 mA
io (2− ms) = 16e−0.5 = 9.7 mA
io (2+ ms) = −6.3 mA
[c] vo (0− ) = 0
vo (0+ ) = 64 V
vo (2− ms) = 320 − 256e−0.5 = 164.73 V
vo (2+ ms) = 100.73
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Problems
7–71
[d]
[e]
P 7.84
t > 0:
vT = 12 × 104 i∆ + 16 × 103 iT
i∆ = −
20
iT = −0.2iT
100
.·. vT = −24 × 103 iT + 16 × 103 iT
RTh =
vT
= −8 kΩ
iT
τ = RC = (−8 × 103 )(2.5 × 10−6 ) = −0.02 1/τ = −50
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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7–72
CHAPTER 7. Response of First-Order RL and RC Circuits
vc = 20e50t V;
50t = ln 1000
P 7.85
20e50t = 20,000
.·.
t = 138.16 ms
Find the Thévenin equivalent with respect to the terminals of the capacitor.
RTh calculation:
iT =
.·.
vT
vT
vT
+
−4
2000 5000
5000
iT
5+2−8
1
=
=−
vT
10,000
10,000
vT
10,000
=−
= −10 kΩ
iT
1
Open circuit voltage calculation:
The node voltage equations:
voc
voc − v1
+
− 4i∆ = 0
2000
1000
v1 − voc
v1
+
− 5 × 10−3 = 0
1000
4000
The constraint equation:
i∆ =
v1
4000
Solving, voc = −80 V,
v1 = −60 V
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Problems
vc(0) = 0;
7–73
vc(∞) = −80 V
τ = RC = (−10,000)(1.6 × 10−6 ) = −16 ms;
1
= −62.5
τ
vc = vc (∞) + [vc(0+ ) − vc (∞)]e−t/τ = −80 + 80e62.5t = 14,400
Solve for the time of the maximum voltage rating:
e62.5t = 181;
62.5t = ln 181;
t = 83.09 ms
P 7.86
vT = 2000iT + 4000(iT − 2 × 10−3 vφ ) = 6000iT − 8vφ
= 6000iT − 8(2000iT )
vT
= −10,000
iT
τ=
10
= −1 ms;
−10,000
1/τ = −1000
i = 25e1000t mA
.·. 25e1000t × 10−3 = 5;
t=
ln 200
= 5.3 ms
1000
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7–74
P 7.87
CHAPTER 7. Response of First-Order RL and RC Circuits
[a]
Using Ohm’s law,
vT = 5000iσ
Using current division,
iσ =
20,000
(iT + βiσ ) = 0.8iT + 0.8βiσ
20,000 + 5000
Solve for iσ :
iσ (1 − 0.8β) = 0.8iT
iσ =
0.8iT
;
1 − 0.8β
vT = 5000iσ =
4000iT
(1 − 0.8β)
Find β such that RTh = −5 kΩ:
RTh =
vT
4000
=
= −5000
iT
1 − 0.8β
.·. β = 2.25
1 − 0.8β = −0.8
[b] Find VTh ;
Write a KCL equation at the top node:
VTh − 40
VTh
+
− 2.25iσ = 0
5000
20,000
The constraint equation is:
(VTh − 40)
=0
5000
Solving,
iσ =
VTh = 50 V
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Problems
7–75
Write a KVL equation around the loop:
50 = −5000i + 0.2
di
dt
Rearranging:
di
= 250 + 25,000i = 25,000(i + 0.01)
dt
Separate the variables and integrate to find i;
di
= 25,000 dt
i + 0.01
Z
0
i
dx
=
x + 0.01
Z
0
t
25,000 dx
.·. i = −10 + 10e25,000t mA
di
= (10 × 10−3 )(25,000)e25,000t = 250e25,000t
dt
Solve for the arc time:
di
v = 0.2 = 50e25,000t = 45,000;
dt
. ·. t =
P 7.88
e25,000t = 900
ln 900
= 272.1 µs
25,000
[a]
τ = (25)(2) × 10−3 = 50 ms;
vc (0+ ) = 80 V;
1/τ = 20
vc (∞) = 0
vc = 80e−20t V
.·. 80e−20t = 5;
e20t = 16;
t=
ln 16
= 138.63 ms
20
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7–76
CHAPTER 7. Response of First-Order RL and RC Circuits
[b] 0+ ≤ t ≤ 138.63− ms:
i = (2 × 10−6 )(−1600e−20t ) = −3.2e−20t mA
t ≥ 138.63+ ms:
τ = (2)(4) × 10−3 = 8 ms;
vc (138.63+ ms) = 5 V;
1/τ = 125
vc (∞) = 80 V
vc = 80 − 75e−125(t−0.13863) V,
t ≥ 138.63 ms
i = 2 × 10−6 (9375)e−125(t−0.13863)
= 18.75e−125(t−0.13863) mA,
t ≥ 138.63+ ms
[c] 80 − 75e−125∆t = 0.85(80) = 68
80 − 68 = 75e−125∆t = 12
e125∆t = 6.25;
P 7.89
∆t =
ln 6.25 ∼
= 14.66 ms
125
[a] RC = (25 × 103 )(0.4 × 10−6 ) = 10 ms;
vo = 0,
1
= 100
RC
t<0
[b] 0 ≤ t ≤ 250 ms :
vo = −100
Z
t
0
−0.20 dx = 20t V
[c] 250 ms ≤ t ≤ 500 ms;
vo (0.25) = 20(0.25) = 5 V
vo (t) = −100
Z
t
0.25
0.20 dx + 5 = −20(t − 0.25) + 5 = −20t + 10 V
[d] t ≥ 500 ms :
vo (0.5) = −10 + 10 = 0 V
vo (t) = 0 V
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Problems
P 7.90
[a] vo = 0,
t<0
1
= 100
RC
RC = (25 × 103 )(0.4 × 10−6 ) = 10 ms
[b] Rf Cf = (5 × 106 )(0.4 × 10−6 ) = 2;
vo =
7–77
1
= 0.5
Rf Cf
−5 × 106
(−0.2)[1 − e−0.5t] = 40(1 − e−0.5t) V,
25 × 103
0 ≤ t ≤ 250 ms
[c] vo (0.25) = 40(1 − e−0.125) ∼
= 4.70 V
−Vm Rf
Vm Rf
+
(2 − e−0.125)e−0.5(t−0.25)
Rs
Rs
= −40 + 40(2 − e−0.125)e−0.5(t−0.25)
= −40 + 44.70e−0.5(t−0.25) V,
250 ms ≤ t ≤ 500 ms
vo =
[d] vo(0.5) = −40 + 44.70e−0.125 ∼
= −0.55 V
vo = −0.55e−0.5(t−0.5) V,
t ≥ 500 ms
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7–78
P 7.91
CHAPTER 7. Response of First-Order RL and RC Circuits
1
vo = −
R(0.5 × 10−6 )
Z
t
4 dx + 0 =
0
−4t
R(0.5 × 10−6 )
−4(15 × 10−3 )
= −10
R(0.5 × 10−6 )
.·.
P 7.92
vo =
.·.
P 7.93
R=
−4(15 × 10−3 )
= 12 kΩ
−10(0.5 × 10−6 )
−4t
−4(40 × 10−3 )
+
6
=
+ 6 = −10
R(0.5 × 10−6 )
R(0.5 × 10−6 )
R=
−4(40 × 10−3 )
= 20 kΩ
−16(0.5 × 10−6 )
1
= 1,250,000
RC
[a] RC = (1000)(800 × 10−12 ) = 800 × 10−9 ;
0 ≤ t ≤ 1 µs:
vg = 2 × 106 t
6
vo = −1.25 × 10
= −2.5 × 1012
t
Z
2 × 106 x dx + 0
0
x2 t
2
= −125 × 1010 t2 V,
0 ≤ t ≤ 1 µs
0
10
vo (1 µs) = −125 × 10 (1 × 10−6 )2 = −1.25 V
1 µs ≤ t ≤ 3 µs:
vg = 4 − 2 × 106 t
4
vo = −125 × 10
Z
t
1×10−6
"
t
(4 − 2 × 106 x) dx − 1.25
x2 t
= −125 × 104 4x
−2 × 106
− 1.25
2 1×10−6
1×10−6
= −5 × 106 t + 5 + 125 × 1010 t2 − 1.25 − 1.25
= 125 × 1010 t2 − 5 × 106 t + 2.5 V, 1 µs ≤ t ≤ 3 µs
#
vo (3 µs) = 125 × 1010 (3 × 10−6 )2 − 5 × 106 (3 × 10−6 ) + 2.5
= −1.25
3 µs ≤ t ≤ 4 µs:
vg = −8 + 2 × 106 t
4
vo = −125 × 10
Z
t
(−8 + 2 × 106 x) dx − 1.25
3×10−6
"
t
x2 t
= −125 × 104 −8x
+2 × 106
− 1.25
2 3×10−6
3×10−6
= 107 t − 30 − 125 × 1010 t2 + 11.25 − 1.25
= −125 × 1010 t2 + 107 t − 20 V, 3 µs ≤ t ≤ 4 µs
#
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Problems
7–79
vo (4 µs) = −125 × 1010 (4 × 10−6 )2 + 107 (4 × 10−6 ) − 20 = 0
[b]
[c] The output voltage will also repeat. This follows from the observation that
at t = 4 µs the output voltage is zero, hence there is no energy stored in
the capacitor. This means the circuit is in the same state at t = 4 µs as it
was at t = 0, thus as vg repeats itself, so will vo .
P 7.94
[a]
Cdvp vp − vb
+
= 0;
dt
R
therefore
dvp
1
vb
+
vp =
dt
RC
RC
vn − va
d(vn − vo )
+C
= 0;
R
dt
dvo
dvn
vn
va
=
+
−
dt
dt
RC
RC
therefore
But vn = vp
Therefore
dvn
vn
dvp
vp
vb
+
=
+
=
dt
RC
dt
RC
RC
Therefore
dvo
1
=
(vb − va );
dt
RC
1
vo =
RC
Z
0
t
(vb − va) dy
[b] The output is the integral of the difference between vb and va and then
scaled by a factor of 1/RC.
Z t
1
[c] vo =
(vb − va ) dx
RC 0
RC = (50 × 103 )(10 × 10−9 ) = 0.5 ms
vb − va = −25 mV
vo =
1
0.0005
Z
0
−50tsat = −6;
t
−25 × 10−3 dx = −50t
tsat = 120 ms
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7–80
P 7.95
CHAPTER 7. Response of First-Order RL and RC Circuits
The equation for an integrating amplifier:
1
vo =
RC
t
Z
(vb − va) dy + vo(0)
0
Find the values and substitute them into the equation:
RC = (100 × 103 )(0.05 × 10−6 ) = 5 ms
1
= 200;
RC
vb − va = −15 − (−7) = −8 V
vo (0) = −4 + 12 = 8 V
vo = 200
Z
t
0
−8 dx + 8 = (−1600t + 8) V,
0 ≤ t ≤ tsat
RC circuit analysis for v2:
v2(0+ ) = −4 V;
v2(∞) = −15 V;
τ = RC = (100 k)(0.05 µ) = 5 ms
v2 = v2(∞) + [v2(0+ ) − v2(∞)]e−t/τ
= −15 + (−4 + 15)e−200t = −15 + 11e−200t V,
vf + v2 = vo
.·.
0 ≤ t ≤ tsat
vf = vo − v2 = 23 − 1600t − 11e−200t V,
0 ≤ t ≤ tsat
Note that
−1600tsat + 8 = −20
. ·.
tsat =
−28
= 17.5 ms
−1600
so the op amp operates in its linear region until it saturates at 17.5 ms.
P 7.96
Use voltage division to find the voltage at the non-inverting terminal:
vp =
80
(−45) = −36 V = vn
100
Write a KCL equation at the inverting terminal:
−36 − 14
d
+ 2.5 × 10−6 (−36 − vo) = 0
80,000
dt
.·.
2.5 × 10−6
dvo
−50
=
dt
80,000
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Problems
7–81
Separate the variables and integrate:
dvo
= −250
dt
Z
vo (t)
vo (0)
.·.
dx = −250
dvo = −250dt
Z
t
. ·.
dy
0
vo (t) − vo(0) = −250t
vo (0) = −36 + 56 = 20 V
vo (t) = −250t + 20
Find the time when the voltage reaches 0:
0 = −250t + 20
P 7.97
.·.
t=
20
= 80 ms
250
[a] T2 is normally ON since its base current ib2 is greater than zero, i.e.,
ib2 = VCC /R when T2 is ON. When T2 is ON, vce2 = 0, therefore ib1 = 0.
When ib1 = 0, T1 is OFF. When T1 is OFF and T2 is ON, the capacitor C
is charged to VCC , positive at the left terminal. This is a stable state;
there is nothing to disturb this condition if the circuit is left to itself.
[b] When S is closed momentarily, vbe2 is changed to −VCC and T2 snaps
OFF. The instant T2 turns OFF, vce2 jumps to VCC R1 /(R1 + RL ) and ib1
jumps to VCC /(R1 + RL ), which turns T1 ON.
[c] As soon as T1 turns ON, the charge on C starts to reverse polarity. Since
vbe2 is the same as the voltage across C, it starts to increase from −VCC
toward +VCC . However, T2 turns ON as soon as vbe2 = 0. The equation
for vbe2 is vbe2 = VCC − 2VCC e−t/RC . vbe2 = 0 when t = RC ln 2, therefore
T2 stays OFF for RC ln 2 seconds.
P 7.98
[a] For t < 0, vce2 = 0. When the switch is momentarily closed, vce2 jumps to
vce2 =
VCC
6(5)
R1 =
= 1.2 V
R1 + RL
25
T2 remains open for (23,083)(250) × 10−12 ln 2 ∼
= 4 µs.
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7–82
CHAPTER 7. Response of First-Order RL and RC Circuits
[b] ib2 =
VCC
= 259.93 µA,
R
ib2 = 0,
ib2
P 7.99
−5 ≤ t ≤ 0 µs
0 < t < RC ln 2
=
VCC
VCC −(t−RC ln 2)/RLC
+
e
R
RL
=
259.93 + 300e−0.2×10
6 (t−4×10−6 )
µA,
RC ln 2 < t
[a] While T2 has been ON, C2 is charged to VCC , positive on the left terminal.
At the instant T1 turns ON the capacitor C2 is connected across b2 − e2 ,
thus vbe2 = −VCC . This negative voltage snaps T2 OFF. Now the polarity
of the voltage on C2 starts to reverse, that is, the right-hand terminal of
C2 starts to charge toward +VCC . At the same time, C1 is charging
toward VCC , positive on the right. At the instant the charge on C2
reaches zero, vbe2 is zero, T2 turns ON. This makes vbe1 = −VCC and T1
snaps OFF. Now the capacitors C1 and C2 start to charge with the
polarities to turn T1 ON and T2 OFF. This switching action repeats itself
over and over as long as the circuit is energized. At the instant T1 turns
ON, the voltage controlling the state of T2 is governed by the following
circuit:
It follows that vbe2 = VCC − 2VCC e−t/R2 C2 .
[b] While T2 is OFF and T1 is ON, the output voltage vce2 is the same as the
voltage across C1 , thus
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Problems
7–83
It follows that vce2 = VCC − VCC e−t/RLC1 .
[c] T2 will be OFF until vbe2 reaches zero. As soon as vbe2 is zero, ib2 will
become positive and turn T2 ON. vbe2 = 0 when VCC − 2VCC e−t/R2C2 = 0,
or when t = R2 C2 ln 2.
[d] When t = R2C2 ln 2,
we have
vce2 = VCC − VCC e−[(R2 C2 ln 2)/(RLC1 )] = VCC − VCC e−10 ln 2 ∼
= VCC
[e] Before T1 turns ON, ib1 is zero. At the instant T1 turns ON, we have
ib1 =
VCC
VCC −t/RLC1
+
e
R1
RL
[f] At the instant T2 turns back ON, t = R2 C2 ln 2; therefore
ib1 =
VCC
VCC −10 ln 2 ∼ VCC
+
e
=
R1
RL
R1
[g]
[h]
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7–84
CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.100 [a] tOFF2 = R2 C2 ln 2 = 18 × 103 (2 × 10−9 ) ln 2 ∼
= 25 µs
[b] tON2 = R1 C1 ln 2 ∼
= 25 µs
[c] tOFF1 = R1 C1 ln 2 ∼
= 25 µs
[d] tON1 = R2 C2 ln 2 ∼
= 25 µs
9
9
+
= 3.5 mA
3 18
9
9
[f] ib1 =
+ e−6 ln 2 ∼
= 0.5469 mA
18 3
[g] vce2 = 9 − 9e−6 ln 2 ∼
= 8.86 V
[e] ib1 =
P 7.101 [a] tOFF2 = R2 C2 ln 2 = (18 × 103 )(2.8 × 10−9 ) ln 2 ∼
= 35 µs
[b] tON2 = R1 C1 ln 2 ∼
= 37.4 µs
[c] tOFF1 = R1 C1 ln 2 ∼
= 37.4 µs
[d] tON1 = R2 C2 ln 2 = 35 µs
[e] ib1 = 3.5 mA
9
[f] ib1 =
+ 3e−5.6 ln 2 ∼
= 0.562 mA
18
[g] vce2 = 9 − 9e−5.6 ln 2 ∼
= 8.81 V
Note in this circuit T2 is OFF 35 µs and ON 37.4 µs of every cycle,
whereas T1 is ON 35 µs and OFF 37.4 µs every cycle.
P 7.102 If R1 = R2 = 50RL = 100 kΩ,
C1 =
48 × 10−6
= 692.49 pF;
100 × 103 ln 2
If R1 = R2 = 6RL = 12 kΩ,
C1 =
then
C2 =
36 × 10−6
= 519.37 pF
100 × 103 ln 2
then
48 × 10−6
= 5.77 nF;
12 × 103 ln 2
C2 =
36 × 10−6
= 4.33 nF
12 × 103 ln 2
Therefore 692.49 pF ≤ C1 ≤ 5.77 nF and
519.37 pF ≤ C2 ≤ 4.33 nF
P 7.103 [a] We want the lamp to be in its nonconducting state for no more than 10 s,
the value of to :
10 = R(10 × 10−6 ) ln
1−6
4−6
and
R = 1.091 MΩ
[b] When the lamp is conducting
VTh =
20 × 103
(6) = 0.108 V
20 × 103 + 1.091 × 106
RTh = 20 kk1.091 M = 19,640 Ω
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Problems
7–85
So,
(tc − to ) = (19,640)(10 × 10−6 ) ln
4 − 0.108
= 0.289 s
1 − 0.108
The flash lasts for 0.289 s.
P 7.104 [a] At t = 0 we have
τ = (800)(25) × 10−3 = 20 sec;
vc (∞) = 40 V;
1/τ = 0.05
vc (0) = 5 V
vc = 40 − 35e−0.05t V,
0 ≤ t ≤ to
.·. e0.05to = 1.4
40 − 35e−0.05to = 15;
to = 20 ln 1.4 s = 6.73 s
At t = to we have
The Thévenin equivalent with respect to the capacitor is
800
20
τ=
(25) × 10−3 =
s;
81
81
vc (to ) = 15 V;
vc(∞) =
1
81
=
= 4.05
τ
20
40
V
81
40
40 −4.05(t−to)
40 1175 −4.05(t−to)
vc (t) =
+ 15 −
e
V=
+
e
81
81
81
81
. ·.
40 1175 −4.05(t−to)
+
e
=5
81
81
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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7–86
CHAPTER 7. Response of First-Order RL and RC Circuits
1175 −4.05(t−to) 365
e
=
81
81
e4.05(t−to) =
t − to =
1175
= 3.22
365
1
ln 3.22 ∼
= 0.29 s
4.05
One cycle = 7.02 seconds.
N = 60/7.02 = 8.55 flashes per minute
[b] At t = 0 we have
τ = 25R × 10−3 ;
1/τ = 40/R
vc = 40 − 35e−(40/R)t
40 − 35e−(40/R)to = 15
. ·. t o =
R
ln 1.4,
40
R in kΩ
At t = to :
vTh =
τ=
10
400
(40) =
;
R + 10
R + 10
RTh =
(25)(10R) × 10−3
0.25R
=
;
R + 10
R + 10
10R
kΩ
R + 10
1
4(R + 10)
=
τ
R
4(R+10)
400
400
vc =
+ 15 −
e− R (t−to )
R + 10
R + 10
. ·.
or
400
15R − 250 − 4(R+10) (t−to )
R
+
e
=5
R + 10
R + 10
15R − 250 − 4(R+10) (t−to ) 5R − 350
R
e
=
R + 10
(R + 10)
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Problems
. ·. e
4(R+10)
(t−to )
R
. ·. t − t o =
=
7–87
3R − 50
R − 70
R
3R − 50
ln
4(R + 10)
R − 70
At 12 flashes per minute to + (t − to ) = 5 s
. ·.
R
3R − 50
R
ln 1.4 +
ln
=5
R − 70
|40 {z } 4(R + 10)
dominant
term
Start the trial-and-error procedure by setting (R/40) ln 1.4 = 5, then
R = 200/(ln 1.4) or 594.40 kΩ. If R = 594.40 kΩ then t − to ∼
= 0.29 s.
Second trial set (R/40) ln 1.4 = 4.7 s or R = 558.74 kΩ.
t − to ∼
= 0.30 s
With R = 558.74 kΩ,
This procedure converges to R = 559.3 kΩ.
P 7.105 [a] to = RC ln
Vmin − Vs
Vmax − Vs
= (3700)(250 × 10−6 ) ln
−700
−100
= 1.80 s
tc − to =
RCRL
Vmax − VTh
ln
R + RL
Vmin − VTh
RL
1.3
=
= 0.26;
R + RL
1.3 + 3.7
VTh =
1000(1.3)
= 260 V;
1.3 + 3.7
RC = (3700)(2501̧0−6 ) = 0.925 s
RTh = 3.7 kk1.3 k = 962 Ω
.·. tc − to = (0.925)(0.26) ln(640/40) = 0.67 s
.·. tc = 1.8 + 0.67 = 2.47 s
flashes/min =
60
= 24.32
2.47
[b] 0 ≤ t ≤ to :
vL = 1000 − 700e−t/τ1
τ1 = RC = 0.925 s
to ≤ t ≤ tc :
vL = 260 + 640e−(t−to )/τ2
τ2 = RTh C = 962(250) × 10−6 = 0.2405 s
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7–88
CHAPTER 7. Response of First-Order RL and RC Circuits
0 ≤ t ≤ to :
i=
1000 − vL
7
= e−t/0.925 A
3700
37
to ≤ t ≤ tc :
i=
1000 − vL
74
64 −(t−to )/0.2405
=
−
e
3700
370 370
Graphically, i versus t is
The average value of i will equal the areas (A1 + A2) divided by tc .
.·. iavg =
A1 + A2
tc
7 to −t/0.925
e
dt
37 0
6.475
=
(1 − e− ln 7 ) = 0.15 A–s
37
Z tc
74 − 64e−(t−to )/0.2405
A2 =
dt
370
to
74
15.392 − ln 16
=
(tc − to ) +
(e
− 1)
370
370
17.797
15.392
=
ln 16 −
(1 − e− ln 16)
370
370
= 0.09436 A–s
A1 =
iavg =
Z
(0.15 + 0.09436)
(1000) = 99.06 mA
0.925 ln 7 + 0.2405 ln 16
[c] Pavg = (1000)(99.06 × 10−3 ) = 99.06 W
No. of kw hrs/yr =
(99.06)(24)(365)
= 867.77
1000
Cost/year = (867.77)(0.05) = 43.39 dollars/year
P 7.106 [a] Replace the circuit attached to the capacitor with its Thévenin equivalent,
where the equivalent resistance is the parallel combination of the two
resistors, and the open-circuit voltage is obtained by voltage division
across the lamp resistance. The resulting circuit is
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Problems
RTh = RkRL =
RRL
;
R + RL
VTh =
7–89
RL
Vs
R + RL
From this circuit,
vC (∞) = VTh ;
vC (0) = Vmax ;
τ = RTh C
Thus,
vC (t) = VTh + (Vmax − VTh )e−(t−to )/τ
where
RRL C
τ=
R + RL
[b] Now, set vC (tc) = Vmin and solve for (tc − to ):
VTh + (Vmax − VTh )e−(tc−to )/τ = Vmin
e−(tc −to )/τ =
Vmin − VTh
Vmax − VTh
−(tc − to )
Vmin − VTh
= ln
τ
Vmax − VTh
(tc − to ) = −
RRL C
Vmin − VTh
RRL C
Vmax − VTh
ln
=
ln
R + RL Vmax − VTh
R + RL Vmin − VTh
P 7.107 [a] 0 ≤ t ≤ 0.5:
21
30 21 −t/τ
i=
+
−
e
60
60 60
where τ = L/R.
i = 0.35 + 0.15e−60t/L
i(0.5) = 0.35 + 0.15e−30/L = 0.40
.·. e30/L = 3;
L=
30
= 27.31 H
ln 3
[b] 0 ≤ t ≤ tr , where tr is the time the relay releases:
30
i=0+
− 0 e−60t/L = 0.5e−60t/L
60
.·. 0.4 = 0.5e−60tr /L ;
e60tr /L = 1.25
tr =
27.31 ln 1.25 ∼
= 0.10 s
60
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8
Natural and Step Responses of
RLC Circuits
Assessment Problems
1
1
=
,
therefore C = 500 nF
(2RC)2
LC
1
[b] α = 5000 =
,
therefore C = 1 µF
2RC
AP 8.1 [a]
s1,2 = −5000 ±
[c] √
s
25 × 106 −
1
= 20,000,
LC
s1,2 = −40 ±
therefore C = 125 nF
q
(40)2 − 202 103 ,
s1 = −5.36 krad/s,
AP 8.2 iL
=
=
1
50 × 10−3
(103 )(106 )
= (−5000 ± j5000) rad/s
20
Z
t
0
s2 = −74.64 krad/s
[−14e−5000x + 26e−20,000x] dx + 30 × 10−3
(
−14e−5000x
20
−5000
t
26e−20,000t
+
−20,000
0
t
0
)
+ 30 × 10−3
=
56 × 10−3 (e−5000t − 1) − 26 × 10−3 (e−20,000t − 1) + 30 × 10−3
=
[56e−5000t − 56 − 26e−20,000t + 26 + 30] mA
=
56e−5000t − 26e−20,000t mA,
t≥0
AP 8.3 From the given values of R, L, and C, s1 = −10 krad/s and s2 = −40 krad/s.
[a] v(0− ) = v(0+ ) = 0,
therefore iR (0+ ) = 0
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8–1 system, or transmission in any form or by any means, electronic,
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8–2
CHAPTER 8. Natural and Step Responses of RLC Circuits
[b] iC (0+ ) = −(iL (0+ ) + iR(0+ )) = −(−4 + 0) = 4 A
dvc (0+ )
dvc (0+ )
4
= ic (0+ ) = 4,
therefore
=
= 4 × 108 V/s
dt
dt
C
[d] v = [A1e−10,000t + A2 e−40,000t] V,
t ≥ 0+
[c] C
dv(0+ )
= −10,000A1 − 40,000A2
dt
v(0+ ) = A1 + A2 ,
Therefore A1 + A2 = 0,
−A1 − 4A2 = 40,000;
[e] A2 = −40,000/3 V
[f] v = [40,000/3][e−10,000t − e−40,000t] V,
A1 = 40,000/3 V
t≥0
1
= 8000,
therefore R = 62.5 Ω
2RC
10 V
[b] iR(0+ ) =
= 160 mA
62.5 Ω
AP 8.4 [a]
iC (0+ ) = −(iL(0+ ) + iR (0+ )) = −80 − 160 = −240 mA = C
Therefore
dv(0+ )
−240 m
=
= −240 kV/s
dt
C
dvc (0+ )
= ωd B2 − αB1
dt
[c] B1 = v(0+ ) = 10 V,
Therefore 6000B2 − 8000B1 = −240,000,
[d] iL = −(iR + iC );
iR = v/R;
v = e−8000t[10 cos 6000t −
iC = C
iC = e−8000t[−240 cos 6000t +
iL = 10e−8000t[8 cos 6000t +
1 2
1
106
AP 8.5 [a]
=
=
,
2RC
LC
4
[b] 0.5CV02 = 12.5 × 10−3 ,
[c] 0.5LI02 = 12.5 × 10−3 ,
B2 = (−80/3) V
dv
dt
80
sin 6000t] V
3
Therefore iR = e−8000t[160 cos 6000t −
dv(0+ )
dt
1280
sin 6000t] mA
3
460
sin 6000t] mA
3
82
sin 6000t] mA,
3
t≥0
1
= 500,
2RC
therefore V0 = 50 V
therefore
R = 100 Ω
I0 = 250 mA
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8–3
Problems
[d] D2 = v(0+ ) = 50,
iR (0+ ) =
dv(0+ )
= D1 − αD2
dt
50
= 500 mA
100
Therefore iC (0+ ) = −(500 + 250) = −750 mA
dv(0+ )
10−3
= −750 ×
= −75,000 V/s
dt
C
1
Therefore D1 − αD2 = −75,000;
α=
= 500,
2RC
Therefore
D1 = −50,000 V/s
[e] v = [50e−500t − 50,000te−500t ] V
iR =
v
= [0.5e−500t − 500te−500t ] A,
R
t ≥ 0+
V0
40
=
= 0.08 A
R
500
iC (0+ ) = I − iR(0+ ) − iL (0+ ) = −1 − 0.08 − 0.5 = −1.58 A
diL (0+ )
Vo
40
=
=
= 62.5 A/s
dt
L
0.64
1
1
α=
= 1000;
= 1,562,500;
s1,2 = −1000 ± j750 rad/s
2RC
LC
iL = if + B10 e−αt cos ωd t + B20 e−αt sin ωd t,
if = I = −1 A
AP 8.6 [a] iR(0+ ) =
[b]
[c]
[d]
[e]
iL (0+ ) = 0.5 = if + B10 ,
therefore B10 = 1.5 A
diL (0+ )
= 62.5 = −αB10 + ωd B20 ,
dt
therefore B20 = (25/12) A
Therefore iL(t) = −1 + e−1000t[1.5 cos 750t + (25/12) sin 750t] A,
[f] v(t) =
LdiL
= 40e−1000t [cos 750t − (154/3) sin 750t]V
dt
t≥0
t≥0
AP 8.7 [a] i(0+ ) = 0, since there is no source connected to L for t < 0.
+
−
[b] vc (0 ) = vC (0 ) =
!
15 k
(80) = 50 V
15 k + 9 k
di(0+ )
di(0+ )
= 100,
= 10,000 A/s
dt
dt
1
[d] α = 8000;
= 100 × 106 ;
s1,2 = −8000 ± j6000 rad/s
LC
[e] i = if + e−αt[B10 cos ωd t + B20 sin ωd t];
if = 0, i(0+ ) = 0
[c] 50 + 80i(0+ ) + L
Therefore B10 = 0;
Therefore B20 = 1.67 A;
di(0+ )
= 10,000 = −αB10 + ωd B20
dt
i = 1.67e−8000t sin 6000t A,
t≥0
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8–4
CHAPTER 8. Natural and Step Responses of RLC Circuits
AP 8.8 vc (t) = vf + e−αt [B10 cos ωd t + B20 sin ωd t],
dvc (0+ )
= 0;
dt
+
vc (0 ) = 50 V;
B10 = −50 V;
B20
Therefore
vf = 100 V
therefore 50 = 100 + B10
0 = −αB10 + ωd B20
α
8000
= B10 =
(−50) = −66.67 V
ωd
6000
Therefore vc (t) = 100 − e−8000t[50 cos 6000t + 66.67 sin 6000t] V,
t≥0
Problems
P 8.1
[a] α =
1
1012
=
= 25,000
2RC
(4000)(10)
1
1012
=
= 4 × 108
LC
(250)(10)
√
= −25,000 ± 625 × 106 − 400 × 106 = −25,000 ± 15,000
ωo2 =
s1,2
s1 = −10,000 rad/s
s2 = −40,000 rad/s
[b] overdamped
[c] ωd =
q
ωo2 − α2
.·. α2 = ωo2 − ωd2 = 4 × 108 − 144 × 106 = 256 × 106
α = 16 × 103 = 16,000
1
= 16,000;
2RC
.·. R =
109
= 3125 Ω
(32,000)(10)
[d] s1 = −16,000 + j12,000 rad/s;
s2 = −16,000 − j12,000 rad/s
1
1
[e] α = 4 × 104 =
;
.·. R =
= 2500 Ω
2RC
2C(4 × 104 )
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Problems
P 8.2
[a] iR(0) =
8–5
15
= 75 mA
200
iL (0) = −45 mA
iC (0) = −iL (0) − iR (0) = 45 − 75 = −30 mA
[b] α =
1
1
=
= 12,500
2RC
2(200)(0.2 × 10−6 )
1
1
=
= 108
−3
LC
(50 × 10 )(0.2 × 10−6 )
√
= −12,500 ± 1.5625 × 108 − 108 = −12,500 ± 7500
ωo2 =
s1,2
s1 = −5000 rad/s;
s2 = −20,000 rad/s
v = A1 e−5000t + A2e−20,000t
v(0) = A1 + A2 = 15
dv
−30 × 10−3
(0) = −5000A1 − 20,000A2 =
= −15 × 104 V/s
dt
0.2 × 10−6
Solving,
A1 = 10;
A2 = 5
v = 10e−5000t + 5e−20,000t V,
[c] iC
t≥0
dv
dt
=
C
=
0.2 × 10−6 [−50,000e−5000t − 100,000e−20,000t]
=
−10e−5000t − 20e−20,000t mA
iR = 50e−5000t + 25e−20,000t mA
iL = −iC − iR = −40e−5000t − 5e−20,000t mA,
P 8.3
t≥0
1
1
=
= 8000
2RC
2(312.5)(0.2 × 10−6 )
1
1
=
= 108
−3
−6
LC
(50 × 10 )(0.2 × 10 )
s1,2 = −8000 ±
√
80002 − 108 = −8000 ± j6000 rad/s
.·. response is underdamped
v(t) = B1 e−8000t cos 6000t + B2 e−8000t sin 6000t
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8–6
CHAPTER 8. Natural and Step Responses of RLC Circuits
v(0+ ) = 15 V = B1 ;
iR (0+ ) =
15
= 48 mA
312.5
iC (0+ ) = [−iL(0+ ) + iR(0+ )] = −[−45 + 48] = −3 mA
dv(0+ )
−3 × 10−3
=
= −15,000 V/s
dt
0.2 × 10−6
dv(0)
= −8000B1 + 6000B2 = −15,000
dt
6000B2 = 8000(15) − 15,000;
.·. B2 = 17.5 V
v(t) = 15e−8000t cos 6000t + 17.5e−8000t sin 6000t V,
P 8.4
α=
t≥0
1
1
=
= 104
2RC
2(250)(0.2 × 10−6 )
α2 = 108 ;
.·. α2 = ωo2
Critical damping:
v = D1 te−αt + D2 e−αt
iR(0+ ) =
15
= 60 mA
250
iC (0+ ) = −[iL(0+ ) + iR(0+ )] = −[−45 + 60] = −15 mA
v(0) = D2 = 15
dv
= D1 [t(−αe−αt) + e−αt ] − αD2 e−αt
dt
dv
iC (0)
−15 × 10−3
(0) = D1 − αD2 =
=
= −75,000
dt
C
0.2 × 10−6
D1 = αD2 − 75,000 = (104 )(15) − 75,000 = 75,000
v = (75,000t + 15)e−10,000t V,
t≥0
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Problems
P 8.5
[a]
8–7
1
= 50002
LC
There are many possible solutions. This one begins by choosing
L = 10 mH. Then,
C=
1
(10 ×
10−3 )(5000)2
= 4 µF
We can achieve this capacitor value using components from Appendix H
by combining four 1 µF capacitors in parallel.
Critically damped:
. ·. R =
α = ω0 = 5000
1
2(4 × 10−6 )(5000)
so
1
= 5000
2RC
= 25 Ω
We can create this resistor value using components from Appendix H by
combining a 10 Ω resistor and a 15 Ω resistor in series. The final circuit:
q
[b] s1,2 = −α ± α2 − ω02 = −5000 ± 0
Therefore there are two repeated real roots at −5000 rad/s.
P 8.6
[a] Underdamped response:
α < ω0
so
α < 5000
Therefore we choose a larger resistor value than the one used in Problem
8.5. Choose R = 100 Ω:
1
α=
= 1250
2(100)(4 × 10−6 )
√
s1,2 = −1250 ± 12502 − 50002 = −1250 ± j4841.23 rad/s
[b] Overdamped response:
α > ω0
so
α > 5000
Therefore we choose a smaller resistor value than the one used in
Problem 8.5. Choose R = 20 Ω:
1
α=
= 6250
2(20)(4 × 10−6 )
√
s1,2 = −1250 ± 62502 − 50002 = −1250 ± 3750
= −2500 rad/s;
and
− 10,000 rad/s
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8–8
P 8.7
CHAPTER 8. Natural and Step Responses of RLC Circuits
[a] α = 8000;
ωd =
q
ωd = 6000
ωo2 − α2
.·. ωo2 = ωd2 + α2 = 36 × 106 + 64 × 106 = 100 × 106
1
= 100 × 106
LC
C=
[b] α =
1
= 25 nF
(100 × 106 )(0.4)
1
2RC
. ·. R =
1
1
=
= 2500 Ω
2αC
(16,000)(25 × 10−9 )
[c] Vo = v(0) = 75 V
[d] Io = iL (0) = −iR(0) − iC (0)
75
= 30 mA
2500
iR (0) =
iC (0) = C
dv
(0) = 25 × 10−9 [6000(−300) − 8000(75)] = −60 mA
dt
.·. Io = −30 + 60 = 30 mA
[e] iC (t) = 25 × 10−9
iR (t) =
dv(t)
= e−8000t(48.75 sin 6000t − 60 cos 6000t) mA
dt
v(t)
= e−8000t(30 cos 6000t − 120 sin 6000t) mA
2500
iL (t) = −iR (t) − iC (t)
= e−8000t(30 cos 6000t + 71.25 sin 6000t) mA,
t≥0
Check:
diL
L
= 0.4 × 10−3 e−8000t[187,000 cos 6000t − 750,000 sin 6000t]
dt
v(t) = e−8000t[75 cos 6000t − 300 sin 6000t] V
P 8.8
[a] −α +
q
−α −
α2 − ωo2 = −250
q
α2 − ωo2 = −1000
Adding the above equations,
− 2α = −1250
α = 625 rad/s
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Problems
8–9
1
1
=
= 625
2RC
2R(0.1 × 10−6 )
R = 8 kΩ
q
2 α2 − ωo2 = 750
4(α2 − ωo2) = 562,500
.·. ωo = 500 rad/s
1
LC
1
. ·. L =
= 40 H
4
(25 × 10 )(0.1 × 10−6 )
ωo2 = 25 × 104 =
[b] iR =
v(t)
= −1e−250t + 4e−1000t mA,
R
iC = C
t ≥ 0+
dv(t)
= 0.2e−250t − 3.2e−1000t mA,
dt
iL = −(iR + iC ) = 0.8e−250t − 0.8e−1000t mA,
P 8.9
[a]
1
2RC
2
. ·. C =
=
t ≥ 0+
t≥0
1
= (500)2
LC
1
= 1 µF
(500)2 (4)
1
= 500
2RC
. ·. R =
1
= 1 kΩ
2(500)(10−6 )
v(0) = D2 = 8 V
iR (0) =
8
= 8mA
1000
iC (0) = −8 + 10 = 2 mA
dv
2 × 10−3
(0) = D1 − 500D2 =
= 2000 V/s
dt
10−6
.·. D1 = 2000 + 500(8) = 6000 V/s
[b] v = 6000te−500t + 8e−500t V,
t≥0
dv
= [−3 × 106 t + 2000]e−500t
dt
dv
iC = C
= (−3000t + 2)e−500t mA,
dt
t ≥ 0+
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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8–10
P 8.10
CHAPTER 8. Natural and Step Responses of RLC Circuits
α = 500/2 = 250
R=
1
106
=
= 1000 Ω
2αC
(500)(18)
v(0+ ) = −11 + 20 = 9 V
9
= 9 mA
1000
iR (0+ ) =
dv
= 1100e−100t − 8000e−400t
dt
dv(0+ )
= 1100 − 8000 = −6900 V/s
dt
iC (0+ ) = 2 × 10−6 (−6900) = −13.8 mA
iL (0+ ) = −[iR(0+ ) + iC (0+ )] = −[9 − 13.8] = 4.8 mA
P 8.11
[a] 2α = 1000;
q
α = 500 rad/s
2 α2 − ωo2 = 600;
C=
L=
ωo = 400 rad/s
1
1
=
= 4 µF
2αR
2(500)(250)
1
ωo2 C
=
1
= 1.5625 H
× 10−6 )
(400)2 (4
iC (0+ ) = A1 + A2 = 45 mA
diC diL diR
+
+
=0
dt
dt
dt
diC (0)
diL (0) diR (0)
=−
−
dt
dt
dt
diL (0)
0
=
= 0 A/s
dt
1.5625
diR (0)
1 dv(0)
1 iC (0)
45 × 10−3
=
=
=
= 45 A/s
dt
R dt
R C
(250)(4 × 10−6 )
. ·.
diC (0)
= 0 − 45 = −45 A/s
dt
.·. 200A1 + 800A2 = 45;
A1 + A2 = 0.045
Solving, A1 = −15 mA;
A2 = 60 mA
.·. iC = −15e−200t + 60e−800t mA,
t ≥ 0+
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Problems
8–11
[b] By hypothesis
v = A3 e−200t + A4 e−800t,
t≥0
v(0) = A3 + A4 = 0
dv(0)
45 × 10−3
=
= 11,250 V/s
dt
4 × 10−6
−200A3 − 800A4 = 11,250;
.·. A3 = 18.75 V;
v = 18.75e−200t − 18.75e−800t V,
t≥0
v
= 75e−200t − 75e−800t mA,
250
[d] iL = −iR − iC
[c] iR (t) =
iL = −60e−200t + 15e−800t mA,
P 8.12
A4 = −18.75 V
t ≥ 0+
t≥0
From the form of the solution we have
v(0) = A1 + A2
dv(0+ )
= −α(A1 + A2 ) + jωd (A1 − A2)
dt
We know both v(0) and dv(0+ )/dt will be real numbers. To facilitate the
algebra we let these numbers be K1 and K2 , respectively. Then our two
simultaneous equations are
K1 = A1 + A2
K2 = (−α + jωd )A1 + (−α − jωd )A2
The characteristic determinant is
1
∆=
1
(−α + jωd ) (−α − jωd )
= −j2ωd
The numerator determinants are
N1 =
K1
1
K2 (−α − jωd )
and N2 =
1
= −(α + jωd )K1 − K2
K1
(−α + jωd ) K2
= K2 + (α − jωd )K1
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8–12
CHAPTER 8. Natural and Step Responses of RLC Circuits
It follows that A1 =
and A2 =
N1
ωd K1 − j(αK1 + K2 )
=
∆
2ωd
N2
ωd K1 + j(αK1 + K2 )
=
∆
2ωd
We see from these expressions that
P 8.13
A1 = A∗2 .
By definition, B1 = A1 + A2. From the solution to Problem 8.12 we have
A1 + A2 =
2ωd K1
= K1
2ωd
But K1 is v(0), therefore, B1 = v(0), which is identical to Eq. (8.30).
By definition, B2 = j(A1 − A2 ). From Problem 8.12 we have
B2 = j(A1 − A2 ) =
j[−2j(αK1 + K2 )]
αK1 + K2
=
2ωd
ωd
It follows that
K2 = −αK1 + ωd B2 ,
but K2 =
dv(0+ )
dt
and
K1 = B1 .
Thus we have
dv +
(0 ) = −αB1 + ωd B2,
dt
which is identical to Eq. (8.31).
P 8.14
1
= 800 rad/s
2RC
1
ωo2 =
= 106
LC
√
ωd = 106 − 8002 = 600 rad/s
[a] α =
.·. v = B1e−800t cos 600t + B2 e−800t sin 600t
v(0) = B1 = 30
30
= 6 mA;
iC (0+ ) = −12 mA
5000
dv +
−0.012
. ·.
(0 ) =
= −96,000 V/s
dt
125 × 10−9
iR (0+ ) =
−96,000 = −αB1 + ωd B2 = −(800)(30) + 600B2
.·. B2 = −120
.·. v = 30e−800t cos 600t − 120e−800t sin 600t V,
t≥0
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Problems
[b]
8–13
dv
= 6000e−800t (13 sin 600t − 16 cos 600t)
dt
dv
= 0 when 16 cos 600t = 13 sin 600t or
dt
.·. 600t1 = 0.8885,
t1 = 1.48 ms
600t2 = 0.8885 + π,
600t3 = 0.8885 + 2π,
tan 600t =
16
13
t2 = 6.72 ms
t3 = 11.95 ms
2π
2π
=
= 10.47 ms
ωd
600
Td
10.48
[d] t2 − t1 = 5.24 ms;
=
= 5.24 ms
2
2
[e] v(t1) = 30e−(1.184)(cos 0.8885 − 4 sin 0.8885) = −22.7 V
[c] t3 − t1 = 10.47 ms;
Td =
v(t2) = 30e−(5.376)(cos 4.032 − 4 sin 4.032) = 0.334 V
v(t3) = 30e−(9.56) (cos 7.17 − 4 sin 7.17) = −5.22 mV
[f]
P 8.15
[a] α = 0;
ωd = ωo =
√
106 = 1000 rad/s
v = B1 cos ωo t + B2 sin ωo t;
C
v(0) = B1 = 30
dv
(0) = −iL (0) = −0.006
dt
−48,000 = −αB1 + ωd B2 = −0 + 1000B2
.·. B2 =
−48,000
= −48 V
1000
v = 30 cos 1000t − 48 sin 1000t V,
[b] 2πf = 1000;
f =
t≥0
1000 ∼
= 159.15 Hz
2π
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8–14
P 8.16
CHAPTER 8. Natural and Step Responses of RLC Circuits
[c]
√
[a]
ωo2
302 + 482 = 56.6 V
1
109
=
=
= 4 × 106
LC
(2.5)(100)
ωo = 2000 rad/s
1
= 2000;
2RC
R=
1
= 2500 Ω
4000C
[b] v(t) = D1 te−5000t + D2 e−5000t
v(0) = −15 V = D2
iC (0) = 5 +
15
= 11 mA
2.5
dv
iC (0)
11 × 10−3
(0) =
=
= 110,000
dt
C
100 × 10−9
D1 − 2000(−15) = 110,000
so
D1 = 80,000 V/s
.·. v(t) = (80,000t − 15)e−2000t V,
[c] iC (t) = 0 when
t≥0
dv
(t) = 0
dt
dv
= (110,000 − 160 × 106 t))e−2000t
dt
dv
= 0 when 160 × 106 t1 = 110,000;
dt
.·. t1 = 687.5 µs
v(687.5µs) = (55 − 15)e−1.375 = 10.1136 V
1
1
[d] w(0) = (100 × 10−9 )(15)2 + (2.5)(0.005)2 = 42.5 µJ
2
2
1
1
10.1136
w(687.5 µs) = (100 × 10−9 )(10.1136)2 + (2.5)
2
2
2500
% remaining =
P 8.17
[a] α =
= 25.571 µJ
25.571
(100) = 60.17%
42.5
1
= 1250,
2RC
s1 = −500,
2
ωo = 103 ,
therefore overdamped
s2 = −2000
therefore v = A1 e−500t + A2 e−2000t
+
v(0 ) = 0 = A1 + A2 ;
"
dv(0+ )
iC (0+ )
=
= 98,000 V/s
dt
C
#
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Problems
Therefore
A1 =
− 500A1 − 2000A2 = 98,000
+980
,
15
A2 =
−980
15
980 −500t
v(t) =
[e
− e−2000t] V,
15
8–15
t≥0
[b]
Example 8.4: vmax ∼
= 74.1 V
at 1.4 ms
Example 8.5: vmax ∼
= 36.1 V
at 1.0 ms
Problem 8.17: vmax ∼
= 30.9 at 0.92 ms
P 8.18
t<0:
Vo = 15 V,
Io = −60 mA
t > 0:
iR(0) =
15
= 150 mA;
100
iL (0) = −60 mA
iC (0) = −150 − (−60) = −90 mA
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8–16
CHAPTER 8. Natural and Step Responses of RLC Circuits
α=
1
1
=
= 5000 rad/s
2RC
2(100)(10−6 )
ωo2 =
1
1
=
= 16 × 106
LC
(62.5 × 10−3 )(10−6 )
s1,2 = −5000 ±
√
25 × 106 − 16 × 106 = −5000 ± 3000
s1 = −2000 rad/s;
s2 = −8000 rad/s
.·. vo = A1e−2000t + A2 e−8000t
A1 + A2 = vo (0) = 15
dvo
−90 × 10−3
(0) = −2000A1 − 8000A2 =
= −90,000
dt
10−6
Solving,
A1 = 5 V,
A2 = 10 V
.·. vo = 5e−2000t + 10e−8000t V,
P 8.19
ωo2 =
α=
t≥0
1
1
=
= 16 × 106
LC
(62.5 × 10−3 )(10−6 )
1
1
=
= 2500
2RC
2(200)(10−6 )
s1,2 = −2500 ±
√
25002 − 16 × 106 = −2500 ± j3122.5rad/s
vo (t) = B1 e−2500t cos 3122.5t + B2 e−2500t sin 3122.5t
vo (0) = B1 = 15 V
iR(0) =
15
= 75 mA
200
iL (0) = −60 mA
iC (0) = −iR(0) − iL(0) = −15 mA
.·.
iC (0)
= −15,000
C
dvo
(0) = −2500B1 + 3122.5B2 = −15,000
dt
.·.
B2 = 7.21
vo (t) = 15e−2500t cos 3122.5t + 7.21e−2500t sin 3122.5t V,
t≥0
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Problems
P 8.20
ωo2 =
α=
8–17
1
1
=
= 16 × 106
−3
−6
LC
(62.5 × 10 )(10 )
1
1
=
= 4000
2RC
2(125)(10−6 )
.·. α2 = ωo2 (critical damping)
vo (t) = D1 te−4000t + D2 e−4000t
vo (0) = D2 = 15 V
iR(0) =
15
= 120 mA
125
iL (0) = −60 mA
iC (0) = −60 mA
dvo
(0) = −4000D2 + D1
dt
iC (0)
−60 × 10−3
=
= −60,000
C
10−6
D1 − 4000D2 = −60,000;
vo (t) = 15e−4000t V,
D1 = 0
t≥0
P 8.21
vT = −16,000iφ + iT (15,000) = −16,000
−iT (40)
+ it(15,000)
64
vT
= 10,000 + 15,000 = 25 kΩ
iT
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8–18
CHAPTER 8. Natural and Step Responses of RLC Circuits
4000
(7.5) = 6 V;
5000
Vo =
Io = 0
iC (0) = −iR(0) − iL(0) = −
6
= −240 µA
25,000
iC (0)
−240 × 10−6
=
= −60,000
C
4 × 10−9
ωo2 =
α=
1
109
=
= 16 × 106 ;
LC
(4)(15.625)
ωo = 4000 rad/s
1
109
=
= 5000 rad/s
2RC
(2)(4)(25 × 103 )
α2 > ω02
so the response is overdamped
s1,2 = −5000 ±
√
50002 − 40002 = −5000 ± 3000 rad/s
vo = A1 e−2000t + A2e−8000t
vo (0) = A1 + A2 = 6 V
dvo
(0) = −2000A1 − 8000A2 = −60,000
dt
.·. A1 = −2 V;
A2 = 8 V
vo = 8e−8000t − 2e−2000t V,
t≥0
P 8.22
vT = −16,000iφ + iT (15,000) = −16,000
−iT (40)
+ it(15,000
64
vT
= 10,000 + 15,000 = 25 kΩ
iT
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Problems
4000
(7.5) = 6 V;
5000
Vo =
8–19
Io = 0
iC (0) = −iR(0) − iL(0) = −
6
= −240 µA
25,000
iC (0)
−240 × 10−6
=
= −60,000
C
4 × 10−9
ωo2 =
α=
1
109
=
= 25 × 106 ;
LC
(4)(10)
ωo = 5000 rad/s
1
109
=
= 5000 rad/s
2RC
(2)(4)(25 × 103 )
α2 = ω02
so the response is critically damped
vo = D1 te−5000t + D2 e−5000t
vo (0) = D2 = 6 V
dvo
(0) = D1 − αD2 = −60,000
dt
.·. D1 = −60,000 + (5000)(6) = −30,000 V/s
vo = −30,000te−5000t + 6e−5000t V,
t≥0
P 8.23
vT = −16,000iφ + iT (15,000) = −16,000
−iT (40)
+ it(15,000
64
vT
= 10,000 + 15,000 = 25 kΩ
iT
Vo =
4000
(7.5) = 6 V;
5000
Io = 0
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8–20
CHAPTER 8. Natural and Step Responses of RLC Circuits
iC (0) = −iR(0) − iL(0) = −
6
= −240 µA
25,000
iC (0)
−240 × 10−6
=
= −60,000
C
4 × 10−9
ωo2 =
α=
109
1
=
= 62502 ;
LC
(4)(6.4)
1
109
=
= 5000 rad/s
2RC
(2)(4)(25 × 103 )
α2 < ω02
ωd =
ωo = 6250 rad/s
√
so the response is underdamped
62502 − 50002 = 3750 rad/s
vo = B1 e−5000t cos 3750t + B2e−5000t sin 3750t
vo (0) = B1 = 6 V
dvo
(0) = −5000B1 + 3750B2 = −60,000
dt
.·. B2 = −8 V
vo = e−5000t(6 cos 3750t − 8 sin 3750t) V,
P 8.24
P 8.25
t≥0
!
diL
[a] v = L
= 16[e−20,000t − e−80,000t] V,
t≥0
dt
v
[b] iR =
= 40[e−20,000t − e−80,000t] mA,
t ≥ 0+
R
[c] iC = I − iL − iR = [−8e−20,000t + 32e−80,000t] mA,
diL
[a] v = L
dt
!
= 40e−32,000t sin 24,000t V,
[b] iC (t) = I − iR − iL = 24 × 10−3 −
t ≥ 0+
t≥0
v
− iL
625
= [24e−32,000t cos 24,000t − 32e−32,000t sin 24,000t] mA,
P 8.26
diL
v=L
dt
!
= 960,000te−40,000t V,
t ≥ 0+
t≥0
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Problems
P 8.27
8–21
t < 0:
vo (0− ) = vo (0+ ) =
625
(25) = 20 V
781.25
iL (0− ) = iL (0+ ) = 0
t > 0:
−160 × 10−3 +
20
+ iC (0+ ) + 0 = 0;
125
.·. iC (0+ ) = 0
1
1
=
= 800 rad/s
2RC
2(125)(5 × 10−6 )
ωo2 =
1
1
=
= 64 × 104
−3
LC
(312.5 × 10 )(5 × 10−6 )
.·. α2 = ωo2
critically damped
[a] vo = Vf + D10 te−800t + D20 e−800t
Vf = 0
dvo (0)
= −800D20 + D10 = 0
dt
vo (0+ ) = 20 = D20
D10 = 800D20 = 16,000 V/s
.·. vo = 16,000te−800t + 20e−800t V,
t ≥ 0+
[b] iL = If + D30 te−800t + D40 e−800t
iL (0+ ) = 0;
If = 160 mA;
.·. 0 = 160 + D40 ;
−800D40 + D30 = 64;
diL (0+ )
20
=
= 64 A/s
dt
312.5 × 10−3
D40 = −160 mA;
D30 = −64 A/s
.·. iL = 160 − 64,000te−800t − 160e−800t mA
t≥0
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8–22
P 8.28
CHAPTER 8. Natural and Step Responses of RLC Circuits
[a] wL =
Z
∞
0
∞
Z
pdt =
voiL dt
0
vo = 16,000te−800t + 20e−800t V
iL = 0.16 − 64te−800t − 0.16e−800t A
p = 3.2e−800t + 2560te−800t − 3840te−1600t
−1,024,000t2 e−1600t − 3.2e−1600t W
Z
wL = 3.2
∞
Z
e−800t dt + 2560
0
Z
− 1,024,000
e−800t
= 3.2
−800
∞
0
∞
0
Z
te−800t dt − 3480
∞
0
Z
t2 e−1600t dt − 3.2
∞
0
te−1600t dt
∞
0
e−1600t dt
∞
2560 −800t
+
e
(−2560t − 1)
(800)2
0
3840 −1600t
−
e
(−1600t − 1)
(1600)2
∞
0
∞
1,024,000 −1600t
−
e
(16002 t2 + 3200t + 2)
3
(−1600)
0
− 3.2
e−1600t
(−1600)
∞
0
All the upper limits evaluate to zero hence
3.2
2560
3840
(1,024,000)(2)
3.2
+
−
−
−
= 4 mJ
2
2
3
800 800
1600
1600
1600
Note this value corresponds to the final energy stored in the inductor, i.e.
1
wL (∞) = (312.5 × 10−3 )(0.16)2 = 4 mJ.
2
[b] v = 16,000te−800t + 20e−800t V
v
iR =
= 128te−800t + 0.16e−800t A
125
wL =
pR = viR = 2,048,000t2 e−1600t + 5120te−1600t + 3.2e−1600t
wR =
Z
0
∞
pR dt
Z
= 2,048,000
=
∞
0
Z
t2 e−1600t dt + 5120
∞
0
2,048,000e−1600t
[16002 t2 + 3200t + 2]
−16003
5120e−1600t
(−1600t − 1)
16002
∞
+
0
Z
te−1600t dt + 3.2
∞
0
e−1600t dt
∞
+
0
3.2e−1600t
(−1600)
∞
0
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Problems
8–23
Since all the upper limits evaluate to zero we have
2,048,000(2)
5120
3.2
+
+
= 5 mJ
3
2
1600
1600
1600
[c] 160 = iR + iC + iL (mA)
wR =
iR + iL = 160 + 64,000te−800t mA
.·. iC = 160 − (iR + iL ) = −64,000te−800t mA = −64te−800t A
pC = viC = [16,000te−800t + 20e−800t][−64te−800t]
= −1,024,000t2 e−1600t − 1280e−1600t
wC = −1,024,000
wC =
Z
∞
2 −1600t
t e
0
dt − 1280
Z
∞
0
te−1600t dt
−1,024,000e−1600t
[16002 t2 + 3200t + 2]
−16003
∞
0
−
1280e−1600t
(−1600t − 1)
16002
∞
0
Since all upper limits evaluate to zero we have
−1,024,000(2) 1280(1)
−
= −1 mJ
16003
16002
Note this 1 mJ corresponds to the initial energy stored in the capacitor,
i.e.,
wC =
1
wC (0) = (5 × 10−6 )(20)2 = 1 mJ.
2
Thus wC (∞) = 0 mJ which agrees with the final value of v = 0.
[d] is = 160 mA
ps (del) = 160v mW
= 0.16[16,000te−800t + 20e−800t ]
= 3.2e−800t + 2560te−800t W
Z
ws = 3.2
∞
0
e
−800t
3.2e−800t
=
−800
=
∞
0
dt +
Z
∞
0
2560te−800t dt
2560e−800t
+
(−800t − 1)
8002
∞
0
3.2
2560
+
= 8 mJ
800
800
[e] wL = 4 mJ (absorbed)
wR = 5 mJ (absorbed)
wC = 1 mJ (delivered)
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8–24
CHAPTER 8. Natural and Step Responses of RLC Circuits
wS = 8 mJ (delivered)
X
P 8.29
ωo2 =
α=
wdel = wabs = 9 mJ.
1
1
=
= 108 ;
−3
LC
(50 × 10 )(0.2 × 10−6 )
ωo = 104 rad/s
1
1
=
= 12,500 rad/s
2RC
2(200)(0.2 × 10−6 )
s1,2 = −12,500 ±
.·. overdamped
q
s1 = −5000 rad/s;
(12,500)2 − 108 = −12,500 ± 7500 rad/s
s2 = −20,000 rad/s
If = 60 mA
iL = 60 × 10−3 + A01e−5000t + A02e−20,000t
.·. −45 × 10−3 = 60 × 10−3 + A01 + A02 ;
A01 + A02 = −105 × 10−3
diL
15
= −5000A01 − 20,000A02 =
= 300
dt
0.05
Solving,
A01 = −120 mA;
A02 = 15 mA
iL = 60 − 120e−5000t + 15e−20,000t mA,
P 8.30
α=
t≥0
1
1
=
= 8000;
2RC
2(312.5)(0.2 × 10−6 )
ωo = 104
α2 = 64 × 106
underdamped
√
s1,2 = −8000 ± j 80002 − 108 = −8000 ± j6000 rad/s
iL = 60 × 10−3 + B10 e−8000t cos 6000t + B20 e−8000t sin 6000t
−45 × 10−3 = 60 × 10−3 + B10
.·. B10 = −105 mA
diL
(0) = −8000B10 + 6000B20 = 300
dt
.·. B20 = −90 mA
iL = 60 − 105e−8000t cos 6000t − 90e−8000t sin 6000t mA,
t≥0
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Problems
P 8.31
α=
8–25
1
1
=
= 104
−6
2RC
2(250)(0.2 × 10 )
α2 = 104 = ωo2
critical damping
4
4
4
iL = If + D10 te−10 t + D20 e−10 t = 60 × 10−3 + D10 te−10 t + D20 e−10
iL (0) = −45 × 10−3 = 60 × 10−3 + D20 ;
4t
.·. D20 = −105 mA
diL
(0) = −104 D20 + D10 = 300 A/s
dt
.·. D10 = 300 + 104 (−105 × 10−3 ) = −750 A/s
4
4
iL = 60 − 750,000te−10 t − 105e−10 t mA,
P 8.32
t≥0
−15
t<0:
iL (0− ) =
= −5 mA;
3000
The circuit reduces to:
vC (0− ) = 0 V
iL (∞) = 4 mA
ωo2 =
α=
1
106
=
= 6400;
LC
(62.5)(2.5)
ωo = 80 rad/s
1
106
=
= 100
2RC
(4000)(2.5)
s1,2 = −100 ±
√
1002 − 802 = −100 ± 60
s1 = −40 rad/s;
s2 = −160 rad/s
iL = If + A01e−40t + A02e−160t
iL (∞) = If = 4mA
iL (0) = A01 + A02 + If = −5 mA
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8–26
CHAPTER 8. Natural and Step Responses of RLC Circuits
.·. A01 + A02 + 4 = −5 so A01 + A02 = −9 mA
diL
(0) = 0 = −40A1 − 160A02
dt
A01 = −12 mA,
Solving,
A02 = 3 mA
iL = 4 − 12e−40t + 3e−160t mA,
P 8.33
t≥0
1
vC (0+ ) = (240) = 120 V
2
iL (0+ ) = 60 mA;
α=
ωo2
iL (∞) =
240
× 10−3 = 48 mA
5
1
106
=
= 40
2RC
2(2500)(5)
1
106
=
=
= 2500
LC
400
α2 = 1600;
α2 < ωo2 ;
.·.
underdamped
√
s1,2 = −40 ± j 2500 − 1600 = −40 ± j30 rad/s
iL
=
If + B10 e−αt cos ωd t + B20 e−αt sin ωd t
=
48 + B10 e−40t cos 30t + B20 e−40t sin 30t
iL (0) = 48 + B10 ;
B10 = 60 − 48 = 12 mA
diL
120
(0) = 30B20 − 40B10 =
= 1.5 = 1500 × 10−3
dt
80
.·. 30B20 = 40(12) × 10−3 + 1500 × 10−3 ;
B20 = 66 mA
.·. iL = 48 + 12e−40t cos 30t + 66e−40t sin 30t mA,
t≥0
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Problems
P 8.34
α=
8–27
1
1
=
= 1000
2RC
2(400)(1.25 × 10−6 )
1
1
=
= 64 × 104
LC
(1.25 × 10−6 )(1.25)
ωo2 =
s1,2 = −1000 ±
√
10002 − 64 × 104 = −1000 ± 600 rad/s
s1 = −400 rad/s;
s2 = −1600 rad/s
vo (∞) = 0 = Vf
.·. vo = A01e−400t + A02e−1600t
vo (0) = 12 = A01 + A02
Note:
iC (0+ ) = 0
dvo
.·.
(0) = 0 = −400A01 − 1600A02
dt
A01 = 16 V,
Solving,
A02 = −4 V
vo (t) = 16e−400t − 4e−1600t V,
P 8.35
t≥0
[a] io = If + A01 e−400t + A02e−1600t
If =
12
= 30mA;
400
io (0) = 0
0 = 30 × 10−3 + A01 + A02 ,
.·. A01 + A02 = −30 × 10−3
dio
12
(0) =
= −400A01 − 1600A02
dt
1.25
Solving,
A01 = −32 mA;
A02 = 2 mA
io = 30 − 32e−400t + 2e−1600t mA,
[b]
t≥0
dio
= [12.8e−400t − 3.2e−1600t]
dt
vo = L
dio
= 16e−400t − 4e−1600t V,
dt
t≥0
This agrees with the solution to Problem 8.34.
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8–28
P 8.36
CHAPTER 8. Natural and Step Responses of RLC Circuits
iL (0− ) = iL (0+ ) =
7.5
= 30 mA
250
For t > 0
iL (0− ) = iL (0+ ) = 30 mA
α=
1
= 80 rad/s;
2RC
ωd =
√
ωo2 =
1
= 104
LC
so ωo = 100 rad/s
1002 − 802 = 60 rad/s
vo (∞) = 0 = Vf ;
B10 = v(0) = 0
vo = e−80tB20 sin 60t
iC (0+ ) = −30 + 30 + 0 = 0
dvo
.·.
=0
dt
dvo
(0) = −αB10 + ωd B20 = 0 + 60B20 = 0
dt
.·. B10 = 0;
B20 = 0
.·. vo = 0 for t ≥ 0
Note:
vo(0) = 0;
vo (∞) = 0;
dvo (0)
=0
dt
Hence, the 30 mA current circulates between the current source and the ideal
inductor in the equivalent circuit. In the original circuit, the 7.5 V source
sustains a current of 30 mA in the inductor. This is an example of a circuit
going directly into steady state when the switch is closed. There is no
transient period, or interval.
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Problems
P 8.37
For
α=
8–29
t>0
1
= 1000;
2RC
1
= 64 × 104
LC
s1,2 = −1000 ± 600 rad/s
s1 = −400 rad/s;
s2 = −1600 rad/s
vo = Vf + A01 e−400t + A02e−1600t
Vf = 0;
vo (0+ ) = 0;
iC (0+ ) = 30 mA
.·. A01 + A02 = 0
dvo (0+ )
iC (0+ )
=
= 24,000 V/s
dt
1.25 × 10−6
dvo (0+ )
= −400A01 − 1600A02 = 24,000
dt
Solving,
A01 = 20 V;
A02 = −20 V
vo = 20e−400t − 20e−1600t V,
P 8.38
t≥0
[a] From the solution to Prob. 8.37 s1 = −400 rad/s and s2 = −1600 rad/s,
therefore
io = If + A01e−400t + A02e−1600t
If = 30 mA;
io (0+ ) = 0;
.·. 0 = 30 × 10−3 + A01 + A02;
dio (0+ )
=0
dt
−400A01 − 1600A02 = 0
Solving
A01 = −40 mA;
A02 = 10 mA
.·. io = 30 − 40e−400t + 10e−1600t mA,
t≥0
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8–30
CHAPTER 8. Natural and Step Responses of RLC Circuits
[b]
dio
= 16e−400t − 16e−1600t
dt
vo = L
dio
= 20e−400t − 20e−1600t V,
dt
t≥0
This agrees with the solution to Problem 8.27.
P 8.39
q
α2
[a] −α +
−
ω02
= −4000;
.·. α = 10,000 rad/s,
α=
R
= 10,000;
2L
ωo2 =
−α −
ω02 = 64 × 106
q
α2 − ω02 = −16,000
R = 20,000L
1
= 64 × 106 ;
LC
L=
109
= 0.5 H
64 × 106 (31.25)
R = 10,000 Ω
[b] i(0) = 0
L
di(0)
= vc (0);
dt
1
(31.25) × 10−9 vc2(0) = 9 × 10−6
2
.·. vc2(0) = 576;
vc (0) = 24 V
di(0)
24
=
= 48 A/s
dt
0.5
[c] i(t) = A1e−4000t + A2e−16,000t
i(0) = A1 + A2 = 0
di(0)
= −4000A1 − 16,000A2 = 48
dt
Solving,
.·. A1 = 4 mA;
A2 = −4 mA
i(t) = 4e−4000t − 4e−16,000t mA,
[d]
t≥0
di(t)
= −16e−4000t + 64e−16,000t
dt
di
= 0 when 64e−16,000t = 16e−4000t
dt
or e12,000t = 4
.·. t =
ln 4
= 115.52 µs
12,000
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Problems
[e] imax = 4e−0.4621 − 4e−1.8484 = 1.89 mA
di
[f] vL(t) = 0.5 = [−8e−1000t + 32e−4000t] V,
dt
P 8.40
[a]
8–31
t ≥ 0+
1
= 20,0002
LC
There are many possible solutions. This one begins by choosing
L = 1 mH. Then,
C=
1
(1 ×
10−3 )(20,000)2
= 2.5 µF
We can achieve this capacitor value using components from Appendix H
by combining four 10 µF capacitors in series.
Critically damped:
α = ω0 = 20,000
so
R
= 20,000
2L
.·. R = 2(10−3 )(20,000) = 40 Ω
We can create this resistor value using components from Appendix H by
combining a 10 Ω resistor and two 15 Ω resistors in series. The final
circuit:
q
[b] s1,2 = −α ± α2 − ω02 = −20,000 ± 0
Therefore there are two repeated real roots at −20,000 rad/s.
P 8.41
[a] Underdamped response:
α < ω0
so
α < 20,000
Therefore we choose a larger resistor value than the one used in Problem
8.40 to give a smaller value of α. For convenience, pick α = 16,000 rad/s:
R
= 16,000
so
R = 2(16,000)(10−3 ) = 32 Ω
2L
We can create a 32 Ω resistance by combining a 10 Ω resistor and a 22 Ω
resistor in series.
α=
s1,2 = −16,000 ±
q
16,0002 − 20,0002 = −16,000 ± j12,000 rad/s
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8–32
CHAPTER 8. Natural and Step Responses of RLC Circuits
[b] Overdamped response:
α > ω0
so
α > 20,000
Therefore we choose a smaller resistor value than the one used in
Problem 8.40. Choose R = 50 Ω, which can be created by combining two
100 Ω resistors in parallel:
α=
R
= 25,000
2L
s1,2 = −25,000 ±
q
25,0002 − 20,0002 = −25,000 ± 15,000
= −10,000 rad/s;
P 8.42
α = 2000 rad/s;
R
= 2000;
2L
− 40,000 rad/s
ωd = 1500 rad/s
ωo2 − α2 = 225 × 104 ;
α=
and
ωo2 = 625 × 104 ;
wo = 25,000 rad/s
R = 4000L
1
= 625 × 104 ;
LC
L=
1
(625 ×
104 )(80
× 10−9 )
= 2H
.·. R = 8 kΩ
i(0+ ) = B1 = 7.5 mA;
at t = 0+
60 + vL(0+ ) − 30 = 0;
. ·.
vL (0+ ) = −30 V
di(0+ )
−30
=
= −15 A/s
dt
2
di(0+ )
.·.
= 1500B2 − 2000B1 = −15
dt
.·. 1500B2 = 2000(7.5 × 10−3 ) − 15;
.·. i = 7.5e−2000t sin 1500t mA,
.·. B2 = 0 A
t≥0
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Problems
P 8.43
8–33
From Prob. 8.42 we know vc will be of the form
vc = B3 e−2000t cos 1500t + B4 e−2000t sin 1500t
From Prob. 8.42 we have
vc (0) = −30 V = B3
and
dvc (0)
iC (0)
7.5 × 10−3
=
=
= 93.75 × 103
−9
dt
C
80 × 10
dvc (0)
= 1500B4 − 2000B3 = 93,750
dt
.·. 1500B4 = 2000(−30) + 93,750;
B4 = 22.5 V
vc (t) = −30e−2000t cos 1500t + 22.5e−2000t sin 1500t V
P 8.44
[a] ωo2 =
α=
t≥0
1
109
=
= 25 × 106
LC
(125)(0.32)
R
= ωo = 5000 rad/s
2L
.·. R = (5000)(2)L = 1250 Ω
[b] i(0) = iL (0) = 6 mA
vL (0) = 15 − (0.006)(1250) = 7.5 V
di
7.5
(0) =
= 60 A/s
dt
0.125
[c] vC = D1 te−5000t + D2 e−5000t
vC (0) = D2 = 15 V
dvC
iC (0)
−iL (0)
(0) = D1 − 5000D2 =
=
= −18,750
dt
C
C
.·. D1 = 56,250 V/s
vC = 56,250te−5000t + 15e−5000t V,
t≥0
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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8–34
P 8.45
CHAPTER 8. Natural and Step Responses of RLC Circuits
ωo2 =
α=
1
1
=
= 25
LC
(10)(4 × 10−3 )
R
80
=
= 4;
2L
2(10)
α2 < ωo2
.·.
α2 = 16
underdamped
√
s1,2 = −4 ± j 9 = −4 ± j3 rad/s
i = B1 e−4t cos 3t + B2 e−4t sin 3t
i(0) = B1 = −240/100 = −2.4 A
di
(0) = 3B2 − 4B1 = 0
dt
.·. B2 = −3.2 A
i = −2.4e−4t cos 3t − 3.2 sin 3t A,
P 8.46
t≥0
[a] For t > 0:
Since i(0− ) = i(0+ ) = 0
va (0+ ) = 75 V
7
[b] va = 2000i + 10
Z
0
t
i dx + 75
dva
di
= 2000 + 107 i
dt
dt
dva (0+ )
di(0+ )
di(0+ )
= 2000
+ 107 i(0+ ) = 2000
dt
dt
dt
−L
di(0+ )
= 75
dt
di(0+ )
= −2.5(75) = −187.5 A/s
dt
. ·.
dva(0+ )
= −375,000 V/s
dt
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Problems
[c] α =
8–35
R
5000
=
= 6250 rad/s
2L
0.8
1
106
=
= 25 × 106
LC
(0.4)(0.1)
√
= −6250 ± 62502 − 25 × 106 = −6250 ± 3750 rad/s
ωo2 =
s1,2
.·. s1 = −2500 rad/s;
s2 = −10,000 rad/s
Overdamped:
va = A1 e−2500t + A2e−10,000t
va (0) = A1 + A2 = 75 V
dva (0)
= −2500A1 − 10,000A2 = −375,000;
dt
A2 = 25 V
t ≥ 0+
va = 50e−2500t + 25e−10,000t V,
P 8.47
.·. A1 = 50 V,
[a] t < 0:
80
= 100 mA;
vo = 500io = (500)(0.01) = 50 V
800
t > 0:
R
500
α=
=
= 105 rad/s
2L
2(2.5 × 10−3 )
io =
ωo2 =
1
1
=
= 100 × 108
−3
−9
LC
(2.5 × 10 )(40 × 10 )
α2 = ωo2
. ·.
critically damped
.·. io(t) = D1 te−10 t + D2 e−10
5
5t
io (0) = D2 = 100 mA
dio
(0) = −αD2 + D1 = 0
dt
. ·.
D1 = 105 (100 × 10−3 ) = 10,000
5
5
io (t) = 10,000te−10 t + 0.1e−10 t A,
5
[b] vo(t) = D3 te−10 t + D4 e−10
t ≥ 0+
5t
vo (0) = D4 = 50
C
dvo
(0) = −0.1
dt
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8–36
CHAPTER 8. Natural and Step Responses of RLC Circuits
dvo
−0.1
(0) =
= −25 × 105 V/s = −αD4 + D3
dt
40 × 10−9
. ·.
D3 = 105 (50) − 25 × 105 = 25 × 105
5
5
vo (t) = 25 × 105 te−10 t + 50e−10 t V,
P 8.48
t ≥ 0+
t < 0:
i(0) =
240
240
=
= 6A
8 + 30k70 + 11
40
vo (0) = 240 − 8(6) −
70
(6)(20) = 108 V
100
t > 0:
α=
R
20
=
= 10,
2L
2(1)
ωo2 =
α2 = 100
1
1
=
= 200
LC
(1)(5 × 10−3 )
ωo2 > α2
underdamped
s1,2 = −100 ±
√
100 − 200 = −10 ± j10 rad/s
vo = B1 e−10t cos 10t + B2e−10t sin 10t
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Problems
8–37
vo (0) = B1 = 108 V
C
dvo
(0) = −6,
dt
dvo
−6
=
= −1200 V/s
dt
5 × 10−3
dvo
(0) = −10B1 + 10B2 = −1200
dt
10B2 = −1200 + 10B1 = −1200 + 1080;
B2 = −120/10 = −12 V
.·. vo = 108e−10t cos 10t − 12e−10t sin 10t V,
P 8.49
iC (0) = 0;
α=
t≥0
vo (0) = 50 V
R
8000
=
= 25,000 rad/s
2L
2(160 × 10−3 )
ωo2 =
1
1
=
= 625 × 106
−3
−9
LC
(160 × 10 )(10 × 10 )
.·. α2 = ωo2 ;
critical damping
