Circuit Variables
Assessment Problems
AP 1.1 Use a product of ratios to convert 95% of the speed of light from meters per
second to miles per second:
(0.95)
1 in
1 ft
1 mile
177,090.79 miles
3 × 108 m 100 cm
·
·
·
·
=
.
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
950 miles:
177,090.79 miles
950 miles
=
.
1s
xs
Therefore,
x=
950
= 0.00536 = 5.36 × 10−3 s = 5.36 ms.
177,090.79
AP 1.2 We begin by expressing $1 trillion in scientific notation:
$1 trillion = $1 × 1012 .
Divide by 100 = 102 to find the number of $100 bills:
$1 trillion =
1012
= 1010 $100 bills.
102
Calculate the height of a stack of 1010 $100 bills:
1010 bills ·
0.11 mm
1m
·
= 1.1 × 106 m.
bill
1000 mm
Now we can convert from meters to miles, again with a product of ratios:
1.1 × 106 m ·
100 cm
1 in
1 ft
1 mi
·
·
·
= 683.51 miles.
1m
2.54 cm 12 in 5280 ft
1–1
1–2
CHAPTER 1. Circuit Variables
AP 1.3 [a] First we use Eq. (1.2) to relate current and charge:
i=
dq
= 0.25te−2000t .
dt
Therefore, dq = 0.25te−2000t 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 = 0.25
Z t
ye−2000y dy.
0
We solve the integral and make the substitutions for the limits of the
integral:
t
e−2000y
q(t) − q(0) = 0.25
(−2000y − 1)
(−2000)2
0
= 62.5 × 10−9 e−2000t (−2000t − 1) + 62.5 × 10−9
= 62.5 × 10−9 (1 − 2000te−2000t − e−2000t ).
But q(0) = 0 by hypothesis, so
q(t) = 62.5(1 − 2000te−2000t − e−2000t ) nC.
[b] q(0.001) = (62.5)[1 − 2000(0.001)e−2000(0.001) − e−2000(0.001) ] = 37.12 nC.
AP 1.4 n =
75 × 10−6 C/s
= 4.681 × 1014 elec/s.
1.6022 × 10−19 C/elec
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:
Problems
1–3
[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 250 mA of current
entering Terminal 2 is the same as 250 mA of current leaving Terminal 1.
We get
(a) v = 50 V,
i = −0.25 A;
(b) v = 50 V,
(c) v = −50 V,
i = −0.25 A;
(d) v = −50 V,
i = 0.25 A;
i = 0.25 A.
[b] Using the reference system in Fig. 1.6(a) and the passive sign convention,
p = vi = (50)(−0.25) = −12.5 W.
[c] Since the power is less than 0, the box is delivering power.
AP 1.6 p = vi;
w=
Z t
p dx.
0
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 60 hr
= 216,000 s = 216 ks.
p(0) = (6)(15 × 10−3 ) = 90 × 10−3 W;
p(216 ks) = (4)(15 × 10−3 ) = 60 × 10−3 W;
1
w = (60 × 10−3 )(216 × 103 ) + (90 × 10−3 − 60 × 10−3 )(216 × 103 ) = 16,200 J.
2
AP 1.7 [a] p = vi = (15e−250t )(0.04e−250t ) = 0.6e−500t W;
p(0.01) = 0.6e−500(0.01) = 0.6e−5 = 0.00404 = 4.04 mW.
[b] wtotal =
Z ∞
0
p(x) dx =
Z ∞
0
0.6e−500x dx =
0.6 −500x ∞
e
−500
0
= −0.0012(e∞ − e0 ) = 0.0012 = 1.2 mJ.
1–4
CHAPTER 1. Circuit Variables
Chapter Problems
P 1.1
[a] Use a product of ratios to convert 40,075 km to inches:
1000 m 100 cm
1 in
·
·
= 1,577,755,905.5 in.
1 km
1m
2.54 cm
Now calculate the number of $20 bills this distance represents:
40,075 km ·
Number of bills =
1,577,755,905.5 in
= 256,963,502.5 ≈ 256,963,503.
6.14 in/bill
The total dollar amount represented by this number of $20 bills is
Total dollars = (256,963,503)($20) ≈ $5.14 billion.
[b] The weight of the number of bills calculated in part (a) is
(256,963,503)(1 g) = 256,963.5 kg. Use a product of ratios to convert kg
to tons:
1 ton
2.2 lbs
·
= 282.66 tons.
256,963.5 kg ·
1 kg 2000 lbs
P 1.2
1s
= 10.48 s.
50 × 106 bits
1s
= 262 ms.
[b] 8(65.5 × 106 ) bits ·
2 × 109 bits
1s
[c] 8(74 × 1012 ) bits ·
= 11.84 × 106 s
6
50 × 10 bits
[a] 8(65.5 × 106 ) bits ·
11.84 × 106 s ·
1 hr
1 day
1 min
·
·
= 137 days!
60 s 60 min 24 hr
[d] 8(74 × 1012 ) bits ·
296 × 103 s ·
P 1.3
1s
= 296 × 103 s.
2 × 109 bits
1 min
1 hr
1 day
·
·
= 3.4 days.
60 s 60 min 24 hr
[a] To begin, we calculate the number of pixels that make up the display:
npixels = (1920)(1080) = 2,073,600 pixels.
Each pixel requires 24 bits of information. Since 8 bits equal one byte,
each pixel requires 3 bytes of information. We can calculate the number
of bytes of information required for the display by multiplying the
number of pixels in the display by 3 bytes per pixel:
nbytes =
2,073,600 pixels 3 bytes
·
= 6,220,800 bytes/display.
1 display
1 pixel
Problems
1–5
Finally, we use the fact that there are 106 bytes per MB:
1 MB
6,220,800 bytes
· 6
= 6.22 MB/display.
1 display
10 bytes
6,220,800 bytes 8 bits
·
= 4.98 Gbps.
10 ms
1 byte
[c] Convert the dimensions of the monitor from inches to mm:
[b]
24 in ·
25.4 mm
= 609.6 mm.
1 in
14 in ·
25.4 mm
= 355.6 mm.
1 in
Calculate the area of the monitor in mm2 :
Screen area = (609.6)(355.6) = 126,773.76 mm2 .
Divide the area of the monitor by the number of pixels in the monitor to
find the area of an indiviual pixel:
126,773.76 mm2
= 0.1045 mm2 /pixel.
2,073,600 pixels
P 1.4
(1024)(600) pixels 3 bytes 60 frames
·
·
= 110,592 × 103 bytes/sec;
1 frame
1 pixel
1 sec
(110,592 × 103 bytes/sec)(x secs) = 128 × 109 bytes;
x=
128 × 109
= 1157 sec = 19.3 min of video.
110,592 × 103
P 1.5
(0.8)(5.4 × 106 )(1.5 × 103 ) + (0.9)(1.4 × 106 )(45 × 103 )
= 63.18 GWh.
109
P 1.6
[a] (0.6)(64 kWh) ·
240 km ·
[b] 200 mi ·
100 km
= 240 km;
16 kWh
0.62 mi
= 148.8 mi.
1 km
1 km
= 322.58 km;
0.62 mi
100 km
322.58 km
=
16 kWh
x kWh
so x =
51.61 kWh
(100) ≈ 81%.
64 kWh
322.58(16)
= 51.61 kWh.
100
1–6
P 1.7
CHAPTER 1. Circuit Variables
Remember from Eq. 1.2, current is the time rate of change of charge, or i = dq
dt
In this problem, we are given the current and asked to find the total charge.
To do this, we must integrate Eq. 1.2 to find an expression for charge in terms
of current:
q(t) =
Z t
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 =
=
P 1.8
Z ∞
20e−5000x dx =
0
20 −5000x ∞
20
=
e
(e−∞ − e0 )
−5000
−5000
0
20
20
(0 − 1) =
= 0.004 C = 4000 µC.
−5000
5000
[a] First we use Eq. 1.2 to relate current and charge:
i=
dq
= 0.025e−1000t .
dt
Therefore, dq = 0.025e−1000t 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 = 0.025
Z t
e−1000y dy.
0
We solve the integral and make the substitutions for the limits of the
integral:
t
e−1000y
q(t) − q(0) = 0.025
= 25 × 10−6 (1 − e−1000t ).
−1000 0
But q(0) = 0 by hypothesis, so
q(t) = 25(1 − e−1000t ) µC.
[b] As t → ∞, qT = 25 µC.
[c] q(1 × 10−3 ) = (25 × 10−6 )(1 − e(−1000)(0.001) ) = 15.8 µC.
P 1.9
First we use Eq. 1.2 to relate current and charge:
i=
dq
= 0.1 cos 2500t.
dt
Therefore, dq = 0.1 cos 2500t dt.
Problems
1–7
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)
dx = 0.1
Z t
q(0)
cos 2500y dy.
0
We solve the integral and make the substitutions for the limits of the integral,
remembering that sin 0 = 0:
q(t) − q(0) = 0.1
t
sin 2500y
2500
=
0
0.1
0.1
0.1
sin 2500t −
sin[2500(0)] =
sin 2500t.
2500
2500
2500
But q(0) = 0 by hypothesis, i.e., the current passes through its maximum
value at t = 0, so q(t) = 40 × 10−6 sin 2500t C = 40 sin 2500t µC.
P 1.10
w = qV = (1.6022 × 10−19 )(9) = 14.42 × 10−19 = 1.442 aJ.
P 1.11
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:
d 1
1
t
dq
i=
=
−
+ 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
= 0−
= −
1
t
1 −αt
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)e−αt = (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
1–8
CHAPTER 1. Circuit Variables
Remember in the problem statement, α = 0.03679. Using this value for α,
i=
P 1.12
1
e−1 ∼
= 10 A.
0.03679
[a]
p = vi = −(80)(−4) = 320 W.
Power is being absorbed by the box.
[b] Entering.
[c] Gain.
P 1.13
[a] p = −vi = −(80)(4) = −320 W, so power is being delivered by the box.
[b] Leaving.
[c] Lose.
P 1.14
[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 = (25)(12) = 300 W.
Since the power is positive, the battery in Car A is absorbing power, so
Car A must have the “dead” battery.
Z t
60 s
[b] w(t) =
p dx;
1 min = 1 ·
= 60 s;
1 min
0
w(60) =
Z 60
300 dx;
0
w = 300(60 − 0) = 300(60) = 18,000 J = 18 kJ.
P 1.15
Assume we are standing at box A looking toward box B. Use the passive sign
convention to get p = −vi, since the current i is flowing out of 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.
[a] p = −(−20)(0.15) = 3 W
3 W from A to B;
[b] p = −(40)(15) = −600 W
600 W from B to A;
[c] p = −(−2000)(−0.05) = −100 W
[d] p = −(80)(−3) = 240 W
100 W from B to A;
240 W from A to B.
Problems
P 1.16
3 hr ·
p = (9)(0.15) = 1.35 W;
w(t) =
Z t
p dt;
w(10,800) =
0
P 1.17
1–9
3600 s
= 10,800 s;
1 hr
Z 10,800
1.35 dt = 1.35(10,800) = 14.58 kJ.
0
At the Building X 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 Building X end of the line is
negative, power is being generated at the Building X end of the line and
transmitted by the line to be delivered to the Building Y end of the line.
P 1.18
[a] Applying the passive sign convention to the power equation using the
voltage and current polarities shown in Fig. 1.5, p = vi.
p(t) = (80,000te−500t )(15te−500t ) = 120 × 104 t2 e−1000t ;
p(0.01) = 120 × 104 (0.01)2 e−1000(0.01) = 5.45 mW.
[b] 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 ∞
120 × 104 x2 e−1000x dx
0
∞
120 × 104 −1000x
=
e
[(−1000)2 x2 − 2(−1000)x + 2)
(−1000)3
0
120 × 104 0
= 0−
e (0 − 0 + 2) = 2.4 mJ.
(−1000)3
P 1.19
[a] p = vi = (100e−500t )(0.02 − 0.02e−500t ) = (2e−500t − 2e−1000t ) W;
dp
= −1000e−500t + 2000e−1000t = 0
dt
2 = e500t
so
ln 2 = 500t
so
thus
2e−1000t = e−500t ;
p is maximum at t = 1.4 ms;
pmax = p(1.4 ms) = 0.5 W.
[b] w =
Z ∞
0
=
−500t
[2e
−1000t
− 2e
∞
2 −500t
2
−1000t
] dt =
e
−
e
−500
−1000
0
4
2
−
= 2 mJ.
1000 1000
1–10
P 1.20
CHAPTER 1. Circuit Variables
[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
=
0 t < 0,
p
dp
dt
dp
dt
= vi = t(3 − t)(6 − 4t) = 18t − 18t2 + 4t3 mW
=
18 − 36t + 12t2 = 12(t2 − 3t + 1.5);
=
0
0 ≤ t ≤ 3 s;
when t2 − 3t + 1.5 = 0;
√
√
3± 9−6
3± 3
=
;
2√
2
√
t2 = 3/2 + 3/2 = 2.366 s.
3/2 − 3/2 = 0.634 s;
t =
t1
p = 0 t > 3 s;
=
p(t1 )
= 18(0.634) − 18(0.634)2 + 4(0.634)3 = 5.196 mW;
p(t2 )
= 18(2.366) − 18(2.366)2 + 4(2.366)3 = −5.196 mW.
Therefore, maximum power is being delivered at t = 0.634 s.
[b] The maximum power was calculated in part (a) to determine the time at
which the power is maximum: pmax = 5.196 mW (delivered).
[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 = 2.366 s.
[d] This maximum extracted power was calculated in part (a) to determine
the time at which power is maximum: pmax = 5.196 mW (extracted).
[e] w =
Z t
pdx =
0
w(0)
Z t
(18x − 18x2 + 4x3 )dx = 9t2 − 6t3 + t4 .
0
=
0 mJ;
w(2)
=
4 mJ;
w(1) = 4 mJ;
w(3) = 0 mJ.
To give you a feel for the quantities of voltage, current, power, and energy
and their relationships among one another, they are plotted below:
Problems
P 1.21
1–11
[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
= 3.2{e−1600t [1280 × 103 t + 1600] − 1600e−1600t [640,000t2 + 1600t + 1]}
dt
= −3.2e−1600t [128 × 104 (800t2 + t)] = −409.6 × 104 e−1600t t(800t + 1).
dp
= 0 when t = 0 so pmax occurs at t = 0.
dt
[b] pmax = 3.2e−0 [0 + 0 + 1] = 3.2 W.
Therefore,
1–12
CHAPTER 1. Circuit Variables
[c] w =
Z t
pdx;
0
Z t
Z t
Z t
w
2 −1600x
−1600x
= 640,000x e
dx + 1600xe
dx + e−1600x dx
3.2
0
0
0
t
640,000e−1600x
=
[256 × 104 x2 + 3200x + 2]
6
−4096 × 10
0
t
+
t
1600e−1600x
e−1600x
+
.
(−1600x
−
1)
256 × 104
−1600
0
0
When t → ∞ all the upper limits evaluate to zero, hence
(640,000)(2)
1600
1
w
=
+
+
;
3.2
4096 × 106
256 × 104 1600
w = 10−3 + 2 × 10−3 + 2 × 10−3 = 5 mJ.
P 1.22
[a] p = vi = 30e−500t − 30e−1500t − 40e−1000t + 50e−2000t − 10e−3000t ;
p(1 ms) = 3.1 mW.
[b]
w(t)
=
Z t
(30e−500x − 30e−1500x − 40e−1000x
0
+ 50e−2000x − 10e−3000x )dx
=
21.67 − 60e−500t + 20e−1500t + 40e−1000t
− 25e−2000t + 3.33e−3000t µJ;
w(1 ms)
= 1.24 µJ.
[c] wtotal = 21.67 µJ.
P 1.23
[a]
p
= vi = (104 t + 5)e−400t ][(40t + 0.05)e−400t ]
=
400 × 103 t2 e−800t + 700te−800t + 0.25e−800t
= e−800t [400,000t2 + 700t + 0.25];
dp
dt
= {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)
=
0.67 W.
Problems
[c] w
w
=
=
Z t
Z0t
pdx;
2 −800x
400,000x e
dx +
0
Z t
700xe
−800x
dx +
0
t
Z t
0.25e−800x dx
0
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
+
+
= 2.97 mJ.
w=
6
4
512 × 10
64 × 10
800
P 1.24
[a] v(10 ms) = 400e−1 sin 2 = 133.8 V;
i(10 ms) = 5e−1 sin 2 = 1.67 A;
p(10 ms) = v(10 ms)i(10 ms) = 223.79 W.
[b]
p
w
= vi = 2000e−200t sin2 200t
1
−200t 1
− cos 400t
= 2000e
2 2
= 1000e−200t − 1000e−200t cos 400t;
=
Z ∞
0
=
−200t
1000e
dt −
∞
e−200t
1000
−200
0
(
Z ∞
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 J.
4 × 104 + 16 × 104
P 1.25
[a] p = vi = 900 sin(200πt) cos(200πt) = 450 sin(400πt) W.
Therefore, pmax = 450 W.
[b] pmax (extracting) = 450 W.
1 Z 5×10−3
[c] pavg =
450 sin(400πt) dt
0.005 0
5×10−3
225
4 − cos 400πt
= 9 × 10
=
[1 − cos 2π] = 0.
400π
π
0
1 Z 6.25×10−3
[d] pavg =
450 sin(400πt) dt
0.00625 0
P 1.26
[a] q
=
180
180
[1 − cos 2.5π] =
= 57.3 W.
π
π
=
area under i vs. t plot
=
1
(10)(15,000) + (20)(15,000) + 21 (20)(5000)
2
=
75,000 + 300,000 + 50,000 = 425,000 C.
0
1–13
1–14
CHAPTER 1. Circuit Variables
[b] w
Z
=
p dt =
Z
vi dt;
v = 250 × 10−6 t + 10,
0 ≤ t ≤ 15,000s:
i
=
p
=
w1
0 ≤ t ≤ 20 ks.
30 − 666.67 × 10−6 t;
300 + 833.33 × 10−6 t − 166.67 × 10−9 t2 ;
Z 15,000
=
(300 + 833.33 × 10−6 t − 166.67 × 10−9 t2 ) dt
0
= (4500 + 93.75 − 187.5)103 = 4406.25 kJ.
15,000 s ≤ t ≤ 20,000 s:
i
=
p
=
w2
80 − 4 × 10−3 t;
800 − 20 × 10−3 t − 10−6 t2 ;
Z 20,000
=
(4000 − 1750 − 1541.67)103 = 708.33 kJ;
=
wT
P 1.27
(800 − 20 × 10−3 t − 10−6 t2 ) dt
15,000
= w1 + w2 = 4406.25 + 08.33 = 5114.58 kJ.
[a] q = area under i vs. t plot
1
1
1
= (6)(5000) + (14)(5000) + (6)(10,000) + (8)(10,000) + (8)(5000)
2
2
2
= 15,000 + 70,000 + 30,000 + 80,000 + 20,000 = 215,000 C.
[b] w
Z
=
p dt =
Z
vi dt;
v = 250 × 10−6 t + 10,
0 ≤ t ≤ 5000s:
i = 20 − 1.2 × 10−3 t;
p
=
w1
=
0 ≤ t ≤ 20 ks.
200 − 7 × 10−3 t − 0.3 × 10−6 t2 ;
Z 5000
(200 + 7 × 10−3 t − 0.3 × 10−6 t2 ) dt
0
= 106 + 87,500 − 12,500 = 900 kJ.
5000 s ≤ t ≤ 15,000 s:
i
=
17 − 0.6 × 10−3 t;
p
=
170 − 1.75 × 10−3 t − 0.15−6 t2 ;
w2
=
Z 15,000
(170 − 1.75 × 10−3 t − 0.15−6 t2 ) dt
5000
=
1.7 × 106 − 175,000 − 162,500 = 1362.5 kJ;
Problems
15,000 s ≤ t ≤ 20,000 s:
i
=
p
=
w3
=
32 − 1.6 × 10−3 t;
320 − 8 × 10−3 t − 0.4 × 10−6 t2 ;
Z 20,000
(320 − 8 × 10−3 t − 0.4 × 10−6 t2 ) dt
15,000
= 1.6 × 106 − 700,000 − 616,666.67 = 283.33 kJ;
wT = w1 + w2 + w3 = 900 + 1362.5 + 283.33 = 2545.83 kJ.
P 1.28
[a]
= 5 + 1 × 10−3 t mA,
0 ≤ t ≤ 5 ks;
i(t)
= 10 mA,
5 ks ≤ t ≤ 10 ks;
i(t)
= 20 − 1 × 10−3 t mA,
10 ks ≤ t ≤ 15 ks;
i(t)
= 0,
t > 15 ks.
[b] i(t)
p = vi = 240i so
p(t)
= 1200 + 0.24t mW,
0 ≤ t ≤ 5 ks;
p(t)
= 2400 mW,
5 ks ≤ t ≤ 10 ks;
p(t)
= 4800 − 0.24t mW,
10 ks ≤ t ≤ 15 ks;
p(t)
= 0,
t > 15 ks.
[c] To find the energy, calculate the area under the plot of the power:
1
w(5 ks) = (1.2)(5000) + (1.2)(5000) = 9 kJ;
2
1–15
1–16
CHAPTER 1. Circuit Variables
w(10 ks) = w(5 ks) + (2.4)(5000) = 21 kJ;
1
w(15 ks) = w(10 ks) + (1.2)(5000) + (1.2)(5000) = 30 kJ.
2
P 1.29
[a] 0 s ≤ t < 10 ms:
v = 8 V;
i = 25t A;
p = 200t W.
i = 0.5 − 25t A;
p = 200t − 4 W.
i = −250 mA;
p = 0 W.
10 ms < t ≤ 30 ms:
v = −8 V;
30 ms ≤ t < 40 ms:
v = 0 V;
40 ms < t ≤ 60 ms:
v = 8 V;
i = 25t − 1.25 A; p = 200t − 10 W.
t > 60 ms:
v = 0 V;
i = 250 mA;
p = 0 W.
[b] Calculate the area under the curve from zero up to the desired time:
P 1.30
1
(2)(0.01) = 10 mJ;
2
w(0.01)
=
w(0.03)
= w(0.01) − 21 (2)(0.01) + 21 (2)(0.01) = 10 mJ;
w(0.08)
= w(0.03) − 21 (2)(0.01) + 21 (2)(0.01) = 10 mJ.
pa = (−8)(7) = −56 W;
pb = −(−2)(−7) = −14 W;
pc = (10)(15) = 150 W;
pd = −(10)(5) = −50 W;
pe = (−6)(3) = −18 W;
pX
f = (−4)(3) = −12 W.
X
Pabs = 150 W;
Pdel = 56 + 14 + 50 + 18 + 12 = 150 W.
Problems
P 1.31
pa
= va ia = (6)(0.5) = 3 W;
pb
= −vb ib = −(10)(0.1) = −1 W;
pc
= −vc ic = −(−8)(−0.4) = −3.2 W;
pd
= −vd id = −(−2)(0.3) = 0.6 W;
pe
= ve ie = (−2)(0.3) = −0.6 W;
pf
= −vf if = −(4)(−0.2) = 0.8 W;
pg
= −vg ig = −(−6)(0.2) = 1.2 W;
1–17
ph = vh ih = (2)(−0.4) = −0.8 W.
Therefore,
X
Pabs = 3 + 0.6 + 0.8 + 1.2 = 5.6 W;
X
Pdel = 1 + 3.2 + 0.6 + 0.8 = 5.6 W;
X
Pabs =
X
Pdel .
Thus, the interconnection satisfies the power check.
P 1.32
pa
= va ia = (−160)(−10) = 1600 W;
pb
= vb ib = (−100)(−20) = 2000 W;
pc
= −vc ic = −(−60)(6) = 360 W;
pd
= vd id = (800)(−50) = −40,000 W;
pe
= −ve ie = −(800)(−20) = 16,000 W;
pf
= −vf if = −(−700)(14) = 9800 W;
pg
= −vg ig = −(640)(−16) = 10,240 W.
X
Pdel = 40,000 W;
Pabs = 1600 + 2000 + 360 + 16,000 + 9800 + 10,240 = 40,000 W;
X
X
Therefore,
Pdel = Pabs = 40,000 W.
X
P 1.33
[a] If the power balances, the sum of the power values should be zero:
ptotal = −0.918 − 0.810 − 0.012 + 0.400 + 0.224 + 1.116 = 0.
Thus, the power balances.
[b] When the power is positive, the element is absorbing power. Since
elements d, e, and f have positive power, these elements are absorbing
power.
1–18
CHAPTER 1. Circuit Variables
[c] The voltage can be calculated using v = p/i or v = −p/i, with proper
application of the passive sign convention:
P 1.34
va
= −pa /ia = −(−0.918)/(−0.051) = −18 V;
vb
= pb /ib = (−0.81)/(0.045) = −18 V;
vc
= pc /ic = (−0.012)/(−0.006) = 2 V;
vd
= −pd /id = −(0.4)/(−0.02) = 20 V;
ve
= −pe /ie = −(0.224)/(−0.014) = 16 V;
vf
= pf /if = (1.116)/(0.031) = 36 V.
[a] From the diagram and the table we have
pa
= −va ia = −(5000)(−0.150) = 750 W;
pb
= vb ib = (2000)(0.250) = 500 W;
pc
= −vc ic = −(3000)(0.200) = −600 W;
pd
= vd id = (−5000)(0.400) = −2000 W;
pe
= −ve ie = −(1000)(−0.050) = 50 W;
pf
= vf if = (4000)(0.350) = 1400 W;
pg
= −vg ig = −(−2000)(0.400) = 800 W;
ph
= −vh ih = −(−6000)(−0.350) = −2100 W.
X
Pdel
=
600 + 2000 + 2100 = 4700 W;
Pabs
=
750 + 500 + 50 + 1400 + 800 = 3500 W.
Therefore,
X
Pdel 6=
X
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
−4700 + 3500 = 1200 W.
One-half of this difference is 600 W, so it is likely that pc is in error.
Either the voltage or the current probably has the wrong sign. (In
Chapter 2, we will discover that using KVL the voltage vc should be
−3.0 kV, not 3.0 kV!) If the sign of pc is changed from negative to
positive, we can recalculate the power delivered and the power absorbed
as follows:
X
Pdel = 2000 + 2100 = 4100 W;
X
Pabs = 750 + 500 + 600 + 50 + 1400 + 800 = 4100 W.
Now the power delivered equals the power absorbed and the power
balances for the circuit.
Problems
P 1.35
1–19
[a] We can add the powers supplied together and the powers absorbed
together
— if the power balances, these power sums should be equal:
X
Psup = 750 + 400 + 800 = 1950 W;
X
Pabs = 1600 + 150 + 200 = 1950 W.
Thus, the power balances.
[b] The current can be calculated using i = p/v or i = −p/v, with proper
application of the passive sign convention. Remember that the power
supplied is negative and the power absorbed is positive.
P 1.36
ia
= −pa /va = −(−750)/(−3000) = −250 mA;
ib
= −pb /vb = −(1600)/(4000) = −400 mA;
ic
= −pc /vc = −(−400)/(1000) = 400 mA;
id
= pd /vd = (150)/(1000) = 150 mA;
ie
= pe /ve = (−800)/(−4000) = 200 mA;
if
= pf /vf = (200)/(4000) = 50 mA;
pa
= va ia = (120)(−10) = −1200 W;
pb
= −vb ib = −(120)(9) = −1080 W;
pc
= vc ic = (10)(10) = 100 W;
pd
= −vd id = −(10)(−1) = 10 W;
pe
= ve ie = (−10)(−9) = 90 W;
pf
= −vf if = −(−100)(5) = 500 W;
pg
= vg ig = (120)(4) = 480 W;
ph
= vh ih = (−220)(−5) = 1100 W.
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.
1–20
P 1.37
CHAPTER 1. Circuit Variables
[a] The revised circuit model is shown below:
[b] The expression for the total power in this circuit is
v a ia − v b ib − v f if + v g ig + v h ih
= (120)(−8) − (120)(8) − (−120)(6) + (120)(6) + (−240)ih = 0.
Therefore,
240ih = −960 − 960 + 720 + 720 = −480
so
−480
= −2 A.
240
Thus, if the power in the modified circuit is balanced the current in
component h is −2 A.
ih =
Circuit Elements
Assessment Problems
AP 2.1 [a] Because both current sources are in the same branch of the circuit, their
values must be the same. Therefore,
v1
= 0.4 → v1 = 0.4(50) = 20 V.
50
[b] p = v1 (0.4) = (20)(0.4) = 8 W (absorbed).
AP 2.2 [a] The voltage drop from the top node to the bottom node in this circuit
must be the same for every path from the top to the bottom. Therefore,
the voltages of the two voltage sources are equal:
−αi∆ = 6.
Also, the current i∆ is in the same branch as the 15 mA current source,
but in the opposite direction, so
i∆ = −0.015A.