vo (t) = Vf + D10 te−25,000t + D20 e−25,000t
Vf = 250 V
vo (0) = 250 + D20 = 50;
D20 = −200 V
dvo
(0) = −25,000D20 + D10 = 0
dt
D10 = 25,000D20 = −5 × 106 V/s
vo = 250 − 5 × 106 te−25,000t − 200e−25,000t V,
P 8.50
α=
t≥0
R
= 2000 rad/s
2L
ωo2 =
1
1
=
= 256 × 104
−3
LC
(62.5 × 10 )(6.25 × 10−6 )
s1,2 = −2000 ±
√
4 × 106 − 256 × 104 = −2000 ± j1200 rad/s
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8–38
CHAPTER 8. Natural and Step Responses of RLC Circuits
vo = Vf + A01 e−800t + A02e−3200t
vo (0) = 0 = Vf + A01 + A02
vo (∞) = 60 V;
.·. A01 + A02 = −60
dvo (0)
= 0 = −800A01 − 3200A02
dt
.·. A01 = −80 V;
A02 = 20 V
vo = 60 − 80e−800t + 20e−3200t V,
P 8.51
α=
t≥0
R
= 2000 rad/s
2L
ωo2 =
1
1
=
= 4 × 106
LC
(62.5 × 10−3 )(4 × 10−6 )
.·. ωo = 2000 rad/s
The response is therefore critically damped
vo = Vf + D10 te−2000t + D20 e−2000t
vo (0) = 0 = Vf + D20
vo (∞) = 60 V;
.·. D20 = −60 V
dvo (0)
= 0 = D10 − αD20
dt
.·. D10 = (2000)(−60) = −120,000 V/s
vo = 60 − 120,000te−2000t − 60e−2000t V,
P 8.52
α=
t≥0
R
= 2000 rad/s
2L
ωo2 =
1
1
=
= 625 × 104
−3
LC
(62.5 × 10 )(2.56 × 10−6 )
.·. ωo = 2500 rad/s
The response is therefore underdamped.
ωd =
√
25002 − 20002 = 1500 rad/s
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Problems
8–39
vo = Vf + B10 e−2000t cos 1500t + B20 e−2000t sin 1500t
vo (0) = 0 = Vf + B10
.·. B10 = −60 V
vo (∞) = 60 V;
dvo (0)
= 0 = −2000B10 + 1500B20
dt
.·. B20 = −80 V
vo = V,
P 8.53
t≥0
[a] t < 0:
io (0− ) =
60
= 6 mA
10,000
vC (0− ) = 20 − (6000)(0.006) = −16 V
t = 0+ :
3 kΩk6 kΩ = 2 kΩ
.·. vo(0+ ) = (0.006)(2000) − 16 = 12 − 16 = −4 V
and vL (0+ ) = 20 − (−4) = 24 V
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8–40
CHAPTER 8. Natural and Step Responses of RLC Circuits
[b] vo(t) = 2000io + vC
dvo
dio dvC
(t) = 2000
+
dt
dt
dt
dvo +
dio
dvC +
(0 ) = 2000 (0+ ) +
(0 )
dt
dt
dt
vL (0+ ) = L
dio +
(0 )
dt
dio +
vL(0+ )
24
(0 ) =
=
= 48 A/s
dt
L
0.5
C
dvc +
(0 ) = io (0+ )
dt
. ·.
dvc +
6 × 10−3
(0 ) =
= 7680
dt
781.25 × 10−9
. ·.
dvo +
(0 ) = 2000(48) + 7680 = 103,680 V/s
dt
[c] ωo2 =
1
= 2.56 × 106 ;
LC
α=
R
= 2000 rad/s
2L
α2 > ωo2
ωo = 1600 rad/s
overdamped
s1,2 = −2000 ± j1200 rad/s
vo (t) = Vf + A01 e−800t + A02 e−3200t
Vf = vo (∞) = 20 V
20 + A01 + A02 = −4;
Solving
A01 = 11.2;
−800A01 − 3200A02 = 103,680
A02 = −35.2
.·. vo(t) = 20 + 11.2e−800t − 35.2e−3200t V,
P 8.54
t ≥ 0+
[a] Let i be the current in the direction of the voltage drop vo (t). Then by
hypothesis
i = if + B10 e−αt cos ωd t + B20 e−αt sin ωd t
if = i(∞) = 0,
i(0) =
Vg
= B10
R
Therefore i = B10 e−αt cos ωd t + B20 e−αt sin ωd t
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Problems
L
di(0)
= 0,
dt
therefore
8–41
di(0)
=0
dt
di
= [(ωd B20 − αB10 ) cos ωd t − (αB20 + ωd B10 ) sin ωd t] e−αt
dt
Therefore ωd B20 − αB10 = 0;
B20 =
α 0
α Vg
B1 =
ωd
ωd R
Therefore
di
α2 Vg ωd Vg
vo = L = − L
+
dt
ωd R
R
(
!
LVg
=−
R
α2 + ωd2 −αt
e sin ωd t
ωd
Vg L
=−
R
ωo2 −αt
e sin ωd t
ωd
vo = −
Vg L
Rωd
sin ωd t e−αt
)
Vg L
=−
R
=−
)
α2
+ ωd sin ωd t e−αt
ωd
(
[b]
!
!
!
1
e−αt sin ωd t
LC
Vg −αt
e sin ωd t V,
RCωd
t≥0
dvo
Vg
=−
{ωd cos ωd t − α sin ωd t}e−αt
dt
ωd RC
dvo
= 0 when
dt
tan ωd t =
ωd
α
Therefore ωd t = tan−1 (ωd /α) (smallest t)
t=
P 8.55
1
ωd
tan−1
ωd
α
[a] From Problem 8.54 we have
vo =
−Vg −αt
e sin ωd t
RCωd
α=
R
4800
=
= 37,500 rad/s
2L
2(64 × 10−3 )
ωo2 =
1
1
=
= 3906.25 × 106
−3
−9
LC
(64 × 10 )(4 × 10 )
ωd =
q
ωo2 − α2 = 50 krad/s
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8–42
CHAPTER 8. Natural and Step Responses of RLC Circuits
−Vg
−(−72)
=
= 75
RCωd
(4800)(4 × 10−9 )(50 × 103 )
.·. vo = 75e−37,500t sin 50,000t V
[b] From Problem 8.54
1
ωd
td =
tan−1
ωd
α
1
50,000
=
tan−1
50,000
37,500
!
td = 18.55 µs
[c] vmax = 75e−0.0375(18.55) sin[(0.05)(18.55)] = 29.93 V
[d] R = 480 Ω;
α = 3750 rad/s
ωd = 62,387.4 rad/s
vo = 601.08e−3750t sin 62,387.4t V,
t≥0
td = 24.22 µs
vmax = 547.92 V
P 8.56
t < 0:
iL (0) =
−150
= −5 A
30
vC (0) = 18iL (0) = −90 V
t > 0:
α=
R
10
=
= 50 rad/s
2L
2(0.1)
ωo2 =
1
1
=
= 5000
LC
(0.1)(2 × 10−3 )
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
ωo > α2
.·.
s1,2 = −50 ±
8–43
underdamped
√
502 − 5000 = −50 ± j50
vc = 60 + B10 e−50t cos 50t + B20 e−50t sin 50t
. ·.
vc (0) = −90 = 60 + B10
C
dvc
(0) = −5;
dt
B10 = −150
dvc
−5
(0) =
= −2500
dt
2 × 10−3
dvc
(0) = −50B10 + 50B2 = −2500
dt
.·.
B20 = −200
vc = 60 − 150e−50t cos 50t − 200e−50t sin 50t V,
P 8.57
t≥0
[a] vc = Vf + [B10 cos ωd t + B20 sin ωd t] e−αt
dvc
= [(ωd B20 − αB10 ) cos ωd t − (αB20 + ωd B10 ) sin ωd t]e−αt
dt
Since the initial stored energy is zero,
dvc (0+ )
=0
dt
vc (0+ ) = 0 and
It follows that B10 = −Vf
and
B20 =
αB10
ωd
When these values are substituted into the expression for [dvc/dt], we get
dvc
=
dt
α2
+ ωd Vf e−αt sin ωd t
ωd
!
But Vf = V
Therefore
[b]
and
dvc
=
dt
dvc
= 0 when
dt
α2
α2 + ωd2
ω2
+ ωd =
= o
ωd
ωd
ωd
ωo2
V e−αt sin ωd t
ωd
!
sin ωd t = 0,
or ωd t = nπ
where n = 0, 1, 2, 3, . . .
Therefore t =
nπ
ωd
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8–44
CHAPTER 8. Natural and Step Responses of RLC Circuits
nπ
,
ωd
[c] When tn =
and
cos ωd tn = cos nπ = (−1)n
sin ωd tn = sin nπ = 0
Therefore vc (tn ) = V [1 − (−1)n e−αnπ/ωd ]
[d] It follows from [c] that
v(t1) = V + V e−(απ/ωd )
P 8.58
vc (t3) = V + V e−(3απ/ωd)
vc (t1) − V
e−(απ/ωd )
= −(3απ/ω ) = e(2απ/ωd)
d
vc (t3) − V
e
Therefore
But
and
2π
= t3 − t1 = Td ,
ωd
(
1
vc (t1) − V
ln
Td
vc (t3) − V
)
;
Td = t3 − t1 =
7000
63.84
α=
ln
= 1000;
2π
26.02
thus α =
ωd =
1
[vc(t1 ) − V ]
ln
Td
[vc(t3 ) − V ]
3π π
2π
− =
ms
7
7
7
2π
= 7000 rad/s
Td
ωo2 = ωd2 + α2 = 49 × 106 + 106 = 50 × 106
L=
1
(50 ×
106 )(0.1
× 10−6 )
= 200 mH;
R = 2αL = 400 Ω
P 8.59
At t = 0 the voltage across each capacitor is zero. It follows that since the
operational amplifiers are ideal, the current in the 500 kΩ is zero. Therefore
there cannot be an instantaneous change in the current in the 1 µF capacitor.
Since the capacitor current equals C(dvo/dt), the derivative must be zero.
P 8.60
[a] From Example 8.13
therefore
dg(t)
= 2,
dt
g(t) − g(0) = 2t;
iR =
d2 vo
=2
dt2
g(t) =
dvo
dt
g(t) = 2t + g(0);
g(0) =
dvo (0)
dt
5
dvo (0)
× 10−3 = 10 µA = iC = −C
500
dt
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Problems
8–45
dvo (0)
−10 × 10−6
=
= −10 = g(0)
dt
1 × 10−6
dvo
= 2t − 10
dt
dvo = 2t dt − 10 dt
vo − vo (0) = t2 − 10t;
vo = t2 − 10t + 8,
vo (0) = 8 V
0 ≤ t ≤ tsat
[b] t2 − 10t + 8 = −9
t2 − 10t + 17 = 0
t∼
= 2.17 s
P 8.61
Part (1) — Example 8.14, with R1 and R2 removed:
[a] Ra = 100 kΩ;
d2 vo
1
=
dt2
Ra C1
C1 = 0.1 µF;
1
vg ;
Rb C2
vg = 250 × 10−3 ;
therefore
Rb = 25 kΩ;
1
= 100
Ra C1
C2 = 1 µF
1
= 40
Rb C2
d2 vo
= 1000
dt2
dvo (0)
, our solution is vo = 500t2
dt
The second op-amp will saturate when
[b] Since vo (0) = 0 =
vo = 6 V,
or
tsat =
q
6/500 ∼
= 0.1095 s
dvo1
1
=−
vg = −25
dt
RaC1
[d] Since vo1 (0) = 0, vo1 = −25t V
[c]
At t = 0.1095 s,
vo1 ∼
= −2.74 V
Therefore the second amplifier saturates before the first amplifier
saturates. Our expressions are valid for 0 ≤ t ≤ 0.1095 s. Once the second
op-amp saturates, our linear model is no longer valid.
Part (2) — Example 8.14 with vo1(0) = −2 V and vo (0) = 4 V:
[a] Initial conditions will not change the differential equation; hence the
equation is the same as Example 8.14.
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8–46
CHAPTER 8. Natural and Step Responses of RLC Circuits
[b] vo = 5 + A01 e−10t + A02e−20t
(from Example 8.14)
vo (0) = 4 = 5 + A01 + A02
4
2
+ iC (0+ ) −
=0
100
25
iC (0+ ) =
4
dvo (0+ )
mA = C
100
dt
dvo (0+ )
0.04 × 10−3
=
= 40 V/s
dt
10−6
dvo
= −10A01 e−10t − 20A02 e−20t
dt
dvo +
(0 ) = −10A01 − 20A02 = 40
dt
Therefore −A01 − 2A02 = 4 and
Thus, A01 = 2 and A02 = −3
A01 + A02 = −1
vo = 5 + 2e−10t − 3e−20t V
[c] Same as Example 8.14:
dvo1
+ 20vo1 = −25
dt
[d] From Example 8.14:
vo1 (∞) = −1.25 V;
v1(0) = −2 V (given)
Therefore
vo1 = −1.25 + (−2 + 1.25)e−20t = −1.25 − 0.75e−20t V
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
P 8.62
8–47
[a]
2C
dva va − vg va
+
+
=0
dt
R
R
(1) Therefore
dva
va
vg
+
=
dt
RC
2RC
0 − va
d(0 − vb)
+C
=0
R
dt
(2) Therefore
dvb
va
+
= 0,
dt
RC
va = −RC
dvb
dt
2vb
dvb
d(vb − vo )
+C
+C
=0
R
dt
dt
(3) Therefore
From (2) we have
dvb
vb
1 dvo
+
=
dt
RC
2 dt
dva
d2 vb
= −RC 2
dt
dt
and va = −RC
dvb
dt
When these are substituted into (1) we get
(4) − RC
d2 vb dvb
vg
−
=
2
dt
dt
2RC
Now differentiate (3) to get
(5)
d2 vb
1 dvb
1 d2 vo
+
=
dt2
RC dt
2 dt2
But from (4) we have
(6)
d2 vb
1 dvb
vg
+
=
−
dt2
RC dt
2R2 C 2
Now substitute (6) into (5)
d2 vo
vg
=− 2 2
2
dt
RC
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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8–48
CHAPTER 8. Natural and Step Responses of RLC Circuits
d2 vo
vg
= 2 2
2
dt
RC
The two equations are the same except for a reversal in algebraic sign.
[b] When R1 C1 = R2 C2 = RC :
[c] Two integrations of the input signal with one operational amplifier.
P 8.63
[a]
d2 vo
1
=
vg
2
dt
R1 C1R2 C2
1
10−6
=
= 250
R1 C1 R2 C2
(100)(400)(0.5)(0.2) × 10−6 × 10−6
d2 vo
= 250vg
dt2
. ·.
0 ≤ t ≤ 0.5− :
vg = 80 mV
d2 vo
= 20
dt2
Let g(t) =
Z
g(t)
g(0)
dvo
,
dt
dx = 20
Z
t
0
then
dy
g(t) − g(0) = 20t,
g(t) =
dg
= 20 or dg = 20 dt
dt
g(0) =
dvo
(0) = 0
dt
dvo
= 20t
dt
dvo = 20t dt
Z
vo (t)
vo (0)
dx = 20
Z
t
0
vo (t) = 10t2 V,
x dx;
vo(t) − vo (0) = 10t2 ,
vo (0) = 0
0 ≤ t ≤ 0.5−
dvo1
1
=−
vg = −20vg = −1.6
dt
R1C1
dvo1 = −1.6 dt
Z
vo1 (t)
vo1 (0)
dx = −1.6
Z
0
t
dy
vo1 (t) − vo1(0) = −1.6t,
vo1 (t) = −1.6t V,
vo1 (0) = 0
0 ≤ t ≤ 0.5−
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
0.5+ ≤ t ≤ tsat:
d2 vo
= −10,
dt2
Z
dg(t) = −10 dt
g(t)
g(0.5+)
dvo
dt
let g(t) =
dg(t)
= −10;
dt
8–49
dx = −10
Z
t
0.5
dy
g(t) − g(0.5+ ) = −10(t − 0.5) = −10t + 5
dvo (0.5+ )
g(0.5 ) =
dt
+
C
dvo
0 − vo1 (0.5+ )
(0.5+ ) =
dt
400 × 103
vo1 (0.5+ ) = vo (0.5− ) = −1.6(0.5) = −0.80 V
. ·. C
dvo1 (0.5+ )
0.80
=
= 2 µA
dt
0.4 × 103
dvo1
2 × 10−6
+
(0.5 ) =
= 10 V/s
dt
0.2 × 10−6
dvo
dt
.·. g(t) = −10t + 5 + 10 = −10t + 15 =
.·. dvo = −10t dt + 15 dt
Z
vo (t)
vo (0.5+)
dx =
Z
t
0.5+
−10y dy +
vo (t) − vo (0.5+ ) = −5y 2
Z
t
0.5+
15 dy
t
t
+ 15y
0.5
0.5
vo (t) = vo (0.5+ ) − 5t2 + 1.25 + 15t − 7.5
vo (0.5+ ) = vo (0.5− ) = 2.5 V
.·. vo(t) = −5t2 + 15t − 3.75 V,
dvo1
= −20(−0.04) = 0.8,
dt
dvo1 = 0.8 dt;
Z
vo1 (t)
vo1 (0.5+ )
0.5+ ≤ t ≤ tsat
dx = 0.8
vo1 (t) − vo1(0.5+ ) = 0.8t − 0.4;
.·. vo1(t) = 0.8t − 1.2 V,
0.5+ ≤ t ≤ tsat
Z
t
0.5+
dy
vo1 (0.5+ ) = vo1(0.5− ) = −0.8 V
0.5+ ≤ t ≤ tsat
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8–50
CHAPTER 8. Natural and Step Responses of RLC Circuits
Summary:
0 ≤ t ≤ 0.5− s :
vo1 = −1.6t V,
0.5+ s ≤ t ≤ tsat :
vo = 10t2 V
vo1 = 0.8t − 1.2 V,
vo = −5t2 + 15t − 3.75 V
[b] −12.5 = −5t2sat + 15tsat − 3.75
.·. 5t2sat − 15tsat − 8.75 = 0
Solving,
tsat = 3.5 sec
vo1 (tsat) = 0.8(3.5) − 1.2 = 1.6 V
P 8.64
τ1 = (106 )(0.5 × 10−6 ) = 0.50 s
1
= 2;
τ1
τ2 = (5 × 106 )(0.2 × 10−6 ) = 1 s;
. ·.
1
=1
τ2
d2 vo
dvo
.·.
+3
+ 2vo = 20
2
dt
dt
s2 + 3s + 2 = 0
(s + 1)(s + 2) = 0;
s1 = −1,
vo = Vf + A01 e−t + A02e−2t;
s2 = −2
Vf =
20
= 10 V
2
vo = 10 + A01e−t + A02 e−2t
vo (0) = 0 = 10 + A01 + A02;
.·. A01 = −20,
dvo
(0) = 0 = −A01 − 2A02
dt
A02 = 10 V
vo (t) = 10 − 20e−t + 10e−2t V,
dvo1
+ 2vo1 = −1.6;
dt
0 ≤ t ≤ 0.5 s
.·. vo1 = −0.8 + 0.8e−2t V,
0 ≤ t ≤ 0.5 s
vo (0.5) = 10 − 20e−0.5 + 10e−1 = 1.55 V
vo1(0.5) = −0.8 + 0.8e−1 = −0.51 V
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Problems
8–51
At t = 0.5 s
iC =
C
0 + 0.51
= 1.26 µA
400 × 103
dvo
= 1.26 µA;
dt
dvo
1.26
=
= 6.32 V/s
dt
0.2
0.5 s ≤ t ≤ ∞:
d2 vo
dvo
+
3
+ 2 = −10
dt2
dt
vo (∞) = −5
.·. vo = −5 + A01e−(t−0.5) + A02e−2(t−0.5)
1.55 = −5 + A01 + A02
dvo
(0.5) = 6.32 = −A01 − 2A02
dt
.·. A01 + A02 = 6.55;
−A01 − 2A02 = 6.32
Solving,
A01 = 19.42;
A02 = −12.87
.·. vo = −5 + 19.42e−(t−0.5) − 12.87e−2(t−0.5) V,
0.5 ≤ t ≤ ∞
dvo1
+ 2vo1 = 0.8
dt
.·. vo1 = 0.4 + (−0.51 − 0.4)e−2(t−0.5) = 0.4 − 0.91e−2(t−0.5) V,
P 8.65
[a] f(t) =
=
0.5 ≤ t ≤ ∞
inertial force + frictional force + spring force
M[d2 x/dt2 ] + D[dx/dt] + Kx
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8–52
CHAPTER 8. Natural and Step Responses of RLC Circuits
d2 x
f
D
[b]
=
−
2
dt
M
M
!
dx
K
−
x
dt
M
Given vA =
d2 x
,
dt2
1
vB = −
R1C1
Z
t
1
vC = −
R2 C2
Z
t
vD = −
0
then
d2 x
dy 2
0
!
vB dy =
dy = −
1 dx
R1 C1 dt
1
x
R1 R2C1 C2
R3
R3 dx
· vB =
R4
R4 R1 C1 dt
R5 + R6
R5 + R6
1
vE =
vC =
·
·x
R6
R6
R1 R2 C1 C2
−R8
vF =
f(t),
R7
Therefore
vA = −(vD + vE + vF )
d2 x
R8
R3
dx
R5 + R6
=
f(t) −
−
x
2
dt
R7
R4R1 C1 dt
R6 R1 R2C1 C2
Therefore M =
R7
,
R8
D=
R3 R7
R8 R4 R1 C1
and
K=
R7 (R5 + R6 )
R8 R6 R1 R2 C1C2
Box Number Function
P 8.66
1
inverting and scaling
2
summing and inverting
3
integrating and scaling
4
integrating and scaling
5
inverting and scaling
6
noninverting and scaling
[a] Given that the current response is underdamped, we know i will be of the
form
i = If + [B10 cos ωd t + B20 sin ωd t]e−αt
where
and
α=
ωd =
R
2L
q
ωo2 − α2 =
s
1
− α2
LC
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Problems
8–53
The capacitor will force the final value of i to be zero, therefore If = 0.
By hypothesis i(0+ ) = Vdc /R; therefore B10 = Vdc /R.
At t = 0+ the voltage across the primary winding is approximately zero;
hence di(0+ )/dt = 0.
From our equation for i we have
di
= [(ωd B20 − αB10 ) cos ωd t − (ωd B10 + αB20 ) sin ωd t]e−αt
dt
Hence
di(0+ )
= ωd B20 − αB10 = 0
dt
Thus
α
αVdc
B20 = B10 =
ωd
ωd R
It follows directly that
i=
Vdc
α
cos ωd t +
sin ωd t e−αt
R
ωd
[b] Since ωd B20 − αB10 = 0, it follows that
di
= −(ωd B10 + αB20 )e−αt sin ωd t
dt
α2 Vdc
=
ωd R
αB20
But
and ωd B10 =
ωd Vdc
R
Therefore
ωd B10
+
αB20
ωd Vdc α2 Vdc
Vdc ωd2 + α2
=
+
=
R
ωd R
R
ωd
"
But ωd2 + α2 = ωo2 =
#
1
LC
Hence
ωd B10 + αB20 =
Vdc
ωd RLC
Now since v1 = L
v1 = −L
we get
Vdc
Vdc −αt
e−αt sin ωd t = −
e
sin ωd t
ωd RLC
ωd RC
[c] vc = Vdc − iR − L
iR = Vdc
di
dt
di
dt
α
cos ωd t +
sin ωd t e−αt
ωd
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8–54
CHAPTER 8. Natural and Step Responses of RLC Circuits
vc = Vdc − Vdc
= Vdc − Vdc e
α
Vdc −αt
cos ωd t +
sin ωd t e−αt +
e sin ωd t
ωd
ωd RC
−αt
Vdc
αVdc −αt
cos ωd t +
−
e sin ωd t
ωd RC
ωd
1
ωd
= Vdc 1 − e−αt cos ωd t +
1
− α e−αt sin ωd t
RC
= Vdc [1 − e−αt cos ωd t + Ke−αt sin ωd t]
P 8.67
vsp = Vdc
a
1−
e−αt sin ωd t
ωd RC
dvsp
−aVdc d −αt
=
[e sin ωd t]
dt
ωd RC dt
=
−aVdc
[−αe−αt sin ωd t + ωd e−αt cos ωd t]
ωd RC
=
aVdc e−αt
[α sin ωd t − ωd cos ωd t]
ωd RC
dvsp
= 0 when α sin ωd t = ωd cos ωd t
dt
or
ωd
tan ωd t =
;
α
−1
ωd t = tan
1
ωd
.·. tmax =
tan−1
ωd
α
ωd
α
Note that because tan θ is periodic, i.e., tan θ = tan(θ ± nπ), where n is an
integer, there are an infinite number of solutions for t where dvsp /dt = 0, that
is
t=
tan−1 (ωd /α) ± nπ
ωd
Because of e−αt in the expression for vsp and knowing t ≥ 0 we know vsp will
be maximum when t has its smallest positive value. Hence
tmax
tan−1 (ωd /α)
=
.
ωd
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Problems
P 8.68
8–55
[a] vc = Vdc [1 − e−αt cos ωd t + Ke−αt sin ωd t]
dvc
d
= Vdc [1 + e−αt(K sin ωd t − cos ωd t)]
dt
dt
= Vdc {(−αe−αt)(K sin ωd t − cos ωd t)+
e−αt[ωd K cos ωd t + ωd sin ωd t]}
= Vdc e−αt[(ωd − αK) sin ωd t + (α + ωd K) cos ωd t]
dvc
= 0 when (ωd − αK) sin ωd t = −(α + ωd K) cos ωd t
dt
or
α + ωd K
tan ωd t =
αK − ωd
.·. ωd t ± nπ = tan−1
α + ωd K
αK − ωd
1
α + ωd K
tc =
tan−1
± nπ
ωd
αK − ωd
α=
R
4 × 103
=
= 666.67 rad/s
2L
6
ωd =
s
109
− (666.67)2 = 28,859.81 rad/s
1.2
1
K=
ωd
tc =
1
− α = 21.63
RC
o
1 n −1
1
tan (−43.29) + nπ = {−1.55 + nπ}
ωd
ωd
The smallest positive value of t occurs when n = 1, therefore
tc max = 55.23 µs
[b] vc (tc max ) = 12[1 − e−αtc max cos ωd tc max + Ke−αtc max sin ωd tc max ]
= 262.42 V
[c] From the text example the voltage across the spark plug reaches its
maximum value in 53.63 µs. If the spark plug does not fire the capacitor
voltage peaks in 55.23 µs. When vsp is maximum the voltage across the
capacitor is 262.15 V. If the spark plug does not fire the capacitor voltage
reaches 262.42 V.
P 8.69
1
1
[a] w = L[i(0+ )]2 = (5)(16) × 10−3 = 40 mJ
2
2
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8–56
CHAPTER 8. Natural and Step Responses of RLC Circuits
[b] α =
R
3 × 103
=
= 300 rad/s
2L
10
ωd =
s
109
− (300)2 = 28,282.68 rad/s
1.25
106
4 × 106
1
=
=
RC
0.75
3
tmax
1
ωd
=
tan−1
ωd
α
vsp (tmax ) = 12 −
= 55.16 µs
12(50)(4 × 106 ) −αtmax
e
sin ωd tmax = −27,808.04 V
3(28,282.68)
[c] vc (tmax) = 12[1 − e−αtmax cos ωd tmax + Ke−αtmax sin ωd tmax]
1
1
− α = 47.13
ωd RC
K=
vc (tmax ) = 568.15 V
P 8.70
[a] vc = Vdc [1 − e−αt cos ωd t + Ke−αt sin ωd t]
dvc
d
= Vdc [1 + e−αt(K sin ωd t − cos ωd t)]
dt
dt
= Vdc {(−αe−αt)(K sin ωd t − cos ωd t)+
e−αt[ωd K cos ωd t + ωd sin ωd t]}
= Vdc e−αt[(ωd − αK) sin ωd t + (α + ωd K) cos ωd t]
dvc
= 0 when (ωd − αK) sin ωd t = −(α + ωd K) cos ωd t
dt
or
α + ωd K
tan ωd t =
αK − ωd
.·. ωd t ± nπ = tan−1
α + ωd K
αK − ωd
1
α + ωd K
tc =
tan−1
± nπ
ωd
αK − ωd
α=
R
3
=
= 300 rad/s
2L
2(5 × 10−3 )
ωd =
s
109
− (300)2 = 28,282.68 rad/s
1.25
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Problems
1
K=
ωd
tc =
8–57
1
− α = 47.13
RC
1
{−1.56 + nπ}
ωd
The smallest positive value of t occurs when n = 1, therefore
tc max = 55.91 µs
[b] vc (tc max ) = 12[1 − e−αtc max cos ωd tc max + Ke−αtc max sin ωd tc max ] = 568.28 V
[c] From Problem 8.69, the voltage across the spark plug reaches its
maximum value in 55.16 µs. If the spark plug does not fire the capacitor
voltage peaks in 55.91 µs. When vsp is maximum the voltage across the
capacitor is 568.15 V. If the spark plug does not fire the capacitor voltage
reaches 568.28 V.
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9
Sinusoidal Steady State Analysis
Assessment Problems
AP 9.1 [a] V = 170/−40◦ V
[b] 10 sin(1000t + 20◦ ) = 10 cos(1000t − 70◦ )
. ·.
I = 10/−70◦ A
[c] I = 5/36.87◦ + 10/−53.13◦
= 4 + j3 + 6 − j8 = 10 − j5 = 11.18/−26.57◦ A
[d] sin(20,000πt + 30◦ ) = cos(20,000πt − 60◦ )
Thus,
V = 300/45◦ − 100/−60◦ = 212.13 + j212.13 − (50 − j86.60)
= 162.13 + j298.73 = 339.90/61.51◦ mV
AP 9.2 [a] v = 18.6 cos(ωt − 54◦ ) V
[b] I = 20/45◦ − 50/ − 30◦ = 14.14 + j14.14 − 43.3 + j25
= −29.16 + j39.14 = 48.81/126.68◦
Therefore i = 48.81 cos(ωt + 126.68◦ ) mA
[c] V = 20 + j80 − 30/15◦ = 20 + j80 − 28.98 − j7.76
= −8.98 + j72.24 = 72.79/97.08◦
v = 72.79 cos(ωt + 97.08◦ ) V
AP 9.3 [a] ωL = (104 )(20 × 10−3 ) = 200 Ω
[b] ZL = jωL = j200 Ω
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
9–1 system, or transmission in any form or by any means, electronic,
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9–2
CHAPTER 9. Sinusoidal Steady State Analysis
[c] VL = IZL = (10/30◦ )(200/90◦ ) × 10−3 = 2/120◦ V
[d] vL = 2 cos(10,000t + 120◦ ) V
−1
−1
=
= −50 Ω
ωC
4000(5 × 10−6 )
[b] ZC = jXC = −j50 Ω
V
30/25◦
= 0.6/115◦ A
[c] I =
=
ZC
50/−90◦
[d] i = 0.6 cos(4000t + 115◦ ) A
AP 9.4 [a] XC =
AP 9.5 I1 = 100/25◦ = 90.63 + j42.26
I2 = 100/145◦ = −81.92 + j57.36
I3 = 100/−95◦ = −8.72 − j99.62
I4 = −(I1 + I2 + I3 ) = (0 + j0) A,
AP 9.6 [a] I =
therefore i4 = 0 A
125/−60◦
125
/(−60 − θZ )◦
=
|Z|/θz
|Z|
But −60 − θZ = −105◦
.·. θZ = 45◦
Z = 90 + j160 + jXC
.·. XC = −70 Ω;
. ·. C =
[b] I =
XC = −
1
= −70
ωC
1
= 2.86 µF
(70)(5000)
Vs
125/−60◦
=
= 0.982/−105◦ A;
Z
(90 + j90)
.·. |I| = 0.982 A
AP 9.7 [a]
ω = 2000 rad/s
ωL = 10 Ω,
−1
= −20 Ω
ωC
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Problems
Zxy = 20kj10 + 5 + j20 =
9–3
20(j10)
+ 5 − j20
(20 + j10)
= 4 + j8 + 5 − j20 = (9 − j12) Ω
[b] ωL = 40 Ω,
Zxy
−1
= −5 Ω
ωC
"
(20)(j40)
= 5 − j5 + 20kj40 = 5 − j5 +
20 + j40
#
= 5 − j5 + 16 + j8 = (21 + j3) Ω
[c] Zxy
20(jωL)
j106
=
+ 5−
20 + jωL
25ω
"
#
!
20ω 2 L2
j400ωL
j106
+
+
5
−
400 + ω 2 L2 400 + ω 2 L2
25ω
The impedance will be purely resistive when the j terms cancel, i.e.,
=
400ωL
106
=
400 + ω 2 L2
25ω
Solving for ω yields ω = 4000 rad/s.
20ω 2 L2
[d] Zxy =
+ 5 = 10 + 5 = 15 Ω
400 + ω 2 L2
AP 9.8 The frequency 4000 rad/s was found to give Zxy = 15 Ω in Assessment
Problem 9.7. Thus,
V = 150/0◦ ,
Is =
V
150/0◦
=
= 10/0◦ A
Zxy
15
Using current division,
IL =
20
(10) = 5 − j5 = 7.07/−45◦ A
20 + j20
iL = 7.07 cos(4000t − 45◦ ) A,
Im = 7.07 A
AP 9.9 After replacing the delta made up of the 50 Ω, 40 Ω, and 10 Ω resistors with its
equivalent wye, the circuit becomes
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9–4
CHAPTER 9. Sinusoidal Steady State Analysis
The circuit is further simplified by combining the parallel branches,
(20 + j40)k(5 − j15) = (12 − j16) Ω
Therefore I =
136/0◦
= 4/28.07◦ A
14 + 12 − j16 + 4
AP 9.10
V1 = 240/53.13◦ = 144 + j192 V
V2 = 96/−90◦ = −j96 V
jωL = j(4000)(15 × 10−3 ) = j60 Ω
1
6 × 106
= −j
= −j60 Ω
jωC
(4000)(25)
Perform a source transformation:
V1
144 + j192
=
= 3.2 − j2.4 A
j60
j60
V2
96
= −j
= −j4.8 A
20
20
Combine the parallel impedances:
Y =
1
1
1
1
j5
1
+
+
+
=
=
j60 30 −j60 20
j60
12
Z=
1
= 12 Ω
Y
Vo = 12(3.2 + j2.4) = 38.4 + j28.8 V = 48/36.87◦ V
vo = 48 cos(4000t + 36.87◦ ) V
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Problems
9–5
AP 9.11 Use the lower node as the reference node. Let V1 = node voltage across the
20 Ω resistor and VTh = node voltage across the capacitor. Writing the node
voltage equations gives us
V1
V1 − 10Ix
− 2/45◦ +
= 0 and
20
j10
VTh =
−j10
(10Ix )
10 − j10
We also have
Ix =
V1
20
Solving these equations for VTh gives VTh = 10/45◦ V. To find the Thévenin
impedance, we remove the independent current source and apply a test
voltage source at the terminals a, b. Thus
It follows from the circuit that
10Ix = (20 + j10)Ix
Therefore
Ix = 0 and IT =
ZTh =
VT
,
IT
VT
VT
+
−j10
10
therefore ZTh = (5 − j5) Ω
AP 9.12 The phasor domain circuit is as shown in the following diagram:
The node voltage equation is
−10 +
V
V
V V − 100/−90◦
+
+
+
=0
5
−j(20/9) j5
20
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9–6
CHAPTER 9. Sinusoidal Steady State Analysis
Therefore V = 10 − j30 = 31.62/−71.57◦
Therefore v = 31.62 cos(50,000t − 71.57◦ ) V
AP 9.13 Let Ia , Ib , and Ic be the three clockwise mesh currents going from left to
right. Summing the voltages around meshes a and b gives
33.8 = (1 + j2)Ia + (3 − j5)(Ia − Ib)
and
0 = (3 − j5)(Ib − Ia) + 2(Ib − Ic ).
But
Vx = −j5(Ia − Ib ),
therefore
Ic = −0.75[−j5(Ia − Ib)].
Solving for I = Ia = 29 + j2 = 29.07/3.95◦ A.
√
AP 9.14 [a] M = 0.4 0.0625 = 0.1 H,
ωM = 80 Ω
Z22 = 40 + j800(0.125) + 360 + j800(0.25) = (400 + j300) Ω
Therefore |Z22| = 500 Ω,
80
Zτ =
500
[b] I1 =
2
∗
Z22
= (400 − j300) Ω
(400 − j300) = (10.24 − j7.68) Ω
245.20
= 0.50/ − 53.13◦ A
184 + 100 + j400 + Zτ
i1 = 0.5 cos(800t − 53.13◦ ) A
[c] I2 =
jωM
j80
I1 =
(0.5/ − 53.13◦ ) = 0.08/0◦ A
Z22
500/36.87◦
i2 = 80 cos 800t mA
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Problems
AP 9.15
I1 =
9–7
Vs
25 × 103 /0◦
=
Z1 + 2s2 Z2
1500 + j6000 + (25)2 (4 − j14.4)
= 4 + j3 = 5/36.87◦ A
V1 = Vs − Z1 I1 = 25,000/0◦ − (4 + j3)(1500 + j6000)
= 37,000 − j28,500
V2 = −
I2 =
1
V1 = −1480 + j1140 = 1868.15/142.39◦ V
25
V2
1868.15/142.39◦
=
= 125/216.87◦ A
Z2
4 − j14.4
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9–8
CHAPTER 9. Sinusoidal Steady State Analysis
Problems
P 9.1
[a] 80 V
[b] 2πf = 1000π;
f = 500 Hz
[c] ω = 1000π = 3141.59 rad/s
−π
[d] θ(rad) =
= −0.5236 rad
6
[e] θ = −30◦
1
1
[f] T = =
= 2 ms
f
500
π
1
[g] 1000πt − = 0;
.·. t =
= 166.67 µs
6
6000
0.002
π
[h] v = 80 cos 1000π t +
−
3
6
= 80 cos[1000πt + (2π/3) − (π/6)]
= 80 cos[1000πt + (π/2)]
= −80 sin 1000πt V
[i] 1000π(t − to ) − (π/6) = 1000πt − (π/2)
.·. 1000πto =
π
;
3
to =
1
= 333.33 µs
3000
[j] 1000π(t + to ) − (π/6) = 1000πt
.·. 1000πto =
P 9.2
[a]
π
;
6
to =
T
= 8 + 2 = 10 ms;
2
f=
1
= 166.67 µs
6000
T = 20 ms
1
1
=
= 50 Hz
T
20 × 10−3
[b] v = Vm sin(ωt + θ)
ω = 2πf = 100π rad/s
100π(−2 × 10−3 ) + θ = 0;
. ·. θ =
π
rad = 36◦
5
v = Vm sin[100πt + 36◦ ]
80.9 = Vm sin 36◦ ;
Vm = 137.64 V
v = 137.64 sin[100πt + 36◦ ] = 137.64 cos[100πt − 54◦ ] V
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Problems
P 9.3
9–9
[a] By hypothesis
i = 20 cos(ωt + θ)
di
= −20ω sin(ωt + θ)
dt
.·. 20ω = 8000π;
ω = 400π rad/s
[b] f =
ω
= 200 Hz;
2π
T =
1
= 5 ms = 5000 µs
f
625
1
1
= ,
.·. θ = − (360) = −45◦
5000
8
8
.·. i = 20 cos(400πt − 45◦ ) A
P 9.4
[a] ω = 2πf = 3769.91 rad/s,
f=
ω
= 600 Hz
2π
[b] T = 1/f = 1.67 ms
[c] Vm = 10 V
[d] v(0) = 10 cos(−53.13◦ ) = 6 V
−53.13◦ (2π)
[e] φ = −53.13◦ ;
φ=
= −0.9273 rad
360◦
[f] V = 0 when 3769.91t − 53.13◦ = 90◦ . Now resolve the units:
(3769.91 rad/s)t =
143.13◦
= 2.498 rad,
57.3◦ /rad
t = 662.64 µs
[g] (dv/dt) = (−10)3769.91 sin(3769.91t − 53.13◦ )
(dv/dt) = 0 when 3769.91t − 53.13◦ = 0◦
or
3769.91t =
53.13◦
= 0.9273 rad
57.3◦ /rad
Therefore t = 245.97 µs
P 9.5
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9–10
CHAPTER 9. Sinusoidal Steady State Analysis
[a] Left as φ becomes more positive
[b] Left
P 9.6
Z
to +T
to
Vm2
2
cos (ωt + φ) dt =
=
Vm2
Z
Vm2
2
Vm2
=
2
Vm2
=
2
P 9.7
P 9.8
Vm =
√
Vrms =
Z
0
T /2
2Vrms =
s
Vm2
1
T
Z
T /2
0
2
sin
√
t
1 1
+ cos(2ωt + 2φ) dt
2 2
(o
R to +T
to
dt +
Z
to +T
cos(2ωt + 2φ) dt
to
)
i
1 h
T+
sin(2ωt + 2φ) |ttoo +T
2ω
1
T+
[sin(2ωto + 4π + 2φ) − sin(2ωto + 2φ)]
2ω
T
1
2
2 T
= Vm
+
(0) = Vm
2
2ω
2
2(240) = 339.41 V
Vm2 sin2
2π
t dt
T
2π
V2
t dt = m
T
2
Therefore Vrms =
P 9.9
to +T
s
Z
0
4π
V 2T
1 − cos t dt = m
T
4
T /2 1 Vm2 T
Vm
=
T 4
2
[a] The numerical values of the terms in Eq. 9.8 are
Vm = 20,
R/L = 1066.67,
√
R2 + ω 2 L2 = 100
ωL = 60
φ = 25◦ ,
θ = 36.87◦
θ = tan−1 60/80,
Substitute these values into Equation 9.9:
h
i
i = −195.72e−1066.67t + 200 cos(800t − 11.87◦ ) mA,
t≥0
[b] Transient component = −195.72e−1066.67t mA
Steady-state component = 200 cos(800t − 11.87◦ ) mA
[c] By direct substitution into Eq 9.9 in part (a), i(1.875 ms) = 28.39 mA
[d] 200 mA,
800 rad/s,
−11.87◦
[e] The current lags the voltage by 36.87◦ .
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Problems
P 9.10
9–11
[a] From Eq. 9.9 we have
L
di
Vm R cos(φ − θ) −(R/L)t ωLVm sin(ωt + φ − θ)
√
= √ 2
e
−
dt
R + ω 2 L2
R2 + ω 2L2
Ri =
−Vm R cos(φ − θ)e−(R/L)t Vm R cos(ωt + φ − θ)
√
√
+
R2 + ω 2 L2
R2 + ω 2 L2
"
di
R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ)
√
L + Ri = Vm
dt
R2 + ω 2 L2
#
But
√
R
= cos θ
R2 + ω 2 L2
and
√
ωL
= sin θ
R2 + ω 2 L2
Therefore the right-hand side reduces to
Vm cos(ωt + φ)
At t = 0, Eq. 9.9 reduces to
i(0) =
−Vm cos(φ − θ) Vm cos(φ − θ)
√
+ √ 2
=0
R2 + ω 2 L2
R + ω 2 L2
Vm
[b] iss = √ 2
cos(ωt + φ − θ)
R + ω 2 L2
Therefore
diss
−ωLVm
L
=√ 2
sin(ωt + φ − θ)
dt
R + ω 2L2
and
Vm R
Riss = √ 2
cos(ωt + φ − θ)
R + ω 2L2
"
diss
R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ)
√
L
+ Riss = Vm
dt
R2 + ω 2 L2
#
= Vm cos(ωt + φ)
P 9.11
[a] Y = 50/60◦ + 100/ − 30◦ = 111.8/ − 3.43◦
y = 111.8 cos(500t − 3.43◦ )
[b] Y = 200/50◦ − 100/60◦ = 102.99/40.29◦
y = 102.99 cos(377t + 40.29◦ )
[c] Y = 80/30◦ − 100/ − 225◦ + 50/ − 90◦ = 161.59/ − 29.96◦
y = 161.59 cos(100t − 29.96◦ )
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9–12
CHAPTER 9. Sinusoidal Steady State Analysis
[d] Y = 250/0◦ + 250/120◦ + 250/ − 120◦ = 0
y=0
P 9.12
[a] Vg = 300/78◦ ;
. ·. Z =
Ig = 6/33◦
Vg
300/78◦
= 50/45◦ Ω
=
Ig
6/33◦
[b] ig lags vg by 45◦ :
2πf = 5000π;
.·. ig lags vg by
P 9.13
f = 2500 Hz;
T = 1/f = 400 µs
45◦
(400 µs) = 50 µs
360◦
[a] ω = 2πf = 160π × 103 = 502.65 krad/s = 502,654.82 rad/s
25 × 10−3 /0◦
[b] I =
= jωC(25 × 10−3 )/0◦ = 25 × 10−3 ωC /90◦
1/jωC
.·. θi = 90◦
[c] 628.32 × 10−6 = 25 × 10−3 ωC
1
25 × 10−3
=
= 39.79 Ω,
ωC
628.32 × 10−6
[d] C =
.·. XC = −39.79 Ω
1
1
=
39.79(ω)
(39.79)(160π × 103 )
C = 0.05 × 10−6 = 0.05 µF
−1
[e] Zc = j
ωC
P 9.14
= −j39.79 Ω
[a] 400 Hz
[b] θv = 0◦
I=
100/0◦
100
/ − 90◦ ;
=
jωL
ωL
θi = −90◦
100
= 20;
ωL = 5 Ω
ωL
5
[d] L =
= 1.99 mH
800π
[e] ZL = jωL = j5 Ω
[c]
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Problems
P 9.15
9–13
[a] ZL = j(8000)(5 × 10−3 ) = j40 Ω
ZC =
−j
= −j100 Ω
(8000)(1.25 × 10−6 )
600/20◦
= 8.32/76.31◦ A
40 + j40 − j100
[c] i = 8.32 cos(8000t + 76.31◦ ) A
[b] I =
P 9.16
[a] jωL = j(2 × 104 )(300 × 10−6 ) = j6 Ω
1
1
= −j
= −j10 Ω;
4
jωC
(2 × 10 )(5 × 10−6 )
Ig = 922/30◦ A
[b] Vo = 922/30◦ Ze
Ze =
1
;
Ye
Ye =
1
1
1
+j +
10
10 8 + j6
Ye = 0.18 + j0.04 S
Ze =
1
= 5.42/ − 12.53◦ Ω
0.18 + j0.04
Vo = (922/30◦ )(5.42/ − 12.53◦ ) = 5000.25/17.47◦ V
[c] vo = 5000.25 cos(2 × 104 t + 17.47◦ ) V
P 9.17
[a] Z1 = R1 + jωL1
Z2 =
R2 (jωL2 )
ω 2 L22 R2 + jωL2 R22
=
R2 + jωL2
R22 + ω 2 L22
Z1 = Z2
when R1 =
ω 2 L22 R2
R22 + ω 2 L22
and L1 =
R22 L2
R22 + ω 2 L22
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9–14
CHAPTER 9. Sinusoidal Steady State Analysis
(4000)2 (1.25)2 (5000)
= 2500 Ω
50002 + 40002 (1.25)2
[b] R1 =
(5000)2 (1.25)
= 625 mH
50002 + 40002 (1.25)2
L1 =
P 9.18
[a] Y2 =
1
j
−
R2 ωL2
1
R1 − jωL1
= 2
R1 + jωL1
R1 + ω 2L21
Y1 =
Therefore
R2 =
R21 + ω 2 L21
R1
and L2 =
[a] Z1 = R1 − j
1
ωC1
R2 /jωC2
R2
R2 − jωR22 C2
=
=
R2 + (1/jωC2 )
1 + jωR2 C2
1 + ω 2 R22 C22
Z2 =
Z1 = Z2
when R1 =
1
ωR22 C2
=
ωC1
1 + ω 2 R22 C22
R2
1 + ω 2R22 C22
or C1 =
and
1 + ω 2 R22 C22
ω 2 R22 C2
1000
[b] R1 =
1 + (40 ×
C1 =
[a] Y2 =
R21 + ω 2 L21
ω 2 L1
80002 + 10002 (4)2
= 20 H
10002 (4)
L2 =
P 9.20
when
80002 + 10002 (4)2
= 10 kΩ
8000
[b] R2 =
P 9.19
Y2 = Y1
103 )2(1000)2 (50
× 10−4 )2
= 200 Ω
1 + (40 × 103 )2 (1000)2 (50 × 10−9 )2
= 62.5 nF
(40 × 103 )2(1000)2 (50 × 10−9 )
1
+ jωC2
R2
Y1 =
1
jωC1
ω 2 R1 C12 + jωC1
=
=
R1 + (1/jωC1 )
1 + jωR1 C1
1 + ω 2 R21 C12
Therefore
R2 =
Y1 = Y2
1 + ω 2 R21 C12
ω 2 R1 C12
and
when
C2 =
C1
1 + ω 2 R21 C12
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Problems
[b] R2 =
1 + (50 × 103 )2(1000)2 (40 × 10−9 )2
= 1250 Ω
(50 × 103 )2 (1000)(40 × 10−9 )2
C2 =
P 9.21
9–15
40 × 10−9
= 8 nF
1 + (50 × 103 )2 (1000)2 (40 × 10−9 )2
[a] R = 300 Ω = 120 Ω + 180 Ω
ωL −
1
1
= −400 so 10,000L −
= −400
ωC
10,000C
Choose L = 10 mH. Then,
1
1
= 100 + 400 so C =
= 0.2 µF
10,000C
10,000(500)
We can achieve the desired capacitance by combining two 0.1 µF
capacitors in parallel. The final circuit is shown here:
[b] 0.01ω =
1
ω(0.2 × 10−6 )
so ω 2 =
1
= 5 × 108
−6
0.01(0.2 × 10 )
.·. ω = 22,360.7 rad/s
P 9.22
[a] Using the notation and results from Problem 9.18:
RkL = 40 + j20 so R1 = 40,
R2 =
L1 =
20
= 4 mH
5000
402 + 50002 (0.004)2
= 50 Ω
40
402 + 50002 (0.004)2
L2 =
= 20 mH
50002 (0.004)
R2 kjωL2 = 50kj100 = 40 + j20 Ω (checks)
The circuit, using combinations of components from Appendix H, is
shown here:
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9–16
CHAPTER 9. Sinusoidal Steady State Analysis
[b] Using the notation and results from Problem 9.22:
RkC = 40 − j20 so R1 = 40,
R2 =
C2 =
C1 = 10 µF
1 + 50002 (40)2 (10 µ)2
= 50 Ω
50002 (40)(10 µ)2
10 µ
1+
50002 (40)2 (10 µ)2
= 2 µF
R2 k(−j/ωC2 ) = 50k(−j100) = 40 − j20 Ω (checks)
The circuit, using combinations of components from Appendix H, is
shown here:
P 9.23
[a] (40 + j20)k(−j/ωC) = 50kj100k(−j/ωC)
To cancel out the j100 Ω impedance, the capacitive impedance must be
−j100 Ω:
−j
1
= −j100 so C =
= 2 µF
5000C
(100)(5000)
Check:
RkjωLk(−j/ωC) = 50kj100k(−j100) = 50 Ω
Create the equivalent of a 2 µF capacitor from components in Appendix
H by combining two 1 µF capacitors in parallel.
[b] (40 − j20)k(jωL) = 50k(−j100)k(jωL)
To cancel out the −j100 Ω impedance, the inductive impedance must be
j100 Ω:
j5000L = j100 so L =
100
= 20 mH
5000
Check:
RkjωLk(−j/ωC) = 50kj100k(−j100) = 50 Ω
Create the equivalent of a 20 mH inductor from components in Appendix
H by combining two 10 mH inductors in series.
P 9.24
[a] Y =
1
1
1
+
+
3 + j4 16 − j12 −j4
= 0.12 − j0.16 + 0.04 + j0.03 + j0.25
= 0.16 + j0.12 = 200/36.87◦ mS
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Problems
9–17
[b] G = 160 mS
[c] B = 120 mS
[d] I = 8/0◦ A,
IC =
V=
I
8
= 40/−36.87◦ V
=
Y
0.2/36.87◦
V
40/−36.87◦
=
= 10/53.13◦ A
ZC
4/−90◦
iC = 10 cos(ωt + 53.13◦ ) A,
P 9.25
[a] jωL = Rk(−j/ωC) = jωL +
jωL +
−jR
ωCR − j
jωL +
−jR(ωCR + j)
ω 2 C 2 R2 + 1
Im(Zab) = ωL −
Im = 10 A
−jR/ωC
R − j/ωC
ωCR2
=0
ω 2 C 2R2 + 1
CR2
ω 2 C 2R2 + 1
. ·.
L=
. ·.
ω 2 C 2 R2 + 1 =
. ·.
(25×10 )(100)
−1
(CR2 /L) − 1
160×10−6
ω =
=
= 900 × 108
2
2
−9
2
C R
(25 × 10 ) (100)2
CR2
L
−9
2
2
ω = 300 krad/s
[b] Zab(300 × 103 ) = j48 +
P 9.26
(100)(−j133.33)
= 64 Ω
100 − j133.33
First find the admittance of the parallel branches
Yp =
1
1
1
1
+
+ +
= 0.375 − j0.125 S
6 − j2 4 + j12 5 j10
Zp =
1
1
=
= 2.4 + j0.8 Ω
Yp
0.375 − j0.125
Zab = −j12.8 + 2.4 + j0.8 + 13.6 = 16 − j12 Ω
Yab =
1
1
=
= 0.04 + j0.03 S
Zab
16 − j12
= 40 + j30 mS = 50/36.87◦ mS
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9–18
P 9.27
CHAPTER 9. Sinusoidal Steady State Analysis
Zab = 1 − j8 + (2 + j4)k(10 − j20) + (40kj20)
= 1 − j8 + 3 + j4 + 8 + j16 = 12 + j12 Ω = 16.97/45◦ Ω
P 9.28
Vg = 40/ − 15◦ V;
Z=
Ig = 40/ − 68.13◦ mA
Vg
= 1000/53.13◦ Ω = 600 + j800 Ω
Ig
0.4 × 106
Z = 600 + j 3.2ω −
ω
!
0.4 × 106
.·. 3.2ω −
= 800
ω
.·. ω 2 − 250ω − 125,000 = 0
Solving,
ω = 500 rad/s
P 9.29
1
1
=
= −j20 Ω
jωC
(1 × 10−6 )(50 × 103 )
jωL = j50 × 103 (1.2 × 10−3 ) = j60 Ω
Vg = 40/0◦ V
Ze = −j20 + 30kj60 = 24 − j8 Ω
Ig =
40/0◦
= 1.5 + j0.5 mA
24 − j8
Vo = (30kj60)Ig =
30(j60)
(1.5 + j0.5) = 30 + j30 = 42.43/45◦ V
30 + j60
vo = 42.43 cos(50,000t + 45◦ ) V
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Problems
P 9.30
[a]
9–19
1
= −j50 Ω
jωC
jωL = j120 Ω
Ze = 100k − j50 = 20 − j40 Ω
Ig = 2/0◦
Vg = Ig Ze = 2(20 − j40) = 40 − j80 V
Vo =
j120
(40 − j80) = 90 − j30 = 94.87/ − 18.43◦ V
80 + j80
vo = 94.87 cos(16 × 105 t − 18.435◦ ) V
[b] ω = 2πf = 16 × 105 ;
T =
. ·.
f=
8 × 105
π
1
π
=
= 1.25π µs
f
8 × 105
18.435
(1.25π µs) = 201.09 ns
360
.·. vo lags ig by 201.09 ns.
P 9.31
Z = 4 + j(50)(0.24) − j
Io =
1
= 5.66/45◦ Ω
(50)(0.0025)
V 0.1/ − 90◦
=
= 17.67/ − 135◦ mA
◦
/
Z
5.66 45
io (t) = 17.67 cos(50t − 135◦ ) mA
P 9.32
ZL = j(2000)(60 × 10−3 ) = j120 Ω
ZC =
−j
= −j40 Ω
(2000)(12.5 × 10−6 )
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9–20
CHAPTER 9. Sinusoidal Steady State Analysis
Construct the phasor domain equivalent circuit:
Using current division:
I=
(120 − j40)
(0.5) = 0.25 − j0.25 A
120 − j40 + 40 + j120
Vo = j120I = 30 + j30 = 42.43/45◦
vo = 42.43 cos(2000t + 45◦ ) V
P 9.33
[a]
Va = (50 + j150)(2/0◦ ) = 100 + j300 V
Ib =
100 + j300
= j2.5 A
120 − j40
Ic = 2/0◦ + j2.5 + 6 + j3.5 = 8 + j6 A
Vg = 5Ic + Va = 5(8 + j6) + 100 + j300 = 140 + j330 V
[b] ib = 2.5 cos(800t + 90◦ ) A
ic = 10 cos(800t + 36.87◦ ) A
vg = 358.47 cos(800t + 67.01◦ ) V
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Problems
P 9.34
9–21
Is = 3/0◦ mA
1
= −j0.4 Ω
jωC
jωL = j0.4 Ω
After source transformation we have
Vo =
−j0.4kj0.4k5
(66 × 10−3 ) = 10 mV
28 + −j0.4kj0.4k5
vo = 10 cos 200t mV
P 9.35
Va − (100 − j50) Va Va − (140 + j30)
+
+
=0
20
j5
12 + j16
Solving,
Va = 40 + j30 V
IZ + (30 + j20) −
140 + j30 (40 + j30) − (140 + j30)
+
=0
−j10
12 + j16
Solving,
IZ = −30 − j10 A
Z=
(100 − j50) − (140 + j30)
= 2 + j2 Ω
−30 − j10
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9–22
CHAPTER 9. Sinusoidal Steady State Analysis
P 9.36
V1 = j5(−j2) = 10 V
−25 + 10 + (4 − j3)I1 = 0
. ·.
I1 =
15
= 2.4 + j1.8 A
4 − j3
Ib = I1 − j5 = (2.4 + j1.8) − j5 = 2.4 − j3.2 A
VZ = −j5I2 + (4 − j3)I1 = −j5(2.4 − j3.2) + (4 − j3)(2.4 + j1.8) = −1 − j12 V
−25 + (1 + j3)I3 + (−1 − j12) = 0
. ·.
I3 = 6.2 − j6.6 A
IZ = I3 − I2 = (6.2 − j6.6) − (2.4 − j3.2) = 3.8 − j3.4 A
Z=
P 9.37
VZ
−1 − j12
=
= 1.42 − j1.88 Ω
IZ
3.8 − j3.4
Simplify the top triangle using series and parallel combinations:
(1 + j1)k(1 − j1) = 1 Ω
Convert the lower left delta to a wye:
Z1 =
(j1)(1)
= j1 Ω
1 + j1 − j1
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Problems
Z2 =
(−j1)(1)
= −j1 Ω
1 + j1 − j1
Z3 =
(j1)(−j1)
= 1Ω
1 + j1 − j1
9–23
Convert the lower right delta to a wye:
Z4 =
(−j1)(1)
= −j1 Ω
1 + j1 − j1
Z5 =
(−j1)(j1)
= 1Ω
1 + j1 − j1
Z6 =
(j1)(1)
= j1 Ω
1 + j1 − j1
The resulting circuit is shown below:
Simplify the middle portion of the circuit by making series and parallel
combinations:
(1 + j1 − j1)k(1 + 1) = 1k2 = 2/3 Ω
Zab = −j1 + 2/3 + j1 = 2/3 Ω
P 9.38
[a] Zg = 500 − j
106
103 (j0.5ω)
+ 3
ω
10 + j0.5ω
= 500 − j
106 500jω(1000 − j0.5ω)
+
ω
106 + 0.25ω 2
= 500 − j
106
250ω 2
5 × 105 ω
+ 6
+
j
ω
10 + 0.25ω 2
106 + 0.25ω 2
.·. If Zg is purely real,
106
5 × 105 ω
= 6
ω
10 + 0.25ω 2
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9–24
CHAPTER 9. Sinusoidal Steady State Analysis
.·.
2(106 + 0.25ω 2 ) = ω 2
. ·.
4 × 106 = ω 2
ω = 2000 rad/s
[b] When ω = 2000 rad/s
Zg = 500 − j500 + (j1000k1000) = 1000 Ω
20/0◦
= 20/0◦ mA
1000
Vo = Vg − Ig Z1
. ·. Ig =
Z1 = 500 − j500 Ω
Vo = 20/0◦ − (0.02/0◦ )(500 − j500) = 10 + j10 = 14.14/45◦ V
vo = 14.14 cos(2000t + 45◦ ) V
P 9.39
50,000 −j20 × 106
[a] Zeq =
+
k(1200 + j0.2ω)
3
ω
=
=
50,000 −j20 × 106
(1200 + j0.2ω)
+
6
3
ω
1200 + j[0.2ω − 20×10
]
ω
50,000
+
3
−j20×106
(1200
ω
h
+ j0.2ω) 1200 − j 0.2ω −
12002 + 0.2ω −
20×106
ω
20×106
ω
2
i
20 × 106
20 × 106
20 × 106
Im(Zeq ) = −
(1200)2 −
0.2ω 0.2ω −
ω
ω
ω
−20 × 106 (1200)2 − 20 × 106
20 × 106
0.2ω 0.2ω −
ω
"
20 × 106
−(1200) = 0.2ω 0.2ω −
ω
2
"
!#
!#
=0
=0
!
0.22 ω 2 − 0.2(20 × 106 ) − 12002 = 0
ω 2 = 64 × 106
. ·.
ω = 8000 rad/s
f = 1273.24 Hz
[b] Zeq =
50,000
+ −j2500k(1200 + j1600)
3
=
Ig =
. ·.
50,000 (−j2500)(1200 + j1600)
+
= 20,000 Ω
3
1200 − j900
30/0◦
= 1.5/0◦ mA
20,000
ig (t) = 1.5 cos 8000t mA
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Problems
P 9.40
9–25
R
R
jωC
[a] Zp =
=
R + (1/jωC)
1 + jωRC
=
10,000
10,000
=
1 + j(5000)(10,000)C
1 + j50 × 106 C
=
10,000(1 − j50 × 106 C)
1 + 25 × 1014 C 2
=
10,000
5 × 1011 C
−
j
1 + 25 × 1014 C 2
1 + 25 × 1014 C 2
jωL = j5000(0.8) = j4000
.·. 4000 =
5 × 1011 C
1 + 25 × 1014 C 2
.·. 1014 C 2 − 125 × 106 C + 1 = 0
.·. C 2 − 5 × 10−8 C + 4 × 10−16 = 0
Solving,
C1 = 40 nF
C2 = 10 nF
10,000
[b] Re =
1 + 25 × 1014 C 2
When C = 40 nF
Re = 2000 Ω;
80/0◦
Ig =
= 40/0◦ mA;
ig = 40 cos 5000t mA
2000
When C = 10 nF
Re = 8000 Ω;
80/0◦
Ig =
= 10/0◦ mA;
8000
P 9.41
ig = 10 cos 5000t mA
109
[a] ZC =
= −j4000 Ω
j(50,000)(5)
Z1 = 10,000kj50,000L =
10,000(j50,000L)
250,000L2 + j50,000L
=
10,000 + j50,000L
1 + 25L2
250,000L2 + j50,000L
ZT = Z1 + ZR + ZC =
− j4000 + 2000
1 + 25L2
ZT is resistive when
50,000L
= 4000
or
1 + 25L2
L2 − 0.5L + 0.04 = 0
Solving, L1 = 0.4 H and L2 = 0.1 H.
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9–26
CHAPTER 9. Sinusoidal Steady State Analysis
[b] When L = 0.4 H:
ZT = 2000 +
250,000(0.16)
= 10,000 Ω
1 + 25(0.16)
50/0◦
= 5/0◦ mA
10,000
Ig =
ig = 5 cos 50,000t mA
When L = 0.1 H:
250,000(0.01)
ZT = 2000 +
= 4000 Ω
1 + 25(0.01)
50/0◦
= 12.5/0◦ mA
4000
Ig =
ig = 12.5 cos 50,000t mA
P 9.42
[a] Y1 =
1
= 0.2 × 10−3 S
5000
Y2 =
=
1
1200 + j0.2ω
1200
0.2ω
−
j
1.44 × 106 + 0.04ω 2
1.44 × 106 + 0.04ω 2
Y3 = jω50 × 10−9
YT = Y1 + Y2 + Y3
For ig and vo to be in phase the j component of YT must be zero; thus,
ω50 × 10−9 =
0.2ω
1.44 × 106 + 0.04ω 2
or
0.04ω 2 + 1.44 × 106 =
0.2 × 109
= 4 × 106
50
.·. 0.04ω 2 = 2.56 × 106
[b] YT = 0.2 × 10−3 +
.·. ω = 8000 rad/s = 8 krad/s
1200
= 0.5 × 10−3 S
1.44 × 106 + 0.04(64) × 106
.·. ZT = 2000 Ω
Vo = (2.5 × 10−3 /0◦ )(2000) = 5/0◦
vo = 5 cos 8000t V
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Problems
9–27
P 9.43
IN =
5 − j15
+ (1 − j3) mA,
ZN
IN =
−18 − j13.5
+ 4.5 − j6 mA,
ZN
ZN in kΩ
ZN in kΩ
5 − j15
−18 − j13.5
+ 1 − j3 =
+ (4.5 − j6)
ZN
ZN
23 − j15
= 3.5 − j3
ZN
IN =
P 9.44
.·.
ZN = 4 + j3 kΩ
5 − j15
+ 1 − j3 = −j6 mA
4 + j3
[a] jωL = j(1000)(100) × 10−3 = j100 Ω
1
106
= −j
= −j100 Ω
jωC
(1000)(10)
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9–28
CHAPTER 9. Sinusoidal Steady State Analysis
Using voltage division,
Vab =
(100 + j100)k(−j100)
(247.49/45◦ ) = 350/0◦
j100 + (100 + j100)k(−j100)
VTh = Vab = 350/0◦ V
[b] Remove the voltage source and combine impedances in parallel to find
ZTh = Zab :
Yab =
1
1
1
+
+
= 5 − j5 mS
j100 100 + j100 −j100
ZTh = Zab =
1
= 100 + j100 Ω
Yab
[c]
P 9.45
Step 1 to Step 2:
240/0◦
= −j20 = 20/ − 90◦ A
j12
Step 2 to Step 3:
(j12)k36 = 3.6 + j10.8 Ω
Step 3 to Step 4:
(20/ − 90◦ )(3.6 + j10.8) = 216 − j72 = 227.68/ − 18.43◦ V
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Problems
P 9.46
9–29
Step 1 to Step 2:
(4/0◦ )(50) = 200/0◦ V
Step 2 to Step 3:
50 + 30 + j60 = (80 + j60) Ω
Step 3 to Step 4:
200/0◦
= 2/ − 36.87◦ A
(80 + j60)
Step 4 to Step 5:
(80 + j60k − j100 = 100 − j50 Ω
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9–30
P 9.47
CHAPTER 9. Sinusoidal Steady State Analysis
Open circuit voltage:
(9 + j4)Ia − Ib = −60/0◦
−Ia + (9 − j4)Ib = 60/0◦
Solving,
Ia = −5 + j2.5 A;
Ib = 5 + j2.5 A
VTh = 4Ia + (4 − j4)Ib = 10/0◦ V
Short circuit current:
(9 + j4)Ia − 1Ib − 4Isc = −60
−1Ia + (9 − j4)Ib − (4 − j4)Isc = 60
−4Ia − (4 − j4)Ib + (8 − j4)Isc = 0
Solving,
Isc = 2.07/0◦
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Problems
ZTh =
9–31
VTh
10/0◦
=
= 4.83 Ω
Isc
2.07/0◦
Alternate calculation for ZTh :
X
Z = 4 + 1 + 4 − j4 = 9 − j4
Z1 =
4
9 − j4
Z2 =
4 − j4
9 − j4
Z3 =
16 − j16
9 − j4
Za = 4 + j4 +
4
56 + j20
=
9 − j4
9 − j4
Zb = 4 +
4 − j4
40 − j20
=
9 − j4
9 − j4
Za kZb =
2640 − j320
884 − j384
Z3 + Za kZb =
16 − j16 2640 − j320
+
= 4.83 Ω
9 − j4
884 − j384
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9–32
P 9.48
CHAPTER 9. Sinusoidal Steady State Analysis
Open circuit voltage:
V1 − 250
V1
− 0.03Vo +
=0
20 + j10
50 − j100
.·. Vo =
−j100
V1
50 − j100
V1
j3V1
V1
250
+
+
=
20 + j10 50 − j100 50 − j100
20 + j10
V1 = 500 − j250 V;
Vo = 300 − j400 V = VTh
Short circuit current:
250/0◦
Isc =
= 3.5 − j0.5 A
70 + j10
ZTh =
VTh
300 − j400
=
= 100 − j100 Ω
Isc
3.5 − j0.5
The Thévenin equivalent circuit:
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Problems
P 9.49
9–33
Short circuit current
Iβ =
−6Iβ
2
2Iβ = −6Iβ ;
I1 = 0;
. ·. Iβ = 0
.·. Isc = 10/−45◦ A = IN
The Norton impedance is the same as the Thévenin impedance. Find it using
a test source
VT = 6Iβ + 2Iβ = 8Iβ ,
ZTh =
P 9.50
Iβ =
j1
IT
2 + j1
VT
8Iβ
j8
=
=
= 1.6 + j3.2 Ω
IT
[(2 + j1)/j1]Iβ
2 + j1
jωL = j100 × 103 (0.6 × 10−3 ) = j60 Ω
1
−j
=
= −j25 Ω
3
jωC
(100 × 10 )(0.4 × 10−6 )
VT = −j25IT + 5I∆ − 30I∆
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9–34
CHAPTER 9. Sinusoidal Steady State Analysis
I∆ =
−j60
IT
30 + j60
VT = −j25IT + 25
j60
IT
30 + j60
VT
= Zab = 20 − j15 = 25/ − 36.87◦ Ω
IT
P 9.51
1
109
=
= 8 kΩ
ωC1
50,000(2.5)
1
109
=
= 4 kΩ
ωC2
50,000(5)
VT = (2400 − j8000)IT + 40IT (90)
ZTh =
P 9.52
VT
= 6000 − j8000 Ω
IT
Open circuit voltage:
V2 − 15 V2
V2
+ 88Iφ +
=0
10
−j50
Iφ =
5 − (V2 /5)
200
Solving,
V2 = −66 + j88 = 110/126.87◦ V = VTh
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Problems
9–35
Find the Thévenin equivalent impedance using a test source:
IT =
VT
0.8Vt
+ 88Iφ +
10
−j50
Iφ =
−VT /5
200
IT = VT
.·.
IN =
1
VT /5
0.8
− 88
+
10
200
−j50
!
VT
= 30 − j40 = ZTh
IT
VTh
−66 + j88
=
= −2.2 + j0 A
ZTh
30 − j40
The Norton equivalent circuit:
P 9.53
[a]
IT =
VT
VT − αVT
+
1000
−j1000
IT
1
(1 − α)
j −1+α
=
−
=
VT
1000
j1000
j1000
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9–36
CHAPTER 9. Sinusoidal Steady State Analysis
.·. ZTh =
VT
j1000
=
IT
α−1+j
ZTh is real when α = 1.
[b] ZTh = 1000 Ω
[c] ZTh = 500 − j500 =
=
j1000
α−1+j
1000(α − 1)
1000
+j
2
(α − 1) + 1
(α − 1)2 + 1
Equate the real parts:
1000
= 500
(α − 1)2 + 1
. ·.
. ·.
(α − 1)2 + 1 = 2
(α − 1)2 = 1 so α = 0
Check the imaginary parts:
(α − 1)1000
(α − 1)2 + 1
= −500
α=1
Thus, α = 0.
1000
1000(α − 1)
[d] ZTh =
+j
2
(α − 1) + 1
(α − 1)2 + 1
For Im(ZTh ) > 0, α must be greater than 1. So ZTh is inductive for
1 < α ≤ 10.
P 9.54
jωL = j(2000)(1 × 10−3 ) = j2 Ω
1
106
= −j
= −j5 Ω
jωC
(2000)(100)
Vg1 = 20/ − 36.87◦ = 16 − j12 V
Vg2 = 50/−106.26◦ = −14 − j48 V
Vo − (16 − j12) Vo Vo − (−14 − j48)
+
+
=0
j2
10
−j5
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Problems
9–37
Solving,
Vo = 36/0◦
vo (t) = 36 cos 2000t V
P 9.55
V1 − 240 V1
V1
+
+
=0
j10
50
30 + j10
Solving for V1 yields
V1 = 198.63/ − 24.44◦ V
Vo =
P 9.56
30
(V1 ) = 188.43/ − 42.88◦ V
30 + j10
Set up the frequency domain circuit to use the node voltage method:
V1 − V2 V1 − 20/90◦
+
=0
−j8
−j4
At V1:
− 5/0◦ +
At V2:
V2 − V1 V2 V2 − 20/90◦
+
+
=0
−j8
j4
12
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9–38
CHAPTER 9. Sinusoidal Steady State Analysis
In standard form:
!
!
V1
1
1
1
+
+ V2 −
−j8 −j4
−j8
V1
1
1
1
1
−
+ V2
+
+
−j8
−j8 j4 12
= 5/0◦ +
!
!
=
20/90◦
−j4
20/90◦
12
Solving on a calculator:
8
4
V1 = − + j
3
3
V2 = −8 + j4
Thus
8
56
Vg = V1 − 20/90◦ = − − j V
3
3
P 9.57
jωL = j106 (10 × 10−6 ) = j10 Ω
1
−j
= 6
= −j10 Ω
jωC
10 (100 × 10−9 )
Va = 50/ − 90◦ = −j50 V
Vb = 25/90◦ = j25 V
V1 V1 + j25 V1 + j50
+
+
=0
10
10
−j10
Solving,
V1 = 25/ − 53.13◦ V = 15 − j20 V
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Problems
Ia =
9–39
V1 + j50 −j25 + j50
+
−j10
j10
= −0.5 + j1.5 = 1.58/108.43◦ A
ia = 1.58 cos(106 t + 108.43◦ ) A
Ib =
−j25 − V1 −j25 + j50
+
10
j10
= 1 − j0.5 = 1.12/ − 26.57◦ A
ib = 1.12 cos(106 t − 26.57◦ ) A
Ic =
−j50 + j25
j10
= −2.5 A
ic = 2.5 cos(106 t + 180◦ ) A
P 9.58
Vo
Vo
+
+ 20Io = 0
50
−j25
(2 + j4)Vo = −2000Io
Vo = (−200 + j400)Io
Io =
V1 − (Vo /10)
j25
.·. V1 = (−20 + j65)Io
0.006 + j0.013 =
V1
+ Io = (−0.4 + j1.3)Io + Io = (0.6 + j1.3)Io
50
0.6 + j1.3(10 × 10−3 )
.·. Io =
= 10/0◦ mA
(0.6 + j1.3)
Vo = (−200 + j400)Io = −2 + j4 = 4.47/116.57◦ V
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9–40
P 9.59
CHAPTER 9. Sinusoidal Steady State Analysis
Write a KCL equation at the top node:
Vo
Vo − 2.4I∆ Vo
+
+
− (10 + j10) = 0
−j8
j4
5
The constraint equation is:
I∆ =
Vo
−j8
Solving,
Vo = j80 = 80/90◦ V
P 9.60
The circuit with the mesh currents identified is shown below:
The mesh current equations are:
−20/ − 36.87◦ + j2I1 + 10(I1 − I2) = 0
50/ − 106.26◦ + 10(I2 − I1 ) − j5I2 = 0
In standard form:
I1 (10 + j2) + I2(−10) = 20/ − 36.87◦
I1 (−10) + I2 (10 − j5) = 50/ − 106.26◦
Solving on a calculator yields:
I1 = −6 + j10 A;
I2 = −9.6 + j10 A
Thus,
Vo = 10(I1 − I2 ) = 36 V
and
vo (t) = 36 cos 2000t V
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Problems
P 9.61
Va = 60/0◦ V;
9–41
Vb = 90/90◦ V
jωL = j(4 × 104 )(125 × 10−6 ) = j5Ω
−j
−j106
=
= −j20 Ω
ωC
40,000(1.25)
60 = (20 + j5)Ia − j5Ib
j90 = −j5Ia − j15Ib
Solving,
Ia = 2.25 − j2.25 A;
Ib = −6.75 + j0.75 A
Io = Ia − Ib = 9 − j3 = 9.49/ − 18.43◦ A
io (t) = 9.49 cos(40,000t − 18.43◦ ) A
P 9.62
(12 − j12)Ia − 12Ig − 5(−j8) = 0
−12Ia + (12 + j4)Ig + j20 − 5(j4) = 0
Solving,
Ig = 4 − j2 = 4.47/ − 26.57◦ A
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9–42
CHAPTER 9. Sinusoidal Steady State Analysis
P 9.63
10/0◦ = (1 − j1)I1 − 1I2 + j1I3
−5/0◦ = −1I1 + (1 + j1)I2 − j1I3
1 = j1I1 − j1I2 + I3
Solving,
I1 = 11 + j10 A;
I2 = 11 + j5 A;
I3 = 6 A
Ia = I3 − 1 = 5 A
Ib = I1 − I3 = 5 + j10 A
Ic = I2 − I3 = 5 + j5 A
Id = I1 − I2 = j5 A
P 9.64
jωL = j10,000(5 × 10−3 ) = j50 Ω
1
−j
=
= −j50 Ω
jωC
(10,000)(2 × 10−6 )
130/0◦ = (40 + j50)Ia − 40I∆ + 30I∆
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Problems
9–43
0 = −40Ia + 30I∆ + (140 − j50)I∆
Solving,
I∆ = (400 − j400) mA
Vo = 100I∆ = 40 − j40 = 56.57/ − 45◦
vo = 56.57 cos(10,000t − 45◦ ) V
P 9.65
1
109
= −j
= −j100 Ω
jωC
(12,500)(800)
jωL = j(12,500)(0.04) = j500 Ω
Let Z1 = 50 − j100 Ω;
Z2 = 250 + j500 Ω
Ig = 125/0◦ mA
Io =
−Ig Z2
−125/0◦ (250 + j500)
=
Z1 + Z2
(300 + j400)
= −137.5 − j25 mA = 139.75/ − 169.7◦ mA
io = 139.75 cos(12,500t − 169.7◦ ) mA
P 9.66
Zo = 12,000 − j
109
= 12,000 − j16,000 Ω
(20,000)(3.125)
ZT = 6000 + j40,000 + 12,000 − j16,000 = 18,000 + j24,000 Ω = 30,000/53.13◦ Ω
Vo = Vg
Zo
(75/0◦ )(20,000/ − 53.13◦ )
=
= 50/ − 106.26◦ V
◦
/
ZT
30,000 53.13
vo = 50 cos(20,000t − 106.26◦ ) V
P 9.67
1
= −j10 kΩ
jωC1
1
= −j100 kΩ
jωC2
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9–44
CHAPTER 9. Sinusoidal Steady State Analysis
Va − 2
Va
Va
Va − Vo
+
+
+
=0
5000
−j10,000 20,000
100,000
20Va − 40 + j10V a + 5Va + Va − Vo = 0
.·.
(26 + j10)Va − Vo = 40
0 − Va
0 − Vo
+
=0
20,000
−j100,000
j5Va − Vo = 0
Solving,
Vo = 1.43 + j7.42 = 7.56/79.09◦ V
vo (t) = 7.56 cos(106 t + 79.09◦ ) V
P 9.68
[a] Vg = 25/0◦ V
Vp =
20
Vg = 5/0◦ ;
100
Vn = Vp = 5/0◦ V
5
5 − Vo
+
=0
80,000
Zp
Zp = −j80,000k40,000 = 32,000 − j16,000 Ω
Vo =
5Zp
+ 5 = 7 − j = 7.07/ − 8.13◦
80,000
vo = 7.07 cos(50,000t − 8.13◦ ) V
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Problems
[b] Vp = 0.2Vm /0◦ ;
9–45
Vn = Vp = 0.2Vm /0◦
0.2Vm
0.2Vm − Vo
+
=0
80,000 32,000 − j16,000
.·. Vo = 0.2Vm +
32,000 − j16,000
Vm (0.2) = Vm (0.28 − j0.04)
80,000
.·. |Vm (0.28 − j0.04)| ≤ 10
.·. Vm ≤ 35.36 V
P 9.69
1
= −j20 kΩ
jωC
Let Va = voltage across the capacitor, positive at upper terminal
Then:
Vg = 4/0◦ V;
Va − 4/0◦
Va
Va
+
+
= 0;
20,000
−j20,000 20,000
0 − Va 0 − Vo
+
= 0;
20,000
10,000
Vo = −
.·. Va = (1.6 − j0.8) V
Va
2
.·. Vo = −0.8 + j0.4 = 0.89/153.43◦ V
vo = 0.89 cos(200t + 153.43◦ ) V
P 9.70
[a]
Va − 4/0◦
Va
+ jωCo Va +
=0
20,000
20,000
Va =
4
2 + j20,000ωCo
Vo = −
Va
2
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9–46
CHAPTER 9. Sinusoidal Steady State Analysis
Vo =
−2
2/180◦
=
2 + j4 × 106 Co
2 + j4 × 106 Co
.·. denominator angle = 45◦
.·.
so 4 × 106 Co = 2
[b] Vo =
C = 0.5 µF
2/180◦
= 0.707/135◦ V
2 + j2
vo = 0.707 cos(200t + 135◦ ) V
P 9.71
[a]
1
−j109
=
= −j100 Ω
jωC
(106 )(10)
Vg = 30/0◦ V
Vp =
Vg (1/jωCo )
30/0◦
=
= Vn
25 + (1/jωCo )
1 + j25ωCo
Vn
Vn − Vo
+
=0
100
−j100
1 + j1
30(1 − j1)
Vn = (1 − j1)Vn =
j
1 + j25ωCo
√
30 2
|Vo | = q
=6
1 + 625ω 2 Co2
Vo =
Solving,
Co = 280 nF
[b] Vo =
30(1 − j1)
= 6/ − 126.87◦
1 + j7
vo = 6 cos(106 t − 126.87◦ ) V
P 9.72
[a]
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Problems
9–47
Because the op-amps are ideal Iin = Io , thus
Zab =
Vab
Vab
=
;
Iin
Io
Vo1 = Vab ;
Vo2
Io =
Vab − Vo
Z
R2
=−
Vo1 = −KVo1 = −KVab
R1
Vo = Vo2 = −KVab
. ·. Io =
Vab − (−KVab)
(1 + K)Vab
=
Z
Z
.·. Zab =
[b] Z =
P 9.73
1
;
jωC
Vab
Z
Z=
(1 + K)Vab
(1 + K)
Zab =
1
;
jωC(1 + K)
.·. Cab = C(1 + K)
[a] Superposition must be used because the frequencies of the two sources are
different.
[b] For ω = 2000 rad/s:
10k − j5 = 2 − j4 Ω
so
Vo1 =
2 − j4
(20/ − 36.87◦ = 31.62/ − 55.3◦ V
2 − j4 + j2
For ω = 5000 rad/s:
j5k10 = 2 + j4 Ω
Vo2 =
2 + j4
(10/16.26◦ ) = 15.81/34.69◦ V
2 + j4 − j2
Thus,
vo (t) = [31.62 cos(2000t − 55.3◦ ) + 15.81 cos(5000t + 34.69◦ )] V,
t≥0
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9–48
P 9.74
CHAPTER 9. Sinusoidal Steady State Analysis
[a] Superposition must be used because the frequencies of the two sources are
different.
[b] For ω = 80,000 rad/s:
V0o − 5 V0o
V0o
+
+
=0
20
j10 −j10
V0o
1
1
1
+
+
20 j10 −j10
!
=
5
20
.·. V0o = 5/0◦ V
I0o
V0o
=
= −j0.5 = 500/ − 90◦ mA
j10
For ω = 320,000 rad/s:
20kj40 = 16 + j8 Ω
V00 =
16 + j8
(2.5/0◦ ) = 2.643/7.59◦ V
16 + j8 − j2.5
V00
00
·
. . Io =
= 66.08/ − 82.4◦ mA
j40
Thus,
io (t) = [500 sin 80,000t + 66.08 cos(320,000t − 82.4◦ )] mA,
P 9.75
t≥0
[a] jωLL = j100 Ω
jωL2 = j500 Ω
Z22 = 300 + 500 + j100 + j500 = 800 + j600 Ω
∗
Z22
= 800 − j600 Ω
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Problems
9–49
ωM = 270 Ω
Zr =
2
270
1000
[800 − j600] = 58.32 − j43.74 Ω
[b] Zab = R1 + jωL1 + Zr = 41.68 + j180 + 58.32 − j43.74 = 100 + j136.26 Ω
P 9.76
[a] jωL1 = j(200 × 103 )(10−3 ) = j200 Ω
jωL2 = j(200 × 103 )(4 × 10−3 ) = j800 Ω
1
−j
=
= −j400 Ω
3
jωC
(200 × 10 )(12.5 × 10−9 )
.·. Z22 = 100 + 200 + j800 − j400 = 300 + j400 Ω
∗
.·. Z22
= 300 − j400 Ω
q
M = k L1 L2 = 2k × 10−3
ωM = (200 × 103 )(2k × 10−3 ) = 400k
"
400k
Zr =
500
#2
(300 − j400) = k 2(192 − j256) Ω
Zin = 200 + j200 + 192k 2 − j256k 2
1
|Zin | = [(200 + 192k)2 + (200 − 256k)2 ] 2
d|Zin |
1
1
= [(200 + 192k)2 + (200 − 256k)2 ]− 2 ×
dk
2
[2(200 + 192k 2 )384k + 2(200 − 256k 2 )(−512k)]
d|Zin |
= 0 when
dk
768k(200 + 192k 2 ) − 1024k(200 − 256k 2 ) = 0
√
.·. k 2 = 0.125;
.·. k = 0.125 = 0.3536
[b] Zin (min) = 200 + 192(0.125) + j[200 − 0.125(256)]
= 224 + j168 = 280/36.87◦ Ω
I1 (max) =
560/0◦
= 2/ − 36.87◦ A
224 + j168
.·. i1 (peak) = 2 A
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9–50
CHAPTER 9. Sinusoidal Steady State Analysis
Note — You can test that the k value obtained from setting d|Zin |/dt = 0
leads to a minimum by noting 0 ≤ k ≤ 1. If k = 1,
Zin = 392 − j56 = 395.98/ − 8.13◦ Ω
Thus,
|Zin |k=1 > |Zin |k=√0.125
If k = 0,
Zin = 200 + j200 = 282.84/45◦ Ω
Thus,
|Zin |k=0 > |Zin |k=√0.125
P 9.77
[a] jωL1 = j(5000)(2 × 10−3 ) = j10 Ω
jωL2 = j(5000)(8 × 10−3 ) = j40 Ω
jωM = j10 Ω
70 = (10 + j10)Ig + j10IL
0 = j10Ig + (30 + j40)IL
Solving,
Ig = 4 − j3 A;
IL = −1 A
ig = 5 cos(5000t − 36.87◦ ) A
iL = 1 cos(5000t − 180◦ ) A
M
2
= √ = 0.5
L1 L2
16
[c] When t = 100π µs,
[b] k = √
5000t = (5000)(100π) × 10−6 = 0.5π = π/2 rad = 90◦
ig (100πµs) = 5 cos(53.13◦ ) = 3 A
iL (100πµs) = 1 cos(−90◦ ) = 0 A
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Problems
9–51
1
1
1
w = L1 i21 + L2 i22 + Mi1 i2 = (2 × 10−3 )(9) + 0 + 0 = 9 mJ
2
2
2
When t = 200π µs,
5000t = π rad = 180◦
ig (200πµs) = 5 cos(180 − 53.13) = −4 A
iL (200πµs) = 1 cos(180 − 180) = 1 A
1
1
w = (2 × 10−3 )(16) + (8 × 10−3 )(1) + 2 × 10−3 (−4)(1) = 12 mJ
2
2
P 9.78
Remove the voltage source to find the equivalent impedance:
ZTh
20
= 45 + j125 +
|5 + j5|
!2
(5 + j5) = 85 + j85 Ω
Using voltage division:
VTh = Vcd
P 9.79
425
= j20I1 = j20
5 + j5
!
= 850 + j850 V
jωL1 = j50 Ω
jωL2 = j32 Ω
1
= −j20 Ω
jωC
q
jωM = j(4 × 103 )k (12.5)(8) × 10−3 = j40k Ω
Z22 = 5 + j32 − j20 = 5 + j12 Ω
∗
Z22
= 5 − j12 Ω
"
40k
Zr =
|5 + j12|
#2
(5 − j12) = 47.337k 2 − j113.609k 2
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9–52
CHAPTER 9. Sinusoidal Steady State Analysis
Zab = 20 + j50 + 47.337k 2 − j113.609k 2 = (20 + 47.337k 2 ) + j(50 − 113.609k 2 )
Zab is resistive when
50 − 113.609k 2 = 0
or
k 2 = 0.44 so k = 0.66
.·. Zab = 20 + (47.337)(0.44) = 40.83 Ω
P 9.80
In Eq. 9.69 replace ω 2 M 2 with k 2ω 2 L1 L2 and then write Xab as
Xab = ωL1 −
= ωL1
k 2 ω 2 L1 L2 (ωL2 + ωLL )
R222 + (ωL2 + ωLL )2
k 2 ωL2 (ωL2 + ωLL )
1− 2
R22 + (ωL2 + ωLL )2
(
)
For Xab to be negative requires
R222 + (ωL2 + ωLL )2 < k 2 ωL2 (ωL2 + ωLL )
or
R222 + (ωL2 + ωLL )2 − k 2ωL2 (ωL2 + ωLL ) < 0
which reduces to
R222 + ω 2 L22 (1 − k 2) + ωL2 ωLL (2 − k 2 ) + ω 2 L2L < 0
But k ≤ 1, so it is impossible to satisfy the inequality. Therefore Xab can
never be negative if XL is an inductive reactance.
P 9.81
[a]
Zab =
Vab
V1 + V2
=
I1
I1
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Problems
V1
V2
=
,
N1
N2
V2 =
N1 I1 = N2 I2 ,
N2
V1
N1
N1
I1
N2
I2 =
V2 = (I1 + I2 )ZL = I1 1 +
V1 + V2 =
.·. Zab =
Zab
9–53
N1
ZL
N2
N1
N1
+ 1 V2 = 1 +
N2
N2
2
ZL I1
(1 + N1 /N2 )2ZL I1
I1
N1
= 1+
N2
2
ZL
Q.E.D.
[b] Assume dot on N2 is moved to the lower terminal, then
V1
−V2
=
,
N1
N2
V1 =
N1 I1 = −N2 I2,
−N1
V2
N2
I2 =
−N1
I1
N2
and
Zab =
As in part [a]
V2 = (I2 + I1 )ZL
Zab =
(1 − N1 /N2 )V2
(1 − N1 /N2 )(1 − N1 /N2 )ZL I1
=
I1
I1
2
Zab = [1 − (N1 /N2 )] ZL
P 9.82
V1 + V2
I1
Q.E.D.
[a]
N1 I1 = N2 I2 ,
Zab =
I2 =
N1
I1
N2
Vab
V2
V2
=
=
I1 + I2
I1 + I2
(1 + N1/N2 )I1
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9–54
CHAPTER 9. Sinusoidal Steady State Analysis
V1
N1
=
,
V2
N2
V1 =
N1
V2
N2
N1
V1 + V2 = ZL I1 =
+ 1 V2
N2
Zab =
I1 ZL
(N1/N2 + 1)(1 + N1/N2 )I1
.·. Zab =
ZL
[1 + (N1 /N2 )]2
Q.E.D.
[b] Assume dot on the N2 coil is moved to the lower terminal. Then
N1
N1
V2 and I2 = − I1
N2
N2
As before
V2
Zab =
and V1 + V2 = ZL I1
I1 + I2
V1 = −
.·. Zab =
Zab =
V2
ZL I1
=
(1 − N1 /N2 )I1
[1 − (N1 /N2 )]2I1
ZL
[1 − (N1 /N2 )]2
Q.E.D.
P 9.83
ZL =
V3
I3
V2
V3
=
;
10
1
V1
V2
=− ;
8
1
Zab =
10I2 = 1I3
8I1 = −1I2
V1
I1
Substituting,
Zab =
V1
−8V2
82 V2
=
=
I1
−I2 /8
I2
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Problems
=
P 9.84
9–55
82 (10V3 )
(8)2 (10)2 V3
=
= (8)2 (10)2 ZL = (8)2 (10)2 (80/60◦ ) = 512,000/60◦ Ω
I3/10
I3
The phasor domain equivalent circuit is
Vo =
Vm
− IRx ;
2
I=
Vm
Rx − jXC
As Rx varies from 0 to ∞, the amplitude of vo remains constant and its phase
angle increases from 0◦ to −180◦ , as shown in the following phasor diagram:
P 9.85
[a] I =
240 240
+
= (10 − j7.5) A
24
j32
Vs = 240/0◦ + (0.1 + j0.8)(10 − j7.5) = 247 + j7.25 = 247.11/1.68◦ V
[b] Use the capacitor to eliminate the j component of I, therefore
Ic = j7.5 A,
Zc =
240
= −j32 Ω
j7.5
Vs = 240 + (0.1 + j0.8)10 = 241 + j8 = 241.13/1.90◦ V
[c] Let Ic denote the magnitude of the current in the capacitor branch. Then
I = (10 − j7.5 + jIc) = 10 + j(Ic − 7.5) A
Vs = 240/α = 240 + (0.1 + j0.8)[10 + j(Ic − 7.5)]
= (247 − 0.8Ic ) + j(7.25 + 0.1Ic)
It follows that
240 cos α = (247 − 0.8Ic ) and 240 sin α = (7.25 + 0.1Ic )
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9–56
CHAPTER 9. Sinusoidal Steady State Analysis
Now square each term and then add to generate the quadratic equation
Ic2 − 605.77Ic + 5325.48 = 0;
Ic = 302.88 ± 293.96
Therefore
Ic = 8.92 A (smallest value) and Zc = 240/j8.92 = −j26.90 Ω.
Therefore, the capacitive reactance is −26.90 Ω.
P 9.86
[a]
I` =
240 240
+
= 30 − j40 A
8
j6
V` = (0.1 + j0.8)(30 − j40) = 35 + j20 = 40.31/29.74◦ V
Vs = 240/0◦ + V` = 275 + j20 = 275.73/4.16◦ V
[b]
[c] I` = 30 − j40 +
240
= 30 + j8 A
−j5
V` = (0.1 + j0.8)(30 + j8) = −3.4 + j24.8 = 25.03/97.81◦
Vs = 240/0◦ + V` = 236.6 + j24.8 = 237.9/5.98◦
P 9.87
[a] I1 =
120
240
+
= 23.29 − j13.71 = 27.02/−30.5◦ A
24
8.4 + j6.3
I2 =
120 120
−
= 5/0◦ A
12
24
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Problems
I3 =
120
240
+
= 28.29 − j13.71 = 31.44/−25.87◦ A
12
8.4 + j6.3
I4 =
120
= 5/0◦ A;
24
I6 =
240
= 18.29 − j13.71 = 22.86/−36.87◦ A
8.4 + j6.3
I5 =
9–57
120
= 10/0◦ A
12
[b] When fuse A is interrupted,
I1 = 0
I3 = 15 A
I5 = 10 A
I2 = 10 + 5 = 15 A
I4 = −5 A
I6 = 5 A
[c] The clock and television set were fed from the uninterrupted side of the
circuit, that is, the 12 Ω load includes the clock and the TV set.
[d] No, the motor current drops to 5 A, well below its normal running value of
22.86 A.
[e] After fuse A opens, the current in fuse B is only 15 A.
P 9.88
[a] The circuit is redrawn, with mesh currents identified:
The mesh current equations are:
120/0◦ = 23Ia − 2Ib − 20Ic
120/0◦ = −2Ia + 43Ib − 40Ic
0 = −20Ia − 40Ib + 70Ic
Solving,
Ia = 24/0◦ A
Ib = 21.96/0◦ A
Ic = 19.40/0◦ A
The branch currents are:
I1 = Ia = 24/0◦ A
I2 = Ia − Ib = 2.04/0◦ A
I3 = Ib = 21.96/0◦ A
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9–58
CHAPTER 9. Sinusoidal Steady State Analysis
I4 = Ic = 19.40/0◦ A
I5 = Ia − Ic = 4.6/0◦ A
I6 = Ib − Ic = 2.55/0◦ A
[b] Let N1 be the number of turns on the primary winding; because the
secondary winding is center-tapped, let 2N2 be the total turns on the
secondary. From Fig. 9.58,
13,200
240
=
N1
2N2
or
N2
1
=
N1
110
The ampere turn balance requires
N1 Ip = N2 I1 + N2 I3
Therefore,
Ip =
P 9.89
N2
1
(I1 + I3 ) =
(24 + 21.96) = 0.42/0◦ A
N1
110
[a]
The three mesh current equations are
120/0◦ = 23Ia − 2Ib − 20Ic
120/0◦ = −2Ia + 23Ib − 20Ic
0 = −20Ia − 20Ib + 50Ic
Solving,
Ia = 24/0◦ A;
Ib = 24/0◦ A;
Ic = 19.2/0◦ A
. ·. I2 = Ia − Ib = 0 A
[b] Ip =
N2
N2
(I1 + I3 ) =
(Ia + Ib )
N1
N1
=
1
(24 + 24) = 0.436/0◦ A
110
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Problems
9–59
[c] Yes; when the two 120 V loads are equal, there is no current in the
“neutral” line, so no power is lost to this line. Since you pay for power,
the cost is lower when the loads are equal.
.
P 9.90
[a]
125 = (R + 0.05 + j0.05)I1 − (0.03 + j0.03)I2 − RI3
125 = −(0.03 + j0.03)I1 + (R + 0.05 + j0.05)I2 − RI3
Subtracting the above two equations gives
0 = (R + 0.08 + j0.08)I1 − (R + 0.08 + j0.08)I2
. ·. I1 = I2
so
In = I1 − I2 = 0 A
[b] V1 = R(I1 − I3 );
V2 = R(I2 − I3)
Since I1 = I2 (from part [a]) V1 = V2
[c]
250 = (440.04 + j0.04)Ia − 440Ib
0 = −440Ia + 448Ib
Solving,
Ia = 31.656207 − j0.160343 A
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9–60
CHAPTER 9. Sinusoidal Steady State Analysis
Ib = 31.090917 − j0.157479 A
I1 = Ia − Ib = 0.56529 − j0.002864 A
V1 = 40I1 = 22.612 − j0.11456 = 22.612/ − 0.290282◦ V
V2 = 400I1 = 226.116 − j1.1456 = 226.1189/ − 0.290282◦ V
[d]
125 = (40.05 + j0.05)I1 − (0.03 + j0.03)I2 − 40I3
125 = −(0.03 + j0.03)I1 + (400.05 + j0.05)I2 − 400I 3
0 = −40I1 − 400I2 + 448I3
Solving,
I1 = 34.19 − j0.182 A
I2 = 31.396 − j0.164 A
I3 = 31.085 − j0.163 A
V1 = 40(I1 − I3 ) = 124.2/ − 0.35◦ V
V2 = 400(I2 − I3 ) = 124.4/ − 0.18◦ V
[e] Because an open neutral can result in severely unbalanced voltages across
the 125 V loads.
P 9.91
[a] Let N1 = primary winding turns and 2N2 = secondary winding turns.
Then
14,000
250
N2
1
=
;
.·.
=
=a
N1
2N2
N1
112
In part c),
Ip = 2aIa
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Problems
. ·. Ip =
=
9–61
2N2 Ia
1
= Ia
N1
56
1
(31.656 − j0.16)
56
Ip = 565.3 − j2.9 mA
In part d),
Ip N1 = I1 N2 + I2 N2
. ·. Ip =
N2
(I1 + I2 )
N1
=
1
(34.19 − j0.182 + 31.396 − j0.164)
112
=
1
(65.586 − j0.346)
112
Ip = 585.6 − j3.1 mA
[b] Yes, because the neutral conductor carries non-zero current whenever the
load is not balanced.
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10
Sinusoidal Steady State Power
Calculations
Assessment Problems
AP 10.1 [a] V = 100/ − 45◦ V,
I = 20/15◦ A
Therefore
1
P = (100)(20) cos[−45 − (15)] = 500 W,
2
Q = 1000 sin −60◦ = −866.03 VAR,
[b] V = 100/ − 45◦ ,
B→A
I = 20/165◦
P = 1000 cos(−210◦ ) = −866.03 W,
B→A
Q = 1000 sin(−210◦ ) = 500 VAR,
[c] V = 100/ − 45◦ ,
A→B
I = 20/ − 105◦
P = 1000 cos(60◦ ) = 500 W,
A→B
Q = 1000 sin(60◦ ) = 866.03 VAR,
[d] V = 100/0◦ ,
A→B
A→B
I = 20/120◦
P = 1000 cos(−120◦ ) = −500 W,
B→A
Q = 1000 sin(−120◦ ) = −866.03 VAR,
B→A
AP 10.2
pf = cos(θv − θi ) = cos[15 − (75)] = cos(−60◦ ) = 0.5 leading
rf = sin(θv − θi) = sin(−60◦ ) = −0.866
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10–1 system, or transmission in any form or by any means, electronic,
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10–2
AP 10.3
CHAPTER 10. Sinusoidal Steady State Power Calculations
Iρ
0.18
From Ex. 9.4 Ieff = √ = √ A
3
3
P =
2
Ieff
R
0.0324
=
(5000) = 54 W
3
AP 10.4 [a] Z = (39 + j26)k(−j52) = 48 − j20 = 52/ − 22.62◦ Ω
Therefore I` =
250/0◦
= 4.85/18.08◦ A (rms)
48 − j20 + 1 + j4
VL = ZI` = (52/ − 22.62◦ )(4.85/18.08◦ ) = 252.20/ − 4.54◦ V (rms)
IL =
VL
= 5.38/ − 38.23◦ A (rms)
39 + j26
[b] SL = VL I∗L = (252.20/ − 4.54◦ )(5.38/ + 38.23◦ ) = 1357/33.69◦
= (1129.09 + j752.73) VA
PL = 1129.09 W;
QL = 752.73 VAR
[c] P` = |I` |2 1 = (4.85)2 · 1 = 23.52 W;
Q` = |I` |24 = 94.09 VAR
[d] Sg (delivering) = 250I∗` = (1152.62 − j376.36) VA
Therefore the source is delivering 1152.62 W and absorbing 376.36
magnetizing VAR.
|VL |2
(252.20)2
[e] Qcap =
=
= −1223.18 VAR
−52
−52
Therefore the capacitor is delivering 1223.18 magnetizing VAR.
Check:
94.09 + 752.73 + 376.36 = 1223.18 VAR and
1129.09 + 23.52 = 1152.62 W
AP 10.5 Series circuit derivation:
S = 250I∗ = (40,000 − j30,000)
Therefore I∗ = 160 − j120 = 200/ − 36.87◦ A (rms)
I = 200/36.87◦ A (rms)
Z=
V
250
=
= 1.25/ − 36.87◦ = (1 − j0.75) Ω
I
200/36.87◦
Therefore R = 1 Ω,
XC = −0.75 Ω
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Problems
10–3
Parallel circuit derivation
P =
(250)2
;
R
therefore R =
(250)2
= 1.5625 Ω
40,000
Q=
(250)2
;
XC
therefore XC =
(250)2
= −2.083 Ω
−30,000
AP 10.6
S1 = 15,000(0.6) + j15,000(0.8) = 9000 + j12,000 VA
S2 = 6000(0.8) − j6000(0.6) = 4800 − j3600 VA
ST = S1 + S2 = 13,800 + j8400 VA
ST = 200I∗ ;
therefore I∗ = 69 + j42
I = 69 − j42 A
Vs = 200 + jI = 200 + j69 + 42 = 242 + j69 = 251.64/15.91◦ V (rms)
AP 10.7 [a] The phasor domain equivalent circuit and the Thévenin equivalent are
shown below:
Phasor domain equivalent circuit:
Thévenin equivalent:
VTh = 3
−j800
= 48 − j24 = 53.67/ − 26.57◦ V
20 − j40
ZTh = 4 + j18 +
−j800
= 20 + j10 = 22.36/26.57◦ Ω
20 − j40
For maximum power transfer, ZL = (20 − j10) Ω
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10–4
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] I =
53.67/ − 26.57◦
= 1.34/ − 26.57◦ A
40
Therefore P =
1.34
√
2
!2
20 = 17.96 W
[c] RL = |ZTh | = 22.36 Ω
53.67/ − 26.57◦
[d] I =
= 1.23/ − 39.85◦ A
42.36 + j10
Therefore P =
1.23
√
2
!2
(22.36) = 17 W
AP 10.8
Mesh current equations:
660 = (34 + j50)I1 + j100(I1 − I2 ) + j40I1 + j40(I1 − I2)
0 = j100(I2 − I1 ) − j40I1 + 100I2
Solving,
I2 = 3.5/0◦ A;
1
.·. P = (3.5)2 (100) = 612.50 W
2
AP 10.9 [a]
248 = j400I1 − j500I2 + 375(I1 − I2 )
0 = 375(I2 − I1) + j1000I 2 − j500I1 + 400I2
Solving,
I1 = 0.80 − j0.62 A;
I2 = 0.4 − j0.3 = 0.5/ − 36.87◦
1
.·. P = (0.25)(400) = 50 W
2
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Problems
10–5
[b] I1 − I2 = 0.4 − j0.32 A
1
P375 = |I1 − I2 |2(375) = 49.20 W
2
1
[c] Pg = (248)(0.8) = 99.20 W
2
X
Pabs = 50 + 49.2 = 99.20 W
AP 10.10 [a] VTh = 210 V;
V2 = 14 V1 ;
Short circuit equations:
(checks)
I1 = 14 I2
840 = 80I1 − 20I2 + V1
0 = 20(I2 − I1 ) − V2
.·. I2 = 14 A;
[b] Pmax =
210
30
2
RTh =
210
= 15 Ω
14
15 = 735 W
AP 10.11 [a] VTh = −4(146/0◦ ) = −584/0◦ V (rms)
V2 = 4V1;
I1 = −4I2
Short circuit equations:
146/0◦ = 80I1 − 20I2 + V1
0 = 20(I2 − I1 ) − V2
.·. I2 = −146/365 = −0.40 A;
[b] P =
−584
2920
2
RTh =
−584
= 1460 Ω
−0.4
1460 = 58.40 W
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10–6
CHAPTER 10. Sinusoidal Steady State Power Calculations
Problems
P 10.1
1
[a] P = (100)(10) cos(50 − 15) = 500 cos 35◦ = 409.58 W
2
(abs)
Q = 500 sin 35◦ = 286.79 VAR (abs)
1
[b] P = (40)(20) cos(−15 − 60) = 400 cos(−75◦ ) = 103.53 W
2
(abs)
Q = 400 sin(−75◦ ) = −386.37 VAR (del)
1
[c] P = (400)(10) cos(30 − 150) = 2000 cos(−120◦ ) = −1000 W
2
(del)
Q = 2000 sin(−120◦ ) = −1732.05 VAR (del)
1
[d] P = (200)(5) cos(160 − 40) = 500 cos(120◦ ) = −250 W
2
(del)
Q = 500 sin(120◦ ) = 433.01 VAR (abs)
P 10.2
[a] hair dryer = 600 W
vacuum = 630 W
sun lamp = 279 W
air conditioner = 860 W
television = 240 W
P
P = 2609 W
2609
= 21.74 A
120
Yes, the breaker will trip.
Therefore Ieff =
[b]
P 10.3
1700
= 14.17 A
120
Yes, the breaker will not trip if the current is reduced to 14.17 A.
X
P = 2609 − 909 = 1700 W;
p = P + P cos 2ωt − Q sin 2ωt;
dp
= 0 when
dt
cos 2ωt = √
dp
= −2ωP sin 2ωt − 2ωQ cos 2ωt
dt
− 2ωP sin 2ωt = 2ωQ cos 2ωt or
P
;
+ Q2
P2
Ieff =
sin 2ωt = − √
tan 2ωt = −
Q
P
Q
+ Q2
P2
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Problems
10–7
Let θ = tan−1 (−Q/P ), then p is maximum when 2ωt = θ and p is minimum
when 2ωt = (θ + π).
q
P
Q(−Q)
√
Therefore pmax = P + P · √ 2
−
=
P
+
P 2 + Q2
P + Q2
P 2 + Q2
q
P
Q
√
−
Q
·
=
P
−
P 2 + Q2
and pmin = P − P · √ 2
P + Q2
P 2 + Q2
P 10.4
[a] P =
−
1 (240)2
= 60 W
2 480
1
−9 × 106
=
= −360 Ω
ωC
(5000)(5)
Q=
1 (240)2
= −80 VAR
2 (−360)
q
q
pmax = P + P 2 + Q2 = 60 + (60)2 + (80)2 = 160 W (del)
√
[b] pmin = 60 − 602 + 802 = −40 W (abs)
[c] P = 60 W
from (a)
[d] Q = −80 VAR from (a)
[e] generates, because Q < 0
[f] pf = cos(θv − θi )
I=
240
240
+
= 0.5 + j0.67 = 0.83/53.13◦ A
480 −j360
.·. pf = cos(0 − 53.13◦ ) = 0.6 leading
[g] rf = sin(−53.13◦ ) = −0.8
P 10.5
Ig = 4/0◦ mA;
1
= −j1250 Ω;
jωC
jωL = j500 Ω
Zeq = 500 + [−j1250k(1000 + j500)] = 1500 − j500 Ω
1
1
Pg = − |I|2Re{Zeq} = − (0.004)2 (1500) = −12 mW
2
2
The source delivers 12 mW of power to the circuit.
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10–8
P 10.6
CHAPTER 10. Sinusoidal Steady State Power Calculations
jωL = j20,000(0.5 × 10−3 ) = j10 Ω;
−6 +
1
106
=
= −j40 Ω
jωC
j20,000(1.25)
Vo
Vo − 30(Vo /j10)
+
=0
j10
30 − j40
.·. Vo
"
#
1
1 + j3
+
=6
j10 30 − j40
.·. Vo = 100/126.87◦ V
Vo
.·. I∆ =
= 10/36.87◦ A
j10
Io = 6/0◦ − I∆ = 6 − 8 − j6 = −2 − j6 = 6.32/ − 108.43◦ A
1
P30Ω = |Io |2 30 = 600 W
2
P 10.7
Zf = −j10,000k20,000 = 4000 − j8000 Ω
Zi = 2000 − j2000 Ω
Zf
4000 − j8000
.·.
=
= 3 − j1
Zi
2000 − j2000
Vo = −
Zf
Vg ;
Zi
Vg = 1/0◦ V
Vo = −(3 − j1)(1) = −3 + j1 = 3.16/161.57◦ V
P =
1 Vm2
1 (10)
=
= 5 × 10−3 = 5 mW
2 R
2 1000
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Problems
P 10.8
10–9
[a] From the solution to Problem 9.59 we have:
Vo = j80 = 80/90◦ V
1
1
Sg = − Vo I∗g = − (j80)(10 − j10) = −400 − j400 VA
2
2
Therefore, the independent current source is delivering 400 W and 400
magnetizing vars.
I1 =
Vo
= j16 A
5
1
P5Ω = (16)2 (5) = 640 W
2
Therefore, the 8 Ω resistor is absorbing 640 W.
I∆ =
Vo
= −10 A
−j8
1
Qcap = (10)2 (−8) = −400 VAR
2
Therefore, the −j8 Ω capacitor is developing 400 magnetizing vars.
2.4I∆ = −24 V
I2 =
Vo − 2.4I∆
−j80 + 24
=
j4
j4
= 20 − j6 A = 20.88/ − 16.7◦ A
1
Qj4 = |I2 |2 (4) = 872 VAR
2
Therefore, the j4 Ω inductor is absorbing 872 magnetizing vars.
Sd.s. = 12 (2.4I∆ )I∗2 = 12 (−24)(20 + j6)
= −240 − j72 VA
Thus the dependent source is delivering 240 W and 72 magnetizing vars.
[b]
X
Pgen = 400 + 240 = 640 W =
X
Pabs
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10–10
CHAPTER 10. Sinusoidal Steady State Power Calculations
[c]
P 10.9
X
Qgen = 400 + 400 + 72 = 872 VAR =
X
Qabs
[a] From the solution to Problem 9.61 we have
Ia = 2.25 − j2.25 A;
Ib = −6.75 + j0.75 A;
Io = 9 − j3 A
1
S60V = − (60)I∗a = −30(2.25 + j2.25) = −67.5 − j67.5 VA
2
Thus, the 60 V source is developing 67.5 W and 67.5 magnetizing vars.
S90V = − 21 (j90)I∗b = −j45(−6.75 − j0.75)
= −33.75 + j303.75 VA
Thus, the 90 V source is delivering 33.75 W and absorbing 303.75
magnetizing vars.
1
P20Ω = |Ia |2(20) = 101.25 W
2
Thus the 20 Ω resistor is absorbing 101.25 W.
1
Q−j20Ω = |Ib |2(−20) = −461.25 VAR
2
Thus the −j20 Ω capacitor is developing 461.25 magnetizing vars.
1
Qj5Ω = |Io |2 (5) = 225 VAR
2
Thus the j5 Ω inductor is absorbing 225 magnetizing vars.
[b]
[c]
X
X
Pdev = 67.5 + 33.75 = 101.25 W =
X
Pabs
Qdev = 67.5 + 461.25 = 528.75 VAR
X
Qabs = 225 + 303.75 = 528.75 VAR =
P 10.10 [a] line loss = 7500 − 2500 = 5 kW
line loss = |Ig |2 20
|Ig | =
√
X
Qdev
.·. |Ig |2 = 250
250 A
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Problems
10–11
.·. RL = 10 Ω
|Ig |2 RL = 2500
|Ig |2 XL = −5000
.·. XL = −20 Ω
Thus,
|Z| =
q
(30)2 + (X` − 20)2
500
|Ig | = q
900 + (X` − 20)2
25 × 104
= 1000
250
(X` − 20) = ±10.
.·. 900 + (X` − 20)2 =
Solving,
Thus, X` = 10 Ω
or
X` = 30 Ω
[b] If X` = 30 Ω:
500
Ig =
= 15 − j5 A
30 + j10
Sg = −500I∗g = −7500 − j2500 VA
Thus, the voltage source is delivering 7500 W and 2500 magnetizing vars.
Qj30 = |Ig |2X` = 250(30) = 7500 VAR
Therefore the line reactance is absorbing 7500 magnetizing vars.
Q−j20 = |Ig |2 XL = 250(−20) = −5000 VAR
Therefore the load reactance is generating 5000 magnetizing vars.
X
Qgen = 7500 VAR =
X
Qabs
If X` = 10 Ω:
500
Ig =
= 15 + j5 A
30 − j10
Sg = −500I∗g = −7500 + j2500 VA
Thus, the voltage source is delivering 7500 W and absorbing 2500
magnetizing vars.
Qj10 = |Ig |2(10) = 250(10) = 2500 VAR
Therefore the line reactance is absorbing 2500 magnetizing vars. The
load continues to generate 5000 magnetizing vars.
X
Qgen = 5000 VAR =
X
Qabs
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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10–12
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.11 [a] Ieff = 40/115 ∼
= 0.35 A
[b] Ieff = 130/115 ∼
= 1.13 A
P 10.12 Wdc =
Vdc2
T;
R
V2
.·. dc T =
R
Z
to +T
to
Vdc2 =
1
T
Z
to +T
Vdc =
s
1
T
Z
to
0.08
0
to
vs2
dt
R
vs2
dt
R
vs2 dt = Vrms = Veff
to
0 ≤ t ≤ 80 ms
i(t) = 100 − 1000t
Z
to +T
vs2 dt
to +T
P 10.13 i(t) = 250t
.·. Irms =
Z
Ws =
s
1
0.1
80 ms ≤ t ≤ 100 ms
Z
0.08
0
(250)2 t2 dt +
(250)2 t2 dt = (250)2
t3
3
0.08
=
0
Z
0.1
0.08
(100 − 1000t)2 dt
32
3
(100 − 1000t)2 = 104 − 2 × 105 t + 106 t2
Z
0.1
Z
0.1
0.08
0.08
6
10
104 dt = 200
0.1
2 × 105 t dt = 105 t2
= 360
0.08
Z
0.1
0.08
t2 dt =
.·. Irms =
2
P 10.14 P = Irms
R
106 3
t
3
0.1
=
0.08
488
3
q
10{(32/3) + 225 − 360 + (488/3)} = 11.55 A
. ·. R =
1280
= 9.6 Ω
(11.55)2
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Problems
10–13
P 10.15 [a] Area under one cycle of vg2:
A = (100)(25 × 10−6 ) + 400(25 × 10−6 ) + 400(25 × 10−6 ) + 100(25 × 10−6 )
= 1000(25 × 10−6 )
Mean value of vg2:
M.V. =
.·. Vrms
[b] P =
A
1000(25 × 10−6 )
=
= 250
100 × 10−6
100 × 10−6
√
= 250 = 15.81 V (rms)
2
Vrms
250
=
= 62.5 W
R
4
P 10.16 [a]
Vo
Vo − 240
Vo
+
+
=0
−j25
12.5
15 + j20
.·. Vo = 183.53 − j14.12 = 184.07/ − 4.4◦ V
Ig =
240 − 183.53 + j14.12
= 4.52 + j1.13 A
12.50
Sg = −Vg I∗g = −(240)(4.52 − j1.13)
= −1084.24 + j271.06 VA
[b] Source is delivering 1084.24 W.
[c] Source is absorbing 271.06 magnetizing VAR.
[d] Qcap =
(184.07)2
= −1355.29 VAR
−25
P12.5Ω = |Ig |2 (12.5) = 271.06 W
|Io | =
184.07
= 7.36 A
25
P15Ω = |Io |2 (15) = 813.18 W
Qind = |Io |2 (20) = 1084.24 VAR
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10–14
CHAPTER 10. Sinusoidal Steady State Power Calculations
[e]
X
Pdel = 1084.24 W
X
. ·.
[f]
X
Pdiss = 271.06 + 813.18 = 1084.24 W
X
Pdel =
X
Pdiss = 1084.24 W
Qabs = 271.06 + 1084.24 = 1355.29 VAR
X
. ·.
Qdev = 1355.29 VAR
X
mag VAR dev =
P 10.17 Ig = 40/0◦ mA
jωL = j10,000 Ω;
Io =
X
mag VAR abs = 1355.29 VAR
1
= −j10,000 Ω
jωC
j10,000
(40/0◦ ) = 80/90◦ mA
5000
1
1
P = |Io |2 (5000) = (0.08)2 (5000) = 16 W
2
2
1
Q = |Io |2 (−10,000) = −32 VAR
2
S = P + jQ = 16 − j32 VA
|S| = 35.78 VA
P 10.18 [a]
1
= −j40 Ω;
jωC
jωL = j80 Ω
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Problems
10–15
Zeq = 40k − j40 + j80 + 60 = 80 + j60 Ω
40/0◦
Ig =
= 0.32 − j0.24 A
80 + j60
1
1
Sg = − Vg I∗g = − 40(0.32 + j0.24) = −6.4 − j4.8 VA
2
2
P = 6.4 W (del);
Q = 4.8 VAR (del)
|S| = |Sg | = 8 VA
[b] I1 =
−j40
Ig = 0.04 − j0.28 A
40 − j40
1
P40Ω = |I1 |2(40) = 1.6 W
2
1
P60Ω = |Ig |2 (60) = 4.8 W
2
X
Pdiss = 1.6 + 4.8 = 6.4 W =
X
Pdev
[c] I−j40Ω = Ig − I1 = 0.28 + j0.04 A
1
Q−j40Ω = |I−j40Ω |2 (−40) = −1.6 VAR (del)
2
1
Qj80Ω = |Ig |2(80) = 6.4 VAR (abs)
2
X
Qabs = 6.4 − 1.6 = 4.8 VAR =
P 10.19 ST = 40,800 + j30,600 VA
X
Qdev
S1 = 20,000(0.96 − j0.28) = 19,200 − j5600 VA
S2 = ST − S1 = 21,600 + j36,200 = 42,154.48/59.176◦ VA
rf = sin(59.176◦ ) = 0.8587
pf = cos(59.176◦ ) = 0.5124 lagging
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10–16
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.20 [a] Let VL = Vm /0◦ :
SL = 2500(0.8 + j0.6) = 2000 + j1500 VA
I∗` =
2000
1500
+j
;
Vm
Vm
I` =
2000
1500
−j
Vm
Vm
2000
1500
250/θ = Vm +
−j
(1 + j2)
Vm
Vm
250Vm /θ = Vm2 + (2000 − j1500)(1 + j2) = Vm2 + 5000 + j2500
250Vm cos θ = Vm2 + 5000;
250Vm sin θ = 2500
(250)2 Vm2 = (Vm2 + 5000)2 + 25002
62,500Vm2 = Vm4 + 10,000Vm2 + 31.25 × 106
or
Vm4 − 52,500Vm2 + 31.25 × 106 = 0
Solving,
Vm2 = 26,250 ± 25,647.86;
Vm = 227.81 V and Vm = 24.54 V
If Vm = 227.81 V:
sin θ =
2500
= 0.044;
(227.81)(250)
If Vm = 24.54 V:
2500
sin θ =
= 0.4075;
(24.54)(250)
.·. θ = 2.52◦
.·. θ = 24.05◦
[b]
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Problems
10–17
P 10.21 [a] S1 = 60,000 − j70,000 VA
S2 =
|VL |2
(2500)2
=
= 240,000 − j70,000 VA
Z2∗
24 − j7
S1 + S2 = 300,000 VA
2500I ∗L = 300,000;
.·. IL = 120 A(rms)
Vg = VL + IL (0.1 + j1) = 2500 + (120)(0.1 + j1)
= 2512 + j120 = 2514.86/2.735◦ Vrms
[b] T =
1
1
=
= 16.67 ms
f
60
2.735◦
t
=
;
360◦
16.67 ms
.·. t = 126.62 µs
[c] VL lags Vg by 2.735◦ or 126.62 µs
P 10.22 [a]
250I∗1 = 7500 + j2500;
.·. I1 = 30 − j10 A(rms)
250I∗2 = 2800 − j9600;
.·. I2 = 11.2 + j38.4 A(rms)
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10–18
CHAPTER 10. Sinusoidal Steady State Power Calculations
I3 =
500
500
+
= 40 − j10 A(rms)
12.5 j50
Ig1 = I1 + I3 = 70 − j20 A
Sg1 = 250(70 + j20) = 17,500 + j5000 VA
Thus the Vg1 source is delivering 17.5 kW and 5000 magnetizing vars.
Ig2 = I2 + I3 = 51.2 + j28.4 A(rms)
Sg2 = 250(51.2 − j28.4) = 12,800 − j7100 VA
Thus the Vg2 source is delivering 12.8 kW and absorbing 7100
magnetizing vars.
[b]
X
Pgen = 17.5 + 12.8 = 30.3 kW
X
(500)2
= 7500 + 2800 +
= 30.3 kW =
Pgen
12.5
X
Pabs
X
Qdel = 9600 + 5000 = 14.6 kVAR
X
Qabs = 2500 + 7100 +
X
(500)2
= 14.6 kVAR =
Qdel
50
P 10.23 S1 = 1200 + 1196 = 2396 + j0 VA
2396
.·. I1 =
= 19.967 A
120
S2 = 860 + 600 + 240 = 1700 + j0 VA
1700
.·. I2 =
= 14.167 A
120
S3 = 4474 + 12,200 = 16,674 + j0 VA
16,674
.·. I3 =
= 69.475 A
240
Ig1 = I1 + I3 = 89.44 A
Ig2 = I2 + I3 = 83.64 A
Breakers will not trip since both feeder currents are less than 100 A.
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Problems
10–19
P 10.24 [a]
I1 =
5000 − j1250
= 40 − j10 A (rms)
125
I2 =
6250 − j2500
= 50 − j20 A (rms)
125
I3 =
8000 + j0
= 32 + j0 A (rms)
250
.·. Ig1 = 72 − j10 A (rms)
In = I1 − I2 = −10 + j10 A (rms)
Ig2 = 82 − j20 A
Vg1 = 0.05Ig1 + 125 + j0 + 0.15In = 127.1 − j1 V(rms)
Vg2 = −0.15In + 125 + j0 + 0.05Ig2 = 130.6 − j2.5 V(rms)
Sg1 = −[(127.1 − j1)(72 + j10)] = −[9141.2 + j1343] VA
Sg2 = −[(130.6 − j2.5)(82 + j20)] = −[10,759.2 + j2407] VA
Note: Both sources are delivering average power and magnetizing VAR to
the circuit.
[b] P0.05 = |Ig1 |2(0.05) = 264.2 W
P0.15 = |In |2 (0.15) = 30 W
P0.05 = |Ig2 |2(0.05) = 356.2 W
X
Pdis = 264.2 + 30 + 356.2 + 5000 + 8000 + 6250 = 19,900.4 W
X
Qabs = 1250 + 2500 = 3750 VAR
X
Pdev = 9141.2 + 10,759.2 = 19,900.4 W =
X
Qdel = 1343 + 2407 = 3750 VAR =
X
X
Pdis
Qabs
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10–20
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.25
480I∗1 = 7500 + j9000
.·. I1 = 15.625 − j18.75 A(rms)
I∗1 = 15.625 + j18.75;
480I∗2 = 2100 − j1800
I∗2 = 4.375 − j3.75;
I3 =
.·. I2 = 4.375 + j3.75 A(rms)
480/0◦
= 10 + j0 A;
48
I4 =
480/0◦
= 0 − j25 A
j19.2
Ig = I1 + I2 + I3 + I4 = 30 − j40 A
Vg = 480 + (30 − j40)(j0.5) = 500 + j15 = 500.22/1.72◦ V (rms)
P 10.26 [a] Z1 = 240 + j70 = 250/16.26◦ Ω
pf = cos(16.26◦ ) = 0.96 lagging
rf = sin(16.26◦ ) = 0.28
Z2 = 160 − j120 = 200/ − 36.87◦ Ω
pf = cos(−36.87◦ ) = 0.8 leading
rf = sin(−36.87◦ ) = −0.6
Z3 = 30 − j40 = 50/ − 53.13◦ Ω
pf = cos(−53.13◦ ) = 0.6 leading
rf = sin(−53.13◦ ) = −0.8
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Problems
10–21
[b] Y = Y1 + Y2 + Y3
1
;
250/16.26◦
Y1 =
Y2 =
1
;
200/ − 36.87◦
Y3 =
1
50/ − 53.13◦
Y = 19.84 + j17.88 mS
1
= 37.44/ − 42.03◦ Ω
Y
Z=
pf = cos(−42.03◦ ) = 0.74 leading
rf = sin(−42.03◦ ) = −0.67
P 10.27 [a] S1 = 16 + j18 kVA;
S2 = 6 − j8 kVA;
S3 = 8 + j0 kVA
ST = S1 + S2 + S3 = 30 + j10 kVA
250I∗ = (30 + j10) × 103 ;
.·. I = 120 − j40 A
250
= 1.875 + j0.625 Ω = 1.98/18.43◦ Ω
120 − j40
Z=
[b] pf = cos(18.43◦ ) = 0.9487 lagging
P 10.28 [a] From the solution to Problem 10.26 we have
IL = 120 − j40 A (rms)
.·. Vs = 250/0◦ + (120 − j40)(0.01 + j0.08) = 254.4 + j9.2
= 254.57/2.07◦ V (rms)
[b] |IL | =
q
16,000
P` = (16,000)(0.01) = 160 W
[c] Ps = 30,000 + 160 = 30.16 kW
30
[d] η =
(100) = 99.47%
30.16
P 10.29 [a] I =
Q` = (16,000)(0.08) = 1280 VAR
Qs = 10,000 + 1280 = 11.28 kVAR
465/0◦
= 2.4 − j1.8 = 3/ − 36.87◦ A(rms)
124 + j93
P = (3)2 (4) = 36 W
[b] YL =
1
= 5.33 − j4 mS
120 + j90
.·. XC =
1
= −250 Ω
−4 × 10−3
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10–22
CHAPTER 10. Sinusoidal Steady State Power Calculations
1
= 187.5 Ω
5.33 × 10−3
465/0◦
[d] I =
= 2.4279/ − 0.9◦ A
191.5 + j3
[c] ZL =
P = (2.4279)2 (4) = 23.58 W
23.58
(100) = 65.5%
36
Thus the power loss after the capacitor is added is 65.5% of the power
loss before the capacitor is added.
[e] % =
P 10.30
IL =
120,000 − j160,000
= 18.75 − j25 A (rms)
6400
IC =
6400
6400
=j
= jIC
−jXC
XC
I` = 18.75 − j25 + jIC = 18.75 + j(IC − 25)
Vs = 6400 + (4 + j24)[18.75 + j(IC − 25)]
= (7075 − 24IC ) + j(350 + 4IC )
|Vs |2 = (7075 − 24IC )2 + (350 + 4IC )2 = (6400)2
.·. 592IC2 − 336,800IC + 9,218,125 = 0
IC = 284.46 ± 255.63 = 28.33 A(rms)∗
*Select the smaller value of IC to minimize the magnitude of I` .
6400
.·. XC = −
= −221.99
28.33
.·. C =
1
= 11.95 µF
(221.99)(120π)
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Problems
10–23
P 10.31 [a] From Problem 9.75,
Zab = 100 + j136.26
so
I1 =
50
50
=
= 160 − j120 mA
100 + j13.74 + 100 + 136.26
200 + j150
I2 =
jωM
j270
I1 =
(0.16 − j0.12) = 51.84 + j15.12 mA
Z22
800 + j600
VL = (300 + j100)(0.05184 + j0.01512) = 14.04 + j9.72
|VL | = 17.08 V
[b] Pg (ideal) = 50(0.16) = 8 W
Pg (practical) = 8 − |I1|2 (100) = 4 W
PL = |I2 |2(300) = 0.8748 W
% delivered =
0.8748
(100) = 21.87%
4
P 10.32 [a] So = original load = 1600 + j
Sf = final load = 1920 + j
1600
(0.6) = 1600 + j1200 kVA
0.8
1920
(0.28) = 1920 + j560 kVA
0.96
.·. Qadded = 560 − 1200 = −640 kVAR
[b] deliver
[c] Sa = added load = 320 − j640 = 715.54/ − 63.43◦ kVA
pf = cos(−63.43) = 0.447 leading
[d] I∗L =
(1600 + j1200) × 103
= 666.67 + j500 A
2400
IL = 666.67 − j500 = 833.33/ − 36.87◦ A(rms)
|IL | = 833.33 A(rms)
[e] I∗L =
(1920 + j560) × 103
= 800 + j233.33
2400
IL = 800 − j233.33 = 833.33/ − 16.26◦ A(rms)
|IL | = 833.33 A(rms)
P 10.33 [a] Pbefore = Pafter = (833.33)2 (0.05) = 34,722.22 W
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10–24
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] Vs (before) = 2400 + (666.67 − j500)(0.05 + j0.4)
= 2633.33 + j241.67 = 2644.4/5.24◦ V(rms)
|Vs (before)| = 2644.4 V(rms)
Vs (after) = 2400 + (800 − j233.33)(0.05 + j0.4)
= 2533.33 + j308.33 = 2552.028/6.94◦ V(rms)
|Vs (after)| = 2552.028 V(rms)
P 10.34 [a] SL = 20,000(0.85 + j0.53) = 17,000 + j10,535.65 VA
125I∗L = (17,000 + j10,535.65);
I∗L = 136 + j84.29 A(rms)
.·. IL = 136 − j84.29 A(rms)
Vs = 125 + (136 − j84.29)(0.01 + j0.08) = 133.10 + j10.04
= 133.48/4.31◦ V(rms)
|Vs | = 133.48 V(rms)
[b] P` = |I` |2 (0.01) = (160)2 (0.01) = 256 W
(125)2
[c]
= −10,535.65;
XC = −1.48306 Ω
XC
−
1
= −1.48306;
ωC
C=
1
= 1788.59 µF
(1.48306)(120π)
[d] I` = 136 + j0 A(rms)
Vs = 125 + 136(0.01 + j0.08) = 126.36 + j10.88
= 126.83/4.92◦ V(rms)
|Vs | = 126.83 V(rms)
[e] P` = (136)2 (0.01) = 184.96 W
P 10.35 [a]
10 = j1(I1 − I2 ) + j1(I3 − I2) − j1(I1 − I3 )
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Problems
10–25
0 = 1I2 + j2(I2 − I3 ) + j1(I2 − I1 ) + j1(I2 − I1 ) + j1(I2 − I3 )
0 = I3 − j1(I3 − I1 ) + j2(I3 − I2 ) + j1(I1 − I2 )
Solving,
I1 = 6.25 + j7.5 A(rms);
I2 = 5 + j2.5 A(rms);
I3 = 5 − j2.5 A(rms)
Ia = I1 = 6.25 + j7.5 A
Ib = I1 − I2 = 1.25 + j5 A
Ic = I2 = 5 + j2.5 A
Id = I3 − I2 = −j5 A
Ie = I1 − I3 = 1.25 + j10 A
If = I3 = 5 − j2.5 A
Va = 10 V
Vb = j1Ib + j1Id = j1.25 V
Vc = 1Ic = 5 + j2.5 V
Vd = j2Id − j1Ib = 5 + j1.25 V
Ve = −j1Ie = 10 − j1.25 V
Vf = 1If = 5 − j2.5 V
[b]
Sa = −10I∗a = −62.5 + j75 VA
Sb = Vb I∗b = 6.25 + j1.5625 VA
Sc = Vc I∗c = 31.25 + j0 VA
Sd = Vd I∗d = −6.25 + j25 VA
Se = Ve I∗e = 0 − j101.5625 VA
Sf = Vf I∗f = 31.25 VA
[c]
X
Pdev = 62.5 W
X
Pabs = 31.25 + 31.25 = 62.5 W
Note that the total power absorbed by the coupled coils is zero:
6.25 − 6.25 = 0 = Pb + Pd
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10–26
CHAPTER 10. Sinusoidal Steady State Power Calculations
[d]
X
Qdev = 101.5625 VAR
Both the source and the capacitor are developing magnetizing vars.
X
P 10.36 [a]
X
Qabs = 75 + 1.5625 + 25 = 101.5625 VAR
Q absorbed by the coupled coils is Qb + Qd = 26.5625
272/0◦ = 2Ig + j10Ig + j14(Ig − I2 ) − j6I2
+j14Ig − j8I2 + j20(Ig − I2 )
0 = j20(I2 − Ig ) − j14Ig + j8I2 + j4I2
+j8(I2 − Ig ) − j6Ig + 8I2
Solving,
Ig = 20 − j4 A(rms);
I2 = 24/0◦ A(rms)
P8Ω = (24)2 (8) = 4608 W
[b] Pg (developed) = (272)(20) = 5440 W
Vg
272
[c] Zab =
−2=
− 2 = 11.08 + j2.62 = 11.38/13.28◦ Ω
Ig
20 − j4
[d] P2Ω = |Ig |2 (2) = 832 W
X
Pdiss = 832 + 4608 = 5440 W =
P 10.37 [a] Zab = 1 +
. ·. I1 =
I2 =
N1
N2
2
X
Pdev
(1 − j2) = 25 − j50 Ω
100/0◦
= 2.5/0◦ A
15 + j50 + 25 − j50
N1
I1 = 10/0◦ A
N2
.·. IL = I1 + I2 = 12.5/0◦ A(rms)
P1Ω = (12.5)2 (1) = 156.25 W
P15Ω = (2.5)2 (15) = 93.75 W
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Problems
10–27
[b] Pg = −100(2.5/0◦ ) = −250 W
X
Pabs = 156.25 + 93.75 = 250 W =
P 10.38 [a] 25a21 + 4a22 = 500
I25 = a1I;
I4 = a2 I;
Pdev
P25 = a21I2 (25)
P4 = a22I2 (4)
a22I2 4 = 100a21 I2
P4 = 4P25 ;
. ·.
X
100a21 = 4a22
25a21 + 100a21 = 500;
25(4) + 4a22 = 500;
a1 = 2
a2 = 10
2000/0◦
[b] I =
= 2/0◦ A(rms)
500 + 500
I25 = a1I = 4 A
P25Ω = (16)(25) = 400 W
[c] I4 = a2I = 10(2) = 20 A(rms)
V4 = (20)(4) = 80/0◦ V(rms)
P 10.39 [a]
300 = 60I1 + V1 + 20(I1 − I2)
0 = 20(I2 − I1 ) + V2 + 40I2
1
V2 = V1 ;
4
I2 = −4I1
Solving,
V1 = 260 V (rms);
V2 = 65 V (rms)
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10–28
CHAPTER 10. Sinusoidal Steady State Power Calculations
I1 = 0.25 A (rms);
I2 = −1.0 A (rms)
V5A = V1 + 20(I1 − I2) = 285 V (rms)
.·. P = −(285)(5) = −1425 W
Thus 1425 W is delivered by the current source to the circuit.
[b] I20Ω = I1 − I2 = 1.25 A(rms)
.·. P20Ω = (1.25)2 (20) = 31.25 W
P 10.40 ZL = |ZL |/θ◦ = |ZL | cos θ◦ + j|ZL | sin θ◦
Thus |I| = q
|VTh |
(RTh + |ZL | cos θ)2 + (XTh + |ZL | sin θ)2
Therefore P =
0.5|VTh |2|ZL | cos θ
(RTh + |ZL | cos θ)2 + (XTh + |ZL | sin θ)2
Let D = demoninator in the expression for P, then
dP
(0.5|VTh |2 cos θ)(D · 1 − |ZL |dD/d|ZL |)
=
d|ZL |
D2
dD
= 2(RTh + |ZL | cos θ) cos θ + 2(XTh + |ZL | sin θ) sin θ
d|ZL |
dP
dD
= 0 when D = |ZL |
d|ZL |
d|ZL |
!
Substituting the expressions for D and (dD/d|ZL |) into this equation gives us
2
the relationship R2Th + XTh
= |ZL |2 or |ZTh | = |ZL |.
P 10.41 [a]
180 = 3I1 + j4I1 + j3(I2 − I1) + j9(I1 − I2 ) − j3I1
0 = 9I2 + j9(I2 − I1 ) + j3I1
Solving,
I1 = 18 − j18 A(rms);
I2 = 12/0◦ A(rms)
.·. Vo = (12)(9) = 108 V(rms)
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Problems
10–29
[b] P = (12)2 (9) = 1296 W
[c] Sg = −(180)(18 + j18) = −3240 − j3240 VA
% delivered =
.·. Pg = −3240 W
1296
(100) = 40%
3240
P 10.42 [a] Open circuit voltage:
180 = 3I1 + j4I1 − j3I1 + j9I1 − j3I1
. ·. I1 =
180
= 9.31 − j21.72 A(rms)
3 + j7
VTh = j9I1 − j3I1 = j6I1 = 130.34 + j55.86 V
Short circuit current:
180 = 3I1 + j4I1 + j3(Isc − I1) + j9(I1 − Isc ) − j3I1
0 = −j9(Isc − I1) + j3I1
Solving,
Isc = 20 − j20 A
ZTh =
IL =
VTh
130.34 + j55.86
=
= 1.86 + j4.66 Ω
Isc
20 − j20
130.34 + j55.86
= 35 + j15 = 38.08/23.20◦ A
3.72
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10–30
CHAPTER 10. Sinusoidal Steady State Power Calculations
PL = (38.08)2 (1.86) = 2700 W
[b] I1 =
Zo + j9
1.86 − j4.66 + j9
I2 =
(35 + j15) = 30/0◦ A(rms)
j6
j6
Pdev = (180)(30) = 5400 W
[c] Begin by choosing the capacitor value from Appendix H that is closest to
the required reactive impedance, assuming the frequency of the source is
60 Hz:
1
1
4.66 =
so
C=
= 569.22 µF
2π(60)C
2π(60)(4.66)
Choose the capacitor value closest to this capacitance from Appendix H,
which is 470 µF. Then,
1
XL = −
= −5.6438 Ω
2π(60)(470 × 10−6 )
Now set RL as close as possible to
RL =
q
q
R2Th + (XL + XTh )2:
1.8562 + (4.66 − 5.6438)2 = 2.11 Ω
The closest single resistor value from Appendix H is 10 Ω. The resulting
real power developed by the source is calculated below, using the
Thévenin equivalent circuit:
130.34 + j55.86
I=
= 11.9157/27.94◦
1.86 + j4.66 + 10 − j5.6438
P = |130.34 + j55.86|(11.9157) = 1689.7 W
(instead of 5400 W)
P 10.43 [a] From Problem 9.78, ZTh = 85 + j85 Ω and VTh = 850 + j850 V. Thus, for
∗
maximum power transfer, ZL = ZTh
= 85 − j85 Ω:
I2 =
850 + j850
= 5 + j5 A
170
425/0◦ = (5 + j5)I1 − j20(5 + j5)
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Problems
. ·. I1 =
10–31
325 + j100
= 42.5 − j22.5 A
5 + j5
Sg (del) = 425(42.5 + j22.5) = 18,062.5 + j9562.5 VA
Pg = 18,062.5 W
[b] Ploss = |I1|2 (5) + |I2 |2(45) = 11,562.5 + 2250 = 13,812.5 W
% loss in transformer =
P 10.44 [a] ZTh = −j40 +
13,812.5
(100) = 76.47%
18,062.5
(40)(j40)
= 20 − j20 Ω
40 + j40
∗
.·. ZL = ZTh
= 20 + j20 Ω
[b] VTh =
√
80/0◦ (40)
= 40(1 − j1) = 40 2/ − 45◦ V
40 + j40
√
40 2/ − 45◦ √
I=
= 2/ − 45◦ A
40
|Irms | = 1 A
Pload = (1)2 (20 × 103 ) = 20 W
[c] The closest resistor value from Appendix H is 22 Ω. Find the inductor
value:
(5000)L = 20
so
L = 4 mH
The closest inductor value is 1 mH.
40/ − 45◦
40/ − 45◦
I=
=
= 0.8969/ − 25.35◦ A(rms)
20 − j20 + 22 + j5
42 − j15
Pload = (0.8969)2 (22) = 17.70 W
P 10.45 [a]
(instead of 20 W)
115.2 − j86.4 − 240 115.2 − j80
+
=0
ZTh
90 − j30
.·. ZTh =
240 − 115.2 + j86.4
= 60 + j80 Ω
1.44 − j0.48
.·. ZL = 60 − j80 Ω
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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10–32
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] I =
240/0◦
= 2/0◦ A(rms)
120/0◦
P = (2)2 (60) = 240 W
[c] Let R = 15 Ω + 15 Ω + 15 Ω + 15 Ω = 60 Ω
1
= 80
2π(60)C
so
C=
1
= 33.16 µF
2π(60)(80)
Let C = 22 µFk10 µFk1 µF = 33 µF
P 10.46 [a] Open circuit voltage:
V1 = 5Iφ = 5
100 − 5Iφ
25 + j10
(25 + j10)Iφ = 100 − 5Iφ
Iφ =
100
= 3−jA
30 + j10
VTh =
j3
(5Iφ ) = 15 V
1 + j3
Short circuit current:
V2 = 5Iφ =
100 − 5Iφ
25 + j10
Iφ = 3 − j1 A
Isc =
5Iφ
= 15 − j5 A
1
ZTh =
15
= 0.9 + j0.3 Ω
15 − j5
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Problems
10–33
∗
ZL = ZTh
= 0.9 − j0.3 Ω
IL =
0.3
= 8.33 A(rms)
1.8
P = |IL |2(0.9) = 62.5 W
[b] VL = (0.9 − j0.3)(8.33) = 7.5 − j2.5 V(rms)
I1 =
VL
= −0.833 − j2.5 A(rms)
j3
I2 = I1 + IL = 7.5 − j2.5 A(rms)
.·.
5Iφ = I2 + VL
Iφ = 3 − j1 A
Id.s. = Iφ − I2 = −4.5 + j1.5 A
Sg = −100(3 + j1) = −300 − j100 VA
Sd.s. = 5(3 − j1)(−4.5 − j1.5) = −75 + j0 VA
Pdev = 300 + 75 = 375 W
% developed =
62.5
(100) = 16.67%
375
Checks:
P25Ω = (10)(25) = 250 W
P1Ω = (67.5)(1) = 67.5 W
P0.9Ω = 62.5 W
X
Pabs = 230 + 62.5 + 67.5 = 375 =
X
Pdev
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10–34
CHAPTER 10. Sinusoidal Steady State Power Calculations
Qj10 = (10)(10) = 100 VAR
Qj3 = (6.94)(3) = 20.82 VAR
Q−j0.3 = (69.4)(−0.3) = −20.82 VAR
Qsource = −100 VAR
X
Q = 100 + 20.82 − 20.82 − 100 = 0
P 10.47 [a] First find the Thévenin equivalent:
jωL = j3000 Ω
ZTh = 6000k12,000 + j3000 = 4000 + j3000 Ω
VTh =
12,000
(180) = 120 V
6000 + 12,000
−j
= −j1000 Ω
ωC
I=
120
= 18 − j6 mA
6000 + j2000
1
P = |I|2 (2000) = 360 mW
2
[b] Set Co = 0.1 µF so −j/ωC = −j2000 Ω
Set Ro as close as possible to
Ro =
q
40002 + (3000 − 2000)2 = 4123.1 Ω
.·. Ro = 4000 Ω
[c] I =
120
= 14.77 − j1.85 mA
8000 + j1000
1
P = |I|2 (4000) = 443.1 mW
2
Yes;
443.1 mW > 360 mW
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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Problems
10–35
120
= 15 mA
8000
1
P = (0.015)2 (4000) = 450 mW
2
[e] Ro = 4000 Ω;
Co = 66.67 nF
[d] I =
[f] Yes;
450 mW > 443.1 mW
P 10.48 [a] Set Co = 0.1 µF, so −j/ωC = −j2000 Ω; also set Ro = 4123.1 Ω
I=
120
= 14.55 − j1.79 mA
8123.1 + j1000
1
P = |I|2 (4123.1) = 443.18 mW
2
[b] Yes;
443.18 mW > 360 mW
[c] Yes;
443.18 mW < 450 mW
P 10.49 [a] ZTh = 20 + j60 +
(j20)(6 − j18)
= 80 + j60 = 100/36.87◦ Ω
6 + j2
.·. R = |ZTh | = 100 Ω
[b] VTh =
I=
j20
(480/0◦ ) = 480 + j1440 V(rms)
6 − j18 + j20
480 + j1440
= 4.8 + j6.4 = 8/53.13◦ A(rms)
180 + j60
P = 82 (100) = 6400 W
[c] Pick the 100 Ω resistor from Appendix H to match exactly.
P 10.50 [a] Open circuit voltage:
Vφ − 100 Vφ
+
− 0.1Vφ = 0
5
j5
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10–36
CHAPTER 10. Sinusoidal Steady State Power Calculations
.·. Vφ = 40 + j80 V(rms)
VTh = Vφ + 0.1Vφ (−j5) = Vφ (1 − j0.5) = 80 + j60 V(rms)
Short circuit current:
Isc = 0.1Vφ +
Vφ
= (0.1 + j0.2)Vφ
−j5
Vφ − 100 Vφ
Vφ
+
+
=0
5
j5
−j5
.·. Vφ = 100 V(rms)
Isc = (0.1 + j0.2)(100) = 10 + j20 A(rms)
ZTh =
VTh
80 + j60
=
= 4 − j2 Ω
Isc
10 + j20
.·. Ro = |ZTh | = 4.47 Ω
[b]
80 + j60
√
= 7.36 + j8.82 A (rms)
4 + 20 − j2
√
P = (11.49)2 ( 20) = 590.17 W
I=
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Problems
10–37
[c]
80 + j60
= 10 + j7.5 A (rms)
8
P = (102 + 7.52 )(4) = 625 W
I=
[d]
Vφ − 100 Vφ Vo − (25 + j50)
+
+
=0
5
j5
−j5
Vφ = 50 + j25 V (rms)
0.1Vφ = 5 + j2.5 V (rms)
5 + j2.5 + IC = 10 + j7.5
IC = 5 + j5 A (rms)
IL =
Vφ
= 5 − j10 A (rms)
j5
IR = IC + IL = 10 − j5 A (rms)
Ig = IR + 0.1Vφ = 15 − j2.5 A (rms)
Sg = −100I∗g = −1500 − j250 VA
100 = 5(5 + j2.5) + Vcs + 25 + j50
. ·.
Vcs = 50 − j62.5 V (rms)
Scs = (50 − j62.5)(5 − j2.5) = 93.75 − j437.5 VA
Thus,
X
Pdev = 1500
% delivered to Ro =
625
(100) = 41.67%
1500
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10–38
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.51 [a]
54 = I1 + j2(I1 − I2 ) + j3I2
0 = 7I2 + j2(I2 − I1 ) − j3I2 + j8I2 + j3(I1 − I2 )
Solving,
I1 = 12 − j21 A (rms);
I2 = −3 A (rms)
Vo = 7I2 = −21/0◦ V(rms)
[b] P = |I2 |2(7) = 63 W
[c] Pg = (54)(12) = 648 W
% delivered =
63
(100) = 9.72%
648
P 10.52 [a]
54 = I1 + j2(I1 − I2 ) + j4kI2
0 = 7I2 + j2(I2 − I1 ) − j4kI2 + j8I2 + j4k(I1 − I2 )
Place the equations in standard form:
54 = (1 + j2)I1 + j(4k − 2)I2
0 = j(4k − 2)I1 + [7 + j(10 − 8k)]I2
I1 =
54 − I2 j(4k − 2)
(1 + j2)
Substituting,
I2 =
j54(4k − 2)
[7 + j(10 − 8k)](1 + j2) − (4k − 2)
For Vo = 0, I2 = 0, so if 4k − 2 = 0, then k = 0.5.
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
10–39
[b] When I2 = 0
54
I1 =
= 10.8 − j21.6 A(rms)
1 + j2
Pg = (54)(10.8) = 583.2 W
Check:
Ploss = |I1 |2 (1) = 583.2 W
P 10.53 [a]
Open circuit:
VTh = −j3I1 + j2I1 = −jI1
I1 =
54
= 10.8 − j21.6
1 + j2
VTh = −21.6 − j10.8 V
Short circuit:
54 = I1 + j2(I1 − Isc ) + j3Isc
0 = j2(Isc − I1 ) − j3Isc + j8Isc + j3(I1 − Isc )
Solving,
Isc = −3.32 + j5.82
ZTh =
VTh
−21.6 − j10.8
=
= 0.2 + j3.6 = 3.6/86.86◦ Ω
Isc
−3.32 + j5.82
.·. RL = |ZTh | = 3.6 Ω
[b]
I=
−21.6 − j10.8
= 4.614/163.1◦
3.8 + j3.6
P = |I|2(3.6) = 76.6 W,
which is greater than when RL = 7 Ω
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10–40
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.54 [a]
1
= 100 Ω;
ωC
C=
1
= 26.53 µF
(60)(200π)
[b] Vswo = 4000 + (40)(1.25 + j10) = 4050 + j400
= 4069.71/5.64◦ V(rms)
Vsw = 4000 + (40 − j40)(1.25 + j10) = 4450 + j350 = 4463.73/4.50◦ V(rms)
4463.73
% increase =
− 1 (100) = 9.68%
4069.71
√
[c] P`wo = (40 2)2(1.25) = 4000 W
P`w = 402 (1.25) = 2000 W
4000
% increase =
− 1 (100) = 100%
2000
P 10.55 Open circuit voltage:
I1 =
10/0◦
= 2 − j4 A
1 + j2
VTh = j2I1 + j1.2I1 = j3.2I1 = 12.8 + j6.4 = 14.31/26.57◦
Short circuit current:
10/0◦ = (1 + j2)I1 − j3.2Isc
0 = −j3.2I1 + j5.4Isc
Solving,
Isc = 5.89/ − 5.92◦ A
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Problems
ZTh =
10–41
14.31/26.57◦
= 2.43/32.49◦ = 2.048 + j1.304 Ω
5.89/ − 5.92◦
14.31/26.57◦
.·. I2 =
= 3.49/26.57◦ A
4.096
10/0◦ = (1 + j2)I1 − j3.2I2
10 + j3.2I2
10 + j3.2(3.49/26.57◦ )
=
= 5A
1 + j2
1 + j2
.·.
I1 =
Zg =
10/0◦
= 2 + j0 = 2/0◦ Ω
5
P 10.56 [a]
Open circuit:
VTh =
120
(j10) = 36 + j48 V
16 + j12
Short circuit:
(16 + j12)I1 − j10Isc = 120
−j10I1 + (11 + j23)Isc = 0
Solving,
Isc = 2.4 A
ZTh =
36 + j48
= 15 + j20 Ω
2.4
∗
.·. ZL = ZTh
= 15 − j20 Ω
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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10–42
CHAPTER 10. Sinusoidal Steady State Power Calculations
IL =
VTh
36 + j48
=
= 1.2 + j1.6 A(rms)
ZTh + ZL
30
PL = |IL |2 (15) = 60 W
[b] I1 =
Z22 I2
26 + j3
=
(1.2 + j1.6) = 5.23/ − 30.29◦ A (rms)
jωM
j10
Ptransformer = (120)(5.23) cos(−30.29◦ ) − (5.23)2 (4) = 432.8 W
% delivered =
60
(100) = 13.86%
432.8
P 10.57 [a] jωL1 = j(10,000)(1 × 10−3 ) = j10 Ω
jωL2 = j(10,000)(1 × 10−3 ) = j10 Ω
jωM = j10 Ω
200 = (5 + j10)Ig + j5IL
0 = j5Ig + (15 + j10)IL
Solving,
Ig = 10 − j15 A;
IL = −5 A
Thus,
ig = 18.03 cos(10,000t − 56.31◦ ) A
iL = 5 cos(10,000t − 180◦ ) A
M
0.5
= √ = 0.5
L1 L2
1
[c] When t = 50π µs:
[b] k = √
10,000t = (10,000)(50π) × 10−6 = 0.5π rad = 90◦
ig (50π µs) = 18.03 cos(90◦ − 56.31◦ ) = 15 A
iL (50π µs) = 5 cos(90◦ + 180◦ ) = 0 A
1
1
1
w = L1 i21 + L2 i22 + Mi1 i2 = (10−3 )(15)2 + 0 + 0 = 112.5 mJ
2
2
2
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Problems
10–43
When t = 100π µs:
10,000t = (104 )(100π) × 10−6 = π = 180◦
ig (100π µs) = 18.03 cos(180 − 56.31◦ ) = −10 A
iL (100π µs) = 5 cos(180 − 180◦ ) = 5 A
1
1
w = (10−3 )(10)2 + (10−3 )(5)2 + 0.5 × 10−3 (−10)(5) = 37.5 mJ
2
2
[d] From (a), Im = 5 A,
1
.·. P = (5)2 (15) = 187.5 W
2
[e] Open circuit:
200
(−j5) = −80 − j40 V
5 + j10
VTh =
Short circuit:
200 = (5 + j10)I1 + j5Isc
0 = j5I1 + j10Isc
Solving,
Isc = −
ZTh =
80
120
+j
13
13
VTh
−80 − j40
=
= 1 + j8 Ω
Isc
−(80/13) + j(120/13)
.·. RL = 8.06 Ω
[f]
I=
−80 − j40
= 7.40/165.12◦ A
1 + j8 + 8.06
1
P = (7.40)2 (8.06) = 223.42 W
2
∗
[g] ZL = ZTh
= 1 − j8 Ω
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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10–44
CHAPTER 10. Sinusoidal Steady State Power Calculations
[h] I =
−80 − j40
= 44.72/ − 153.43◦
2
1
P = (44.72)2 (1) = 1000 W
2
P 10.58 [a] Replace the circuit to the left of the primary winding with a Thévenin
equivalent:
VTh = (15)(20kj10) = 60 + j120 V
ZTh = 2 + 20kj10 = 6 + j8 Ω
Transfer the secondary impedance to the primary side:
Zp =
1
XC
(100 − jXC ) = 4 − j
Ω
25
25
Now maximize I by setting (XC /25) = 8 Ω:
. ·. C =
[b] I =
1
= 0.25 µF
200(20 × 103 )
60 + j120
= 6 + j12 A
10
P = |I|2(4) = 720 W
Ro
= 6 Ω;
.·. Ro = 150 Ω
25
60 + j120
[d] I =
= 5 + j10 A
12
[c]
P = |I|2(6) = 750 W
P 10.59 [a]
For maximum power transfer, Zab = 90 kΩ
Zab
N1
= 1−
N2
2
ZL
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Problems
N1
N2
2
. ·.
1−
N1
= ±15;
N2
1−
=
10–45
90,000
= 225
400
N1
= 15 + 1 = 16
N2
2
[b] P = |Ii | (90,000) =
180
180,000
!2
180
[c] V1 = Ri Ii = (90,000)
180,000
(90,000) = 90 mW
!
= 90 V
[d]
Vg = (2.25 × 10−3 )(100,000k80,000) = 100 V
Pg (del) = (2.25 × 10−3 )(100) = 225 mW
% delivered =
90
(100) = 40%
225
P 10.60 [a] Zab = 50 − j400 = 1 −
.·. ZL =
N1
N2
2
ZL
1
(50 − j400) = 2 − j16 Ω
(1 − 6)2
[b]
I1 =
24
= 240/0◦ mA
100
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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10–46
CHAPTER 10. Sinusoidal Steady State Power Calculations
N1 I1 = −N2 I2
I2 = −6I1 = −1.44/0◦ A
IL = I1 + I2 = −1.68/0◦ A
VL = (2 − j16)IL = −3.36 + j26.88 = 27.1/97.13◦ V(rms)
P 10.61 [a] ZTh = 720 + j1500 +
200
50
2
(40 − j30) = 1360 + j1020 = 1700/36.87◦ Ω
.·. Zab = 1700 Ω
Zab =
ZL