Substituting,
6
= 400.
0.015
The interconnection is valid if α = 400 V/A.
−α(−0.015) = 6
→
α=
[b] The voltage across the current source must equal the voltage across the 6
V source, since both are connected between the top and bottom nodes.
Using the passive sign convention,
p = vi = (6)(0.015) = 0.09 = 90 mW.
[c] Since the power is positive, the current source is absorbing power.
2–1
2–2
CHAPTER 2. Circuit Elements
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 = 100 V.
Note from the circuit that the current through the resistor is ig = 2 A.
Use Ohm’s law to calculate the value of the resistor:
vR
100 V
R=
=
= 50 Ω.
ig
2A
Using the passive sign convention to calculate the power in the resistor,
pR = (vR )(ig ) = (100 V)(2 A) = 200 W.
The resistor is dissipating 200 W of power.
[b] Note from part (a) the vR = vg and iR = ig . The power delivered by the
source is thus
psource
−125 W
psource = −vg ig
so
vg = −
=−
= 500 V.
ig
250 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
500 V
=
= 2 kΩ.
ig
250 mA
The power absorbed by the resistor must equal the power generated by
the source. Thus,
pR = −psource = −(−125 W) = 125 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 ,
i2g =
pr
20 mW
=
= 25 × 10−6
R
800 Ω
so
ig =
√
25 × 10−6 = 5 × 10−3 A = 5 mA.
Problems
2–3
Then, since vR = vg
vR = RiR = Rig = (800 Ω)(5 mA) = 4 V
so
vg = 4 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
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,
20 mA
ig
=
= 4 V.
G
5 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 = −(4)(0.02) = −80 mW.
Thus the current source delivers 80 mW 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=
400
pg
= 2 = 0.16 S = 160 mS;
2
vg
50
ig = Gvg = (160 mS)(50 V) = 8 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 =
√
vg2 =
pg
50 mW
=
= 2500.
G
20 µS
2500 = 50 V;
ig = Gvg = (20 µS)(50 V) = 0.001 A = 1 mA.
2–4
CHAPTER 2. Circuit Elements
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:
−180 V + v2 + v3 − v1 = 0.
Next, use Ohm’s law to find the three unknown voltages in terms of the
three currents:
v2 = 15i2 ;
v3 = 20i3 ;
v1 = 25i1 .
A KCL equation at the lower right node gives i3 = −i1 ; a KCL equation
at the upper right node gives i2 = i3 ; a KCL equation at the lower left
node gives is = i1 . Now replace the currents i2 and i3 in the Ohm’s law
equations with −i1 :
v2 = 15i2 = −15i1 ;
v3 = 20i3 = −20i1 ;
v1 = 25i1 .
Now substitute these expressions for the three voltages into the first
equation:
180 = v2 + v3 − v1 = −15i1 − 20i1 − 25i1 = −60i1 .
Therefore i1 = 180/(−60) = −3 A.
[b] v1 = 25i1 = 25(−3) = −75 V.
[c] v2 = −15i1 = −15(−3) = 45 V.
[d] v3 = −20i1 = −20(−3) = 60 V.
[e] Since is = i1 , is = −3 A. We can now compute the power associated with
the voltage source:
p180 = (180)is = (180)(−3) = −540 W.
Therefore 180 V source is delivering 540 W.
AP 2.6
i1 = 100/40 = 2.5 A (Ohm’s Law);
Problems
3 = i1 + i2
so
2–5
i2 = 3 − i1 = 3 − 2.5 = 0.5 A (KCL middle top) ;
i2 = i3 (KCL right) ;
−100 + v2 + v3 = 0
so
v2 + v3 = 100 (KVL right) ;
v2 = 80i2 = 80(0.5) = 40 V (Ohm’s Law);
v3 = Ri3 = Ri2 = 0.5i2 (Ohm’s Law).
Substitute the equations for v2 and v3 into the KVL equation and solving for
R,
40 + 0.5R = 100
so
R=
100 − 40
= 120 Ω.
0.5
AP 2.7 [a] Plotting a graph of vt versus it gives
Note that when it = 0, vt = 40 V; therefore the voltage source must be
40 V. Since the plot is a straight line, its slope can be used to calculate
the value of resistance:
40 − 0
40
∆v
=
=
= 800 Ω.
R=
∆i
0.05 − 0
0.05
A circuit model having the same v − i characteristic is a 40 V source in
series with a 800Ω resistor, as shown below:
[b] Draw the circuit model from part (a) and attach a 200 Ω resistor:
2–6
CHAPTER 2. Circuit Elements
To find the power delivered to the 200 Ω resistor we must calculate the
current through the 200 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:
−40 V + 800it + 200it = 0
so
1000it = 40
so
it =
40
= 40 mA.
1000
Thus, the power delivered to the 200 Ω resistor is
p200 = (200)i2t = (200)(0.04)2 = 320 mW.
AP 2.8 [a] From the graph in Assessment Problem 2.7(a), we see that when vt = 0,
it = 50 mA. Therefore the current source must be 50 mA. Since the plot
is a straight line, its slope can be used to calculate the value of resistance:
∆v
40 − 0
40
=
=
= 800 Ω.
∆i
0.05 − 0
0.05
A circuit model having the same v − i characteristic is a 50 mA current
source in parallel with a 800Ω resistor, as shown below:
R=
[b] Draw the circuit model from part (a) and attach a 200 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
−0.05 +
+
= 0,
so
5vt = 40,
thus
vt = 8 V;
800 200
p200 =
vt2
= 320 mW.
200
Problems
2–7
AP 2.9 Label unknown current:
−20 + 450i + 150i = 0
so
→
600i = 20
(KVL and Ohm’s law);
i = 33.33 mA.
vx = 150i = 150(0.0333) = 5 V
vo = 300
(Ohm’s law);
vx
= 300(5/100) = 15 V
100
(Ohm’s law).
Calculate the power for all components:
p20V = −20i = −20(0.0333) = −0.667 W;
pd.s. = −vo
vx
= −(15)(5/100) = −0.75 W;
100
p450 = 450i2 = 450(0.033)2 = 0.5 W;
p150 = 150i2 = 150(0.033)2 = 0.1667 W;
p300 =
vo2
152
=
= 0.75 W.
300
300
Thus the total power absorbed is
pabs = 0.5 + 0.1667 + 0.75 = 1.4167 W.
AP 2.10 [a] io = 0 because no current can exist in a single conductor connecting two
parts of a circuit.
[b]
2–8
CHAPTER 2. Circuit Elements
−200 + 8000ig + 12,000ig = 0
so
ig = 200/20,000 = 10 mA;
v∆ = (12 × 103 )(10 × 10−3 ) = 120 V;
5 × 10−3 v∆ = 0.6 A;
9000i1 = 3000i2
so
i2 = 3i1 ;
0.6 + i1 + i2 = 0
so
0.6 + i1 + 3i1 = 0
thus
i1 = −0.15 A.
[c] i2 = 3i1 = −0.45 A.
[d] pv.s. = −200ig = −200(0.01) = −2 W.
v1 = 9000i1 = 9000(−0.15) = −1350 V.
pc.s. = 5 × 10−3 v∆ v1 = 5 × 10−3 (120)(−1350) = −810 W.
pT = −2 − 810 = −812 W.
Problems
2–9
Problems
P 2.1
[a] Yes, independent voltage sources can carry the 8 A current required by the
connection; independent current source can support any voltage required
by the connection, in this case 20 V, positive at the top.
[b] 30 V source:
absorbing;
10 V source:
delivering;
8 A source:
delivering.
[c] P30V
=
(30)(8) = 240 W (abs);
P10V
= −(10)(8) = −80 W (del);
P8A
= −(20)(8) = −160 W
X
Pabs =
X
(del).
Pdel = 240 W.
[d] The interconnection is valid, but in this circuit the voltage drop across the
8 A current source is 40 V, positive at the top; 30 V source is absorbing,
the 10 V source is absorbing, and the 8 A source is delivering.
P30V
=
(30)(8) = 240 W (abs);
P10V
=
(10)(8) = 80 W (abs);
P8A
= −(40)(8) = −320 W
X
P 2.2
Pabs =
X
(del).
Pdel = 320 W.
The interconnect is valid since the voltage sources can all carry the 50 mA
current supplied by the current source, and the current source can carry the
voltage drop required by the interconnection. Note that the voltage drop
across the branch with the current source must be the same as the voltage
drop across the branch with the three voltage sources. This voltage drop, from
left to right, is 12 + 8 − 30 = −10 V. The voltages and currents are
summarized in the circuit below:
2–10
CHAPTER 2. Circuit Elements
P12V
=
(12)(0.05) = 0.6 W (abs);
P8V
=
(8)(0.05) = 0.4 W
P30V
= −(30)(0.05) = −1.5 W (dev);
P50mA
= −(−10)(0.05) = 0.5 W
X
(abs);
(abs).
Pdev = 1.5 W.
P 2.3
Note that v5 , v20 , and v25 are not specified. The current sources can carry any
amount of current and the voltage sources can carry the 5 A of current. Thus,
the interconnection is valid.
We can now calculate the power developed by the two voltage sources:
pv−sources = p60 + p100 = −(60)(5) + (100)(5) = 200 W.
Since the power is positive, the sources are absorbing 200 W of power, or
developing −200 W of power.
P 2.4
The voltage from the top node to the bottom node is 50 V, established by the
50 V source. The voltage from the top node to the bottom node using the
right hand branch must also be 50 V, so the voltage drop across the 5 A
source must be 20 V.
Now we can calculate the power developed by the current sources.
p15A = −(15)(50) = −750 W;
p5A = −(5)(20) = −100 W.
X
Pdev = 850 W.
Problems
P 2.5
2–11
The interconnection is valid, since the voltage sources can carry the 10 A
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 20 A current source in the middle
branch, since the middle and right hand branch are connected between the
same two terminals. In particular, this means that
v1 (the voltage drop across the middle branch)
= 100V − 50V − v2 (the voltage drop across the right hand branch).
Hence any combination of v1 and v2 such that v1 + v2 = 50 V is a valid
solution.
P 2.6
[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, ix . Note that this current is in the same branch of the circuit as
the dependent current source and these two currents have the same
direction. Therefore, the current ix is the same as the current of the
dependent source:
ix = αvx = (0.6)(−25) = −15 A.
Using the passive sign convention,
ps = −(ix )(25 V) = −(−15 A)(25 V) = 375 W.
Thus the voltage source dissipates 375 W.
2–12
CHAPTER 2. Circuit Elements
P 2.7
[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 connected between the same two nodes and have the
same polarity. Therefore, their voltages are equal, and
ib
−8
=
= −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 connected between the same two nodes, and that the voltages vg and
vi have the same polarities, so these voltages are equal:
vg =
vi = vg = −2 V.
Using the passive sign convention,
ps = (8 A)(vi ) = (8 A)(−2 V) = −16 W.
Thus the current source generates 16 W of power.
P 2.8
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.
Problems
2–13
P 2.9
We assume the interconnection is valid, and therefore the total power in the
circuit balances. Note that the voltage drop from the top node to the bottom
node must be the same in each branch, so
v∆ = 50 + v75
so v75 = v∆ − 50
and
v∆ = vdep − 20 so vdep = v∆ + 20.
The total power in the circuit is
vtotal = 50(75) + 75v75 − 5v∆ vdep + 5v∆ (20) − 25v∆
= 50(75) + 75(v∆ − 50) − 5v∆ (v∆ + 20) + 5v∆ (20) − 25v∆ = 0.
Solving for v∆ gives us 10 V, so v75 = −40 V and vdep = 30 V. Use the values
for v∆ , v75 , and vdep to calculate the power associated with each source:
p50V = (75)(50) = 3750 W
p20V = [5(10)](20) = 1000 W
p75A = (75)(−40) = −3000 W;
pds = −(50)(30) = −1500 W;
p25A = −(25)(10) = −250 W.
X
P 2.10
Pdev = 250 + 3000 + 1500 = 4750 W =
X
Pabs .
[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 i1 = 50 mA.)
2–14
CHAPTER 2. Circuit Elements
[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
v1 − 20 = v2 = v3 + 30.
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 2.11
[a] Using the passive sign convention and Ohm’s law,
i=
v
120
=
= 0.012 = 12 mA.
R
10,000
[b] pv.s. = pR = Ri2 = (10,000)(0.012)2 = 1.44 W.
[c] Using the passive sign convention with the voltage polarity reversed,
i=−
120
v
=−
= −0.012 = −12 mA;
R
10,000
pv.s. = pR = Ri2 = (10,000)(−0.012)2 = 1.44 W.
P 2.12
[a] Using the passive sign convention and Ohm’s law,
v = −Ri = −(8000)(0.075) = −600 V.
(−600)2
v2
=
= 45 W.
R
8000
[c] Using the passive sign convention with the current direction reversed,
[b] pc.s. = pR =
v = Ri = (8000)(0.075) = 600 V;
pc.s. = pR =
v2
6002
=
= 45 W.
R
8000
Problems
P 2.13
[a]
[b]
P 2.14
2–15
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.
Since we know the device is a resistor, we can use the power equation. From
Fig. P2.14(a),
p = vi =
v2
R
so
R=
v2
.
p
Using the values in the table of Fig. P2.14(b)
R=
(5)2
(10)2
(15)2
(20)2
=
=
=
16.67 × 10−3
66.67 × 10−3
150 × 10−3
266.67 × 10−3
=
(25)2
(30)2
=
= 1.5 kΩ.
416.67 × 10−3
600 × 10−3
Note that this value is found in Appendix H.
P 2.15
Since we know the device is a resistor, we can use Ohm’s law to calculate the
resistance. From Fig. P2.15(a),
v = Ri
so
v
R= .
i
2–16
CHAPTER 2. Circuit Elements
Using the values in the table of Fig. P2.15(b),
R=
−22.4
−11.2
11.2
22.4
33.6
=
=
=
=
= 560 Ω.
−0.04
−0.02
0.02
0.04
0.06
Note that this value is found in Appendix H.
P 2.16
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. P2.16(b),
i
0.072
0.162
0.288
0.018
=
=
=
2
2
2
(0.001)
(0.002)
(0.003)
(0.004)2
=
0.648
0.45
=
= 18 kΩ.
2
(0.005)
(0.006)2
Note that this is a value from Appendix H.
P 2.17
[a] Write a KVL equation clockwise aroud the right loop, starting below the
vertical 10 Ω resistor:
−va + vb = 0
so
va = vb .
Using Ohm’s law,
va = 10ia
and
vb = 15ib .
Substituting,
2
ib = ia .
3
Write a KCL equation at the top middle node, summing the currents
leaving:
10ia = 15ib
so
2
5
ix = ia + ib = ia + ia = ia .
3
3
Write a KVL equation clockwise around the left loop, starting below the
voltage source:
−ix + ia + ib = 0
−80 V + vx + va = 0.
so
Problems
2–17
From Ohm’s law,
vx = 10ix
and
va = 10ia .
Substituting,
−80 V + 10ix + 10ia = 0
Substituting for ix :
−80 V + 10
50
80
5
ia + 10ia = −80 V + ia + 10ia = −80 V + ia = 0.
3
3
3
Thus,
80
ia = 80 V
3
so
ia =
80 V
= 3 A.
80/3
[b] From part (a), ib = 32 ia = 23 (3 A) = 2 A.
[c] From the circuit, vb = 15 Ω(ib ) = 15 Ω(2 A) = 30 V.
[d] Use the formula pR = Ri2R to calculate the power absorbed by each
resistor:
p10Ω(top) = i2x (10 Ω) =
2
5 2
5
ia (10 Ω) = (3) (10 Ω) = (5)2 (10 Ω) = 250 W;
3
3
p10Ω(mid) = i2a (10 Ω) = (3)2 (10 Ω) = 90 W;
p15Ω = i2b (15 Ω) = (2)2 (15 Ω) = 60 W.
[e] Using the passive sign convention,
5
5
psource = −(80 V)ix = −(80 V) ia = −(80 V) (3 A)
3
3
= −(80 V)(5 A) = −400 W.
Thus the voltage source delivers 400 W of power to the circuit. Check:
P 2.18
X
Pdis = 250 + 90 + 60 = 400 W;
X
Pdel = 400 W.
Label the unknown resistor voltages and currents:
2–18
CHAPTER 2. Circuit Elements
[a] ib =
25
= 0.005 A (Ohm’s law);
5000
i1 = 0.025 − ib = 0.025 − 0.005 = 0.02 A (KCL).
[b] v1 = 250i1 = 250(0.02) = 5 V (Ohm’s law);
[c] va = 0.025(2000) = 50 V (Ohm’s law);
−vg + va + 25 V = 0
so
vg = va + 25 = 50 + 25 = 75 V (KVL).
[d] pg = −vg (0.025) = −75(0.025) = −1.875 W
so 1.875 W delivered.
P 2.19
[a] Write a KCL equation at the top node:
−4 + i1 + i2 = 0
so
i1 + i2 = 4.
Write a KVL equation around the right loop:
−v80 + v30 + v90 = 0.
From Ohm’s law,
v80 = 80i1 ,
v30 = 30i2 ,
v90 = 90i2 .
Substituting,
−80i1 + 30i2 + 90i2 = 0
so
− 80i1 + 120i2 = 0.
Solving the two equations for i1 and i2 simultaneously,
i1 = 2.4 A
and
i2 = 1.6 A.
[b] Write a KVL equation clockwise around the left loop:
−vo + v80 = 0
So
but
v80 = 80i1 = 80(2.4) = 192 A.
vo = v80 = 192 V.
[c] Calculate power using p = vi for the source and p = Ri2 for the resistors:
psource = −vo (4) = −(192)(4) = −768 W;
p80Ω = 2.42 (80) = 460.8 W;
p30Ω = 1.62 (30) = 76.8 W;
p90Ω = 1.62 (90) = 230.4 W;
X
Pdev = 768 W
X
Pabs = 460.8 + 76.8 + 230.4 = 768 W.
Problems
P 2.20
2–19
[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 vo , 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.
P 2.21
Label the unknown resistor currents and voltages:
[a] KCL at the top node:
i1 = i2 + 0.0025;
KVL around the left loop:
−100 + v1 + vo = 0.
Use Ohm’s law to write the resistor voltages in the previous equation in
terms of the resistor currents:
−100 + 5000i1 + 2000i2 = 0
→
5000i1 + 2000i2 = 100.
Use the KCL equation for i1 to eliminate i1 in the KVL equation:
5000(i2 + 0.0025) + 2000i2 = 100
→
7000i2 = 87.5.
Solving,
87.5
= 0.0125 = 12.5 mA.
7000
Therefore,
i2 =
vo = Ri2 = (2000)(0.0125) = 25 V.
[b] i1 = i2 + 0.0025 = 0.0125 + 0.0025 = 0.015 A;
p100V = −(100)i1 = −(100)(0.015) = −1.5 W;
p25mA = (0.0025)vo = (0.0025)(25) = 0.0625 W;
p5k = 5000i21 = 5000(0.015)2 = 1.125 W;
2–20
CHAPTER 2. Circuit Elements
p2k = 2000i22 = 2000(0.0125)2 = 0.3125 W;
ptotal = p25mA + p100V + p5k + p2k = 0.0625 − 1.5 + 1.125 + 0.3125 = 0.
Thus the power in the circuit balances.
P 2.22
[a]
iCD = 500/15,000 = 33.33 mA;
iBD + iCD = 0.1
so
iBD = 0.1 − 0.033 = 66.67 mA;
4000iBC + 500 − 7500iBD = 0
so
iBC = (500 − 500)/4000 = 0;
iAC = iCD − iBC = 33.33 − 0 = 33.33 mA;
100 = iAB + iAC
so
iAB = 100 − 33.33 = 66.67 mA.
Calculate the power dissipated by the resistors using the equation
pR = Ri2R :
p5kΩ = (5000)(0.0667)2 = 22.22 W
p7.5kΩ = (7500)(0.0667)2 = 33.33 W;
p10kΩ = (10,000)(0.03333)2 = 11.11 W
p15kΩ = (15,000)(0.0333)2 = 16.67 W;
p4kΩ = (4000)(0)2 = 0 W.
[b] Calculate the voltage drop across the current source:
vAD = 5000iAB + 7500iBD = 5000(0.0667) + 7500(0.0667) = 833.33 V.
Now that we have both the voltage and the current for the source, we can
calculate the power supplied by the source:
pg = −833.33(0.1) = −83.33 W
[c]
X
thus
pg (supplied) = 83.33 W.
Pdis = 22.22 + 33.33 + 11.11 + 16.67 + 0 = 83.33 W.
Therefore,
X
Psupp =
X
Pdis .
Problems
P 2.23
2–21
[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 2.24
[a] Start by calculating the voltage drops due to the currents i1 and i2 . Then
use KVL to calculate the voltage drop across and 100 Ω resistor, and
Ohm’s law to find the current in the 100 Ω resistor. Finally, KCL at each
of the middle three nodes yields the currents in the two sources and the
current in the middle 10 Ω resistor. These calculations are summarized in
the figure below:
2–22
CHAPTER 2. Circuit Elements
[b]
p130
= −(130)(15) = −1950 W;
p460
= −(460)(30) = −13,800 W.
X
Pdis = (15)2 (2) + (15)2 (10) + (30)2 (2) + (10)2 (25) + (25)2 (10) + (5)2 (100)
= 450 + 2250 + 1800 + 2500 + 6250 + 2500 = 15,750 W.
X
Psup = 1950 + 13,800 = 15,750 W.
Therefore,
P 2.25
X
Pdis =
X
Psup = 15,750 W.
[a]
v2 = 80 + 4(12) = 128 V;
i1 =
v1
80
=
= 5 A;
6 + 10
16
v1 = 128 − (8 + 12 + 4)(2) = 80 V;
i3 = i1 − 2 = 5 − 2 = 3 A;
vg = v1 + 24i3 = 80 + 24(3) = 152 V;
i4 = 2 + 4 = 6 A;
ig = −i4 − i3 = −6 − 3 = −9 A.
[b] Calculate power using the formula p = Ri2 :
p8 Ω = (8)(2)2 = 32 W;
p12 Ω = (12)(2)2 = 48 W;
p4 Ω(left) = (4)(2)2 = 16 W;
p4 Ω(right) = (4)(6)2 = 144 W;
p24 Ω = (24)(3)2 = 216 W;
p6 Ω = (6)(5)2 = 150 W;
p10 Ω = (10)(5)2 = 250 W;
p12 Ω = (12)(4)2 = 192 W.
Problems
2–23
[c] vg = 152 V.
[d] Sum the power dissipated by the resistors:
X
pdiss = 32 + 48 + 16 + 144 + 216 + 150 + 250 + 192 = 1048 W.
The power associated with the sources is
pvolt−source = (80)(4) = 320 W;
pcurr−source = −vg ig = −(152)(9) = −1368 W.
Thus the total power dissipated is 1048 + 320 = 1368 W and the total
power developed is 1368 W, so the power balances.
P 2.26
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)
= 1440 W (CHECKS).
P 2.27
[a]
2–24
CHAPTER 2. Circuit Elements
ib = 60 V/30 Ω = 2 A;
va = (30 + 60)(2) = 180 V;
−500 + va + vb = 0;
so
vb = 500 − va = 500 − 180 = 320 V.
ie = vb /(60 + 36) = 320/96 = (10/3) A;
id = ie − ib = (10/3) − 2 = (4/3) A;
vc = 30id + vb = 40 + 320 = 360 V;
ic = vc /180 = 360/180 = 2 A;
vd = 500 − vc = 500 − 360 = 140 V;
ia = id + ic = 4/3 + 2 = (10/3) A;
R = vd /ia = 140/(10/3) = 42 Ω.
[b] ig = ia + ib = (10/3) + 2 = (16/3) A.
pg (supplied) = (500)(16/3) = 2666.67 W.
P 2.28
[a] Plot the v − i characteristic
From the plot:
R=
∆v
(82 − 50)
=
= 4 Ω.
∆i
(8 − 0)
When it = 0, vt = 50 V; therefore the ideal voltage source has a voltage
of 50 V.
[b]
When vt = 0,
it =
−50
= −12.5A.
4
Note that this result can also be obtained by extrapolating the v − i
characteristic to vt = 0.
Problems
P 2.29
2–25
[a] Plot the v—i characteristic:
From the plot:
R=
(420 − 100)
∆v
=
= 20 Ω.
∆i
(16 − 0)
When it = 0, vt = 100 V; therefore the ideal current source must have a
current of 100/20 = 5 A.
[b] We attach a 5 Ω resistor to the device model developed in part (a):
Write a KCL equation at the top node:
5 + it = i1 .
Write a KVL equation for the right loop, in the direction of the two
currents, using Ohm’s law:
20i1 + 5it = 0.
Combining the two equations and solving,
20(5 + it ) + 5it = 0
so
25it = −100;
thus
Now calculate the power dissipated by the resistor:
p5 Ω = 5i2t = 5(−4)2 = 80 W.
it = −4 A.
2–26
P 2.30
CHAPTER 2. Circuit Elements
[a] Begin by constructing a plot of voltage versus current:
[b] Since the plot is linear for 0 ≤ is ≤ 225 mA and since R = ∆v/∆i, we can
calculate R from the plotted values as follows:
∆v
75 − 30
45
=
=
= 200 Ω.
∆i
0.225 − 0
0.225
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 75 V.
The model, valid for 0 ≤ is ≤ 225 mA, is shown below:
R=
[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:
−75 V + 200i + 400i = 0
Thus,
i=
so
75 V
= 125 mA.
600 Ω
[d] The circuit is shown below:
600i = 75 V.
Problems
2–27
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:
−75 V + 200i = 0
Thus,
i=
so
200i = 75 V.
75 V
= 375 mA.
200 Ω
[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 = 500 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 2.31
[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.
2–28
CHAPTER 2. Circuit Elements
[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 2.32
Label unknown voltage and current:
io = ix + 3ix = 4ix
(KCL);
−280 + 6ix + 2(4ix ) = 0
(KVL left).
Therefore
14ix = 280 so ix =
280
= 20 A.
14
Thus
vo = 2(4ix ) = 8ix = 8(20) = 160 V.
The only two circuit elements that could supply power are the two sources, so
calculate the power for each source:
p280V = −280ix = −280(20) = −5600 W;
pd.s. = −(3ix )vo = −3(20)(160) = −9600 W.
Both sources are supplying power, so the total power supplied is
5600 + 9600 = 15,200 W.
P 2.33
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
Problems
2–29
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 5 V 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.
[c] We now know the values of voltage and current for every circuit element.
2–30
CHAPTER 2. Circuit Elements
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
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.
P 2.34
[a] The circuit:
v1 = (3)(8) = 24 V
v1
= 12io + 4io = 16io
6
(Ohm’s law);
(KVL).
Thus,
io =
24/6
v1 /6
=
= 0.25 = 250 mA.
16
16
Problems
2–31
[b] Calculate the power for all components:
p8A = −(8)v1 = −(8)(24) = −192 W;
pd.s. = −(v1 /6)io = −(24/6)(0.25) = −1 W;
p3 =
242
v12
=
= 192 W;
3
3
p12 = 12i2o = 12(0.25)2 = 0.75 W;
p4 = 4i2o = 4(0.25)2 = 0.25 W.
Therefore,
ptotal = −192 − 1 + 192 + 0.75 + 0.25 = 0.
Thus the power in the circuit balances.
P 2.35
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, from right to left, as seen in the
circuit below.
[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, from 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.
2–32
CHAPTER 2. Circuit Elements
[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
v4A = −10 V.
so
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.
P 2.36
[a] −50 − 20iσ + 18i∆ = 0;
−18i∆ + 5iσ + 40iσ = 0,
so
18i∆ = 45iσ .
− 50 − 20iσ + 45iσ = 0,
Therefore,
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)
X
Pgen
=
4230 W; Therefore,
=
X
Pdiss = 4230 W.
Problems
P 2.37
50i2 +
0.250 0.250
+
= 0;
50
12.5
i2 = −0.5 mA;
v1 = 100i2 = −50 mV;
20i1 +
(−0.050)
+ (−0.0005) = 0;
25
i1 = 125 µA;
vg = 10i1 + 40i1 = 50i1 ;
Therefore, vg = 6.25 mV.
P 2.38
iE − iB − iC = 0;
iC = βiB
therefore iE = (1 + β)iB ;
i2 = −iB + i1 ;
Vo + iE RE − (i1 − iB )R2 = 0;
−i1 R1 + 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 =
P 2.39
[VCC R2 /(R1 + R2 )] − Vo
.