(1 + N1 /N2 )2
(1 + N1/N2 )2 = 6800/1700 = 4
.·. N1/N2 = 1
[b] VTh
or
N2 = N1 = 1000 turns
255/0◦
=
(j200) = 1020/53.13◦ V
40 + j30
IL =
1020/53.13◦
= 0.316/34.7◦ A(rms)
3060 + j1020
Since the transformer is ideal, P6800 = P1700.
P = |I|2(1700) = 170 W
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
10–47
[c]
255/0◦ = (40 + j30)I1 − j200(0.26 + j0.18)
.·. I1 = 4.13 − j1.80 A(rms)
Pgen = (255)(4.13) = 1053 W
Pdiss = 1053 − 170 = 883 W
% dissipated =
883
(100) = 83.85%
1053
P 10.62 [a] Open circuit voltage:
500 = 100I1 + V1
V2 = 400I2
V1
V2
=
1
2
. ·.
V2 = 2V1
I1 = 2I2
Substitute and solve:
2V1 = 400I1 /2 = 200I1
500 = 100I1 + 100I1
. ·.
. ·.
.·.
V1 = 100I1
I1 = 500/200 = 2.5 A
1
I2 = I1 = 1.25 A
2
V1 = 100(2.5) = 250 V;
V2 = 2V1 = 500 V
VTh = 20I1 + V1 − V2 + 40I2 = −150 V(rms)
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10–48
CHAPTER 10. Sinusoidal Steady State Power Calculations
Short circuit current:
500 = 80(Isc + I1 ) + 360(Isc + 0.5I1 )
2V1 = 40
I1
+ 360(Isc + 0.5I1)
2
500 = 80(I1 + Isc ) + 20I1 + V1
Solving,
Isc = −1.47 A
RTh =
P =
VTh
−150
=
= 102 Ω
Isc
−1.47
752
= 55.15 W
102
[b]
500 = 80[I1 − (75/102)] − 75 + 360[I2 − (75/102)]
575 +
6000 27,000
+
= 80I1 + 180I2
102
102
. ·.
I1 = 3.456 A
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
10–49
Psource = (500)[3.456 − (75/102)] = 1360.35 W
% delivered =
55.15
(100) = 4.05%
1360.35
[c] P80Ω = 80(I1 + IL )2 = 592.13 W
P20Ω = 20I21 = 238.86 W
P40Ω = 40I22 = 119.43 W
P102Ω = 102I2L = 55.15 W
P360Ω = 360(I2 + IL )2 = 354.73 W
X
Pabs = 592.13 + 238.86 + 119.43 + 55.15 + 354.73 = 1360.3 W =
P 10.63 [a] Open circuit voltage:
X
Pdev
40/0◦ = 4(I1 + I3 ) + 12I3 + VTh
I1
= −I3 ;
4
I1 = −4I3
Solving,
VTh = 40/0◦ V
Short circuit current:
40/0◦ = 4I1 + 4I3 + I1 + V1
4V1 = 16(I1 /4) = 4I1 ;
.·. V1 = I1
.·. 40/0◦ = 6I1 + 4I3
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10–50
CHAPTER 10. Sinusoidal Steady State Power Calculations
Also,
40/0◦ = 4(I1 + I3 ) + 12I3
Solving,
I1 = 6 A;
RTh =
I=
I3 = 1 A;
Isc = I1 /4 + I3 = 2.5 A
VTh
40
=
= 16 Ω
Isc
2.5
40/0◦
= 1.25/0◦ A(rms)
32
P = (1.25)2 (16) = 25 W
[b]
40 = 4(I1 + I3) + 12I3 + 20
4V1 = 4I1 + 16(I1 /4 + I3 );
.·. V1 = 2I1 + 4I3
40 = 4I1 + 4I3 + I1 + V1
.·. I1 = 6 A;
I3 = −0.25 A;
I1 + I3 = 5.75/0◦ A
P40V (developed) = 40(5.75) = 230 W
.·. % delivered =
[c] PRL = 25 W;
25
(100) = 10.87%
230
P16Ω = (1.5)2 (16) = 36 W
P4Ω = (5.75)2 (4) = 132.25 W;
P1Ω = (6)2 (1) = 36 W
P12Ω = (−0.25)2 (12) = 0.75 W
X
Pabs = 25 + 36 + 132.25 + 36 + 0.75 = 230 W =
X
Pdev
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Problems
10–51
P 10.64
30Vo = Va ;
Io
= Ia ;
30
Vb
−Va
=
;
1
20
Ib = −20Ia ;
therefore
Va
= 9 kΩ
Ia
therefore
Vb
9000
=
= 22.5 Ω
Ib
400
Therefore Ib = [50/(2.5 + 22.5)] = 2 A (rms); since the ideal transformers are
lossless, P10Ω = P22.5Ω, and the power delivered to the 22.5 Ω resistor is
22 (22.5) or 90 W.
Vb
a2 10
=
= 2.5 Ω;
therefore a2 = 100,
Ib
400
50
[b] Ib =
= 10 A;
P = (100)(2.5) = 250 W
5
P 10.65 [a]
a = 10
P 10.66 [a] Begin with the MEDIUM setting, as shown in Fig. 10.31, as it involves
only the resistor R2. Then,
Pmed = 500 W =
V2
1202
=
R2
R2
Thus,
1202
R2 =
= 28.8 Ω
500
[b] Now move to the LOW setting, as shown in Fig. 10.30, which involves the
resistors R1 and R2 connected in series:
Plow =
V2
V2
=
= 250 W
R1 + R2
R1 + 28.8
Thus,
1202
− 28.8 = 28.8 Ω
250
[c] Note that the HIGH setting has R1 and R2 in parallel:
R1 =
Phigh =
V2
1202
=
= 1000 W
R1 kR2
28.8k28.8
If the HIGH setting has required power other than 1000 W, this problem
could not have been solved. In other words, the HIGH power setting was
chosen in such a way that it would be satisfied once the two resistor
values were calculated to satisfy the LOW and MEDIUM power settings.
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10–52
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.67 [a] PL =
V2
;
R1 + R2
PM =
R1 + R2 =
V2
;
R2
V2
PL
V2
PM
R2 =
V 2 (R1 + R2)
PH =
R1 R2
R1 + R2 =
PH = V 2
[b] PH =
R1 =
V 2V 2/PL
PL
PH =
V2
;
PL
−
V2
PM
2
PM
PM − PL
V2
PM
=
V2
V2
−
PL
PM
PM PL PM
PL (PM − PL )
(750)2
= 1125 W
(750 − 250)
P 10.68 First solve the expression derived in P10.67 for PM as a function of PL and PH .
Thus
PM − PL =
2
PM
PH
2
PM
− PM + PL = 0
PH
or
2
PM
− PM PH + PL PH = 0
PH
.·. PM =
±
2
s
PH
2
s
2
− PL PH
PH
1
PL
=
± PH
−
2
4
PH
For the specified values of PL and PH
√
PM = 500 ± 1000 0.25 − 0.24 = 500 ± 100
.·. PM 1 = 600 W;
PM 2 = 400 W
Note in this case we design for two medium power ratings
If PM 1 = 600 W
R2 =
(120)2
= 24 Ω
600
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Problems
R1 + R2 =
10–53
(120)2
= 60 Ω
240
R1 = 60 − 24 = 36 Ω
(120)2 (60)
= 1000 W
(36)(24)
CHECK: PH =
If PM 2 = 400 W
R2 =
(120)2
= 36 Ω
400
R1 + R2 = 60 Ω (as before)
R1 = 24 Ω
CHECK: PH = 1000 W
P 10.69 R1 + R2 + R3 =
(120)2
= 24 Ω
600
(120)2
R2 + R3 =
= 16 Ω
900
.·. R1 = 24 − 16 = 8 Ω
R3 + R1 kR2 =
(120)2
= 12 Ω
1200
8R2
.·. 16 − R2 +
= 12
8 + R2
R2 −
8R2
=4
8 + R2
8R2 + R22 − 8R2 = 32 + 4R2
R22 − 4R2 − 32 = 0
R2 = 2 ±
√
4 + 32 = 2 ± 6
.·. R2 = 8 Ω;
.·. R3 = 8 Ω
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10–54
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.70 R2 =
(220)2
= 96.8 Ω
500
R1 + R2 =
(220)2
= 193.6 Ω
250
.·. R1 = 96.8 Ω
CHECK: R1 kR2 = 48.4 Ω
PH =
(220)2
= 1000 W
48.4
P 10.71 Choose R1 = 22 Ω and R2 = 33 Ω:
PL =
PM
1202
= 262 W
22 + 33
1202
=
= 436 W
33
PH =
(instead of 240 W)
(instead of 400 W)
1202 (55)
= 1091 W
(22)(33)
(instead of 1000 W)
P 10.72 Choose R1 = R2 = 100 Ω :
PL =
2202
= 242 W
100 + 100
PM =
2202
= 484 W
100
PH =
2202 (200)
= 968 W
(100)(100)
(instead of 250 W)
(instead of 500 W)
(instead of 1000 W)
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11
Balanced Three-Phase Circuits
Assessment Problems
AP 11.1 Make a sketch:
We know VAN and wish to find VBC . To do this, write a KVL equation to
find VAB , and use the known phase angle relationship between VAB and VBC
to find VBC .
VAB = VAN + VNB = VAN − VBN
Since VAN , VBN , and VCN form a balanced set, and VAN = 240/ − 30◦ V, and
the phase sequence is positive,
VBN = |VAN |//VAN − 120◦ = 240/ − 30◦ − 120◦ = 240/ − 150◦ V
Then,
VAB = VAN − VBN = (240/ − 30◦ ) − (240/ − 150◦ ) = 415.46/0◦ V
Since VAB , VBC , and VCA form a balanced set with a positive phase sequence,
we can find VBC from VAB :
VBC = |VAB |/(/VAB − 120◦ ) = 415.69/0◦ − 120◦ = 415.69/ − 120◦ V
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11–1 system, or transmission in any form or by any means, electronic,
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11–2
CHAPTER 11. Balanced Three-Phase Circuits
Thus,
VBC = 415.69/ − 120◦ V
AP 11.2 Make a sketch:
We know VCN and wish to find VAB . To do this, write a KVL equation to
find VBC , and use the known phase angle relationship between VAB and VBC
to find VAB .
VBC = VBN + VNC = VBN − VCN
Since VAN , VBN , and VCN form a balanced set, and VCN = 450/ − 25◦ V, and
the phase sequence is negative,
VBN = |VCN |//VCN − 120◦ = 450/ − 23◦ − 120◦ = 450/ − 145◦ V
Then,
VBC = VBN − VCN = (450/ − 145◦ ) − (450/ − 25◦ ) = 779.42/ − 175◦ V
Since VAB , VBC , and VCA form a balanced set with a negative phase
sequence, we can find VAB from VBC :
VAB = |VBC |//VBC − 120◦ = 779.42/ − 295◦ V
But we normally want phase angle values between +180◦ and −180◦ . We add
360◦ to the phase angle computed above. Thus,
VAB = 779.42/65◦ V
AP 11.3 Sketch the a-phase circuit:
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Problems
11–3
[a] We can find the line current using Ohm’s law, since the a-phase line
current is the current in the a-phase load. Then we can use the fact that
IaA , IbB , and IcC form a balanced set to find the remaining line currents.
Note that since we were not given any phase angles in the problem
statement, we can assume that the phase voltage given, VAN , has a phase
angle of 0◦ .
2400/0◦ = IaA (16 + j12)
so
IaA =
2400/0◦
= 96 − j72 = 120/ − 36.87◦ A
16 + j12
With an acb phase sequence,
/IbB = /IaA + 120◦
and /IcC = /IaA − 120◦
so
IaA = 120/ − 36.87◦ A
IbB = 120/83.13◦ A
IcC = 120/ − 156.87◦ A
[b] The line voltages at the source are Vab Vbc, and Vca. They form a
balanced set. To find Vab , use the a-phase circuit to find VAN , and use
the relationship between phase voltages and line voltages for a
y-connection (see Fig. 11.9[b]). From the a-phase circuit, use KVL:
Van = VaA + VAN = (0.1 + j0.8)IaA + 2400/0◦
= (0.1 + j0.8)(96 − j72) + 2400/0◦ = 2467.2 + j69.6
2468.18/1.62◦ V
From Fig. 11.9(b),
√
Vab = Van( 3/ − 30◦ ) = 4275.02/ − 28.38◦ V
With an acb phase sequence,
/Vbc = /Vab + 120◦
and /Vca = /Vab − 120◦
so
Vab = 4275.02/ − 28.38◦ V
Vbc = 4275.02/91.62◦ V
Vca = 4275.02/ − 148.38◦ V
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11–4
CHAPTER 11. Balanced Three-Phase Circuits
[c] Using KVL on the a-phase circuit
Va0n = Va0 a + Van = (0.2 + j0.16)IaA + Van
= (0.02 + j0.16)(96 − j72) + (2467.2 + j69.9)
= 2480.64 + j83.52 = 2482.05/1.93◦ V
With an acb phase sequence,
/Vb0 n = /Va0 n + 120◦
and /Vc0 n = /Va0 n − 120◦
so
Va0n = 2482.05/1.93◦ V
Vb0n = 2482.05/121.93◦ V
Vc0n = 2482.05/ − 118.07◦ V
AP 11.4
√
√
IcC = ( 3/ − 30◦ )ICA = ( 3/ − 30◦ ) · 8/ − 15◦ = 13.86/ − 45◦ A
AP 11.5
IaA = 12/(65◦ − 120◦ ) = 12/ − 55◦
"
!
#
!
/ − 30◦
1
◦
√ / − 30 IaA =
√
IAB =
· 12/ − 55◦
3
3
= 6.93/ − 85◦ A
AP 11.6 [a] IAB =
"
!
#
1
√ /30◦ [69.28/ − 10◦ ] = 40/20◦ A
3
Therefore Zφ =
[b] IAB =
"
!
4160/0◦
= 104/ − 20◦ Ω
40/20◦
#
1
√ / − 30◦ [69.28/ − 10◦ ] = 40/ − 40◦ A
3
Therefore Zφ = 104/40◦ Ω
AP 11.7
Iφ =
110
110
+
= 30 − j40 = 50/ − 53.13◦ A
3.667 j2.75
Therefore |IaA | =
AP 11.8 [a] |S| =
Q=
√
√
3Iφ = 3(50) = 86.60 A
√
3(208)(73.8) = 26,587.67 VA
q
(26,587.67)2 − (22,659)2 = 13,909.50 VAR
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Problems
[b] pf =
11–5
22,659
= 0.8522 lagging
26,587.67
!
2450
/0◦ V;
AP 11.9 [a] VAN = √
VAN I∗aA = Sφ = 144 + j192 kVA
3
Therefore
(144 + j192)1000
√
= (101.8 + j135.7) A
I∗aA =
2450/ 3
IaA = 101.8 − j135.7 = 169.67/ − 53.13◦ A
|IaA | = 169.67 A
[b] P =
(2450)2
;
R
Q=
(2450)2
;
X
therefore R =
(2450)2
= 41.68 Ω
144,000
therefore X =
(2450)2
= 31.26 Ω
192,000
√
VAN
2450/ 3
[c] Zφ =
=
= 8.34/53.13◦ = (5 + j6.67) Ω
◦
/
IaA
169.67 − 53.13
.·. R = 5 Ω,
X = 6.67 Ω
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11–6
CHAPTER 11. Balanced Three-Phase Circuits
Problems
P 11.1
Va = Vm /0◦ = Vm + j0
Vb = Vm / − 120◦ = −Vm (0.5 + j0.866)
Vc = Vm /120◦ = Vm (−0.5 + j0.866)
Va + Vb + Vc = (Vm )(1 + j0 − 0.5 − j0.866 − 0.5 + j0.866)
= Vm (0) = 0
P 11.2
[a] First, convert the cosine waveforms to phasors:
Va = 208/27◦ ;
Vb = 208/ − 147◦ ;
Vc = 208/ − 93◦
Subtract the phase angle of the a-phase from all phase angles:
/V0a = 27◦ − 27◦ = 0◦
/V0b = 147◦ − 27◦ = 120◦
/V0c = −93◦ − 27◦ = −120◦
Compare the result to Eqs. 11.1 and 11.2:
Therefore acb
[b] First, convert the cosine waveforms to phasors:
Va = 4160/ − 18◦ ;
Vb = 4160/ − 138◦ ;
Vc = 4160/ + 102◦
Subtract the phase angle of the a-phase from all phase angles:
/V0a = −18◦ + 18◦ = 0◦
/V0b = −138◦ + 18◦ = −120◦
/V0c = 102◦ + 18◦ = 120◦
Compare the result to Eqs. 11.1 and 11.2:
Therefore abc
P 11.3
[a] Va = 139/0◦ V
Vb = 139/120◦ V
Vc = 139/ − 120◦ V
Balanced, negative phase sequence
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Problems
11–7
[b] Va = 381/0◦ V
Vb = 381/240◦ V = 622/ − 120◦ V
Vc = 381/120◦ V
Balanced, positive phase sequence
[c] Va = 2771/ − 120◦ V
Vb = 2771/0◦ V
Vc = 2771/120◦ V
Balanced, negative phase sequence
[d] Va = 170/ − 60◦ V
Vb = 170/180◦ V
Vc = 170/60◦ V
Balanced, positive phase sequence
[e] Unbalanced, due to unequal amplitudes
[f] Unbalanced, due to unequal phase angle separation
P 11.4
P 11.5
Va + Vb + Vc
=0
3(RW + jXW )
√
[a] Van = 1/ 3/30◦ Vab = 240/120◦ V (rms)
The a-phase circuit is
I=
240/120◦
= 2.4/83.13◦ A (rms)
80 + j60
[c] VAN = (76 + j55)IaA = 225.15/119.02◦ V (rms)
√
VAB = 3/ − 30◦ VAN = 389.98/89.02◦ A (rms)
[b] IaA =
P 11.6
Zga + Zla + ZLa = 60 + j80 Ω
Zgb + Zlb + ZLb = 40 + j30Ω
Zgc + Zlc + ZLc = 20 + j15Ω
VN − 240 VN − 240/120◦ VN − 240/ − 120◦ VN
+
+
+
=0
60 + j80
40 + j30
20 + j15
10
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11–8
CHAPTER 11. Balanced Three-Phase Circuits
Solving for VN yields
VN = 42.94/ − 156.32◦ V (rms)
Io =
P 11.7
VN
= 4.29/ − 156.32◦ A (rms)
10
VAN = 7620/30◦ V
VBN = 7620/150◦ V
VCN = 7620/ − 90◦ V
VAB = VAN − VBN = 13,198.23/0◦ V
VBC = VBN − VCN = 13,198.23/120◦ V
VCA = VCN − VAN = 13,198.23/ − 120◦ V
vAB = 13,198.23 cos ωt V
vBC = 13,198.23 cos(ωt + 120◦ ) V
vCA = 13,198.23 cos(ωt − 120◦ ) V
P 11.8
200
= 8 A (rms)
25
200/ − 120◦
=
= 4/ − 66.87◦ A (rms)
30 − j40
[a] IaA =
IbB
IcC =
200/120◦
= 2/83.13◦ A (rms)
80 + j60
The magnitudes are unequal and the phase angles are not 120◦ apart.
b] Io = IaA + IbB + IcC = 9.96/ − 9.79◦ A (rms)
P 11.9
[a]
6600
IaA = √
= 15.24/16.26◦ A (rms)
3(240 − j70)
|IaA | = |IL | = 15.24 A (rms)
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Problems
11–9
[b] Van = (15.24/16.26◦ )(240 − j66) = 3801.24/0.91◦
√
|Vab | = 3(3801.24) = 6583.94 V (rms)
P 11.10 [a] IaA
277/0◦
=
= 2.77/ − 36.87◦ A (rms)
80 + j60
IbB
277/ − 120◦
=
= 2.77/ − 156.87◦ A (rms)
80 + j60
IcC =
277/120◦
= 2.77/83.13◦ A (rms)
80 + j60
Io = IaA + IbB + IcC = 0
[b] VAN = (78 + j54)IaA = 262.79/ − 2.17◦ V (rms)
[c] VAB = VAN − VBN
VBN = (77 + j56)IbB = 263.73/ − 120.84◦ V (rms)
VAB = 262.79/ − 2.17◦ − 263.73/ − 120.84◦ = 452.89/28.55◦ V (rms)
[d] Unbalanced — see conditions for a balanced circuit on p. 504 of the text!
P 11.11 Make a sketch of the a-phase:
[a] Find the a-phase line current from the a-phase circuit:
IaA
125/0◦
125/0◦
=
=
0.1 + j0.8 + 19.9 + j14.2
20 + j15
= 4 − j3 = 5/ − 36.87◦ A (rms)
Find the other line currents using the acb phase sequence:
IbB = 5/ − 36.87◦ + 120◦ = 5/83.13◦ A (rms)
IcC = 5/ − 36.87◦ − 120◦ = 5/ − 156.87◦ A (rms)
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11–10
CHAPTER 11. Balanced Three-Phase Circuits
[b] The phase voltage at the source is Van = 125/0◦ V. Use Fig. 11.9(b) to
find the line voltage, Van, from the phase voltage:
√
Vab = Van( 3/ − 30◦ ) = 216.51/ − 30◦ V (rms)
Find the other line voltages using the acb phase sequence:
Vbc = 216.51/ − 30◦ + 120◦ = 216.51/90◦ V (rms)
Vca = 216.51/ − 30◦ − 120◦ = 216.51/ − 150◦ V (rms)
[c] The phase voltage at the load in the a-phase is VAN . Calculate its value
using IaA and the load impedance:
VAN = IaA ZL = (4 − j3)(19.9 + j14.2) = 122.2 − j2.9 = 122.23/ − 1.36◦ V (rms)
Find the phase voltage at the load for the b- and c-phases using the acb
sequence:
VBN = 122.23/ − 1.36◦ + 120◦ = 122.23/118.64◦ V (rms)
VCN = 122.23/ − 1.36◦ − 120◦ = 122.23/ − 121.36◦ V (rms)
[d] The line voltage at the load in the a-phase is VAB . Find this line voltage
from the phase voltage at the load in the a-phase, VAN , using Fig,
11.9(b):
√
VAB = VAN ( 3/ − 30◦ ) = 211.72/ − 31.36◦ V (rms)
Find the line voltage at the load for the b- and c-phases using the acb
sequence:
VBC = 211.72/ − 31.36◦ + 120◦ = 211.72/88.64◦ V (rms)
VCA = 211.72/ − 31.36◦ − 120◦ = 211.72/ − 151.36◦ V (rms)
P 11.12 [a] Van = Vcn − /120◦ = 20/ − 210◦ = 20/150◦ V (rms)
Zy = Z∆ /3 = 39 − j33 Ω
The a-phase circuit is
20/150◦
[b] IaA =
= 0.4/ − 173.13◦ A (rms)
40 − j30
[c] VAN = (39 + j33)IaA = 20.44/146.63◦ V (rms)
√
VAB = 3/30◦ VAN = 35.39/176.63◦ A (rms)
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Problems
11–11
P 11.13 Zy = Z∆ /3 = 4 + j3 Ω
The a-phase circuit is
IaA =
240/ − 170◦
= 37.48/151.34◦ A (rms)
(1 + j1) + (4 + j3)
1
IAB = √ / − 30◦ IaA = 21.64/121.34◦ A (rms)
3
P 11.14 [a] IAB =
69,000
= 76.67/16.26◦ A (rms)
864 − j252
IBC = 76.67/ − 103.74◦ A (rms)
ICA = 76.67/136.26◦ A (rms)
√
[b] IaA = 3/ − 30◦ IAB = 132.79/ − 13.74◦ A (rms)
IbB = 132.79/ − 133.74◦ A (rms)
IcC = 132.79/106.26◦ A (rms)
[c]
13,000
√ / − 30◦ + (0.5 + j4)(132.79/ − 13.74◦ )
3
= 39,755.70/ − 29.24◦ V (rms)
Van =
Vab =
√
3/30◦ Van = 68,858.88/0.76◦ V (rms)
Vbc = 68,858.88/ − 119.24◦ V (rms)
Vca = 68,858.88/120.76◦ V (rms)
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11–12
CHAPTER 11. Balanced Three-Phase Circuits
P 11.15 [a]
IaA =
7650
7650
+
= 252.54/ − 6.49◦ A (rms)
72 + j21
50
|IaA | = 252.54 A (rms)
√
7650 3/30◦
[b] IAB =
= 88.33/30◦ A (rms)
150
|IAB | = 88.33 A (rms)
[c] IAN
7650/0◦
=
= 102/ − 16.26◦ A (rms)
72 + j21
|IAN | = 102 A (rms)
[d] Van = (252.54/ − 6.49◦ )(j1) + 7650/0◦ = 7682.66/1.87◦ V (rms)
√
|Vab | = 3(7682.66) = 13,306.76 V (rms)
√
208
P 11.16 Van = 1/ 3/ − 30◦ Vab = √ /20◦ V (rms)
3
Zy = Z∆ /3 = 1 − j3 Ω
The a-phase circuit is
Zeq = (4 + j3)k(1 − j3) = 2.6 − j1.8 Ω
VAN
2.6 − j1.8
=
(1.4 + j0.8) + (2.6 − j1.8)
VAB =
√
!
208
√ /20◦ = 92.1/ − 0.66◦ V (rms)
3
3/30◦ VAN = 159.5/29.34◦ V (rms)
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Problems
P 11.17 [a] IAB =
11–13
13,200/0◦
= 105.6/36.87◦ A (rms)
100 − j75
IBC = 105.6/156.87◦ A (rms)
ICA = 105.6/ − 83.13◦ A (rms)
√
[b] IaA = 3/ − 30◦ IAB = 182.9/66.87◦ A (rms)
IbB = 182.9/ − 173.13◦ A (rms)
IcC = 182.9/ − 53.13◦ A (rms)
[c] Iba = IAB = 105.6/36.87◦ A (rms)
Icb = IBC = 105.6/156.87◦ A (rms)
Iac = ICA = 105.6/ − 83.13◦ A (rms)
P 11.18 [a] IAB =
480/0◦
= 192/16.26◦ A (rms)
2.4 − j0.7
IBC =
480/120◦
= 48/83.13◦ A (rms)
8 + j6
ICA =
480/ − 120◦
= 24/ − 120◦ A (rms)
20
[b] IaA = IAB − ICA
= 210/20.79◦
IbB = IBC − IAB
= 178.68/ − 178.04◦
IcC = ICA − IBC
= 70.7/ − 104.53◦
P 11.19 [a]
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11–14
CHAPTER 11. Balanced Three-Phase Circuits
[b] IaA = √
13,800
= 2917/ − 29.6◦ A (rms)
3(2.375 + j1.349)
|IaA | = 2917 A (rms)
[c] VAN = (2.352 + j1.139)(2917/ − 29.6◦ ) = 7622.93/ − 3.76◦ V (rms)
√
|VAB | = 3|VAN | = 13,203.31 V (rms)
[d] Van = (2.372 + j1.319)(2917/ − 29.6◦ ) = 7616.93/ − 0.52◦ V (rms)
√
|Vab | = 3|Van | = 13,712.52 V (rms)
|IaA |
[e] |IAB | = √ = 1684.13 A (rms)
3
[f] |Iab | = |IAB | = 1684.13 A (rms)
P 11.20 [a] Since the phase sequence is acb (negative) we have:
Van = 2399.47/30◦ V (rms)
Vbn = 2399.47/150◦ V (rms)
Vcn = 2399.47/ − 90◦ V (rms)
1
ZY = Z∆ = 0.9 + j4.5 Ω/φ
3
√
[b] Vab = 2399.47/30◦ − 2399.47/150◦ = 2399.47 3/0◦ = 4156/0◦ V (rms)
Since the phase sequence is negative, it follows that
Vbc = 4156/120◦ V (rms)
Vca = 4156/ − 120◦ V (rms)
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Problems
11–15
[c]
Iba =
4156
= 301.87/ − 78.69◦ A (rms)
2.7 + j13.5
Iac = 301.87/ − 198.69◦ A (rms)
IaA = Iba − Iac = 522.86/ − 48.69◦ A (rms)
Since we have a balanced three-phase circuit and a negative phase
sequence we have:
IbB = 522.86/71.31◦ A (rms)
IcC = 522.86/ − 168.69◦ A (rms)
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11–16
CHAPTER 11. Balanced Three-Phase Circuits
[d]
IaA =
2399.47/30◦
= 522.86/ − 48.69◦ A (rms)
0.9 + j4.5
Since we have a balanced three-phase circuit and a negative phase
sequence we have:
IbB = 522.86/71.31◦ A (rms)
IcC = 522.86/ − 168.69◦ A (rms)
P 11.21 [a]
[b] IaA
2399.47/30◦
=
= 1.2/46.26◦ A (rms)
1920 − j560
VAN = (1910 − j636)(1.2/46.26◦ ) = 2415.19/27.84◦ V (rms)
√
|VAB | = 3(2415.19) = 4183.24 V (rms)
1.2
[c] |Iab| = √ = 0.69 A (rms)
3
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Problems
11–17
[d] Van = (1919.1 − j564.5)(1.2/46.26◦ ) = 2400.48/29.87◦ V (rms)
√
|Vab | = 3(2400.48) = 4157.76 V (rms)
P 11.22 The a-phase of the circuit is shown below:
I1 =
120/20◦
= 12/ − 16.87◦ A (rms)
8 + j6
I∗2 =
600/36◦
= 5/16◦ A (rms)
120/20◦
I = I1 + I2 = 12/ − 16.87◦ + 5/ − 16◦ = 17/ − 16.61◦ A (rms)
Sa = VI∗ = (120/20◦ )(17/16.61◦ ) = 2040/36.61◦ VA
ST = 3Sa = 6120/36.61◦ VA
P 11.23 The complex power of the source per phase is
Ss = 20,000/( cos−1 0.6) = 20,000/53.13◦ = 12,000 + j16,000 kVA. This
complex power per phase must equal the sum of the per-phase impedances of
the two loads:
Ss = S1 + S2
so
12,000 + j16,000 = 10,000 + S2
.·. S2 = 2000 + j16,000 VA
|Vrms |2
Z2∗
Also,
S2 =
|Vrms | =
|Vload|
√
= 120 V (rms)
3
Thus,
Z2∗
|Vrms |2
(120)2
=
=
= 0.11 − j0.89 Ω
S2
2000 + j16,000
.·. Z2 = 0.11 + j0.89 Ω
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11–18
CHAPTER 11. Balanced Three-Phase Circuits
P 11.24 [a] ST ∆ = 14,000/41.41◦ − 9000/53.13◦ = 5.5/22◦ kVA
S∆ = ST ∆ /3 = 1833.46/22◦ VA
3000/53.13◦
= 300 V (rms)
10/ − 30◦
√
√
|Vline| = |Vab| = 3|Van| = 300 3 = 519.62 V (rms)
[b] |Van| =
P 11.25 |Iline| =
1600
√ = 11.547 A (rms)
240/ 3
√
|V |
240/ 3
|Zy | =
=
= 12
|I|
11.547
Zy = 12/ − 50◦ Ω
Z∆ = 3Zy = 36/ − 50◦ = 23.14 − j27.58 Ω/φ
P 11.26 Let pa , pb , and pc represent the instantaneous power of phases a, b, and c,
respectively. Then assuming a positive phase sequence, we have
pa = vaniaA = [Vm cos ωt][Im cos(ωt − θφ )]
pb = vbnibB = [Vm cos(ωt − 120◦ )][Im cos(ωt − θφ − 120◦ )]
pc = vcnicC = [Vm cos(ωt + 120◦ )][Im cos(ωt − θφ + 120◦ )]
The total instantaneous power is pT = pa + pb + pc , so
pT = Vm Im [cos ωt cos(ωt − θφ ) + cos(ωt − 120◦ ) cos(ωt − θφ − 120◦ )
+ cos(ωt + 120◦ ) cos(ωt − θφ + 120◦ )]
Now simplify using trigonometric identities. In simplifying, collect the
coefficients of cos(ωt − θφ ) and sin(ωt − θφ). We get
pT = Vm Im [cos ωt(1 + 2 cos2 120◦ ) cos(ωt − θφ )
+2 sin ωt sin2 120◦ sin(ωt − θφ )]
= 1.5Vm Im [cos ωt cos(ωt − θφ ) + sin ωt sin(ωt − θφ )]
= 1.5Vm Im cos θφ
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Problems
11–19
P 11.27 [a] S1 = (4.864 + j3.775) kVA
S2 = 17.636(0.96) + j17.636(0.28) = (16.931 + j4.938) kVA
√
13,853
= 0.521
3VL IL sin θ3 = 13,853;
sin θ3 = √
3(208)(73.8)
Therefore
cos θ3 = 0.854
Therefore
13,853
P3 =
× 0.854 = 22,693.584 W
0.521
S3 = 22.694 + j13.853 kVA
ST = S1 + S2 + S3 = 44.49 + j22.57 kVA
1
ST /φ = ST = 14.83 + j7.52 kVA
3
208 ∗
√ IaA = (14.83 + j7.52)103 ;
I∗aA = 123.49 + j62.64 A
3
IaA = 123.49 − j62.64 = 138.46/ − 26.90◦ A
(rms)
[b] pf = cos(0◦ − 26.90◦ ) = 0.892 lagging
P 11.28 From the solution to Problem 11.18 we have:
SAB = (480/0◦ )(192/ − 16.26◦ ) = 88,473.7 − j25,804.5 VA
SBC = (480/120◦ )(48/ − 83.13◦ ) = 18,431.98 + j13,824.03 VA
SCA = (480/ − 120◦ )(24/120◦ ) = 11,520 + j0 VA
P 11.29 [a] S1/φ = 40,000(0.96) − j40,000(0.28) = 38,400 − j11,200 VA
S2/φ = 60,000(0.8) + j60,000(0.6) = 48,000 + j36,000 VA
S3/φ = 33,600 + j5200 VA
ST /φ = S1 + S2 + S3 = 120,000 + j30,000 VA
.·. I∗aA =
120,000 + j30,000
= 50 + j12.5
2400
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11–20
CHAPTER 11. Balanced Three-Phase Circuits
.·. IaA = 50 − j12.5 A
Van = 2400 + (50 − j12.5)(1 + j8) = 2550 + j387.5 = 2579.27/8.64◦ V (rms)
√
|Vab | = 3(2579.27) = 4467.43 V (rms)
[b] Sg/φ = (2550 + j387.5)(50 + j12.5) = 122,656.25 + j51,250 VA
% efficiency =
P 11.30 [a] I∗aA =
120,000
(100) = 97.83%
122,656.25
(160 + j46.67)103
= 133.3 + j38.9
1200
IaA = 133.3 − j38.9 A (rms)
Van = 1200 + (133.3 − j38.9)(0.18 + j1.44) = 1280 + j185 V (rms)
IC =
1280 + j185
= −3.1 + j21.3 A (rms)
−j60
Ina = IaA + IC = 130.25 − j17.556 = 131.4/ − 7.68◦ A (rms)
[b] Sg/φ = (1280 + j185)(130.25 + j17.556) = 163,472 + j46,567 VA
SgT = 3Sg/φ = −490.4 − j139.7 kVA
Therefore, the source is delivering 490.4 kW and 139.7 kvars.
[c] Pdel = 490.4 kW
Pabs = 3(160,000) + 3|IaA |2(0.18)
= 490.4 kW = Pdel
[d] Qdel = 3|IC |2(60) + 139.7 × 103 = 223.3 kVAR
Qabs = 3(46,667) + 3|IaA |2(1.44)
= 223.4 kVAR = Qdel (roundoff)
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Problems
11–21
P 11.31 [a]
IaA =
1365/0◦
= 27.3/ − 53.13◦ A (rms)
30 + j40
IaA
ICA = √ /150◦ = 15.76/96.87◦ A (rms)
3
[b] Sg/φ = −1365I∗aA = −22,358.75 − j29,811.56 VA
.·. Pdeveloped/phase = 22.359 kW
Pabsorbed/phase = |IaA |228.5 = 21.241 kW
% delivered =
21.241
(100) = 95%
22.359
P 11.32 [a] POUT = 746 × 100 = 74,600 W
PIN = 74,600/(0.97) = 76,907.22 W
√
3VL IL cos θ = 76,907.22
76,907.22
= 242.58 A (rms)
3(208)(0.88)
√
√
[b] Q = 3VL IL sin φ = 3(208)(242.58)(0.475) = 41,511.90 VAR
IL = √
P 11.33 [a]
√
24,000 3/0◦
I1 =
= 66.5 − j49.9 A (rms)
400 + j300
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11–22
CHAPTER 11. Balanced Three-Phase Circuits
√
24,000 3/0◦
I2 =
= 33.3 + j24.9 A (rms)
800 − j600
I∗3 =
57,600 + j734,400
√
= 1.4 + j17.7
24,000 3
I3 = 1.4 − j17.7 A (rms)
IaA = I1 + I2 + I3 = 101.2 − j42.7 A = 109.8/ − 22.9◦ A (rms)
√
Van = (2 + j16)(101.2 − j42.7) + 24,000 3 = 42,454.8 + j1533.8 V (rms)
Sφ = Van I∗aA = (42,454.8 + j1533.8)(101.2 + j42.7)
= 4,230,932.5 + j1,968,040.5 VA
ST = 3Sφ = 12,692.8 + j5904.1 kVA
√
[b] S1/φ = 24,000 3(66.5 + j49.9) = 2764.4 + j2074.3 kVA
√
S2/φ = 24,000 3(33.3 − j24.9) = 1384.3 − j1035.1 kVA
S3/φ = 57.6 + j734.4 kVA
Sφ (load) = 4206.3 + j1773.6 kVA
% delivered =
4206.3
(100) = 99.4%
4230.9
P 11.34
4000I∗1 = (210 + j280)103
I∗1 =
210
280
+j
= 52.5 + j70 A (rms)
4
4
I1 = 52.5 − j70 A (rms)
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Problems
I2 =
11–23
4000/0◦
= 240 + j70 A (rms)
15.36 − j4.48
.·. IaA = I1 + I2 = 292.5 + j0 A (rms)
Van = 4000 + j0 + 292.5(0.1 + j0.8) = 4036.04/3.32◦ V (rms)
|Vab| =
√
3|Van| = 6990.62 V (rms)
P 11.35 Assume a ∆-connect load (series):
1
Sφ = (96 × 103 )(0.8 + j0.6) = 25,600 + j19,200 VA
3
∗
Z∆φ
=
|480|2
= 5.76 − j4.32 Ω/φ
25,600 + j19,200
Z∆φ = 5.76 + 4.32 Ω
Now assume a Y-connected load (series):
1
ZY φ = Z∆φ = 1.92 + j1.44 Ω/φ
3
Now assume a ∆-connected load (parallel):
Pφ =
|480|2
R∆
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11–24
CHAPTER 11. Balanced Three-Phase Circuits
R∆φ =
Qφ =
|480|2
= 9Ω
25,600
|480|2
X∆
X∆ φ =
|480|2
= 12 Ω
19,200
Now assume a Y-connected load (parallel):
1
RY φ = R∆φ = 3 Ω
3
1
XY φ = X∆φ = 4 Ω
3
P 11.36 [a]
SL/φ
1
720
=
720 + j
(0.6) 103 = 240,000 + j180,000 VA
3
0.8
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Problems
I∗aA =
11–25
240,000 + j180,000
= 100 + j75 A (rms)
2400
IaA = 100 − j75 A (rms)
Van = 2400 + (0.8 + j0.6)(100 − j75)
= 2960 + j580 = 3016.29/11.09◦ V (rms)
|Vab | =
√
3(3016.29) = 5224.37 V (rms)
[b]
I1 = 100 − j75 A
(from part [a])
1
S2 = 0 − j (576) × 103 = −j192,000 VAR
3
I∗2 =
−j192,000
= −j80 A (rms)
2400
.·. I2 = j80 A (rms)
IaA = 100 − j75 + j80 = 100 + j5 A (rms)
Van = 2400 + (100 + j5)(0.8 + j6.4)
= 2448 + j644 = 2531.29/14.74◦ V (rms)
|Vab | =
√
3(2531.29) = 4384.33 V (rms)
[c] |IaA | = 125 A (rms)
Ploss/φ = (125)2 (0.8) = 12,500 W
Pg/φ = 240,000 + 12,500 = 252.5 kW
%η =
240
(100) = 95.05%
252.5
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11–26
CHAPTER 11. Balanced Three-Phase Circuits
[d] |IaA | = 100.125 A (rms)
P`/φ = (100.125)2 (0.8) = 8020 W
%η =
240,000
(100) = 96.77%
248,200
[e] Zcap/Y = −j
24002
= −j30 Ω
−192,000
Zcap/∆ = 3Zcap/Y = −j90 Ω
. ·.
1
= 90;
ωC
C=
1
= 29.47 µF
(90)(120π)
P 11.37 [a] From Assessment Problem 11.9, IaA = (101.8 − j135.7) A (rms)
Therefore Icap = j135.7 A (rms)
√
2450/ 3
Therefore ZCY =
= −j10.42 Ω
j135.7
Therefore CY =
1
= 254.5 µF
(10.42)(2π)(60)
ZC∆ = (−j10.42)(3) = −j31.26 Ω
Therefore C∆ =
254.5
= 84.84 µF
3
[b] CY = 254.5 µF
[c] |IaA | = 101.8 A (rms)
P 11.38 [a]
1
Sg = (150)(0.8 − j0.6) = 40 − j30 kVA
3
1
S1 = (30 + j30) = 10 + j10 kVA
3
S2 = Sg − S1 = 30 − j40 kVA
.·. I∗aA =
(30 − j40)103
= 12 − j16
2500
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Problems
11–27
IaA = 12 + j16 A (rms)
Z=
[b] R =
2500
= 75 − j100 Ω
12 + j16
(2500)2
= 208.33 Ω
30 × 103
XL =
(2500)2
= −156.25 Ω
−40 × 103
1
P 11.39 [a] Sg/φ = (41.6)(0.707 + j0.707) × 103 = 9803.73 + j9803.73 VA
3
I∗aA =
9803.73 + j9803.73
√
= 70.76 + j70.76 A (rms)
240/ 3
IaA = 70.76 − j70.76 A (rms)
240
VAN = √ − (0.04 + j0.03)(70.76 − j70.76)
3
= 133.61 + j0.71 = 133.61/0.30◦ V (rms)
|VAB | =
√
3(133.61) = 231.42 V (rms)
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11–28
CHAPTER 11. Balanced Three-Phase Circuits
[b] SL/φ = (133.61 + j0.71)(70.76 + j70.76) = 9404 + j9504.5 VA
SL = 3SL/φ = 28,212 + j28,513 VA
Check:
Sg = 41,600(0.7071 + j0.7071) = 29,415 + j29,415 VA
P` = 3|IaA |2 (0.04) = 1202 W
Pg = PL + P` = 28,212 + 1202 = 29,414 W
(checks)
Q` = 3|IaA |2 (0.03) = 901 VAR
Qg = QL + Q` = 28,513 + 901 = 29,414 VAR (checks)
P 11.40 Zφ = |Z|/θ =
VAN
IaA
θ = /VAN − /IaA
θ1 = /VAB − /IaA
For a positive phase sequence,
/VAB = /VAN + 30◦
Thus,
θ1 = /VAN + 30◦ − /IaA = θ + 30◦
Similarly,
Zφ = |Z|/θ =
VCN
IcC
θ = /VCN − /IcC
θ2 = /VCB − /IcC
For a positive phase sequence,
/VCB = /VBA − 120◦ = /VAB + 60◦
/IcC = /IaA + 120◦
Thus,
θ2 = /VAB + 60◦ − (/IaA + 120◦ ) = θ1 − 60◦
= θ + 30◦ − 60◦ = θ − 30◦
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Problems
P 11.41 IaA =
11–29
VAN
= |IL |/−θφ A,
Zφ
Zφ = |Z|/θφ ,
VBC = |VL |/ − 90◦ V,
Wm = |VL | |IL | cos[−90◦ − (−θφ)]
= |VL | |IL | cos(θφ − 90◦ )
= |VL | |IL | sin θφ ,
therefore
√
√
3Wm = 3|VL | |IL | sin θφ = Qtotal
P 11.42 [a] Z = 16 − j12 = 20/ − 36.87◦ Ω
VAN = 680/0◦ V;
.·. IaA = 34/36.87◦ A
√
= 680 3/ − 90◦ V
VBC = VBN − VCN
√
Wm = (680 3)(34) cos(−90 − 36.87◦ ) = −24,027.07 W
√
3Wm = −41,616.1 W
[b] Qφ = (342 )(−12) = −13,872 VAR
√
QT = 3Qφ = −41,616 VAR = 3Wm
P 11.43 [a] W2 − W1 = VL IL [cos(θ − 30◦ ) − cos(θ + 30◦ )]
= VL IL [cos θ cos 30◦ + sin θ sin 30◦
− cos θ cos 30◦ + sin θ sin 30◦ ]
= 2VL IL sin θ sin 30◦ = VL IL sin θ,
therefore
√
√
3(W2 − W1 ) = 3VL IL sin θ = QT
[b] Zφ = (8 + j6) Ω
√
QT = 3[2476.25 − 979.75] = 2592 VAR,
QT = 3(12)2 (6) = 2592 VAR;
Zφ = (8 − j6) Ω
√
QT = 3[979.75 − 2476.25] = −2592 VAR,
QT = 3(12)2 (−6) = −2592 VAR;
√
Zφ = 5(1 + j 3) Ω
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11–30
CHAPTER 11. Balanced Three-Phase Circuits
QT =
√
3[2160 − 0] = 3741.23 VAR,
√
QT = 3(12)2 (5 3) = 3741.23 VAR;
Zφ = 10/75◦ Ω
√
QT = 3[−645.53 − 1763.63] = −4172.80 VAR,
QT = 3(12)2 [−10 sin 75◦ ] = −4172.80 VAR
P 11.44 Wm1 = |VAB ||IaA | cos(/VAB − /IaA ) = (199.58)(2.4) cos(65.68◦ ) = 197.26 W
Wm2 = |VCB ||IcC | cos(/VCB − /IcC ) = (199.58)(2.4) cos(5.68◦ ) = 476.64 W
CHECK: W1 + W2 = 673.9 = (2.4)2 (39)(3) = 673.9 W
P 11.45 tan φ =
√
3(W2 − W1)
= 0.75
W1 + W2
.·. φ = 36.87◦
√
.·. 2400 3|IL | cos 66.87◦ = 40,823.09
|IL | = 25 A
|Z| =
2400
= 96 Ω
25
.·. Z = 96/36.87◦ Ω
P 11.46 [a] W1 = |VBA ||IbB | cos θ
Negative phase sequence:
√
VBA = 240 3/150◦ V
IaA =
240/0◦
= 18/30◦ A
13.33/ − 30◦
IbB = 18/150◦ A
√
W1 = (18)(240) 3 cos 0◦ = 7482.46 W
W2 = |VCA ||IcC | cos θ
√
VCA = 240 3/ − 150◦ V
IcC = 18/ − 90◦ A
√
W2 = (18)(240) 3 cos(−60◦ ) = 3741.23 W
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Problems
11–31
[b] Pφ = (18)2 (40/3) cos(−30◦ ) = 3741.23 W
PT = 3Pφ = 11,223.69 W
W1 + W2 = 7482.46 + 3741.23 = 11,223.69 W
.·. W1 + W2 = PT
(checks)
P 11.47 From the solution to Prob. 11.18 we have
IaA = 210/20.79◦ A
and
IbB = 178.68/ − 178.04◦ A
[a] W1 = |Vac| |IaA | cos(θac − θaA )
= 480(210) cos(60◦ − 20.79◦ ) = 78,103.2 W
[b] W2 = |Vbc| |IbB | cos(θbc − θbB )
= 480(178.68) cos(120◦ + 178.04◦ ) = 40,317.7 W
[c] W1 + W2 = 118,421 W
PAB = (192)2 (2.4) = 88,473.6 W
PBC = (48)2 (8) = 18,432 W
PCA = (24)2 (20) = 11,520 W
PAB + PBC + PCA = 118,425.7
therefore W1 + W2 ≈ Ptotal
(round-off differences)
1
P 11.48 [a] Z = Z∆ = 4.48 + j15.36 = 16/73.74◦ Ω
3
IaA
600/0◦
=
= 37.5/ − 73.74◦ A
◦
/
16 73.74
IbB = 37.5/ − 193.74◦ A
√
VAC = 600 3/ − 30◦ V
√
VBC = 600 3/ − 90◦ V
√
W1 = (600 3)(37.5) cos(−30 + 73.74◦ ) = 28,156.15 W
√
W2 = (600 3)(37.5) cos(−90 + 193.74◦ ) = −9256.15 W
[b] W1 + W2 = 18,900 W
PT = 3(37.5)2 (13.44/3) = 18,900 W
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11–32
CHAPTER 11. Balanced Three-Phase Circuits
[c]
√
3(W1 − W2 ) = 64,800 VAR
QT = 3(37.5)2 (46.08/3) = 64,800 VAR
P 11.49 [a] Zφ = 100 − j75 = 125/ − 36.87◦ Ω
Sφ =
[b]
(13,200)2
= 1,115,136 + j836,352 VA
125/36.87◦
13,200 ◦ ∗
√ /30 IaA = Sφ
3
so IaA = 182.9/66.87◦
Wm1 = (13,200)(182.9) cos(0 − 66.87◦ ) = 948,401.92 W
Wm2 = (13,200)(182.9) cos(−60◦ + 53.13◦ ) = 2,397,006.08 W
Check:
PT = 3(1,115,136) W = Wm1 + Wm2 .
P 11.50 [a] Negative phase sequence:
√
VAB = 240 3/ − 30◦ V
√
VBC = 240 3/90◦ V
√
VCA = 240 3/ − 150◦ V
√
240 3/ − 30◦
IAB =
= 20.78/ − 60◦ A
20/30◦
√
240 3/90◦
IBC =
= 6.93/90◦ A
◦
/
60 0
√
240 3/ − 150◦
ICA =
= 10.39/ − 120◦ A
◦
/
40 − 30
IaA = IAB + IAC = 18/ − 30◦ A
IcC = ICB + ICA = ICA + IBC = 16.75/ − 108.06◦
√
Wm1 = 240 3(18) cos(−30 + 30◦ ) = 7482.46 W
√
Wm2 = 240 3(16.75) cos(−90 + 108.07◦ ) = 6621.23 W
[b] Wm1 + Wm2 = 14,103.69 W
√
PA = (12 3)2 (20 cos 30◦ ) = 7482.46 W
√
PB = (4 3)2(60) = 2880 W
√
PC = (6 3)2 [40 cos(−30◦ )] = 3741.23 W
PA + PB + PC = 14,103.69 = Wm1 + Wm2
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Problems
P 11.51 [a] I∗aA =
11–33
144(0.96 − j0.28)103
= 20/ − 16.26◦ A
7200
VBN = 7200/ − 120◦ V;
VBC = VBN − VCN
VCN = 7200/120◦ V
√
= 7200 3/ − 90◦ V
IbB = 20/ − 103.74◦ A
√
Wm1 = (7200 3)(20) cos(−90◦ + 103.74◦ ) = 242,278.14 W
[b] Current coil in line aA, measure IaA .
Voltage coil across AC, measure VAC .
[c] IaA = 20/16.76◦ A
√
VCA = VAN − VCN = 7200 3/ − 30◦ V
√
Wm2 = (7200 3)(20) cos(−30◦ − 16.26◦ ) = 172,441.86 W
[d] Wm1 + Wm2 = 414.72kW
PT = 432,000(0.96) = 414.72 kW = Wm1 + Wm2
P 11.52 [a]
[b]
[c]
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11–34
CHAPTER 11. Balanced Three-Phase Circuits
[d]
P 11.53 [a] Q =
|V|2
XC
.·. |XC | =
(13,800)2
= 158.70 Ω
1.2 × 106
1
1
= 158.70;
C=
= 16.71 µF
ωC
2π(60)(158.70)
√
(13,800/ 3)2
1
[b] |XC | =
= (158.70)
6
1.2 × 10
3
. ·.
.·. C = 3(16.71) = 50.14 µF
P 11.54 [a] The capacitor from Appendix H whose value is closest to 16.71 µF is 22 µF.
|XC | =
Q=
[b] I∗aA =
1
1
=
= 120.57 Ω
ωC
2π(60)(22 × 10−6 )
|V |2
(13,800)2
=
= 1,579,497 VAR/φ
XC
120.57
1,200,000 − j379,497
√
= 50.2 − j15.9 A
13,800/ 3
13,800 ◦
√ /0 + (0.6 + j4.8)(50.2 + j15.9) = 7897.8/1.76◦
3
√
|Vab | = 3(7897.8) = 13,679.4 V
Van =
This voltage falls within the allowable range of 13 kV to 14.6 kV.
P 11.55 [a] The capacitor from Appendix H whose value is closest to 50.14 µF is 47 µF.
|XC | =
Q=
1
1
=
= 56.4 Ω
ωC
2π(60)(47 × 10−6 )
|V |2
(13,800)2
=
= 1,124,775.6 VAR
3XC
3(56.4)
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Problems
[b] I∗aA =
11–35
1,200,000 + j75,224
√
= 150.6 + j9.4 A
13,800/ 3
13,800 ◦
√ /0 + (0.6 + j4.8)(150.6 − j9.4) = 8134.8/5.06◦
3
√
|Vab | = 3(8134.8) = 14,089.9 V
Van =
This voltage falls within the allowable range of 13 kV to 14.6 kV.
P 11.56 If the capacitors remain connected when the substation drops its load, the
expression for the line current becomes
13,800 ∗
√ IaA = −j1.2 × 106
3
or
I∗aA = −j150.61 A
Hence
IaA = j150.61 A
Now,
Van =
13,800 ◦
√ /0 + (0.6 + j4.8)(j150.61) = 7244.49 + j90.37 = 7245.05/0.71◦ V
3
The magnitude of the line-to-line voltage at the generating plant is
|Vab| =
√
3(7245.05) = 12,548.80 V.
This is a problem because the voltage is below the acceptable minimum of 13
kV. Thus when the load at the substation drops off, the capacitors must be
switched off.
P 11.57 Before the capacitors are added the total line loss is
PL = 3|150.61 + j150.61|2 (0.6) = 81.66 kW
After the capacitors are added the total line loss is
PL = 3|150.61|2 (0.6) = 40.83 kW
Note that adding the capacitors to control the voltage level also reduces the
amount of power loss in the lines, which in this example is cut in half.
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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11–36
CHAPTER 11. Balanced Three-Phase Circuits
P 11.58 [a]
13,800 ∗
√ IaA = 80 × 103 + j200 × 103 − j1200 × 103
3
√
√
80
3
−
j1000
3
I∗aA =
= 10.04 − j125.51 A
13.8
.·. IaA = 10.04 + j125.51 A
13,800 ◦
√ /0 + (0.6 + j4.8)(10.04 + j125.51)
3
= 7371.01 + j123.50 = 7372.04/0.96◦ V
Van =
.·. |Vab| =
√
3(7372.04) = 12,768.75 V
[b] Yes, the magnitude of the line-to-line voltage at the power plant is less
than the allowable minimum of 13 kV.
P 11.59 [a]
13,800 ∗
√ IaA = (80 + j200) × 103
3
√
√
80 3 + j200 3
∗
IaA =
= 10.04 + j25.1 A
13.8
.·. IaA = 10.04 − j25.1 A
13,800 ◦
√ /0 + (0.6 + j4.8)(10.04 − j25.1)
3
= 8093.95 + j33.13 = 8094.02/0.23◦ V
Van =
.·. |Vab| =
[b] Yes:
√
3(8094.02) = 14,019.25 V
13 kV < 14,019.25 < 14.6 kV
[c] Ploss = 3|10.04 + j125.51|2 (0.6) = 28.54 kW
[d] Ploss = 3|10.04 + j25.1|2 (0.6) = 1.32 kW
[e] Yes, the voltage at the generating plant is at an acceptable level and the
line loss is greatly reduced.
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12
Introduction to the Laplace
Transform
Assessment Problems
eβt + e−βt
2
Therefore,
AP 12.1 [a] cosh βt =
1
L{cosh βt} =
2
Z
∞
0−
[e(s−β)t + e−(s−β)t]dt
"
1 e−(s−β)t
=
2 −(s − β)
1
=
2
∞
0−
e−(s+β)t
+
−(s + β)
1
1
+
s−β s+β
!
=
s2
∞
0−
#
s
− β2
eβt − e−βt
2
Therefore,
[b] sinh βt =
L{sinh βt} =
1
2
Z
∞
0−
h
i
e−(s−β)t − e−(s+β)t dt
"
1 e−(s−β)t
=
2 −(s − β)
=
1
2
#∞
0−
"
1 e−(s+β)t
−
2 −(s + β)
1
1
−
s−β s+β
!
=
(s2
#∞
0−
β
− β 2)
AP 12.2 [a] Let f(t) = te−at :
F (s) = L{te−at } =
Now,
1
(s + a)2
L{tf(t)} = −
dF (s)
ds
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12–1 system, or transmission in any form or by any means, electronic,
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12–2
CHAPTER 12. Introduction to the Laplace Transform
So,
L{t · te
−at
"
#
d
1
2
}=−
=
2
ds (s + a)
(s + a)3
[b] Let f(t) = e−at sinh βt, then
L{f(t)} = F (s) =
(
df(t)
L
dt
)
β
(s + a)2 − β 2
= sF (s) − f(0− ) =
s(β)
βs
−
0
=
(s + a)2 − β 2
(s + a)2 − β 2
[c] Let f(t) = cos ωt. Then
F (s) =
s
2
(s + ω 2 )
and
dF (s)
−(s2 − ω 2 )
= 2
ds
(s + ω 2)2
Therefore L{t cos ωt} = −
AP 12.3
F (s) =
dF (s)
s2 − ω 2
= 2
ds
(s + ω 2 )2
6s2 + 26s + 26
K1
K2
K3
=
+
+
(s + 1)(s + 2)(s + 3)
s+1 s+2 s+3
K1 =
6 − 26 + 26
= 3;
(1)(2)
K3 =
54 − 78 + 26
=1
(−2)(−1)
K2 =
24 − 52 + 26
=2
(−1)(1)
Therefore f(t) = [3e−t + 2e−2t + e−3t] u(t)
AP 12.4
7s2 + 63s + 134
K1
K2
K3
F (s) =
=
+
+
(s + 3)(s + 4)(s + 5)
s+3 s+4 s+5
K1 =
63 − 189 − 134
= 4;
1(2)
K3 =
175 − 315 + 134
= −3
(−2)(−1)
K2 =
112 − 252 + 134
=6
(−1)(1)
f(t) = [4e−3t + 6e−4t − 3e−5t ]u(t)
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Problems
AP 12.5
F (s) =
10(s2 + 119)
(s + 5)(s2 + 10s + 169)
s1,2 = −5 ±
F (s) =
12–3
√
25 − 169 = −5 ± j12
K1
K2
K2∗
+
+
s + 5 s + 5 − j12 s + 5 + j12
K1 =
10(25 + 119)
= 10
25 − 50 + 169
K2 =
10[(−5 + j12)2 + 119]
= j4.17 = 4.17/90◦
(j12)(j24)
Therefore
f(t) = [10e−5t + 8.33e−5t cos(12t + 90◦ )] u(t)
= [10e−5t − 8.33e−5t sin 12t] u(t)
AP 12.6
F (s) =
K0 =
4s2 + 7s + 1
K0
K1
K2
=
+
+
2
2
s(s + 1)
s
(s + 1)
s+1
1
= 1;
(1)2
K1 =
d 4s2 + 7s + 1
K2 =
ds
s
"
=
#
4−7+1
=2
−1
=
s=−1
s(8s + 7) − (4s2 + 7s + 1)
s2
s=−1
1+2
=3
1
Therefore f(t) = [1 + 2te−t + 3e−t ] u(t)
AP 12.7
F (s) =
=
(s2
K1
K2
K1∗
+
+
(s + 2 − j1)2 (s + 2 − j1) (s + 2 + j1)2
+
K1 =
40
40
=
2
+ 4s + 5)
(s + 2 − j1)2 (s + 2 + j1)2
K2∗
(s + 2 + j1)
40
= −10 = 10/180◦
(j2)2
and
K1∗ = −10
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12–4
CHAPTER 12. Introduction to the Laplace Transform
"
d
40
K2 =
ds (s + 2 + j1)2
#
=
s=−2+j1
−80(j2)
= −j10 = 10/ − 90◦
(j2)4
K2∗ = j10
Therefore
f(t) = [20te−2t cos(t + 180◦ ) + 20e−2t cos(t − 90◦ )] u(t)
= 20e−2t [sin t − t cos t] u(t)
AP 12.8
F (s) =
5s2 + 29s + 32
5s2 + 29s + 32
s+8
=
=5−
2
(s + 2)(s + 4)
s + 6s + 8
(s + 2)(s + 4)
s+8
K1
K2
=
+
(s + 2)(s + 4)
s+2 s+4
K1 =
−2 + 8
= 3;
2
K2 =
−4 + 8
= −2
−2
Therefore,
F (s) = 5 −
3
2
+
s+2 s+4
f(t) = 5δ(t) + [−3e−2t + 2e−4t ]u(t)
AP 12.9
F (s) =
2s3 + 8s2 + 2s − 4
4(s + 1)
4
= 2s − 2 +
= 2s − 2 +
2
s + 5s + 4
(s + 1)(s + 4)
s+4
f(t) = 2
dδ(t)
− 2δ(t) + 4e−4t u(t)
dt
AP 12.10
7s3 [1 + (9/s) + (134/(7s2 ))]
lim sF (s) = lim 3
=7
s→∞
s→∞ s [1 + (3/s)][1 + (4/s)][1 + (5/s)]
"
#
.·. f(0+ ) = 7
7s3 + 63s2 + 134s
lim sF (s) = lim
=0
s→0
s→0 (s + 3)(s + 4)(s + 5)
"
#
.·. f(∞) = 0
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Problems
12–5
s3 [4 + (7/s) + (1/s)2 ]
lim sF (s) = lim
=4
s→∞
s→∞
s3 [1 + (1/s)]2
"
#
.·. f(0+ ) = 4
4s2 + 7s + 1
lim sF (s) = lim
=1
s→0
s→0
(s + 1)2
"
#
.·. f(∞) = 1
"
#
40s
lim sF (s) = lim 4
=0
s→∞
s→∞ s [1 + (4/s) + (5/s2 )]2
.·. f(0+ ) = 0
"
#
40s
lim sF (s) = lim
=0
s→0
s→0 (s2 + 4s + 5)2
.·. f(∞) = 0
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12–6
CHAPTER 12. Introduction to the Laplace Transform
Problems
P 12.1
P 12.2
[a] (10 + t)[u(t + 10) − u(t)] + (10 − t)[u(t) − u(t − 10)]
= (t + 10)u(t + 10) − 2tu(t) + (t − 10)u(t − 10)
[b] (−24 − 8t)[u(t + 3) − u(t + 2)] − 8[u(t + 2) − u(t + 1)] + 8t[u(t + 1) − u(t − 1)]
+8[u(t − 1) − u(t − 2)] + (24 − 8t)[u(t − 2) − u(t − 3)]
= −8(t + 3)u(t + 3) + 8(t + 2)u(t + 2) + 8(t + 1)u(t + 1) − 8(t − 1)u(t − 1)
−8(t − 2)u(t − 2) + 8(t − 3)u(t − 3)
P 12.3
[a] f(t) = 5t[u(t) − u(t − 2)] + 10[u(t − 2) − u(t − 6)]+
(−5t + 40)[u(t − 6) − u(t − 8)]
[b] f(t) = 10 sin πt[u(t) − u(t − 2)]
[c] f(t) = 4t[u(t) − u(t − 5)]
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Problems
P 12.4
12–7
[a]
[b] f(t) = −20t[u(t) − u(t − 1)] − 20[u(t − 1) − u(t − 2)]
+20 cos( π2 t)[u(t − 2) − u(t − 4)]
+(100 − 20t)[u(t − 4) − u(t − 5)]
P 12.5
As ε → 0 the amplitude → ∞; the duration → 0; and the area is independent
of ε, i.e.,
A=
Z
ε 1
dt = 1
π ε2 + t 2
∞
−∞
P 12.6
F (s) =
Z
−ε/2
−ε
Z ε/2
Z ε
4 −st
−4 −st
4 −st
e dt +
e dt +
e dt
3
3
ε
ε
−ε/2
ε/2 ε3
Therefore F (s) =
4 sε
[e − 2esε/2 + 2e−sε/2 − e−sε ]
sε3
L{δ 00(t)} = lim F (s)
ε→0
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12–8
CHAPTER 12. Introduction to the Laplace Transform
After applying L’Hopital’s rule three times, we have
2s
s
s
2s 3s
lim
sesε − esε/2 − e−sε/2 + se−sε =
ε→0 3
4
4
3 2
Therefore L{δ 00(t)} = s2
P 12.7
P 12.8
1
1
1
[a] A =
bh =
(2ε)
=1
2
2
ε
[b] 0;
[c] ∞
1 −st
esε − e−sε
e dt =
2ε
2εs
ε
Z
F (s) =
−ε
1
sesε + se−sε
1 2s
F (s) =
lim
=
·
=1
2s ε→0
1
2s 1
"
P 12.9
[a] I =
3
Z
#
3
(t + 2)δ(t) dt +
−1
Z
3
8(t3 + 2)δ(t − 1) dt
−1
= (03 + 2) + 8(13 + 2) = 2 + 8(3) = 26
[b] I =
2
Z
2
t δ(t) dt +
−2
Z
2
2
t δ(t + 1.5) dt +
−2
Z
2
−2
δ(t − 3) dt
= 02 + (−1.5)2 + 0 = 2.25
1
P 12.10 f(t) =
2π
Z
dn f(t)
P 12.11 L
dtn
Therefore
)
(
∞
−∞
(4 + jω)
1
· πδ(ω) · ejtω dω =
(9 + jω)
2π
!
4 + j0 −jt0
2
πe
=
9 + j0
9
= sn F (s) − sn−1 f(0− ) − sn−2 f 0 (0− ) − · · · ,
L{δ n (t)} = sn (1) − sn−1 δ(0− ) − sn−2 δ 0(0− ) − sn−3 δ 00(0− ) − · · · = sn
P 12.12 [a] Let dv = δ 0(t − a) dt,
v = δ(t − a)
du = f 0 (t) dt
u = f(t),
Therefore
Z
∞
∞
f(t)δ (t − a) dt = f(t)δ(t − a)
0
−∞
−∞
−
Z
∞
−∞
δ(t − a)f 0 (t) dt
= 0 − f 0 (a)
0
[b] L{δ (t)} =
Z
∞
0−
0
δ (t)e
−st
"
d(e−st )
dt = −
dt
#
t=0
h
= − −se−st
i
t=0
=s
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Problems
P 12.13 L{e−at f(t)} =
∞
Z
0−
[e−atf(t)]e−st dt =
(
Z
12–9
∞
0−
f(t)e−(s+a)t dt = F (s + a)
)
d sin ωt
sω
sω
P 12.14 [a] L
u(t) = 2
−
sin(0)
=
dt
s + ω2
s2 + ω 2
d cos ωt
s2
s2
−ω 2
u(t) = 2
−
cos(0)
=
−
1
=
[b] L
dt
s + ω2
s2 + ω 2
s2 + ω 2
(
)
d3 (t2)
2
[c] L
u(t) = s3 3 − s2(0) − s(0) − 2(0) = 2
3
dt
s
d sin ωt
ωs
[d]
= (cos ωt) · ω,
L{ω cos ωt} = 2
dt
s + ω2
(
)
d cos ωt
= −ω sin ωt
dt
L{−ω sin ωt} = −
ω2
s2 + ω 2
d3 (t2 u(t))
= 2δ(t);
dt3
P 12.15 [a]
t
Z
0−
x dx =
t2
L
2
(
)
L{2δ(t)} = 2
t2
2
1
2
=
Z
∞
0−
t2e−st dt
∞
"
1 e−st 2 2
=
(s t + 2st + 2)
2 −s3
0−
#
1
1
(2) = 3
3
2s
s
Z t
1
. ·. L
x dx = 3
s
0−
=
L{t}
1/s2
1
=
= 3
s
s
s
0−
Z t
1
. ·. L
x dx = 3
CHECKS
s
0−
[b] L
Z
t
x dx =
P 12.16 L{f(at)} =
Z
Let u = at,
∞
0−
f(at)e−st dt
u = 0−
du = a dt,
when t = 0−
and u = ∞ when t = ∞
Therefore L{f(at)} =
Z
∞
0−
f(u)e−(u/a)s
du
1
= F (s/a)
a
a
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12–10
CHAPTER 12. Introduction to the Laplace Transform
P 12.17 [a] L{t} =
1
;
s2
1
(s + a)2
therefore L{te−at } =
ejωt − e−jωt
j2
Therefore
[b] sin ωt =
1
j2
L{sin ωt} =
=
s2
!
ω
+ ω2
1
1
−
s − jω s + jω
!
=
1
j2
!
2jω
s2 + ω 2
[c] sin(ωt + θ) = (sin ωt cos θ + cos ωt sin θ)
Therefore
L{sin(ωt + θ)} = cos θL{sin ωt} + sin θL{cos ωt}
ω cos θ + s sin θ
=
s2 + ω 2
e−st
(−st − 1)
s2
0
[e] f(t) = cosh t cosh θ + sinh t sinh θ
From Assessment Problem 12.1(a)
s
L{cosh t} = 2
s −1
[d] L{t} =
Z
∞
∞
te−st dt =
0
=0−
1
1
(0 − 1) = 2
2
s
s
From Assessment Problem 12.1(b)
L{sinh t} =
s2
1
−1
.·. L{cosh(t + θ)} = cosh θ
=
P 12.18 [a] L{f 0 (t)} =
Z
#
s
1
+ sinh θ 2
2
(s − 1)
s −1
sinh θ + s[cosh θ]
(s2 − 1)
e−st
dt +
−ε 2ε
ε
"
Z
∞
ε
− ae−a(t−ε) e−st dt
1 sε
a
=
(e − e−sε ) −
e−sε = F (s)
2sε
s+a
a
s
lim F (s) = 1 −
=
ε→0
s+a
s+a
[b] L{e−at} =
1
s+a
Therefore L{f 0 (t)} = sF (s) − f(0− ) =
s
s
−0 =
s+a
s+a
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Problems
P 12.19 [a] L{40e−8(t−3) u(t − 3)} =
12–11
40e−3s
(s + 8)
[b] First rewrite f(t) as
f(t) = (5t − 10)u(t − 2) + (40 − 10t)u(t − 4)
+(10t − 80)u(t − 8) + (50 − 5t)u(t − 10)
= 5(t − 2)u(t − 2) − 10(t − 4)u(t − 4)
+10(t − 8)u(t − 8) − 5(t − 10)u(t − 10)
.·. F (s) =
P 12.20 [a] L{te−at } =
5[e−2s − 2e−4s + 2e−8s − e−10s ]
s2
Z
∞
0−
te−(s+a)t dt
e−(s+a)t
=
(s + a)2
− (s + a)t − 1
= 0+
1
(s + a)2
.·. L{te−at} =
1
(s + a)2
(
∞
0−
)
d −at
s
[b] L
(te )u(t) =
−0
dt
(s + a)2
(
)
d −at
s
L
(te )u(t) =
dt
(s + a)2
[c]
d −at
(te ) = −ate−at + e−at
dt
−a
1
−a
s+a
L{−ate−at + e−at } =
+
=
+
2
2
(s + a)
(s + a)
(s + a)
(s + a)2
. ·. L
(
)
d −at
s
(te ) =
dt
(s + a)2
CHECKS
P 12.21 [a] f(t) = 5t[u(t) − u(t − 2)]
+(20 − 5t)[u(t − 2) − u(t − 6)]
+(5t − 40)[u(t − 6) − u(t − 8)]
= 5tu(t) − 10(t − 2)u(t − 2)
+10(t − 6)u(t − 6) − 5(t − 8)u(t − 8)
.·. F (s) =
5[1 − 2e−2s + 2e−6s − e−8s ]
s2
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12–12
CHAPTER 12. Introduction to the Laplace Transform
[b]
f 0 (t) = 5[u(t) − u(t − 2)] − 5[u(t − 2) − u(t − 6)]
+5[u(t − 6) − u(t − 8)]
= 5u(t) − 10u(t − 2) + 10u(t − 6) − 5u(t − 8)
L{f 0 (t)} =
5[1 − 2e−2s + 2e−6s − e−8s ]
s
[c]
f 00 (t) = 5δ(t) − 10δ(t − 2) + 10δ(t − 6) − 5δ(t − 8)
L{f 00 (t)} = 5[1 − 2e−2s + 2e−6s − e−8s ]
P 12.22 [a] L
[b]
Z
Z
t
0−
t
0−
e−ax dx =
(
e−ax dx =
1 e−at
L
−
a
a
F (s)
1
=
s
s(s + a)
1 e−at
−
a
a
)
=
1 1
1
1
−
=
a s s+a
s(s + a)
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Problems
dF (s)
d
P 12.23 [a]
=
ds
ds
∞
Z
f(t)e
0−
−st
dt = −
Therefore L{tf(t)} = −
[b]
d2 F (s)
=
ds2
Z
Therefore
dn F (s)
= (−1)n
dsn
d4
ds4
d
L{t sin βt} = (−1)
ds
Z
∞
1
s2
Z
∞
−t3f(t)e−st dt
0−
tn f(t)e−st dt = (−1)n L{tn f(t)}
0−
β
2
s + β2
1
tf(t)e−st dt
0−
d3 F (s)
=
ds3
t2f(t)e−st dt;
[c] L{t5 } = L{t4 t} = (−1)4
∞
dF (s)
ds
∞
0−
Z
12–13
=
!
120
s6
=
(s2
2βs
+ β 2 )2
L{te−t cosh t}:
From Assessment Problem 12.1(a),
s
F (s) = L{cosh t} = 2
s −1
dF
(s2 − 1)1 − s(2s)
s2 + 1
=
=
−
ds
(s2 − 1)2
(s2 − 1)2
Therefore
−
dF
s2 + 1
= 2
ds
(s − 1)2
Thus
L{t cosh t} =
s2 + 1
(s2 − 1)2
(s + 1)2 + 1
s2 + 2s + 2
L{e t cosh t} =
= 2
[(s + 1)2 − 1]2
s (s + 2)2
−t
P 12.24 [a]
Z
∞
F (u)du =
s
=
Z
Z
∞ Z ∞
s
0−
∞
0−
f(t)
f(t)e−ut dt du =
Z
∞
e
−ut
du dt =
s
"
#
Z
Z
∞
0−
"
(
∞
(
t sin βt
therefore L
t
)
=
Z
∞
s
"
∞
f(t)e−ut du dt
s
e−tu
f(t)
−t
0−
∞
−e−st
f(t)
=
f(t)
dt = L
−
−t
t
0
2βs
[b] L{t sin βt} = 2
(s + β 2)2
Z
Z
∞
s
#
dt
)
#
2βu
du
2
(u + β 2)2
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12–14
CHAPTER 12. Introduction to the Laplace Transform
Let ω = u2 + β 2, then ω = s2 + β 2 when u = s, and ω = ∞ when u = ∞;
also dω = 2u du, thus
(
t sin βt
L
t
)
=β
∞
Z
s2 +β 2
P 12.25 [a] f1(t) = e−at sin ωt;
"
F1(s) =
F (s) = sF1(s) − f1 (0− ) =
[b] f1(t) = e−at cos ωt;
F (s) =
[c]
#
dω
−1
=
β
ω2
ω
∞
=
s2 +β 2
β
s2 + β 2
ω
(s + a)2 + ω 2
sω
−0
(s + a)2 + ω 2
F1(s) =
s+a
(s + a)2 + ω 2
F1(s)
s+a
=
s
s[(s + a)2 + ω 2 ]
d −at
[e sin ωt] = ωe−at cos ωt − ae−at sin ωt
dt
Therefore F (s) =
Z
t
0−
e−ax cos ωx dx =
ω(s + a) − ωa
ωs
=
2
2
(s + a) + ω
(s + a)2 + ω 2
−ae−at cos ωt + ωe−at sin ωt + a
a2 + ω 2
Therefore
1
−a(s + a)
ω2
a
F (s) = 2
+
+
2
2
2
2
2
a + ω (s + a) + ω
(s + a) + ω
s
"
=
P 12.26 Ig (s) =
#
s+a
s[(s + a)2 + ω 2 ]
1.2s
;
s2 + 1
1
= 1.6;
RC
1
= 1;
LC
1
= 1.6
C
V (s)
1 V (s)
+
+ C[sV (s) − v(0− )] = Ig (s)
R
L s
1
1
V (s)
+
+ sC = Ig (s)
R Ls
V (s) =
=
1
R
1
sIg (s)
Ig (s)
LsIg (s)
C
=
= 2 R
1
1
2
+ Ls + sC
RLs + 1 + s LC
s + C s + LC
(1.6)(1.2)s2
1.92s2
=
(s2 + 1.6s + 1)(s2 + 1)
(s2 + 1.6s + 1)(s2 + 1)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
1
P 12.27 [a]
L
Z
t
0
v1 dτ +
12–15
v1 − v2
= ig u(t)
R
and
dv2 v2 v1
C
+
−
=0
dt
R
R
V1
V1 − V2
[b]
+
= Ig
sL
R
V2 − V1
+ sCV2 = 0
R
or
(R + sL)V1 (s) − sLV2 (s) = RLsIg (s)
−V1 (s) + (RCs + 1)V2 (s) = 0
Solving,
V2 (s) =
C[s2
vo − Vdc
1
P 12.28 [a]
+
R
L
. ·.
[b] Vo +
sIg (s)
+ (R/L)s + (1/LC)]
Z
R
vo +
L
t
vo dx + C
0
Z
dvo
=0
dt
t
0
vo dx + RC
dvo
= Vdc
dt
R Vo
Vdc
+ RCsVo =
L s
s
. ·.
sLVo + RVo + RCLs2 Vo = LVdc
. ·.
Vo (s) =
[c] io =
s2
(1/RC)Vdc
+ (1/RC)s + (1/LC)
1Zt
vo dx
L 0
Vo
Vdc /RLC
=
sL
s[s2 + (1/RC)s + (1/LC)]
Io (s) =
P 12.29 [a] For t ≥ 0+ :
Rio + L
io = C
dio
+ vo = 0
dt
dvo
dt
.·. RC
dio
d2 vo
=C 2
dt
dt
dvo
d2 vo
+ LC 2 + vo = 0
dt
dt
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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12–16
CHAPTER 12. Introduction to the Laplace Transform
or
d2 vo R dvo
1
+
+
vo = 0
2
dt
L dt
LC
R
1
[b] s2Vo (s) − sVdc − 0 + [sVo (s) − Vdc ] +
Vo (s) = 0
L
LC
Vo (s) s2 +
R
1
s+
= Vdc (s + R/L)
L
LC
Vo (s) =
Vdc [s + (R/L)]