(1 + β)RE + R1 R2 /(R1 + R2 )
Here is Equation 2.21:
iB =
(VCC R2 )/(R1 + R2 ) − V0
;
(R1 R2 )/(R1 + R2 ) + (1 + β)RE
VCC R2
(15)(80)
=
= 12 V;
R1 + R2
100
R1 R2
(20)(80)
=
= 16 kΩ;
R1 + R2
100
iB =
12 − 0.2
11.8
=
= 0.59 mA;
16 + 40(0.1)
20
2–33
2–34
CHAPTER 2. Circuit Elements
iC = βiB = (39)(0.59) = 23.01 mA;
iE = iC + iB = 23.01 + 0.59 = 23.6 mA;
v3d = (23.6)(0.1) = 2.36 V;
vbd = Vo + v3d = 2.56 V;
i2 =
vbd
2.56
=
× 10−3 = 32 µA;
R2
80
i1 = i2 + iB = 32 + 590 = 622 µA;
vab = 20(0.622) = 12.44 V;
iCC = iC + i1 = 23.01 + 0.622 = 23.632 mA;
v13 + 23.01(0.5) + 2.36 = 15;
v13 = 1.135 V.
P 2.40
[a]
[b]
P 2.41
Each radiator is modeled as a 48 Ω resistor:
Problems
2–35
Write a KVL equation for each of the three loops:
240
= 5 A;
48
−240 + 48i1 = 0
→
i1 =
−48i1 + 48i2 = 0
→
i2 = i1 = 5 A;
−48i2 + 48i3 = 0
→
i3 = i2 = 5 A.
Therefore, the current through each radiator is 5 A and the power for each
radiator is
prad = Ri2 = 48(5)2 = 1200 W.
There are three radiators, so the total power for this heating system is
ptotal = 3prad = 3(1200) = 3600 W.
P 2.42
Each radiator is modeled as a 48 Ω resistor:
Write a KVL equation for this loop:
−240 + 48i + 48i + 48i = 0
→
i=
240
= 1.67 A.
3(48)
Calculate the power for each radiator:
prad = 48i2 = 48(1.67)2 = 133.33 W.
Calculate the total power for this heating system:
ptotal = 3prad = 3(133.33) = 400 W.
Each radiator has much less power than the radiators in Fig. P2.41, and the
total power of this configuration is just 1/9th of the total power in Fig. P2.41.
2–36
P 2.43
CHAPTER 2. Circuit Elements
Each radiator is modeled as a 48 Ω resistor:
Write a KVL equation for the left and right loops:
−240 + 48i1 = 0
→
−48i1 + 48i2 + 48i2 = 0
i1 =
240
= 5 A;
48
→
i2 =
i1
5
= = 2.5 A.
2
2
The power for the center radiator is
pcen = 48i21 = 48(5)2 = 1200 W.
The power for each of the radiators on the right is
pright = 48i22 = 48(2.5)2 = 300 W.
Thus the total power for this heating system is
ptotal = pcen + 2pright = 1200 + 2(300) = 1800 W.
The center radiator produces 1200 W, just like each of the three radiators in
Problem 2.41. But the other two radiators produce only 300 W each, which is
1/4th of the power of the radiators in Problem 2.41. The total power of this
configuration is 1/2 of the total power in Fig. P2.41.
P 2.44
Each radiator is modeled as a 48 Ω resistor:
Write a KVL equation for the left and right loops:
−240 + 48i1 + 48i2 = 0;
Problems
−48i2 + 48i3 = 0
→
i2 = i3 .
Write a KCL equation at the top node:
i1 = i2 + i3
→
i1 = i2 + i2 = 2i2 .
Substituting into the first KVL equation gives
−240 + 48(2i2 ) + 48i2 = 0
→
i2 =
240
= 1.67 A.
3(48)
Solve for the currents i1 and i3 :
i3 = i2 = 1.67 A;
i1 = 2i2 = 2(1.67) = 3.33 A.
Calculate the power for each radiator using the current for each radiator:
pleft = 48i21 = 48(3.33)2 = 533.33 W;
pmiddle = pright = 48i22 = 48(1.67)2 = 133.33 W.
Thus the total power for this heating system is
ptotal = pleft + pmiddle + pright = 533.33 + 133.33 + 133.33 = 800 W.
All radiators in this configuration have much less power than their
counterparts in Fig. P2.41. The total power for this configuration is only
22.2% of the total power for the heating system in Fig. P2.41.
2–37
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 35 Ω, the 75 Ω and the 40 Ω resistors in series:
35 Ω + 75 Ω + 40 Ω = 150 Ω.
Now combine this 150 Ω resistor in parallel with the 100 Ω resistor:
150 Ωk100 Ω =
(150)(100)
15,000
=
= 60 Ω.
150 + 100
250
This equivalent 60 Ω resistor is in series with the 140 Ω resistor:
60 Ω + 140 Ω = 200 Ω.
Finally, this equivalent 200 Ω resistor is in parallel with the 300 Ω resistor:
200 Ωk300 Ω =
(200)(300)
60,000
=
= 120 Ω.
200 + 300
500
Thus, the simplified circuit is as shown:
3–1
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 120 Ω equivalent resistor:
v = (120 Ω)(0.03 A) = 3.6 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 = −(3.6 V)(0.03 A) = −108 mW.
Thus, the source delivers 108 mW of power to the circuit.
[c] We now can return to the original circuit, shown in the first figure. In this
circuit, v = 3.6 V, as calculated in part (a). This is also the voltage drop
across the 300 Ω resistor, so we can use Ohm’s law to calculate the
current through this resistor:
3.6 V
= 12 mA.
300 Ω
Now write a KCL equation at the upper left node to find the current iB :
iA =
−30 mA + iA + iB = 0
so
iB = 30 mA − iA = 30 mA − 12 mA = 18 mA.
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 + 140iB + 35iC + 75iC + 40iC = 0.
So
Thus
150iC = v − 140iB = 3.6 V − (140)(0.018) = 1.08 V.
iC =
1.08
= 7.2 mA.
150
Now that we have the current through the 75 Ω resistor we can use the
formula p = Ri2 to find the power:
p75 Ω = (75)(7.2 × 10−3 )2 = 3.888 mW = 3888 µW.
Problems
3–3
AP 3.2
[a] We can use voltage division to calculate the voltage vo across the 4 kΩ
resistor:
4000
(60 V) = 20 V.
vo (no load) =
4000 + 8000
[b] When we have a load resistance of 4 kΩ then the voltage vo is across the
parallel combination of two 4 kΩ resistors. First, calculate the equivalent
resistance of the parallel combination:
(4000)(4000)
= 2000 Ω = 2 kΩ.
4000 + 4000
Now use voltage division to find vo across this equivalent resistance:
4 kΩk4 kΩ =
vo =
2000
(60 V) = 12 V.
2000 + 8000
[c] If the load terminals are short-circuited, the 4 kΩ resistor is effectively
removed from the circuit, leaving only the voltage source and the 8 kΩ
resistor. We can calculate the current in the resistor using Ohm’s law:
i=
60 V
= 7.5 mA.
8 kΩ
Now we can use the formula p = Ri2 to find the power dissipated in the
8 kΩ resistor:
p8k = (8000)(0.0075)2 = 0.45 W.
[d] The power dissipated in the 4 kΩ resistor will be maximum at no load
since vo is maximum. In part (a) we determined that the no-load voltage
is 20 V, so we can use the formula p = v 2 /R to calculate the power:
p4k (max) =
(20)2
= 0.1 W.
4000
3–4
CHAPTER 3. Simple Resistive Circuits
AP 3.3
[a] We will write a current division equation for the current in the branch
containing the 5 kΩ resistor and use this equation to solve for R:
i5k =
Rk(3000 + 5000 + 7000)
(0.03) = 0.012;
3000 + 5000 + 7000
Rk(3000 + 5000 + 7000) =
(3000 + 5000 + 7000)(0.012)
= 6000;
0.03
15,000R
= 6000;
R + 15,000
15,000R = 6000R + 6000(15,000) so 9000R = 6000(15,000);
Thus, R =
6000(15,000)
= 10,000 = 10 kΩ.
9000
[b] With R = 10 kΩ we can calculate the current through R and then use this
current to find the power dissipated by R, using the formula p = Ri2 :
iR = 0.03 − i5k = 0.03 − 0.012 = 0.018 = 18 mA,
so
pR = (10,000)(0.018)2 = 3.24 W.
[c] The equivalent resistance seen by the current source is
Req = 4000 + 10,000k15,000 = 4000 + 6000 = 10,000 = 10 kΩ.
The power supplied by the current source equals the power dissipated by
the equivalent resistance (whose current is 30 mA) so
p = 10,000(0.03)2 = 9 W.
Thus, the current source generates 9 W of power.
AP 3.4
Problems
3–5
[a] First we need to determine the equivalent resistance of the 80 Ω resistor
and the resistors to its right:
Req = 80 Ωk(50 Ω + 30 Ω) = 80 Ωk80 Ω = 40 Ω
Now we can use voltage division to find the voltage vo :
vo =
10
(120 V) = 10 V.
10 + 40 + 70
[b] The current through the 10 Ω resistor can be found using Ohm’s law:
10 V
vo
=
= 1 A.
10
10 Ω
Now we use current division to find the current in the 80 Ω branch:
80k(50 + 30)
40
i80Ω =
(1 A) =
= 0.5 A.
80
80
[c] We can find the power dissipated by the 50 Ω resistor if we can find the
current in this resistor. The current in the 10 Ω resistor divides between
the 80 Ω branch and the branch with the 50 Ω and 30 Ω resistors. Since
0.5 A is in the 80 Ω branch, the remaining 0.5 A current must be in the
other branch. Therefore,
i40Ω =
i50Ω = 0.5 A.
Thus,
p50Ω = (50)(0.5)2 = 12.5 W.
AP 3.5 [a]
We can find the current i using Ohm’s law:
i=
5V
= 0.2 A = 200 mA.
25 Ω
[b]
Rm = 50 Ωk5.555 Ω = 5 Ω.
3–6
CHAPTER 3. Simple Resistive Circuits
We can use the meter resistance to find the current using Ohm’s law:
imeas =
5V
= 0.1667 = 166.67 mA.
25 Ω + 5 Ω
AP 3.6 [a]
Use voltage division to find the voltage v:
v=
50,000
(150 V) = 100 V.
50,000 + 25,000
[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 50 kΩ resistor and
the voltmeter:
(50,000)(150,000)
50,000 Ωk150,000 Ω =
= 37.5 kΩ.
50,000 + 150,000
Thus,
vmeas =
37,500
(150 V) = 90 V.
37,500 + 25,000
AP 3.7 [a] Using the condition for a balanced bridge, the products of the opposite
resistors must be equal. Therefore,
1000Rx = (2000)(500)
so
Rx =
(2000)(500)
= 1000 Ω = 1 kΩ.
1000
Problems
3–7
[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:
iR1 ,R3 =
30 V
= 10 mA;
1000 Ω + 2000 Ω
iR2 ,Rx =
30 V
= 20 mA.
500 Ω + 1000 Ω
We can calculate the power dissipated by each resistor using the formula
p = Ri2 :
pR1 = (1000 Ω)(0.01 A)2 = 100 mW;
pR2 = (500 Ω)(0.02 A)2 = 200 mW;
pR3 = (2000 Ω)(0.01 A)2 = 200 mW;
pRx = (1000 Ω)(0.02 A)2 = 400 mW.
Thus, the Rx resistor must dissipate the most power.
AP 3.8 Convert the three Y-connected resistors, 1 Ω, 10 Ω, and 40 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:
(1)(10) + (1)(40) + (10)(40)
= 450 Ω;
1
(1)(10) + (1)(40) + (10)(40)
Rb =
= 45 Ω;
10
(1)(10) + (1)(40) + (10)(40)
Rc =
= 11.25 Ω.
40
Ra =
The circuit with these new ∆-connected resistors is shown below:
3–8
CHAPTER 3. Simple Resistive Circuits
From this circuit we see that the 450 Ω resistor is parallel to the 50 Ω resistor:
450 Ωk50 Ω =
(450)(50)
= 45 Ω.
450 + 50
Also, the 11.25 Ω resistor is parallel to the 15 Ω resistor:
11.25 Ωk15 Ω =
(11.25)(15)
45
=
Ω.
11.25 + 15
7
Once the parallel combinations are made, we can see that the equivalent 45 Ω
resistor is in series with the equivalent 45/7 Ω resistor, and that this series
combination is in parallel with the voltage source. Therefore, we can use
voltage division to find v:
v=
45/7
(24) = 3 V.
45/7 + 45
Problems
3–9
Problems
P 3.1
[a] The 3 kΩ and 8 kΩ resistors are in series, as are the 5 kΩ and 7 kΩ
resistors. The simplified circuit is shown below:
[b] The 180 Ω and 300 Ω resistors are in series, as are the 140 Ω and 200 Ω
resistors. The simplified circuit is shown below:
[c] The 40 Ω, 50 Ω, and 60 Ω resistors are in series, as are the 45 Ω and 30 Ω
resistors. The simplified circuit is shown below:
[d] The 6 Ω and 1.2 Ω resistors are in series, as are the 6 Ω and 10 Ω resistors.
The simplified circuit is shown below:
3–10
P 3.2
CHAPTER 3. Simple Resistive Circuits
[a] The 4 kΩ and 12 kΩ resistors are in parallel, as are the 6 kΩ and 3 kΩ
resistors. The simplified circuit is shown below:
[b] The 12 Ω and 20 Ω resistors are in parallel, as are the 28 Ω and 21 Ω
resistors. The simplified circuit is shown below:
[c] The 30 Ω and 5 Ω resistors are in parallel, as are the 9 Ω and 18 Ω resistors.
The simplified circuit is shown below:
[d] The 100 kΩ and 300 kΩ resistors are in parallel, as are the 75 kΩ, 50 kΩ,
and 150 kΩ resistors. 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] Circuit in Fig. P3.1(a):
Req = {[(5 k + 7 k)k6 k] + 3 k + 8 k}k10 k = [(12 kk6 k) + 11 k]k10 k
= (4 k + 11 k)k10 k = 15 kk10 k = 6 kΩ.
Circuit in Fig. P3.1(b):
Req = [240k(180 + 300)] + 140 + 200 = (240k480) + 340 = 160 + 340 = 500 Ω.
Problems
3–11
Circuit in Fig. P3.1(c):
Req = (40 + 50 + 60)k(30 + 45) = 150k75 = 50 Ω.
Circuit in Fig. P3.1(d):
Req = ([(6 + 10)k64] + 6 + 1.2)k30 = [(16k64) + 6 + 1.2]k30
= (12.8 + 6 + 1.2)k30 = 20k30 = 12 Ω.
[b] Note that in every case, the power delivered by the source must equal the
power absorbed by the equivalent resistance in the circuit. For the circuit
in Fig. P3.1(a):
P = i2s Req = 0.0022 (6000) = 0.024 = 24 mW.
For the circuit in Fig. P3.1(b):
P =
102
Vs2
=
= 0.2 W.
Req
500
For the circuit in Fig. P3.1(c):
vs2
502
P =
=
= 50 W.
Req
50
For the circuit in Fig. P3.1(d):
P = i2s (Req ) = (5)2 (12) = 300 W.
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] Circuit in Fig. P3.2(a):
400 + (6000k3000k(5000 + (4000k12,000))) = 400 + (2000k(5000 + 3000))
= 400 + 1600 = 2000 = 2 kΩ.
Circuit in Fig. P3.2(b):
Req = 12k20k[18 + (28k21)] = 12k20k(18 + 12) = 12k20k30 = 6 Ω.
Circuit in Fig. P3.2(c):
Req = 4 + (9k18) + [5k30k(20 + 40)] = 4 + 6 + (5k30k60) = 4 + 6 + 4 = 14 Ω.
Circuit in Fig. P3.2(d):
Req = (100 kk300 k) + (75 kk50 kk150 k) + 25 k = 75 k + 25 k + 25 k = 125 kΩ.
3–12
CHAPTER 3. Simple Resistive Circuits
[b] Note that in every case, the power delivered by the source must equal the
power absorbed by the equivalent resistance in the circuit. For the circuit
in Fig. P3.2(a):
P =
vs2
602
=
= 1.8 W.
Req
2000
For the circuit in Fig. P3.2(b):
P = i2s (Req ) = (0.2)2 (6) = 0.24 W.
For the circuit in Fig. P3.2(c):
P =
82
vs2
=
= 4.57 W.
Req
14
For the circuit in Fig. P3.2(d):
P =
P 3.5
vs2
0.52
=
= 2 µW.
Req
125,000
[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.
[c] Rab = 2 k + [(14 k + 10 k)k12 k] + 6 k = (24 kk12 k) + 8 k
= 8 k + 8 k = 16 kΩ.
[d] Rab = 600k200k300k(250 + 150) = 600k200k300k400 = 80 Ω.
P 3.6
Write an expression for the resistors in series and parallel from the right side
of the circuit to the left. Then simplify the resulting expression from left to
right to find the equivalent resistance.
[a] Rab = [(18 + 12)k30 + 15]k60 = (30k30 + 15)k60 = (15 + 15)k60 = 30k60 = 20 Ω.
[b] 60k20 = 1200/80 = 15 Ω;
15 + 8 + 7 = 30 Ω;
12k24 = 288/36 = 8 Ω;
30k120 = 3600/150 = 24 Ω;
Rab = 15 + 24 + 25 = 64 Ω.
[c] 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.
Problems
[d] 50 + 30 = 80 Ω;
P 3.7
3–13
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] Circuit in Fig. P3.7(a):
Req = 15k(18 + 48k16) = 10 Ω.
Circuit in Fig. P3.7(b):
Req = ([(750 + 250)k1000] + 250)k([(300 + 500)k2400] + 900)
= [(1000k1000) + 250]k[(800k2400) + 900] = (500 + 250)k(600 + 900)
= 750k1500 = 500 Ω.
Circuit in Fig. P3.7(c):
5k10k15k10k(12 + 18) = 2 Ω;
16k(14 + 2) = 8 Ω;
Req = 4 + 8 + 12 = 24 Ω.
Circuit in Fig. P3.7(d):
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;
Req = 30k60 = 20 Ω.
[b] Note that in every case, the power delivered by the source must equal the
power absorbed by the equivalent resistance in the circuit. For the circuit
in Fig. P3.7(a):
P =
vs2
202
=
= 40 W.
Req
10
For the circuit in Fig. P3.7(b):
P = i2s (Req ) = (0.05)2 (500) = 1.25 W.
3–14
CHAPTER 3. Simple Resistive Circuits
For the circuit in Fig. P3.7(c):
P =
vs2
1442
=
= 864 W.
Req
24
For the circuit in Fig. P3.7(d):
P = i2s (Req ) = 52 (20) = 500 W.
P 3.8
[a] R + R = 2R.
[b] R + R + R + · · · + R = nR.
[c] R + R = 2R = 2000 so R = 1000 = 1 kΩ.
This is a resistor from Appendix H.
[d] nR = 6000;
so if n = 4, R = 1.5 kΩ.
This is a resistor from Appendix H. So connect four 1.5k resistors in
series to get 6 kΩ.
P 3.9
[a] Req = RkR =
[b] Req
R
R2
= .
2R
2
= RkRkRk · · · kR
(n R’s)
−1
1
1
1
=
+ + ···
R R
R
−1
n
R
=
= .
R
n
R
= 6000 so R = 12 kΩ.
2
This is a resistor from Appendix H.
R
[d]
= 900 so R = 900n.
n
If n = 2
R = 900(2) = 1.8 kΩ.
This is a resistor from Appendix H. So connect two 1.8k resistors in
parallel to get 900Ω.
[c]
P 3.10
[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.
Problems
[b] v1 = 4is = 48 V;
3–15
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 3.11
[a] p4Ω = i2s 4 = (12)2 4 = 576 W;
p3Ω = (8)2 3 = 192 W;
p18Ω = (4)2 18 = 288 W;
p6Ω = (8)2 6 = 384 W.
[b] p120V (delivered) = 120is = 120(12) = 1440 W.
[c] pdiss = 576 + 288 + 192 + 384 = 1440 W.
P 3.12
vo (no load) = 25 =
R2
(75),
100 + R2
P 3.13
2500 + 25R2 = 75R2 ,
Req
(75),
100 + Req
→
50RL
,
50 + RL
1250 + 25RL = 50RL ,
vo (with load) = 15 =
Req = 25 = 50kRL =
→
→
→
1500 + 15Req = 75Req ,
→
R2 = 50 Ω.
→
Req = 25 Ω.
RL = 50 Ω.
160(3300)
= 66 V.
(4700 + 3300)
[b] i = 160/8000 = 20 mA;
[a] vo =
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:
iR 1 =
160 − 66
,
R1
Thus, R1 ≥
942
0.5
Therefore,
or
94
R1
Thus, R2 = 12,408 Ω.
R1 ≤ 0.5;
R1 ≥ 17,672 Ω.
Now use the voltage specification:
R2
(160) = 66;
R2 + 17,672
2
3–16
P 3.14
CHAPTER 3. Simple Resistive Circuits
[a]
40 kΩ + 60 kΩ = 100 kΩ;
25 kΩk100 kΩ = 20 kΩ;
vo1 =
20,000
(380) = 80 V;
(75,000 + 20,000)
vo =
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
[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 3.15
[a] vo =
100R2
= 20
R1 + R2
Let Re = R2 kRL =
vo =
100Re
= 16
R1 + Re
so
R1 = 4R2
R2 RL
R2 + RL
so
R1 = 5.25Re
Problems
Then, 4R2 = 5.25Re =
Thus, R2 = 15 kΩ
3–17
5.25(48R2 )
48 + R2
and
R1 = 4(15k) = 60 kΩ
[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 = 100 − 16 = 84 V
p R1 =
842
= 117.6 mW
60k
Thus the minimum power rating for all resistors should be 1/8 W.
P 3.16
Refer to the solution to Problem 3.15. The voltage divider will reach the
maximum power it can safely dissipate when the power dissipated in R1 equals
0.15 W. Thus,
vR2 1
= 0.15
60,000
so
vR1 = 94.87 V.
vo = 100 − 94.87 = 5.13 V.
So,
100Re
= 5.13
60,000 + Re
Thus,
and
Re = 3245.55 Ω.
(15,000)RL
= 3245.55
15,000 + RL
and
RL = 4141.69 Ω.
The minimum value for RL from Appendix H is 4.7 kΩ.
P 3.17
[a] At no load:
vo = kvs =
R2
vs .
R1 + R2
At full load:
vo = αvs =
Therefore k
=
α
=
1−α
α
Thus
Re
vs ,
R1 + Re
R2
R1 + R2
Re
R1 + Re
and
and
where Re =
(1 − k)
R2 .
k
(1 − α)
R1 =
Re .
α
R1 =
R2 Ro
(1 − k)
=
R2 .
Ro + R2
k
Ro R2
.
Ro + R2
3–18
CHAPTER 3. Simple Resistive Circuits
Solving for R2 yields
Also,
[b] R1
R2
R2 =
(k − α)
Ro .
α(1 − k)
(1 − k)
(k − α)
R1 =
R2 .·.
Ro .
k
αk
0.25
Ro = 6.667 kΩ;
=
0.375
0.25
=
Ro = 20 kΩ.
0.125
R1 =
[c]
Maximum dissipation in R2 occurs at no load, therefore,
PR2 (max) =
752
= 281.25 mW.
20,000
Maximum dissipation in R1 occurs at full load.
PR1 (max) =
[100 − 0.5(100)]2
= 375 mW.
6666.67
[d]
P 3.18
PR1
=
PR2
=
(100)2
= 1.5 W = 1500 mW;
6666.67
(0)2
= 0 W.
20,000
(30)2
= 30,
R1 + R2 + R3
Therefore, R1 + R2 + R3 = 30 Ω.
(R1 + R2 )30
= 15;
(R1 + R2 + R3 )
Therefore, 2(R1 + R2 ) = R1 + R2 + R3 .
Problems
Thus, R1 + R2 = R3 ;
2R3 = 30;
R3 = 15 Ω.
R2 (30)
= 5;
R1 + R2 + R3
6R2 = R1 + R2 + R3 = 30;
Thus, R2 = 5 Ω;
P 3.19
i1200Ω =
R1 = 30 − R2 − R3 = 10 Ω.
800k2400k1200
400
(0.32) =
(0.32) = 106.667 mA;
1200
1200
P1200Ω = 1200i21200Ω = 1200(106.667 × 10−3 )2 = 13.653 W.
P 3.20
[a]
Req = (10 + 40 + 20)k[12 + (20k180)] = 70k30 = 21 Ω;
v12A = 12(21) = 252 V;
vo = v40Ω =
v20Ω =
io =
40
(252) = 144 V;
10 + 40 + 20
20k180
18
(252) = (252) = 151.2 V;
12 + (20k180)
30
151.2
= 7.56 A.
20
(252 − 151.2)2
= 846.72 W.
12
[c] p12A = −(252)(12) = −3024 W.
Thus the power developed by the current source is 3024 W.
[b] p12Ω =
P 3.21
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 );
3–19
3–20
CHAPTER 3. Simple Resistive Circuits
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
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 3.22
[a] Let vo be the voltage across the parallel branches, positive at the upper
terminal, then
ig = vo G1 + vo G2 + · · · + vo GN = vo (G1 + G2 + · · · + GN ).
It follows that
vo =
ig
.
(G1 + G2 + · · · + GN )
The current in the k th branch is
ig G k
ik =
.
[G1 + G2 + · + GN ]
ik = v o G k ;
Thus,
Problems
[b] i6.25 =
P 3.23
3–21
1.142(0.16)
= 0.032 = 32 mA.
[4 + 0.4 + 1 + 0.16 + 0.1 + 0.05]
[a] The equivalent resistance to the right of the 10 kΩ resistor is
3k + 8k + [6kk(5k + 7k)] = 15 kΩ. Therefore,
15kk10k
6k
(0.002) =
(0.002) = 1.2 mA.
10k
10k
[b] The voltage drop across the 10 kΩ resistor can be found using Ohm’s law:
i10k =
v10k = (10, 000)i10k = (10, 000)(0.0012) = 12 V.
[c] The voltage v10k drops across the 3 kΩ resistor, the 8 kΩ resistor and the
equivalent resistance of the 6 kΩ and the parallel branch containing the
5 kΩ and 7 kΩ resistors. Thus, using voltage division,
v6k =
6kk(5k + 7k)
4
(12) = (12) = 3.2 V.
3k + 8k + [6kk(5k + 7k)]
15
[d] The voltage v6k drops across the branch containing the 5 kΩ and 7 kΩ
resistors. Thus, using voltage division,
5k
(3.2) = 1.33 V.
v5k =
5k + 7k
P 3.24
[a] The equivalent resistance of the resistors to the right of the 30 Ω resistor is
Req = 6 + 1.2 + [64k(6 + 10)] = 7.2 + (64k16) = 7.2 + 12.8 = 20 Ω.
Using current division,
30k20
(5) = 3 A.
20
[b] The current into the top right node is 3 A and divides to give
iReq =
i10 = i10+6 =
P 3.25
64k(10 + 6)
(3) = 2.4 A.
10 + 6
[a] The equivalent resistance of the circuit to the right of the 60 kΩ resistor is
20 k + (200 kk50 k) = 20 k + 40 k = 60 kΩ.
Thus by current division,
30 kk60 kk60 k
15 k
(0.02) =
(0.02) = 5 mA.
60 k
60 k
[b] Using Ohm’s law:
i60k =
v60 k = 60,000i60 k = 60,000(0.005) = 300 V.
[c] The voltage across the 60 kΩ resistor divides between two resistors – the
20 kΩ resistor and the 200 kk50 k = 40 kΩ equivalent resistance. Using
voltage division,
20 k
20 k
(v60 k ) =
(300) = 100 V.
v20 k =
20 k + 40 k
60 k
3–22
CHAPTER 3. Simple Resistive Circuits
[d] The current in the 20 kΩ resistor can be found using Ohm’s law:
i20 k =
v20 k
100
=
= 5 mA.
20 k
20,000
[e] The current i20 k divides between the 200 kΩ branch and the 50 kΩ branch.
Using current division,
i200 k =
P 3.26
200 kk50 k
40 k
i20 k =
(0.005) = 1 mA.
200 k
200 k
[a] Begin by finding the equivalent resistance of the 120 Ω resistor and all
resistors to its right:
[(12k24) + (60k20) + 7]k120 = (8 + 15 + 7)k120 = 30k120 = 24 Ω.
Now use voltage division to find the voltage across the 15 Ω resistor:
v15 =
15
(128) = 30 V.
15 + 24 + 25
[b] Use Ohm’s law to find the current in the 15 Ω resistor:
i15 = v15 /15 = 30/15 = 2 A.
[c] From part (a), the equivalent resistance of all resistors to the right of the
120 Ω resistor is 30 Ω. Now use current division:
24
120k30
(2) =
(2) = 0.4 A.
i120 =
120
120
[d] Using Ohm’s law,
v120 = 120i120 = 120(0.4) = 48 V.