+ (R/L)s + (1/LC)]
[s2
1 t
vo
dvo
vo dx +
+C
L 0
R
dt
Idc
Vo (s) Vo (s)
[b]
=
+
+ sCVo (s)
s
sL
R
Z
P 12.30 [a] Idc =
.·. Vo (s) =
[c] io = C
s2
Idc/C
+ (1/RC)s + (1/LC)
dvo
dt
.·. Io(s) = sCVo (s) =
s2
sIdc
+ (1/RC)s + (1/LC)
P 12.31 [a] For t ≥ 0+ :
vo
dvo
+C
+ io = 0
R
dt
vo = L
dio
;
dt
dvo
d2 io
=L 2
dt
dt
. ·.
L dio
d2 io
+ LC 2 + io = 0
R dt
dt
or
d2 io
1 dio
1
+
+
io = 0
dt2
RC dt
LC
[b] s2Io (s) − sIdc − 0 +
1
1
[sIo(s) − Idc] +
Io (s) = 0
RC
LC
Io (s) s2 +
1
1
s+
= Idc(s + 1/RC)
RC
LC
Io (s) =
Idc[s + (1/RC)]
+ (1/RC)s + (1/LC)]
[s2
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
P 12.32 [a] 300 = 60i1 + 25
0=5
12–17
di1
d
d
di1
+ 10 (i2 − i1) + 5 (i1 − i2) − 10
dt
dt
dt
dt
d
di1
(i2 − i1) + 10
+ 40i2
dt
dt
Simplifying the above equations gives:
300 = 60i1 + 10
0 = 40i2 + 5
[b]
di1
di2
+5
dt
dt
di1
di2
+5
dt
dt
300
= (10s + 60)I1 (s) + 5sI2 (s)
s
0 = 5sI1 (s) + (5s + 40)I2 (s)
[c] Solving the equations in (b),
I1(s) =
60(s + 8)
s(s + 4)(s + 24)
I2(s) =
−60
(s + 4)(s + 24)
P 12.33 From Problem 12.26:
V (s) =
1.92s2
(s2 + 1.6s + 1)(s2 + 1)
s2 + 1.6s + 1 = (s + 0.8 + j0.6)(s + 0.8 − j0.6);
s2 + 100 = (s − j1)(s + j1)
Therefore
V (s) =
=
K1 =
K2 =
1.92s2
(s + 0.8 + j0.6)(s + 0.8 − j0.6)(s − j1)(s + j1)
K1
K1∗
K2
K2∗
+
+
+
s + 0.8 − j0.6 s + 0.8 + j0.6 s − j1 s + j1
1.92s2
(s + 0.8 + j0.6)(s2 + 1)
1.92s2
(s + j1)(s2 + 1.6s + 1)
s=−0.8+j0.6
= 1/ − 126.87◦
= 0.6/0◦
s=−j1
Therefore
v(t) = [2e−0.8t cos(0.6t − 126.87◦ ) + 1.2 cos(t)]u(t) V
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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12–18
P 12.34
CHAPTER 12. Introduction to the Laplace Transform
1
= 2 × 106 ;
C
V2 (s) =
s2
1
= 4 × 106 ;
LC
R
= 5000;
L
Ig =
0.015
s
30,000
+ 5000s + 4 × 106
s1 = −1000;
s2 = −4000
V2 (s) =
30,000
(s + 1000)(s + 4000)
=
10
10
−
s + 1000 s + 4000
v2(t) = [10e−1000t − 10e−4000t]u(t) V
P 12.35 [a]
1
1
=
= 50 × 106
−3
LC
(200 × 10 )(100 × 10−9 )
1
1
=
= 2000
RC
(5000)(100 × 10−9 )
Vo (s) =
s2
70,000
+ 2000s + 50 × 106
s1,2 = −1000 ± j7000 rad/s
Vo (s) =
70,000
(s + 1000 − j7000)(s + 1000 + j7000)
=
K1
K1∗
+
s + 1000 − j7000 s + 1000 + j7000
K1 =
70,000
= 5/ − 90◦
j14,000
vo (t) = 10e−1000t cos(7000t − 90◦ )]u(t) V
= [10e−1000t sin 7000t]u(t) V
[b] Io (s) =
=
K1 =
K2 =
35(10,000)
s(s + 1000 − j7000)(s + 1000 + j7000)
K1
K2
K2∗
+
+
s
s + 1000 − j7000 s + 1000 + j7000
35(10,000)
= 7 mA
50 × 106
35(10,000)
= 3.54/171.87◦ mA
(−1000 + j7000)(j14,000)
io (t) = [7 + 7.07e−1000t cos(7000t + 171.87◦ )]u(t) mA
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
P 12.36
R
= 5000;
L
Vo (s) =
s2
1
= 4 × 106
LC
48(s + 5000)
+ 5000s + 4 × 106
s1,2 = −2500 ±
√
6.25 × 106 − 4 × 106
s1 = −1000 rad/s;
Vo (s) =
K1 =
s2 = −4000 rad/s
48(s + 5000)
K1
K2
=
+
(s + 1000)(s + 4000)
s + 1000 s + 4000
48(4000)
= 64 V;
3000
Vo (s) =
12–19
K2 =
48(1000)
= −16 V
−3000
64
16
−
s + 1000 s + 4000
vo (t) = [64e−1000t − 16e−4000t ]u(t) V
P 12.37 [a]
1
1
=
= 500
3
RC
(1 × 10 )(2 × 10−6 )
1
1
=
= 40,000
LC
(12.5)(2 × 10−6 )
Vo (s) =
500,000Idc
s + 500s + 40,000
=
500,000Idc
(s + 100)(s + 400)
=
15,000
(s + 100)(s + 400)
=
K1
K2
+
s + 100 s + 400
K1 =
15,000
= 50;
300
Vo (s) =
K2 =
50
50
−
s + 100 s + 400
15,000
= −50
−300
vo (t) = [50e−100t − 50e−400t ]u(t) V
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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12–20
CHAPTER 12. Introduction to the Laplace Transform
[b] Io (s) =
0.03s
(s + 100)(s + 400)
=
K1
K2
+
s + 100 s + 400
K1 =
0.03(−100)
= −0.01
300
K2 =
0.03(−400)
= 0.04
−300
Io (s) =
−0.01
0.04
+
s + 100 s + 400
io (t) = (40e−400t − 10e−100t)u(t) mA
[c] io (0) = 40 − 10 = 30 mA
Yes. The initial inductor current is zero by hypothesis, the initial resistor
current is zero because the initial capacitor voltage is zero by hypothesis.
Thus at t = 0 the source current appears in the capacitor.
P 12.38
1
= 8000;
RC
Io (s) =
s2
1
= 16 × 106
LC
0.005(s + 8000)
+ 8000s + 16 × 106
s1,2 = −4000
Io (s) =
0.005(s + 8000)
K1
K2
=
+
2
2
(s + 4000)
(s + 4000)
s + 4000
K1 = 0.005(s + 8000)
= 20
s=−4000
K2 =
d
[0.005(s + 8000)]s=−4000 = 0.005
ds
Io (s) =
20
0.005
+
2
(s + 4000)
s + 4000
io (t) = [20te−4000t + 0.005e−4000t ]u(t) V
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
P 12.39 [a] I1(s) =
12–21
K1
K2
K3
+
+
s
s + 4 s + 24
K1 =
(60)(8)
= 5;
(4)(24)
K3 =
(60)(−16)
= −2
(−24)(−20)
I1(s) =
(60)(4)
= −3
(−4)(20)
K2 =
5
3
2
−
−
s s + 4 s + 24
i1 (t) = (5 − 3e−4t − 2e−24t)u(t) A
I2(s) =
K1 =
K1
K2
+
s + 4 s + 24
−60
= −3;
20
I2(s) =
K2 =
−3
3
+
s + 4 s + 24
−60
=3
−20
i2 (t) = (3e−24t − 3e−4t )u(t) A
[b] i1(∞) = 5 A;
i2(∞) = 0 A
[c] Yes, at t = ∞
300
= 5A
60
Since i1 is a dc current at t = ∞ there is no voltage induced in the 10 H
inductor; hence, i2 = 0. Also note that i1(0) = 0 and i2 (0) = 0. Thus our
solutions satisfy the condition of no initial energy stored in the circuit.
i1 =
P 12.40 [a] F (s) =
K1 =
K2 =
K3 =
K1
K2
K3
+
+
s+1 s+2 s+4
8s2 + 37s + 32
(s + 2)(s + 4)
s=−1
8s2 + 37s + 32
(s + 1)(s + 4)
s=−2
8s2 + 37s + 32
(s + 1)(s + 2)
s=−4
=1
=5
=2
f(t) = [e−t + 5e−2t + 2e−4t ]u(t)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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12–22
CHAPTER 12. Introduction to the Laplace Transform
[b] F (s) =
K1 =
K2 =
K3 =
K4 =
K1
K2
K3
K4
+
+
+
s
s+2 s+4 s+6
13s3 + 134s2 + 392s + 288
(s + 2)(s + 4)(s + 6)
s=0
13s3 + 134s2 + 392s + 288
s(s + 4)(s + 6)
s=−2
13s3 + 134s2 + 392s + 288
s(s + 2)(s + 6)
s=−4
13s3 + 134s2 + 392s + 288
s(s + 2)(s + 4)
s=−6
=6
=4
=2
=1
f(t) = [6 + 4e−2t + 2e−4t + e−6t]u(t)
[c] F (s) =
K1 =
K1
K2
K2∗
+
+
s + 1 s + 1 − 2j s + 1 + 2j
20s2 + 16s + 12
s2 + 2s + 5
=4
s=−1
20s2 + 16s + 12
K2 =
(s + 1)(s + 1 + 2j)
= 8 + j6 = 10/36.87◦
s=−1+2j
f(t) = [4e−t + 20e−t cos(2t + 36.87◦ )]u(t)
[d] F (s) =
K1 =
K2 =
K1
K2
K2∗
+
+
s
s+7−j s+7+j
250(s + 7)(s + 14)
s2 + 14s + 50
250(s + 7)(s + 14)
s(s + 7 + j)
= 490
s=0
s=−7+j
= 125/ − 163.74◦
f(t) = [490 + 250e−7t cos(t − 163.74◦ )]u(t)
P 12.41 [a] F (s) =
K1 =
K1 K2
K3
+
+
2
s
s
s+5
100
s+5
= 20
s=0
d 100
−100
K2 =
=
ds s + 5
(s + 5)2
K3 =
100
s2
s=0
= −4
=4
s=−5
f(t) = [20t − 4 + 4e−5t ]u(t)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
[b] F (s) =
K1 =
K2 =
12–23
K1
K2
K3
+
+
2
s
(s + 1)
s+1
50(s + 5)
(s + 1)2
s=0
50(s + 5)
s
s=−1
= 250
"
= −200
#
"
d 50(s + 5)
50 50(s + 5)
K3 =
=
−
ds
s
s
s2
#
s=−1
= −250
f(t) = [250 − 200te−t − 250e−t ]u(t)
[c] F (s) =
K1 =
K1 K2
K3
K3∗
+
+
+
s2
s
s+3−j s+3+j
100(s + 3)
s2 + 6s + 10
= 30
s=0
"
d 100(s + 3)
K2 =
ds s2 + 6s + 10
#
"
100
100(s + 3)(2s + 6)
= 2
−
s + 6s + 10
(s2 + 6s + 10)2
K3 =
100(s + 3)
+ 3 + j)
s2 (s
#
s=0
= 10 − 18 = −8
= 4 + j3 = 5/36.87◦
s=−3+j
f(t) = [30t − 8 + 10e−3t cos(t + 36.87◦ )]u(t)
[d] F (s) =
K1 =
K2 =
K1
K2
K3
K4
+
+
+
s
(s + 1)3 (s + 1)2 s + 1
5(s + 2)2
(s + 1)3
s=0
5(s + 2)2
s
s=−1
= 20
= −5
d 5(s + 2)2
10(s + 2) 5(s + 2)2
K3 =
=
−
ds
s
s
s2
"
#
"
#
s=−1
= −10 − 5 = −15
1 d 10(s + 2) 5(s + 2)2
K4 =
−
2 ds
s
s2
"
#
1 10 10(s + 2) 10(s + 2) 10(s + 2)2
=
−
−
+
2 s
s2
s2
s3
"
#
s=−1
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12–24
CHAPTER 12. Introduction to the Laplace Transform
1
= (−10 − 10 − 10 − 10) = −20
2
f(t) = [20 − 2.5t2 e−t − 15te−t − 20e−t ]u(t)
[e] F (s) =
K1 =
K2 =
K1
K2
K2∗
K3
K3∗
+
+
+
+
s
(s + 2 − j)2 (s + 2 + j)2 s + 2 − j s + 2 − j
400
(s2 + 4s + 5)2
400
s(s + 2 + j)2
= 16
s=0
= 44.72/26.57◦
s=−2+j
"
#
"
d
400
400
−800
K3 =
= 2
+
2
2
ds s(s + 2 + j)
s (s + 2 + j)
s(s + 2 + j)3
#
s=−2+j
= 12 + j16 − 20 + j40 = −8 + j56 = 56.57/98.13◦
f(t) = [16 + 89.44te−2t cos(t + 26.57◦ ) + 113.14e−2t cos(t + 98.13◦ )]u(t)
P 12.42 [a]
5
F (s) = s2 + 6s + 8
5s2 + 38s + 80
5s2 + 30s + 40
8s + 40
F (s) = 5 +
K1 =
K2 =
8s + 40
K1
K2
= 10 +
+
+ 6s + 8
s+2 s+4
s2
8s + 40
s+4
s=−2
8s + 40
s+2
s=−4
= 12
= −4
f(t) = 5δ(t) + [12e−2t − 4e−4t ]u(t)
[b]
10
F (s) = s2 + 48s + 625
10s2 + 512s + 7186
10s2 + 480s + 6250
32s + 936
F (s) = 10 +
K1 =
32s + 936
K1
K2∗
=
10
+
+
s2 + 48s + 625
s + 24 − j7 s + 24 + j7
32s + 936
s + 24 + j7
s=−24+j7
= 16 − j12 = 20/ − 36.87◦
f(t) = 10δ(t) + [40e−24t cos(7t − 36.87◦ )]u(t)
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Problems
[c]
12–25
s − 10
F (s) =
s2 + 15s + 50
s3 + 5s2 − 50s − 100
s3 + 15s2 + 50s
−10s2 − 100s − 100
−10s2 − 150s − 500
50s + 400
F (s) = s − 10 +
K1 =
K2 =
K1
K2
+
s + 5 s + 10
50s + 400
s + 10
s=−5
50s + 400
s+5
s=−10
= 30
= 20
f(t) = δ 0(t) − 10δ(t) + [30e−5t + 20e−10t ]u(t)
K1 K2
K3
K3∗
P 12.43 [a] F (s) = 2 +
+
+
s
s
s + 1 − j2 s + 1 + j2
K1 =
100(s + 1)
s2 + 2s + 5
= 20
s=0
"
#
"
d 100(s + 1)
100
100(s + 1)(2s + 2)
K2 =
= 2
−
2
ds s + 2s + 5
s + 2s + 5
(s2 + 2s + 5)2
#
s=0
= 20 − 8 = 12
K3 =
100(s + 1)
+ 1 + j2)
s2 (s
s=−1+j2
= −6 + j8 = 10/126.87◦
f(t) = [20t + 12 + 20e−t cos(2t + 126.87◦ )]u(t)
[b] F (s) =
K1 =
K2 =
K1
K2
K3
K4
+
+
+
3
2
s
(s + 5)
(s + 5)
s+5
500
(s + 5)3
s=0
500
s
= −100
=4
s=−5
d 500
−500
K3 =
=
ds s
s2
s=−5
1 d −500
1 1000
K4 =
=
2
2 ds
s
2 s3
= −20
s=−5
= −4
f(t) = [4 − 50t2 e−5t − 20te−5t − 4e−5t ]u(t)
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12–26
CHAPTER 12. Introduction to the Laplace Transform
[c] F (s) =
K1 =
K2 =
K1
K2
K3
K4
+
+
+
3
2
s
(s + 1)
(s + 1)
s+1
40(s + 2)
(s + 1)3
s=0
40(s + 2)
s
s=−1
= 80
"
= −40
#
"
d 40(s + 2)
40 40(s + 2)
K3 =
=
−
ds
s
s
s2
"
1 d 40 40(s + 2)
K4 =
−
2 ds s
s2
#
s=−1
= −40 − 40 = −80
#
"
1 −40 40 80(s + 2)
=
− 2 +
2 s2
s
s3
#
s=−1
1
= (−40 − 40 − 80) = −80
2
f(t) = [80 − 20t2 e−t − 80te−t − 80e−t ]u(t)
[d] F (s) =
K1 =
K2 =
K1
K2
K3
K4
K5
+
+
+
+
4
3
2
s
(s + 1)
(s + 1)
(s + 1)
s+1
(s + 5)2
(s + 1)4
s=0
(s + 5)2
s
s=−1
= 25
= −16
d (s + 5)2
2(s + 5) (s + 5)2
K3 =
=
−
ds
s
s
s2
"
#
= −8 − 16 = −24
"
1 d 2(s + 5) (s + 5)2
K4 =
−
2 ds
s
s2
"
#
s=−1
#
1 2 2(s + 5) 2(s + 5) 3(s + 5)2
=
−
−
+
2 s
s2
s2
s3
"
#
s=−1
1
= (−2 − 8 − 8 − 32) = −25
2
1 d 2 2(s + 5) 2(s + 5) 3(s + 5)2
K5 =
−
−
+
6 ds s
s2
s2
s3
"
#
1 −2
2
4(s + 5)
2
4(s + 5) 4(s + 5) 6(s + 5)2
=
−
+
−
+
+
−
6 s2
s2
s3
s2
s3
s3
s4
"
#
s=−1
1
= (−2 − 2 − 16 − 2 − 16 − 16 − 96) = −25
6
f(t) = [25 − (8/3)t3 e−t − 12t2 e−t − 25te−t − 25e−t ]u(t)
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Problems
P 12.44 f(t) = L
−1
(
K
K∗
+
s + α − jβ s + α + jβ
12–27
)
= Ke−αt ejβt + K ∗ e−αte−jβt
= |K|e−αt [ejθ ejβt + e−jθ e−jβt ]
= |K|e−αt [ej(βt+θ) + e−j(βt+θ)]
= 2|K|e−αt cos(βt + θ)
n
P 12.45 [a] L{t f(t)} = (−1)
Let f(t) = 1,
n
"
dn F (s)
dsn
#
1
then F (s) = ,
s
Therefore L{tn } = (−1)n
"
It follows that L{t(r−1) } =
and
L{t(r−1)e−at } =
[b] f(t) = L
Therefore
f(t) =
=
(−1)n n!
n!
= (n+1)
(n+1)
s
s
#
(r − 1)!
sr
(r − 1)!
(s + a)r
K
K
Ktr−1 e−at
L{tr−1 e−at} =
=
L
(r − 1)!
(s + a)r
(r − 1)!
(
Therefore
−1
dn F (s)
(−1)n n!
=
dsn
s(n+1)
thus
(
K
K∗
+
(s + α − jβ)r (s + α + jβ)r
)
)
Ktr−1 −(α−jβ)t K ∗tr−1 −(α+jβ)t
e
+
e
(r − 1)!
(r − 1)!
i
|K|tr−1e−αt h jθ jβt
e e + e−jθ e−jβt
(r − 1)!
"
#
2|K|tr−1 e−αt
=
cos(βt + θ)
(r − 1)!
1.92s3
P 12.46 [a] lim sV (s) = lim 4
=0
s→∞
s→∞ s [1 + (1.6/s) + (1/s2 )][1 + (1/s2 )]
"
#
Therefore v(0+ ) = 0
[b] No, V has a pair of poles on the imaginary axis.
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12–28
CHAPTER 12. Introduction to the Laplace Transform
P 12.47 sVo (s) =
sVdc /RC
+ (1/RC)s + (1/LC)
s2
lim sVo (s) = 0,
s→0
.·. vo(∞) = 0
.·. vo (0+ ) = 0
lim sVo (s) = 0,
s→∞
sIo (s) =
s2
Vdc /RC)
+ (1/RC)s + (1/LC)
Vdc/RLC
Vdc
=
,
1/LC
R
lim sIo (s) =
s→0
.·. io (0+ ) = 0
lim sIo (s) = 0,
s→∞
P 12.48 sVo (s) =
s2
(Idc/C)s
+ (1/RC)s + (1/LC)
lim sVo (s) = 0,
s→0
.·. vo(∞) = 0
.·. vo (0+ ) = 0
lim sVo (s) = 0,
s→∞
sIo (s) =
s2Idc
s2 + (1/RC)s + (1/LC)
lim sIo (s) = 0,
s→0
.·. io (∞) = 0
.·. io (0+ ) = Idc
lim sIo (s) = Idc,
s→∞
P 12.49 sIo (s) =
Idcs[s + (1/RC)]
s2 + (1/RC)s + (1/LC)
lim sIo (s) = 0,
s→0
.·. io (∞) = 0
lim sIo (s) = Idc,
s→∞
P 12.50 [a] sF (s) =
.·. io (0+ ) = Idc
8s3 + 37s2 + 32s
(s + 1)(s + 2)(s + 4)
lim sF (s) = 0,
s→0
lim sF (s) = 8,
s→∞
Vdc
.·. io (∞) =
R
.·. f(∞) = 0
.·. f(0+ ) = 8
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Problems
[b] sF (s) =
12–29
13s3 + 134s2 + 392s + 288
(s + 2)(s2 + 10s + 24)
.·. f(∞) = 6
lim sF (s) = 6;
s→0
lim sF (s) = 13,
s→∞
.·. f(0+ ) = 13
20s3 + 16s2 + 12s
[c] sF (s) =
(s + 1)(s2 + 2s + 5)
.·. f(∞) = 0
lim sF (s) = 0,
s→0
lim sF (s) = 20,
s→∞
[d] sF (s) =
.·. f(0+ ) = 20
250(s + 7)(s + 14)
(s2 + 14s + 50)
250(7)(14)
= 490,
.·. f(∞) = 490
s→0
50
lim sF (s) = 250,
.·. f(0+ ) = 250
lim sF (s) =
s→∞
100
s(s + 5)
F (s) has a second-order pole at the origin so we cannot use the final
value theorem.
P 12.51 [a] sF (s) =
lim sF (s) = 0,
s→∞
[b] sF (s) =
.·. f(0+ ) = 0
50(s + 5)
(s + 1)2
lim sF (s) = 250,
s→0
lim sF (s) = 0,
s→∞
.·. f(∞) = 250
.·. f(0+ ) = 0
100(s + 3)
s(s2 + 6s + 10)
F (s) has a second-order pole at the origin so we cannot use the final
value theorem.
[c] sF (s) =
lim sF (s) = 0,
s→∞
[d] sF (s) =
.·. f(0+ ) = 0
5(s + 2)2
(s + 1)3
lim sF (s) = 20,
.·. f(∞) = 20
lim sF (s) = 0,
.·. f(0+ ) = 0
s→0
s→∞
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12–30
CHAPTER 12. Introduction to the Laplace Transform
[e] sF (s) =
(s2
400
+ 4s + 5)2
lim sF (s) = 16,
.·. f(∞) = 16
lim sF (s) = 0,
.·. f(0+ ) = 0
s→0
s→∞
P 12.52 All of the F (s) functions referenced in this problem are improper rational
functions, and thus the corresponding f(t) functions contain impulses (δ(t)).
Thus, neither the initial value theorem nor the final value theorem may be
applied to these F (s) functions!
100(s + 1)
s(s2 + 2s + 5)
F (s) has a second-order pole at the origin, so we cannot use the final
value theorem here.
P 12.53 [a] sF (s) =
lim sF (s) = 0,
s→∞
[b] sF (s) =
500
(s + 5)3
lim sF (s) = 4,
s→0
lim sF (s) = 0,
s→∞
[c] sF (s) =
.·. f(0+ ) = 0
.·. f(∞) = 4
.·. f(0+ ) = 0
40(s + 2)
(s + 1)3
lim sF (s) = 80,
.·. f(∞) = 80
lim sF (s) = 0,
.·. f(0+ ) = 0
s→0
s→∞
[d] sF (s) =
(s + 5)2
(s + 1)4
lim sF (s) = 25,
.·. f(∞) = 25
lim sF (s) = 0,
.·. f(0+ ) = 0
s→0
s→∞
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Problems
P 12.54 [a] ZL = j120π(0.01) = j3.77 Ω;
ZC =
The phasor-transformed circuit is
IL =
12–31
−j
= −j26.526 Ω
120π(100 × 10−6 )
1
= 36.69/56.61◦ mA
15 + j3.77 − j26.526
.·. iL−ss (t) = 36.69 cos(120πt + 56.61◦ ) mA
[b] The steady-state response is the second term in Eq. 12.109, which
matches the steady-state response just derived in part (a).
P 12.55 The transient and steady-state components are both proportional to the
magnitude of the input voltage. Therefore,
K=
40
= 0.947
42.26
So if we make the amplitude of the sinusoidal source 0.947 instead of 1, the
current will not exceed the 40 mA limit. A plot of the current through the
inductor is shown below with the amplitude of the sinusoidal source set at
0.947.
P 12.56 We begin by using Eq. 12.105, and changing the right-hand side so it is the
Laplace transform of Kte−100t:
15IL (s) + 0.01sIL (s) + 104
IL (s)
A
=
s
(s + 100)2
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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12–32
CHAPTER 12. Introduction to the Laplace Transform
Solving for IL(s),
IL (s) =
100Ks
K1
K1∗
=
+
(s2 + 1500s + 106 )(s + 100)2
s + 750 − j661.44 s + 750 + j661.44
+
K1 =
K2 =
K3
K2
+
2
(s + 100)
s + 100
100Ks
(s + 750 + j661.44)(s + 100)2
(s2
100Ks
+ 1500s + 106 )
"
s=−100
d
100Ks
K3 =
ds (s2 + 1500s + 106 )
#
= 87.9K /139.59◦ µA
s=−750+j661.44
= −11.63K mA
= 133.86K µA
s=−100
Therefore,
iL (t) = K[0.176e−750t cos(661.44t + 139.59◦ ) − 11.63te−100t + 0.134e−100t ]u(t) mA
Plot the expression above with K = 1:
The maximum value of the inductor current is 0.068K mA. Therefore,
K=
40
= 588
0.068
So the inductor current rating will not be exceeded if the input to the RLC
circuit is 588te−100t V.
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13
The Laplace Transform in Circuit
Analysis
Assessment Problems
AP 13.1 [a] Y =
1
1
C[s2 + (1/RC)s + (1/LC)
+
+ sC =
R sL
s
1
106
=
= 80,000;
RC
(500)(0.025)
1
= 25 × 108
LC
25 × 10−9 (s2 + 80,000s + 25 × 108 )
s
√
[b] z1,2 = −40,000 ± 16 × 108 − 25 × 108 = −40,000 ± j30,000 rad/s
Therefore Y =
−z1 = −40,000 − j30,000 rad/s
−z2 = −40,000 + j30,000 rad/s
p1 = 0 rad/s
AP 13.2 [a] Z = 2000 +
1
4 × 107 s
= 2000 + 2
Y
s + 80,000s + 25 × 108
2000(s2 + 105 s + 25 × 108 )
2000(s + 50,000)2
=
s2 + 80,000s + 25 × 108
s2 + 80,000s + 25 × 108
[b] −z1 = −z2 = −50,000 rad/s
=
−p1 = −40,000 − j30,000 rad/s
−p2 = −40,000 + j30,000 rad/s
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
13–1 system, or transmission in any form or by any means, electronic,
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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13–2
CHAPTER 13. The Laplace Transform in Circuit Analysis
AP 13.3 [a] At t = 0− ,
0.2v1 = (0.8)v2;
v1 = 4v2 ;
Therefore v1(0− ) = 80V = v1(0+ );
I=
v1 + v2 = 100 V
v2(0− ) = 20V = v2(0+ )
(80/s) + (20/s)
20 × 10−3
=
5000 + [(5 × 106 )/s] + (1.25 × 106 /s)
s + 1250
80 5 × 106
V1 =
−
s
s
20 × 10−3
s + 1250
20 1.25 × 106
V2 =
−
s
s
[b] i = 20e−1250tu(t) mA;
!
=
20 × 10−3
s + 1250
!
80
s + 1250
=
20
s + 1250
v1 = 80e−1250tu(t) V
v2 = 20e−1250tu(t) V
AP 13.4 [a]
I=
Vdc/s
Vdc /L
= 2
R + sL + (1/sC)
s + (R/L)s + (1/LC)
Vdc
= 40;
L
I=
R
= 1.2;
L
1
= 1.0
LC
40
K1
K1∗
=
+
(s + 0.6 − j0.8)(s + 0.6 + j0.8)
s + 0.6 − j0.8 s + 0.6 + j0.8
K1 =
40
= −j25 = 25/ − 90◦ ;
j1.6
K1∗ = 25/90◦
[b] i = 50e−0.6t cos(0.8t − 90◦ ) = [50e−0.6t sin 0.8t]u(t) A
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
[c] V = sLI =
160s
(s + 0.6 − j0.8)(s + 0.6 + j0.8)
=
K1
K1∗
+
s + 0.6 − j0.8 s + 0.6 + j0.8
K1 =
13–3
160(−0.6 + j0.8)
= 100/36.87◦
j1.6
[d] v(t) = [200e−0.6t cos(0.8t + 36.87◦ )]u(t) V
AP 13.5 [a]
The two node voltage equations are
V1 − V2
5
V2 V2 − V1 V2 − (15/s)
+ V1 s =
and
+
+
=0
s
s
3
s
15
Solving for V1 and V2 yields
V1 =
5(s + 3)
,
2
s(s + 2.5s + 1)
V2 =
2.5(s2 + 6)
s(s2 + 2.5s + 1)
[b] The partial fraction expansions of V1 and V2 are
15
50/3
5/3
15
125/6
25/3
−
+
and V2 =
−
+
s
s + 0.5 s + 2
s
s + 0.5 s + 2
It follows that
50
5
v1 (t) = 15 − e−0.5t + e−2t u(t) V and
3
3
V1 =
125 −0.5t 25 −2t
v2 (t) = 15 −
e
+ e
u(t) V
6
3
[c] v1 (0+ ) = 15 −
50 5
+ =0
3
3
v2 (0+ ) = 15 −
[d] v1(∞) = 15 V;
125 25
+
= 2.5 V
6
3
v2 (∞) = 15 V
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13–4
CHAPTER 13. The Laplace Transform in Circuit Analysis
AP 13.6 [a]
With no load across terminals a − b Vx = 20/s:
1 20
20
− VTh s + 1.2
− VTh = 0
2 s
s
therefore VTh =
Vx = 5IT
and
20(s + 2.4)
s(s + 2)
ZTh =
VT
IT
Solving for IT gives
(VT − 5IT )s
+ VT − 6IT
2
Therefore
IT =
14IT = VT s + 5sIT + 2VT ;
therefore ZTh =
5(s + 2.8)
s+2
[b]
I=
VTh
20(s + 2.4)
=
ZTh + 2 + s
s(s + 3)(s + 6)
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Problems
AP 13.7 [a] i2 = 1.25e−t − 1.25e−3t ;
Therefore
13–5
di2
= −1.25e−t + 3.75e−3t
dt
therefore
di2
= 0 when
dt
1.25e−t = 3.75e−3t
or
e2t = 3,
t = 0.5(ln 3) = 549.31 ms
i2 (max) = 1.25[e−0.549 − e−3(0.549)] = 481.13 mA
[b] From Eqs. 13.68 and 13.69, we have
∆ = 12(s2 + 4s + 3) = 12(s + 1)(s + 3) and N1 = 60(s + 2)
Therefore I1 =
N1
5(s + 2)
=
∆
(s + 1)(s + 3)
A partial fraction expansion leads to the expression
2.5
2.5
I1 =
+
s+1 s+3
Therefore we get
i1 = 2.5[e−t + e−3t ]u(t) A
di1
di1 (0.54931)
= −2.5[e−t + 3e−3t ];
= −2.89 A/s
dt
dt
[d] When i2 is at its peak value,
[c]
di2
=0
dt
Therefore L2
[e] i2(max) =
di2
dt
!
M
= 0 and i2 = −
12
di1
dt
!
−2(−2.89)
= 481.13 mA (checks)
12
AP 13.8 [a] The s-domain circuit with the voltage source acting alone is
V 0 − (20/s)
V0
V 0s
+
+
=0
2
1.25s
20
V0 =
200
100/3 100/3
=
−
(s + 2)(s + 8)
s+2
s+8
v0 =
100 −2t
[e − e−8t]u(t) V
3
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13–6
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] With the current source acting alone,
V 00
V 00s
5
V 00
+
+
=
2
1.25s
20
s
V 00 =
100
50/3
50/3
=
−
(s + 2)(s + 8)
s+2 s+8
v 00 =
50 −2t
[e − e−8t ]u(t) V
3
[c] v = v 0 + v 00 = [50e−2t − 50e−8t ]u(t) V
Vo
Vo s
+
= Ig ;
s+2
10
[b] −z1 = −2 rad/s;
Vo
10(s + 2)
= H(s) = 2
Ig
s + 2s + 10
−p1 = −1 + j3 rad/s;
−p2 = −1 − j3 rad/s
AP 13.9 [a]
therefore
AP 13.10 [a]
Vo =
10(s + 2)
1
Ko
K1
K1∗
·
=
+
+
s2 + 2s + 10 s
s
s + 1 − j3 s + 1 + j3
Ko = 2;
K1 = 5/3/ − 126.87◦ ;
K1∗ = 5/3/126.87◦
vo = [2 + (10/3)e−t cos(3t − 126.87◦ )]u(t) V
[b] Vo =
10(s + 2)
K2
K2∗
·
1
=
+
s2 + 2s + 10
s + 1 − j3 s + 1 + j3
K2 = 5.27/ − 18.43◦ ;
K2∗ = 5.27/18.43◦
vo = [10.54e−t cos(3t − 18.43◦ )]u(t) V
AP 13.11 [a]
H(s) = L{h(t)} = L{vo (t)}
vo (t) = 10,000 cos θe−70t cos 240t − 10,000 sin θe−70t sin 240t
= 9600e−70t cos 240t − 2800e−70t sin 240t
Therefore H(s) =
=
9600(s + 70)
2800(240)
−
2
2
(s + 70) + (240)
(s + 70)2 + (240)2
s2
9600s
+ 140s + 62,500
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Problems
[b] Vo (s) = H(s) ·
1
9600
= 2
s
s + 140s + 62,500
=
K1 =
13–7
K1
K1∗
+
s + 70 − j240 s + 70 + j240
9600
= −j20 = 20/ − 90◦
j480
Therefore
vo (t) = [40e−70t cos(240t − 90◦ )]u(t) V = [40e−70t sin 240t]u(t) V
AP 13.12 From Assessment Problem 13.9:
H(s) =
10(s + 2)
+ 2s + 10
s2
Therefore H(j4) =
10(2 + j4)
= 4.47/ − 63.43◦
10 − 16 + j8
Thus,
vo = (10)(4.47) cos(4t − 63.43◦ ) = 44.7 cos(4t − 63.43◦ ) V
AP 13.13 [a]
Let R1 = 10 kΩ,
then V1 = V2 =
Also
R2 = 50 kΩ,
C = 400 pF,
Vg R2
R2 + (1/sC)
R2 C = 2 × 10−5
V1 − Vg V1 − Vo
+
=0
R1
R1
therefore Vo = 2V1 − Vg
Now solving for Vo /Vg , we get H(s) =
It follows that H(j50,000) =
R2 Cs − 1
R2 Cs + 1
j−1
= j1 = 1/90◦
j+1
Therefore vo = 10 cos(50,000t + 90◦ ) V
[b] Replacing R2 by Rx gives us H(s) =
Rx Cs − 1
Rx Cs + 1
Therefore
H(j50,000) =
j20 × 10−6 Rx − 1
Rx + j50,000
=
j20 × 10−6 Rx + 1
Rx − j50,000
Thus,
50,000
= tan 60◦ = 1.7321,
Rx
Rx = 28,867.51 Ω
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13–8
CHAPTER 13. The Laplace Transform in Circuit Analysis
Problems
P 13.1
1
i=
L
Z
t
0−
vdτ + I0 ;
P 13.2
VTh = Vab = CV0
P 13.3
Iscab = IN =
P 13.4
[a] Y =
therefore I =
1
sC
=
V0
;
s
1
L
ZTh =
V
s
+
I0
V
I0
=
+
s
sL
s
1
sC
−LI0
−I0
=
;
ZN = sL
sL
s
Therefore, the Norton equivalent is the same as the circuit in Fig. 13.4.
1
1
C[s2 + (1/RC)s + (1/LC)]
+
+ sC =
R sL
s
Z=
1
s/C
8 × 107 s
= 2
= 2
Y
s + (1/RC)s + (1/LC)
s + 40,000s + 256 × 106
[b] zero at z1 = 0
poles at −p1 = −8000 rad/s and −p2 = −32,000 rad/s
P 13.5
[a]
Z=
(R + 1/sC)(sL)
(Rs)(s + 1/RC)
= 2
R + sL + (1/sC)
s + (R/L)s + (1/LC)
R
= 10,000;
L
Z=
[b] Z =
s2
1
= 1600;
RC
1000s(s + 1600)
+ 10,000s + 16 × 106
1000s(s + 1600)
(s + 2000)(s + 8000)
z1 = 0;
−z2 = −1600 rad/s
−p1 = −2000 rad/s;
P 13.6
1
= 16 × 106
LC
−p2 = −8000 rad/s
1
L[s2 + (R/L)s + (1/LC)]
[a] Z = R + sL +
=
sC
s
=
[s2 + 8000s + 25 × 106 ]
s
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Problems
13–9
[b] s1,2 = −4000 ± j3000 rad/s
Zeros at −4000 + j3000 rad/s and −4000 − j3000 rad/s
Pole at 0.
P 13.7
Zab = 1k[s + (1/sk1)] = 1k[s + (1/(s + 1))] =
=
s + (1/(s + 1))
1 + s + (1/(s + 1))
s2 + s + 1
(s + 0.5 + j0.866)(s + 0.5 − j0.866)
=
2
s + 2s + 2
(s + 1 + j1)(s + 1 − j1)
Zeros at −0.5 + j0.866 rad/s and −0.5 − j0.866 rad/s; poles at −1 + j1 rad/s
and −1 − j1 rad/s.
P 13.8
Transform the Y-connection of the two resistors and the inductor into the
equivalent delta-connection:
where
Za =
(s)(1) + (1)(s) + (1)(1)
2s + 1
=
s
s
Zb = Zc =
(s)(1) + (1)(s) + (1)(1)
= 2s + 1
1
Then
Zab = Za k[(1/skZc ) + (1/skZb )] = Za k2(1/skZb )
1/skZb =
Zab =
=
1
(2s + 1)
s
1
+ 2s + 1
s
=
2s + 1
+s+1
2s2
2s + 1
2(2s + 1)
k 2
s
2s + s + 1
2(2s + 1)2
2
=
2
(2s + 1)(2s + s + 1) + 2s(2s + 1)
s+1
No zeros; one pole at −1 rad/s.
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13–10
P 13.9
CHAPTER 13. The Laplace Transform in Circuit Analysis
[a] For t > 0:
2.5s
−150
[b] Vo =
5
(16 × 10 )/s + 5000 + 2.5s
s
=
s2
−150s
+ 2000s + 64 × 104
−150s
(s + 400)(s + 1600)
K1
K2
[c] Vo =
+
s + 400 s + 1600
=
K1 =
−150s
s + 1600
K2 =
−150s
s + 400
Vo =
= 50
s=−400
s=−1600
= −200
50
200
−
s + 400 s + 1600
vo (t) = (50e−400t − 200e−1600t)u(t) V
P 13.10 [a] For t < 0:
Vc − 50
Vc
Vc − 137.5
+
+
=0
400
1200
500
Vc
1
1
1
50
137.5
+
+
=
+
400 1200 500
400
500
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Problems
13–11
Vc = 75 V
iL (0− ) =
75 − 137.5
= −0.125 A
500
For t > 0:
[b] Vo =
5 × 105
75
I+
s
5
0=−
137.5
5 × 105
75
+ 100I +
I+
− 1.25 × 10−3 + 0.01sI
s
s
s
5 × 105
62.5
I 100 +
+ 0.01s =
+ 1.25 × 10−3
s
s
!
. ·. I =
s2
6250 + 0.125s
+ 104 s + 5 × 107
5 × 105
Vo =
s
=
[c] Vo =
6250 + 0.125s
75
+
2
4
7
s + 10 s + 5 × 10
s
75s2 + 812,500s + 6875 × 106
s(s2 + 104 s + 5 × 107 )
K1
K2
K2∗
+
+
s
s + 5000 − j5000 s + 5000 + j5000
K1 =
K2 =
75s2 + 812,500s + 6875 × 106
s2 + 104 s + 5 × 107
75s2 + 812,500s + 6875 × 106
s(s + 5000 + j5000
= 137.5
s=0
= 40.02/141.34◦
s=−5000+j5000
vo (t) = [137.5 + 80.04e−5000t cos(5000t + 141.34◦ )]u(t) V
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13–12
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.11 [a] For t < 0:
iL (0− ) =
−100
−100
=
= −5 A
4 + 10k40 + 8
20
10
(5) = 1 A
50
i1 =
vC (0− ) = 10(1) + 4(5) − 100 = −70 V
For t > 0:
[b] (20 + 2s + 100/s)I = 10 +
5(s + 7)
+ 10s + 50
. ·.
I=
Vo =
100
70
I−
s
s
s2
70
s
−70s2 − 200s
−70(s + 20/7)
=
= 2
2
s(s + 10s + 50)
s + 10s + 50
=
K1 =
K1
K1∗
+
s + 5 − j5 s + 5 + j5
−70(s + 20/7)
s + 5 + j5
s=−5+j5
= 38.1/ − 156.8◦
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Problems
13–13
[c] vo (t) = 76.2e−5t cos(5t − 156.8◦ )u(t) V
P 13.12 [a] For t < 0:
1
1
1
1
= +
+
= 0.1875;
Re
8 80 20
Re = 5.33 Ω
v1 = (9)(5.33) = 48 V
iL (0− ) =
48
= 2.4 A
20
vC (0− ) = −v1 = −48 V
For t = 0+ :
s-domain circuit:
where
R = 20 Ω;
C = 6.25 µF;
L = 6.4 mH;
[b]
and
γ = −48 V;
ρ = −2.4 A
Vo
Vo
ρ
+ Vo sC − γC +
− =0
R
sL s
.·. Vo =
s2
γ[s + (ρ/γC)]
+ (1/RC)s + (1/LC)
ρ
−2.4
=
= 8000
γC
(−48)(6.25 × 10−6 )
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13–14
CHAPTER 13. The Laplace Transform in Circuit Analysis
1
1
=
= 8000
RC
(20)(6.25 × 10−6 )
1
1
=
= 25 × 106
−3
LC
(6.4 × 10 )(6.25 × 10−6 )
Vo =
[c] IL =
s2
−48(s + 8000)
+ 8000s + 25 × 106
ρ
Vo
2.4
Vo
− =
+
sL s
0.0064s
s
=
[d] Vo =
=
s(s2
−7500(s + 8000)
2.4
2.4(s + 4875)
−
= 2
6
+ 8000s + 25 × 10 )
s
(s + 8000s + 25 × 106 )
−48(s + 8000)
s2 + 8000s + 25 × 106
K1
K1∗
+
s + +4000 − j3000 s + 4000 + j3000
K1 =
−48(s + 8000)
s + 4000 + j3000
= 40/126.87◦
s=−4000+j3000
vo (t) = [80e−4000t cos(3000t + 126.87◦ )]u(t) V
[e] IL =
=
s2
2.4(s + 4875)
+ 8000s + 25 × 106
K1
K1∗
+
s + 4000 − j3000 s + 4000 + j3000
K1 =
2.4(s + 4875)
s + 4000 + j3000
s=−4000+j3000
= 1.25/ − 16.26◦
iL (t) = [2.5e−4000t cos(3000t − 16.26◦ )]u(t) A
P 13.13 [a] io(0− ) =
Io =
20
= 5 mA
4000
20/s + Lρ
sC(20/s + Lρ)
= 2
R + sL + 1/sC
s LC + RsC + 1
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Problems
=
s2
13–15
20/L + sρ
40 + s(0.005)
= 2
+ sR/L + 1/LC
s + 8000s + 16 × 106
4000(40 + 0.005s)
0.0025s(s + 8000)
− 0.0025 + 2
6
+ 8000s + 16 × 10
s + 8000s + 16 × 106
20
40,000
20s + 120,000
=
+
=
(s + 4000)2
(s + 4000)2 s + 4000
Vo = RIo − Lρ + sLIo =
s2
vo (t) = [20te−4000t + 40,000e−4000t]u(t) V
[b] Io =
0.005(s + 8000)
+ 8000s + 16 × 106
K1
K2
=
+
(s + 4000)2 s + 4000
s2
K1 = 20
K2 = 0.005
io (t) = [20te−4000t + 0.005e−4000t]u(t) A
P 13.14 For t < 0:
vo (0− ) − 500 vo(0− ) vo (0− )
+
+
=0
5
25
100
25vo (0− ) = 10,000
iL (0− ) =
.·.
vo (0− ) = 400 V
vo (0− )
400
=
= 16 A
25
25
For t > 0 :
Vo + 400
Vo
Vo − (400/s)
+
+
=0
25 + 25s 100
100/s
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13–16
CHAPTER 13. The Laplace Transform in Circuit Analysis
Vo
1
1
s
+
+
25 + 25s 100 100
.·.
Vo =
=4−
400
25 + 25s
400(s − 3)
s2 + 2s + 5
Vo − (400/s)
−20s − 20
= 2
100/s
s + 2s + 5
Io =
=
K1 =
K1
K1∗
+
s + 1 − j2 s + 1 + j2
−20(s + 1)
s + 1 + j2
s=−1+j2
= −10
io (t) = [−20e−t cos 2t]u(t) A
P 13.15
Vo =
(18/s)(8 × 106 /s)
2800 + 0.2s + (8 × 106 /s)
=
720 × 106
s(s2 + 14,000s + 40 × 106 )
=
720 × 106
s(s + 4000)(s + 10,000)
=
K1
K2
K3
+
+
s
s + 4000 s + 10,000
K1 =
720 × 106
= 18
4 × 107
720 × 106
K2 =
= −30
(−4000)(6000)
K3 =
720 × 106
= 12
(−6000)(−10,000)
Vo =
18
30
12
−
+
s
s + 4000 s + 10,000
vo (t) = [18 − 30e−4000t + 12e−10,000t]u(t) V
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Problems
13–17
P 13.16 With a non-zero initial voltage on the capacitor, the s-domain circuit becomes:
(Vo − 30/s)s
Vo − 18/s
+
=0
0.2s + 2800
8 × 106
Vo
"
#
5
s
30
90
+
=
+
s + 14,000 8 × 106
80 × 106 s(s + 14,000)
30s2 + 420,000s + 720 × 106
·
. . Vo =
s(s + 4000)(s + 10,000)
=
K1
K2
K3
+
+
s
s + 4000 s + 10,000
720 × 106
K1 =
= 18
40 × 106
K2 =
30s2 + 420,000s + 720 × 106
s(s + 10,000)
s=−4000
K3 =
30s2 + 420,000s + 720 × 106
s(s + 4000)
s=−10,000
Vo =
18
20
8
+
−
s
s + 4000 s + 10,000
= 20
= −8
vo (t) = [18 + 20e−4000t − 8e−10,000t]u(t) V
P 13.17 [a] For t < 0:
V2 =
10
(450) = 90 V
10 + 40
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13–18
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] V1 =
25(450/s)
(125,000/s) + 25 + 1.25 × 10−3 s
=
9 × 106
9 × 106
=
s2 + 20, 000s + 108
(s + 10,000)2
v1 (t) = (9 × 106 te−10,000t)u(t) V
[c] V2 =
90
(25,000/s)(450/s)
−
s
(125,000/s) + 1.25 × 10−3 s + 25
=
90(s + 20,000)
s2 + 20,000s + 108
=
900,000
90
+
2
(s + 10,000)
s + 10,000
v2 (t) = [9 × 105 te−10,000t + 90e−10,000t]u(t) V
P 13.18 [a] iL(0− ) = iL (0+ ) =
"
24
= 8A
3
directed upward
#
20(10/s)
25IT (10/s)
200
VT = 25Iφ +
IT =
+
IT
20 + (10/s)
20 + (10/s)
10 + 20s
VT
250 + 200
45
=Z=
=
IT
20s + 10
2s + 1
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Problems
13–19
Vo Vo (2s + 1)
Vo
8
+
+
=
5
45
5.625s
s
[9s + (2s + 1)s + 8]Vo
8
=
45s
s
Vo [2s2 + 10s + 8] = 360
Vo =
[b] Vo =
2s2
360
180
= 2
+ 10s + 8
s + 5s + 4
180
K1
K2
=
+
(s + 1)(s + 4)
s+1 s+4
K1 =
180
= 60;
3
Vo =
60
60
−
s+1 s+4
K2 =
180
= −60
−3
vo (t) = [60e−t − 60e−4t ]u(t) V
P 13.19 vC (0− ) = vC (0+ ) = 0
0.01
Vo
Vo
Vo s
6Vo
=
+
+
−
6
s
50,000 5000 50 × 10
50,000
500 × 103
= (1000 + 10,000 + s − 6000)Vo
s
Vo =
=
500 × 103
K1
K2
=
+
s(s + 5000)
s
s + 5000
100
100
−
s
s + 5000
vo (t) = [100 − 100e−5000t ]u(t) V
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13–20
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.20
5 × 10−3
Vo
Vo
=
+ 3.75 × 10−3 Vφ +
6
s
200 + 4 × 10 /s
0.04s
Vφ =
4 × 106 /s
4 × 106 Vo
V
=
o
200 + 4 × 106 /s
200s + 4 × 106
.·.
5 × 10−3
Vo s
15,000Vo
25Vo
=
+
+
6
6
s
200s + 4 × 10
200s + 4 × 10
s
.·. Vo =
s2
s + 20,000
K1
K2
=
+
8
2
+ 20,000s + 10
(s + 10,000)
s + 10,000
K1 = 10,000;
Vo =
K2 = 1
10,000
1
+
2
(s + 10,000)
s + 10,000
vo (t) = [10,000te−10,000t + e−10,000t]u(t) V
P 13.21 [a]
Vo − 35/s
Vo − 8Iφ
+ 0.4V∆ +
=0
2
s + (250/s)
"
#
Vo − 8Iφ
V∆ =
s;
s + (250/s)
Iφ =
(35/s) − Vo
2
Solving for Vo yields:
Vo =
29.4s2 + 56s + 1750
29.4s2 + 56s + 1750
=
s(s2 + 2s + 50)
s(s + 1 − j7)(s + 1 + j7)
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Problems
Vo =
K1
K2
K2∗
+
+
s
s + 1 − j7 s + 1 + j7
K1 =
29.4s2 + 56s + 1750
s2 + 2s + 50
K2 =
29.4s2 + 56s + 1750
s(s + 1 + j7)
13–21
= 35
s=0
s=−1+j7
= −2.8 + j0.6 = 2.86/167.91◦
.·. vo(t) = [35 + 5.73e−t cos(7t + 167.91◦ )]u(t) V
[b] At t = 0+
vo = 35 + 5.73 cos(167.91◦ ) = 29.4 V
vo − 35
+ 0.4v∆ = 0;
2
vo − 35 + 0.8v∆ = 0
vo = v∆ + 8iφ = v∆ + 8(0.4v∆ ) = 4.2 V
vo + (0.8)
vo
= 35;
4.2
.·. vo (0+ ) = 29.4 V(checks)
At t = ∞, the circuit is
v∆ = 0,
.·. vo = 35 V(checks)
iφ = 0
P 13.22 [a]
I1 + s(I1 − I2) =
10
s
and
1
I2 + I2 + s(I2 − I1 ) = 0
s
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13–22
CHAPTER 13. The Laplace Transform in Circuit Analysis
Solving the second equation for I1 :
s2 + s + 1
I2
s2
Substituting into the first equation and solving for I2:
I1 =
s2 + s + 1
10
(s + 1)
− s I2 =
2
s
s
"
#
. ·.
I2 =
. ·.
I1 =
2s2
s2 + s + 1
10s
10(s2 + s + 1)
·
=
s2
2s2 + 2s + 1
s(2s2 + 2s + 1)
Io = I1 − I2 =
=
10(s2 + s + 1)
10s
5(s + 1)
− 2
=
2
2
s(2s + 2s + 1) 2s + 2s + 1
s(s + s + 0.5)
K1
K2
K2∗
+
+
s
s + 0.5 − j0.5 s + 0.5 + j0.5
K1 = 10;
. ·.
10s
+ 2s + 1
K2 = 5/ − 180◦
io (t) = [10 − 10e−0.5t cos 0.5t]u(t) A
[b] Vo = sIo =
5(s + 1)
K1
K1∗
=
+
s2 + s + 0.5
s + 0.5 − j0.5 s + 0.5 + j0.5
K1 = 3.54/ − 45◦
. ·.
vo (t) = 7.07e−0.5t cos(0.5t − 45◦ )u(t) V
[c] At t = 0+ the circuit is
.·. vo(0+ ) = 5 V = 7.07 cos(−45◦ );
Io(0+ ) = 0
Both values agree with our solutions for vo and io .
At t = ∞ the circuit is
.·. vo(∞) = 0;
io(∞) = 10 A
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
13–23
Both values agree with our solutions for vo and io .
P 13.23 [a]
Vo =
(1/sC)(sL)(Ig /s)
Ig /C
= 2
R + sL + (1/sC)
s + (R/L)s + (1/LC)
Ig
0.015
=
= 0.15
C
0.1
R
= 7;
L
1
= 10
LC
Vo =
0.15
s2 + 7s + 10
[b] sVo =
0.15s
s2 + 7s + 10
lim sVo = 0;
s→0
lim sVo = 0;
s→∞
[c] Vo =
.·. vo (∞) = 0
.·. vo (0+ ) = 0
0.15
0.05
−0.05
=
+
(s + 2)(s + 5)
s+2
s+5
vo = [50e−2t − 50e−5t ]u(t) mV
P 13.24 IL =
IL =
Ig
Vo
Ig
+
=
− sCVo
s
1/sC
s
15
15s
15
−10
25
−
=
−
+
s
(s + 2)(s + 5)
s
s+2 s+5
iL (t) = [15 + 10e−2t − 25e−5t ]u(t) mA
Check:
iL (0+ ) = 0 (ok);
iL(∞) = 15 mA (ok)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–24
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.25 [a]
[b] Zeq = 50,000 +
107 20 × 1012 /s2
+
3s
12 × 106 /s
= 50,000 +
=
107
20 × 1012
+
3s
12 × 106 s
100,000s + 107
2s
I1 =
20/s
0.4 × 10−3
=
Zeq
s + 100
V1 =
107
4000/3
I1 =
3s
s(s + 100)
V2 =
107 0.4 × 10−4
2000/3
·
=
6s
s + 100
s(s + 100)
[c] i1 (t) = 0.4e−100tu(t) mA
V1 =
40/3
40/3
−
;
s
s + 100
v1(t) = (40/3)(1 − 1e−100t)u(t) V
V2 =
20/3
20/3
−
;
s
s + 100
v2(t) = (20/3)(1 − 1e−100t)u(t) V
[d] i1(0+ ) = 0.4 mA
i1 (0+ ) =
20
× 10−3 = 0.44 mA(checks)
50
v1 (0+ ) = 0;
v2(0+ ) = 0(checks)
v1 (∞) = 40/3 V;
v2(∞) = 20/3 V(checks)
v1 (∞) + v2 (∞) = 20 V(checks)
(0.3 × 10−6 )v1 (∞) = 4 µC
(0.6 × 10−6 )v2 (∞) = 4 µC(checks)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
13–25
P 13.26 [a]
V1 − 75/s V1 V1 − V2
+
+
=0
10
20
10
V2 V2 − V1 (V2 − 75/s)s
+
+
=0
5s
10
250
Thus,
5V1 − 2V2 =
150
s
−25sV1 + (s2 + 25s + 50)V2 = 75s
∆=
N2 =
5
−2
−25s s2 + 25s + 50
5
150/s
= 5(s + 5)(s + 10)
= 375(s + 10)
−25s 75s
V2 =
N2
375(s + 10)
75
=
=
∆
5(s + 5)(s + 10)
s+5
Vo =
75
75
375
−
=
s
s+5
s(s + 5)
Io =
V2
15
3
3
=
= −
5s
s(s + 5)
s s+5
[b] vo(t) = (75 − 75e−5t )u(t) V
io (t) = (3 − 3e−5t )u(t) A
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–26
CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] At t = 0+ the circuit is
vo (0+ ) = 0;
io (0+ ) = 0
Checks
At t = ∞ the circuit is
vo (∞) = 75 V;
io (∞) =
75
20
·
= 3 A Checks
10 + (200/30) 30
P 13.27 [a]
10
10
I1 + (I1 − I2) + 10(I1 − 9/s) = 0
s
s
10
10
(I2 − 9/s) + (I2 − I1) + 10I2 = 0
s
s
Simplifying,
(s + 2)I1 − I2 = 9
−I1 + (s + 2)I2 =
9
s
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
∆=
N1 =
(s + 2)
−1
−1
(s + 2)
9
−1
13–27
= s2 + 4s + 3 = (s + 1)(s + 3)
=
9/s (s + 2)
9s2 + 18s + 9
9
= (s + 1)2
s
s
9
(s + 1)2
9(s + 1)
N1
I1 =
=
=
∆
s (s + 1)(s + 3)
s(s + 3)
"
N2 =
#
(s + 2) 9
−1
=
9/s
18
(s + 1)
s
N2
18(s + 1)
18
=
=
∆
s(s + 1)(s + 3)
s(s + 3)
I2 =
Ia = I1 =
9(s + 1)
3
6
= +
s(s + 3)
s s+3
9
9 9(s + 1)
6
6
− I1 = −
= −
s
s s(s + 3)
s s+3
Ib =
[b] ia(t) = 3(1 + 2e−3t )u(t) A
ib (t) = 6(1 − e−3t )u(t) A
10
10 3
6
[c] Va = Ib =
+
s
s s s+3
=
30
60
30 20
20
+
= 2 +
−
2
s
s(s + 3)
s
s
s+3
Vb =
10
10
(I2 − I1) =
s
s
6
6
3
6
−
−
+
s s+3
s s+3
=
10 3
12
30 40
40
−
= 2 −
+
s s s+3
s
s
s+3
Vc =
10
10 9 6
6
(9/s − I2) =
− +
s
s s s s+3
=
30 20
20
+
−
s2
s
s+3
[d] va(t) = [30t + 20 − 20e−3t ]u(t) V
vb (t) = [30t − 40 + 40e−3t ]u(t) V
vc (t) = [30t + 20 − 20e−3t ]u(t) V
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–28
CHAPTER 13. The Laplace Transform in Circuit Analysis
[e] Calculating the time when the capacitor voltage drop first reaches 1000 V:
30t + 20 − 20e−3t = 1000
or 30t − 40 + 40e−3t = 1000
Note that in either of these expressions the exponential tem is negligible
when compared to the other terms. Thus,
30t + 20 = 1000 or
30t − 40 = 1000
Thus,
980
1040
= 32.67 s or t =
= 34.67 s
30
30
Therefore, the breakdown will occur at t = 32.67 s.
t=
P 13.28 [a]
V1 V1 − 50/s V1 − Vo
+
+
=0
10
25/s
4s
−5 Vo − V1 Vo − 50/s
+
+
=0
s
4s
30
Simplfying,
(4s2 + 10s + 25)V1 − 25Vo = 200s
−15V1 + (2s + 15)Vo = 400
∆=
No =
(4s2 + 10s + 25)
−25
−15
(2s + 15)
(4s2 + 10s + 25) 200s
−15
= 8s(s + 5)2
= 200(8s2 + 35s + 50)
400
Vo =
No
200(8s2 + 35s + 50)
25(8s2 + 35s + 50)
K1
K2
K3
=
=
=
+
+
2
2
2
∆
8s(s + 5)
s(s + 5)
s
(s + 5)
s+5
K1 =
(25)(50)
= 50;
25
K2 =
25(200 − 175 + 50)
= −375
−5
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
d 8s2 + 35s + 50
K3 = 25
ds
s
"
#
s=−5
13–29
s(16s + 35) − (8s2 + 35s + 50)
= 25
s2
"
#
s=−5
= −5(−45) − 75 = 150
.·. Vo =
50
375
150
−
+
2
s
(s + 5)
s+5
[b] vo(t) = [50 − 375te−5t + 150e−5t ]u(t) V
[c] At t = 0+ :
vo (0+ ) = 50 + 150 = 200 V(checks)
At t = ∞:
vo (∞)
vo (∞) − 50
−5+
=0
10
30
.·. 3vo (∞) − 150 + vo(∞) − 50 = 0;
. ·.
.·. 4vo (∞) = 200
vo (∞) = 50 V(checks)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–30
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.29 [a]
5
0 = 2.5s(I1 − 6/s) + (I1 − I2) + 10I1
s
−75
5
= (I2 − I1) + 5(I2 − 6/s)
s
s
or
(s2 + 4s + 2)I1 − 2I2 = 6s
−I1 + (s + 1)I2 = −9
∆=
−1
(s + 1)
−2
9 (s + 1)
= 5(s + 2)(s + 3)
= 6(s2 + s − 3)
N1
6(s2 + s − 3)
=
∆
s(s + 2)(s + 3)
(s2 + 4s + 2) 6s
N2 =
I2 =
−2
6s
N1 =
I1 =
(s2 + 4s + 2)
−1
9
= −9s2 − 30s − 18
N2
−9s2 − 30s − 18
=
∆
s(s + 2)(s + 3)
6(s2 + s − 3)
[b] sI1 =
(s + 2)(s + 3)
lim sI1 = i1 (0+ ) = 6 A;
s→∞
sI2 =
lim sI1 = i1(∞) = −3 A
s→0
−9s2 − 30s − 18
(s + 2)(s + 3)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
lim sI2 = i2 (0+ ) = −9 A;
lim sI2 = i2(∞) = −3 A
s→∞
[c] I1 =
13–31
s→0
6(s2 + s − 3)
K1
K2
K3
=
+
+
s(s + 2)(s + 3)
s
s+2 s+3
K1 =
6(−3)
= −3;
6
K3 =
6(9 − 3 − 3)
=6
(−3)(−1)
K2 =
6(4 − 2 − 3)
=3
(−2)(1)
i1 (t) = [−3 + 3e−2t + 6e−3t ]u(t) A
I2 =
−9s2 − 30s − 18
K1
K2
K3
=
+
+
s(s + 2)(s + 3)
s
s+2 s+3
K1 =
−18
= −3;
6
K3 =
−81 + 90 − 18
= −3
(−3)(−1)
K2 =
−36 + 60 − 18
= −3
(−2)(1)
i2 (t) = [−3 − 3e−2t − 3e−3t ]u(t) A
P 13.30 [a]
At Vo :
Vo − V1 50 Vo − V2
−
+
=0
10s
s
5
. ·.
Vo (2s + 1) − 2sV2 − V1 = 500
Supernode:
V1 s V1 − Vo V2 V2 − V1
+
+
+
=0
5
10s
1
5
. ·.
−Vo (2s + 1) + 12sV2 + (2s2 + 1)V1 = 0
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13–32
CHAPTER 13. The Laplace Transform in Circuit Analysis
Constraint:
V1 s
V1 − V2 = 4I∆ = 4 −
5
. ·.
V2 = (0.8s + 1)V1
Simplifying:
Vo (2s + 1) − V1 (1.6s2 + 2s + 1) = 500
−Vo (2s + 1) − V1 (11.6s2 + 12s + 1) = 0
∆=
No =
Vo =
2s + 1
−(1.6s2 + 2s + 1)
−(2s + 1) (11.6s2 + 12s + 1)
500 −(1.6s2 + 2s + 1)
2
= 20(s2 + 1.5s + 0.5)
= 500(11.6s2 + 12s + 1)
0 (11.6s + 12s + 1)
No
25(11.6s2 + 12s + 1)
=
∆
s(s + 0.5)(s + 1)
[b] vo(0+ ) = lim sVo = 25(11.6) = 290 V
s→∞
vo (∞) = lim sVo =
s→0
25
= 50 V
0.5
[c] At t = 0+ the circuit is
4I∆ + 1I1 = 0;
I1 − I∆ = 50
.·. 4Iφ + 50 + I∆ = 0;
5I∆ = −50
.·. I∆ = Io (0+ ) = −10 A
Also I1 = 50 − 10 = 40 A
Vo (0+ ) = 5(I1 − I∆ ) + 1I1 = 6I1 − 5I∆ = 240 − 5(−10) = 290 V (checks)
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Problems
13–33
At t = ∞ the circuit is
Vo (∞) = 50(1) = 50 V(checks)
[d] Vo =
25(11.6s2 + 12s + 1)
K1
K2
K3
=
+
+
s(s + 0.5)(s + 1)
s
s + 0.5 s + 1
K1 =
25
= 50;
(0.5)(1)
K3 =
15
= 30
(−1)(−0.5)
Vo =
50
210
30
+
+
s
s + 0.5 s + 1
K2 =
−52.5
= 210
(−0.5)(0.5)
vo (t) = (50 + 210e−0.5t + 30e−t )u(t) V
vo (∞) = 50 V(checks)
vo (0+ ) = 50 + 210 + 30 = 290 V(checks)
P 13.31 [a]
120
250
= 50(I1 − 0.05Vφ ) +
(I1 − I2 )
s
s
250
250
250
250
= 50I1 − 2.5
(I2 − I1) +
I1 −
I2
s
s
s
s
Simplfying,
(50s + 875)I1 − 875I2 = 120
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13–34
CHAPTER 13. The Laplace Transform in Circuit Analysis
250(s − 1)I1 + (20s2 + 450s + 250)I2 = 0
∆=
(50s + 875)
2
−875
= 1000s(s2 + 40s + 625)
250(s − 1) (20s + 450s + 250)
N1 =
N2 =
120
−875
= 1200(2s2 + 45s + 25)
2
0 (20s + 450s + 250)
(50s + 875) 120
250(s − 1)
0
= −30,000(s − 1)
I1 =
N1
1200(2s2 + 45s + 25)
=
∆
s(s2 + 40s + 625)
I2 =
N2
−30,000(s − 1)
=
∆
s(s2 + 40s + 625)
Io = I2 − 0.05Vφ = I2 − 0.05
I2 − I1 =
250
(I2 − I1)
s
−2400(s + 35)
s(s2 + 40s + 625)
250
−600,000(s + 35)
(I2 − I1 ) =
s
s(s2 + 40s + 625)
. ·.
[b] sIo =
Io =
(s2
−30,000(s − 1)
30,000(s + 35)
1080
+
=
2
2
2
s(s + 40s + 625) s(s + 40s + 625)
s(s + 40s + 625)
1080
+ 40s + 625)
io (0+ ) = lim sIo = 0
s→∞
io (∞) = lim sVo =
s→0
1080
= 1728 mA
625
[c] At t = 0+ the circuit is
i(0+ ) = 0 (checks)
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Problems
13–35
At t = ∞ the circuit is
120 = 50(ia − i1) + 700ia
= 50(ia − 0.05vφ ) + 700ia = 750ia − 2.5vφ
. ·.
vφ = −700ia
120 = 750ia + 1750ia = 2500ia
120
= 48 mA
2500
vφ = −700ia = −33.60 V
ia =
io (∞) = 48 × 10−3 − 0.05(−33.60) = 48 × 10−3 + 1.68 = 1728 mA (checks)
[d] Io =
1080
K1
K2
K2∗
=
+
+
s(s2 + 40s + 625)
s
s + 20 − j15 s + 20 + j15
1080
= 1.728
625
1080
K2 =
= 1.44/126.87◦
(−20 + j15)(j30)
K1 =
io (t) = [1728 + 2880e−20t cos(15t + 126.87◦ )]u(t) mA
Check:
io (0+ ) = 0 mA;
100k5s =
500s
100s
=
5s + 100
s + 20
io (∞) = 1728 mA
P 13.32 [a]
"
#
100s
50
5000s
Vo =
=
2
s + 20 (s + 25)
(s + 20)(s + 25)2
Io =
Vo
50s
=
100
(s + 20)(s + 25)2
IL =
Vo
1000
=
5s
(s + 20)(s + 25)2
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13–36
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] Vo =
K1
K2
K3
+
+
2
s + 20 (s + 25)
s + 25
K1 =
K2 =
5000s
(s + 25)2
5000s
(s + 20)
s=−20
= −4000
= 25,000
s=−25
d 5000s
K3 =
ds s + 20
s=−25
"
5000
5000s
=
−
s + 20 (s + 20)2
#
= 4000
s=−25
vo (t) = [−4000e−20t + 25,000te−25t + 4000e−25t ]u(t) V
Io =
K1
K2
K3
+
+
2
s + 20 (s + 25)
s + 25
K1 =
50s
(s + 25)2
K2 =
50s
(s + 20)
s=−20
= −40
= 250
s=−25
d
50s
K3 =
ds s + 20
s=−25
"
50
50s
=
−
s + 20 (s + 20)2
#
= 40
s=−25
io (t) = [−40e−20t + 250te−25t + 40e−25t ]u(t) V
IL =
K1
K2
K3
+
+
2
s + 20 (s + 25)
s + 25
K1 =
1000
(s + 25)2
K2 =
1000
(s + 20)
= 40
s=−20
s=−25
d 1000
K3 =
ds s + 20
= −200
s=−25
"
1000
= −
(s + 20)2
#
s=−25
= −40
iL (t) = [40e−20t − 200te−25t − 40e−25t ]u(t) V
P 13.33 vC = 12 × 105 te−5000t V,
dvC
iC = C
dt
!
C = 5 µF;
therefore
= 6e−5000t(1 − 5000t) A
iC > 0 when 1 > 5000t
or iC > 0 when 0 < t < 200 µs
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Problems
13–37
and iC < 0 when t > 200 µs
iC = 0 when 1 − 5000t = 0,
or t = 200 µs
dvC
= 12 × 105 e−5000t[1 − 5000t]
dt
.·. iC = 0 when
dvC
=0
dt
P 13.34
10s
40
400
40
·
=
=
10s + 1000 s
10s + 1000
s + 100
VTh =
ZTh = 1000 + 1000k10s = 1000 +
I=
(5 ×
=
K1 =
105 )/s
10,000s
2000(s + 50)
=
10s + 1000
s + 100
40/(s + 100)
40s
=
2
+ 2000(s + 50)/(s + 100)
2000s + 600,000s + 5 × 107
0.02s
K1
K1∗
=
+
s2 + 300s + 25,000
s + 150 − j50 s + 150 + j50
0.02s
s + 150 + j50
s=−150+j50
= 31.62 × 10−3 /71.57◦
i(t) = 63.25e−150t cos(50t + 71.57◦ )u(t) mA
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13–38
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.35 [a] The s-domain equivalent circuit is
I=
Vg
Vg /L
=
,
R + sL
s + (R/L)
I=
K0
K1
K1∗
+
+
s + R/L s − jω s + jω
K0 =
Vg =
Vm (ωL cos φ − R sin φ)
,
R2 + ω 2 L2
Vm (ω cos φ + s sin φ)
s2 + ω 2
K1 =
Vm /φ − 90◦ − θ(ω)
√
2 R2 + ω 2 L2
where tan θ(ω) = ωL/R. Therefore, we have
i(t) =
Vm (ωL cos φ − R sin φ) −(R/L)t Vm sin[ωt + φ − θ(ω)]
√
e
+
R2 + ω 2 L2
R2 + ω 2 L2
Vm
sin[ωt + φ − θ(ω)]
+ ω 2 L2
Vm (ωL cos φ − R sin φ) −(R/L)t
[c] itr =
e
R2 + ω 2 L2
Vg
[d] I =
,
Vg = Vm /φ − 90◦
R + jωL
[b] iss (t) = √
R2
Therefore I = √
Vm /φ − 90◦
Vm
√
/φ − θ(ω) − 90◦
=
2
2
2
2
2
2
/
R + ω L θ(ω)
R +ω L
Vm
Therefore iss = √ 2
sin[ωt + φ − θ(ω)]
R + ω 2 L2
[e] The transient component vanishes when
ωL cos φ = R sin φ or
tan φ =
ωL
R
or
φ = θ(ω)
1
1
P 13.36 [a] W = L1 i21 + L2 i22 + Mi1i2
2
2
W = 4(15)2 + 9(100) + 150(6) = 2700 J
[b] 120i1 + 8
di1
di2
−6
=0
dt
dt
270i2 + 18
di2
di1
−6
=0
dt
dt
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Problems
13–39
Laplace transform the equations to get
120I1 + 8(sI1 − 15) − 6(sI2 + 10) = 0
270I2 + 18(sI2 + 10) − 6(sI1 − 15) = 0
In standard form,
(8s + 120)I1 − 6sI2 = 180
−6sI1 + (18s + 270)I2 = −270
∆=
N1 =
N2 =
8s + 120
−6s
−6s
18s + 270
180
= 108(s + 10)(s + 30)
−6s
= 1620(s + 30)
−270 18s + 270
8s + 120 180
−6s
−270
= −1080(s + 30)
I1 =
N1
1620(s + 30)
15
=
=
∆
108(s + 10)(s + 30)
s + 10
I2 =
N2
−1080(s + 30)
−10
=
=
∆
108(s + 10)(s + 30)
s + 10
[c] i1 (t) = 15e−10t u(t) A;
i2 (t) = −10e−10t u(t) A
Z ∞
e−20t ∞
[d] W120Ω =
(225e−20t )(120) dt = 27,000
= 1350 J
−20 0
0
W270Ω =
Z
0
∞
(100e−20t )(270) dt = 27,000
e−20t
−20
∞
= 1350 J
0
W120Ω + W270Ω = 2700 J
1
1
[e] W = L1 i21 + L2 i22 + Mi1 i2 = 900 + 900 − 900 = 900 J
2
2
With the dot reversed the s-domain equations are
(8s + 120)I1 + 6sI2 = 60
6sI1 + (18s + 270)I2 = −90
As before,
N1 =
N2 =
∆ = 108(s + 10)(s + 30). Now,
60
−6s
= 1620(s + 10)
−90 18s + 270
8s + 120 60
−6s
−90
= −1080(s + 10)
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13–40
CHAPTER 13. The Laplace Transform in Circuit Analysis
I1 =
N1
15
=
;
∆
s + 30
I2 =
i1 (t) = 15e−30t u(t) A;
Z
∞
W270Ω =
Z
∞
W120Ω =
i2 (t) = −10e−30tu(t) A
(100e−60t )(270) dt = 450 J
0
0
N2
−10
=
∆
s + 30
(225e−60t )(120) dt = 450 J
W120Ω + W270Ω = 900 J
P 13.37 The s-domain equivalent circuit is
V1 − 48/s
V1 + 9.6
V1
+
+
=0
4 + (100/s)
0.8s
0.8s + 20
V1 =
Vo =
s2
20
−30,000
V1 =
0.8s + 20
(s + 25)(s + 5 − j10)(s + 5 + j10)
=
K1 =
K2 =
−1200
+ 10s + 125
K1
K2
K2∗
+
+
s + 25 s + 5 − j10 s + 5 + j10
s2
−30,000
+ 10s + 125
s=−25
−30,000
(s + 25)(s + 5 + j10)
= −60
= 67.08/63.43◦
s=−5+j10
vo (t) = [−60e−25t + 134.16e−5t cos(10t + 63.43◦ )]u(t) V
P 13.38 For t < 0:
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Problems
13–41
For t > 0+ :
Note that because of the dot locations on the coils, the sign of the mutual
inductance is negative! (See Example C.1 in Appendix C.)
L1 − M = 3 + 1 = 4 H;
18 × 4 = 72;
L2 − M = 2 + 1 = 3 H
18 × 3 = 54
V − 72
V
V + 54
+
+
=0
4s + 20 −s + 10
3s
V
1
1
1
+
+
4s + 20 −s + 10 3s
=
72
54
−
4s + 20 3s
"
#
3s(−s + 10) + 3s(4s + 20) + (4s + 20)(−s + 10)
72(3s) − 54(4s + 20)
V
=
3s(−s + 10)(4s + 20)
3s(4s + 20)
V =
[72(3s) − 54(4s + 20)](−s + 10)
5s2 + 110s + 200
Io =
V
−108
−1.2
1.2
=
=
+
−s + 10
(s + 2)(s + 20)
s + 2 s + 20
io (t) = 1.2[e−20t − e−2t]u(t) A
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–42
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.39 [a]
150
= (25 + 0.9375s)I1 + 0.625sI2
s
0 = 0.625sI2 + (50 − 1.25s)I1
∆=
0.625s
0.625s
1.25s + 50
150
N1 =
I1 =
0.9375s + 25
0.625s
=
0 1.25s + 50
= 0.78125(s2 + 100s + 1600)
187.5(s + 40)
s
N1
240(s + 40)
=
∆
s(s + 20)(s + 80)
240(s + 40)
(s + 20)(s + 80)
[b] sI1 =
lim sI1 = i1(∞) = 6 A
s→0
lim sI1 = i1 (0) = 0
s→∞
[c] I1 =
K1
K2
K3
+
+
s
s + 20 s + 80
K1 = 6;
K2 = −4;
K3 = −2
i1 (t) = (6 − 4e−20t − 2e−80t )u(t) A
P 13.40 [a] From the solution to Problem 13.39 we have
0.9375s + 25 150
N2 =
I2 =
0.625s
0
= −93.75
−120
K1
K2
=
+
(s + 20)(s + 80)
s + 20 s + 80
K1 =
−120
= −2;
60
K2 =
−120
=2
−60
i2 (t) = (−2e−20t + 2e−80t )u(t) A
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
13–43
[b] Reversing the dot on the 1.25 H coil will reverse the sign of M, thus the
circuit becomes
The two simulanteous equations are
150
= (25 + 0.9375s)I1 − 0.625sI2
s
0 = −0.625sI1 + (1.25s + 50)I2
When these equations are compared to those derived in Problem 13.39
we see the only difference is the algebraic sign of the 0.625s term. Thus
reversing the dot will have no effect on I1 and will reverse the sign of I2 .
Hence,
i2 (t) = (2e−20t − 2e−80t )u(t) A
P 13.41 [a] s-domain equivalent circuit is
i2(0+ ) = −
Note:
[b]
20
= −2 A
10
24
= (120 + 3s)I1 + 3sI2 + 6
s
0 = −6 + 3sI1 + (360 + 15s)I2 + 36
In standard form,
(s + 40)I1 + sI2 = (8/s) − 2
sI1 + (5s + 120)I2 = −10
∆=
N1 =
s + 40
s
s
5s + 120
= 4(s + 20)(s + 60)
(8/s) − 2
s
−10
5s + 120
=
−200(s − 4.8)
s
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13–44
CHAPTER 13. The Laplace Transform in Circuit Analysis
I1 =
[c] sI1 =
N1
−50(s − 4.8)
=
∆
s(s + 20)(s + 60)
−50(s − 4.8)
(s + 20)(s + 60)
lim sI1 = i1 (0+ ) = 0 A
s→∞
lim sI1 = i1(∞) =
s→0
[d] I1 =
(−50)(−4.8)
= 0.2 A
(20)(60)
K1
K2
K3
+
+
s
s + 20 s + 60
K1 =
240
= 0.2;
1200
K3 =
−50(−60) + 240
= 1.35
(−60)(−40)
K2 =
−50(−20) + 240
= −1.55
(−20)(40)
i1 (t) = [0.2 − 1.55e−20t + 1.35e−60t ]u(t) A
P 13.42 [a] Voltage source acting alone:
Vo1 − 60/s V01 s
V01
+
+
=0
10
80
20 + 10s
.·. V01 =
480(s + 2)
s(s + 4)(s + 6)
Vo2 V02 s V02 − 30/s
+
+
=0
10
80
10(s + 2)
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Problems
.·. V02 =
240
s(s + 4)(s + 6)
Vo = Vo1 + Vo2 =
[b] Vo =
13–45
480(s + 2) + 240
480(s + 2.5)
=
s(s + 4)(s + 6)
s(s + 4)(s + 6)
K2
K3
K1
+
+
s
s+4 s+6
(480)(2.5)
= 50;
(4)(6)
K1 =
K2 =
480(−1.5)
= 90;
(−4)(2)
K3 =
480(−3.5)
= −140
(−6)(−2)
vo (t) = [50 + 90e−4t − 140e−6t ]u(t) V
Y11 Y12
P 13.43 ∆ =
Y12 Y22
N2 =
V2 =
= Y11Y22 − Y122
Y11 [(Vg /R1 ) + γC − (ρ/s)]
Y12
(Ig − γC)
N2
∆
Substitution and simplification lead directly to Eq. 13.90.
P 13.44 [a] Vo = −
Zf =
Zf
Vg
Zi
104 (80 × 106 /s)
80 × 106
=
104 + 80 × 106 /s
s + 8000
Zi = 4000 +
Vg =
16,000
s2