[e] The 48 V divides across the branch to the right of the 120 Ω resistor. This
branch has a 12k24 = 8 Ω resistance, a 60k20 = 15 Ω resistance and a 7 Ω
resistance. Use voltage division to find the voltage across the 60k20 Ω
resistance:
60k20
15
v60k20 =
(48) =
(48) = 24 V.
(12k24) + (60k20) + 7
8 + 15 + 7
[f ] i20 =
P 3.27
v60k20
24
=
= 1.2 A.
20
20
[a] The equivalent resistance to the right of the 16 Ω resistor is
14 + [5k10k15k10k(12 + 18)] = 14 + 2 = 16 Ω.
By voltage division,
v4 =
4
4
(144) = (144) = 24 V.
4 + (16k16) + 12
24
Problems
3–23
[b] Using Ohm’s law,
v4
24
=
= 6 A.
4
4
[c] From part (a), the resistance to the right of the 16 Ω resistor is also 16 Ω.
Therefore, the current in the 4 Ω resistor divides equally between the two
16 Ω branches. The current in the 14 Ω resistor is i14 = 6/2 = 3 A.
i4 =
[d] Using current division,
5k10k15k10k(12 + 18)
2
i14 = (3) = 0.2 A.
12 + 18
30
i18 =
P 3.28
60k30 = 20 Ω;
i30Ω =
50 + 25
(25) = 15 A;
30 + 20 + 50 + 75
v2 = (15)(20) = 300 V;
v2 + 30i30 = 750 V;
v1 − 12(25) = 750;
v1 = 1050 V.
P 3.29
[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.
6
[b] v6k = (Vs ) = 0.75Vs ;
8
v3k =
3
(Vs ) = 0.25Vs ;
12
vx = (0.75Vs ) − (0.25Vs ) = 0.5Vs .
P 3.30
Use current division to find the current in the branch containing the 24 kΩ
and 16 kΩ resistors, from top to bottom:
i24k+16k =
(24 k + 16 k)k(20 k + 40 k)
(0.05) = 0.03 = 30 mA.
24 k + 16 k
3–24
CHAPTER 3. Simple Resistive Circuits
Use Ohm’s law to find the voltage drop across the 16 kΩ resistor, positive at
the top:
v16k = (0.03)(16 k) = 480 V.
Find the current in the branch containing the 20 kΩ and 40 kΩ resistors, from
top to bottom:
i20k+40k = 0.05 − i24k+16k = 0.05 − 0.03 = 0.02 = 20 mA.
Use Ohm’s law to find the voltage drop across the 40 kΩ resistor, positive at
the top:
v40k = (40 k)(0.02) = 800 V;
vo = v16k − v40k = 480 − 800 = −320 V.
P 3.31
54 Ωk27 Ω = 18 Ω;
Therefore, ig =
i2Ω =
P 3.32
18 Ω + 2 Ω = 20 Ω;
20k(10 + 15 + 35) = 15 Ω;
675
= 15 A.
30 + 15
20k60
(15) = 11.25 A;
20
io =
27k54
(11.25) = 7.5 A.
27
[a] The equivalent resistance seen by the voltage source is
60k[8 + 30k(4 + 80k20)] = 60k[8 + 30k20] = 60k20 = 15 Ω.
Thus,
300
= 20 A.
15
[b] Use current division to find the current in the 8 Ω division:
ig =
15
(20) = 15 A.
20
Use current division again to find the current in the 30 Ω resistor:
i30 =
12
(15) = 6 A.
30
Thus,
p30 = (6)2 (30) = 1080 W.
Problems
P 3.33
3–25
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 3.34
Original meter:
Rs =
50 × 10−3
= 0.005 Ω;
10
Modified meter:
Re =
(0.015)(0.005)
= 0.00375 Ω;
0.02
.·. (Ifs )(0.00375) = 50 × 10−3 ;
.·. Ifs = 13.33 A.
P 3.35
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,
50 mV
= 50 Ω.
1 mA
Rv = 1000 (full-scale reading) − 50.
3–26
CHAPTER 3. Simple Resistive Circuits
[a] Rv = 1000(100) − 50 = 99, 950 Ω;
[b] Rv = 1000(5) − 50 = 4950 Ω;
[c] Rv = 1000(0.1) − 50 = 50 Ω;
60k5000
(5) = 3.73878 V.
20 + (60k5000)
60
(5) = 3.75 V;
[b] vtrue =
20 + 60
3.73878
% error =
− 1 × 100 = −0.299%.
3.75
P 3.36
[a] vmeas =
P 3.37
[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 =
100 mV
100
=
Ω.
4998 mA
4998
im =
1
(100/4998)
(imeas ) =
(imeas ).
50 + (100/4998)
2500
[c] Yes
P 3.38
The current in the shunt resistor at full-scale deflection is
iA = ifullscale − 5 × 10−3 A. The voltage across RA at full-scale deflection is
always 200 mV; therefore,
RA =
200 × 10−3
200
=
.
−3
ifullscale − 5 × 10
1000ifullscale − 5
200
= 20.01 mΩ.
10,000 − 5
[b] Let Rm be the equivalent ammeter resistance:
[a] RA =
0.2
= 0.02 = 20 mΩ.
10
200
[c] RA =
= 25.0156 mΩ.
8000 − 5
Rm =
Problems
[d] Rm =
P 3.39
0.2
= 25 mΩ.
8
Rmeter = Rm + Rmovement =
750 V
= 500 kΩ;
1.5 mA
vmeas = (25 kΩk125 kΩk500 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 3.40
[a]
40 × 103 i1 + 60 × 103 (i1 − iB ) = 10;
60 × 103 (i1 − iB ) = 0.6 + 50iB (120);
.·.
100i1 − 60iB = 10 × 10−3 ;
60i1 − 66iB = 0.6 × 10−3 .
Calculator solution yields iB = 180 µA.
[b] With the insertion of the ammeter the equations become
100i1 − 60iB = 10 × 10−3
(no change);
60 × 103 (i1 − iB ) = 103 iB + 0.6 + 50iB (120);
60i1 − 67iB = 0.6 × 10−3 ;
Calculator solution yields iB = 174.2 µA.
174.2
[c] % error =
− 1 100 = −3.22%.
180
3–27
3–28
P 3.41
CHAPTER 3. Simple Resistive Circuits
The measured value is
ig =
180
= 5.96 A;
(20.22 + 10)
The true value is
ig =
60k30.5 = 20.22 Ω.
imeas =
60
(5.96) = 3.95 A.
90.5
60k30 = 20 Ω.
180
= 6 A;
(20 + 10)
itrue =
60
(6) = 4 A;
90
3.95
− 1 × 100 = −1.28%.
%error =
4
P 3.42
Begin by using Ohm’s law to find the actual value of the current io :
itrue =
5
= 62.5 mA;
20 + 60
imeas =
5
= 62.112 mA;
20 + 60 + 0.5
62.112
% error =
− 1 100 = −0.621% ≈ −0.6%.
62.5
P 3.43
From the problem statement we have
0.080 =
vs (10 × 106 )
;
10 × 106 + Rs
0.072 =
vs (5 × 106 )
.
5 × 106 + Rs
In standard form,
10 × 106 vs − 0.08Rs = 0.08(10 × 106 );
5 × 106 vs − 0.072Rs = 0.072(5 × 106 ).
[a] Solving simultaneously on a calculator,
vs = 90 mV.
[b] Rs = 1250 kΩ.
Problems
P 3.44
3–29
[a] vmeter = 150V.
[b] Rmeter = (200)(500) = 100 kΩ;
100 kk60 k = 37.5 kΩ;
vmeter =
37.5 k
(150) = 83.33 V.
30 k + 37.5 k
[c] 30 kk100 k = 23.077 kΩ;
vmeter =
23.077 k
(150) = 41.67 V.
23.077 k + 60 k
[d] vmeter a = 150 V;
vmeter b + vmeter c = 125 V.
No, because of the loading effect.
P 3.45
[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.
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;
vm2 = (0.711)(180) = 128 V.
3–30
P 3.46
CHAPTER 3. Simple Resistive Circuits
The current in the series-connected voltmeters is
im =
205.2
136.8
=
= 0.76 mA.
270,000
180,000
v50 kΩ = (0.76 × 10−3 )(50,000) = 38 V;
Vpower supply = 205.2 + 136.8 + 38 = 380 V.
P 3.47
[a] Rmovement = 5 Ω;
R1 + Rmovement =
50
= 25 kΩ
2 × 10−3
R2 + R1 + Rmovement =
.·. R1 = 24,995 Ω.
100
= 50 kΩ;
2 × 10−3
R3 + R2 + R1 + Rmovement =
.·. R2 = 25 kΩ.
200
= 100 kΩ;
2 × 10−3
.·. R3 = 50 kΩ.
[b]
imove =
188
(2) = 1.88 mA;
200
v1 = (1.88)(50) = 94 V;
i1 =
94
= 0.188 mA;
500
i2 = imove + i1 = 1.88 + 0.188 = 2.068 mA;
vmeas = vx = 94 + 50i2 = 197.4 V.
[c] v1
=
100 V;
i2 = 2 + 0.20 = 2.20 mA;
i1
=
100/500 = 0.20 mA;
vmeas = vx = 100 + 50(2.20) = 210 V.
Problems
P 3.48
[a] R1
=
(50)/10−3 = 50 kΩ;
R2
=
(20)/10−3 = 20 kΩ;
R3
=
(2)/10−3 = 2 kΩ.
[b] Let ia
=
actual current in the movement;
id
=
design current in the movement;
ia
Then % error =
− 1 100.
id
For the 50 V scale:
50
50
ia =
=
,
50,000 + 50
50,050
ia
50,000
=
= 0.9990;
id
50,050
P 3.49
50
;
50,000
% error = (0.9990 − 1)100 = −0.10%.
For the 20 V scale:
ia
20,000
=
= 0.9975;
id
20,050
For the 2 V scale:
ia
2000
=
= 0.9756;
id
2050
id =
% error = (0.9975 − 1.0)100 = −0.25%.
% error = (0.9756 − 1.0)100 = −2.44%.
[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
3–31
3–32
P 3.50
CHAPTER 3. Simple Resistive Circuits
Since the bridge is balanced, we can remove the detector without disturbing
the voltages and currents in the circuit.
It follows that
i1 =
ig (R2 + Rx )
ig (R2 + Rx )
X
;
=
R1 + R2 + R3 + Rx
R
i2 =
ig (R1 + R3 )
ig (R1 + R3 )
X
;
=
R1 + R2 + R3 + Rx
R
v 3 = R 3 i1 = v x = i2 R x ;
R3 ig (R2 + Rx )
Rx ig (R1 + R3 )
X
X
.·.
=
;
R
R
.·. R3 (R2 + Rx ) = Rx (R1 + R3 ).
From which Rx =
P 3.51
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.
Problems
P 3.52
3–33
[a]
The condition for a balanced bridge is that the product of the opposite
resistors must be equal:
(1000)(750)
= 1500 Ω.
500
[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:
(500)(Rx ) = (1000)(750)
is =
so
Rx =
24V
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;
p750 = (750)(0.0192)2 = 276.18 mW;
p1000 = (1000)(0.0096)2 = 92.16 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.
3–34
P 3.53
CHAPTER 3. Simple Resistive Circuits
Redraw the circuit, replacing the detector branch with a short circuit.
15 kΩk3 kΩ = 2.5 kΩ;
9 kΩk45 kΩ = 7.5 kΩ;
is =
50
= 5 mA;
10
v1 = 5(2.5) = 12.5 V;
v2 = 5(7.5) = 37.5 V;
i1 =
12.5
= 833.3 µA;
15
i2 =
37.5
= 4166.7 µA;
9
id = i1 − i2 = −3333.4 µA.
P 3.54
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 3.55
Use the figure below to transform the Y to an equivalent ∆:
Ra =
(25)(100) + (25)(40) + (40)(100)
7500
=
= 300 Ω;
25
25
Problems
Rb =
(25)(100) + (25)(40) + (40)(100)
7500
=
= 187.5 Ω;
40
40
Rc =
(25)(100) + (25)(40) + (40)(100)
7500
=
= 75 Ω.
100
100
3–35
Replace the Y with its equivalent ∆ in the circuit to get the figure below:
Find the equivalent resistance to the right of the 5 Ω resistor:
300k[(125k187.5) + (37.5k75)] = 75 Ω.
The equivalent resistance seen by the source is thus 5 + 75 = 80 Ω. Use Ohm’s
law to find the current provided by the source:
is =
40
= 0.5 A.
80
Thus, the power associated with the source is
Ps = −(40)(0.5) = −20 W.
P 3.56
Use the figure below to transform the Y to an equivalent ∆:
Ra =
(25)(125) + (25)(37.5) + (37.5)(125)
8750
=
= 233.33 Ω;
37.5
37.5
Rb =
(25)(125) + (25)(37.5) + (37.5)(125)
8750
=
= 350 Ω;
25
25
3–36
CHAPTER 3. Simple Resistive Circuits
Rc =
(25)(125) + (25)(37.5) + (37.5)(125)
8750
=
= 70 Ω.
125
125
Replace the Y with its equivalent ∆ in the circuit to get the figure below:
Find the equivalent resistance to the right of the 5 Ω resistor:
350k[(100k233.33) + (40k70)] = 75 Ω.
The equivalent resistance seen by the source is thus 5 + 75 = 80 Ω. Use Ohm’s
law to find the current provided by the source:
is =
40
= 0.5 A.
80
Thus, the power associated with the source is
Ps = −(40)(0.5) = −20 W.
P 3.57
Use the figure below to transform the ∆ to an equivalent Y:
R1 =
(40)(25)
= 9.756 Ω;
40 + 25 + 37.5
R2 =
(25)(37.5)
= 9.1463 Ω;
40 + 25 + 37.5
R3 =
(40)(37.5)
= 14.634 Ω.
40 + 25 + 37.5
Problems
3–37
Replace the ∆ with its equivalent Y in the circuit to get the figure below:
Find the equivalent resistance to the right of the 5 Ω resistor:
(100 + 9.756)k(125 + 9.1463) + 14.634 = 75 Ω.
The equivalent resistance seen by the source is thus 5 + 75 = 80 Ω. Use Ohm’s
law to find the current provided by the source:
is =
40
= 0.5 A.
80
Thus, the power associated with the source is
Ps = −(40)(0.5) = −20 W.
P 3.58
Begin by transforming the ∆-connected resistors (5 Ω, 10 Ω, 105 Ω) to
Y-connected resistors. Both the Y-connected and ∆-connected resistors are
shown below to assist in using Eqs. 3.15 – 3.17:
Now use Eqs. 3.15 – 3.17 to calculate the values of the Y-connected resistors:
R1 =
(5)(10)
= 5/12 Ω;
5 + 10 + 105
R3 =
(10)(105)
= 8.75 Ω.
5 + 10 + 105
R2 =
(5)(105)
= 4.375 Ω;
5 + 10 + 105
3–38
CHAPTER 3. Simple Resistive Circuits
The transformed circuit is shown below:
The equivalent resistance seen by the 2 A source can be calculated by making
series and parallel combinations of the resistors to the right of the 2 A source:
Req = (28 + 8.75)k(20 + 5/12) + 4.375 = 36.75k(245/12) + 4.375
= 13.125 + 4.375 = 17.5 Ω.
Therefore, the voltage v across the 2 A source is given by
v = (17.5)(2) = 35 V.
P 3.59
[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
i=
96
(1000) = 240 mA;
400
io =
400
(240) = 96 mA.
1000
80
(240) = 48 mA,
400
[c] Now that io and i1 are known return to the original circuit
[b] i1 =
Problems
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 3.60
[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 =
15
(24 + 36)k(14 + 6)
(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 +
P 3.61
[a] Convert the upper delta to a wye.
R1 =
(80)(200)
= 40 Ω;
400
R2 =
(80)(120)
= 24 Ω;
400
3–39
3–40
CHAPTER 3. Simple Resistive Circuits
R3 =
(120)(200)
= 60 Ω.
400
Convert the lower delta to a wye.
R4 =
(60)(90)
= 18 Ω;
300
R5 =
(60)(150)
= 30 Ω;
300
R6 =
(90)(150)
= 45 Ω.
300
Now redraw the circuit using the wye equivalents.
Rab = 2 + 40 +
(80)(120)
+ 30 = 42 + 48 + 30 = 120 Ω.
200
[b] When vab = 600 V:
600
= 5 A;
ig =
120
(5)(80)
i15 =
= 2 A;
200
p15Ω = (4)(15) = 60 W.
P 3.62
[a] Calculate the values of the Y-connected resistors that are equivalent to the
10 Ω, 40 Ω, and 50 Ω ∆-connected resistors:
RX =
(10)(50)
= 5 Ω;
10 + 40 + 50
RZ =
(10)(40)
= 4 Ω.
10 + 40 + 50
RY =
(50)(40)
= 20 Ω;
10 + 40 + 50
Replacing the R2 —R3 —R4 delta with its equivalent Y gives
Problems
3–41
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
800
(10)(8) + (8)(40) + (10)(40)
=
= 80 Ω;
RY =
10
10
(10)(8) + (8)(40) + (10)(40)
800
RZ =
=
= 20 Ω
40
40
.
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 3.63
Replace the upper and lower deltas with the equivalent wyes:
R1U =
(10)(50)
(50)(40)
(10)(40)
= 5 Ω; R2U =
= 20 Ω; R3U =
= 4 Ω;
100
100
100
3–42
CHAPTER 3. Simple Resistive Circuits
R1L =
(10)(60)
(60)(30)
(10)(30)
= 6 Ω; R2L =
= 18 Ω; R3L =
= 3 Ω.
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.
P 3.64
Subtracting Eq. 3.13 from Eq. 3.14 gives
R1 − R2 = (Rc Rb − Rc Ra )/(Ra + Rb + Rc ).
Adding this expression to Eq. 3.12 and solving for R1 gives
R1 = Rc Rb /(Ra + Rb + Rc ).
To find R2 , subtract Eq. 3.14 from Eq. 3.12 and add this result to Eq. 3.13.
To find R3 , subtract Eq. 3.12 from Eq. 3.13 and add this result to Eq. 3.14.
Using the hint, Eq. 3.14 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 + R2 R3 + R3 R1 )/R2 . To find Ra : First use
Eqs. 3.15–3.17 to obtain the ratios (R1 /R3 ) = (Rc /Ra ) or 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. 3.13. To find Rc , use Eqs. 3.15–3.17 to obtain
the ratios Rb = (R3 /R2 )Rc and Ra = (R3 /R1 )Rc . Now use the relationships to
eliminate Rb and Ra from Eq. 3.12.
P 3.65
1
R1
=
Ra
R1 R2 + R2 R3 + R3 R1
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
=
Problems
P 3.66
[a]
va =
vin R4
;
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,
R3
R4
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
=
[b] ∆R = 0.02Ro ;
Ro =
(750)(1000)
R2 R4
=
= 3000 Ω;
R3
250
∆R = (0.02)(3000) = 60 Ω;
−60(1000)(10)
.·. vo ≈
= −37.5 mV.
(3000 + 1000)2
[c] vo
=
=
−(∆R)R4 vin
(Ro + R4 + ∆R)(Ro + R4 )
−60(1000)(10)
(3000 + 1000 + 60)(3000 + 1000)
= −36.946 mV.
3–43
3–44
P 3.67
CHAPTER 3. Simple Resistive Circuits
[a] approx value =
true value =
−(∆R)R4 vin
;
(Ro + R4 )2
−(∆R)R4 vin
;
(Ro + R4 + ∆R)(Ro + R4 )
approx value
(Ro + R4 + ∆R)
.·.
=
;
true value
(Ro + R4 )
.·. % error =
−∆R
Ro + R4
− 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 3.68
−R3 ∆R
.
R4 (R2 + R3 )
−(250)(60)
× 100 = −1.5%.
(1000)(1000)
∆R(R3 )(100)
= 1;
(R2 + R3 )R4
∆R(250)(100)
= 1;
(1000)(1000)
.·. ∆R = 40 Ω;
% change =
P 3.69
40
× 100 = 1.33%.
3000
[a] Rab = 2R1 +
Therefore
Thus
R2 (2R1 + RL )
= RL ;
2R1 + R2 + RL
2R1 − RL +
R2 (2R1 + RL )
= 0;
2R1 + R2 + RL
RL2 = 4R12 + 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
Problems
Therefore
and
vo =
3–45
R2
vi
·
RL
RL 2R1 + R2 + RL
vo
R2
=
.
vi
2R1 + R2 + RL
[b] (600)2 = 4(R1 + R2 )R1 ;
9 × 104 = R12 + R1 R2 ;
R2
vo
= 0.6 =
;
vi
2R1 + R2 + 600
.·. 1.2R1 + 0.6R2 + 360 = R2 ;
0.4R2 = 1.2R1 + 360;
R2 = 3R1 + 900;
.·. 9 × 104 = R12 + R1 (3R1 + 900) = 4R12 + 900R1 ;
.·. R12 + 225R1 − 22,500 = 0.
R1 = −112.5 ±
q
(112.5)2 + 22,500 = −112.5 ± 187.5;
.·. R1 = 75 Ω;
.·. R2 = 3(75) + 900 = 1125 Ω.
[c] From Appendix H, choose R1 = 68 Ω and R2 = 1.2 kΩ. For these values,
Rab = RL =
% error =
q
(4)(68)(68 + 1200) = 587.3 Ω;
587.3
− 1 100 = −2.1%.
600
vo
1200
=
= 0.624;
vi
2(68) + 1200 + 587.3
0.624
% error =
− 1 100 = 4%.
0.6
P 3.70
[a] After making the Y-to-∆ transformation, the circuit reduces to
3–46
CHAPTER 3. Simple Resistive Circuits
Combining the parallel resistors reduces the circuit to
3RRL
2.25R2 + 3.75RRL
=
3R + RL
3R + RL .
Now note:
0.75R +
Therefore
2.25R2 + 3.75RRL
3R
3R + RL
3R(3R + 5RL )
! =
.
Rab =
2
15R + 9RL
2.25R + 3.75RRL
3R +
3R + RL
!
If R = RL , we have
Therefore
Rab =
3RL (8RL )
= RL .
24RL
Rab = RL .
[b] When R = RL , the circuit reduces to
io =
ii (3RL )
1
1 vi
=
ii =
,
4.5RL
1.5
1.5 RL
Therefore
P 3.71
vo
= 0.5.
vi
[a] 3(3R − RL ) = 3R + RL ;
9R − 1800 = 3R + 600;
6R = 2400,
R = 400 Ω;
2(400)(600)2
R2 =
= 2400 Ω.
3(400)2 − (600)2
1
vo = 0.75RL io = vi .
2
Problems
[b]
vo =
vi
180
=
= 60 V;
3
3
io =
60
= 100 mA;
600
i1 =
180 − 60
120
=
= 50 mA;
2400
2400
ig =
180
= 300 mA;
600
i2 = 300 − 50 = 250 mA;
i3 = 100 − 50 = 50 mA;
i4 = 250 − 50 = 200 mA;
p2400 top = (50 × 10−3 )2 (2400) = 6 W;
p400 left = (250 × 10−3 )2 (400) = 25 W;
p400 right = (50 × 10−3 )2 (400) = 1 W;
p400 vertical = (200 × 10−3 )2 (400) = 16 W;
p600 load = (100 × 10−3 )2 (600) = 6 W;
The 400 Ω resistor carrying i2 .
[c] p400 left = 25 W.
[d] The 400 Ω resistor carrying i3 .
[e] p400 right = 1 W.
P 3.72
[a] Using the equation for voltage division,
Vy =
βRy
βRy
VS =
VS = βVS .
βRy + (1 − β)Ry
Ry
3–47
3–48
CHAPTER 3. Simple Resistive Circuits
[b] Since β represents the touch point with respect to the bottom of the
screen, (1 − β) represents the location of the touch point with respect to
the top of the screen. Therefore, the y-coordinate of the pixel
corresponding to the touch point is
y = (1 − β)py .
Remember that the value of y is capped at (py − 1).
P 3.73
[a] Use the equations developed in the Practical Perspective and in
Problem 3.72:
Vx
4.5
=
Vx = αVS
so
α=
= 0.75;
VS
6
Vy = βVS
so
β=
Vy
1.5
=
= 0.25.
VS
6
[b] Use the equations developed in the Practical Perspective and in
Problem 3.72:
x = (1 − α)px = (1 − 0.75)(720) = 180;
y = (1 − β)py = (1 − 0.25)(1260) = 945.
Therefore, the touch occurred in the lower left corner of the screen.
P 3.74
Use the equations developed in the Practical Perspective and in Problem 3.72:
x = (1 − α)px
so
α=1−
108
x
=1−
= 0.8;
px
540
Vx = αVS = (0.8)(4) = 3.2 V;
y = (1 − β)py
so
β =1−
y
602
=1−
= 0.3;
py
860
Vy = βVS = (0.3)(4) = 1.2 V.
P 3.75
From the results of Problem 3.74, the voltages corresponding to the touch
point (480, 192) are
Vx1 = 3.2 V;
Vy1 = 1.2 V.
Now calculate the voltages corresponding to the touch point (270, 430):
x = (1 − α)px
so
α=1−
270
x
=1−
= 0.5;
px
540
Problems
3–49
Vx2 = αVS = (0.5)(4) = 2V;
y = (1 − β)py
so
β =1−
430
y
=1−
= 0.5;
py
860
Vy2 = βVS = (0.5)(4) = 2V.
When the screen is touched at two points simultaneously, only the smaller of
the two voltages in the x direction is sensed. The same is true in the y
direction. Therefore, the voltages actually sensed are
Vx = 2V;
Vy = 1.2V.
These two voltages identify the touch point as (270, 602), which does not
correspond to either of the points actually touched! Therefore, the resistive
touch screen is appropriate only for single point touches.
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 − v2
−5 +
+
= 0;
5
120
v2
v2 − v1
0.04 +
+
= 0.
25
120
Place
instandard
these equations
form:
1
1
1
v1
+
+ v2 −
=
5;
5 120
120
1
1
1
+
v1 −
+ v2
= −0.04.
120
25 120
Solving, v1 = 24.133 V and v2 = 3.33 V;
[b] p5A = −(5 A)v1 = −(5 A)(24.133 V) = −120.67 W = 120.67 W(delivered).
[c] p40mA = (0.04 A)v2 = (0.04 A)(3.33 V) = 0.133 W
= 133.33 mW(absorbed)= −133.33 mW(delivered).
4–1
4–2
CHAPTER 4. Techniques of Circuit Analysis
AP 4.2
v1 − 144 v1
v1 − v2
+
+
=0
4
10
80
v2 − v1 v2
+
=0
−3 +
80
5
Solving, v1 = 100 V;
so
29v1 − v2 = 2880;
so
−v1 + 17v2 = 240.
v2 = 20 V.
AP 4.3 [a]
The node voltage equation is:
vo − 6.25i∆ vo − 45
vo
+
+
= 0.
100
5
25
The dependent source constraint equation is:
−0.45 +
45 − vo
.
25
Place these equations in standard form:
1
1
6.25
45
1
vo
+ +
+ i∆ −
=
+ 0.45;
100 5 25
5
25
1
45
vo
+ i∆ (1)
=
.
25
25
Solving,
vo = 15 V;
i∆ = 1.2 A.
vo − 6.25i∆
15 − 7.5
ids =
=
= 1.5 A;
5
5
pds = [6.25(1.2)](1.5) = 11.25 W.
Thus, the dependent source absorbs 11.25 W.
p450mA = −(0.45)(15) = −6.75 W;
pX
45V = −(1.2)(45) = −54 W;
pdev = 6.75 + 54 = 60.75 W ;
Thus the independent sources develop 60.75 W.
i∆ =
Problems
Also,
X
pdis = pds + p100Ω + p5Ω + p25Ω
= 11.25 + (15)2 /100 + (1.5)2 (5) + (1.2)2 (25)
= 11.25 + 2.25 + 11.25 + 36 = 60.75 W (checks!).
[b] The independent sources are delivering power to the circuit, while the
dependent source is extracting power from the circuit.
AP 4.4
The two node voltage equations are:
vb vb − vc
7+
+
= 0;
3
1
vc − vb vc − 4
−2vx +
+
= 0.
1
2
The constraint equation for the dependent source is:
vx = vc − 4.
Place these equations in standard form:
1
+1
+ vc (−1)
+ vx (0)
= −7;
vb
3
1
4
vb (−1)
+ vc 1 +
+ vx (−2) =
;
2
2
vb (0)
+ vc (1)
+ vx (−1) =
4.
Solving, vc = 9 V, vx = 5 V, and vo = vb = 1.5 V.
AP 4.5 Place 4v∆ inside a supernode and use the lower node as a reference. The
resulting circuit is
4–3
4–4
CHAPTER 4. Techniques of Circuit Analysis
The supernode KCL equation is
v2
v2
v1 − 100 v1
+
+
+
= 0.
10
60 20 30
The dependent source constraint equation is
v∆ = v2 .
The supernode constraint equation is
v1 − v2 = 4v∆ = 4v2 .
Solving,
v1 = 75 V;
v2 = 15 V.
vo = 100 − v1 = 25 V.
AP 4.6 Redraw the circuit identifying the reference node and the two unknown node
voltages. Note that the left-most node voltage is 80 V.
The node voltages v1 and v2 , together with the dependent voltage source, form
a supernode. The supernode equation is
v1 − 80
v1
v2 − 80 v2
+
+
+
= 0.
6
12.5
15
4
The supernode constraint equation is
v1 − v2 = 7.5iφ .
The constraint equation due to the dependent source is
iφ =
v1
.
12.5
Place these three equations in standard form:
Problems
v1
1
1
+
6 12.5
v1 (1)
v1
1
12.5
+ v2
1
1
+
15 4
+ iφ (0)
=
+ v2 (−1)
+ iφ (−7.5) =
+ v2 (0)
+ iφ (−1)
18.67;
0;
Solving, v1 = 50 V and v2 = 20 V and
Therefore, v = 80 − v2 = 80 − 20 = 60 V.
= 0;
iφ = 4 A.
AP 4.7 [a]
The three mesh current equations are:
−44 + 4i1 + 6(i1 − i2 )
= 0;
6(i2 − i1 ) + i2 + 3(i2 − i3 )
= 0;
−2 + 3(i3 − i2 ) + 2i3
=
0.
Place these equations in standard form:
i1 (10)
+ i2 (−6) + i3 (0)
i1 (−6) + i2 (10)
i1 (0)
= 44;
+ i3 (−3) =
+ i2 (−3) + i3 (5)
0;
= 2.
Solving, i1 = 8 A;
i2 = 6 A;
i3 = 4 A.
Therefore p44V = −(44)i1 = −(44)(8) = −352 W, so the 44 V source
delivers 352 W to the circuit.
[b] p3Ω = 3(i2 − i3 )2 = 3(2)2 = 12 W.
Thus, the 3 Ω resistor dissipates 12 W of power.
AP 4.8 [a] b = 8,
n = 6,
b − n + 1 = 3.
[b] Redraw the circuit identifying the three mesh currents:
4–5
4–6
CHAPTER 4. Techniques of Circuit Analysis
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 delivering 46,640 W.
CHECK:
p30V = −30i2 = −1560 W (left source);
p30V = −30i3 = −1800 W (right source);
X
pdev = 46,640 + 1560 + 1800 = 50 kW;
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.
Problems
4–7
AP 4.9 Redraw the circuit identifying the three mesh currents:
The mesh current equations are:
−80 + 6(ia − ib ) + 12.5(ia − ic )
= 0;
15ib − 7.5iφ + 6(ib − ia )
= 0;
7.5iφ + 4ic + 12.5(ic − ia )
= 0.
The dependent source constraint equation is iφ = ia − ic . We can substitute
this simple expression for iφ into the second and third mesh equations and
place the equations in standard form:
18.5ia − 6ib − 12.5ic
=
80;
−13.5ia + 21ib + 7.5ic
=
0;
−5ia + 0ib + 9ic
=
0.
Solving, ia = 9 A;
ib = 4 A;
ic = 5 A.
Thus, vo = 15ib = 15(4) = 60 V.
AP 4.10 Redraw the circuit identifying the three mesh currents:
The mesh current equations are:
10, 000i1 + 4000(i1 − i2 ) + 1500(i1 − 0.025)
= 0;
−20 + 3000i2 + 2000(i2 − 0.025) + 4000(i2 − i1 )
= 0;
Now place the equations in standard form:
15, 500i1 − 4000i2
=
37.5;
−4000i1 + 9000i2
=
20 + 50;
4–8
CHAPTER 4. Techniques of Circuit Analysis
Solving, i1 = 5 mA;
i2 = 10 mA.
Thus, p10k = 10, 000i21 = 10, 000(0.005)2 = 0.25 W.
AP 4.11 Redraw the circuit and identify the mesh currents:
There are current sources on the perimeters of two meshes, so we know the
currents in those meshes, as seen in the figure. The remaining mesh current
equation is
40 + 500(ia − 0.3) + 25(ia + vφ /50) = 0.
The dependent source requires the following constraint equation:
vφ = 25(ia + vφ /50).
Place the mesh current equation and the dependent source equation is
standard form:
525ia + 0.5vφ
=
110;
25ia − 0.5vφ
=
0.
Solving:
ia = 0.2 A;
vφ = 10 V.
Thus, ia = 0.2 A.
AP 4.12 Redraw the circuit and identify the mesh currents:
The 150 mA current source is shared by the meshes i2 and i3 . Thus we
combine these meshes to form a supermesh and write the following equation:
50i2 + 32i3 + 20i2 + 100(i2 − i1 ) = 0.
Problems
4–9
The other mesh current equation is
−30 + 60i1 + 100(i1 − i2 ) = 0.
The supermesh constraint equation is
i3 − i2 = 0.15.
Place these three equations in standard form:
160i1 − 100i2 + 0i3
=
30;
−100i1 + 170i2 + 32i3
=
0;
0i1 − 1i2 + 1i3
=
0.15.
Solving,
i1 = 0.25 A;
i2 = 0.1 A;
i3 = 0.25 A.
Thus,
p100 Ω = 100(i1 − i2 )2 = 100(0.15)2 = 2.25 W.
AP 4.13 To find the power delivered by the voltage source, calculate the current in
that source using the mesh current method. Redraw the circuit and identify
the mesh currents:
The 7.5 A current source is shared by the two meshes, so combine them to
create a supermesh whose KVL equation is
25 + 30i1 + 5i2 + 100 = 0.
The supermesh constraint equation is
i1 − i2 = 7.5.
In standard form,
30i1 + 5i2 = −125;
1i1 − 1i2 = 7.5.
Solving,
i1 = −2.5 A;
i2 = −10 A.
p100V = 100i2 = (100)(−10) = −1000 = −1 kW.
The 100 V source delivers 1 kW of power to the circuit.
4–10
CHAPTER 4. Techniques of Circuit Analysis
AP 4.14 To find the power delivered by the voltage-controlled current source, calculate
the voltage drop across that source. Redraw the circuit, label the reference
node and identify the node voltages:
The three KCL equations are
v1 − v2
+ 30 = 0;
−v∆ +
2.5
v2 − v1 v2 v2 − v3
+
+
= 0;
2.5
7
2
v 3 − v 2 v 3 + ix
+
− 30 = 0;
2
1.5
The constraint equations due to the dependent sources are
v∆ = v1 − v2 ;
ix =
v2
.
7
Place the five equations in standard form:
v1 (1/2.5) + v2 (−1/2.5) + v3 (0) + v∆ (−1) + ix (0)
v1 (−1/2.5) + v2 [(1/2.5) + (1/7) + (1/2)] + v3 (−1/2) + v∆ (0) + ix (0)
= −30;
=
0;
v1 (0) + v2 (−1/2) + v3 [(1/2) + (1/1.5)] + v∆ (0) + ix (1/1.5) =
30;
v1 (1) + v2 (−1) + v3 (0) + v∆ (−1) + ix (0)
=
0;
v1 (0) + v2 (1/7) + v3 (0) + v∆ (0) + ix (−1) = −30;
Solving,
v1 = 120 V;
v2 = 70 V;
v3 = 50 V;
v∆ = 50 V;
pVCCS = −v1 v∆ = −(120)(50) = −6000 = −6 kW.
Thus, the voltage-controlled current source delivers 6 kW.
ix = 10 A;
Problems
4–11
AP 4.15 [a] The circuit:
Source-transform the branch with the 7.5 V source and the branch with
the 15 V source. The resulting circuit, shown to the left of the 160 Ω
resistor is
Combine the three parallel current sources into a single current source
with the value 10 m − 15 m + 25 m = 20 mA. Combine the three parallel
resistors into a single equivalent resistor with the value
2000k500k600 = 240 Ω. The simplified circuit is
Source-transform the parallel 20 mA current source and 240 Ω resistor to
get
Combine the series-connected resistors to give 240 + 160 = 400 Ω. Then
source-transform the 4.8 V source and the 400 Ω resistor to give
Combine the two parallel 400 Ω resistors into an equivalent 200 Ω resistor.
Then source-transform the 12 mA source and parallel 200 Ω resistor to
give the equivalent circuit to the left of the voltage v:
4–12
CHAPTER 4. Techniques of Circuit Analysis
Use voltage division to find v:
v=
1000
(2.4) = 2 V.
1000 + 200
[b] Return to the original circuit. Calculate the current in the 160 Ω resistor
from left to right using KCL at the node to the right of the 160 Ω resistor:
v
2
2
v
+
=
+
= 0.007 = 7 mA.
400 1000
400 1000
Use Ohm’s law to find the voltage across the 160 Ω resistor from left to
right:
i160 =
v160 = 160i160 = (160)(0.007) = 1.12 V.
Write a KVL equation around the loop that includes the 10 mA source,
the 160 Ω resistor, and the 1 kΩ resistor to find the voltage across the
10 mA source:
v10mA = v160 + v = 1.12 + 2 = 3.12 V.
The power delivered by the 10 mA source is thus,
p10mA = −(0.01)v10mA = −(0.01)(3.12) = −31.2 mW.
The 10 mA source delivers 31.2 mW to the circuit.
AP 4.16 Find the open-circuit voltage:
−0.075 +
v1
v1
+
= 0;
4000 5000
v1 = 166.67 V;
so
voc =
3000
v1 = 100 V.
5000
Problems
4–13
Find the short-circuit current:
isc =
4000k2000
(0.075) = 50 mA.
2000
Thus,
IN = isc = 50 mA;
RN =
100
voc
=
= 2 kΩ.
isc
0.05
AP 4.17 Find the open-circuit voltage.
−12 + 12(i1 − 8) + 6i1 = 0;
so
vTh = 2(8) + 6i1 = 16 + 36 = 52V.
Find the short-circuit current.
18i1 = 12 + 96;
thus
i1 = 6A.
4–14
CHAPTER 4. Techniques of Circuit Analysis
−12 + 12(i1 − 8) + 6(i1 − isc ) = 0;
6(isc − i1 ) + 2(isc − 8) = 0.
Solving,
isc = 8.67 A
so
RTh =
52
= 6 Ω.
8.67
The Thévenin equivalent is
AP 4.18 Find the Norton equivalent to the left of the 15 kΩ resistor 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.
The Norton equivalent is thus a 15 mA source in parallel with a 10 kΩ
resistor, as seen to the left of the 15 kΩ resistor in the circuit below.
Problems
4–15
Now attach the ammeter, modeled as a 50 Ω resistor, to the Norton equivalent
and use current division to calculate the meter reading io :
vo =
10,000k15,050
(0.15) = 5.988 mA.
15,050
AP 4.19
The node voltage equations and dependant source equation are:
v1
v1 − v2
v1 − 280
+
+
+ 0.2i∆ = 0;
2000
2000
2000
v2 − v1
v2
+
− 0.2i∆
= 0;
2000
5600
280 − v1
i∆
=
.
2000
In standard form:
1
1
1
1
280
+
+
+ v2 −
+ i∆ (0.2) =
;
v1
2000 2000 2000
2000
2000
1
1
1
v1 −
+ v2
+
+ i∆ (−0.2) = 0;
2000
2000 5600
280
1
.
+ v2 (0) + i∆ (1) =
v1
2000
2000
Solving,
v1 = 120 V; v2 = 112 V; i∆ = 0.08 A.
VTh = v2 = 112 V.
4–16
CHAPTER 4. Techniques of Circuit Analysis
The mesh current equations are:
−280 + 2000i∆ + 2000(i∆ − isc )
=
2000(isc − 0.2i∆ ) + 2000(isc − i∆ )
= 0.
Put these equations in standard form:
i∆ (4000) + isc (−2000)
=
280;
i∆ (−2400) + isc (4000)
= 0.
Solving, i∆ = 0.1 A;
isc = 0.06 A;
112
RTh =
= 1866.67 Ω.
0.06
AP 4.20
OPEN CIRCUIT
100 = 2500i1 + 625(i1 + 10−3 v2 );
v2 =
6000
(5000i1 ).
10,000
Solving,
i1 = 0.02 A;
v2 = voc = 60 V.
0;
Problems
SHORT CIRCUIT
v2 = 0;
i1 =
isc =
5000
i1 ;
4000
100
= 0.032 A.
2500 + 625
Thus,
5
isc = i1 = 0.04 A;
4
RTh =
60
= 1.5 kΩ.
0.04
AP 4.21 [a]
RTh = 5000k(1600 + 2400k4800 + 1800) = 2.5 kΩ;
R = RTh = 2.5 kΩ.
[b]
7200i1 − 4800i2 = 60;
−4800i1 + 4800i2 + 8400i3 = 0;
4–17
4–18
CHAPTER 4. Techniques of Circuit Analysis
i2 − i3 = 0.015.
Solving,
i1 = 19.4 mA;
i2 = 16.6 mA;
i3 = 1.6 mA;
vTh = 5000i3 = 8 V.
pmax = (1.6 × 10−3 )2 (2500) = 6.4 mW.
AP 4.22 [a] Write KVL equations for the left mesh and the supermesh, place them in
standard form, and solve:
At i1 :
− 60 + 2400i1 + 4800(i1 − i2 ) = 0;
Supermesh:
4800(i2 − i1 ) + 1600i3 + (5000k2500)i3 + 1800i3 = 0;
Constraint:
i2 − i3 = 0.015 = 0;
Standard form:
i1 (7200) + i2 (−4800) + i3 (0) = 60;
i1 (−4800) + i2 (4800) + i3 (5066.67) = 0;
i1 (0) + i2 (1) + i3 (−1) = 0.015.
Calculator solution:
i1 = 19.933 mA;
i2 = 17.4 mA;
i3 = 2.4 mA.
Calculate the power delivered by the voltage source:
p60V = −60i1 = −60(0.019933) = −1.196 W (del);
Problems
4–19
[b] Calculate voltage across the current source and use it to calculate the
power of the current source:
v15mA = 4800(i1 − i2 ) = 12.16 V;
p15mA = (0.015)(12.16) = 182.4 mW (abs).
[c] Calculate the power absorbed by the 2.5 kΩ resistor and the percentage
power:
i2.5k =
5000k2500
i3 = 1.6 mA;
2500
p2.5k = (0.0016)2 (2500) = 6.4 mW;
% delivered to R:
0.0064
(100) = 0.535%.
1.196
4–20
CHAPTER 4. Techniques of Circuit Analysis
Problems
P 4.1
[a] 11 branches, 7 branches with resistors, 2 branches with independent
sources, 2 branches with dependent sources.
[b] The current is unknown in every branch except the one containing the 5
mA current source, so the current is unknown in 10 branches.
[c] 11 essential branches each containing a single element.
[d] The current is known only in the essential branch containing the current
source, and is unknown in the remaining 10 essential branches.
[e] From the figure there are 5 nodes – four identified by rectangular boxes
and one identified by a triangle.
[f ] There are 5 essential nodes, four 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
7 window panes or meshes.
P 4.2 [a] From Problem 4.1(d) there are 10 essential branches where the current is
unknown, so we need 10 simultaneous equations to describe the circuit.
[b] From Problem 4.1(f), there are 5 essential nodes, so we can apply KCL at
(5 − 1) = 4 of these essential nodes. There would also be two dependent
source constraint equations.
[c] The remaining 4 equations needed to describe the circuit will be derived
from KVL equations.
[d] We must avoid using the meshes containing current sources, as we have no
way of determining the voltage drop across a current source.
P 4.3 [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.
Problems
4–21
[b] There are four essential nodes in this circuit, identified by the dark black
dots in Fig. P4.3. 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.
[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 + R3 i3 − R2 i2 = 0;
R3 i3 + R5 i5 − R4 i4 = 0.
P 4.4 [a] At node g,
i4 + i6 − i5 − I = 0.
[b] From Example 4.2, the equations at nodes b, c, and e are
At node b:
− i1 + i2 + i6 − I = 0;
At node c:
i1 − i3 − i5 = 0;
4–22
CHAPTER 4. Techniques of Circuit Analysis
At node e:
i3 + i4 − i2 = 0.
Add the equations at nodes b and c to give
i2 + i6 − I − i3 − i5 = 0.
From the equation at node e,
i2 − i3 = i4 .
Combine the two equations above and rearrange to give the equation at
node g.
P 4.5
[a] At node 1:
− ig + i1 + i2 = 0;
At node 2:
− i2 + i3 + i4 = 0;
At node 3:
ig − i1 − i3 − i4 = 0.
[b] There are many possible solutions. For example, solve the equation at
node 1 for ig :
ig = i1 + i2 .
Substitute this expression for ig into the equation at node 3:
(i1 + i2 ) − i1 − i3 − i4 = 0
so
i2 − i3 − i4 = 0.
Multiply this last equation by −1 to get the equation at node 2:
−(i2 − i3 − i4 ) = −0
so
P 4.6
2+
vo vo − 55
+
= 0;
4
5
vo = 20 V;
p2A = (20)(2) = 40 W (absorbing).
− i2 + i3 + i4 = 0.
Problems
4–23
P 4.7 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 − v1 v2 − 30
v2
+
+
= 0.
12
6+2
4
Place
form:
theseequationsin standard
1
1
v1 1 +
+ v2 −
= 4.5;
8
8
1
1
1 1
+ +
+ v2
= 7.5.
v1 −
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.
P 4.8
Note that we have chosen the lower node as the reference node, and that the
voltage at the upper node with respect to the reference node is vo . Write a
KCL equation (node voltage equation) by summing the currents leaving the
upper node:
vo − 180 vo
+
+ 5 = 0.
12 + 5
2
Solve by multiplying both sides of the KCL equation by 17 and collecting the
terms involving vo on one side of the equation and the constants on the other
side of the equation:
vo − 180 + 8.5vo + 85 = 0
.·.
9.5vo = 95
so
vo = 95/9.5 = 10 V.
4–24
CHAPTER 4. Techniques of Circuit Analysis
P 4.9 [a] From the solution to Problem 4.8 we know vo = 10 V; therefore
p5A = (10)(5) = 50 W.
The power developed by the 5 A source is −50 W; that is, the 5 mA
source is absorbing 50 W.
[b] The current into the positive terminal of the 180 V source in the figure of
Problem 4.8 is
ig = (vo − 180)/(12 + 5) = (10 − 180)/17 = −170/17 = −10 A.
The power in the 180 V source is
p180V = (180)(−10) = −1800 W.
The power developed by the 180 V source is 1800 W.
[c] p12Ω
=
12i2g = 12(−10)2 = 1200 W;
p5Ω
=
5i2g = 5(−10)2 = 500 W;
p2Ω
= vo2 /2 = 102 /2 = 50 W.
X
X
pdis = 50 + 1200 + 500 + 50 = 1800 W;
pdev = 1800 W. (checks!)
P 4.10
[a] The node voltage equation is:
vo − 180 vo
+
+ 5 = 0.
17
2
Solving,
vo − 180 + 8.5vo + 85 = 0
.·.
9.5vo = 95
so
[b] Let vx = voltage drop across 5 A source:
vx = vo − (5)(10) = 10 − 50 = −40 V;
p5A = (5)(−40) = −200 W.
The power developed by the 5 A source is 200 W.
[c] Let ig = current into positive terminal of 180 V source:
ig = (vo − 180)/17 = (10 − 180)/17 = −10 A;
p180V = (180)(−10) = −1800 W.
The power developed by the 180 V source is 1800 W.
[d] p12Ω
=
(−10)2 (12) = 1200 W;
p5Ω
=
(−10)2 (5) = 500 W;
p2Ω
=
(10)2 /2 = 50 W;
p10Ω
=
(5)2 (10) = 250 W.
vo = 10 V.
Problems
X
X
4–25
pdis = 1200 + 500 + 50 + 250 = 2000 W;
pdev = 200 + 1800 = 2000 W. (checks!)
[e] vo is independent of any finite resistance connected in series with the 5 A
current source.
P 4.11 [a]
The three node voltage equations are:
v1 − 80 v1 − v2 v1 − v3
+
+
= 0;
800
500
1000
v2
v2 − v3
v2 − v1
+
+
= 0;
500
3000
2000
v3 − v1 v3 − v2
v3
+
+
= 0.
1000
2000
250
Place these equations in standard form:
1
1
80
1
1
1
+ v2 −
+ v3 −
=
+
+
;
800 500 1000
500
1000
800
1
1
1
1
1
+
+
+ v2
+ v3 −
= 0;
500
500 3000 2000
2000
1
1
1
1
1
+ v2 −
+ v3
+
+
= 0.
1000
2000
1000 2000 250
v1
v1 −
v1 −
Solving, v1 = 40 V;
v2 = 30 V;
v3 = 10 V.
Now calculate the branch currents from the node voltage values:
ia =
40 − 10
= 30 mA;
1000
ib =
80 − 40
= 50 mA;
800
ic =
40 − 30
= 20 mA;
500
id =
30 − 10
= 10 mA;
2000
ie =
30
= 10 mA;
3000
if =
10
= 40 mA.
250
[b] p80V = −80ib = −80(0.05) = −4 W.
Thus, the power developed in the circuit is 4 W. Note that the resistors
cannot develop power!
4–26
CHAPTER 4. Techniques of Circuit Analysis
P 4.12 [a]
v1
v1 − 640 v1 − v2
+
+
=0
so 31v1 − 20v2 + 0v3 = 6400;
50
5
2.5
v2 − v1 v2 − v3
+
+ 12.8 = 0
so −2v1 + 3v2 − v3 = −64;
2.5
5
v3
v3 − v2
+
− 12.8 = 0
so 0v1 − v2 + 3v3 = 64.
2.5
5
Solving, v1 = 380 V; v2 = 269 V; v3 = 111 V.
640 − 380
= 52 A;
5
[b] ig =
pg = (640)(52) = 33,280 W (del).
P 4.13 2.4 +
v1 − v2
v1
+
= 0;
125
25
v2 − v1
v2
v2
+
+
− 3.2 = 0.
25
250 375
Solving, v1 = 25 V;
v2 = 90 V.
CHECK:
p125Ω =
(25)2
= 5 W;
125
p250Ω =
(90)2
= 32.4 W;
250
p25Ω =
(90 − 25)2
= 169 W;
25
p375Ω =
(90)2
= 21.6 W;
375
p2.4A = (25)(2.4) = 60 W.
X
pabs = 5 + 169 + 32.4 + 21.6 + 60 = 288 W.
X
pdev = (90)(3.2) = 288 W (CHECKS).
Problems
4–27
P 4.14
v1 − 30
v1
v1 − v2
+
+
− 4 = 0;
15
31.25
25
v2 − v3
v2 − v1
+4+
− 1 = 0;
25
50
v3 − v2
v3
+
+ 1 = 0.
50
50
Solving, v1 = 30 V;
v2 = −46 V;
v3 = −48 V.
p4A = 4(v2 − v1 ) = 4(−76) = −304 W (del);
p1A = 1(v3 − v2 ) = −2 W (del);
p30V = (30)
v1 − 30
= 0 W;
15
Thus, the total power delivered in the circuit, which equals the total power
dissipated in the circuit, is 304 + 2 = 306 W.
Check:
(30 − 30)2
p15Ω =
= 0 W;
15
p25Ω =
(30 + 46)2
= 231.04 W;
25
p31.25Ω =
302
= 28.8 W;
31.25
p50Ω (lower) =
(−48)2
= 46.08 W;
50
p50Ω (right) =
(−46 + 48)2
= 0.08 W.
50
X
pdiss = 0 + 231.04 + 28.8 + 46.08 + 0.08 = 306 W
(CHECKS).
4–28
CHAPTER 4. Techniques of Circuit Analysis
P 4.15 [a]
v1 − 120 v1 − v2 v1 − v3
+
+
0.5
20
50
v2 − v1 v2 v2 − v3
+
+
20
1
10
v3 + 120 v3 − v2 v3 − v1
+
+
0.5
10
50
In standard form:
=
0;
=
0;
=
0.
1
1
1
1
1
v1
+
+
+ v2 −
+ v3 −
0.5 20 50
20
50
1
1
1
1
1
v1 −
+ v2
+ +
+ v3 −
20
20 1 10
10
1
1
1
1
1
v1 −
+ v2 −
+ v3
+
+
50
10
0.5 10 50
Solving, v1 = 114.74 V;
v2 = −4.78 V;
120 − v1
= 10.52 A
0.5
v2
i2 =
= −4.78 A
1
v3 + 120
i3 =
= 15.3 A
0.5
Thus, i1 =
[b]
X
=
240;
=
0;
= −240.
v3 = −112.35 V.
v1 − v2
= 5.976 A;
20
v2 − v3
i5 =
= 10.757 A;
10
v1 − v3
i6 =
= 4.542 A.
50
i4 =
Pdev = 120i1 + 120i3 = 3098 W;
X
P 4.16 [a]
Pdis = i21 (0.5) + i22 (1) + i23 (0.5) + i24 (20) + i25 (10) + i26 (50) = 3098 W.
vo − v1 vo − v2 vo − v3
vo − vn
+
+
+ ··· +
= 0;
R
R
R
R
.·. nvo = v1 + v2 + v3 + · · · + vn ;
1
1
.·. vo = [v1 + v2 + v3 + · · · + vn ] =
n
n
Xn
v .
k=1 k
Problems
4–29
1
[b] vo = (150 + 200 − 50) = 100 V.
3
P 4.17 [a] Identify the reference node and label the node voltages:
Write the KCL equations and the dependent source constraint equation:
v1
v1 − v2
−5 +
+
= 0 so 4v1 − 3v2 + 0i∆ = 75;
15
5
v2
v2
v2 + 5i∆
v2 − v1
+
+
+
= 0 so
5
30 10
30
i∆ =
v1 − v2
5
− 6v1 + 11v2 + 5i∆ = 0;
so v1 − v2 − 5i∆ = 0.
Solving, v1 = 30 V;
v2 = 15 V;
i∆ = 3 A;
io =
15 + 15
= 1 A.
30
p5i∆ = (−15)(1) = −15 W(del);
p5A = −5(30) = −150 W(del)
.·. pdev = 165 W.
[b]
X
pabs =
.·.
X
(30)2 (15)2 (15)2
+
+
+ (3)2 (5) + (1)2 (30) = 165 W;
15
30
10
pdev =
X
pabs = 165 W.
P 4.18
vo − 80 vo
vo + 75iσ
+
+
= 0;
5
50
25
Solving, vo = 50 V;
io =
iσ = 1 A.
50 − (−75)(1)
= 5 A.
25
iσ =
vo
;
50
4–30
CHAPTER 4. Techniques of Circuit Analysis
p75iσ = −75iσ io = −375 W.
.·. The dependent voltage source delivers 375 W to the circuit.
P 4.19
[a]
v − 120
v∆
v − v∆
+
+
= 0;
800
3000
1250
v∆
v∆ − v
+
− 0.02 = 0.
1250
500
Solving,
v = 80 V;
v∆ = 30 V;
vo = v − v∆ = 50 V.
120 − v
120 − 80
[b] i120V =
=
= 0.05 A;
800
800
P120V = −120i120V = −120(0.05) = −6 W
(6 W supplied);
v∆
= (80)(30/3000) = 0.8 W
3000
(0.8 W absorbed);
P20mA = −v∆ (0.02) = −30(0.02) = −0.6 W
(0.6 W supplied).
Pds = v
Ptotal = 6 + 0.6 = 6.6 W supplied.
P 4.20 [a]
v2 − v3
;
50
v1
v1 − v2
−2io +
+
=0
100
25
v2 − v1
v2
v2 − v3
+
+
=0
25
200
50
v3 − v2 v3 − 5io v3 − 38.5
+
+
=0
50
5
20
io =
Solving, v1 = −50 V;
v2 = −30 V;
so
5v1 − 8v2 + 4v3 = 0;
so
−8v1 + 13v2 − 4v3 = 0;
so
0v1 − 4v2 + 29v3 = 192.5.
v3 = 2.5 V.
Problems
4–31
v2 − v3
−30 − 2.5
=
= −0.65 A;
50
50
[b] io =
v3 − 5io
2.5 − 5(−0.65)
=
= 1.15 A;
5
5
38.5 − 2.5
i38.5V =
= 1.8 A.
20
i3 =
p2io = −2io v1 = −2(−0.65)(−50) = −65 W(dev);
p5io = 5io i3 = 5(−0.65)(1.15) = −3.7375 W(dev);
p38.5V = −38.5(1.8) = −69.30 W(dev);
X
pdev = 69.3 + 65 + 3.7375 = 138.0375 W.
CHECK
X
.·.
pdis
X
=
(−50)2 (−30)2 202
+
+
+ (0.65)2 (50) + (1.15)2 5 + (1.8)2 (20)
100
200
25
=
138.0375 W;
pdev =
X
pdis = 138.0375 W.