.·. Vo =
[b] Vo =
109
4000(s + 4000)
=
62.5s
s
−320 × 106
s(s + 4000)(s + 8000)
K1
K2
K3
+
+
s
s + 4000 s + 8000
K1 =
−20,000(16,000)
= −10
(4000)(8000)
K2 =
−320 × 106
= 20
(−4000)(4000)
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13–46
CHAPTER 13. The Laplace Transform in Circuit Analysis
K3 =
−320 × 106
= −10
(−8000)(−4000)
.·. vo(t) = (−10 + 20e−4000t − 10e−8000t)u(t) V
[c] −10 + 20e−4000ts − 10e−8000ts = −5
.·. 20e−4000ts − 10e−8000ts = 5
Let x = e−4000ts . Then
20x − 10x2 = 5;
Solving,
x = 1±
√
0.5
or x2 − 2x + 0.5 = 0
so
x = 0.2929
.·. e−4000ts = 0.2929;
[d] vg = m tu(t);
Vg =
.·. ts = 306.99 µs
m
s2
Vo =
−20,000m
s(s + 4000)(s + 8000)
K1 =
−20,000m
−20,000m
=
(4000)(8000)
32 × 106
.·. −5 =
−20,000m
32 × 106
.·. m = 8000 V/s
Thus, m must be less than or equal to 8000 V/s to avoid saturation.
P 13.45 [a] Let va be the voltage across the 0.5 µF capacitor, positive at the upper
terminal.
Let vb be the voltage across the 100 kΩ resistor, positive at the upper
terminal.
Also note
106
2 × 106
106
4 × 106
0.5
=
and
=
;
Vg =
0.5s
s
0.25s
s
s
sVa
Va − (0.5/s)
Va
+
+
=0
6
s × 10
200,000
200,000
sVa + 10Va −
Va =
5
+ 10Va = 0
s
5
s(s + 20)
0 − Va
(0 − Vb )s
+
=0
200,000
4 × 106
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Problems
. ·.
Vb = −
13–47
20
−100
Va = 2
s
s (s + 20)
Vb
(Vb − 0)s (Vb − Vo )s
+
+
=0
100,000
4 × 106
4 × 106
40Vb + sVb + sVb = sVo
. ·.
2(s + 20)Vb
Vo =
;
s
−100
−200
Vo = 2
=
3
s
s3
[b] vo(t) = −100t2 u(t) V
[c] −100t2 = −4;
t = 0.2 s = 200 ms
P 13.46
Va − 0.016/s
Va s
(Va − Vo )s
+
+
=0
6
2000
50 × 10
50 × 106
(0 − Va )s (0 − Vo )
+
=0
50 × 106
10,000
Va =
−5000Vo
s
−5000Vo
0.016
.·.
(2s + 25,000) − sVo = 25,000
s
s
Vo =
−4000
(s + 5000 − j10,000)(s + 5000 + j10,000)
K1 =
−400
= j0.02 = 0.02/90◦
j10,000
vo (t) = 40e−5000t cos(10,000t + 90◦ ) = −40e−5000t sin(10,000t)u(t) mV
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13–48
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.47 [a]
Vp =
50/s
50
Vg2 =
Vg2
5 + 50/s
5s + 50
Vp − 40/s Vp − Vo Vp − Vo
+
+
=0
20
5
100/s
Vp
1
1
s
1
s
+ +
− Vo
+
20 5 100
5 100
=
2
s
s + 25
50
16 2
1
s
− = Vo
+
100
5s + 50 s
s
5 100
"
= Vo
s + 20
100
#
100
16(s + 25)
2
−40s + 2000
Vo =
−
=
s + 20 10(s + 10)(s) s
s(s + 10)(s + 20)
=
K1
K2
K3
+
+
s
s + 10 s + 20
K1 = 10;
. ·.
K2 = −24;
K3 = 14
vo (t) = [10 − 24e−10t + 14e−20t]u(t) V
[b] 10 − 24x + 14x2 = 5
14x2 − 24x + 5 = 0
x = 0 or 0.242691
e−10t = 0.242691
.·.
t = 141.60 ms
P 13.48 Let vo1 equal the output voltage of the first op amp. Then
Vo1 =
−Zf 1
Vg
ZA1
ZA1 = 25,000 +
where Zf 1 = 25 × 103 Ω
25,000(20 × 104 /s)
25,000 + (20 × 104 /s)
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Problems
=
.·.
13–49
25,000(s + 16)
Ω
(s + 8)
Vo1 =
−(s + 8)
Vg
(s + 16)
.·.
vg (t) = 16u(t) mV;
Vg =
16 × 10−3
s
Vo1 =
−16 × 10−3 (s + 8)
−0.008 −0.008
=
+
s(s + 16)
s
s + 16
.·.
vo1 = −0.008(1 + e−16t ) V
The op amp will saturate when vo1 = ±6 V. Hence, saturation will occur when
−0.008(1 + e−16t) = −6
Thus
t=
e−16t = 749
so
ln 749
= −0.414 s
−16
Thus, the first op amp never saturates. We must investigate the output of the
second op amp:
2 × 108
Ω and ZA2 = 25,000 Ω
s
Vo =
−Zf 2
Vo1
ZA2
.·.
−8000
−8000 −(s + 8)
Vo =
Vo1 =
Vg
s
s
(s + 16)
where Zf 2 =
"
=
#
8000(s + 8)
Vg
s(s + 16)
.·.
vg (t) = 16u(t) mV;
Vg =
16 × 10−3
s
Vo =
128(s + 8)
K1 K2
K3
= 2 +
+
2
s (s + 16)
s
s
s + 16
K1 =
128(8)
= 64
16
d s+8
K2 = 128
ds s + 16
=4
s=0
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13–50
CHAPTER 13. The Laplace Transform in Circuit Analysis
K3 =
128(−8)
= −4
256
vo (t) = [64t + 4 − 4e−16t ]u(t) V
The op amp will saturate when vo = ±6 V. Hence, saturation will occur when
64t + 4 − 4e−16t = 6 or 16t − 0.5 = e−16t
This equation can be solved by trial and error. First note that t > 0.5/16 or
t > 31.25 ms.
Try 40 ms:
0.64 − 0.5 = 0.14;
e−0.64 = 0.53
Try 50 ms:
0.80 − 0.5 = 0.30;
e−0.80 = 0.45
Try 60 ms:
0.96 − 0.5 = 0.46;
e−0.96 = 0.38
Further trial and error gives
tsat ∼
= 56.5 ms
P 13.49 [a]
4
100,000Vi
Vi = Vo +
20
100,000 + (4 × 108 /s)
0.2Vi −
. ·.
sVi
= Vo
s + 4000
Vo
−0.8(s − 1000)
= H(s) =
Vi
(s + 4000)
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Problems
13–51
[b] −z1 = 1000 rad/s
−p1 = −4000 rad/s
P 13.50 [a]
Vo
1/sC
1
=
=
Vi
R + 1/sC
RCs + 1
H(s) =
[b]
(1/RC)
250
=
;
s + (1/RC)
s + 250
−p1 = −250 rad/s
Vo
R
RCs
s
=
=
=
Vi
R + 1/sC
RCs + 1
s + (1/RC)
s
;
z1 = 0, −p1 = −250 rad/s
s + 250
Vo
sL
s
s
[c]
=
=
=
Vi
R + sL
s + R/L
s + 8000
=
z1 = 0;
[d]
−p1 = −8000 rad/s
Vo
R
R/L
8000
=
=
=
Vi
R + sL
s + (R/L)
s + 8000
−p1 = −8000 rad/s
[e]
Vo s
Vo
Vo − Vi
+
+
=0
6
4 × 10
10,000
40,000
sVo + 400Vo + 100Vo = 100Vi
H(s) =
Vo
100
=
Vi
s + 500
−p1 = −500 rad/s
P 13.51 [a]
[b]
1/sC
1
1/RC
=
=
R + 1/sC
RsC + 1
s + 1/RC
There are no zeros, and a single pole at −1/RC rad/sec.
R
R/L
=
R + sL
s + R/L
There are no zeros, and a single pole at −R/L rad/sec.
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13–52
CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] There are several possible solutions. One is
R = 10 Ω;
L = 10 mH;
C = 100 µF
R
RsC
s
=
=
R + 1/sC
RsC + 1
s + 1/RC
There is a single zero at 0 rad/sec, and a single pole at −1/RC rad/sec.
sL
s
[b]
=
R + sL
s + R/L
There is a single zero at 0 rad/sec, and a single pole at −R/L rad/sec.
P 13.52 [a]
[c] There are several possible solutions. One is
R = 100 Ω;
P 13.53 [a]
L = 10 mH;
C = 1 µF
R
(R/L)s
= 2
1/sC + sL + R
s + (R/L)s + 1/LC
There is a single zero at 0 rad/sec, and two poles:
p1 = −(R/2L) +
q
(R/2L)2 − (1/LC);
p2 = −(R/2L) −
[b] There are several possible solutions. One is
R = 250 Ω;
L = 10 mH;
q
(R/2L)2 − (1/LC)
C = 1 µF
These component values yield the following poles:
−p1 = −5000 rad/sec
and
− p2 = −20,000 rad/sec
[c] There are several possible solutions. One is
R = 200 Ω;
L = 10 mH;
C = 1 µF
These component values yield the following poles:
−p1 = −10,000 rad/sec
and
− p2 = −10,000 rad/sec
[d] There are several possible solutions. One is
R = 120 Ω;
L = 10 mH;
C = 1 µF
These component values yield the following poles:
−p1 = −6000 + j8000 rad/sec
P 13.54 [a] Zi = 1000 +
Zf =
and
− p2 = −6000 − j8000 rad/sec
5 × 106
1000(s + 5000)
=
s
s
40 × 106
40 × 106
k40,000 =
s
s + 1000
H(s) = −
Zf
−40 × 106 /(s + 1000)
−40,000s
=
=
Zi
1000(s + 5000)/s
(s + 1000)(s + 5000)
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Problems
[b] Zero at z1 = 0;
13–53
Poles at −p1 = −1000 rad/s and −p2 = −5000 rad/s
P 13.55 [a] Let R1 = 250 kΩ; R2 = 125 kΩ; C2 = 1.6 nF; and Cf = 0.4 nF. Then
(R2 + 1/sC2 )1/sCf
(s + 1/R2 C2 )
=
Zf = C +C
1
1
R2 + sC2 + sCf
Cf s s + C22Cf Rf2
1
= 2.5 × 109
Cf
1
62.5 × 107
=
= 5000 rad/s
R2 C2
125 × 103
C2 + Cf
2 × 10−9
=
= 25,000 rad/s
C2Cf R2
(0.64 × 10−18 )(125 × 103 )
.·. Zf =
2.5 × 109 (s + 5000)
Ω
s(s + 25,000)
Zi = R1 = 250 × 103 Ω
H(s) =
Vo
−Zf
−104 (s + 5000)
=
=
Vg
Zi
s(s + 25,000)
[b] −z1 = −5000 rad/s
−p1 = 0;
−p2 = −25,000 rad/s
P 13.56 [a]
Va − Vg
sVa
(Va − Vo )s
+
+
=0
6
1000
5 × 10
5 × 106
5000Va − 5000Vg + 2sVa − sVo = 0
(5000 + 2s)Va − sVo = 5000Vg
(0 − Va )s 0 − Vo
+
=0
5 × 106
5000
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13–54
CHAPTER 13. The Laplace Transform in Circuit Analysis
.·.
−sVa − 1000Vo = 0;
Va −
−1000
Vo
s
−1000
(2s + 5000)
Vo − sVo = 5000Vg
s
1000Vo (2s + 5000) + s2 Vo = −5000sVg
Vo (s2 + 2000s + 5 × 106 ) = −5000sVg
Vo
−5000s
= 2
Vg
s + 2000s + 5 × 106
√
s1,2 = −1000 ± 106 − 5 × 106 = −1000 ± j2000
Vo
−5000s
=
Vg
(s + 1000 − j2000)(s + 1000 + j2000)
[b] z1 = 0;
−p1 = −1000 + j2000;
−p2 = −1000 − j2000
P 13.57 [a]
Vo
Vo
+
+ Vo (10−7 )s = Ig
5000 0.2s
.·. Vo =
Ig =
10 × 106 s
· Ig
s2 + 2000s + 50 × 106
0.1s
;
s2 + 108
.·. H(s) =
Io = 10−7 sVo
s2
s2 + 2000s + 50 × 106
(s2 )(0.1s)
[b] Io =
(s + 1000 − j7000)(s + 1000 + j7000)(s2 + 108 )
Io =
0.1s3
(s + 1000 − j7000)(s + 1000 + j7000)(s + j104 )(s − j104 )
[c] Damped sinusoid of the form
Me−1000t cos(7000t + θ1)
[d] Steady-state sinusoid of the form
N cos(104 t + θ2 )
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
[e] Io =
13–55
K1
K1∗
K2
K2∗
+
+
+
s + 1000 − j7000 s + 1000 + j7000 s − j104 s + j104
K1 =
0.1(−1000 + j7000)3
= 46.9 × 10−3 / − 140.54◦
(j14,000)(−1000 − j5000)(−1000 + j17,000)
K2 =
0.1(j104 )3
= 92.85 × 10−3 /21.8◦
(j20,000)(1000 + j3000)(1000 + j17,000)
io (t) = [93.8e−1000t cos(7000t − 140.54◦ ) + 185.7 cos(104 t + 21.8◦ )] mA
Test:
Z=
1
;
Y
. ·. Z =
Y =
1
1
1
2 + j5
+
+
=
5000 j2000 −j1000
10,000
10,000
= 1856.95/ − 68.2◦ Ω
2 + j5
Vo = Ig Z = (0.1/0◦ )(1856.95/ − 68.2◦ ) = 185.695/ − 68.2◦ V
Io = (10−7 )(j104 )Vo = 185.7/21.8◦ mA
ioss = 185.7 cos(104 t + 21.8◦ ) mA(checks)
P 13.58
Vg = 25sI1 − 35sI2
16 × 106
0 = −35sI1 + 50s + 10,000 +
I2
s
!
∆=
N2 =
25s
−35s
−35s 50s + 10,000 + 16 × 106 /s
25s Vg
−35s 0
= 25(s + 2000)(s + 8000)
= 35sVg
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–56
CHAPTER 13. The Laplace Transform in Circuit Analysis
I2 =
N2
35sVg
=
∆
25(s + 2000)(s + 8000)
H(s) =
.·.
I2
1.4s
=
Vg
(s + 2000)(s + 8000)
z1 = 0;
−p1 = −2000 rad/s;
−p2 = −8000 rad/s
P 13.59 [a]
2000(Io − Ig ) + 8000Io + µ(Ig − Io )(2000) + 2sIo = 0
. ·. I o =
1000(1 − µ)
Ig
s + 1000(5 − µ)
.·. H(s) =
1000(1 − µ)
s + 1000(5 − µ)
[b] µ < 5
[c]
µ H(s)
Io
−3 4000/(s + 8000)
20,000/s(s + 8000)
0 1000/(s + 5000)
5000/s(s + 5000)
4 −3000/(s + 1000)
−15,000/s(s + 1000)
5 −4000/s
−20,000/s2
6 −5000/(s − 1000)
µ = −3:
Io =
2.5
2.5
−
;
s
(s + 8000)
−25,000/s(s − 1000)
io = [2.5 − 2.5e−8000t]u(t) A
µ = 0:
1
1
−
;
s s + 5000
µ = 4:
Io =
Io =
−15
15
−
;
s
s + 1000
io = [1 − e−5000t]u(t) A
io = [−15 + 15e−1000t ]u(t) A
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
13–57
µ = 5:
−20,000
;
s2
µ = 6:
Io =
Io =
io = −20,000t u(t) A
25
25
−
;
s
s − 1000
io = 25[1 − e1000t]u(t) A
P 13.60 [a]
y(t) = 0
t<0
0 ≤ t ≤ 10 :
y(t) =
Z
t
625 dλ = 625t
0
Z
10 ≤ t ≤ 20 :
y(t) =
20 ≤ t < ∞ :
y(t) = 0
10
t−10
625 dλ = 625(10 − t + 10) = 625(20 − t)
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13–58
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b]
y(t) = 0
t<0
0 ≤ t ≤ 10 :
t
Z
y(t) =
312.5 dλ = 312.5t
0
10 ≤ t ≤ 20 :
y(t) =
Z
10
y(t) =
Z
10
20 ≤ t ≤ 30 :
30 ≤ t < ∞ :
y(t) = 0
0
312.5 dλ = 3125
t−20
312.5 dλ = 312.5(30 − t)
[c]
y(t) = 0
0≤t≤1:
t<0
y(t) =
Z
0
t
625 dλ = 625t
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
1 ≤ t ≤ 10 :
y(t) =
Z
1
0
Z
10 ≤ t ≤ 11 :
y(t) =
11 ≤ t < ∞ :
y(t) = 0
13–59
625 dλ = 625
1
t−10
625 dλ = 625(11 − t)
P 13.61 [a] 0 ≤ t ≤ 40:
y(t) =
Z
0
t
t
(10)(1)(dλ) = 10λ = 10t
0
40 ≤ t ≤ 80:
y(t) =
Z
t ≥ 80 :
40
40
t−40
(10)(1)(dλ) = 10λ
t−40
= 10(80 − t)
y(t) = 0
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–60
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] 0 ≤ t ≤ 10:
y(t) =
Z
t
t
40 dλ = 40λ = 40t
0
0
10 ≤ t ≤ 40:
y(t) =
Z
t
t
t−10
40 dλ = 40λ
= 400
t−10
40 ≤ t ≤ 50:
y(t) =
Z
t ≥ 50 :
40
40
t−10
40 dλ = 40λ
t−10
= 40(50 − t)
y(t) = 0
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
13–61
[c] The expressions are
0≤t≤1:
y(t) =
1 ≤ t ≤ 40 :
y(t) =
t
t
Z
400 dλ = 400λ = 400t
0
0
Z
t
t
t−1
Z
40 ≤ t ≤ 41 :
y(t) =
41 ≤ t < ∞ :
y(t) = 0
400 dλ = 400λ
40
40
t−1
= 400
t−1
400 dλ = 400λ
t−1
= 400(41 − t)
[d]
[e] Yes, note that h(t) is approaching 40δ(t), therefore y(t) must approach
40x(t), i.e.
y(t) =
Z
0
t
h(t − λ)x(λ) dλ →
Z
0
t
40δ(t − λ)x(λ) dλ
→ 40x(t)
This can be seen in the plot, e.g., in part (c), y(t) ∼
= 40x(t).
Vo
1
=
;
Vi
s+1
For 0 ≤ t ≤ 1:
P 13.62 H(s) =
vo =
Z
0
h(t) = e−t
t
e−λ dλ = (1 − e−t) V
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13–62
CHAPTER 13. The Laplace Transform in Circuit Analysis
For 1 ≤ t ≤ ∞:
vo =
Z
t
t−1
P 13.63 H(s) =
e−λ dλ = (e − 1)e−t V
Vo
s
1
=
= 1−
;
Vi
s+1
s+1
h(t) = δ(t) − e−t
h(λ) = δ(λ) − e−λ
For 0 ≤ t ≤ 1:
vo =
Z
t
0
[δ(λ) − e−λ ] dλ = [1 + e−λ ] |t0= e−t V
For 1 ≤ t ≤ ∞:
vo =
Z
t
t
t−1
(−e−λ ) dλ = e−λ
t−1
= (1 − e)e−t V
P 13.64 [a] From Problem 13.50(a)
H(s) =
250
s + 250
h(λ) = 250e−250λ
0 ≤ t ≤ 4 ms:
vo =
Z
t
16(250)e−250λ dλ = 16(1 − e−250t) V
0
4 ms ≤ t ≤ ∞:
vo =
Z
t
t−0.004
16(250)e−250λ dλ = 16(e − 1)e−250t V
[b]
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Problems
2500
s + 2500
0 ≤ t ≤ 4 ms:
.·. h(λ) = 2500e−2500λ
P 13.65 [a] H(s) =
vo =
13–63
t
Z
16(2500)e−2500λ dλ = 16(1 − e−2500t) V
0
4 ms ≤ t ≤ ∞:
vo =
t
Z
t−0.004
16(2500)e−2500λ dλ = 16(e10 − 1)e−2500t V
[b] decrease
[c] The circuit with R = 10 kΩ.
P 13.66 [a]
vo =
Z
0
t
10(10e−4λ ) dλ
= 100
e−4λ
−4
t
0
= −25[e−4t − 1]
= 25(1 − e−4t ) V,
0≤t≤∞
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13–64
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b]
0 ≤ t ≤ 0.5:
vo =
Z
t
0
t
100(1 − 2λ) dλ = 100(λ − λ2 ) = 100t(1 − t)
0
0.5 ≤ t ≤ ∞:
vo =
Z
0.5
0
100(1 − 2λ) dλ = 100(λ − λ2 )
0.5
= 25
0
[c]
P 13.67 [a] −1 ≤ t ≤ 4:
vo =
Z
t+1
0
10λ dλ = 5λ2
Z
= 5t2 + 10t + 5 V
0
4 ≤ t ≤ 9:
vo =
t+1
t+1
t−4
10λ dλ = 5λ2
t−4
9 ≤ t ≤ 14:
vo = 10
Z
t+1
10
t−4
λ dλ + 10
Z
= 50t − 75 V
t+1
10
10 dλ
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
10
= 5λ2
t+1
+100λ
10
t−4
13–65
= −5t2 + 140t − 480 V
14 ≤ t ≤ 19:
vo = 100
Z
t+1
t−4
dλ = 500 V
19 ≤ t ≤ 24:
vo =
Z
20
t−4
100λ dλ +
Z
t+2
10(30 − λ) dλ
20
20
t+1
= 100λ
+300λ
20
t−2
t+2
−5λ2
20
= −5t2 + 190t − 1305 V
24 ≤ t ≤ 29:
vo = 10
Z
t+1
t+1
(30 − λ) dλ = 300λ
t−4
t−4
−5λ2
t+1
t−4
= 1575 − 50t V
29 ≤ t ≤ 34:
vo = 10
Z
30
30
t−4
(30 − λ) dλ = 300λ
t−4
−5λ2
30
t−2
= 5t2 − 340t + 5780 V
Summary:
vo = 0
− ∞ ≤ t ≤ −1
vo = 5t2 + 10t + 5 V
vo = 50t − 75 V
−1≤t≤4
4≤t≤9
vo = −5t2 + 140t − 480 V
vo = 500 V
9 ≤ t ≤ 14
14 ≤ t ≤ 19
vo = −5t2 + 190t − 1305 V
vo = 1575 − 50t V
19 ≤ t ≤ 24
24 ≤ t ≤ 29
vo = 5t2 − 340t + 5780 V
vo = 0
29 ≤ t ≤ 34
34 ≤ t ≤ ∞
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–66
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b]
2
P 13.68 [a] h(λ) = λ
5
0≤λ≤5
2
h(λ) = 4 − λ
5
5 ≤ λ ≤ 10
0 ≤ t ≤ 5:
2
λ dλ = 2t2
0 5
5 ≤ t ≤ 10:
Z
t
vo = 10
vo = 10
Z
5
0
4λ2
=
2
2
λ dλ + 10
5
5
t
+40λ
0
5
Z
t
5
2
4 − λ dλ
5
4λ2
−
2
t
5
= −100 + 40t − 2t2
10 ≤ t ≤ ∞:
vo = 10
Z
5
0
2
λ dλ + 10
5
Z
5
10 2
4 − λ dλ
5
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
=
4λ2
2
5
10
+40λ
0
5
−
4λ2
2
13–67
10
5
= 50 + 200 − 150 = 100
vo = 2t2 V
0≤t≤5
vo = 40t − 100 − 2t2 V
vo = 100 V
5 ≤ t ≤ 10
10 ≤ t ≤ ∞
[b]
1
[c] Area = (10)(2) = 10
2
5
h(λ) = λ
2
0 ≤ t ≤ 2:
vo = 10
Z
t
0
1
(4)h = 10 so h = 5
2
0≤λ≤2
5
h(λ) = 10 − λ
2
. ·.
2≤λ≤4
5
λ dλ = 12.5t2
2
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–68
CHAPTER 13. The Laplace Transform in Circuit Analysis
2 ≤ t ≤ 4:
vo = 10
=
Z
2
0
5
λ dλ + 10
2
25λ2
2
2
Z
5
10 − λ dλ
2
t
2
t
+100λ
0
2
−
25λ2
2
t
2
= −100 + 100t − 12.5t2
4 ≤ t ≤ ∞:
vo = 10
Z
2
0
Z 4
5
5
λ dλ + 10
10 − λ dλ
2
2
2
25λ2
=
2
2
4
+100λ
0
2
25λ2
−
2
4
2
= 50 + 200 − 150 = 100
vo = 12.5t2 V
0≤t≤2
vo = 100t − 100 − 12.5t2 V
vo = 100 V
2≤t≤4
4≤t≤∞
[d] The waveform in part (c) is closer to replicating the input waveform
because in part (c) h(λ) is closer to being an ideal impulse response.
That is, the area was preserved as the base was shortened.
P 13.69 [a] Vo =
. ·.
16
Vg
20
H(s) =
Vo
4
=
Vg
5
h(λ) = 0.8δ(λ)
[b]
0 < t < 0.5 s :
vo =
Z
0
t
75[0.8δ(λ)] dλ = 60 V
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
13–69
0.5 s ≤ t ≤ 1.0 s:
vo =
Z
t−0.5
0
−75[0.8δ(λ)] dλ = −60 V
1s < t < ∞ :
vo = 0
[c]
Yes, because the circuit has no memory.
P 13.70 [a]
Vo − Vg Vo s Vo
+
+
=0
5
4
20
(5s + 5)Vo = 4Vg
H(s) =
Vo
0.8
=
;
Vg
s+1
h(λ) = 0.8e−λ u(λ)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–70
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b]
0 ≤ t ≤ 0.5 s;
vo =
Z
0
t
75(0.8e−λ ) dλ = 60
vo = 60 − 60e−t V,
e−λ
−1
t
0
0 ≤ t ≤ 0.5 s
0.5 s ≤ t ≤ 1 s:
vo =
Z
t−0.5
(−75)(0.8e
0
= −60
e−λ
−1
−λ
t−0.5
+60
0
) dλ +
e−λ
−1
Z
t
t−0.5
t
t−0.5
= 120e−(t−0.5) − 60e−t − 60 V,
1 s ≤ t ≤ ∞;
vo =
Z
t−0.5
t−1
= −60
(−75)(0.8e−λ ) dλ +
e−λ
−1
t−0.5
+60
t−1
e−λ
−1
75(0.8e−λ ) dλ
Z
0.5 s ≤ t ≤ 1 s
t
t−0.5
75(0.8e−λ ) dλ
t
t−0.5
= 120e−(t−0.5) − 60e−(t−1) − 60e−t V,
1s ≤ t ≤ ∞
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
13–71
[c]
[d] No, the circuit has memory because of the capacitive storage element.
P 13.71 vi = 25 sin 10λ [u(λ) − u(λ − π/10)]
H(s) =
32
s + 32
h(λ) = 32e−32λ
h(t − λ) = 32e−32(t − λ) = 32e−32t e32λ
vo = 800e
−32t
= 800e
−32t
Z
t
"
e32λ
(32 sin 10λ − 10 cos 10λ
322 + 102
0
e32λ sin 10λ dλ
=
800e−32t 32t
[e (32 sin 10t − 10 cos 10t) + 10]
1124
=
800
[32 sin 10t − 10 cos 10t + 10e−32t ]
1124
t
0
#
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–72
CHAPTER 13. The Laplace Transform in Circuit Analysis
vo (0.075) = 10.96 V
16s
0.8s
2
1.6
P 13.72 H(s) =
=
= 0.8 1 −
= 0.8 −
40 + 4s + 16s
s+2
s+2
s+2
h(λ) = 0.8δ(λ) − 1.6e−2λ u(λ)
Z
vo =
t
0
75[0.8δ(λ) − 1.6e
= 60 − 120
e−2λ
−2
−2λ
] dλ =
Z
0
t
60δ(λ) dλ − 120
Z
t
0
e−2λ dλ
t
0
= 60 + 60(e−2t − 1)
= 60e−2t u(t) V
P 13.73
Vo =
5 × 103 Ig
(20 × 103 )
25 × 103 + 2.5 × 106 /s
Vo
4000s
= H(s) =
Ig
s + 100
H(s) = 4000 1 −
100
4 × 105
= 4000 −
s + 100
s + 100
h(t) = 4000δ(t) − 4 × 105 e−100t
vo =
Z
+
10−3
0
Z
(−20 × 10−3 )[4000δ(λ) − 4 × 105 e−100λ] dλ
5×10−3
10−3
(10 × 10−3 )[−4 × 105 e−100λ] dλ
Z
= −80 + 8000
10−3
0
Z
e−100λ dλ − 4000
5×10−3
10−3
e−100λ dλ
= −80 − 80(e−0.1 − 1) + 40(e−0.5 − e−0.1)
= 40e−0.5 − 120e−0.1 = −84.32 V
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
13–73
Alternate:
Ig =
Z
4×10−3
(10 × 10−3 )e−st dt +
0
Z
6×10−3
4×10−3
(−20 × 10−3 )e−st dt
10 30 −4×10−3 s 20 −6×10−3 s
=
− e
+ e
× 10−3
s
s
s
40
−3
−3
[1 − 3e−4×10 s + 2e−6×10 s ]
s + 100
Vo = Ig H(s) =
−3 s
40
120e−4×10
=
−
s + 100
s + 100
80e−6×10 s
+
]
s + 100
vo (t) = 40e−100t − 120e−100(t−4×10
+80e−100(t−6×10
−3 )
−3
−3 )
u(t − 4 × 10−3 )
u(t − 6 × 10−3 )
vo (5 × 10−3 ) = 40e−0.5 − 120e−0.1 + 80(0) = −84.32 V
P 13.74 [a] H(s) =
=
(checks)
Vo
1/LC
= 2
Vi
s + (R/L)s + (1/LC)
s2
100
100
=
+ 20s + 100
(s + 10)2
h(λ) = 100λe−10λ u(λ)
0 ≤ t ≤ 0.5:
Z
vo = 500
t
0
λe−10λ dλ
(
e−10λ
= 500
(−10λ − 1)
100
t
0
)
= 5[1 − e−10t(10t + 1)]
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–74
CHAPTER 13. The Laplace Transform in Circuit Analysis
0.5 ≤ t ≤ ∞:
Z
vo = 500
t
t−0.5
λe−10λ dλ
(
e−10λ
= 500
(−10λ − 1)
100
t
t−0.5
)
= 5e−10t [e5(10t − 4) − 10t − 1]
[b]
P 13.75 [a] Io =
Vo
Vo s
Vo (s + 50)
+
=
5
6
10
5 × 10
5 × 106
Vo
5 × 106
= H(s) =
Ig
s + 50
h(λ) = 5 × 106 e−50λu(λ)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
0 ≤ t ≤ 0.1 s:
vo =
t
Z
0
(50 × 10−6 )(5 × 106 )e−50λ dλ = 250
e−50λ
−50
13–75
t
0
= 5(1 − e−50t) V
0.1 s ≤ t ≤ 0.2 s:
vo =
t−0.1
Z
0
+
Z
t
t−0.1
= −250
h
(−50 × 10−6 )(5 × 106 e−50λ dλ)
(50 × 10−6 )(5 × 106 e−50λ dλ)
e−50λ
−50
t−0.1
+250
0
i
e−50λ
−50
t
t−0.1
h
= 5 e−50(t−0.1) − 1 − 5 e−50t − e−50(t−0.1)
vo = [10e−50(t−0.1) − 5e−50t − 5] V
i
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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13–76
CHAPTER 13. The Laplace Transform in Circuit Analysis
0.2 s ≤ t ≤ ∞:
vo =
Z
t−0.1
t−0.2
"
− 250e−50λ dλ +
t−0.1
= 5e
−50λ
t−0.2
Z
t
t−0.1
t
−5e
−50λ
t−0.1
250e−50λ dλ
#
vo = [10e−50(t−0.1) − 5e−50(t−0.2) − 5e−50t ] V
[b] Io =
Vo s
s
5 × 106 Ig
=
·
5 × 106
5 × 106
s + 50
Io
s
50
= H(s) =
= 1−
Ig
s + 50
s + 50
h(λ) = δ(λ) − 50e−50λ
0 < t < 0.1 s:
io =
Z
t
0
(50 × 10−6 )[δ(λ) − 50e−50λ ] dλ
= 50 × 10−6 − 25 × 10−3
e−50λ
−50
t
0
= 50 × 10−6 + 50 × 10−6 [e−50t − 1] = 50e−50t µA
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Problems
13–77
0.1 s < t < 0.2 s:
io =
Z
+
t−0.1
(−50 × 10−6 )[δ(λ) − 50e−50λ ] dλ
0
Z
t
t−0.1
(50 × 10−6 )(−50e−50λ) dλ
= −50 × 10−6 + 2.5 × 10−3
e−50λ
−50
t−0.1
0
−2.5 × 10−3
e−50λ
−50
t
t−0.1
= −50 × 10−6 − 50 × 10−6 e−50(t−0.1) + 50 × 10−6
+50 × 10−6 e−50t − 50 × 10−6 e−50(t−0.1)
= 50e−50t − 100e−50(t−0.1) µA
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13–78
CHAPTER 13. The Laplace Transform in Circuit Analysis
0.2 s < t < ∞:
io =
Z
+
t−0.1
t−0.2
Z
(−50 × 10−6 )(−50e−50λ) dλ
t
t−0.1
(50 × 10−6 )(−50e−50λ) dλ
= 50e−50t − 100e−50(t−0.1) + 50e−50(t−0.2) µA
[c] At t = 0.1− :
vo = 5(1 − e−5 ) = 4.97 V;
i100kΩ =
4.97
= 49.66 µA
0.1
.·. io = 50 − 49.66 = 0.34 µA
From the solution for io we have
io (0.1− ) = 50e−5 = 0.34 µA (checks)
At t = 0.1+ :
vo (0.1+ ) = vo (0.1− ) = 4.97 V
i100kΩ = 49.66 µA
.·. io(0.1+ ) = −(50 + 49.66) = −99.66 µA
From the solution for io we have
io (0.1+ ) = 50e−5 − 100 = 99.66 µA
(checks)
At t = 0.2− :
vo = 10e−5 − 5e−10 − 5 = −4.93 V
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Problems
13–79
i100kΩ = 49.33 µA
io = −50 + 49.33 = −0.67 µA
From the solution for io ,
vo (0.2− ) = 50e−10 − 100e−5 = −0.67 µA (checks)
At t = 0.2+ :
vo (0.2+ ) = vo (0.2− ) = −4.93 V;
i100kΩ = −49.33 µA
io = 0 + 49.33 = 49.33 µA
From the solution for io ,
io (0.2+ ) = 50e−10 − 100e−5 + 50 = 49.33 µA(checks)
P 13.76 [a] Y (s) =
Z
∞
0
y(t)e−st dt
Z
∞
Y (s) =
=
Z
∞Z
∞
=
Z
0
0
0
e−st
∞
Z
h(λ)x(t − λ) dλ dt
0
∞
0
e−st h(λ)x(t − λ) dλ dt
h(λ)
Z
∞
e−st x(t − λ) dt dλ
0
But x(t − λ) = 0 when t < λ.
Therefore Y (s) =
Let u = t − λ;
Z
∞
Y (s) =
Z
∞
=
Z
∞
=
0
0
0
h(λ)
Z
∞
h(λ)
0
du = dt;
Z
Z
∞
λ
e−st x(t − λ) dt dλ
u = 0 when t = λ;
u = ∞ when t = ∞.
∞
0
e−s(u+λ) x(u) du dλ
h(λ)e−sλ
Z
∞
0
e−su x(u) du dλ
h(λ)e−sλ X(s) dλ = H(s) X(s)
Note on x(t − λ) = 0,
t<λ
We are using one-sided Laplace transforms; therefore h(t) and x(t) are
assumed zero for t < 0.
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13–80
CHAPTER 13. The Laplace Transform in Circuit Analysis
a
1
a
= ·
= H(s)X(s)
2
s(s + a)
s (s + a)2
[b] F (s) =
.·. h(t) = u(t),
.·. f(t) =
=
Z
x(t) = at e−atu(t)
t
0
(1)aλe
−aλ
t
"
e−aλ
dλ = a
(−aλ − 1)
a2
0
1 −at
1
[e (−at − 1) − 1(−1)] = [1 − e−at − ate−at]
a
a
1 1 −at
=
− e − te−at u(t)
a a
Check:
F (s) =
K0 =
1
;
a
f(t) =
P 13.77 H(j3) =
a
K0
K1
K2
=
+
+
2
2
s(s + a)
s
(s + a)
s+a
K1 = −1;
K2 =
d a
ds s
s=−a
=−
1
a
1
1
− te−at − e−at u(t)
a
a
4(3 + j3)
= 0.42/8.13◦
−9 + j24 + 41
.·. vo (t) = 16.97 cos(3t + 8.13◦ ) V
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Problems
P 13.78 Vo =
13–81
50
20
30(s + 3000)
−
=
s + 8000 s + 5000
(s + 5000)(s + 8000)
30
Vo = H(s)Vg = H(s)
s
.·. H(s) =
s(s + 3000)
(s + 5000)(s + 8000)
H(j6000) =
(j6000)(3000 + j6000)
= 0.52/66.37◦
(5000 + j6000)(8000 + j6000)
.·. vo (t) = 61.84 cos(6000t + 66.37◦ ) V
P 13.79 [a]
Vp =
0.01s
s
Vg =
Vg
80 + 0.01s
s + 8000
Vn = Vp
Vn
Vn − Vo
+
+ (Vn − Vo )8 × 10−9 s = 0
5000
25,000
5Vn + Vn − Vo + (Vn − Vo )2 × 10−4 s = 0
6Vn + 2 × 10−4 sVn = Vo + 2 × 10−4 sVo
2 × 10−4 Vn (s + 30,000) = 2 × 10−4 Vo (s + 5000)
s + 30,000
Vo =
Vi =
s + 5000
H(s) =
s + 30,000
s + 5000
sVg
s + 8000
Vo
s(s + 30,000)
=
Vg
(s + 5000)(s + 8000)
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13–82
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] vg = 0.6u(t);
Vg =
0.6
s
Vo =
0.6(s + 30,000)
K1
K2
=
+
(s + 5000)(s + 8000)
s + 5000 s + 8000
K1 =
0.6(25,000)
= 5;
3000
K2 =
0.6(22,000)
= −4.4
−3000
.·. vo(t) = (5e−5000t − 4.4e−8000t)u(t) V
[c] Vg = 2 cos 10,000t V
H(jω) =
. ·.
j10,000(30,000 + j10,000)
= 2.21/ − 6.34◦
(5000 + j10,000)(8000 + j10,000)
vo = 4.42 cos(10,000t − 6.34◦ ) V
P 13.80 [a] H(s) =
−Zf
Zi
Zf =
(1/Cf )
108
=
s + (1/Rf Cf )
s + 1000
Zi =
Ri [s + (1/Ri Ci )]
10,000(s + 400)
=
s
s
H(s) =
−104 s
(s + 400)(s + 1000)
[b] H(j400) =
−104 (j400)
= 6.565/ − 156.8◦
(400 + j400)(1000 + j400)
vo (t) = 13.13 cos(400t − 156.8◦ ) V
P 13.81 Original charge on C1 ; q1 = V0 C1
The charge transferred to C2 ; q2 = V0 Ce =
V0 C1 C2
C1 + C2
The charge remaining on C1; q10 = q1 − q2 =
Therefore V2 =
q2
V0 C1
=
C2
C1 + C2
and
V1 =
V0 C12
C1 + C2
q10
V0 C1
=
C1
C1 + C2
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Problems
13–83
P 13.82 [a] The s-domain circuit is
The node-voltage equation is
Therefore V =
ρR
s + (R/Le )
V
V
V
ρ
+ +
=
sL1 R sL2
s
where Le =
L1 L2
L1 + L2
Therefore v = ρRe−(R/Le)t u(t) V
[b] I1 =
V
V
ρ[s + (R/L2 )]
K0
K1
+
=
=
+
R sL2
s[s + (R/Le )]
s
s + (R/Le )
K0 =
ρL1
;
L1 + L2
K1 =
Thus we have i1 =
[c] I2 =
ρL2
L1 + L2
ρ
[L1 + L2 e−(R/Le)t]u(t) A
L1 + L2
V
(ρR/L2 )
K2
K3
=
=
+
sL2
s[s + (R/Le )]
s
s + (R/Le )
K2 =
ρL1
;
L1 + L2
K3 =
Therefore i2 =
−ρL1
L1 + L2
ρL1
[1 − e−(R/Le)t ]u(t)
L1 + L2
[d] λ(t) = L1 i1 + L2i2 = ρL1
P 13.83 [a] As R → ∞, v(t) → ρLe δ(t) since the area under the impulse generating
function is ρLe .
ρL1
i1 (t) →
u(t) A as R → ∞
L1 + L2
i2 (t) →
ρL1
u(t) A as R → ∞
L1 + L2
[b] The s-domain circuit is
V
V
ρ
+
= ;
sL1 sL2
s
therefore V =
ρL1 L2
= ρLe
L1 + L2
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13–84
CHAPTER 13. The Laplace Transform in Circuit Analysis
Therefore v(t) = ρLe δ(t)
I1 = I2 =
V
ρL1
=
sL2
L1 + L2
Therefore i1 = i2 =
1
s
ρL1
u(t) A
L1 + L2
P 13.84 [a]
Vo =
=
5
· 16 × 10−3 s
200 + 20 × 10−3 s
4s
40,000
=4−
s + 10,000
s + 10,000
vo (t) = 4δ(t) − 40,000e−10,000tu(t) V
[b] At t = 0 the voltage impulse establishes a current in the inductors; thus
103
iL (0) =
20
Z
0+
0−
5δ(t) dt = 250 A
It follows that since iL (0− ) = 0 that
diL
(0) = 250δ(t)
dt
.·. vo(0) = (16 × 10−3 )(250δ(t)) = 4δ(t)
This agrees with our solution.
At t = 0+ our circuit is
.·. iL(t) = 250e−t/τ A,
t ≥ 0+
τ = L/R = 0.1 ms
.·. iL(t) = 250e−10,000t A,
t ≥ 0+
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Problems
vo (t) = 16 × 10−3
diL
= −40,000e−10,000t V,
dt
13–85
t ≥ 0+
which agrees with our solution.
1/C1
25 × 1010
P 13.85 [a] Z1 =
=
Ω
s + 1/R1 C1
s + 20 × 104
1/C2
6.25 × 1010
Z2 =
=
Ω
s + 1/R2 C2
s + 12,500
V0
V0 − 10/s
+
=0
Z2
Z1
V0 (s + 12,500) V0 (s + 20 × 104 )
10 (s + 20 × 104 )
+
=
6.25 × 1010
25 × 1010
s
25 × 1010
V0 =
2(s + 200,000)
K1
K2
=
+
s(s + 50,000)
s
s + 50,000
K1 =
2(200,000)
=8
50,000
K2 =
2(150,000)
= −6
−50,000
.·. vo = [8 − 6e−50,000t]u(t) V
[b] I0 =
V0
2(s + 200,000)(s + 12,500)
=
Z2
s(s + 50,000)6.25 × 1010
162,500s + 25 × 108
1+
s(s + 50,000)
= 32 × 10
"
= 32 × 10−12
"
−12
K1 = 50,000;
K1
K2
1+
+
s
s + 50,000
#
#
K2 = 112,500
io = 32δ(t) + [1.6 × 106 + 3.6 × 106 e−50,000t]u(t) pA
[c] When
Z1 =
C1 = 64 pF
156.25 × 108
Ω
s + 12,500
V0 (s + 12,500) V0 (s + 12,500)
10 (s + 12,500)
+
=
8
8
625 × 10
156.25 × 10
s 156.25 × 108
.·. V0 + 4V0 =
40
s
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13–86
CHAPTER 13. The Laplace Transform in Circuit Analysis
V0 =
8
s
vo = 8u(t) V
I0 =
V0
8 (s + 12,500)
12,500
−12
=
=
128
×
10
1
+
Z2
s 6.25 × 1010
s
io (t) = 128δ(t) + 1.6 × 106 u(t) pA
P 13.86 Let a =
1
1
=
R1C1
R2C2
Then Z1 =
1
C1 (s + a)
and
Z2 =
1
C2(s + a)
Vo
Vo
10/s
+
=
Z2 Z1
Z1
Vo C2(s + a) + V0 C1 (s + a) = (10/s)C1 (s + a)
10
C1
Vo =
s C1 + C2
Thus, vo is the input scaled by the factor
C1
.
C1 + C2
P 13.87 [a] For t < 0:
Req = 0.8 kΩk4 kΩk16 kΩ = 0.64 kΩ;
i1 (0− ) =
3200
= 0.8 A;
4000
i2(0− ) =
v = 5(640) = 3200 V
3200
= 0.2 A
1600
[b] For t > 0:
i1 + i2 = 0
8(∆i1 ) = 2(∆i2 )
i1 (0− ) + ∆i1 + i2(0− ) + ∆i2 = 0;
∆i2 = −0.8 A;
therefore ∆i1 = −0.2 A
i1 (0+ ) = 0.8 − 0.2 = 0.6 A
[c] i2 (0− ) = 0.2 A
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
13–87
[d] i2(0+ ) = 0.2 − 0.8 = −0.6 A
[e] The s-domain equivalent circuit for t > 0 is
I1 =
0.6
0.006
=
0.01s + 20,000
s + 2 × 106
6
i1 (t) = 0.6e−2×10 t u(t) A
6
[f] i2(t) = −i1(t) = −0.6e−2×10 t u(t) A
[g] V = −0.0064 + (0.008s + 4000)I1 =
= −1.6 × 10−3 −
7200
s + 2 × 106
−0.0016(s + 6.5 × 106 )
s + 2 × 106
6
v(t) = [−1.6 × 10−3 δ(t)] − [7200e−2×10 t u(t)] V
P 13.88 [a] For t < 0,
0.5v1 = 2v2 ;
v1 + v2 = 100;
therefore v1 = 4v2
therefore v1(0− ) = 80 V
[b] v2(0− ) = 20 V
[c] v3 (0− ) = 0 V
[d] For t > 0:
I=
100/s
× 10−6 = 32 × 10−6
3.125/s
i(t) = 32δ(t) µA
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13–88
CHAPTER 13. The Laplace Transform in Circuit Analysis
106 0+
32 × 10−6 δ(t) dt + 80 = −64 + 80 = 16 V
0.5 0−
106 Z 0+
[f] v2(0+ ) = −
32 × 10−6 δ(t) dt + 20 = −16 + 20 = 4 V
2 0−
20
0.625 × 106
[g] V3 =
· 32 × 10−6 =
s
s
Z
[e] v1(0+ ) = −
v3(0+ ) = 20 V
v3 (t) = 20u(t) V;
Check: v1(0+ ) + v2(0+ ) = v3(0+ )
P 13.89 [a]
0.5
106
·
50,000 + 5 × 106 /s s
Vo =
500,000
10
=
50,000s + 5 × 106
s + 100
vo = 10e−100tu(t) V
[b] At t = 0 the current in the 1 µF capacitor is 10δ(t) µA
. ·.
+
6
vo (0 ) = 10
Z
0+
0−
10 × 10−6 δ(t) dt = 10 V
After the impulsive current has charged the 1 µF capacitor to 10 V it
discharges through the 50 kΩ resistor.
Ce =
C1 C2
0.25
=
= 0.2 µF
C1 + C2
1.25
τ = (50,000)(0.2 × 10−6 ) = 10−2
1
= 100 (checks)
τ
Note – after the impulsive current passes the circuit becomes
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
13–89
The solution for vo in this circuit is also
vo = 10e−100tu(t) V
P 13.90 [a] After making a source transformation, the circuit is as shown. The
impulse current will pass through the capacitive branch since it appears
as a short circuit to the impulsive current,
+
6
Therefore vo(0 ) = 10
Z
0+
0−
"
#
δ(t)
dt = 1000 V
1000
Therefore wC = (0.5)Cv 2 = 0.5 J
[b] iL(0+ ) = 0;
therefore wL = 0 J
Vo
Vo
[c] Vo (10−6 )s +
+
= 10−3
250 + 0.05s 1000
Therefore
1000(s + 5000)
Vo = 2
s + 6000s + 25 × 106
=
K1
K1∗
+
s + 3000 − j4000 s + 3000 + j4000
K1 = 559.02/ − 26.57◦ ;
K1∗ = 559.02/26.57◦
vo = [1118.03e−3000t cos(4000t − 26.57◦ )]u(t) V
[d] The s-domain circuit is
Vo s
Vo
Vo
+
+
= 10−3
106 250 + 0.05s 1000
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13–90
CHAPTER 13. The Laplace Transform in Circuit Analysis
Note that this equation is identical to that derived in part [c], therefore
the solution for Vo will be the same.
P 13.91 [a]
20 = sI1 − 0.5sI2
0 = −0.5sI1 + s +
s
∆=
3
I2
s
−0.5s
−0.5s (s + 3/s)
20 −0.5s
N1 =
I1 =
=
= s2 + 3 − 0.25s2 = 0.75(s2 + 4)
= 20s +
0 (s + 3/s)
60
20s2 + 60
20(s2 + 3)
=
=
s
s
s
N1
20(s2 + 3)
80
s2 + 3
=
=
·
∆
s(0.75)(s2 + 4)
3 s(s2 + 4)
K0
K1
K1∗
+
+
s
s − j2 s + j2
"
80 3
K0 =
= 20;
3 4
20
cos 2t u(t) A
3
.·. i1 = 20 +
s
[b] N2 =
#
80 −4 + 3
10
K1 =
= /0◦
3 (j2)(j4)
3
20
= 10s
−0.5s 0
N2
10s
40
s
K1
K1∗
I2 =
=
=
=
+
∆
0.75(s2 + 4)
3 s2 + 4
s − j2 s + j2
40
K1 =
3
j2
j4
!
=
20 ◦
/0
3
40
i2 =
cos 2t u(t) A
3
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Problems
13–91
3
3 40
s
40
K1
K1∗
[c] V0 = I2 =
=
=
+
s
s 3 s2 + 4
s2 + 4
s − j2 s + j2
K1 =
40
= −j10 = 10/90◦
j4
vo = 20 cos(2t − 90◦ ) = 20 sin 2t
vo = [20 sin 2t]u(t) V
[d] Let us begin by noting i1 jumps from 0 to (80/3) A between 0− and 0+
and in this same interval i2 jumps from 0 to (40/3) A. Therefore in the
derivatives of i1 and i2 there will be impulses of (80/3)δ(t) and
(40/3)δ(t), respectively. Thus
di1
80
40
= δ(t) −
sin 2t A/s
dt
3
3
di2
40
80
= δ(t) −
sin 2t A/s
dt
3
3
From the circuit diagram we have
20δ(t) = 1
=
di1
di2
− 0.5
dt
dt
80
40
20δ(t) 40
δ(t) −
sin 2t −
+
sin 2t
3
3
3
3
= 20δ(t)
Thus our solutions for i1 and i2 are in agreement with known circuit
behavior.
Let us also note the impulsive voltage will impart energy into the circuit.
Since there is no resistance in the circuit, the energy will not dissipate.
Thus the fact that i1, i2, and vo exist for all time is consistent with
known circuit behavior.
Also note that although i1 has a dc component, i2 does not. This follows
from known transformer behavior.
Finally we note the flux linkage prior to the appearance of the impulsive
voltage is zero. Now since v = dλ/dt, the impulsive voltage source must
be matched to an instantaneous change in flux linkage at t = 0+ of 20.
For the given polarity dots and reference directions of i1 and i2 we have
λ(0+ ) = L1 i1(0+ ) + Mi1 (0+ ) − L2 i2(0+ ) − Mi2 (0+ )
80
80
40
40
λ(0 ) = 1
+ 0.5
−1
− 0.5
3
3
3
3
+
=
120 60
−
= 20 (checks)
3
3
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13–92
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.92 [a] The circuit parameters are
1202
1202
1202
288
= 12 Ω
Rb =
= 8Ω
Xa =
=
Ω
1200
1800
350
7
The branch currents are
120/0◦
120/0◦
35
35
I1 =
= 10/0◦ A(rms)
I2 =
= −j
= / − 90◦ A(rms)
12
j1440/35
12
12
Ra =
I3 =
120/0◦
= 15/0◦ A(rms)
8
.·. Io = I1 + I2 + I3 = 25 − j
35
= 25.17/ − 6.65◦ A(rms)
12
Therefore,
i2 =
35 √
2 cos(ωt − 90◦ ) A
12
Thus,
i2 (0− ) = i2(0+ ) = 0 A
and
and
√
iL = 25.17 2 cos(ωt − 6.65◦ ) A
√
iL (0− ) = iL (0+ ) = 25 2 A
[b] Begin by using the s-domain circuit in Fig. 13.60 to solve for V0
symbolically. Write a single node voltage equation:
V0 − (Vg + L` Io )
V0
V0
+
+
=0
sL`
Ra sLa
.·. V0 =
(Ra /L` )Vg + Io Ra
s + [Ra (La + L` )]/La L`
√
where L` = 1/120π H, La = 12/35π H, Ra = 12 Ω, and I0Ra = 300 2 V.
Thus,
√
√
√
1440π(122.92 2s − 3000π 2)
300 2
V0 =
+
(s + 1475π)(s2 + 14,400π 2 )
s + 1475π
√
K1
K2
K2∗
300 2
=
+
+
+
s + 1475π s − j120π s + j120π s + 1475π
The coefficients are
√
K1 = −121.18 2 V
√
K2 = 61.03 2/6.85◦ V
√
K2∗ = 61.03 2/ − 6.85◦
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Problems
13–93
√
√
Note that K1 + 300 2 = 178.82 2 V. Thus, the inverse transform of V0 is
√
√
v0 = 178.82 2e−1475πt + 122.06 2 cos(120πt + 6.85◦ ) V
Initially,
√
√
√
v0 (0+ ) = 178.82 2 + 122.06 2 cos 6.85◦ = 300 2 V
√
Note that at t = 0+ the initial value of iL , which is √
25 2 A, exists√in the
12 Ω resistor Ra . Thus, the initial value of V0 is (25 2)(12) = 300 2 V.
[c] The phasor domain equivalent circuit has a j1 Ω inductive impedance in
series with the parallel combination of a 12 Ω resistive impedance and a
j1440/35 Ω inductive impedance (remember that ω = 120π rad/s). Note
that Vg = 120/0◦ + (25.17/ − 6.65◦ )(j1) = 125.43/11.50◦ V(rms). The
node voltage equation in the phasor domain circuit is
V0 − 125.43/11.50◦ V0 35V0
+
+
=0
j1
12
1440
.·. V0 = 122.06/6.85◦ V(rms)
√
Therefore, v0 = 122.06 2 cos(120πt + 6.85◦ ) V, agreeing with the
steady-state component of the result in part (b).
[d] A plot of v0 , generated in Excel, is shown below.
P 13.93 [a] At t = 0− the phasor domain equivalent circuit is
I1 =
−j120
= −j10 = 10/ − 90◦ A (rms)
12
I2 =
−j120(35)
35
35
= − = /180◦ A (rms)
j1440
12
12
I3 =
−j120
= −j15 = 15/ − 90◦ A (rms)
8
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13–94
CHAPTER 13. The Laplace Transform in Circuit Analysis
IL = I1 + I2 + I3 = −
35
− j25 = 25.17/ − 96.65◦ A (rms)
12
√
iL = 25.17 2 cos(120πt − 96.65◦ )A
√
iL (0− ) = iL (0+ ) = −2.92 2A
35 √
2 cos(120πt + 180◦ )A
12
√
35 √
2 = −2.92 2A
i2 (0− ) = i2(0+ ) = −
12
i2 =
Vg = Vo + j1IL
35
12
25 − j122.92 = 125.43/ − 78.50◦ V (rms)
√
125.43 2 cos(120πt − 78.50◦ )V
√
125.43 2[cos 120πt cos 78.50◦ + sin 120πt sin 78.50◦ ]
√
√
25 2 cos 120πt + 122.92 2 sin 120πt
Vg = −j120 + 25 − j
=
vg =
=
=
√
√
25 2s + 122.92 2(120π)
·
. . Vg =
s2 + (120π)2
s-domain circuit:
where
Ll =
1
H;
120π
La =
√
iL (0) = −2.92 2 A;
12
H;
35π
Ra = 12 Ω
√
i2(0) = −2.92 2 A
The node voltage equation is
0=
Vo − (Vg + iL (0)Ll )
Vo
Vo + i2(0)La
+
+
sLl
Ra
sLa
Solving for Vo yields
Vo =
Vg Ra /Ll
Ra [iL(0) − i2 (0)]
+
[s + Ra (Ll + La )/La Ll ] [s + Ra (Ll + La )/Ll La ]
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Problems
13–95
Ra
= 1440π
Ll
1
12
12( 120π
+ 35π
)
Ra (Ll + La )
=
= 1475π
12
1
Ll La
( 35π )( 120π )
√
√
iL (0) − i2 (0) = −2.92 2 + 2.92 2 = 0
√
√
1440π[25 2s + 122.92 2(120π)]
·
. . Vo =
(s + 1475π)[s2 + (120π)2 ]
K1
K2
K2∗
=
+
+
s + 1475π s − j120π s + j120π
√
√
K1 = −14.55 2
K2 = 61.03 2/ − 83.15◦
√
√
.·. vo(t) = −14.55 2e−1475πt + 122.06 2 cos(120πt − 83.15◦ )V
Check:
√
vo (0) = (−14.55 + 14.55) 2 = 0
[b]
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13–96
CHAPTER 13. The Laplace Transform in Circuit Analysis
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Problems
13–97
√
[c] In Problem 13.92, the line-to-neutral voltage spikes at 300 2 V. Here the
line-to-neutral voltage has no spike. Thus the amount of voltage
disturbance depends on what part of the cycle the sinusoidal steady-state
voltage is switched.
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13–98
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.94 [a] First find Vg before Rb is disconnected. The phasor domain circuit is
120/θ◦ 120/θ◦ 120/θ◦
+
+
Ra
Rb
jXa
120/θ◦
=
[(Ra + Rb )Xa = jRa Rb ]
Ra Rb Xa
IL =
Since Xl = 1 Ω we have
Vg = 120/θ◦ +
120/θ◦
[Ra Rb + j(Ra + Rb )Xa ]
Ra Rb Xa
Ra = 12 Ω;
Rb = 8 Ω;
Vg =
Xa =
1440
Ω
35
120/θ◦
(1475 + j300)
1400
25 ◦
/θ (59 + j12) = 125.43/(θ + 11.50)◦
12
√
vg = 125.43 2 cos(120πt + θ + 11.50◦ )V
=
Let β = θ + 11.50◦ . Then
√
vg = 125.43 2(cos 120πt cos β − sin 120πt sin β)V
Therefore
√
125.43 2(s cos β − 120π sin β)
Vg =
s2 + (120π)2
The s-domain circuit becomes
where ρ1 = iL(0+ ) and ρ2 = i2(0+ ).
The s-domain node voltage equation is
Vo − (Vg + ρ1 Ll )
Vo
Vo + ρ2 La
+
+
=0
sLl
Ra
sLa
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Problems
13–99
Solving for Vo yields
Vo =
Vg Ra /Ll + (ρ1 − ρ2 )Ra
[s +
(La +Ll )Ra
]
La Ll
Substituting the numerical values
Ll =
1
H;
120π
La =
12
H;
35π
Ra = 12 Ω;
Rb = 8 Ω;
gives
Vo =
1440πVg + 12(ρ1 − ρ2 )
(s + 1475π)
Now determine the values of ρ1 and ρ2 .
ρ1 = iL(0+ )
IL =
ρ2 = i2 (0+ )
and
120/θ◦
[(Ra + Rb )Xa − jRa Rb ]
Ra Rb Xa
"
#
120/θ◦
(20)(1440)
=
− j96
96(1440/35)
35
= 25.17/(θ − 6.65)◦ A(rms)
√
.·. iL = 25.17 2 cos(120πt + θ − 6.65◦ )A
√
iL (0+ ) = ρ1 = 25.17 2 cos(θ − 6.65◦ )A
√
√
.·. ρ1 = 25 2 cos θ + 2.92 2 sin θA
120/θ◦
35
I2 =
= /(θ − 90)◦
j(1440/35)
12
35 √
2 cos(120πt + θ − 90◦ )A
12
√
35 √
ρ2 = i2(0+ ) =
2 sin θ = 2.92 2 sin θA
12
√
.·. ρ1 = ρ2 = 25 2 cos θ
√
(ρ1 − ρ2 )Ra = 300 2 cos θ
√
1440π
300
2 cos θ
. ·. V o =
· Vg +
s + 1475π
s + 1475π
√
√
"
#
1440π
125.43 2(s cos β − 120π sin β)
300 2 cos θ
=
+
s + 1475π
s2 + 14,400π 2
s + 1475π
√
K1 + 300 2 cos θ
K2
K2∗
=
+
+
s + 1475π
s − j120π s + j120π
i2 =
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13–100
CHAPTER 13. The Laplace Transform in Circuit Analysis
Now
√
(1440π)(125.43 2)[−1475π cos β − 120π sin β]
K1 =
14752 π 2 + 14,400π 2
√
−1440(125.43 2)[1475 cos β + 120 sin β]
=
14752 + 14,000
Since β = θ + 11.50◦ , K1 reduces to
√
√
K1 = −121.18 2 cos θ + 14.55 2 sin θ
From the partial fraction expansion√for Vo we see vo (t) will go directly
into steady state when K1 = −300 2 cos θ. It follow that
√
√
14.55 2 sin θ = −178.82 2 cos θ
or
tan θ = −12.29
Therefore,
θ = −85.35◦
[b] When θ = −85.35◦ , β = −73.85◦
√
1440π(125.43
2)[−120π sin(−73.85◦ ) + j120π cos(−73.85◦ )
.·. K2 =
(1475π + j120π)(j240π)
√
720 2(120.48 + j34.88)
=
−120 + j1475
√
= 61.03 2/ − 78.50◦
√
.·. vo = 122.06 2 cos(120πt − 78.50◦ ) V t > 0
= 172.61 cos(120πt − 78.50◦ ) V t > 0
[c] vo1 = 169.71 cos(120πt − 85.35◦ )V
vo2 = 172.61 cos(120πt − 78.50◦ )V
t<0
t>0
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Problems
13–101
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13–102
CHAPTER 13. The Laplace Transform in Circuit Analysis
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14
Introduction to
Frequency-Selective Circuits
Assessment Problems
AP 14.1
fc = 8 kHz,
ωc =
1
;
RC
ωc = 2πfc = 16π krad/s
R = 10 kΩ;
1
1
.·. C =
=
= 1.99 nF
ωc R
(16π × 103 )(104 )
AP 14.2 [a] ωc = 2πfc = 2π(2000) = 4π krad/s
L=
R
5000
=
= 0.40 H
ωc
4000π
[b] H(jω) =
ωc
4000π
=
ωc + jω
4000π + jω
When ω = 2πf = 2π(50,000) = 100,000π rad/s
H(j100,000π) =
4000π
1
=
= 0.04/87.71◦
4000π + j100,000π
1 + j25
.·. |H(j100,000π)| = 0.04
[c] .·. θ(100,000π) = −87.71◦
AP 14.3
ωc =
R
5000
=
= 1.43 Mrad/s
L
3.5 × 10−3
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
14–1 system, or transmission in any form or by any means, electronic,
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14–2
CHAPTER 14. Introduction to Frequency-Selective Circuits
1
106
106
=
=
= 10 krad/s
RC
R
100
106
[b] ωc =
= 200 rad/s
5000
106
[c] ωc =
= 33.33 rad/s
3 × 104
AP 14.4 [a] ωc =
AP 14.5 Let Z represent the parallel combination of (1/SC) and RL . Then
RL
(RL Cs + 1)
Z=
Thus H(s) =
=
where K =
AP 14.6
Z
RL
=
R+Z
R(RL Cs + 1) + RL
(1/RC)
s+
R+RL
RL
1
RC
=
(1/RC)
s+
1
K
1
RC
RL
R + RL
ωo2 =
1
LC
Q=
ωo
ωo
=
β
R/L
so L =
1
ωo2 C
=
so R =
1
(24π ×
103 )2 (0.1
× 10−6 )
= 1.76 mH
ωo L
(24π × 103 )(1.76 × 10−3 )
=
= 22.10 Ω
Q
6
AP 14.7
ωo = 2π(2000) = 4000π rad/s;
β = 2π(500) = 1000π rad/s;
β=
1
RC
so C =
ωo2 =
1
LC
so L =
ωo2 =
1
LC
so L =
AP 14.8
β=
1
RC
so R =
R = 250 Ω
1
1
=
= 1.27 µF
βR
(1000π)(250)
1
106
=
= 4.97 mH
ωo2 C
(4000π)2 (1.27)
1
ωo2 C
=
1
(104 π)2 (0.2
× 10−6 )
= 5.07 mH
1
1
=
= 3.98 kΩ
βC
400π(0.2 × 10−6 )
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Problems
AP 14.9
ωo2 =
1
LC
Q=
fo
5 × 103
=
= 25 = ωo RC
β
200
so L =
1
ωo2 C
=
1
(400π)2 (0.2
× 10−6 )
14–3
= 31.66 mH
25
Q
.·. R =
=
= 9.95 kΩ
ωo C
(400π)(0.2 × 10−6 )
AP 14.10
ωo = 8000π rad/s
C = 500 nF
ωo2 =
1
LC
Q=
ωo
ωo L
1
=
=
β
R
ωo CR
.·. R =
so L =
1
ωo2 C
= 3.17 mH
1
1
=
= 15.92 Ω
ωo CQ
(8000π)(500)(5 × 10−9 )
AP 14.11
ωo = 2πfo = 2π(20,000) = 40π krad/s;
Q=
ωo
ωo
=
β
(R/L)
ωo2 =
1
LC
so L =
so C =
R = 100 Ω;
Q=5
QR
5(100)
=
= 3.98 mH
ωo
(40π × 103 )
1
1
=
= 15.92 nF
ωo2 L
(40π × 103 )2 (3.98 × 10−3 )
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14–4
CHAPTER 14. Introduction to Frequency-Selective Circuits
Problems
P 14.1
[a] ωc =
R
127
=
= 12.7 krad/s
L
10 × 10−3
.·. fc =
[b] H(s) =
ωc
12,700
=
= 2021.27 Hz
2π
2π
ωc
12,700
=
s + ωc
s + 12,700
12,700
12,700 + jω
H(jω) =
H(jωc ) =
12,700
= 0.7071/ − 45◦
12,700 + j12,700
H(j0.2ωc ) =
H(j5ωc ) =
12,700
= 0.981/ − 11.31◦
12,700 + j2540
12,700
= 0.196/ − 78.69◦
12,700 + j63,500
[c] vo (t)|ωc = 7.07 cos(12,700t − 45◦ ) V
vo (t)|0.2ωc = 9.81 cos(2540t − 11.31◦ ) V
vo (t)|5ωc = 1.96 cos(63,500t − 78.69◦ ) V
P 14.2
[a]
R
= 10,000π rad/s
L
R = (0.001)(10,000)(π) = 31.42 Ω
[b] Re = 31.42k68 = 21.49 Ω
ωloaded =
Re
= 21,488.34 rad/s
L
.·. floaded = 3419.98 Hz
[c] The 33 Ω resistor in Appendix H is closest to the desired value of 31.42 Ω.
Therefore,
ωc = 33 krad/s so fc = 5252.11 Hz
P 14.3
[a] H(s) =
Vo
R
(R/L)
=
=
Vi
sL + R + Rl
s + (R + Rl )/L
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Problems
14–5
(R/L)
[b] H(jω) = R+R l
+ jω
L
|H(jω)| = r
(R/L)
R+Rl 2
L
+ ω2
|H(jω)|max occurs when ω = 0
R
R + Rl
R
R/L
[d] |H(jωc )| = √
= r
2(R + Rl )
R+Rl 2
[c] |H(jω)|max =
L
+ ωc2
R + Rl 2
;
.·. ωc = (R + Rl )/L
L
127 + 75
[e] ωc =
= 20,200 rad/s
0.01
12,700
H(jω) =
20,200 + jω
.·. ωc2 =
H(j0) = 0.6287
0.6287
√ / − 45◦ = 0.4446/ − 45◦
2
12,700
H(j6060) =
= 0.6022/ − 16.70◦
20,200 + j6060
H(j20,200) =
H(j60,600) =
P 14.4
12,700
= 0.1988/ − 71.57◦
20,200 + j60,600
1
1
=
= 10 krad/s
3
RC
(10 )(100 × 10− 9)
ωc
fc =
= 1591.55 Hz
2π
ωc
10,000
[b] H(jω) =
=
s + ωc
s + 10,000
[a] ωc =
H(jω) =
H(jωc ) =
10,000
10,000 + jω
10,000
= 0.7071/ − 45◦
10,000 + j10,000
H(j0.1ωc ) =
10,000
= 0.9950/ − 5.71◦
10,000 + j1000
H(j10ωc ) =
10,000
= 0.0995/ − 84.29◦
10,000 + j100,000
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14–6
CHAPTER 14. Introduction to Frequency-Selective Circuits
[c] vo (t)|ωc = 200(0.7071) cos(10,000t − 45◦ )
= 141.42 cos(10,000t − 45◦ ) mV
vo (t)|0.1ωc = 200(0.9950) cos(1000t − 5.71◦ )
= 199.01 cos(1000t − 5.71◦ ) mV
vo (t)|10ωc = 200(0.0995) cos(100,000t − 84.29◦ )
= 19.90 cos(100,000t − 84.29◦ ) mV
P 14.5
[a] fc =
[b]
ωc
50,000
50
=
=
× 103 = 7957.75 Hz
2π
2π
2π
1
= 50 × 103
RC
R=
1
(50 ×
103 )(0.5
× 10−6 )
= 40 Ω
[c] With a load resistor added in parallel with the capacitor the transfer
function becomes
RL k(1/sC)
RL /sC
H(s) =
=
R + RL k(1/sC)
R[RL + (1/sC)] + RL /sC
=
RL
1/RC
=
RRL sC + R + RL
s + [(R + RL )/RRL C]
This transfer function is in the form of a low-pass filter, with a cutoff
frequency equal to the quantity added to s in the denominator.
Therefore,
R + RL
1
R
ωc =
=
1+
RRL C
RC
RL
. ·.
R
= 0.05
RL
[d] H(j0) =
P 14.6
.·. RL = 20R = 800 Ω
RL
800
=
= 0.9524
R + RL
840
[a] ωc = 2π(100) = 628.32 rad/s
1
1
1
[b] ωc =
so R =
=
= 338.63 Ω
RC
ωc C
(628.32)(4.7 × 10−6 )
[c]
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Problems
14–7
Vo
1/sC
1/RC
628.32
=
=
=
Vi
R + 1/sC
s + 1/RC
s + 628.32
Vo
(1/sC)kRL
1/RC
628.32
[e] H(s) =
=
=
=
R + RL
Vi
R + (1/sC)kRL
s + 2(628.32)
s+
1/RC
RL
[f] ωc = 2(628.32) = 1256.64 rad/s
[d] H(s) =
[g] H(0) = 1/2
P 14.7
[a] Let Z =
RL (1/SC)
RL
=
RL + 1/SC
RL Cs + 1
Then H(s) =
=
=
[b] |H(jω)| = q
Z
Z +R
RL
RRL Cs + R + RL
(1/RC)
R + RL
s+
RRL C
(1/RC)
ω 2 + [(R + RL )/RRL C]2
|H(jω)| is maximum at ω = 0.
RL
R + RL
RL
(1/RC)
[d] |H(jωc )| = √
=q
2(R + RL )
ωC2 + [(R + RL )/RRL C]2
[c] |H(jω)|max =
.·. ωc =
[e] ωc =
R + RL
1
=
(1 + (R/RL ))
RRL C
RC
1
(103 )(10−7 )
H(j0) =
[1 + (103 /104 )] = 10,000(1 + 0.1) = 11,000 rad/s
10,000
= 0.9091/0◦
11,000
H(jωc ) =
10,000
= 0.6428/ − 45◦
11,000 + j11,000
H(j0.1ωc ) =
10,000
= 0.9046/ − 5.71◦
11,000 + j1100
H(j10ωc ) =
10,000
= 0.0905/ − 84.29◦
11,000 + j110,000
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14–8
P 14.8
CHAPTER 14. Introduction to Frequency-Selective Circuits
[a] ZL = jωL = j0L = 0
At ω = 0,
so it is a short circuit.
Vo = Vi
[b] ZL = jωL = j∞L = ∞
At ω = ∞,
so it is an open circuit.
Vo = 0
[c] This is a low pass filter, with a gain of 1 at low frequencies and a gain of 0
at high frequencies.
Vo
R
R/L
[d] H(s) =
=
=
Vi
R + sL
s + R/L
R
330
[e] ωc =
=
= 33 krad/s
L
0.01
P 14.9
R
RL
Vo
RkRL
L R + RL
[a] H(s) =
=
=
R
RL
Vi
RkRL + sL
s+
L R + RL
R
R
RL
[b] ωc(U L) = ;
ωc(L) =
so the cutoff frequencies are different.
L
L R + RL
H(0)(U L) = 1;
H(0)(L) = 1
so the passband gains are the same.
[c] ωc(U L) = 33,000 rad/s
ωc(L) = 33,000 − 0.05(33,000) = 31,350 rad/s
31,350 =
330
RL
0.01 330 + RL
so
RL
= 0.95
330 + RL
.·. 0.05RL = 313.5 so RL ≥ 6270 Ω
P 14.10 [a]
1
1
=
= 4000 rad/s
3
RC
(50 × 10 )(5 × 10−9 )
fc =
4000
= 636.62 Hz
2π
[b] H(s) =
s
s + ωc
.·.
H(jωc ) = H(j4000) =
H(jω) =
jω
4000 + jω
j4000
= 0.7071/45◦
4000 + j4000
H(j0.2ωc ) = H(j800) =
H(j5ωc ) = H(j20ωc ) =
j800
= 0.1961/78.69◦
4000 + j800
j20,000
= 0.9806/11.31◦
4000 + j20,000
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Problems
14–9
[c] vo (t)|ωc = (0.7071)(500) cos(4000t + 45◦ )
= 353.55 cos(4000t + 45◦ ) mV
vo (t)|0.2ωc = (0.1961)(500) cos(800t + 78.60◦ )
= 98.06 cos(800t + 78.69◦ ) mV
vo (t)|5ωc = (0.9806)(500) cos(20,000t + 11.31◦ )
= 490.29 cos(20,000t + 11.31◦ ) mV
P 14.11 [a] H(s) =
Vo
R
=
Vi
R + Rc + (1/sC)
R
s
·
R + Rc [s + (1/(R + Rc )C)]
R
jω
[b] H(jω) =
·
R + Rc jω + (1/(R + Rc )C)
=
|H(jω)| =
R
ω
·q
R + Rc
ω 2 + (R+R1c )2 C 2
The magnitude will be maximum when ω = ∞.
R
[c] |H(jω)|max =
R + Rc
Rωc
q
[d] |H(jωc )| =
(R + Rc ) ωc2 + [1/(R + Rc )C]2
.·. |H(jω)| = √
.·. ωc2 =
or ωc =
[e] ωc =
R
2(R + Rc )
when
1
(R + Rc )2 C 2
1
(R + Rc )C
1
(62.5 ×
103 )(5
× 10−9 )
= 3200 rad/s
R
50
=
= 0.8
R + Rc
62.5
. ·.
H(jω) =
H(jωc ) =
0.8jω
3200 + jω
(0.8)j3200
= 0.5657/45◦
3200 + j3200
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14–10
CHAPTER 14. Introduction to Frequency-Selective Circuits
H(j0.2ωc ) =
H(j5ωc ) =
(0.8)j640
= 0.1569/78.69◦
3200 + j640
(0.8)j16,000
= 0.7845/11.31◦
3200 + j16,000
P 14.12 [a] ωc = 2π(500) = 3141.59 rad/s
1
1
1
[b] ωc =
so R =
=
= 1.45 MΩ
RC
ωc C
(3141.59)(220 × 10−12 )
[c]
Vo
R
s
s
=
=
=
Vi
R + 1/sC
s + 1/RC
s + 3141.59
Vo
RkRL
s
s
[e] H(s) =
=
=
=
R + RL
Vi
RkRL + (1/sC)
s + 2(3141.59)
s+
1/RC
RL
[f] ωc = 2(3141.59) = 6283.19 rad/s
[d] H(s) =
[g] H(∞) = 1
1
= 2π(300) = 600π rad/s
RC
1
1
. ·. R =
=
= 5305.16 Ω = 5.305 kΩ
ωc C
(600π)(100 × 10−9 )
P 14.13 [a] ωc =
[b] Re = 5305.16k47,000 = 4767.08 Ω
ωc =
fc =
1
1
=
= 2097.7 rad/s
Re C
(4767.08)(100 × 10−9 )
ωc
2097.7
=
= 333.86 Hz
2π
2π
P 14.14 [a] R = ωc L = (1500 × 103 )(100 × 10−6 ) = 150 Ω (a value from Appendix H)
[b] With a load resistor in parallel with the inductor, the transfer function
becomes
sLkRL
sLRL
s[RL /(R + RL )]
H(s) =
=
=
R + sLkRL
R(sL + RL ) + sLRL
s + [RRL /(R + RL )]
This transfer function is in the form of a high-pass filter whose cutoff
frequency is the quantity added to s in the denominator. Thus,
ωc =
RRL
L(R + RL )
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Problems
14–11
Substituting in the values of R and L from part (a), we can solve for the
value of load resistance that gives a cutoff frequency of 1200 krad/s:
150RL
= 1200 × 103
100 × 10−6 (150 + RL )
so
RL = 600 Ω
The smallest resistor from Appendix H that is larger than 600 Ω is 680 Ω.
P 14.15 [a] For ω = 0, the inductor behaves as a short circuit, so Vo = 0.
For ω = ∞, the inductor behaves as an open circuit, so Vo = Vi .
Thus, the circuit is a high-pass filter.
sL
s
s
[b] H(s) =
=
=
R + sL
s + R/L
s + 15,000
R
[c] ωc =
= 15,000 rad/s
L
jR/L
j
1
[d] |H(jR/L)| =
=
=√
jR/L + R/L
j +1
2
RL
s
Vo
RL ksL
R + RL
P 14.16 [a] H(s) =
=
=
R
RL
Vi
R + RL ksL
s+
L R + RL
=
1
s
2
1
s + (15,000)
2
R
RL
1
[b] ωc =
= (15,000) = 7500 rad/s
L R + RL
2
1
[c] ωc(L) = ωc(U L)
2
[d] The gain in the passband is also reduced by a factor of 1/2 for the loaded
filter.
P 14.17 By definition Q = ωo /β therefore β = ωo /Q. Substituting this expression into
Eqs. 14.34 and 14.35 yields
ωo
ωc1 = −
+
2Q
v
u
u
t
v
u
ωo
2Q
ωo u
ωo
ωc2 =
+t
2Q
2Q
!2
!2
+ ωo2
+ ωo2
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14–12
CHAPTER 14. Introduction to Frequency-Selective Circuits
Now factor ωo out to get

ωc1 = ωo −


1
+
2Q
+
v
u
u
t
ωc1 ωc2 =
q
 1
ωc2 = ωo 
√
P 14.18 ωo =
v
u
u
t
2Q
1
2Q
1+
1
2Q
1+
!2
!2






(121)(100) = 110 krad/s
ωo
= 17.51 kHz
2π
fo =
β = 121 − 100 = 21 krad/s or 2.79 kHz
Q=
P 14.19 β =
ωo
110
=
= 5.24
β
21
ωo
50,000
=
= 12.5 krad/s;
Q
4

1
ωc2 = 50,000  +
8
fc2 =
s
1
8
1+

 = 56.64 krad/s
56.64 k
= 9.01 kHz
2π

1
ωc1 = 50,000 − +
8
fc1 =
2
12,500
= 1.99 kHz
2π
s
2
1
1+
8

 = 44.14 krad/s
44.14 k
= 7.02 kHz
2π
P 14.20 [a] ωo2 =
R=
1
LC
1
= 79.16 mH
[8000(2π)]2 (5 × 10−9 )
so L =
ωo L
8000(2π)(79.16 × 10−3 )
=
= 1.99 kΩ
Q
2

1
[b] fc1 = 8000 − +
4
s

1 
1+
= 6.25 kHz
16
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems


s
1
[c] fc2 = 8000  +
4
14–13
1 
1+
= 10.25 kHz
16
[d] β = fc2 − fc1 = 4 kHz
or
fo
8000
β=
=
= 4 kHz
Q
2
P 14.21 [a] We need ωc close to 2π(8000) = 50,265.48 rad/s. There are several
possible approaches – this one starts by choosing L = 10 mH. Then,
1
= 39.58 nF
[2π(8000)]2 (0.01)
C=
Use the closest value from Appendix H, which is 0.047 µF to give
ωc =
s
1
= 46,126.56 rad/s or fc = 7341.27 Hz
(0.01)(47 × 10−9 )
Then, R =
ωo L
(46,126.56)(0.01)
=
= 230 Ω
Q
2
Use the closest value from Appendix H, which is 220 Ω to give
(46,126.56)(0.01)
= 2.1
220
7341.27 − 8000
[b] % error in fc =
(100) = −8.23%
8000
Q=
% error in Q =
P 14.22 [a] ωo2 =
2.1 − 2
(100) = 5%
2
1
1
=
= 1010
−3
LC
(10 × 10 )(10 × 10−9 )
ωo = 105 rad/s = 100 krad/s
ωo
105
=
= 15.9 kHz
2π
2π
[c] Q = ωo RC = (100 × 103 )(8000)(10 × 10−9 ) = 8
[b] fo =

[d] ωc1 = ωo −

1
+
2Q
v
u
u
t
1+
1
2Q
!2
ωc1
[e] .·. fc1 =
= 14.95 kHz
2π

v
u
u
1
 1
[f] ωc2 = ωo 
+ t1 +
2Q
2Q
!2



1

5
+
 = 10 −

16
1

5
+
 = 10 
16
s
s
1+

1 
= 93.95 krad/s
256

1 
1+
= 106.45 krad/s
256
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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14–14
CHAPTER 14. Introduction to Frequency-Selective Circuits
ωc2
[g] .·. fc2 =
= 16.94 kHz
2π
ωo
105
[h] β =
=
= 12.5 krad/s or 1.99 kHz
Q
8
P 14.23 [a] L =
1
ωo2 C
1
(50 ×
10−9 )(20

 1
= ωo 
+
2Q
v
u
u
t
1
1+
2Q
. ·.
= 22.10 krad/s

1

ωc1 = ωo −
+
2Q
!2
= 50 mH



s
1
1


+ 1+
 = 20,000 
1
1+
2Q
. ·.
10
fc2 =
v
u
u
t
= 18.10 krad/s
[c] β =
× 103 )2
Q
5
=
= 5 kΩ
3
ωo C
(20 × 10 )(50 × 10−9 )
R=
[b] ωc2
=
!2
ωc2
= 3.52 kHz
2π


1

+
 = 20,000 −
10
fc1 =
100
ωc1
= 2.88 kHz
2π
s

1 
1+
100
ωo
20,000
=
= 4000 rad/s or 636.62 Hz
Q
5
P 14.24 [a] We need ωc = 20,000 rad/s. There are several possible approaches – this
one starts by choosing L = 1 mH. Then,
C=
1
20,0002 (0.001)
= 2.5 µF
Use the closest value from Appendix H, which is 2.2 µF to give
ωc =
s
1
= 21,320 rad/s
(0.001)(2.2 × 10−6 )
Then, R =
Q
5
=
= 106.6 Ω
ωo C
(21320)(2.2 × 10−6 )
Use the closest value from Appendix H, which is 100 Ω to give
Q = 100(21,320)(2.2 × 10−6 ) = 4.69
[b] % error in ωc =
% error in Q =
21,320 − 20,000
(100) = 6.6%
20,000
4.69 − 5
(100) = −6.2%
5
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Problems
14–15
1
1
=
= 625 × 106
−3
−9
LC
(40 × 10 )(40 × 10 )
P 14.25 [a] ωo2 =
ωo = 25 × 103 rad/s = 25 krad/s
fo =
[b] Q =
[c] ωc1
25,000
= 3978.87 Hz
2π
(25 × 103 )(40 × 10−3 )
ωo L
=
=5
R + Ri
200

v
u
u
1

= ωo −
+ t1 +
1
2Q
2Q
!2



s
1
1


+ 1+
 = 25,000 −
10
100
= 22.62 krad/s or 3.60 kHz
[d] wc2

 1
= ωo 
+
2Q
v
u
u
t
1
1+
2Q
!2



s
1
1


+ 1+
 = 25,000 
10
100
= 27.62 krad/s or 4.4 kHz
[e] β = ωc2 − ωc1 = 27.62 − 22.62 = 5 krad/s
or
ωo
25,000
β=
=
= 5 krad/s or 795.77 Hz
Q
5
P 14.26 [a] H(s) =
(R/L)s
(R+Ri )
s
L
1
+
+ LC
For the numerical values in Problem 14.25 we have
4500s
H(s) = 2
s + 5000s + 625 × 106
s2
.·. H(jω) =
H(jωo ) =
(625 ×
4500jω
− ω 2 ) + j5000ω
106
j4500(25 × 103 )
= 0.9/0◦
j5000(25 × 103 )
.·. vo(t) = 500(0.9) cos 25,000 = 450 cos 25,000t mV
[b] From the solution to Problem 14.25,
ωc1 = 22.62 krad/s
H(j22.62 k) =
j4500(22.62 × 103 )
= 0.6364/45◦
(113.12 + j113.12) × 106
.·. vo(t) = 500(0.6364) cos(22,620t + 45◦ ) = 318.2 cos(22,620t + 45◦ ) mV
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14–16
CHAPTER 14. Introduction to Frequency-Selective Circuits
[c] From the solution to Problem 14.25,
ωc2 = 27.62 krad/s
H(j27.62 k) =
j4500(27.62 × 103 )
= 0.6364/ − 45◦
6
(−138.12 + j138.12) × 10
.·. vo(t) = 500(0.6364) cos(27,620t − 45◦ ) = 318.2 cos(27,620t − 45◦ ) mV
P 14.27 [a] ωo =
q
1/LC
Q=
ωo
β
β=
R
L
so L =
so β =
1
ωo2 C
=
1
(50,000)2 (0.01
× 10−6 )
= 40 mH
ωo
50,000
=
= 6250 rad/s
Q
8
so R = Lβ = (40 × 10−3 )(6250) = 250 Ω
[b] From part (a), β = 6250 rad/s. Then,
ωc1,2
β
=± +
2
s
β
6250
+ ωo2 = ±
+
2
2
ωc1 = 46,972.56 rad/s
P 14.28 H(jω) =
50,0002
6250
2
2
+ 50,0002 = ±3125 + 50,097.56
ωc2 = 53,222.56 rad/s
jω(6250)
− ω 2 + jω(6250)
[a] H(j50,000) =
Vo = (1)Vi
s
50,0002
j50,000(6250)
=1
− 50,0002 + j(50,000)(6250)
.·. vo (t) = 50 cos 50,000t mV
[b] H(j46,972.56) =
50,0002
1
Vo = √ /45◦ Vi
2
[c] H(j53,222.56) =
j46,972.56(6250)
1
= √ /45◦
2
− 46,972.56 + j(46,972.56)(6250)
2
.·. vo(t) = 35.36 cos(46,972.56t + 45◦ ) mV
50,0002
1
Vo = √ / − 45◦ Vi
2
j53,222.56(6250)
1
= √ / − 45◦
2
− 53,222.56 + j(53,222.56)(6250)
2
.·. vo(t) = 35.36 cos(53,222.56t − 45◦ ) mV
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Problems
[d] H(j10,000) =
j10,000(6250)
= 0.026/88.5◦
− 10,0002 + j(10,000)(6250)
50,0002
.·. vo (t) = 1.3 cos(10,000t + 88.5◦ ) mV
Vo = 0.026/88.5◦ Vi
[e] H(j250,000) =
50,0002
j250,000(6250)
= 0.026/ − 88.5◦
− 250,0002 + j(250,000)(6250)
Vo = 0.026/ − 88.5◦ Vi
P 14.29 H(s) = 1 −
14–17
.·. vo (t) = 1.3 cos(250,000t − 88.5◦ ) mV
s2 + (1/LC)
(R/L)s
=
s2 + (R/L)s + (1/LC)
s2 + (R/L)s + (1/LC)
50,0002 − ω 2
H(jω) =
50,0002 − ω 2 + jω(6250)
[a] H(j50,000) =
Vo = (0)Vi
50,0002 − 50,0002
=0
50,0002 − 50,0002 + j(50,000)(6250)
.·. vo (t) = 0 mV
[b] H(j46,972.56) =
50,0002 − 46,972.562
1
= √ / − 45◦
2
2
50,000 − 46,972.56 + j(46,972.56)(6250)
2
1
Vo = √ / − 45◦ Vi
2
[c] H(j53,222.56) =
1
Vo = √ /45◦ Vi
2
[d] H(j10,000) =
.·. vo(t) = 35.36 cos(46,972.56t − 45◦ ) mV
50,0002 − 53,222.562
1
= √ /45◦
2
2
50,000 − 53,222.56 + j(53,222.56)(6250)
2
.·. vo(t) = 35.36 cos(53,222.56t + 45◦ ) mV
50,0002 − 10,0002
= 0.9997/ − 1.49◦
50,0002 − 10,0002 + j(10,000)(6250)
Vo = 0.9997/ − 1.49◦ Vi
[e] H(j250,000) =
.·. vo (t) = 49.98 cos(10,000t − 1.49◦ ) mV
50,0002 − 250,0002
= 0.9997/1.49◦
2
2
50,000 − 250,000 + j(250,000)(6250)
Vo = 0.9997/1.49◦ Vi
.·. vo (t) = 49.98 cos(250,000t + 1.49◦ ) mV
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14–18
CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.30 [a]
[b] L =
1
ωo2 C
R=
1
=
(50 ×
103 )2 (20
× 10−4 )
= 20 mH
ωo L
(50 × 103 )(20 × 10−3 )
=
= 160 Ω
Q
6.25
[c] Re = 160k480 = 120 Ω
Re + Ri = 120 + 80 = 200 Ω
Qsystem =
[d] βsystem =
ωo L
(50 × 103 )(20 × 10−3 )
=
=5
Re + Ri
200
ωo
Qsystem
βsystem(Hz) =
P 14.31 [a]
50 × 103
= 10 krad/s
5
=
10,000
= 1591.55 Hz
2π
Vo
Z
1
=
where Z =
Vi
Z +R
Y
and Y = sC +
H(s) =
=
=
=
RL
RLCs2
RL Ls
+ (R + RL )Ls + RRL
(1/RC)s
s2 +
h
R+RL
RL
1
RC
i
s+
1
LC
RL
R+RL
1
s
R+RL
RL
RC
h
i
1
1
L
s2 + R+R
s + LC
RL
RC
Kβs
,
s2 + βs + ωo2
R + RL
[b] β =
RL
1
1
LCRL s2 + sL + RL
+
=
sL RL
RL Ls
K=
RL
,
R + RL
β=
1
(RkRL )C
1
RC
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
[c] βU =
14–19
1
RC
R + RL
R
. ·. β L =
βU = 1 +
βU
RL
RL
[d] Q =
ωo
ωo RC
= R+R L
β
RL
[e] QU = ωo RC
. ·. Q L =
[f] H(jω) =
RL
1
QU =
QU
R + RL
[1 + (R/RL )]
Kjωβ
ωo2 − ω 2 + jωβ
H(jωo ) = K
Let ωc represent a corner frequency. Then
K
Kωc β
|H(jωc )| = √ = q
2
(ωo2 − ωc2 )2 + ωc2 β 2
1
ωc β
. ·. √ = q
2
(ωo2 − ωc2 )2 + ωc2 β 2
Squaring both sides leads to
(ωo2 − ωc2 )2 = ωc2 β 2 or (ωo2 − ωc2 ) = ±ωc β
.·. ωc2 ± ωc β − ωo2 = 0
or
β
ωc = ∓ ±
2
s
β2
+ ωo2
4
The two positive roots are
β
ωc1 = − +
2
s
β2
+ ωo2
4
β
and ωc2 = +
2
s
β2
+ ωo2
4
where
R
β = 1+
RL
P 14.32 [a] ωo2 =
1
1
and ωo2 =
RC
LC
1
1
=
= 1012
−3
LC
(5 × 10 )(200 × 10−12 )
ωo = 1 Mrad/s
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14–20
CHAPTER 14. Introduction to Frequency-Selective Circuits
R + RL 1
[b] β =
·
=
RL
RC
500 × 103
400 × 103
ωo
106
=
= 16
β
62.5 × 103
RL
[d] H(jωo ) =
= 0.8/0◦
R + RL
!
1
3
(100 × 10 )(200 × 10−12 )
!
= 62.5 krad/s
[c] Q =
.·. vo(t) = 250(0.8) cos(106 t) = 200 cos 106 t mV
R
[e] β = 1 +
RL
1
100
= 1+
(50 × 103 ) rad/s
RC
RL
ωo = 106 rad/s
Q=
ωo
20
=
β
1 + (100/RL )
where RL is in kilohms
[f]
P 14.33 ωo2 =
1
1
=
= 1016
−6
LC
(2 × 10 )(50 × 10−12 )
ωo = 100 Mrad/s
QU = ωo RC = (100 × 106 )(2.4 × 103 )(50 × 10−12 ) = 12
.·.
RL
12 = 7.5;
R + RL
7.5
.·. RL =
R = 4 kΩ
4.5
P 14.34 [a] In analyzing the circuit qualitatively we visualize vi as a sinusoidal voltage
and we seek the steady-state nature of the output voltage vo .
At zero frequency the inductor provides a direct connection between the
input and the output, hence vo = vi when ω = 0.
At infinite frequency the capacitor provides the direct connection, hence
vo = vi when ω = ∞.
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Problems
14–21
At the resonant frequency of the parallel combination of L and C the
impedance of the combination is infinite and hence the output voltage
will be zero when ω = ωo .
At frequencies on either side of ωo the amplitude of the output voltage
will be nonzero but less than the amplitude of the input voltage.
Thus the circuit behaves like a band-reject filter.
[b] Let Z represent the impedance of the parallel branches L and C, thus
Z=
sL(1/sC)
sL
= 2
sL + 1/sC
s LC + 1
Then
H(s) =
=
H(s) =
Vo
R
R(s2 LC + 1)
=
=
Vi
Z +R
sL + R(s2 LC + 1)
[s2 + (1/LC)]
s2 +
1
RC
s+
s2 + ωo2
s2 + βs + ωo2
1
LC
[c] From part (b) we have
H(jω) =
ωo2 − ω 2
ωo2 − ω 2 + jωβ
It follows that H(jω) = 0 when ω = ωo .
1
.·. ωo = √
LC
[d] |H(jω)| = q
ωo2 − ω 2
(ωo2 − ω 2 )2 + ω 2 β 2
1
|H(jω)| = √ when ω 2 β 2 = (ωo2 − ω 2)2
2
or ± ωβ = ωo2 − ω 2, thus
ω 2 ± βω − ωo2 = 0
The two positive roots of this quadratic are
ωc1 =
ωc2 =
−β
+
2
β
+
2
v
!
u
u β 2
t
2
v
!
u
u β 2
t
2
+ ωo2
+ ωo2
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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14–22
CHAPTER 14. Introduction to Frequency-Selective Circuits
Also note that since β = ωo /Q

 −1
ωc1 = ωo 

2Q
 1
ωc2 = ωo 
2Q
+
v
u
u
t
+
v
u
u
t
1+
1+
1
2Q
!2

1
2Q
!2





[e] It follows from the equations derived in part (d) that
β = ωc2 − ωc1 = 1/RC
[f] By definition Q = ωo /β = ωo RC.
P 14.35 [a] ωo2 =
1
1
=
= 1012
−6
LC
(50 × 10 )(20 × 10−9 )
.·. ωo = 1 Mrad/s
ωo
= 159.15 kHz
2π
[c] Q = ωo RC = (106 )(750)(20 × 10−9 ) = 15
[b] fo =

1

[d] ωc1 = ωo −
+
2Q
v
u
u
t
1
1+
2Q
!2
= 967.22 krad/s
[e] fc1 =


1

6
+
 = 10 −
30

s
1 
1+
900
ωc1
= 153.94 kHz
2π

 1
[f] ωc2 = ωo 
+
2Q
v
u
u
t
1
1+
2Q
= 1.03 Mrad/s
!2


1

6
+
 = 10 
30
s

1 
1+
900
ωc1
= 164.55 kHz
2π
[h] β = fc2 − fc1 = 10.61 kHz
[g] fc2 =
P 14.36 [a] ωo = 2πfo = 8π krad/s
L=
R=
1
ωo2 C
=
1
(8000π)2 (0.5
× 10−6 )
= 3.17 mH
Q
5
=
= 397.89 Ω
ωo C
(8000π)(0.5 × 10−6 )
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Problems
[b] fc2

 1
= fo 
+
2Q
v
u
u
t
1
1+
2Q
!2
= 4.42 kHz

v
u

14–23


s
1
1


+ 1+
 = 4000 
u
1
1

fc1 = fo −
+ t1 +
2Q
2Q
10
!2
= 3.62 kHz


1

+
 = 4000 −
10
100
s

1 
1+
100
[c] β = fc2 − fc1 = 800 Hz
or
fo
4000
β=
=
= 800 Hz
Q
5
P 14.37 [a] Re = 397.89k1000 = 284.63 Ω
Q = ωo Re C = (8000π)(284.63)(0.5 × 10−6 ) = 3.58
[b] β =
fo
4000
=
= 1.12 kHz
Q
3.58

1
[c] fc2 = 4000 
+
7.16
s

1
[d] fc1 = 4000 −
+
7.16

1 
1+
= 4.60 kHz
7.162
s

1 
1+
= 3.48 kHz
7.162
P 14.38 [a] We need ωc = 2π(4000) = 25,132.74 rad/s. There are several possible
approaches – this one starts by choosing L = 100 µH. Then,
1
C=
= 15.83 µF
2
[2π(4000)] (100 × 10−6 )
Use the closest value from Appendix H, which is 22 µF, to give
ωc =
s
1
100 ×
Then, R =
10−6 )(22
× 10−6 )
= 21,320.07 rad/s so fc = 3393.19 Hz
Q
5
=
= 10.66 Ω
ωo C
(21320.07)(22 × 10−6 )
Use the closest value from Appendix H, which is 10 Ω, to give
Q = 10(21,320.07)(22 × 10−6 ) = 4.69
3393.19 − 4000
(100) = −15.2%
4000
[b] % error in fc =
% error in Q =
4.69 − 5
(100) = −6.2%
5
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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14–24
CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.39 [a] ωo =
q
1/LC
Q=
ωo
β
β=
R
L
so L =
so β =
1
ωo2 C
=
1
(25,000)2 (200
× 10−9 )
= 8 mH
ωo
25,000
=
= 10,000 rad/s
Q
2.5
so R = Lβ = (8 × 10−3 )(10,000) = 80 Ω
[b] From part (a), β = 10,000 rad/s.
ωc1,2
β
=± +
2
s
β
10,000
+ ωo2 = ±
+
2
2
ωc1 = 20,495.1 rad/s
P 14.40 H(jω) =
s
10,000
2
2
+ 25,0002 = ±5000 + 25,495.1
ωc2 = 30,495.1 rad/s
ωo2 − ω 2
25,0002 − ω 2
=
ωo2 − ω 2 + jωβ
25,0002 − ω 2 + jω(10,000)
[a] H(j25,000) =
Vo = (0)Vi
25,0002 − 25,0002
=0
25,0002 − 25,0002 + j(25,000)(10,000)
.·. vo (t) = 0 mV
[b] H(j20,495.1) =
25,0002 − 20,495.12
1
= √ / − 45◦
2
2
25,000 − 20,495.1 + j(20,495.1)(10,000)
2
1
Vo = √ / − 45◦ Vi
2
[c] H(j30,495.1) =
1
Vo = √ /45◦ Vi
2
[d] H(j5000) =
.·. vo(t) = 176.78 cos(20,495.1t − 45◦ ) mV
25,0002 − 30,495.12
1
= √ /45◦
2
2
25,000 − 30,495.1 + j(30,495.1)(10,000)
2
.·. vo(t) = 176.78 cos(30,495.1t + 45◦ ) mV
25,0002 − 50002
= 0.9965/ − 4.76◦
25,0002 − 50002 + j(5000)(10,000)
Vo = 0.9965/ − 4.76◦ Vi
.·. vo (t) = 249.1 cos(5000t − 4.76◦ ) mV
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Problems
25,0002 − 125,0002
= 0.9965/4.76◦
25,0002 − 125,0002 + j(125,000)(10,000)
[e] H(j125,000) =
.·. vo (t) = 249.1 cos(125,000t + 4.76◦ ) mV
Vo = 0.9965/4.76◦ Vi
P 14.41 H(jω) =
ωo2
jωβ
jω(10,000)
=
2
2
− ω + jωβ
25,000 − ω 2 + jω(10,000)
[a] H(j25,000) =
j(25,000)(10,000)
=1
− 25,0002 + j(25,000)(10,000)
25,0002
.·. vo (t) = 250 cos 25,000t mV
Vo = (1)Vi
[b] H(j20,495.1) =
1
Vo = √ /45◦ Vi
2
[c] H(j30,495.1) =
j(20,495.1)(10,000)
1
/45◦
√
=
2
2
25,000 − 20,495.1 + j(20,495.1)(10,000)
2
.·. vo(t) = 176.78 cos(20,495.1t + 45◦ ) mV
[d] H(j5000) =
j(30,495.1)(10,000)
1
= √ / − 45◦
2
− 30,495.1 + j(30,495.1)(10,000)
2
25,0002
1
Vo = √ / − 45◦ Vi
2
.·. vo(t) = 176.78 cos(30,495.1t − 45◦ ) mV
j(5000)(10,000)
= 0.083/85.24◦
2
2
25,000 − 5000 + j(5000)(10,000)
.·. vo (t) = 20.75 cos(5000t + 85.24◦ ) mV
Vo = 0.083/85.24◦ Vi
[e] H(j125,000) =
25,0002
j(125,000)(10,000)
= 0.083/ − 85.24◦
− 125,0002 + j(125,000)(10,000)
Vo = 0.083/ − 85.24◦ Vi
P 14.42 [a] Let Z =
Z=
14–25
.·. vo (t) = 20.75 cos(125,000t − 85.24◦ ) mV
RL (sL + (1/sC))
RL + sL + (1/sC)
RL (s2LC + 1)
s2LC + RL Cs + 1
Then H(s) =
Vo
s2 RL CL + RL
=
Vi
(R + RL )LCs2 + RRL Cs + R + RL
Therefore
H(s) =
=
RL
[s2 + (1/LC)]
i
·h
RRL
s
1
R + RL
s2 + R+R
+
L
LC
L
K(s2 + ωo2 )
s2 + βs + ωo2
where K =
RL
;
R + RL
ωo2 =
1
;
LC
β=
RRL
R + RL
1
L
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14–26
CHAPTER 14. Introduction to Frequency-Selective Circuits
1
[b] ωo = √
LC
RRL
1
[c] β =
R + RL L
ωo
ωo L
[d] Q =
=
β
[RRL /(R + RL )]
K(ωo2 − ω 2 )
[e] H(jω) = 2
(ωo − ω 2 ) + jβω
H(jωo ) = 0
[f] H(j0) =
Kωo2
=K
ωo2
[g] H(jω) = nh
H(j∞) =
[h] H(jω) =
h
i
K (ωo /ω)2 − 1
i
(ωo /ω)2 − 1 + jβ/ω
−K
=K
−1
o
K(ωo2 − ω 2)
(ωo2 − ω 2 ) + jβω
H(j0) = H(j∞) = K
Let ωc represent a corner frequency. Then
K
|H(jωc )| = √
2
K
K(ωo2 − ωc2 )
. ·. √ = q
2
(ωo2 − ωc2 )2 + ωc2 β 2
Squaring both sides leads to
(ωo2 − ωc2 )2 = ωc2 β 2 or (ωo2 − ωc2 ) = ±ωc β
.·. ωc2 ± ωc β − ωo2 = 0
or
s
β
β2
ωc = ∓ ±
+ ωo2
2
4
The two positive roots are
s
β
β2
β
ωc1 = − +
+ ωo2 and ωc2 = +
2
4
2
where
RRL
1
1
β=
· and ωo2 =
R + RL L
LC
s
β2
+ ωo2
4
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Problems
P 14.43 [a] ωo2 =
14–27
1
1
=
= 0.25 × 1018 = 25 × 1016
−6
−12
LC
(10 )(4 × 10 )
ωo = 5 × 108 = 500 Mrad/s
β=
RRL
1
(30)(150)
1
· =
· −6 = 25 Mrad/s = 3.98 MHz
R + RL L
180
10
Q=
ωo
500 M
=
= 20
β
25 M
[b] H(j0) =
RL
150
=
= 0.8333
R + RL
180
RL
= 0.8333
R + RL
H(j∞) =