P 4.21 [a] Redraw the circuit, choosing the node voltages and reference node as
shown:
The node voltage equations are:
v1 − 50 v1 v1 − vo
+
+
− 3i1 = 0
6
8
2
vo vo − v1
−5 +
+
+ 3i1 = 0.
4
2
The dependent source requires the following constraint equation:
50 − v1
i1 =
.
6
Place these equations in standard form:
1 1 1
1
50
v1
+ +
+ vo −
+ i1 (−3) =
6 8 2
2
6
1
1 1
v1 −
+ vo
+
+ i1 (3)
= 5
2
4 2
50
1
+ vo (0)
+ i1 (1)
=
.
v1
6
6
4–32
CHAPTER 4. Techniques of Circuit Analysis
Solving, v1 = 32 V;
Therefore, vo = 16 V.
vo = 16 V;
i1 = 3 A.
[b] p3i1 = 3i1 (vo − v1 ) = −144 W.
Therefore the dependent source delivers 144 W to the circuit.
[c] p50V = −50i1 = −150 W;
p5A = −5(vo ) = −80 W.
Therefore the independent sources deliver 150 + 80 = 230 W to the
circuit.
P 4.22
v1
v1 − v2 v1 − 80
+
+
=0
70
10
5
v2 − v1 v2 − 80
v2
+
+
=0
12
10
5
Solving, v1 = 70 V;
Thus, io =
so
22v1 − 7v2 = 1120;
so
−6v1 + 23v2 = 960.
v2 = 60 V.
v1 − v2
= 1 A.
10
P 4.23 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
Problems
Place these equations in standard form:
1
1
1
+
+
+ i∆ (1)
= 1;
vo
40 10 20
1
20
10
vo
+ i∆ 1 −
= 1+ .
10
30
30
Solving,
i∆ = −3.2 A
and
vo = 24 V.
P 4.24 [a]
v1 − 20
v1
v1 + 10
+
+ 11 × 10−3 +
= 0;
3
3
25 × 10
0.25 × 10
0.5 × 103
v1 = −5 V.
i1 =
20 + 5
= 1 mA;
25,000
i2 =
v1
−5
=
= −20 mA;
250
250
i5 =
−10 + 5
= −10 mA;
500
i4 =
−10
= −10 mA;
1000
i4 + i3 − 11 + i5 = 0;
.·. i3 = 11 − i4 − i5 = 11 + 10 + 10 = 31 mA.
[b] p20V = −20i1 = −20(1 × 10−3 ) = −20 mW;
p10V = −10i3 = −10(31 × 10−3 ) = −310 mW;
p11mA = (−5 + 10)(.011) = 55 mW;
X
pdev = 20 + 310 = 330 mW.
p25k = 25 × 103 i21 = 25 mW;
p0.25k = 0.25 × 103 i22 = 100 mW;
p0.5k = 0.5 × 103 i25 = 50 mW;
4–33
4–34
CHAPTER 4. Techniques of Circuit Analysis
p1k = 1 × 103 i24 = 100 mW;
X
pdiss = 25 + 100 + 50 + 100 + 55 = 330 mW.
X
pdiss =
X
pdev = 330 mW.
P 4.25
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
=
50
;
80
=
0.75 +
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 mA;
p50V
= −(50)(0.196) = −9.8 W;
p80Ω
=
(50 − 34)2 /80 = 3.2 W;
p800Ω
=
(50 − 53.2)2 /800 = 12.8 mW;
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
=
X
pdel = 9.8 + 39.9 = 49.7 W.
50
.
800
Problems
4–35
P 4.26 [a]
v2 − 500 v2 − v4 v2 − v3
+
+
=0
4
2
3
v3 − v2 v3 v3 − v5
+
+
=0
3
6
2
v4 − v2 v4 − 500 v4 − v5
+
+
=0
2
11
4
v5 − v3 v5 v5 − v4
+
+
=0
2
4
4
Solving, v2 = 300 V;
i5Ω =
v3 = 180 V;
so
13v2 − 4v3 − 6v4 + 0v5 = 1500;
so
−2v2 + 6v3 + 0v4 − 3v5 = 0;
so
−22v2 + 0v3 + 37v4 − 11v5 = 2000;
so
0v2 − 2v3 − v4 + 4v5 = 0.
v4 = 280 V;
v5 = 160 V.
500 − 280
500 − v4
=
= 20 A;
11
11
p5Ω = (20)2 (5) = 2000 W.
[b]
i500V
=
=
p500V =
Check:
500 − v2 500 − v4
+
4
11
500 − 300 500 − 280
+
= 50 + 20 = 70 A;
4
11
35,000 W.
4–36
CHAPTER 4. Techniques of Circuit Analysis
X
Pdis
=
(50)2 (4) + (40)2 (3) + (30)2 (6) + (20)2 (11) + (10)2 (2)
+(30)2 (4) + (10)2 (2) + (40)2 (4) = 35,000 W.
P 4.27 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
v2
v2 − 12
v − v1
+
+
+
= 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 = v 2 .
Place this set of equations in standard form:
1
1
1
v1
+
+ v −
+ v2 (0)
+ ix (0) = 4.8;
7.5 2.5
2.5
1
1
1
1
v1 −
+ v
+
+ v2
+1
+ ix (0) = 12;
2.5
2.5 10
2.5
1
v1 −
+ v(0)
+ v2 (0)
+ ix (1) = 0;
7.5
v1 (0)
+ v(1)
+ v2 (−1)
+ ix (1)
=
Solving this set of equations gives v1 = 15 V, v2 = 10 V, ix = 2 A, and
v = 8 V.
0.
Problems
4–37
P 4.28 A supernode exists that includes the nodes labeled v1 and v3 as well as the
current-controlled voltage source between those two nodes:
Note that v∆ = v1 . The supernode equation is
v1 − v2 v3 − 150 v3
v1
v1
+
+
+
−
= 0.
15
30
10
20 4.5
The KCL equation at the node labeled v2 is
v2 − v1
v2
v2 − 150
+
+
= 0.
30
60
10
The supernode constraint equation is
v3 − v1 = 11iφ = 11v3 /20.
In standard form,
1
1
1
1
1
1
+
−
+ v2 −
+ v3
+
= 15;
15 30 4.5
30
10 20
1
1
1
1
−
+ v2
+
+
+ v3 (0) = 15;
30
30 60 10
v1
v1
v1 (1) + v2 (0) + v3
11
− 1 = 0.
20
Solving,
v1 = 90 V;
v2 = 120 V;
v3 = 200 V.
Thus,
v∆
90
pVCCS = −
v3 = −
(200) = −4000 W.
4.5
4.5
The voltage-controlled current source develops 4 kW.
4–38
CHAPTER 4. Techniques of Circuit Analysis
P 4.29 [a] The voltage at node 1 is v4 − 30iφ . The voltage at node 2 is 250 V.
iφ =
v3 − v4
235 − 222
=
= 3.25 A;
4
4
30iφ = 30(3.25) = 97.5 V;
v1 = v4 − 30iφ = 222 − 97.5 = 124.5 V.
[b] v3 + v∆ = 250;
.·. v∆ = 250 − 235 = 15 V.
3.2v∆ = (3.2)(15) = 48 A;
250 − 124.5 250 − 235
+
= 77.75 A;
2
1
i250V =
p250V = −250ig = −250(77.75) = −19,437.5 W(del).
i30iφ = iφ − 222/40 − 48 = 3.25 − 5.55 − 48 = −50.3 A;
p30iφ = (30iφ )i30iφ = (97.5)(−50.3) = −4904.25 W(dev);
p3.2v∆ = (3.2v∆ )(v4 ) = (48)(222) = 10,656 W(abs);
.·.
X
pdev = 19,437.5 + 4904.25 = 24,341.75 W.
p10Ω =
(124.5)2
v12
=
= 1550.025 W;
10
10
p2Ω =
(250 − 124.5)2
= 7875.125 W;
2
p1Ω =
(250 − 235)2
= 225 W;
1
p20Ω =
(235)2
= 2761.25 W;
20
p4Ω = (3.25)2 (4) = 42.25 W;
p40Ω =
(222)2
= 1232.10 W.
40
.·.
pdiss
X
=
10,656 + 1550.025 + 7875.125 + 225+
2761.250 + 42.25 + 1232.1 = 24,341.75 W.
Thus,
X
pdev =
X
[c] Agree with analyst.
pdiss .
Problems
4–39
P 4.30 [a]
This circuit has a supernode that includes the nodes v1 , v2 and the 12 V
source. The supernode equation is
v1
v2
v2
−0.5 +
+
+
= 0.
40 100 200
The supernode constraint equation is
v2 − v1 = 12.
Place these two equations in standard form:
1
1
1
v1
+
+ v2
= 0.5;
40
100 200
v1 (−1)
+ v2 (1)
=
12.
Solving, v1 = 8 V and v2 = 20 V, so vo = v2 = 20 V.
p0.5A = −(0.5)v1 = −(0.5)(8) = −4 W.
The 0.5 A source delivers 4 W.
[b]
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 v is
v
v − 12 v − 12
+
+
= 0.
40
100
200
Solving, v = −8 V so vo = 12 − v = 20 V.
0.5 +
p0.5A = (0.5)v = (0.5)(−8) = −4 W.
The 0.5 A source delivers 4 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.
4–40
CHAPTER 4. Techniques of Circuit Analysis
Both methods give the same result but the choice of reference node in
part (b) yielded fewer equations to solve, so the preferred method is the
one used in part (b).
P 4.31 From Eq. 4.13, iB = vc /(1 + β)RE .
Substitute vc from Eq. 4.14 into Eq. 4.13 to give, iB = (vb − Vo )/(1 + β)RE .
Now substitute vb from
# to give,
" Eq. 4.16 into the above expression
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 )
[R1 R2 /(R1 + R2 )] + (1 + β)RE
P 4.32 [a]
The three mesh current equations are:
1000ia + 2000(ia − if ) + 500(ia − ib )
= 0;
−80 + 800ib + 500(ib − ia ) + 3000(ib − if )
= 0;
250if + 3000(if − ib ) + 2000(if − ia )
= 0.
Place these equations in standard form:
ia (3500) + ib (−500) + if (−2000)
= 0;
ia (−500) + ib (4300) + if (−3000)
= 80;
ia (−2000) + ib (−3000) + if (5250)
= 0.
Solving, ia = 30 mA;
ib = 50 mA;
if = 40 mA.
Now calculate the remaining branch currents:
ic
= ib − ia = 20 mA;
id
= if − ia = 10 mA.
ie
= ib − if = 10 mA.
Problems
4–41
[b] p80V = −(80)ib = −80(0.05) = −4 W (dev).
Therefore, the total power developed is 4 W.
P 4.33 [a]
The three mesh current equations are:
−120 + 0.5i1 + 20(i1 − i6 ) + 1(i1 − i3 )
= 0;
−120 + 1(i3 − i1 ) + 10(i3 − i6 ) + 0.5i3
=
50i6 + 10(i6 − i3 ) + 20(i6 − i1 )
0;
= 0,
Place these equations in standard form:
i1 (21.5) + i3 (−1) + i6 (−20)
= 120;
i1 (−1) + i3 (11.5) + i6 (−10)
= 120;
i1 (−20) + i3 (−10) + i6 (80)
= 0.
Solving, i1 = 10.52 A;
i3 = 15.3 A;
i6 = 4.54 A.
Now calculate the remaining branch currents:
[b]
i2
= i1 − i3 = −4.78 A;
i4
= i1 − i6 = 5.98 A;
i5
= i3 − i6 = 10.76 A.
X
Pdev = 120i1 + 120i3 = 3098 W;
X
Pdis = i21 (0.5) + i22 (1) + i23 (0.5) + i24 (20) + i25 (10) + i26 (50) = 3098 W.
4–42
CHAPTER 4. Techniques of Circuit Analysis
P 4.34
The three mesh current equations are:
−80 + 5(i1 − i2 ) + 70(i1 − i3 )
= 0;
5i2 + 10(i2 − i3 ) + 5(i2 − i1 )
= 0;
12i3 + 70(i3 − i1 ) + 10(i3 − i2 )
= 0.
Place these equations in standard form:
i1 (5 + 70) + i2 (−5) + i3 (−70)
=
i1 (−5) + i2 (5 + 10 + 5) + i3 (−10)
= 0;
i1 (−70) + i2 (−10) + i3 (12 + 70 + 10)
=
Solving, i1 = 6 A;
i2 = 4 A;
Thus, io = i3 − i2 = 1 A.
80;
0.
i3 = 5 A.
P 4.35 [a]
The four mesh current equations are:
Problems
−500 + 4(i1 − i2 ) + 3(i1 − i3 ) + 6(i1 − i4 )
= 0;
5i2 + 6i2 + 2(i2 − i3 ) + 4(i2 − i1 )
= 0;
4i3 + 2(i3 − i4 ) + 3(i3 − i1 ) + 2(i3 − i2 )
= 0;
3i4 + 1i4 + 6(i4 − i1 ) + 2(i4 − i3 )
= 0.
Place these equations in standard form:
i1 (4 + 3 + 6) + i2 (−4) + i3 (−3) + i4 (−6)
=
i1 (−4) + i2 (5 + 6 + 2 + 4) + i3 (−2) + i4 (0)
= 0;
i1 (−3) + i2 (−2) + i3 (2 + 4 + 2 + 3) + i4 (−2) =
i1 (−6) + i2 (0) + i3 (−2) + i4 (2 + 3 + 1 + 6)
[b] p500 = −(500)i1 = −(500)(70) = −35 kW.
P 4.36 [a]
110 + 12 = 17i1 − 10i2 − 3i3 ;
0 = −10i1 + 28i2 − 12i3 ;
−12 − 70 = −3i1 − 12i2 + 17i3 .
Solving, i1 = 8 A;
i2 = 2 A;
i3 = −2 A.
p110 = −110i1 = −880 W(del);
p12 = −12(i1 − i3 ) = −120 W(del);
p70 = 70i3 = −140 W(del);
.·.
X
pdev = 1140 W.
0;
= 0.
Solving, i1 = 70 A;
i2 = 20 A;
i3 = 30 A;
The power absorbed by the 5 Ω resistor is
p5 = i22 (5) = (20)2 (5) = 2000 W.
500;
i4 = 40 A.
4–43
4–44
CHAPTER 4. Techniques of Circuit Analysis
[b] p4Ω = (8)2 (4) = 256 W;
p10Ω = (6)2 (10) = 360 W;
p12Ω = (−4)2 (12) = 192 W;
p2Ω = (−2)2 (2) = 8 W;
p6Ω = (2)2 (6) = 24 W;
p3Ω = (10)2 (3) = 300 W;
.·.
X
pabs = 1140 W.
P 4.37 [a]
40 = 50i1 − 45i2 ;
64 = −45i1 + 50.5i2 .
Solving, i1 = 9.8 A;
ia = i1 = 9.8 A;
i2 = 10 A.
ib = i1 − i2 = −0.2 A;
ic = −i2 = −10 A.
[b] If the polarity of the 64 V source is reversed, we have
40 = 50i1 − 45i2 ;
−64 = −45i1 + 50.5i2 .
i1 = −1.72 A and i2 = −2.8 A.
ia = i1 = −1.72 A;
ib = i1 − i2 = 1.08 A;
P 4.38
The two mesh current equations are:
ic = −i2 = 2.8 A.
Problems
−80 + 5i1 + 50(i1 − i2 )
= 0;
10i2 − 75iσ + 15i2 + 50(i2 − i1 )
= 0.
4–45
The dependent source constraint equation is
iσ = i1 − i2 .
Place these equations in standard form:
i1 (5 + 50) + i2 (−50) + iσ (0)
=
80;
i1 (−50) + i2 (10 + 15 + 50) + iσ (−75)
=
0;
i1 (1) + i2 (−1) + iσ (−1)
=
0.
Solving, i1 = 6 A;
i2 = 5 A;
iσ = 1 A.
Thus, pd.s. = −75iσ i2 = −75(1)(5) = −375 W. The dependent source delivers
375 W.
P 4.39 [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.
P 4.40 Redraw the circuit identifying the three mesh currents:
4–46
CHAPTER 4. Techniques of Circuit Analysis
The three mesh-current equations are
−400 + 7500i1 + 20,000(i1 − i3 ) + 15,000(i1 − i2 )
15,000(i2 − i1 ) + 10,000(i2 − i3 ) + 20,000i2
vx /2 + 10,000(i3 − i2 ) + 20,000(i3 − i1 )
= 0;
=
0;
= 0.
The dependent source constraint equation is
vx = 20,000(i1 − i3 ).
Use the dependent source constraint equation to eliminate vx in the third
mesh-current equation and place the resulting three equations in standard
form:
42,500i1 − 15,000i2 − 20,000i3
=
400;
−15,000i1 + 45,000i2 − 10,000i3
=
0;
−10,000i1 − 10,000i2 + 20,000i3
=
0.
Solving,
i1 = 20 mA;
i2 = 10 mA;
i3 = 15 mA.
p400V = −400i1 = −8 W.
Thus, the independent source is delivering 8 W.
P 4.41 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 ).
Problems
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.
P 4.42
Two mesh current equations and a dependent source constraint equation:
−60 + 4i1 + 10(i1 − i2 ) + 2i1 = 0;
2.5i2 −
v∆
+ 20i2 + 10(i2 − i1 ) = 0;
2
v∆ = 10(i1 − i2 ).
In standard form:
16i1 − 10i2 + 0v∆ = 60;
−10i1 + 32.5i2 − 0.5v∆ = 0;
10i1 − 10i2 − 1v∆ = 0.
Solving, i1 = 5 A;
i2 = 2 A;
Thus,
p10Ω =
2
v∆
302
=
= 90 W.
10
10
v∆ = 30 V.
4–47
4–48
CHAPTER 4. Techniques of Circuit Analysis
P 4.43
Two mesh current equations:
400i1 − 250i2 = −300;
−250i1 + 625i2 = −1200.
Solving,
i1 = −2.6 A;
i2 = −2.96 A.
Therefore,
v1 = −125(−2.6 + 2.4) = 25 V;
v2 = 375(−2.96 + 3.2) = 90 V.
P 4.44
600 = 64i1 − 40i2 − 14i3 ;
−400 = −40i1 + 50i2 − 2i3 ;
−12 = i3 .
Solving, i1 = 2.9 A;
i2 = −6.16 A;
i3 = −12 A.
[a] v12A = 2(12 − 6.16) + 14(12 + 2.9)
= 220.28 V;
p12A = −12v12A = −12(220.28) = −2643.36 W.
Therefore, the 12 A source delivers 2643.36 W.
Problems
4–49
[b] p400V = 400(−6.16) = −2464 W;
p600V = −600i1 = −600(2.9) = −1740 W.
Therefore, the total power delivered is 2643.36 + 2464 + 1740 = 6847.36 W
[c]
X
presistors = (2.9)2 (10) + (6.16)2 (8) + (9.06)2 (40) + (14.9)2 (14) + (5.84)2 (2) = 6847.36 W;
X
pabs = 6847.36W =
X
pdel (CHECKS).
P 4.45 [a] Identify and label the mesh currents:
Write the KVL equations, which already use the controlling current for
the dependent source:
5i∆ + 30(i∆ − i1 ) + 15(i∆ − 5) = 0 so i∆ (50) + i1 (−30) + i2 (0) = 75;
10(i1 − i2 ) + 30(i1 − i∆ ) = 0 so i∆ (−30) + i1 (40) + i2 (−10) = 0;
30i2 − 5i∆ + 10(i2 − i1 ) = 0 so i∆ (−5) + i1 (−10) + i2 (40) = 0;
Solving, i∆ = 3 A;
i1 = 2.5 A;
i2 = 1 A.
p5i∆ = −5i∆ i2 = −5(3)(1) = −15 W(del);
v5A = 15(5 − i∆ ) = 15(5 − 3) = 30 V;
p5A = −5v5A = −5(30) = −150 W(del);
.·. pdev = 165 W.
[b]
X
pabs = 15(5 − i∆ )2 + 5i2∆ + 30(i∆ − i1 )2 + 10(i1 − i2 )2 + 30i22
= 15(2)2 + 5(3)2 + 30(0.5)2 + 10(1.5)2 + 30(1)2 = 165 W;
.·.
X
pdev =
X
pabs = 165 W.
4–50
CHAPTER 4. Techniques of Circuit Analysis
P 4.46
Mesh equations:
50i1 − 20i2 − 25ig = 0;
−20i1 + 120i2 − 30i∆ − 100ig = 0.
Constraint equations:
ig = 4;
i∆ = i1 .
Solving, i1 = 4 A;
i2 = 5 A.
i25Ω = 4 − i1 = 0 A;
i20Ω = i2 − i1 = 1 A;
i100Ω = 4 − i2 = −1 A;
i5Ω = i1 = 4 A;
v4A = 100(4 − i2 ) = −100 V;
p4A = −v4A ig = −(−100)(4) = 400 W (abs);
v30i∆ = 30i∆ = 30i1 = 120 V;
p30i∆ = −30i∆ i2 = −120(5) = −600 W.
Therefore, the dependent source is developing 600 W, all other elements are
absorbing power, and the total power developed is thus 600 W.
CHECK:
p5Ω = 16(5) = 80 W;
Problems
p25Ω = 0 W;
p20Ω = 1(20) = 20 W;
p100Ω = 1(100) = 100 W;
p4A = 400 W;
X
pabs = 80 + 0 + 20 + 100 + 400 = 600 W (CHECKS).
P 4.47 [a]
The mesh current equation for the right mesh is:
3300(i1 − 0.008) + 6500i1 + 200(i1 − 0.008) = 0.
Solving,
10,000i1 = 28
.·. i1 = 2.8 mA.
Then,
i∆ = i1 − 0.008 = −5.2 mA.
[b] vo = (0.008)(980) − (−0.0052)(3300) = 25 V;
p8mA = −(25)(0.008) = −200 mW.
Thus, the 8 mA source delivers 200 mW.
[c] 200i∆ = 200(−0.0052) = −1.04 V;
pdep source = 200i∆ i1 = (−1.04)(0.0028) = −2.912 mW.
The dependent source delivers 2.912 mW.
P 4.48
Mesh current equations:
−120 + 500i1 − 500i∆ + 4000(i1 − i2 ) = 0;
625i2 + 1000(i2 − 3iφ ) + 4000(i2 − i1 ) + 500i∆ = 0.
Dependent source constraint equations:
i∆ = i2 − 3iφ ;
4–51
4–52
CHAPTER 4. Techniques of Circuit Analysis
iφ = i1 − i2 .
In standard form:
4500i1 − 4000i2 − 500i∆ + 0iφ = 120;
−4000i1 + 5625i2 + 500i∆ − 3000iφ = 0;
0i1 − 1i2 + 1i∆ + 3iφ = 0;
−1i1 + 1i2 + 0i∆ + 1iφ = 0.
Solving,
i1 = 100 mA;
i2 = 80 mA;
i∆ = 20 mA;
iφ = 20 mA.
[a] Source power:
p120V = −120i1 = −120(0.1) = −12 W;
pCCVS = −500i∆ iφ = −500(0.02)(0.02) = −0.2 W;
vCCCS = 1000i∆ = 1000(0.02) = 20 V;
pCCCS = 3iφ vCCCS = 3(0.02)(20) = 1.2 W.
Thus the two voltage sources are generating power and
pgenerated = 12 + .2 = 12.2 W.
[b] Resistor power:
p500Ω = 500i21 = 500(0.1)2 = 5 W;
p4000Ω = 4000i2φ = 4000(0.02)2 = 1.6 W;
p625Ω = 625i22 = 625(0.08)2 = 4 W;
p1000Ω = 1000i2∆ = 1000(0.02)2 = 0.4 W.
Thus,
pdissipated = 1.2 + 5 + 1.6 + 4 + 0.4 = 12.2 W.
P 4.49
−18 + 3i1 + 9i2 − 15 + 6i2 + 2i1 = 0;
i2 − i1 = 3;
Problems
Solving, i1 = −0.6 A;
i2 = 2.4 A.
p18V = −18i1 = 10.8 W (diss);
p3 Ω = (−0.6)2 (3) = 1.08 W;
p2 Ω = (−0.6)2 (2) = 0.72 W;
p9 Ω = (2.4)2 (9) = 51.84 W;
p6 Ω = (2.4)2 (6) = 34.56 W;
X
pdiss = 99 W.
vo = 15i2 − 15 = 36 − 15 = 21 V;
p3A = −3vo = −63 W (dev);
p15V = −15i2 = −36 W (dev);
X
pdev = 99 W =
X
pdiss .
P 4.50 [a]
200 = 85i1 − 25i2 − 50i3 ;
0 = −75i1 + 35i2 + 150i3
(supermesh);
i3 − i2 = 4.3(i1 − i2 ).
Solving, i1 = 4.6 A;
i2 = 5.7 A;
i3 = 0.97 A.
ia = i2 = 5.7 A;
ib = i1 = 4.6 A;
ic = i3 = 0.97 A;
id = i1 − i2 = −1.1 A;
ie = i1 − i3 = 3.63 A.
4–53
4–54
CHAPTER 4. Techniques of Circuit Analysis
[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 4.51 [a]
The meshes labeled ib and ic share a current source, so combine them to
form a supermesh whose KVL equation is
500ib + 1000(ic − id ) + 250(ic − ia ) = 0.
The supermesh constraint equation is
ic − ib = 0.011.
The KVL equations for the remaining two meshes are
−20 + 25,000ia + 250(ia − ic ) = 0;
−10 + 1000(id − ic ) = 0.
In standard form, the four equations are
ia (−250) + ib (500) + ic (1250) + id (−1000) = 0;
ia (0) + ib (−1) + ic (1) + id (0) = 0.011;
ia (25,250) + ib (0) + ic (−250) + id (0) = 20;
ia (0) + ib (0) + ic (−1000) + id (1000) = 10.
Problems
Solving,
ia = 1 mA;
ib = 10 mA;
ic = 21 mA;
i2 = ia − ic = −20 mA;
i1 = ia = 1 mA;
id = 31 mA.
i3 = id = 31 mA.
[b] p20V = −20i1 = 20(0.001) = −20 mW;
p10V = −10i3 = 10(0.031) = −310 mW;
v11mA = 1000(ic − id ) + 250(ic − ia )
= 1000(0.021 − 0.031) + 250(0.021 − 0.001) = −5 V;
p11mA = −0.011v11mA = 55 mW;
X
pdev = 20 + 310 = 330 mW.
p25k = 25,000i21 = 25 mW;
p250 = 250i22 = 100 mW;
p500 = 500i2b = 50 mW;
p1000 = 1000(id − ic )2 = 100 mW;
X
pdiss = 25 + 100 + 50 + 100 + 55 = 330 mW;
X
pdiss =
X
pdev = 330 mW.
P 4.52 [a]
The supermesh equation is:
10 + 3i1 + 9i2 − 15 + 6i2 + 2i1 = 0.
The supermesh constraint equation is :
i2 − i1 = 3.
Place these equations in standard form:
i1 (5) + i2 (15)
= 5;
i1 (−1) + i2 (1)
=
3.
Solving,
i1 = −2 A;
Now find the power:
i2 = 1 A.
4–55
4–56
CHAPTER 4. Techniques of Circuit Analysis
p3Ω
=
(−2)2 (3) = 12 W;
p2Ω
=
(−2)2 (2) = 8 W;
p9Ω
=
12 (9) = 9 W;
p6Ω
=
12 (6) = 6 W;
p10V
=
(10)(−2) = −20 W;
v3A
=
9i2 − 15 + 6i2 = 9(1) − 15 + 6(1) = 0 V
(the 3 A source acts like a short circuit carrying 3 A of current);
p3A
=
(0)(3) = 0 W;
p15V
= −(15)(1) = −15 W.
In
Xsummary:
pdiss = 12 + 8 + 9 + 5 = 35 W;
X
pdev = 15 + 20 = 35 W. (note that the power of the 3 A source is zero)
Thus the power dissipated in the circuit is 35 W.
[b]
Now there is no longer a supermesh. The two simple mesh current
equations are:
10 + 3i1 + 2i1
=
0;
−15 + 6i2 + 9i2
=
0.
Since these equations are uncoupled, each can be solved separately:
5i1 = −10 .·.
i1 = −2 A;
15i2 = 15
.·.
i2 = 1 A.
Since the currents are the same as in part (a), the power will be the same
as calculated in part (a). Thus, the power dissipated in the circuit is
again 35 W.
[c] As noted in part (a), the 3 A source has zero voltage drop, so is equivalent
to a short circuit (which has no voltage drop by definition) carrying 3 A
of current, as in the circuit of part (b).
Problems
4–57
[d]
The supermesh equation is:
−18 + 3i1 + 9i2 − V + 6i2 + 2i1 = 0.
The supermesh constraint equation is :
i2 − i1 = 3.