250  1
[c] fc2 =
+
π 40
fc1 =

s

1 
1+
= 81.59 MHz
1600
250  1
− +
π
40
s
1+

1 
= 77.61 MHz
1600
Check:
β = fc2 − fc1 = 3.98 MHz.
ωo
500 × 106
[d] Q =
= RRL 1
β
·L
R+R
L
500(R + RL )
50
30
=
1+
RRL
3
RL
where RL is in ohms.
=
[e]
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14–28
CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.44 [a] ωo2 =
1
= 625 × 106
LC
. ·. L =
1
= 64 mH
(625 × 106 )(25 × 10−9 )
RL
= 0.9;
R + RL
. ·.
.·. 0.1RL = 0.9R
500
.·. R =
= 55.6 Ω
9
RL = 9R
RL
1
[b] β =
R · = 781.25 rad/s
R + RL
L
ωo
25,000
=
= 32
β
781.25
Q=
P 14.45 [a] |H(jω)| = q
. ·.
1010
(1010 − ω 2 )2 + (50,000ω)2
=1
1020 = (1010 − ω 2 )2 + (50,000ω)2
= −2 × 1010 ω 2 + ω 4 + 25 × 108 ω 2
. ·.
ω 2 = 175 × 108
so
ω = 132,287.57 rad/s
[b] From the equation for |H(jω)| in part (a), the frequency for which the
magnitude is maximum is the frequency for which the denominator is
minimum. This is the frequency at which
√
(1010 − ω 2 )2 = 0 so ω = 1010 = 100,000 rad/s
1010
[c] |H(j100,000)| = q
(1010 − 100,0002 )2 + [50,000(100,000)]2
=2
P 14.46 [a] Use the cutoff frequencies to calculate the bandwidth:
ωc1 = 2π(697) = 4379.38 rad/s
Thus
ωc2 = 2π(941) = 5912.48 rad/s
β = ωc2 − ωc1 = 1533.10 rad/s
Calculate inductance using Eq. (14.32) and capacitance using Eq.
(14.31):
L=
C=
R
600
=
= 0.39 H
β
1533.10
1
Lωc1 ωc2
=
1
= 0.10 µF
(0.39)(4379.38)(5912.48)
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Problems
14–29
[b] At the outermost two frequencies in the low-frequency group (687 Hz and
941 Hz) the amplitudes are
|V697Hz| = |V941Hz | =
|Vpeak|
√
= 0.707|Vpeak |
2
because these are cutoff frequencies. We calculate the amplitudes at the
other two low frequencies using Eq. (14.32):
ωβ
|V | = (|Vpeak|)(|H(jω)|) = |Vpeak| q
(ωo2 − ω 2 )2 + (ωβ)2
Therefore
|V770Hz | = |Vpeak| = q
(4838.05)(1533.10)
(5088.522 − 4838.052 )2 + [(4838.05)(1533.10)]2
= 0.948|Vpeak |
and
|V852Hz | = |Vpeak| = q
(5353.27)(1533.10)
(5088.522 − 5353.272 )2 + [(5353.27)(1533.10)]2
= 0.948|Vpeak |
It is not a coincidence that these two magnitudes are the same. The
frequencies in both bands of the DTMF system were carefully chosen to
produce this type of predictable behavior with linear filters. In other
words, the frequencies were chosen to be equally far apart with respect to
the response produced by a linear filter. Most musical scales consist of
tones designed with this dame property – note intervals are selected to
place the notes equally far apart. That is why the DTMF tones remind
us of musical notes! Unlike musical scales, DTMF frequencies were
selected to be harmonically unrelated, to lower the risk of misidentifying
a tone’s frequency if the circuit elements are not perfectly linear.
[c] The high-band frequency closest to the low-frequency band is 1209 Hz.
The amplitude of a tone with this frequency is
|V1209Hz| = |Vpeak| = q
(7596.37)(1533.10)
(5088.522 − 7596.372 )2 + [(7596.37)(1533.10)]2
= 0.344|Vpeak |
This is less than one half the amplitude of the signals with the low-band
cutoff frequencies, ensuring adequate separation of the bands.
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14–30
CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.47 The cutoff frequencies and bandwidth are
ωc1 = 2π(1209) = 7596 rad/s
ωc2 = 2π(1633) = 10.26 krad/s
β = ωc2 − ωc1 = 2664 rad/s
Telephone circuits always have R = 600 Ω. Therefore, the filters inductance
and capacitance values are
L=
R
600
=
= 0.225 H
β
2664
C=
1
= 0.057 µF
ωc1 ωc2 L
At the highest of the low-band frequencies, 941 Hz, the amplitude is
ωβ
|Vω | = |Vpeak| q
(ωo2 − ω 2 )2 + ω 2 β 2
where
ωo =
√
ωc1 ωc2 . Thus,
|Vpeak|(5912)(2664)
|Vω | = q
[(8828)2 − (5912)2 ]2 + [(5912)(2664)]2
= 0.344 |Vpeak|
Again it is not coincidental that this result is the same as the response of the
low-band filter to the lowest of the high-band frequencies.
P 14.48 From Problem 14.46 the response to the largest of the DTMF low-band tones
is 0.948|Vpeak |. The response to the 20 Hz tone is
|V20Hz | =
[(50892
|Vpeak|(125.6)(1533)
− 125.62 )2 + [(125.6)(1533)]2 ]1/2
= 0.00744|Vpeak |
.·.
|V20Hz |
|V20Hz |
0.00744|Vpeak |
=
=
= 0.5
|V770Hz |
|V852Hz |
0.948|Vpeak |
.·. |V20Hz | = 63.7|V770Hz |
Thus, the 20Hz signal can be 63.7 times as large as the DTMF tones.
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15
Active Filter Circuits
Assessment Problems
AP 15.1
H(s) =
−(R2 /R1 )s
s + (1/R1 C)
1
= 1 rad/s;
R1 C
R2
= 1,
R1
.·.
R1 = 1 Ω,
.·. R2 = R1 = 1 Ω
Hprototype(s) =
AP 15.2
H(s) =
.·. C = 1 F
−s
s+1
−(1/R1 C)
−20,000
=
s + (1/R2 C)
s + 5000
1
= 20,000;
R1 C
.·. R1 =
C = 5 µF
1
= 10 Ω
(20,000)(5 × 10−6 )
1
= 5000
R2 C
.·. R2 =
1
= 40 Ω
(5000)(5 × 10−6 )
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
15–1 system, or transmission in any form or by any means, electronic,
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15–2
CHAPTER 15. Active Filter Circuits
AP 15.3
ωc = 2πfc = 2π × 104 = 20,000π rad/s
.·. kf = 20,000π = 62,831.85
C0 =
C
kf km
.·. km =
.·.
0.5 × 10−6 =
1
kf km
1
= 31.83
(0.5 × 10−6 )(62,831.85)
AP 15.4 For a 2nd order Butterworth high pass filter
H(s) =
s2
√
s2 + 2s + 1
For the circuit in Fig. 15.25
H(s) =
s2
s2 +
2
R2 C
s+
1
R1 R2 C 2
Equate the transfer functions. For C = 1F,
√
2
= 2,
R2 C
√
.·. R2 =
1
= 1,
R1 R2 C 2
2 = 0.707 Ω
1
.·. R1 = √ = 1.414 Ω
2
AP 15.5
Q = 8, K = 5, ωo = 1000 rad/s, C = 1 µF
For the circuit in Fig 15.26
1
−
s
R1 C
!
H(s) =
R
+
R
2
2
1
s2 +
s+
R3 C
R1 R2 R3C 2
Kβs
= 2
s + βs + ωo2
β=
2
,
R3 C
β=
ωo
1000
=
= 125 rad/s
Q
8
.·.
R3 =
2
βC
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Problems
15–3
2 × 106
.·. R3 =
= 16 kΩ
(125)(1)
Kβ =
1
R1 C
.·. R1 =
1
1
=
= 1.6 kΩ
KβC
5(125)(1 × 10−6 )
R1 + R2
R1 R2 R3 C 2
ωo2 =
106 =
(1600 + R2 )
(1600)(R2 )(16,000)(10−6 )2
Solving for R2,
R2 =
(1600 + R2 )106
,
256 × 105
246R2 = 16,000,
R2 = 65.04 Ω
AP 15.6
ωo = 1000 rad/s;
Q = 4;
C = 2 µF
s2 + (1/R2 C 2)
#
4(1
−
σ)
1
2
s +
s+
RC
R2 C 2
2
2
s + ωo
1
= 2
;
ωo =
;
2
s + βs + ωo
RC
H(s) =
"
R=
1
1
=
= 500 Ω
ωo C
(1000)(2 × 10−6 )
β=
ωo
1000
=
= 250
Q
4
.·.
β=
4(1 − σ)
RC
4(1 − σ)
= 250
RC
4(1 − σ) = 250RC = 250(500)(2 × 10−6 ) = 0.25
1−σ =
0.25
= 0.0625;
4
.·.
σ = 0.9375
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15–4
CHAPTER 15. Active Filter Circuits
Problems
P 15.1
Summing the currents at the inverting input node yields
0 − Vi 0 − Vo
+
=0
Zi
Zf
.·.
Vo
Vi
=−
Zf
Zi
Vo
Zf
.·. H(s) =
=−
Vi
Zi
P 15.2
[a] Zf =
R2 (1/sC2 )
R2
=
[R2 + (1/sC2 )]
R2 C2 s + 1
(1/C2 )
s + (1/R2 C2)
Likewise
(1/C1 )
Zi =
s + (1/R1 C1 )
=
.·. H(s) =
−(1/C2 )[s + (1/R1 C1)]
[s + (1/R2 C2)](1/C1 )
=−
C1 [s + (1/R1 C1)]
C2 [s + (1/R2 C2)]
"
−C1 jω + (1/R1 C1 )
[b] H(jω) =
C2 jω + (1/R2 C2 )
H(j0) =
−C1
C2
R2C2
R1C1
=
#
−R2
R1
!
C1 j
−C1
[c] H(j∞) = −
=
C2 j
C2
[d] As ω → 0 the two capacitor branches become open and the circuit reduces
to a resistive inverting amplifier having a gain of −R2/R1 .
As ω → ∞ the two capacitor branches approach a short circuit and in
this case we encounter an indeterminate situation; namely vn → vi but
vn = 0 because of the ideal op amp. At the same time the gain of the
ideal op amp is infinite so we have the indeterminate form 0 · ∞.
Although ω = ∞ is indeterminate we can reason that for finite large
values of ω H(jω) will approach −C1 /C2 in value. In other words, the
circuit approaches a purely capacitive inverting amplifier with a gain of
(−1/jωC2 )/(1/jωC1 ) or −C1 /C2 .
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Problems
P 15.3
15–5
(1/C2 )
s + (1/R2 C2 )
[a] Zf =
Zi = R1 +
1
R1
=
[s + (1/R1 C1 )]
sC1
s
H(s) = −
(1/C2 )
s
·
[s + (1/R2 C2)] R1 [s + (1/R1 C1)]
=−
[b] H(jω) = −
H(j0) = 0
1
s
R1C2 [s + (1/R1 C1 )][s + (1/R2 C2 )]
1
R1C2 jω +
jω
1
R1 C1
jω +
1
R2 C2
[c] H(j∞) = 0
[d] As ω → 0 the capacitor C1 disconnects vi from the circuit. Therefore
vo = vn = 0.
As ω → ∞ the capacitor short circuits the feedback network, thus
ZF = 0 and therefore vo = 0.
P 15.4
[a] ωc =
1
R2 C
K=
R2
R1
so R2 =
so R1 =
1
1
=
= 6366 Ω
ωc C
2π(2500)(10 × 10−9 )
R2
6366
=
= 1273 Ω
K
5
[b] Both the cutoff frequency and the passband gain are changed.
P 15.5
[a] 5(2) = 10 V so Vcc ≥ 10 V
[b] H(jω) =
−5(2π)(2500)
jω + 2π(2500)
H(j5000π) =
−5(5000π)
5
= −2.5 + j2.5 = √ /135◦
5000π + j5000π
2
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15–6
CHAPTER 15. Active Filter Circuits
10
Vo = √ /135◦ Vi
2
[c] H(j1000π) =
−5(5000π)
= 4.9/168.7◦
5000π + j1000π
Vo = 4.9/168.7◦ Vi
[d] H(j25,000π) =
so vo(t) = 9.8 cos(1000πt + 168.7◦ ) V
−5(5000π)
= 0.98/101.3◦
5000π + j25,000π
Vo = 0.98/101.3◦ Vi
P 15.6
so vo (t) = 7.07 cos(5000πt + 135◦ ) V
so vo (t) = 1.96 cos(25,000πt + 101.3◦ ) V
[a] K = 10(10/20) = 3.16 =
R2
R1
R2 =
1
1
=
= 212.21 Ω
3
ωc C
(2π)(10 )(750 × 10−9 )
R1 =
R2
212.21
=
= 67.16 Ω
K
3.16
[b]
P 15.7
[a]
1
= 2π(1000) so RC = 1.5915 × 10−4
RC
There are several possible approaches. Here, choose Rf = 150 Ω. Then
1.5915 × 10−4
= 1.06 × 10−6
150
Choose C = 1 µF. This gives
C=
ωc =
1
= 6.67 × 103 rad/s so fc = 1061 Hz
(150)(10−6 )
To get a passband gain of 10 dB, choose
Ri =
Rf
150
=
= 47.47 Ω
3.16
3.16
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Problems
15–7
Choose Ri = 47 Ω to give K = 20 log 10(150/47) = 10.08 dB. The resulting
circuit is
1061 − 1000
(100) = 6.1%
1000
10.08 − 10
% error in passband gain =
(100) = 0.8%
10
[b] % error in fc =
P 15.8
[a] ωc =
1
R1 C
K=
R2
R1
so R1 =
1
1
=
= 159 Ω
ωc C
2π(4000)(250 × 10−9 )
so R2 = KR1 = (8)(159) = 1273 Ω
[b] The passband gain changes but the cutoff frequency is unchanged.
P 15.9
[a] 8(0.25) = 2 V so Vcc ≥ 2 V
−8jω
[b] H(jω) =
jω + 8000π
H(j600π) =
−8(j8000π)
8
= √ / − 135◦
8000π + j8000π
2
8
Vo = √ / − 135◦ Vi
2
[c] H(j1600π) =
so vo (t) = 1.41 cos(8000πt − 135◦ ) V
−8(j1600π)
= 1.57/ − 101.3◦
8000π + j1600π
Vo = 1.57/ − 101.3◦ Vi
so vo (t) = 392.2 cos(1600πt − 101.3◦ ) mV
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15–8
CHAPTER 15. Active Filter Circuits
[d] H(j40,000π) =
−8(j40,000π)
= 7.84/ − 168.7◦
8000π + j40,000π
Vo = 7.84/ − 168.7◦ Vi
P 15.10 [a] R1 =
so vo (t) = 1.96 cos(40,000πt − 168.7◦ ) V
1
1
=
= 5.10 kΩ
3
ωc C
(2π)(8 × 10 )(3.9 × 10−9 )
K = 10(14/20) = 5.01 =
R2
R1
.·. R2 = 5.01R1 = 25.55 kΩ
[b]
P 15.11 [a]
1
= 2π(8000) so RC = 19.89 × 10−6
RC
There are several possible approaches. Here, choose C = 0.047 µF. Then
19.89 × 10−6
= 423
0.047 × 10−6
Choose Ri = 390 Ω. This gives
Ri =
ωc =
1
= 54.56 krad/s so fc = 8.68 kHz
(0.047 × 10−6 )(390)
To get a passband gain of 14 dB, choose
Rf = 5Ri = 5(390) = 1950 Ω
Choose Rf = 1.8 kΩ to give a passband gain of 20 log 10(1800/390) = 13.3
dB. The resulting circuit is
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Problems
15–9
8683.76 − 8000
(100) = 8.5%
8000
[b] % error in fc =
% error in passband gain =
13.3 − 14
(100) = −5.1%
14
P 15.12 For the RC circuit
H(s) =
Vo
s
=
Vi
s + (1/RC )
R0 = km R;
C0 =
C
km kf
RC
1
.·. R0 C 0 =
= ;
kf
kf
H 0 (s) =
1
= kf
R0 C 0
s
s
(s/kf )
=
=
s + (1/R0 C 0)
s + kf
(s/kf ) + 1
For the RL circuit
H(s) =
s
s + (R/L)
R0 = km R;
L0 =
km L
kf
R0
= kf
L0
R
= kf
L
H 0 (s) =
s
(s/kf )
=
s + kf
(s/kf ) + 1
P 15.13 For the RC circuit
H(s) =
Vo
(1/RC)
=
Vi
s + (1/RC )
R0 = km R;
C0 =
C
km kf
C
1
1
.·. R0 C 0 = km R
= RC =
km kf
kf
kf
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15–10
CHAPTER 15. Active Filter Circuits
1
= kf
R0 C 0
H 0 (s) =
(1/R0 C 0)
kf
=
s + (1/R0 C 0)
s + kf
H 0 (s) =
1
(s/kf ) + 1
For the RL circuit
R0 = km R;
L0 =
R0
km R
= km = kf
0
L
L
kf
R
L
H(s) =
R/L
s + R/L
so
km
L
kf
= kf
H 0 (s) =
(R0 /L0 )
kf
=
0
0
s + (R /L )
s + kf
H 0 (s) =
1
(s/kf ) + 1
(R/L)s
βs
= 2
+ (R/L)s + (1/LC)
s + βs + ωo2
For the prototype circuit ωo = 1 and β = ωo /Q = 1/Q.
For the scaled circuit
P 15.14 H(s) =
s2
(R0 /L0 )s
H (s) = 2
s + (R0 /L0 )s + (1/L0 C 0)
0
where R0 = km R; L0 =
km
C
L; and C 0 =
kf
kf km
R0
km R
.·.
= km = kf
0
L
L
kf
R
L
= kf β
kf2
1
kf km
=
=
= kf2
km
L0 C 0
LC
LC
kf
Q0 =
ωo0
kf ωo
=
=Q
β0
kf β
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Problems
15–11
therefore the Q of the scaled circuit is the same as the Q of the unscaled
circuit. Also note β 0 = kf β.
.·. H 0 (s) =
s2 +
H 0 (s) = 2
s
kf
P 15.15 [a] L = 1 H;
R=
kf
s
Q
kf
s+
Q
1
Q
+
1
Q
s
kf
s
kf
kf2
+1
C = 1F
1
1
=
= 0.05 Ω
Q
20
ωo0
= 40,000;
ωo
Thus,
[b] kf =
km =
R0
5000
=
= 100,000
R
0.05
R0 = km R = (0.05)(100,000) = 5 kΩ
L0 =
km
100,000
L=
(1) = 2.5 H
kf
40,000
C0 =
C
1
=
= 250 pF
km kf
(40,000)(100,000)
[c]
P 15.16 [a] Since ωo2 = 1/LC and ωo = 1 rad/s,
C=
1
1
= F
L
Q
[b] H(s) =
s2
H(s) =
(R/L)s
+ (R/L)s + (1/LC)
s2
(1/Q)s
+ (1/Q)s + 1
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15–12
CHAPTER 15. Active Filter Circuits
[c] In the prototype circuit
R = 1 Ω;
L = 16 H;
.·. km =
1
= 0.0625 F
L
C=
R0
= 10,000;
R
kf =
ωo0
= 25,000
ωo
Thus
R0 = km R = 10 kΩ
L0 =
km
10,000
L=
(16) = 6.4 H
kf
25,000
C0 =
C
0.0625
=
= 250 pF
km kf
(10,000)(25,000)
[d]
[e] H 0 (s) = H 0 (s) =
s
25,000
s2
1
16
2
+
s
25,000
1
16
s
25,000
+1
1562.5s
+ 1562.5s + 625 × 106
P 15.17 [a] Using the first prototype
ωo = 1 rad/s;
km =
C = 1 F;
L = 1 H;
R0
40,000
=
= 1600;
R
25
kf =
R = 25 Ω
ωo0
= 50,000
ωo
Thus,
R0 = km R = 40 kΩ;
C0 =
L0 =
km
1600
L=
(1) = 32 mH;
kf
50,000
C
1
=
= 12.5 nF
km kf
(1600)(50,000)
Using the second prototype
ωo = 1 rad/s;
L=
1
= 40 mH;
25
C = 25 F
R = 1Ω
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Problems
R0
= 40,000;
R
kf =
ωo0
= 50,000
ωo
R0 = km R = 40 kΩ;
L0 =
km
40,000
L=
(0.04) = 32 mH;
kf
50,000
km =
15–13
Thus,
C0 =
C
25
=
= 12.5 nF
km kf
(40,000)(50,000)
[b]
P 15.18 For the scaled circuit
H 0 (s) =
L0 =
.·.
s2 +
s2 +
km
L;
kf
R0
L0
1
L0 C 0
s+
C0 =
R
L
1
L0 C 0
C
km kf
kf2
1
=
;
L0 C 0
LC
R0
.·.
= kf
L0
R0 = km R
It follows then that
2
s +
H 0 (s) =
s2 +
R
L
= 2
s
kf
kf s +
s
kf
+
kf2
LC
2
+
R
L
= H(s)|s=s/kf
kf2
LC
1
LC
s
kf
+
1
LC
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15–14
CHAPTER 15. Active Filter Circuits
P 15.19 For the circuit in Fig. 15.31
H(s) =
s2 +
s
RC
s2 +
1
LC
+
It follows that
1
LC
s2 + L01C 0
H (s) = 2
s + R0sC 0 + L01C 0
0
where R0 = km R;
C0 =
.·.
km
L;
kf
L0 =
C
km kf
kf2
1
=
L0 C 0
LC
1
kf
=
0
0
RC
RC
2
s +
H 0 (s) =
s2 +
= 2
s
kf
kf
RC
+
kf2
LC
s
kf
kf2
LC
s+
2
1
RC
= H(s)|s=s/kf
1
LC
+
s
kf
+
1
LC
P 15.20 [a] For the circuit in Fig. P15.20(a)
1
s+
Vo
s2 + 1
s
H(s) =
=
=
1
1
2 + 1 s+1
Vi
s
+s+
Q
Q
s
For the circuit in Fig. P15.18(b)
H(s) =
Qs + Qs
Vo
=
Vi
1 + Qs + Qs
=
H(s) =
Q(s2 + 1)
Qs2 + s + Q
s2 + 1
s2 +
1
Q
s+1
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Problems
2
s
50,000
2
s
+ 15
50,000
2
[b] H 0 (s) =
15–15
+1
s
50,000
8
+1
s + 25 × 10
s2 + 10,000s + 25 × 108
=
P 15.21 For prototype circuit (a):
H(s) =
Vo
Q
Q
=
1 =
Vi
Q + s+ 1
Q + s2s+1
s
H(s) =
Q(s2 + 1)
s2 + 1
=
Q(s2 + 1) + s
s2 + Q1 s + 1
For prototype circuit (b):
H(s) =
=
Vo
1
=
Vi
1 + (s(s/Q)
2 +1)
s2 + 1
s2 +
P 15.22 [a] km =
L0 =
[b]
1
Q
s+1
R0
1000
=
= 1000;
R
1
kf =
C
1
=
= 5000
0
km C
(1000)(200 × 10−9 )
km
1000
(L) =
(1) = 200 mH
kf
5000
V − 10/s
V
V
+
+
=0
1000
0.2s 1000 + (5 × 106 /s)
V
1
5
s
+ +
1000 s 1000s + 5 × 106
=
1
100s
V =
10(s + 5000)
5(s + 5000)
=
2s2 + 10,000s + 25 × 106
s2 + 5000s + 12.5 × 106
Io =
V
25(s + 5000)
=
2
0.2s
s(s + 5000s + 12.5 × 106 )
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15–16
CHAPTER 15. Active Filter Circuits
=
K1
K2
K2∗
+
+
s
s + 2500 − j2500 s + 2500 + j2500
K1 = 0.01;
K2 = −0.005
io (t) = 10 − 10e−2500t cos 2500t mA
Since km = 1000 and the source voltage didn’t change, the amplitude of
the current is reduced by a factor of 1000. Since kf = 5000 the
coefficients of t are multiplied by 5000.
P 15.23 km =
C0 =
R0
5000
=
= 100;
R
50
C
4 × 10−3
=
= 8 nF
km kf
(100)(5000)
50 Ω → 5 kΩ;
L0 =
ωo0
kf =
= 5000
ωo
700 Ω → 70 kΩ
km
100
L=
(20) = 0.4 H
kf
5000
0.05vφ →
0.05
vφ = 5 × 10−4 vφ
100
The original expression for the current:
io (t) = 1728 + 2880e−20t cos(15t − 233.13◦ ) mA
The frequency components will be multiplied by kf = 5000:
20 → 20(5000) = 105 ;
15 → 15(5000) = 75,000
The magnitudes will be reduced by km = 100:
1728 → 1728/100 = 17.28;
2880 → 2880/100 = 28.80
The expression for the current in the scaled circuit is thus,
5
io (t) = 17.28 + 28.80e−10 t cos(75,000t − 233.13◦ ) mA
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Problems
15–17
P 15.24 From the solution to Problem 14.22, ωo = 100 krad/s and β = 12.5 krad/s.
Compute the two scale factors:
kf =
ωo0
2π(200 × 103 )
=
= 4π
ωo
100 × 103
km =
1 C
1 10 × 10−9
1
=
=
0
−9
kf C
4π 2.5 × 10
π
Thus,
R0 = km R =
8000
= 2546.48 Ω
π
L0 =
km
1/π
L=
(10 × 10−3 ) = 253.3 µH
kf
4π
Calculate the cutoff frequencies:
0
ωc1
= kf ωc1 = 4π(93.95 × 103 ) = 1180.6 krad/s
0
ωc2
= kf ωc2 = 4π(106.45 × 103 ) = 1337.7 krad/s
To check, calculate the bandwidth:
0
0
β 0 = ωc2
− ωc1
= 157.1 krad/s = 4πβ (checks!)
P 15.25 From the solution to Problem 14.35, ωo = 106 rad/s and β = 2π(10.61) krad/s.
Calculate the scale factors:
kf =
ωo0
50 × 103
=
= 0.05
ωo
106
kf L0
0.05(200 × 10−6 )
km =
=
= 0.2
L
50 × 10−6
Thus,
R = km R = (0.2)(750) = 150 Ω
0
C
20 × 10−9
C =
=
= 2 µF
km kf
(0.2)(0.05)
0
Calculate the bandwidth:
β 0 = kf β = (0.05)[2π(10.6 × 103 )] = 3330 rad/s
To check, calculate the quality factor:
Q=
ωo
106
=
= 15
β
2π(10.61 × 103 )
Q0 =
ωo0
50 × 103
=
= 15 (checks)
β0
3330
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15–18
CHAPTER 15. Active Filter Circuits
P 15.26 [a] From Eq 15.1 we have
H(s) =
−Kωc
s + ωc
where K =
R2
,
R1
.·. H 0 (s) =
−K 0 ωc0
s + ωc0
where K 0 =
R02
R01
ωc =
1
R2C
ωc0 =
1
R02 C 0
By hypothesis R01 = km R1 ;
and C 0 =
R02 = km R2,
C
. It follows that
kf km
K 0 = K and ωc0 = kf ωc , therefore
H 0 (s) =
−Kkf ωc
−Kωc
=s
s + kf ωc
+ ωc
kf
[b] H(s) =
−K
s+1
−K
[c] H 0 (s) = s kf
+1
=
−Kkf
s + kf
P 15.27 [a] From Eq. 15.4
H(s) =
ωc =
−Ks
R2
where K =
and
s + ωc
R1
1
R1C
.·. H 0 (s) =
and ωc0 =
−K 0 s
R02
0
where
K
=
s + ωc0
R01
1
R01 C 0
By hypothesis
R01 = km R1 ;
R02 = km R2 ;
C0 =
C
km kf
It follows that
K 0 = K and ωc0 = kf ωc
.·. H 0 (s) =
−Ks
−K(s/kf )
= s
s + kf ωc
+ ωc
kf
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Problems
−Ks
s+1
−K(s/kf )
[b] H(s) =
[c] H 0 (s) = P 15.28 [a] Hhp =
s
kf
+1
s
;
s+1
0
.·. Hhp
=
=
.·. RH =
kf =
=
1
= 1.59 kΩ
(2000π)(0.1 × 10−6 )
ωo0
5000(2π)
=
= 10,000π
ω
1
10,000π
s + 10,000π
1
= 10,000π;
RL CL
[b] H 0(s) =
ωo0
1000(2π)
=
= 2000π
ω
1
s
s + 2000π
1
;
s+1
0
.·. Hlp
=
−Ks
s + kf
kf =
1
= 2000π;
RH CH
Hlp =
15–19
.·. RL =
1
= 318.3 Ω
(10,000π)(0.1 × 10−6 )
s
10,000π
·
s + 2000π s + 10,000π
10,000πs
(s + 2000π)(s + 10,000π)
√
(2000π)(10,000π) = 1000π 20 rad/s
√
(10,000π)(j1000π
20)
√
√
H 0 (jωo ) =
(2000π + j1000π 20)(10,000π + j1000π 20)
√
j10 20
√
√
=
= 0.8333/0◦
(2 + j 20)(10 + j 20)
[c] ωo =
√
ωc1 ωc1 =
q
[d] G = 20 log 10(0.8333) = −1.58 dB
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15–20
CHAPTER 15. Active Filter Circuits
[e]
P 15.29 [a] For the high-pass section:
ωo0
4000(2π)
=
= 8000π
ω
1
s
H 0 (s) =
s + 8000π
kf =
. ·.
1
= 8000π;
R1 (10 × 10−9 )
R1 = 3.98 kΩ
.·.
R2 = 3.98 kΩ
For the low-pass section:
kf =
ωo0
400(2π)
=
== 800π
ω
1
H 0 (s) =
. ·.
800π
s + 800π
1
= 800π;
R2 (10 × 10−9 )
R2 = 39.8 kΩ
. ·.
R1 = 39.8 kΩ
0 dB gain corresponds to K = 1. In the summing amplifier we are free to
choose Rf and Ri so long as Rf /Ri = 1. To keep from having many
different resistance values in the circuit we opt for Rf = Ri = 39.8 kΩ.
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Problems
15–21
[b]
[c] H 0 (s) =
s
800π
+
s + 8000π s + 800π
s2 + 1600πs + 64 × 105 π 2
(s + 800π)(s + 8000π)
q
√
[d] ωo = (8000π)(800π) = 800π 10
√
√
√
−(800π 10)2 + 1600π(j800π 10) + 64 × 105 π 2
0
√
√
H (j800π 10) =
(800π + j800π 10)(8000π + j800π 10)
√
j128 × 104 10π 2
√
√
=
(800π)2 (1 + j 10)(10 + j 10)
√
j2 10
√
√
=
(1 + j 10)(10 + j 10)
=
= 0.1818/0◦
[e] G = 20 log 10 0.1818 = −14.81 dB
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15–22
CHAPTER 15. Active Filter Circuits
[f]
P 15.30 ωo = 2πfo = 400π rad/s
β = 2π(1000) = 2000π rad/s
.·. ωc2 − ωc1 = 2000π
√
ωc1 ωc2 = ωo = 400π
Solve for the cutoff frequencies:
ωc1 ωc2 = 16 × 104 π 2
ωc2 =
16 × 104 π 2
ωc1
16 × 104 π 2
·
..
− ωc1 = 2000π
ωc1
or ωc21 + 2000πωc1 − 16 × 104 π 2 = 0
ωc1 = −1000π ±
√
ωc1 = 1000π(−1 ±
106 π 2 + 0.16 × 106 π 2
√
1.16) = 242.01 rad/s
.·. ωc2 = 2000π + 242.01 = 6525.19 rad/s
Thus, fc1 = 38.52 Hz
Check:
and
fc2 = 1038.52 Hz
β = fc2 − fc1 = 1000Hz
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Problems
ωc2 =
1
= 6525.19
RL CL
RL =
1
= 30.65 Ω
(6525.19)(5 × 10−6 )
ωc1 =
1
= 242.01
RH CH
RH =
1
= 826.43 Ω
(242.01)(5 × 10−6 )
P 15.31 ωo = 1000 rad/s;
GAIN = 6
β = 4000 rad/s;
C = 0.2 µF
15–23
β = ωc2 − ωc1 = 4000
ωo =
√
ωc1 ωc2 = 1000
Solve for the cutoff frequencies:
.·. ωc21 + 4000ωc1 − 106 = 0
√
ωc1 = −2000 ± 1000 5 = 236.07 rad/s
ωc2 = 4000 + ωc1 = 4236.07 rad/s
Check:
ωc1 =
β = ωc2 − ωc1 = 4000 rad/s
1
RL CL
.·. RL =
1
(0.2 ×
10−6 )(236.07)
= 21.18 kΩ
1
= 4236.07
RH CH
RH =
1
(0.2 ×
10−6 )(4236.07)
= 1.18 kΩ
Rf
=6
Ri
If Ri = 1 kΩ
Rf = 6Ri = 6 kΩ
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15–24
CHAPTER 15. Active Filter Circuits
P 15.32 H(s) =
Zf =
Vo
−Zf
=
Vi
Zi
1
(1/C2 )
kR2 =
;
sC2
s + (1/R2 C2 )
Zi = R1 +
1
sR1 C1 + 1
=
sC1
sC1
−1/C2
−(1/R1 C2)s
s + (1/R2 C2)
.·. H(s) =
=
s + (1/R1 C1)
[s + (1/R1 C1 )][s + (1/R2 C2 )]
s/R1
=
−Kβs
s2 + βs + ωo2
−250s
−250s
−3.57(70s)
√
= 2
= 2
(s + 50)(s + 20)
s + 70s + 1000
s + 70s + ( 1000)2
√
ωo = 1000 = 31.6 rad/s
[a] H(s) =
β = 70 rad/s
K = −3.57
ωo
[b] Q =
= 0.45
β
ωc1,2 = ±
β
+
2
v
!
u
u β 2
t
ωc1 = 12.17 rad/s
P 15.33 [a] H(s) =
2
+ ωo2 = ±35 +
√
352 + 1000 = ±35 + 47.17
ωc2 = 82.17 rad/s
(1/sC)
(1/RC)
=
R + (1/sC)
s + (1/RC)
H(jω) =
(1/RC)
jω + (1/RC)
(1/RC)
|H(jω)| = q
ω 2 + (1/RC)2
|H(jω)|2 =
(1/RC)2
ω 2 + (1/RC)2
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Problems
15–25
[b] Let Va be the voltage across the capacitor, positive at the upper terminal.
Then
Va − Vin
Va
+ sCVa +
=0
R1
R2 + sL
Solving for Va yields
Va =
R1
LCs2
(R2 + sL)Vin
+ (R1 R2 C + L)s + (R1 + R2)
But
vo =
sLVa
R2 + sL
Therefore
Vo =
R1
H(s) =
LCs2
R1
H(jω) =
sLVin
+ (L + R1 R2 C)s + (R1 + R2 )
LCs2
sL
+ (L + R1R2 C)s + (R1 + R2)
jωL
[(R1 + R2) − R1 LCω 2 ] + jω(L + R1 R2C)
ωL
|H(jω)| = q
[R1 + R2 − R1LCω 2 ]2 + ω 2 (L + R1 R2 C)2
|H(jω)|2 =
=
ω 2 L2
(R1 + R2 − R1 LCω 2)2 + ω 2 (L + R1 R2 C)2
ω 2 L2
R21 L2 C 2ω 4 + (L2 + R21 R22 C 2 − 2R21 LC)ω 2 + (R1 + R2 )2
[c] Let Va be the voltage across R2 positive at the upper terminal. Then
Va − Vin
Va
+
+ Va sC + Va sC = 0
R1
R2
(0 − Va )sC + (0 − Va )sC +
.·. Va =
0 − Vo
=0
R3
R2 Vin
2R1 R2 Cs + R1 + R2
and Va = −
Vo
2R3 Cs
It follows directly that
H(s) =
Vo
−2R2 R3 Cs
=
Vin
2R1 R2 Cs + (R1 + R2 )
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15–26
CHAPTER 15. Active Filter Circuits
H(jω) =
−2R2 R3 C(jω)
(R1 + R2 ) + jω(2R1 R2 C)
2R2 R3 Cω
|H(jω)| = q
(R1 + R2 )2 + ω 2 4R21 R22 C 2
|H(jω)|2 =
4R22 R23 C 2ω 2
(R1 + R2 )2 + 4R21 R22 C 2ω 2
P 15.34 For the scaled circuit
H 0 (s) =
1/(R0 )2C10 C20
s2 + R02C 0 s + (R0)21C 0 C 0
1
1
2
where
R0 = km R;
C10 = C1/kf km ;
C20 = C2/kf km
It follows that
kf2
1
=
(R0 )2 C10 C20
R2 C1C2
2
R0 C10
=
2kf
RC1
kf2 /RC1 C2
.·. H 0 (s) =
s2 +
k2
2kf
s
RC1
= 2
s
kf
+ R2 Cf1 C2
1/RC1 C2
+
2
RC1
s
kf
+
1
R2 C1 C2
1
P 15.35 [a] y = 20 log 10 √
= −10 log 10 (1 + ω 2n )
1 + ω 2n
From the laws of logarithms we have
−10
y=
ln(1 + ω 2n )
ln 10
Thus
dy
−10 2nω 2n−1
=
dω
ln 10 (1 + ω 2n )
x = log10 ω =
ln ω
ln 10
.·. ln ω = x ln 10
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Problems
1 dω
= ln 10,
ω dx
dy
=
dx
dy
dω
!
15–27
dω
= ω ln 10
dx
dω
dx
!
=
−20nω 2n
dB/decade
1 + ω 2n
at ω = ωc = 1 rad/s
dy
= −10n dB/decade.
dx
1
[b] y = 20 log 10 √
= −10n log 10 (1 + ω 2 )
[ 1 + ω 2 ]n
=
−10n
ln(1 + ω 2)
ln 10
dy
−10n
=
dω
ln 10
1
−20nω
2ω =
2
1+ω
(ln 10)(1 + ω 2 )
As before
dω
= ω(ln 10);
dx
dy
−20nω 2
.·.
=
dx
(1 + ω 2 )
√
At the corner ωc = 21/n − 1
.·. ωc2 = 21/n − 1
dy
−20n[21/n − 1]
=
dB/decade.
dx
21/n
[c] For the Butterworth Filter
For the cascade of identical sections
n
dy/dx (dB/decade)
n
dy/dx (dB/decade)
1
−10
1
−10
2
−20
2
−11.72
3
−30
3
−12.38
4
−40
4
−12.73
∞
−∞
∞ −13.86
[d] It is apparent from the calculations in part (c) that as n increases the
amplitude characteristic at the cutoff frequency decreases at a much
faster rate for the Butterworth filter.
Hence the transition region of the Butterworth filter will be much
narrower than that of the cascaded sections.
P 15.36 [a] n ∼
=
(−0.05)(−30) ∼
= 2.76
log10 (7000/2000)
. ·. n = 3
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15–28
CHAPTER 15. Active Filter Circuits
[b] Gain = 20 log 10 q
1
1 + (7000/2000)6
= −32.65 dB
1
(s + 1)(s2 + s + 1)
[b] fc = 2000 Hz;
ωc = 4000π rad/s;
P 15.37 [a] H(s) =
H 0 (s) =
kf = 4000π
1
( ksf
+
1)[( ksf )2
+
s
kf
+ 1]
=
kf3
(s + kf )(s2 + kf s + kf2 )
=
(4000π)3
(s + 4000π)[s2 + 4000πs + (4000π)2 ]
[c] H 0 (j14,000π) =
64
(4 + j14)(−180 + j52)
= 0.02332/ − 236.77◦
Gain = 20 log 10 (0.02332) = −32.65 dB
P 15.38 [a] In the first-order circuit R = 1 Ω and C = 1 F.
km =
R0
1000
=
= 1000;
R
1
R0 = km R = 1000 Ω;
kf =
C0 =
ωo0
2π(2000)
=
= 4000π
ωo
1
C
1
=
= 79.58 nF
km kf
(1000)(4000π)
In the second-order circuit R = 1 Ω, 2/C1 = 1 so C1 = 2 F, and
C2 = 1/C1 = 0.5 F. Therefore in the scaled second-order circuit
R0 = km R = 1000 Ω;
C20 =
C10 =
C1
2
=
= 159.15 nF
km kf
(1000)(4000π)
C2
0.5
=
= 39.79 nF
km kf
(1000)(4000π)
[b]
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Problems
15–29
(−0.05)(−48)
= 3.99 .·. n = 4
log10 (2000/500)
From Table 15.1 the transfer function of the first section is
P 15.39 [a] n =
s2
s2 + 0.765s + 1
For the prototype circuit
H1 (s) =
2
= 0.765;
R2
R2 = 2.61 Ω;
R1 =
1
= 0.383 Ω
R2
The transfer function of the second section is
s2
H2 (s) = 2
s + 1.848s + 1
For the prototype circuit
2
= 1.848;
R2
R2 = 1.082 Ω;
R1 =
1
= 0.9240 Ω
R2
The scaling factors are:
kf =
ωo0
2π(2000)
=
= 4000π
ωo
1
C0 =
C
km kf
. ·.
km =
. ·.
10 × 10−9 =
1
4000πkm
1
= 7957.75
4000π(10 × 10−9 )
Therefore in the first section
R01 = km R1 = 3.05 kΩ;
R02 = km R2 = 20.77 kΩ
In the second section
R01 = km R1 = 7.35 kΩ;
R02 = km R2 = 8.61 kΩ
[b]
P 15.40 n = 5: 1 + (−1)5 s10 = 0;
s10 = 1/(0 + 360k)◦
so
s10 = 1
s = 1/36k ◦
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15–30
CHAPTER 15. Active Filter Circuits
k sk+1
k sk+1
0 1/0◦
5 1/180◦
1 1/36◦
6 1/216◦
2 1/72◦
7 1/252◦
3 1/108◦
8 1/288◦
4 1/144◦
9 1/324◦
Group by conjugate pairs to form denominator polynomial.
(s + 1)[s − (cos 108◦ + j sin 108◦ )][(s − (cos 252◦ + j sin 252◦ )]
· [(s − (cos 144◦ + j sin 144◦ )][(s − (cos 216◦ + j sin 216◦ )]
= (s + 1)(s + 0.309 − j0.951)(s + 0.309 + j0.951)·
(s + 0.809 − j0.588)(s + 0.809 + j0.588)
which reduces to
(s + 1)(s2 + 0.618s + 1)(s2 + 1.618s + 1)
n = 6: 1 + (−1)6 s12 = 0
s12 = −1
s12 = 1/180◦ + 360k
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Problems
k sk+1
k sk+1
0 1/15◦
6 1/195◦
1 1/45◦
7 1/225◦
2 1/75◦
8 1/255◦
3 1/105◦
9 1/285◦
4 1/135◦
10 1/315◦
5 1/165◦
11 1/345◦
15–31
Grouping by conjugate pairs yields
(s + 0.2588 − j0.9659)(s + 0.2588 + j0.9659)×
(s + 0.7071 − j0.7071)(s + 0.7071 + j0.7071)×
(s + 0.9659 − j0.2588)(s + 0.9659 + j0.2588)
or (s2 + 0.5176s + 1)(s2 + 1.4142s + 1)(s2 + 1.9318s + 1)
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15–32
CHAPTER 15. Active Filter Circuits
s2
P 15.41 H 0 (s) =
s2 +
s2
H 0 (s) =
s2
=
1
2
s+
2 k2 )
km R2 (C/km kf )
km R1 km R2 (C 2/km
f
kf2
2kf
+
s+
R2 C
R1 R2 C 2
(s/kf )2
!
2
s
1
2
(s/kf ) +
+
R2 C kf
R1 R2C 2
(−0.05)(−48)
= 3.99 .·.
n=4
log10(32/8)
From Table 15.1 the transfer function is
1
H(s) = 2
(s + 0.765s + 1)(s2 + 1.848s + 1)
P 15.42 [a] n =
The capacitor values for the first stage prototype circuit are
2
= 0.765
C1
C2 =
. ·.
C1 = 2.61 F
1
= 0.38 F
C1
The values for the second stage prototype circuit are
2
= 1.848
C1
C2 =
. ·.
C1 = 1.08 F
1
= 0.92 F
C1
The scaling factors are
km =
R0
= 1000;
R
kf =
ωo0
= 16,000π
ωo
Therefore the scaled values for the components in the first stage are
R1 = R2 = R = 1000 Ω
C1 =
2.61
= 52.01 nF
(16,000π)(1000)
C2 =
0.38
= 7.61 nF
(16,000π)(1000)
The scaled values for the second stage are
R1 = R2 = R = 1000 Ω
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Problems
C1 =
1.08
= 21.53 nF
(16,000π)(1000)
C2 =
0.92
= 18.38 nF
(16,000π)(1000)
15–33
[b]
P 15.43 [a] The cascade connection is a bandpass filter.
[b] The cutoff frequencies areq2 kHz and 8 kHz.
The center frequency is (2)(8) = 4 kHz.
The Q is 4/(8 − 2) = 2/3 = 0.67
[c] For the high pass section kf = 4000π. The prototype transfer function is
Hhp (s) =
. ·.
s4
(s2 + 0.765s + 1)(s2 + 1.848s + 1)
0
Hhp
(s) =
·
=
(s/4000π)4
[(s/4000π)2 + 0.765(s/4000π) + 1]
[(s/4000π)2
1
+ 1.848(s/4000π) + 1]
s4
(s2 + 3060πs + 16 × 106 π 2)(s2 + 7392πs + 16 × 106 π 2)
For the low pass section kf = 16,000π
Hlp (s) =
. ·.
(s2
1
+ 0.765s + 1)(s2 + 1.848s + 1)
0
Hlp
(s) =
·
=
[(s/16,000π)2
1
+ 0.765(s/16,000π) + 1]
1
[(s/16,000π)2 + 1.848(s/16,000π) + 1]
(16,000π)4
([s2 + 12,240πs + (16,000π)2 )][s2 + 29,568πs + (16,000π)2 ]
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15–34
CHAPTER 15. Active Filter Circuits
The cascaded transfer function is
0
0
H 0 (s) = Hhp
(s)Hlp
(s)
For convenience let
D1 = s2 + 3060πs + 16 × 106 π 2
D2 = s2 + 7392πs + 16 × 106 π 2
D3 = s2 + 12,240πs + 256 × 106 π 2
D4 = s2 + 29,568πs + 256 × 106 π 2
Then
H 0 (s) =
65,536 × 1012 π 4s4
D1 D2 D3 D4
[d] ωo = 2π(4000) = 8000π rad/s
s = j8000π
s4 = 4096 × 1012 π 4
D1 = (16 × 106 π 2 − 64 × 106 π 2 ) + j(8000π)(3060π)
= 106 π 2(−48 − j24.48) = 106 π 2(53.88/152.98◦ )
D2 = (16 × 106 π 2 − 64 × 106 π 2 ) + j(8000π)(7392π)
= 106 π 2(−48 + j59.136) = 106 π 2 (76.16/129.07◦ )
D1 = (256 × 106 π 2 − 64 × 106 π 2) + j(8000π)(12,240π)
= 106 π 2(192 + j97.92) = 106 π 2(215.53/27.02◦ )
D1 = (256 × 106 π 2 − 64 × 106 π 2) + j(8000π)(29,568π)
= 106 π 2(192 + j236.544) = 106 π 2 (304.66/50.93◦ )
(65,536)(4096)π 8 × 1024
H (jωo ) = 8
(π × 1024 )[(53.88)(76.16)(215.53)(304.66)/360◦ ]
0
= 0.996/ − 360◦ = 0.996/0◦
P 15.44 [a] From the statement of the problem, K = 10 ( = 20 dB). Therefore for the
prototype bandpass circuit
R1 =
R2 =
Q
16
=
= 1.6 Ω
K
10
Q
16
=
Ω
−K
502
2Q2
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Problems
15–35
R3 = 2Q = 32 Ω
The scaling factors are
kf =
ωo0
= 2π(6400) = 12,800π
ωo
km =
C
1
=
= 1243.40
0
−9
C kf
(20 × 10 )(12,800π)
Therefore,
R01 = km R1 = (1.6)(1243.30) = 1.99 kΩ
R02 = km R2 = (16/502)(1243.40) = 39.63 Ω
R03 = km R3 = 32(1243.40) = 39.79 kΩ
[b]
P 15.45 From Eq 15.56 we can write
H(s) =
2
−
R3 C
1
s
R1 C
2
R1 + R2
s2 +
s+
R3C
R1 R2 R3C 2
R3 C
2
or
R3
2
−
s
2R1
R3 C
H(s) =
2
R1 + R2
s2 +
s+
R3 C
R1 R2R3 C 2
Therefore
2
ωo
=β= ;
R3 C
Q
and K =
R1 + R2
= ωo2 ;
R1R2 R3 C 2
R3
2R1
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15–36
CHAPTER 15. Active Filter Circuits
By hypothesis C = 1 F and ωo = 1 rad/s
.·.
2
1
=
or R3 = 2Q
R3
Q
R1 =
R3
Q
=
2K
K
R1 + R2
=1
R1R2 R3
Q
Q
+ R2 =
(2Q)R2
K
K
.·. R2 =
Q
2Q2 − K
P 15.46 [a] First we will design a unity gain filter and then provide the passband gain
with an inverting amplifier. For the high pass section the cut-off
frequency is 500 Hz. The order of the Butterworth is
n=
(−0.05)(−20)
= 2.51
log10(500/200)
. ·. n = 3
Hhp (s) =
s3
(s + 1)(s2 + s + 1)
For the prototype first-order section
R1 = R2 = 1 Ω,
C = 1F
For the prototype second-order section
R1 = 0.5 Ω,
R2 = 2 Ω,
C = 1F
The scaling factors are
ωo0
kf =
= 2π(500) = 1000π
ωo
km =
C
1
106
=
=
C 0 kf
(15 × 10−9 )(1000π)
15π
In the scaled first-order section
106
0
0
R1 = R2 = km R1 =
(1) = 21.22 kΩ
15π
C 0 = 15 nF
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Problems
15–37
In the scaled second-order section
R01 = 0.5km = 10.61 kΩ
R02 = 2km = 42.44 kΩ
C 0 = 15 nF
For the low-pass section the cut-off frequency is 4500 Hz. The order of
the Butterworth filter is
(−0.05)(−20)
n=
= 2.51;
. ·. n = 3
log10(11,250/4500)
Hlp (s) =
1
(s + 1)(s2 + s + 1)
For the prototype first-order section
R1 = R2 = 1 Ω,
C = 1F
For the prototype second-order section
R1 = R2 = 1 Ω;
C1 = 2 F;
C2 = 0.5 F
The low-pass scaling factors are
km =
R0
= 104 ;
R
kf =
ωo0
= (4500)(2π) = 9000π
ωo
For the scaled first-order section
C
1
R01 = R02 = 10 kΩ;
C0 =
=
= 3.54 nF
kf km
(9000π)(104 )
For the scaled second-order section
R01 = R02 = 10 kΩ
C10 =
C1
2
=
= 7.07 nF
kf km
(9000π)(104 )
C20 =
C2
0.5
=
= 1.77 nF
kf km
(9000π)(104 )
GAIN AMPLIFIER
20 log 10 K = 20 dB,
.·. K = 10
Since we are using 10 kΩ resistors in the low-pass stage, we will use
Rf = 100 kΩ and Ri = 10 kΩ in the inverting amplifier stage.
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15–38
CHAPTER 15. Active Filter Circuits
[b]
P 15.47 [a] Unscaled high-pass stage
Hhp (s) =
s3
(s + 1)(s2 + s + 1)
The frequency scaling factor is kf = (ωo0 /ωo ) = 1000π. Therefore the
scaled transfer function is
0
Hhp
(s) = =
(s/1000π)3
s
1000π
(s +
+1
s
1000π
3
1000π)[s2
3
+
s
1000π
+1
s
+ 1000πs + 106 π 2]
Unscaled low-pass stage
Hlp (s) =
1
(s + 1)(s2 + s + 1)
The frequency scaling factor is kf = (ωo0 /ωo ) = 9000π. Therefore the
scaled transfer function is
1
Hlp0 (s) = 2 s
s
s
+1
+ 9000π + 1
9000π
9000π
=
(9000π)3
(s + 9000π)(s2 + 9000πs + 81 × 106 π 2)
Thus the transfer function for the filter is
729 × 1010 π 3s3
0
H 0 (s) = 10Hhp
(s)Hlp0 (s) =
D1 D2 D3 D4
where
D1 = s + 1000π
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Problems
15–39
D2 = s + 9000π
D3 = s2 + 1000πs + 106 π 2
D4 = s2 + 9000πs + 81 × 106 π 2
[b] At 200 Hz
ω = 400π rad/s
D1 (j400π) = 400π(2.5 + j1)
D2 (j400π) = 400π(22.5 + j1)
D3 (j400π) = 4 × 105 π 2 (2.1 + j1.0)
D4 (j400π) = 4 × 105 π 2 (202.1 + j9)
Therefore
D1 D2 D3 D4 (j400π) = 256π 6 1014 (28,534.82/52.36◦ )
H 0 (j400π) =
(729π 3 × 1010 )(64 × 106 π 3)
256π 6 × 1014 (28,534.82/52.36◦ )
= 0.639/ − 52.36◦
.·. 20 log 10 |H 0 (j400π)| = 20 log 10 (0.639) = −3.89 dB
At f = 1500 Hz,
ω = 3000π rad/s
Then
D1 (j3000π) = 1000π(1 + j3)
D2 (j3000π) = 3000π(3 + j1)
D3 (j3000π) = 106 π 2 (−8 + j3)
D4 (j3000π) = 106 π 2 (8 + j3)
H 0 (j3000π) =
(729 × π 3 × 1010 )(27 × 109 π 3)
27 × 1018 π 6(730/270◦ )
= 9.99/90◦
.·. 20 log 10 |H 0 (j3000π)| = 19.99 dB
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15–40
CHAPTER 15. Active Filter Circuits
[c] From the transfer function the gain is down 19.99 + 3.89 or 23.88 dB at
200 Hz. Because the upper cut-off frequency is nine times the lower
cut-off frequency we would expect the high-pass stage of the filter to
predict the loss in gain at 200 Hz. For a 3nd order Butterworth
1
GAIN = 20 log 10 q
= −23.89 dB.
1 + (500/200)6
1500 Hz is in the passband for this bandpass filter. Hence we expect the
gain at 1500 Hz to nearly equal 20 dB as specified in Problem 15.39.
Thus our scaled transfer function confirms that the filter meets the
specifications.
P 15.48 [a] From Table 15.1
Hlp (s) =
1
√
(s2 + 0.518s + 1)(s2 + 2s + 1)(s2 + 1.932s + 1)
Hhp (s) = 1
s2
+ 0.518
1
s
+1
1
s2
1
√ 1
+ 2 s + 1 s12 + 1.932 1s + 1
s6
√
Hhp (s) = 2
(s + 0.518s + 1)(s2 + 2s + 1)(s2 + 1.932s + 1)
P 15.49 [a] kf = 25,000
(s/25,000)6
[(s/25,000)2 + 0.518(s/25,000) + 1]
0
Hhp
(s) =
·
=
[(s/25,000)2
s6
(s2 + 12,950s + 625 × 106 )(s2 + 35,355s + 625 × 106 )
·
(s2
[b] H 0(j25,000) =
=
1
√
+ 2s/25,000 + 1][(s/25,000)2 + 1.932s/25,000 + 1]
1
+ 48,300s + 625 × 106 )
−(25,000)6
[12,950(j25,000)][35,355(j25,000)][48,300(j25,000)]
−(25,000)3
(12,950)(35,355)(48,300)j 3
= 0.7066/ − 90◦
20 log 10 |H 0 (j25,000)| = −3.02 dB
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Problems
15–41
P 15.50 [a] At very low frequencies the two capacitor branches are open and because
the op amp is ideal the current in R3 is zero. Therefore at low frequencies
the circuit behaves as an inverting amplifier with a gain of R2 /R1 . At
very high frequencies the capacitor branches are short circuits and hence
the output voltage is zero.
[b] Let the node where R1 , R2 , R3 , and C2 join be denoted as a, then
(Va − Vi )G1 + Va sC2 + (Va − Vo )G2 + Va G3 = 0
−Va G3 − Vo sC1 = 0
or
(G1 + G2 + G3 + sC2 )Va − G2 Vo = G1 Vi
Va =
−sC1
Vo
G3
Solving for Vo /Vi yields
H(s) =
=
=
=
=
−G1 G3
(G1 + G2 + G3 + sC2)sC1 + G2 G3
s2 C
−G1 G3
1 C2 + (G1 + G2 + G3 )C1 s + G2 G3
−G1 G3 /C1 C2
s2 +
s2 +
s2
where K =
and b1 =
h
i
(G1 +G2 +G3 )
s
C2
1 G2 G3
−G
G2 C1 C2
h
i
(G1 +G2 +G3 )
s
C2
+
G2 G3
C1 C2
+
G2 G3
C1 C2
−Kbo
+ b1 s + bo
G1
;
G2
bo =
G2 G3
C1 C2
G1 + G2 + G3
C2
[c] Rearranging we see that
G1 = KG2
G3 =
b o C1 C2
b o C1
=
G2
G2
since by hypothesis C2 = 1 F
b1 =
G1 + G2 + G3
= G1 + G2 + G3
C2
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15–42
CHAPTER 15. Active Filter Circuits
b o C1
.·. b1 = KG2 + G2 +
G2
b1 = G2 (1 + K) +
b o C1
G2
Solving this quadratic equation for G2 we get
b1
G2 =
±
2(1 + K)
=
b1 ±
q
s
b21 − bo C1 4(1 + K)
4(1 + K)2
b21 − 4bo (1 + K)C1
2(1 + K)
For G2 to be realizable
b21
4bo (1 + K)
C1 <
[d] 1. Select C2 = 1 F
2. Select C1 such that C1 <
b21
4bo (1 + K)
3. Calculate G2 (R2 )
4. Calculate G1 (R1 ); G1 = KG2
5. Calculate G3 (R3 ); G3 = bo C1 /G2
P 15.51 [a] In the second order section of a third order Butterworth filter bo = b1 = 1
Therefore,
C1 ≤
b21
1
=
= 0.05 F
4bo (1 + K)
(4)(1)(5)
. ·.
C1 = 0.05 F (limiting value)
[b] G2 =
1
= 0.1 S
2(1 + 4)
G3 =
1
(0.05) = 0.5 S
0.1
G1 = 4(0.1) = 0.4 S
Therefore,
R1 =
1
= 2.5 Ω;
G1
R2 =
1
= 10 Ω;
G2
R3 =
1
= 2Ω
G3
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Problems
[c] kf =
15–43
ωo0
= 2π(2500) = 5000π
ωo
km =
C2
1
=
= 6366.2
0
C2 kf
(10 × 10−9 )kf
C10 =
0.05
= 0.5 × 10−9 = 500 pF
kf km
R01 = (2.5)(6366.2) = 15.92 kΩ
R02 = (10)(6366.2) = 63.66 kΩ
R03 = (2)(6366.2) = 12.73 kΩ
[d] R01 = R02 = (6366.2)(1) = 6.37 kΩ
C0 =
C
1
= 8 = 10 nF
kf km
10
[e]
P 15.52 [a] By hypothesis the circuit becomes:
For very small frequencies the capacitors behave as open circuits and
therefore vo is zero. As the frequency increases, the capacitive branch
impedances become small compared to the resistive branches. When this
happens the circuit becomes an inverting amplifier with the capacitor C2
dominating the feedback path. Hence the gain of the amplifier
approaches (1/jωC2 )/(1/jωC1 ) or C1 /C2 . Therefore the circuit behaves
like a high-pass filter with a passband gain of C1 /C2 .
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15–44
CHAPTER 15. Active Filter Circuits
[b] Summing the currents away from the upper terminal of R2 yields
Va G2 + (Va − Vi )sC1 + (Va − Vo )sC2 + Va sC3 = 0
or
Va [G2 + s(C1 + C2 + C3)] − Vo sC2 = sC1 Vi
Summing the currents away from the inverting input terminal gives
(0 − Va )sC3 + (0 − Vo )G1 = 0
or
sC3 Va = −G1 Vo ;
Va =
−G1 Vo
sC3
Therefore we can write
−G1 Vo
[G2 + s(C1 + C2 + C3 )] − sC2 Vo = sC1 Vi
sC3
Solving for Vo /Vi gives
H(s) =
Vo
−C1 C3s2
=
Vi
C2 C3s2 + G1 (C1 + C2 + C3 )s + G1 G2 ]
= h
s2 +
=
s2
−C1 2
s
C2
G1
(C1
C2 C3
2
+ C2 + C3 )s +
−Ks
+ b1 s + bo
G1 G2
C2 C3
i
Therefore the circuit implements a second-order high-pass filter with a
passband gain of C1 /C2 .
[c] C1 = K:
b1 =
G1
(K + 2) = G1 (K + 2)
(1)(1)
.·. G1 =
bo =
b1
;
K +2
R1 =
K +2
b1
G1 G2
= G1 G2
(1)(1)
.·. G2 =
bo
bo
= (K + 2)
G1
b1
.·. R2 =
b1
bo (K + 2)
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Problems
15–45
[d] From Table 15.1 the transfer function of the second-order section of a
third-order high-pass Butterworth filter is
H(s) =
Ks2
s2 + s + 1
Therefore b1 = bo = 1
Thus
C1 = K = 8 F
R1 =
8+2
= 10 Ω
1
R2 =
1
= 0.1 Ω
1(8 + 2)
P 15.53 [a] Low-pass filter:
n=
(−0.05)(−30)
= 3.77;
log10(1000/400)
.·. n = 4
In the first prototype second-order section: b1 = 0.765, bo = 1, C2 = 1 F
C1 ≤
b21
(0.765)2
≤
≤ 0.0732
4bo (1 + K)
(4)(2)
choose C1 = 0.03 F
G2 =
0.765 ±
q
(0.765)2 − 4(2)(0.03)
4
=
0.765 ± 0.588
4
Arbitrarily select the larger value for G2 , then
G2 = 0.338 S;
. ·.
G1 = KG2 = 0.338 S;
G3 =
R2 =
. ·.
1
= 2.96 Ω
G2
R1 =
b o C1
(1)(0.03)
=
= 0.089
G2
0.338
. ·.
1
= 2.96 Ω
G1
R3 = 1/G3 = 11.3 Ω
Therefore in the first second-order prototype circuit
R1 = R2 = 2.96 Ω;
C1 = 0.03 F;
R3 = 11.3 Ω
C2 = 1 F
In the second second-order prototype circuit:
b1 = 1.848, b0 = 1, C2 = 1 F
. · . C1 ≤
(1.848)2
≤ 0.427
8
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15–46
CHAPTER 15. Active Filter Circuits
choose C1 = 0.30 F
G2 =
1.848 ±
q
(1.848)2 − 8(0.3)
4
=
1.848 ± 1.008
4
Arbitrarily select the larger value, then
G2 = 0.7139 S;
.·.
R2 =
G1 = KG2 = 0.7139 S;
G3 =
.·.
1
= 1.4008 Ω
G2
R1 =
b o C1
(1)(0.30)
=
= 0.4202 S
G2
0.7139
1
= 1.4008 Ω
G1
. ·.
R3 = 1/G3 = 2.3796 Ω
In the low-pass section of the filter
kf =
ωo0
= 2π(400) = 800π
ωo
km =
C2
1
125,000
=
=
02
−9
C kf
(10 × 10 )kf
π
Therefore in the first scaled second-order section
R01 = R02 = 2.96km = 118 kΩ
R03 = 11.3km = 448 kΩ
C10 =
0.03
= 300 pF
kf km
C20 = 10 nF
In the second scaled second-order section
R01 = R02 = 1.4008km = 55.74 kΩ
R03 = 2.38km = 94.68 kΩ
C10 =
0.3
= 3 nF
kf km
C20 = 10 nF
High-pass filter section
n=
(−0.05)(−30)
= 3.77;
log10(6400/2560)
n=4
In the first prototype second-order section:
b1 = 0.765; bo = 1; C2 = C3 = 1 F
C1 = K = 1 F
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Problems
R1 =
K +2
3
=
= 3.92 Ω
b1
0.765
R2 =
b1
0.765
=
= 0.255 Ω
bo (K + 2)
3
15–47
In the second prototype second-order section: b1 = 1.848; bo = 1;
C2 = C3 = 1 F
C1 = K = 1 F
R1 =
K +2
3
=
= 1.623 Ω
b1
1.848
R2 =
b1
1.848
=
= 0.616 Ω
bo (K + 2)
3
In the high-pass section of the filter
ωo0
kf =
= 2π(6400) = 12,800π
ωo
km =
C
1
7812.5
=
=
0
−9
C kf
(10 × 10 )(12,800π)
π
In the first scaled second-order section
R01 = 3.92km = 9.75 kΩ
R02 = 0.255km = 634 Ω
C10 = C20 = C30 = 10 nF
In the second scaled second-order section
R01 = 1.623km = 4.04 kΩ
R02 = 0.616km = 1.53 kΩ
C10 = C20 = C30 = 10 nF
In the gain section, let Ri = 10 kΩ and Rf = 10 kΩ.
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15–48
CHAPTER 15. Active Filter Circuits
[b]
P 15.54 [a] The prototype low-pass transfer function is
Hlp (s) =
1
+ 0.765s + 1)(s2 + 1.848s + 1)
(s2
The low-pass frequency scaling factor is
kflp = 2π(400) = 800π
The scaled transfer function for the low-pass filter is
Hlp0 (s) = =
1
s
800π
[s2
2
+
0.765s
800π
+1
2
s
800π
8
+
1.848s
800π
4
+1
4096 × 10 π
+ 612πs + (800π)2 ] [s2 + 1478.4πs + (800π)2 ]
The prototype high-pass transfer function is
Hhp (s) =
s4
(s2 + 0.765s + 1)(s2 + 1.848s + 1)
The high-pass frequency scaling factor is
kfhp = 2π(6400) = 12,800π
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Problems
15–49
The scaled transfer function for the high-pass filter is
0
Hhp
(s) = =
(s/12,800π)4
s
12,800π
[s2
2
+
0.765s
12,800π
+ 9792πs +
+1
s
12,800π
4
s
(12,800π)2 ][s2
2
+
1.848s
12,800π
+1
+ 23,654.4πs + (12,800π)2 ]
The transfer function for the filter is
h
0
H 0 (s) = Hlp0 (s) + Hhp
(s)
[b] fo =
q
fc1 fc2 =
q
i
400)(6400) = 1600 Hz
ωo = 2πfo = 3200π rad/s
(jωo )2 = −1024 × 104 π 2
(jωo )4 = 1,048,576 × 108 π 4
Hlp0 (jωo ) =
4096 × 108 π 4
×
[−960 × 104 π 2 + j612(3200π 2 )]
[−960 ×
=
104 π 2
1
+ j1478.4(3200π 2 )]
40,000
(−3000 + j612)(−3000 + j1478.4)
= 3906.2 × 10−6 / − 322.24◦
1,048,576 × 108 π 4
0
Hhp
(jωo ) =
[15,360 × 104 π 2 + j9792(3200π 2 )]
1
[15,360 × 104 π 2 + j23,654.4(3200π 2 )]
=
10.24 × 106
(48,000 + j9792)(48,000 + j23,654.4)
= 3906.2 × 10−6 / − 37.76◦
.·. H 0 (jωo ) = −3906.2 × 10−6 (1/ − 322.24◦ + 1/ − 37.76◦ )
= −3906.2 × 10−6 (1.58/0◦ ) = −6176.35 × 10−6 /0◦
G = 20 log 10 |H 0 (jωo )| = 20 log 10(6176.35 × 10−6 ) = −44.19 dB
P 15.55 [a] At low frequencies the capacitor branches are open; vo = vi . At high
frequencies the capacitor branches are short circuits and the output
voltage is zero. Hence the circuit behaves like a unity-gain low-pass filter.
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15–50
CHAPTER 15. Active Filter Circuits
[b] Let va represent the voltage-to-ground at the right-hand terminal of R1 .
Observe this will also be the voltage at the left-hand terminal of R2 . The
s-domain equations are
(Va − Vi )G1 + (Va − Vo )sC1 = 0
(Vo − Va )G2 + sC2 Vo = 0
or
(G1 + sC1 )Va − sC1 Vo = G1 Vi
−G2 Va + (G2 + sC2)Vo = 0
.·. Va =
G2 + sC2Vo
G2
. ·.
"
#
. ·.
Vo
G1 G2
=
Vi
(G1 + sC1 )(G2 + sC2) − C1 G2 s
(G2 + sC2)
(G1 + sC1 )
− sC1 Vo = G1 Vi
G2
which reduces to
Vo
G1 G2 /C1 C2
bo
= 2 G1
G1 G2 = 2
Vi
s + b1 s + bo
s + C1 s + C1 C2
[c] There are four circuit components and two restraints imposed by H(s);
therefore there are two free choices.
G1 ·
[d] b1 =
. . G1 = b1 C1
C1
bo =
G1 G2 ·
bo
. . G2 = C2
C1 C2
b1
[e] No, all physically realizeable capacitors will yield physically realizeable
resistors.
[f] From Table 15.1 we know the transfer function of the prototype 4th order
Butterworth filter is
1
H(s) = 2
(s + 0.765s + 1)(s2 + 1.848s + 1)
In the first section bo = 1,
b1 = 0.765
.·. G1 = (0.765)(1) = 0.765 S
R1 = 1/G1 = 1.307 Ω
G2 =
1
(1) = 1.307 S
0.765
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Problems
15–51
R2 = 1/G2 = 0.765 Ω
In the second section bo = 1,
b1 = 1.848
.·. G1 = 1.848 S
R1 = 1/G1 = 0.541 Ω
G2 =
1
(1) = 0.541 S
1.848
R2 = 1/G2 = 1.848 Ω
P 15.56 [a] kf =
ωo0
= 2π(3000) = 6000π
ωo
km =
C
1
106
=
=
C 0 kf
(4.7 × 10−9 )(6000π)
28.2π
In the first section
R01 = 1.307km = 14.75 kΩ
R02 = 0.765km = 8.64 kΩ
In the second section
R01 = 0.541km = 6.1 kΩ
R02 = 1.848km = 20.86 kΩ
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15–52
CHAPTER 15. Active Filter Circuits
[b]
P 15.57 [a] Interchanging the Rs and Cs yields the following circuit.
At low frequencies the capacitors appear as open circuits and hence the
output voltage is zero. As the frequency increases the capacitor branches
approach short circuits and va = vi = vo . Thus the circuit is a unity-gain,
high-pass filter.
[b] The s-domain equations are
(Va − Vi )sC1 + (Va − Vo )G1 = 0
(Vo − Va )sC2 + Vo G2 = 0
It follows that
Va (G1 + sC1) − G1 Vo = sC1Vi
and Va =
(G2 + sC2 )Vo
sC2
Thus
("
#
)
(G2 + sC2 )
(G1 + sC1 ) − G1 Vo = sC1 Vi
sC2
Vo {s2C1 C2 + sC1G2 + G1 G2 } = s2 C1 C2Vi
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Problems
H(s) =
Vo
=
Vi
15–53
s2
G2
G1 G2
+
s+
C2
C1 C2
Vo
s2
=
= 2
Vi
s + b1 s + bo
s2
[c] There are 4 circuit components: R1 , R2 , C1 and C2 .
There are two transfer function constraints: b1 and bo .
Therefore there are two free choices.
G2
G1 G2
[d] bo =
;
b1 =
C1 C2
C2
.·. G2 = b1 C2 ;
G1 =
R2 =
1
b 1 C2
bo
b1
C1 .·. R1 =
b1
b o C1
[e] No, all realizeable capacitors will produce realizeable resistors.
[f] The second-order section in a 3rd-order Butterworth high-pass filter is
s2 /(s2 + s + 1). Therefore bo = b1 = 1 and
R1 =
1
= 1 Ω.
(1)(1)
R2 =
1
= 1 Ω.
(1)(1)
P 15.58 [a] kf =
ωo0
= 104 π
ωo
C
1
105
km = 0 =
=
C kf
(75 × 10−9 )(104 π)
75π
C10 = C20 = 75 nF;
[b] R = 424.4 Ω;
R01 = R02 = km R = 424.4 Ω
C = 75 nF
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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15–54
CHAPTER 15. Active Filter Circuits
[c]
s3
(s + 1)(s2 + s + 1)
[d] Hhp(s) =
0
Hhp
(s) =
=
(s/104 π)3
[(s/104 π) + 1][(s/104 π)2 + (s/104 π) + 1]
s3
(s + 104 π)(s2 + 104 πs + 108 π 2)
0
[e] Hhp
(j104 π) =
. ·.
(j104 π)3
= 0.7071/135◦
4
4
4
2
4
4
8
2
(j10 π + 10 π)[(j10 π) + 10 π(j10 π) + 10 π ]
0
|Hhp
| = 0.7071 = −3.01 dB
P 15.59 [a] It follows directly from Eqs 15.64 and 15.65 that
H(s) =
s2 + 1
s2 + 4(1 − σ)s + 1
Now note from Eq 15.69 that (1 − σ) equals 1/4Q, hence
H(s) =
s2 + 1
s2 + Q1 s + 1
[b] For Example 15.13 ωo = 5000 rad/s and Q = 5. Therefore kf = 5000 and
(s/5000)2 + 1
1
s
(s/5000)2 +
+1
5 5000
s2 + 25 × 106
= 2
s + 1000s + 25 × 106
H 0 (s) =
P 15.60 [a] ωo = 2000π rad/s
. ·.
kf =
ωo0
= 2000π
ωo
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Problems
km =
15–55
C
1
105
=
=
C 0 kf
(15 × 10−9 )(2000π)
3π
R0 = km R =
σ =1−
105
(1) = 10,610 Ω
3π
so
R0 /2 = 5305 Ω
1
1
=1−
= 0.9875
4Q
4(20)
σR0 = 10,478 Ω;
(1 − σ)R0 = 133 Ω
C 0 = 15 nF
2C 0 = 30 nF
[b]
[c] kf = 2000π
H(s) =
=
(s/2000π)2 + 1
1
(s/2000π)2 + 20
(s/2000π) + 1
s2 + 4 × 106 π 2
s2 + 100πs + 4 × 106 π 2
P 15.61 To satisfy the gain specification of 20 dB at ω = 0 and α = 1 requires
R1 + R2
= 10
R1
or
R2 = 9R1
Use the specified resistor of 11.1 kΩ for R1 and a 100 kΩ potentiometer for R2 .
Since (R1 + R2 )/R1 1 the value of C1 is
C1 =
1
= 39.79 nF
2π(40)(105 )
Choose a capacitor value of 40 nF. Using the selected values of R1 and R2 the
maximum gain for α = 1 is
20 log 10
111.1
11.1
= 20.01 dB
α=1
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15–56
CHAPTER 15. Active Filter Circuits
When C1 = 40 nF the frequency 1/R2 C1 is
1
109
= 5
= 250 rad/s = 39.79 Hz
R2 C1
10 (40)
The magnitude of the transfer function at 250 rad/s is
|H(j250)|α=1 =
111.1 × 103 + j250(11.1)(100)(40)10−3
= 7.11
11.1 × 103 + j250(11.1)(100)(40)10−3
Therefore the gain at 39.79 Hz is
20 log 10(7.11)α=1 = 17.04 dB
P 15.62 20 log 10
R1 + R2
R1
= 13.98
R1 + R2
.·.
= 5;
R1
Choose
R1 = 100 kΩ. Then
1
= 100π rad/s;
R2 C1
P 15.63 |H(j0)| =
.·. R2 = 4R1
R2 = 400 kΩ
. · . C1 =
1
= 7.96 nF
(100π)(400 × 103 )
R1 + αR2
11.1 + α(100)
=
R1 + (1 − α)R2
11.1 + (1 − α)100
P 15.64 [a] Combine the impedances of the capacitors in series in Fig. P15.64(b) to
get
1
1−α
α
1
=
+
=
sCeq
sC1
sC1
sC1
which is identical to the impedance of the capacitor in Fig. P15.60(a).
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Problems
15–57
[b]
Vx =
α/sC1
V = αV
(1 − α)/sC1 + α/sC1
Vy =
αR2
= αV = Vx
(1 − α)R2 + αR2
[c] Since x and y are both at the same potential, they can be shorted together,
and the circuit in Fig. 15.34 can thus be drawn as shown in Fig. 15.53(c).
[d] The feedback path between Vo and Vs containing the resistance R4 + 2R3
has no effect on the ratio Vo /Vs , as this feedback path is not involved in
the nodal equation that defines the voltage ratio. Thus, the circuit in
Fig. P15.64(c) can be simplified into the form of Fig. 15.2, where the
input impedance is the equivalent impedance of R1 in series with the
parallel combination of (1 − α)/sC1 and (1 − α)R2 , and the feedback
impedance is the equivalent impedance of R1 in series with the parallel
combination of α/sC1 and αR2 :
Zi = R1 +
=
· (1 − α)R2
(1 − α)R2 +
(1−α)
sC1
R1 + (1 − α)R2 + R1R2 C1 s
1 + R2 C1s
Zf = R1 +
=
(1−α)
sC1
α
sC1
· αR2
αR2 + sCα1
R1 + αR2 + R1 R2 C1s
1 + R2 C1s
P 15.65 As ω → 0
|H(jω)| →
2R3 + R4
=1
2R3 + R4
Therefore the circuit would have no effect on low frequency signals. As ω → ∞
|H(jω)| →
[(1 − β)R4 + Ro ](βR4 + R3 )
[(1 − β)R4 + R3 ](βR4 + Ro )
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