To meet the problem specifications, the voltage across the 3 A current
source must be zero:
9i2 − V + 6i2 = 0.
Place these equations in standard form:
i1 (5) + i2 (15) + V (−1) =
i1 (−1) + i2 (1) + V (0)
18;
= 3.
i1 (0) + i2 (15) + V (−1) =
0.
Solving,
i1 = 3.6 A;
i2 = 6.6 A;
V = 99 V.
Note that the power dissipated in this circuit has changed:
p18V
= −(18)(3.6) = −64.8 W;
v3A
=
9i2 − 99 + 6i2 = 9(6.6) − 99 + 6(6.6) = 0 V
(the 3 A source acts like a short circuit carrying 3 A of current);
p3A
=
(0)(3) = 0 W;
p99V
= −(99)(6.6) = −653.4 W.
In
Xsummary:
pdiss = 64.8 + 653.4 = 718.2 W.
Thus the power dissipated in the circuit is 718.2 W.
4–58
CHAPTER 4. Techniques of Circuit Analysis
P 4.53 [a]
40(i3 − i1 ) + 10(i3 − i2 ) + 35(i4 − i2 ) + 150 = 0;
35(i2 − i4 ) + 10(i2 − i3 ) + 15id = 0;
3ia = i3 − i4 ;
id = i4 ;
ia = i1 − i3 ;
i1 = 30 A;
Solving, i1 = 30 A;
i2 = 8 A;
ia = 30 − 24 = 6 A;
ib = 8 − 24 = −16 A;
id = 6 A;
i3 = 24 A;
i4 = 6 A.
ic = 8 − 6 = 2 A;
ie = ic + id = 6 + 2 = 8 A.
[b] va = 40ia = 240 V;
vb = 150 − 35ic = 80 V;
p30A = −30va = −30(240) = −7200 W (gen);
p15id = 15id ie = 15(6)(8) = 720 W (diss);
p3ia = 3ia vb = 3(6)(80) = 1440 W (diss);
p150V = 150id = 150(6) = 900 W (diss);
p40Ω = (6)2 (40) = 1440 W (diss);
p10Ω = (−16)2 (10) = 2560 W (diss);
p35Ω = (2)2 (35) = 140 W (diss);
X
Pgen = 7200 W;
X
Pdiss = 720 + 1440 + 900 + 1440 + 2560 + 140 = 7200 W.
P 4.54 [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.
Problems
4–59
[b]
The mesh current equations:
2i1 + 10(i1 − i2 ) + 8(i1 − 4) =
4i2 + 1(i2 − 4) + 10(i2 − i1 )
0;
= 0.
Place the equations in standard form:
i1 (2 + 10 + 8) + i2 (−10)
= 32;
i1 (−10) + i2 (4 + 1 + 10)
= 4.
Solving, i1 = 2.6 A;
i2 = 2 A.
Find the power in the 10 Ω resistor:
i10Ω = i1 − i2 = 0.6 A;
p10Ω = (0.6)2 (10) = 3.6 W.
[c] No, the voltage across the 4 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 = 2i1 + 4i2 = 2(2.6) + 4(2) = 13.2 V;
p4A = −(13.2)(4) = −52.8 W;
Thus the 4 A source develops 52.8 W.
P 4.55 [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]
The node voltage equations are:
4–60
CHAPTER 4. Techniques of Circuit Analysis
v1 v1 − v2 v1 − v3
+
+
= 0;
1
8
10
v2
v2 − v1 v2 − v3
−4 +
+
+
= 0;
20
8
2
v3 − v1 v3 − v2 v3
+
+
= 0.
10
2
4
Putthe equations
in standard
form:
1
1
1
1
+ v2 −
+ v3 −
= 0;
v1 1 + +
8 10
8
10
1
1
1 1
1
v1 −
+ v2
+ +
+ v3 −
= 4;
8
20 8 2
2
1
1
1
1
1
v1 −
+ v2 −
+ v3
+
+
= 0.
10
2
2 10 4
Solving,
v1 = 1.72 V; v2 = 11.33 V; v3 = 6.87 V.
p4A = −(11.33)(4) = −45.32 W;
Therefore, the 4 A source is developing 45.32 W.
P 4.56 [a] Choose the reference node so that a node voltage is identical to the
voltage across the 15 A source; thus:
Since the 15 A source is developing 3750 W, v1 must be 250 V. We can
then use the node-voltage method to write KCL equations in succession
at v1 (solving for v2 ), v2 (solving for v3 ) and v3 (solving for idc ).
[b] Since v1 is known, we can sum the currents away from node 1 to find v2 ;
thus:
250 − (420 + v2 ) 250 − v2 250
+
+
− 15 = 0;
7.2
20
15
.·. v2 = −50 V.
Now that we know v2 we sum the currents away from node 2 to find v3 ;
thus:
v2 + 420 − 250 v2 − 250 v2 − v3
+
+
= 0;
7.2
20
40
.·. v3 = 50/3 V.
Problems
4–61
Now that we know v3 we sum the currents away from node 3 to find idc ;
thus:
v3
v3 + 50
+
= idc ;
50
40
.·. idc = 2 A.
P 4.57 [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 10 V source is obtained by summing the currents at either terminal of
the source.
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 10 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:
vx v2
v1
−
+
+ 10 = 0;
25
2
5
v3 v4 vx
−10 +
+
+
= 0.
4
2
2
Constraints:
v2 = vx ;
−
v3
= ix ;
4
v4 − v3 = 2ix ;
v1 − v2 = 10.
4–62
CHAPTER 4. Techniques of Circuit Analysis
Solving, v1 = 50 V; v2 = 40 V;
v1
vx
io =
−
= −18 A;
25
2
v3 = −20 V;
v4 = −10 V;
ix = 5 A.
p10V = −10io = 180 W.
Thus, the 10 V source absorbs 180 W.
P 4.58 [a] There are three meshes, but if io = 0, the current in the upper mesh is
known. Therefore we would need to write two KVL equations, but there
would be three unknowns: the two mesh currents and the voltage vdc .
There is no obvious way to write a third equation that we need to solve
for three unknowns. There are also four essential nodes, one of which will
be chosen as the reference node. But if io = 0 there is no voltage drop
across the 30 Ω resistor so the voltages at the left and right nodes are the
same. Make the bottom node the reference and write a KCL equation at
the left and middle nodes to solve for the two unknown node voltages;
note that these KVL equations do not involve vdc . Then write a KCL
equation at the right node to solve for vdc . Thus, use the node-voltage
method.
[b]
If io = 0 then v1 = v3 ; therefore,
v1 − v2 v1 − 230
+
= 0;
5
20
v2 − v1 v2 − v3 v2 − 115
+
+
= 0.
5
15
10
Solving, v1 = 170 V = v3 ;
At the node labeled v3 ,
Solving, vdc = 195 V.
230 − 170
[c] ia =
= 3 A;
20
ib =
115 − 155
= −4 A;
10
v2 = 155 V.
170 − 155 170 − vdc
+
= 0.
15
25
Problems
ic =
195 − 170
= 1 A;
25
id =
170 − 155
= 3 A;
5
ie =
155 − 170
= −1 A;
15
p230V = −230ia = −690 W (dev);
p115V = −115ib = 460 W (abs);
pvdc = −vdc ic = −195 W (dev);
p20Ω = i2a (20) = 180 W;
p5Ω = i2d (5) = 45 W;
p10Ω = i2b (10) = 160 W;
p15Ω = i2e (15) = 15 W;
p25Ω = i2c (25) = 25 W;
X
pdiss = 460 + 180 + 45 + 160 + 15 + 25 = 885 W;
X
pdev = 690 + 195 = 885 W (CHECKS),
P 4.59 [a] Apply source transformations to both current sources to get
io =
13.8 + 4.2
= 3 mA.
2700 + 2300 + 1000
[b]
The node voltage equations:
v1
v1 − v2
6 × 10−3 +
+
2300
2700
v2
v2 − v1
+
− 4.2 × 10−3
1000
2700
=
0;
=
0.
4–63
4–64
CHAPTER 4. Techniques of Circuit Analysis
Place
standard form:
in these equations
1
1
1
v1
+
+ v2 −
= −6 × 10−3 ;
2700 2300
2700
1
1
1
v1 −
+ v2
+
= 4.2 × 10−3 .
2700
1000 2700
Solving, v1 = −6.9 V;
v2 = 1.2 V.
v2 − v1
.·. io =
= 3 mA.
2700
P 4.60 [a] Applying a source transformation to each current source yields
Now combine the 20 V and 10 V sources into a single voltage source and
the 5 Ω, 4 Ω and 1 Ω resistors into a single resistor to get
Now use a source transformation on each voltage source, thus
which can be reduced to
2k8
.·. io =
(1.25) = 1 A.
2
Problems
[b]
50ia − 40ib = 20 − 10 − 10 = 0;
−40ia + 42ib = 10;
Solving, ia = 0.8 A,
P 4.61 [a]
io =
120
= 4 mA.
30,000
ib = 1 A,
io = ib = 1 A.
4–65
4–66
CHAPTER 4. Techniques of Circuit Analysis
[b]
va
=
ib
(15,000)(0.004) = 60 V;
va
=
= 1 mA;
60,000
= 12 − 1 − 4 = 7 mA;
vb
=
ia
p100V
60 − (0.007)(4000) = 32 V;
32
= 0.007 −
= 6.6 mA;
80,000
= −(100)(6.6 × 10−3 ) = −660 mW.
Check:
p12mA
= −(60)(12 × 10−3 ) = −720 mW;
ig
X
Pdev
=
660 + 720 = 1380 mW;
X
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] Begin by source-transforming both ends of the circuit to give
Now simplify the right side of the circuit by combining the
series-connected voltage sources and the series-connected resistors to give
Source-transform the right side of the circuit to give
Combine the parallel 80 kΩ and 20 kΩ resistors to give a 16 kΩ and
Problems
4–67
source-transform the right side one final time:
Combine the series-connected voltage sources and use voltage division to
find the voltage v0 :
v0 =
24,000
(−20 − 80) = −50 V.
8000 + 24,000 + 16,000
[b] Identify the four meshes and label their currents:
The two mesh current equations are
24,000i1 + 80,000(i1 − i2 ) + 8000(i1 + 0.0025) = 0;
15,000i2 + 40 + 5000(i2 + 0.012) + 80,000(i2 − i1 ) = 0.
Solving,
i1 = −2.0833 mA;
i2 = −2.667 mA.
Therefore,
v0 = 24,000i1 = −50 V.
P 4.63 [a] First remove the 8 Ω and 80 Ω resistors:
Next use a source transformation to convert the 5 A current source and
20 Ω resistor:
which simplifies to
4–68
CHAPTER 4. Techniques of Circuit Analysis
45
.·. vo = (−240) = −135V.
80
[b] Return to the original circuit with vo = −135 V and io = vo /45 = −3 A:
ig =
340
− (−3) = 7.25 A;
80
p340V = −(340)(7.25) = −2465 W.
Therefore, the 340 V source is developing 2465 W.
[c] v1 = 340 + 60io = 340 − 180 = 160 V;
vg = v1 + 5(8) = 160 + 40 = 200 V;
p5A = −(5)(200) = −1000 W.
Therefore the 5 A source is developing 1000 W.
[d]
X
pdev = 2465 + 1000 = 3465 W;
X
pdiss = (5)2 (8) +
.·.
X
pdiss =
X
1602 3402
+
+ (−3)2 (5 + 45 + 10) = 3465 W;
20
80
pdev .
P 4.64 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.
Problems
4–69
Finally, combine the 20 Ω and 12 Ω parallel resistors to give
RN = 20k12 = 7.5 Ω and the Norton equivalent circuit:
P 4.65 vTh =
30
(80) = 60 V;
40
RTh = 2.5 +
(30)(10)
= 10 Ω.
40
P 4.66 To find RN , 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 RN = 12k12 = 6 Ω.
Create a short circuit between the terminals a and b and use mesh current
analysis to find iN = isc . Begin by redrawing the circuit and labeling the mesh
currents:
4–70
CHAPTER 4. Techniques of Circuit Analysis
The mesh current equations are
25i1 − 5i2 − 20isc = 72;
−5i1 + 25i2 − 8isc = 0;
−20i1 − 8i2 + 28isc = 0.
Solving, i1 = 12.72 A, i2 = 6 A, and iN = isc = 10.8 A. Therefore, the Norton
equivalent circuit is a 10.8 A source in parallel with a 6 Ω resistor.
P 4.67 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
Therefore the Norton equivalent is
Problems
P 4.68 After making a source transformation the circuit becomes
300 = 48i1 − 40i2 ;
−450 = −40i1 + 200i2 ;
.·. i1 = 5.25A and i2 = −1.2 A.
vTh = 8i1 + 10i2 = 30 V;
RTh = (40k8 + 10)k150 = 15 Ω.
P 4.69
i1 = 45/15,000 = 3 mA;
45 = vTh − 0.003RTh ,
i2 = 25/5000 = 5 mA;
vTh = 45 + 0.003RTh .
4–71
4–72
CHAPTER 4. Techniques of Circuit Analysis
25 = vTh − 0.005RTh ,
vTh = 25 + 0.005RTh ;
.·. 45 + 0.003RTh = 25 + 0.005RTh
so
vTh = 45 + 30 = 75 V.
P 4.70
v1 =
600
(120) = 48 V;
1500
v2 =
800
(120) = 47.9042 V;
2004
vTh = v1 − v2 = 48 − 47.9042 = 95.8 mV.
RTh =
igal =
(900)(600) (1204)(800)
+
= 840.64 Ω.
1500
2004
0.0958
= 110.034 µA.
840.64 + 30
RTh = 10 kΩ;
Problems
P 4.71
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 4.72 First, find the Thévenin equivalent with respect to Ro .
Ro (Ω)
10
io (A) vo (V )
Ro (Ω)
io (A)
vo (V )
1.2
12
33 0.762 25.143
15 1.067
16
47 0.623
29.3
22 0.923
20.31
68
33.31
0.49
4–73
4–74
CHAPTER 4. Techniques of Circuit Analysis
P 4.73 [a] First, find the Thévenin equivalent with respect to a,b using a succession
of source transformations.
.·. VTh = 48 V
vmeas =
RTh = 6 kΩ.
100
(48) = 45.283 V.
106
45.283 − 48
× 100 = −5.66%.
[b] %error =
48
P 4.74 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
Problems
4–75
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 an open 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
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.
4–76
CHAPTER 4. Techniques of Circuit Analysis
P 4.75 Begin by calculating the short circuit current, which is also iN , from the
circuit below:
The short circuit eliminates both the 8 Ω resistor and the dependent current
source from the circuit, resulting in the simplified circuit shown below.
Perform a source transformation to get a circuit with only parallel branches.
In the source transformed circuit we can combine the two current sources to
calculate the short circuit current, since the short circuit eliminates the 2 Ω
resistor from the circuit:
iN = isc = 12 − 4 = 8 A.
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.
Problems
4–77
Substitute the second equation into the first equation:
iT = 4(vT /8) + vT /2 = vT .
Thus,
RTh = vT /iT = 1 Ω.
The Norton is an 8 A source in parallel with a 1 Ω resistor.
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 − 24
+
+
+
= 0;
100
25
20
2
Solving, vTh = 20 V.
Short-circuit current:
isc = 12 + 2isc ,
RTh =
.·. isc = −12 A;
20
= −(5/3) Ω.
−12
4–78
CHAPTER 4. Techniques of Circuit Analysis
Rtotal =
20
= 3.33 Ω;
6
Rmeter = 3.33 − 3.13 = 0.2 Ω.
[b] Actual current:
iactual =
20
= 6.38 A;
3.13
% error =
6 − 6.38
× 100 = −6%.
6.38
P 4.77 [a] Open circuit:
The node voltage equations are:
v1 − 17.4 v1
+
+ 0.1
= 0;
40
15
v2
v2 − 17.4
−0.1 +
+
= 0.
14
26
The above equations are decoupled, so just solve the second equation for
v2 and use v2 to solve for voc :
−36.4 + 26v2 + 14v2 − 243.6 = 0; .·.
v2 = 280/40 = 7 V.
10
voc =
(7) = 5 V.
10 + 4
Problems
4–79
Short circuit:
Write a node voltage equation at v2 :
v2 − 17.4 v2
−0.1 +
+
= 0.
26
4
Solving,
−5.2 + 2v2 − 34.8 + 13v2 = 0; .·.
v2 = 40/15 V.
Calculate the short circuit current:
isc = (40/15)/4 = 2/3 A.
Therefore, RTh = 5/(2/3) = 7.5 Ω.
[b]
RTh = 10k(26 + 4) = 7.5 Ω (CHECKS).
P 4.78 [a] Replace the current source with an open circuit and find the equivalent
resistance looking into the terminals a,b:
RTh = 12k(2 + 8 + 10) = 12k20 = 7.5 Ω.
[b] Replace the voltage source with a short circuit and find the equivalent
resistance from the terminals a,b:
RN = 2.5 + (10k30) = 10 Ω.
4–80
CHAPTER 4. Techniques of Circuit Analysis
P 4.79 IN = 0 since there are no independent sources in the circuit. Thus we need
only find RN .
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
.·. RN = 12.5 Ω.
P 4.80 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
Problems
i∆ =
4–81
vT − v1
.
12
In standard form:
1
1
1
1
+
+
+ vT −
+ i∆ (−1) = 0;
10 2.5 12
12
1
1
1
10
+ vT
+
+ i∆ −
= 1;
12
12 6
6
v1
v1 −
v1 (1) + vT (−1) + i∆ (12) = 0.
Solving,
v1 = 2 V;
vT = 8 V;
i∆ = 0.5 A.
vT
= 8 Ω.
.·. RTh =
1A
P 4.81 VTh = 0 since there are no independent sources in the circuit. To find RTh we
first find Ra0 b0 .
iT =
vT vT − v1
+
;
25
10
v1
v1 + 6.5ix v1 − vT
+
+
= 0 so 16v1 + 65ix = 5vT ;
50
5
10
ix =
vT
,
25
65ix = 2.6vT ;
16v1 + 2.6vT = 5vT ;
4–82
CHAPTER 4. Techniques of Circuit Analysis
.·. v1 = 0.15vT .
iT =
vT vT − 0.15vT
6.25
+
=
vT ;
25
10
50
vT
= 50/6.25 = 8 Ω = Ra0 b0 ;
iT
.·. RTh = 12 + 8 = 20 Ω.
P 4.82 [a] From the solution to Problem 4.72 we have
Ro (Ω) po (W)
10
14.4
15
17.07
22
18.74
33
19.16
47
18.25
68 16.32
The 33 Ω resistor dissipates the most power, because its value is closest to
the Thévenin equivalent resistance of the circuit.
[b]
[c] From the solution to Problem 4.72, VTh = 48 V and RTh = 30 Ω. Thus, the
maximum power delivered to the load occurs when Ro = RTh = 30 Ω, so
pmax = 242 /30 = 19.2 W.
Problems
P 4.83 [a] From the solution of Problem 4.75 we have RN = 1 Ω and IN = 8 A.
Therefore
Ro = RN = 1 Ω.
[b] p =
(4)2
= 16 W.
1
[c]
The node voltage equation is:
v v
v − 24
+ 3ix + 4 + + = 0.
2
8 1
The dependent source constraint equation is:
v
ix = .
8
Solving, v = 4 V;
ix = 0.5 A.
Calculate the power:
vdep source = 4 − 24 = −20 V
so pdep source = 3ix vdep source = 3(0.5)(−20) = −30 W.
i24V = 3ix +
4 − 24
= 1.5 − 10 = −8.5;
2
p24V = 24i24V = 24(−8.5) = −204 W.
p4A = 4v = 4(4) = 16 W.
X
pdev = 30 + 204 = 234 W.
16
× 100 = 6.84%.
234
[d] The 10 Ω resistor in Appendix H is closest to the Thévenin equivalent
resistance.
% delivered =
[e] Substitute the 10 Ω resistor into the original circuit and calculate the
power developed by the sources in this circuit:
4–83
4–84
CHAPTER 4. Techniques of Circuit Analysis
The node voltage equation is:
v − 24
v
v
+ 3ix + 4 + +
= 0.
2
8 10
The dependent source constraint equation is:
v
ix = .
8
Solving, v = 7.273 V;
ix = 0.909 A.
Calculate the power:
p10Ω =
v2
= 5.3 W;
10
vdep source = 7.273 − 24 = −16.727 V
so pdep source = 3ix vdep source = 3(0.909)(−16.727) = −45.6 W.
i24V = 3ix +
7.273 − 24
= −5.636 A;
2
p24V = 24i24V = 24(−5.636) = −135.27 W.
p4A = 4v = 4(7.273) = 29.09 W.
X
pdev = 45.6 + 135.27 = 180.87 W.
% delivered =
5.3
× 100 = 2.93%.
180.87
P 4.84 [a] Since 0 ≤ Ro ≤ ∞ maximum power will be delivered to the 6 Ω resistor
when Ro = 0.
302
[b] P =
= 150 W.
6
P 4.85 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:
Problems
4–85
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 −
= 25;
4 4
4
4
1
1 1 1
vTh −
+ v1
+ +
+ vφ (0)
= 30;
4
4 4 4
vTh (0)
+ v1 (1)
+ vφ (−1)
= 20.
Solving, vTh = 120 V, v1 = 80 V, and vφ = 60 V.
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.
4–86
CHAPTER 4. Techniques of Circuit Analysis
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
=
= 3 Ω.
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
vload =
= 60 V.
2
Therefore,
2
vload
602
=
= 1200 W.
Rload
3
[c] The resistor from Appendix H closest in value to RTh is 10 Ω. Attach this
resistor to the Thévenin equivalent circuit:
pmax =
v10Ω =
10
(120) = 92.31 V,
13
so p10Ω =
92.312
= 852.1 W.
10
P 4.86 Substituting the value R = 3 Ω into the circuit and identifying three mesh
currents we have the circuit below:
Problems
4–87
The mesh current equations are:
−100 + 4(i1 − i2 ) + vφ + 20
−vφ + 4i2 + 4(i2 − i3 ) + 4(i2 − i1 )
−20 − vφ + 4(i3 − i2 ) + 3i3
=
0;
= 0;
=
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.
p100V = −(100)i1 = −(100)(30) = −3000 W;
p20V = 20(i1 − i2 ) = 20(10) = 200 W;
pdep source = −vφ i2 = −(40)(20) = −800 W;
.·.
pgen = 3000 + 800 = 3800 W.
From Problem 4.85(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.
P 4.87 [a] Open circuit voltage
Node voltage equation:
v1 − 100 v1 − 13i∆ v1 − v2
+
+
= 0.
2
5
4
4–88
CHAPTER 4. Techniques of Circuit Analysis
Constraint equations:
100 − v1
v2 − v1
;
− v∆ = 0;
2
4
Solving, v2 = 90 V = vTh ; v1 = 90 V;
Short circuit current:
i∆ =
v1 − 100 v1 − 13i∆ v1
+
+
= 0;
2
5
4
i∆ =
100 − v1
.
2
Solving, v1 = 80 V = v∆ ; i∆ = 10 A.
v1
+ v∆ = 20 + 80 = 100 A;
isc =
4
RTh =
vTh
90
=
= 0.9 Ω;
isc
100
.·. Ro = RTh = 0.9 Ω.
[b]
pmax =
(45)2
= 2250 W.
0.9
v∆ = v1 − v2 .
v∆ = 0 V;
i∆ = 5 A.
Problems
4–89
[c]
v1 − 100 v1 − 13i∆ v1 − 45
+
+
= 0;
2
5
4
i∆ =
100 − v1
.
2
Solving, v1 = 85 V;
i∆ = 7.5 A;
v∆ = v1 − v2 = 85 − 45 = 40 V.
i100V = i∆ = 7.5 A;
p100V (dev) = 100(7.5) = 750 W.
i12 = v∆ /4 = 40/4 = 10 A;
i1 = i12 − i∆ = 10 − 7.5 = 2.5 A;
p13i∆ (dev) = (97.5)(2.5) = 243.75 W.
pv∆ (dev) = (45)(40) = 1800 W.
X
pdev = 750 + 243.75 + 1800 = 2793.75 W.
% delivered =
2250
× 100 = 80.54%.
2793.75
P 4.88 We begin by finding the Thévenin equivalent with respect to Ro . After making
a couple of source transformations the circuit simplifies to
ix =
80 + 60ix
;
140
ix = 1 A;
VTh = 40ix − 60ix = −20ix = −20 V.
4–90
CHAPTER 4. Techniques of Circuit Analysis
Using the test-source method to find the Thévenin resistance gives
Use the node voltage method:
v
v − vT
v
+
+
= 0;
50
60
200
vT − v vT + 60ix
+
− 1 = 0;
60
40
ix =
v − vT
;
60
Solving,
RTh =
vT = 50 V.
vT
= 50 Ω.
1A
Thus our problem is reduced to analyzing the circuit shown below.
−20
50 + Ro
2
Ro = 1.5;
400Ro
= 1.5;
Ro2 + 100Ro + 2500
1.5Ro2 − 250Ro + 3750 = 0;
.·. Ro = 16.67 Ω;
Ro = 150 Ω.
P 4.89 [a] We begin by finding the Thévenin equivalent with respect to the terminals
of Ro .
Problems
4–91
Open circuit voltage
−200 = 25i1 − 1i2 − 20i3 ;
0 = −i1 + 3i2 − 2i3 + 14i∆ ;
100 = −20i1 − 2i2 + 25i3 ;
i∆ = i1 − i3 .
Solving, i1 = −2.5 A;
i2 = 37.5 A;
i3 = 5 A;
i∆ = −7.5 A.
vTh = 20(i1 − i3 ) = 20(−7.5) = −150 V.
Now find the short-circuit current.
Note with the short circuit from a to b that i∆ is zero, hence 14i∆ is also
zero.
−200 = 5i1 − 1i2 + 0i3 ;
0 = −1i1 + 3i2 − 2i3 ;
100 = 0i1 − 2i2 + 5i3 ;
Solving, i1 = −40 A;
i2 = 0 A;
isc = i1 − i3 = −60 A.
RTh = (−150)/(−60) = 2.5 Ω.
i3 = 20 A.
4–92
CHAPTER 4. Techniques of Circuit Analysis
For maximum power transfer Ro = RTh = 2.5 Ω.
752
[b] pmax =
= 2250 W.
2.5
[c] Now use the mesh current method to analyze the original circuit with the
variable resistor set to 2.5 Ω.
25i1 − 20i2 − i4 = −200;
−20i1 + 22.5i2 − 2.5i3 = 0;
−2.5i2 + 7.5i3 − 2i4 = 100;
−i1 − 2i3 + 3i4 + 14i∆ = 0;
i∆ = i1 − i2 .
Solving, i1 = −21.25 A;
i2 = −17.5 A;
i3 = 12.5 A;
i4 = 18.75 A.
p200V = 200i1 = 200(−21.25) = −4250 W;
p100V = −100i3 = −100(12.5) = −1250 W;
pdepsource = 14i∆ i4 = 14(−21.25 + 17.5)(18.75) = −984.375 W;
X
pdev = 4250 + 1250 + 984.375 = 6484.375 W;
.·. % delivered =
2250
(100) = 34.7%.
6484.375
[d] 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:
150
i10 =
= 12 A;
2.5 + 10
Problems
4–93
p10 = 10(12)2 = 1440 W.
Thus, using a 10 Ω resistor selected from Appendix H will cause 1440 W
of power to be delivered to the load, compared to the maximum power of
2250 W that will be delivered if a 2.5 Ω resistor is used.
P 4.90 [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
v∆ =
(280 − v1 )
5 = 56 − 0.2v1 ;
25
v1 = 210 V;
v∆ = 14 V;
VTh = 280 − v∆ = 280 − 14 = 266 V.
Short circuit current
v1 − 280 v2
v2
v1
+
+
+
+ 0.5125(280) = 0;
100
10
20 400
v∆ = 280 V;
v2 + 50iφ = v1 ;
iφ =
280 v2
+
= 56 + 0.05v2 ;
5
20
4–94
CHAPTER 4. Techniques of Circuit Analysis
v2 = −968V;
v1 = −588 V;
iφ = isc = 56 + 0.05(−968) = 7.6 A;
RTh = VTh /isc = 266/7.6 = 35 Ω;
.·. Ro = 35 Ω.
[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 = −189V and v2 = −379 V; thus,
280 − 133 280 + 189
ig =
+
= 76.30 A.
5
10
p280V (dev) = (280)(76.3) = 21,364 W.
P 4.91 [a] Find the Thévenin equivalent with respect to the terminals of RL .
Open circuit voltage:
Problems
4–95
The mesh current equations are:
480 + 6(i1 − i2 ) + 40(i1 − i3 ) + 4i1
=
4i2 + 8(i2 − i3 ) + 6(i2 − i1 )
= 0;
−20iβ + 2i3 + 40(i3 − i1 ) + 8(i3 − i2 )
= 0.
0;
The dependent source constraint equation is:
iβ = i1 − i2 .
Place these equations in standard form:
−480;
i1 (6 + 40 + 4) + i2 (−6) + i3 (−40) + iβ (0)
=
i1 (−6) + i2 (4 + 8 + 6) + i3 (−8) + iβ (0)
= 0;
i1 (−40) + i2 (−8) + i3 (8 + 2 + 40) + iβ (−20)
= 0;
i1 (−1) + i2 (1) + i3 (0) + iβ (1)
= 0.
Solving, i1 = −99.6 A; i2 = −78 A;
VTh = 40(i1 − i3 ) = 48 V.
Short-circuit current:
i3 = −100.8 A;
iβ = −21.6 A.
The mesh current equations are:
480 + 6(i1 − i2 ) + 4i1
=
4i2 + 8(i2 − i3 ) + 6(i2 − i1 )
= 0;
−20iβ + 2i3 + 8(i3 − i2 )
= 0.
0;
The dependent source constraint equation is:
iβ = i1 − i2 .
Place these equations in standard form:
i1 (6 + 4) + i2 (−6) + i3 (0) + iβ (0)
=
−480;
i1 (−6) + i2 (4 + 8 + 6) + i3 (−8) + iβ (0)
=
0;
i1 (0) + i2 (−8) + i3 (8 + 2) + iβ (−20)
=
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;
VTh
48
RTh =
=
= 12 Ω.
isc
4
iβ = −18.67 A.
4–96
CHAPTER 4. Techniques of Circuit Analysis
RL = RTh = 12 Ω.
[b] pmax =
242
= 48 W.
12
P 4.92 [a] Begin by replacing the voltage source with a short circuit. This places the
12 Ω and 4 Ω resistors in parallel, so replace them with the equivalent 3 Ω
resistor. The resulting circuit is
Write a single KCL equation at the node labeled vx :
vx
vx
−4.5 +
+
= 0.
1
6+2+3
Solving,
vx = 4.125 V.
Use voltage division to find v1 :
2+3
v1 =
(4.125) = 1.875 V.
6+2+3
Return to the original circuit and replace the current source with an open
circuit to give
Write a single KCL equation at the node labeled vy :
vy
vy − 30
vy
+
+
= 0.
1 + 6 + 2 12
4
Solving,
vy = 16.875 V.
Use voltage division to find v2 :
6+1
v2 =
(16.875) = 13.125 V.
2+6+1
Problems
4–97
Thus,
v = v1 + v2 = 1.875 + 13.125 = 15 V.
[b] Begin by replacing the voltage source with a short circuit. The resulting
circuit is
Write a single KCL equation at the upper node:
v1
v1
+
+ 5 = 0.
12 + 5
2
Solving,
v1 = −8.9474 V.
Return to the original circuit and replace the current source with an open
circuit to give
Use voltage division to find v2 :
2
(180) = 18.9474 V.
12 + 5 + 2
Thus,
vo = v1 + v2 = −8.9474 + 18.9474 = 10 V.
P 4.93 45 V source acting alone:
io1 = 45/40 = 1.125 A.
4–98
CHAPTER 4. Techniques of Circuit Analysis
10 V source acting alone:
is =
30
10
=
A;
40 + 40/3
160
io2 = −
30 40
·
= −0.125 A.
160 60
8 A current source acting alone. Begin by performing a delta-to-wye transform
on the right side:
io3 =
(8)(30)
= 6 A;
40
io = io1 + io2 + io3 = 1.125 − 0.125 + 6 = 7 A.
P 4.94 6 A source:
30 Ωk5 Ωk60 Ω = 4 Ω;
.·. io1 =
20
(6) = 4.8 A.
20 + 5
Problems
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 Voltage source acting alone:
io1 =
180
180
+
= 2 + 3 = 5 A;
90
40 + 100k25
i80Ω =
25
(3) = 0.6 A;
125
so vo1 = 80i80Ω = 48 V.
4–99
4–100
CHAPTER 4. Techniques of Circuit Analysis
Current source acting alone:
v2
v2 − v3
+ 7.5 +
= 0;
80
20
v3 − v2
v3
v3
+
+
= 0.
25
20
40
Solving, v2 = −184 V = vo2 ;
i40 =
v3
= −2 A;
40
i30 =
7.5(60)
= 5 A;
90
v3 = −80 V;
io2 = −i30 − i40 = −5 + 2 = −3 A;
.·. vo = vo1 + vo2 = 48 − 184 = −136 V;
io = io1 + io2 = 5 − 3 = 2 A.
P 4.96 [a] By hypothesis i0o + i00o = 3.5 mA.
i000
o = −5
(2)
= −1.25mA;
(8)
.·. io = 3.5 − 1.25 = 2.25mA.
Problems
4–101
[b] With all three sources in the circuit write a single node voltage equation.
vb
vb − 8
+
+ 0.005 − 0.01 = 0;
6000
2000
.·. vb = 13.5V.
io =
vb
= 2.25mA.
6000
P 4.97 [a] Begin by replacing the voltage source with a short circuit. Then identify
the meshes and label their currents:
Write two KVL equations for the meshes whose currents are unknown:
6i1 + 8(i1 − i2 ) = 0;
2(i2 + 3i1 ) + 4(i2 + 5) + 8(i2 − i1 ) = 0.
Solving,
i1 = −0.889 A;
i2 = −1.556 A.
Use Ohm’s law to find v1 :
v1 = 4(5 + i2 ) = 13.778 V.
Return to the original circuit and replace the current source with an open
circuit. Identify and label essential nodes:
Write two KCL equations and a dependent source constraint equation:
vx − 50 vx vx − v2
+
+
− 3i1 = 0;
6
8
2
v2 − vx v2
+
+ 3i1 = 0;
2
4
4–102
CHAPTER 4. Techniques of Circuit Analysis
i1 =
50 − vx
.
6
Solving,
vx = 26.667 V;
v2 = 2.222 V.
Thus,
v = v1 + v2 = 13.778 + 2.222 = 16 V.
[b] Begin by replacing the voltage source with a short circuit. Identify and
label essential nodes:
Write a KCL equation, a supernode equation, and a supernode constraint
equation:
va − vb
va
+
= 0;
−4.8 +
7.5
2.5
vb − va
vb
vc
vc
+
+
+
= 0;
2.5
10 2.5
1
va
v c − v b = ix =
.
7.5
Solving,
va = 9.8276 V;
vb = 1.10345 V;
vc = 2.4138 V.
So,
v1 = vb = 1.10345 V.
Return to the original circuit and replace the current source with an open
circuit. Identify and label essential nodes:
Write a supernode equation and a supernode constraint equation:
vx
vx
vy
vy − 12
+
+
+
= 0;
7.5 + 2.5 10 2.5
1
vx
v y − v x = ix =
.
2.5 + 7.5
Problems
Solving,
vx = 6.89655 V;
vy = 7.5862 V.
So,
v2 = vx = 6.89655 V.
Therefore,
vo = v1 + v2 = 1.10345 + 6.89655 = 8 V.
P 4.98 Voltage source acting alone:
vo1 − 25
vo1
vo1 − 25
+
− 2.2
= 0;
4000
20,000
4000
Simplifying
5vo1 − 125 + vo1 − 11vo1 + 275 = 0;
.·. vo1 = 30V.
Current source acting alone:
vo2
vo2
vo2
+
+ 0.005 − 2.2
= 0;
4000 20,000
4000
Simplifying
5vo2 + vo2 + 100 − 11vo2 = 0;
.·. vo2 = 20V.
vo = vo1 + vo2 = 30 + 20 = 50V.
4–103
4–104
CHAPTER 4. Techniques of Circuit Analysis
P 4.99 [a]
voc = vTh = 120 V;
Therefore
[b] iL =
RTh =
iL =
40
= 1 A;
40
iL =
120 − 40
80
=
.
RTh
RTh
80
= 80 Ω.
1
vo
vTh − vo
=
.
RL
RTh
vTh
vTh − vo
RTh =
=
− 1 RL .
vo /RL
vo
Therefore
P 4.100 [a]
v − v1
v
v − v2
+ +
= 0;
2xr
R 2r(L − x)
"
#
v=
v1 RL + xR(v2 − v1 )
.
RL + 2rLx − 2rx2
1
1
v1
v2
1
+ +
=
+
;
v
2xr R 2r(L − x)
2xr 2r(L − x)
[b] Let D = RL + 2rLx − 2rx2 .
(RL + 2rLx − 2rx2 )R(v2 − v1 ) − [v1 RL + xR(v2 − v1 )]2r(L − 2x)
dv
=
;
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 )
Problems
4–105
Solving for the roots of the quadratic yields
x=
[c] x =


s


s
L
−v1 ±
v2 − v1 
L
−v1 ±
v2 − v1 
v2 = 2000 V,
v1 v2 −
v1 v2 −
R
(v2 − v1 )2 .

2rL


R
(v1 − v2 )2 ;

2rL
v1 = 1800 V,
r = 4 × 10−5 Ω/m;


L = 20 km;
R = 7.1 Ω;
20,000
L
=
= 100;
v2 − v1
2000 − 1800
v1 v2 = 3.6 × 106 ;
7.1(200)2
R
2
(v2 − v1 ) =
= 177,500;
2rL
(8 × 10−5 )(20 × 103 )
x = 100{−1800 ±
q
3.6 × 106 − 177,500};
= 100{−1800 ± 1850} = 100(50) = 5000 m.
[d] vmin
=
v1 RL + R(v2 − v1 )x
;
RL + 2rLx − 2rx2
=
(1800)(7.1)(20 × 103 ) + 7.1(200)(5000)
(7.1)(20,000) + 8 × 10−5 (20,000)(5000) − 8 × 10−5 (25 × 106 )
=
1775 V .
P 4.101
100 = 6ia − 1ib + 0ic − 2id − 2ie + 0if − 1ig
0 = −1ia + 4ib − 2ic + 0id + 0ie + 0if + 0ig
4–106
CHAPTER 4. Techniques of Circuit Analysis
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
A calculator 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
pabs
=
100 + 400 + 50 + 200 + 300 + 450 + 0 + 450 + 300
+ 200 + 50 + 400 + 100 = 3000 W
X
pgen
=
100ia = 100(30) = 3000 W (CHECKS)
P 4.102
110 = 27.5ia − 0.3ib − 27ic ;
110 = −0.3ia + 36.5ib − 36id ;
0 = −27ia + 54.5ic − 0.3id − 27ie ;
0 = −36ib − 0.3ic + 72.5id − 36ie ;
0 = −27ic − 36id + 81ie .
Problems
4–107
Solving,
ia = 19.187 A; ib = 17.363 A; ic = 15.275 A;
id = 14.388 A; ie = 11.487 A.
So,
v1 = 27(15.275 − 11.487) = 102.276 V;
v2 = 36(14.388 − 11.487) = 104.436 V;
v3 = 18(11.487) = 206.766 V.
P 4.103 [a] In studying the circuit in Fig. P4.103 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
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
16 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
2
3
−8vy + ia + ib − 16 + 2ib + 2(ic − id ) + 1ic = 0.
4
3
Note that vy = 2ib /3; make that substitution and multiply the equation
by 12:
2
−96 ib + 9ia + 8ib − 192 + 24ib + 24(ic − id ) + 12ic = 0
3
or
9ia − 32ib + 36ic − 24id = 192.
Now note:
id = 2ix ;
and
ix = ic − id
4–108
CHAPTER 4. Techniques of Circuit Analysis
so
.·. 3id = 2ic .
id = 2(ic − id )
Now use the following constraints:
vx
= ic − ib
3
Therefore
and
3
v x = ia .
4
ia = 4ic − 4ib .
Finally,
ia − ic = 2.
In standard form:
9ia − 32ib + 36ic − 24id = 192;
0ia + 0ib + 2ic − 3id = 0;
1ia + 4ib − 4ic + 0id = 0;
1ia + 0ib − 1ic + 0id = 2;
Solving,
ia = 33.6 A;
ib = 23.2 A;
ic = 31.6 A;
id = 21.067 A.
p16V = −16ib = −16(23.2) = −371.2 W.
Therefore, the 16 V source delivers 371.2 W of power.
P 4.104 v1 =
R1 R3 R4
R1 [R2 (R3 + R4 ) + R3 R4 ]
Ig2 −
Ig1 ;
(R1 + R2 )(R3 + R4 ) + R3 R4
(R1 + R2 )(R3 + R4 ) + R3 R4
−R1 [R2 (R3 + R4 ) + R3 R4 ]
dv1
=
;
dIg1
(R1 + R2 )(R3 + R4 ) + R3 R4
dv1
R1 R3 R4
=
.
dIg2
(R1 + R2 )(R3 + R4 ) + R3 R4
v2 =
R3 R4 (R1 + R2 )
R1 R3 R4
Ig2 −
Ig1 ;
(R1 + R2 )(R3 + R4 ) + R3 R4
(R1 + R2 )(R3 + R4 ) + R3 R4
dv2
−R1 R3 R4
=
;
dIg1
(R1 + R2 )(R3 + R4 ) + R3 R4
dv2
R3 R4 (R1 + R2 )
=
.
dIg2
(R1 + R2 )(R3 + R4 ) + R3 R4
Problems
P 4.105 From the solution to Problem 4.104 we have
−25[5(125) + 3750]
175
dv1
=
=−
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 = 13 − 12 = 1 A;
175
175
.·. ∆v1 = (−
)(1) = −
= −14.583 V.
12
12
Thus, v1 = 25 − 14.583 = 10.417 V.
Also,
∆v2 = (−12.5)(1) = −12.5 V.
Thus, v2 = 90 − 12.5 = 77.5 V.
The PSpice solution is
v1 = 10.417 V
and
v2 = 77.5 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
(50)(75)(30)
dv2
=
= 15 V/A.
dIg2
30(125) + 3750
By hypothesis, ∆Ig2 = 15 − 16 = −1 A;
.·. ∆v1 = (12.5)(−1) = −12.5 V.
Thus, v1 = 25 − 12.5 = 12.5 V.
4–109
4–110
CHAPTER 4. Techniques of Circuit Analysis
Also,
∆v2 = (15)(−1) = −15 V.
Thus, v2 = 90 − 15 = 75 V .
The PSpice solution is
v1 = 12.5 V
and
v2 = 75 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 = 13 − 12 = 1 A;
∆Ig2 = 15 − 16 = −1 A.
Therefore,
∆v1 = −
175
− 12.5 = −27.0833 V;
12
∆v2 = −12.5 − 15 = −27.5 V.
Hence
v1 = 25 − 27.0833 = −2.0833 V;
v2 = 90 − 27.5 = 62.5 V.
The PSpice solution is
v1 = −2.0833 V
and
v2 = 62.5 V.
These values are in agreement with our predicted values.
Problems
4–111
P 4.108 By hypothesis,
∆R1 = 22.5 − 25 = −2.5 Ω;
∆R2 = 5.5 − 5 = 0.5 Ω;
∆R3 = 45 − 50 = −5 Ω;
∆R4 = 82.5 − 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 = 20.083 V.
∆v2 = 0.5(−2.5) + 6.5(0.5) + 0.54(−5) + 0.24(7.5) = 1.1 V;
.·. v2 = 90 + 1.1 = 91.1 V.
The PSpice solution is
v1 = 19.882 V
and
v2 = 90.7415 V.
Note our predicted values are within a fraction of a volt of the actual values.
The Operational Amplifier
Assessment Problems
AP 5.1 [a] Derive the expression for the output voltage using circuit analysis:
vn = vp = 0;
.·.
0 − vs
0 − vo
+
= 0;
10,000 40,000
KCL at vn :
vo = (−40/10)vs ,
vs ( V)
−6
vo ( V)
24(sat. at 15)
so
vo = −4vs .
−3.5 −1.25
1
14
−4 −9.6 −21.6(sat. at −15)
5
2.4
5.4
Two of the values, −6 V and 5.4 V, cause the op amp to saturate.
[b] Use the negative power supply value to determine the largest input
voltage:
−15 = −4vs ,
vs = 3.75 V.
Use the positive power supply value to determine the smallest input
voltage:
15 = −4vs ,
Therefore
vs = −3.75 V;
− 3.75 V ≤ vs ≤ 3.75 V.
AP 5.2 From Eq. 5.4 and Assessment Problem 5.1
vo = (−Rf /Ri )vs = (−Rx /10,000)vs = (−Rx /10,000)(0.8)
= −0.8Rx /10,000 = −8 × 10−5 Rx .
Use the negative power supply value to determine one limit on the value of Rx :
−8 × 10−5 Rx = −15
so
Rx = 15/8 × 10−5 = 187.5 kΩ.
5–1
5–2
CHAPTER 5. The Operational Amplifier
Now use the positive power supply value to determine the upper limit on the
value of Rx :
−8 × 10−5 Rx = 15
so
Rx = −15/8 × 10−5 = −187.5 kΩ.
Since we cannot have negative resistor values, the lower limit for Rx is 0,
which is a short circuit. Therefore,
0 ≤ Rx ≤ 187.5 kΩ.
AP 5.3 [a] This is an inverting summing amplifier so
vo = (−Rf /Ra )va + (−Rf /Rb )vb = −(400/25)va − (400/40)vb = −16va − 10vb
Substituting the values for va and vb :
vo = −16(0.25) − 10(0.1) = −4 − 1 = −5 V.
[b] Substitute the value for vb into the equation for vo from part (a) and use
the negative power supply value:
vo = −16va − 10(0.1) = −16va − 1 = −12 V.
Therefore 16va = 11,
so va = 0.6875 V.
[c] Substitute the value for va into the equation for vo from part (a) and use
the negative power supply value:
vo = −16(0.25) − 10vb = −4 − 10vb = −12 V;
Therefore 10vb = 8,
so vb = 0.8 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 = −16va + 10vb = −4 + 1 = −3 V.
Repeat part (b):
vo = −16va + 1 = −12 V;
16va = 13,
va = 0.8125 V.
Repeat part (c), using the value of the positive power supply:
vo = −4 + 10vb = 16 V;
AP 5.4 [a] vp = vn =
10vb = 20;
45
vg = 0.6vg ;
75
0.6vg 0.6vg − vo
.·.
+
= 0;
15
48
.·. vo = 2.52vg = 2.52(3), vo = 7.56 V.
vb = 2.0 V.
Problems
5–3
[b] vo = 2.52vg = ±10.
vg = ±3.97 V,
[c]
−3.97 V ≤ vg ≤ 3.97 V.
0.6vg 0.6vg − vo
+
= 0;
15
Rf
0.6Rf
+ 0.6 vg = vo = ±10;
15
.·. 3Rf + 45 = ±150;
3Rf = 150 − 45;
Rf = 35 kΩ.
AP 5.5 [a] Since this is a difference amplifier with Ra /Rb = Rc /Rd , we can use the
expression for the output voltage in terms of the input voltages and the
resistor values given in Eq. 5.10:
vo =
Rb
36,000
(vb − va ) =
(vb − va ) = 3(vb − va ).
Ra
12,000
Substitute in the value for va :
vo = 3(vb − va ) = 3vb − 6.
Set this expression for vo to the positive power supply value:
3vb − 6 = 12 V so vb = 6 V.
Now set the expression for vo to the negative power supply value:
3vb − 6 = −12 V so vb = −2 V;
Therefore
− 2 V ≤ vb ≤ 6 V.
[b] Now Ra /Rb 6= Rc /Rd so we need to use Eq. 5.8:
vo =
18(48)
vb − 3va = 2vb − 3va .
12(36)
Now substitute the value for va :
vo = 2vb − 6.
Set this expression for vo to the positive power supply value:
2vb − 6 = 12 V so vb = 9 V.
Now set the expression for vo to the negative power supply value:
2vb − 6 = −12 V so vb = −3 V;
Therefore
− 3 V ≤ vb ≤ 9 V.
5–4
CHAPTER 5. The Operational Amplifier
AP 5.6 Acm =
Adm =
(20)(50) − (50)Rx
;
20(50 + Rx )
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;
If we use −1000 Rx = 20.07 kΩ;
19.93 kΩ ≤ Rx ≤ 20.07 kΩ.
AP 5.7 [a] Replace the op amp with the more realistic model of the op amp from Fig.
5.18:
Write the node voltage equation at the left hand node:
vn
vn − vg vn − vo
+
+
= 0.
480,000
16,000
24,000
Multiply both sides by 480,000 and simplify:
vn + 30vn − 30vg + 20vn − 20vo = 0 so 51vn − 20vo = 30vg .
Write the node voltage equation at the right hand node:
vo − 100,000(−vn ) vo − vn
+
= 0.
2000
24,000
Multiply through by 24,000 and simplify:
12vo + 12 × 105 vn + vo − vn = 0 so 12 × 105 vn + 13vo = 0.
Problems
5–5
Use Cramer’s method to solve for vo :
−20
51
∆=
12 × 105 13
No =
51
30vg
5
12 × 10
= 24,000,663;
= −36 × 106 vg ;
0
No
= −1.49996vg ;
∆
vo =
so
vo
= −1.49996.
vg
[b] Use Cramer’s method again to solve for vn :
N1 =
30vg −20
0
= 390vg ;
13
N1
= 1.65 × 10−5 vg ;
∆
vn =
vg = 1 V,
vn = 16.25 µV.
[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:
ig =
vg − vn
vg − 1.625 × 10−5 vg
=
.
16,000
16,000
Solve for the ratio of vg to ig to get the input resistance:
Rg =
16,000
vg
=
= 16,000.26 Ω.
ig
1 − 1.625 × 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
24,000
=−
= −1.5.
vg
16,000
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 = 16,000 Ω.
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 − 2.5 vn − vo
+
+ 0 = 0.
16,000
80,000
But vp = 0 and vn = vp = 0. Thus,
vo
−2.5
−
= 0 so vo = −12.5 V.
16,000 80,000
Note that vo is between the two power supply voltages, so our
assumption that the op amp is in its linear operating region is confirmed.
P 5.2
[a] Let the value of the voltage source be vs :
vn − vs vn − vo
+
+ 0 = 0.
16,000
80,000
But vn = vp = 0. Therefore,
vo = −
80,000
vs = −5vs .
16,000
When vs = 0.4 V, vo = −5(0.4) = −2 V.
When vs = 2.0 V, vo = −5(2.0) = −10 V.
When vs = 3.5 V, vo = −5(3.5) = −17.5 V; saturates at vo = −15 V.
When vs = −0.6 V, vo = −5(−0.6) = 3 V.
When vs = −1.6 V, vo = −5(−1.6) = 8 V.
When vs = −2.4 V, vo = −5(−2.4) = 12 V; saturates at vo = 10 V.
10
[b] −5vs = 10
so
vs =
= −2 V;
−5
−15
−5vs = −15
so
vs =
= 3 V.
−5
Problems
5–7
The range of source voltages that avoids saturation is −2 V ≤ vs ≤ 3 V.
P 5.3
vb − va
vb − vo
+
= 0,
20,000
160,000
therefore vo = 9vb − 8va .
[a] va = 1.5 V,
vb = 0 V,
vo = −12 V.
[b] va = 3.0 V,
vb = 0 V,
vo = −24 V (sat).
[c] va = 1.0 V,
vb = 2 V,
vo = 10 V.
[d] va = 4.0 V,
vb = 6 V,
vo = 22 V (sat).
[e] If vb = 4.5 V,
vo = 40.5 − 8va = ±18;
.·. 2.8125 ≤ va ≤ 7.3125 V.
P 5.4
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 5.5
vo = (0.001)(9000) = 9 V;
i6kΩ =
9
= 1.5 mA;
6000
i15kΩ =
i9kΩ =
9
= 0.6 mA;
15,000
9
= 1 mA;
9000
.·. io = −0.6 − 1.5 − 1 = −3.1 mA,
P 5.6
[a] ia =
0.120
= 20 µA;
6000
va = −20 × 103 ia = −400 mV.
[b]
va
va
va − vo
+
+
= 0;
60,000 20,000 240,000
.·. vo = 17va = −6.8 V.
[c] ia = 20 µA.
−vo
va − vo
[d] io =
+
= 111.67 µA.
80,000 240,000
P 5.7
vp =
3000
(3) = 1 V = vn ;
3000 + 6000
vn − 5 vn − vo
+
= 0;
10,000
5000
(1 − 5) + 2(1 − vo ) = 0;
5–8
CHAPTER 5. The Operational Amplifier
vo = −1.0 V;
iL =
vo
1
=−
= −250 × 10−6 ;
4000
4000
iL = −250 µA.
P 5.8
[a] Replace the combination of vg , 3.2 kΩ, and the 4.8 kΩ resistors with its
Thévenin equivalent.
Then vo =
−[30,000 + σ170,000]
(0.30).
1920
At saturation vo = −10 V;
−
therefore
30,000 + σ170,000
(0.3) = −10,
1920
or σ = 0.2.
Thus for 0 ≤ σ < 0.20 the operational amplifier will not saturate.
−(30,000 + 20,400)
[b] When σ = 0.12,
vo =
(0.30) = −7.875 V.
1920
vo
vo
Also
+
+ io = 0;
180,000 50,400
.·. io = −
P 5.9
vo
vo
7.875
7.875
−
=
+
mA = 200 µA.
180,000 50,400
180,000 50,400
[a] Let v∆ be the voltage from the potentiometer contact to ground. Then
0 − vg 0 − v∆
+
= 0;
4000
20,000
−5vg − v∆ = 0,
.·. v∆ = −5(40 × 10−3 ) = −0.2 V;
v∆
v∆ − 0
v∆ − vo
+
+
= 0;
αR∆ 20,000 (1 − α)R∆
v∆
v∆ − vo
+ 6v∆ +
= 0;
α
1−α
v∆
1
1
vo
+6+
=
;
α
1−α
1−α
Problems
5–9
#
"
(1 − α)
.
.·. vo = −0.2 1 + 6(1 − α) +
α
When α = 0.25,
vo = −0.2(1 + 4.5 + 3) = −1.7 V.
When α = 0.8,
vo = −0.2(1 + 1.2 + 0.25) = −0.49 V;
.·. −1.7 V ≤ vo ≤ −0.49 V.
"
#
(1 − α)
[b] −0.2 1 + 6(1 − α) +
= −12;
α
α + 6α(1 − α) + (1 − α) = 60α;
α + 6α − 6α2 + 1 − α = 60α;
.·. 6α2 + 54α − 1 = 0
P 5.10
α∼
= 0.0185.
so
30,000
=6
Rin
so
Rin =
30,000
= 5000 = 5 kΩ,
6
[b] −6vin = 15
so
vin =
15
= −2.5 V;
−6
[a]
−6vin = −15
.·.
[c] −
so
vin =
15
= 2.5 V;
6
−2.5 V ≤ vin ≤ 2.5 V.
Rf
(5) = −15
5000
so
Rf = 15 kΩ;
vo
Rf
15,000
=
=
= 3.
vin
Rin
5000
The amplifier has a gain of 3.
P 5.11
[a] The gain of an inverting amplifier is the negative of the ratio of the
feedback resistor to the input resistor. If the gain of the inverting
amplifier is to be 4, the feedback resistor must be 4 times as large as the
input resistor. There are many possible designs that use resistor values
chosen from Appendix H. We present one here that uses 18 kΩ and 22 kΩ
5–10
CHAPTER 5. The Operational Amplifier
resistors in the feedback path and a 10 kΩ in the forward path. The
resulting circuit is shown here:
[b] To amplify signals in the range −5 V to 4 V 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.
−4(−5) = 20 V
− 4(4) = −16 V.
and
Thus, the power supplies should have values of −16 V and 20 V.
P 5.12
[a] This circuit is an example of an inverting summing amplifier.
180
180
180
[b] vo = −
va −
vb −
vc = −4.5 − 9 + 7.5 = −6 V.
20
30
60
[c] vo = −13.5 − 3vc = ±9;
.·. vc = −7.5 V when vo = 9 V;
vc = −1.5 V when vo = −9 V;
.·. −7.5 V ≤ vc ≤ −1.5 V.
P 5.13
[a] Write a KCL equation at the inverting input to the op amp:
vd − vc
vd
vd − vo
vd − va vd − vb
+
+
+
+
= 0;
55,000
66,000
220,000 550,000 330,000
vo = 14.1vd − 6va − 5vb − 1.5vc = 141 − 96 − 60 + 9 = −6 V.
[b] vo = 141 − 96 − 5vb + 9 = 54 − 5vb
54 − 5vb = −12
54 − 5vb = 12
so
so
vb = 13.2 V;
vb = 8.4 V;
.·. 8.4 V ≤ vb ≤ 13.2 V.
P 5.14
[a]
vd − va vd − vb
vd − vc
vd
vd − vo
+
+
+
+
= 0;
55,000
66,000
220,000 550,000
Rf
12vd − 12va + 10vd − 10vb + 3vd − 3vc + 1.2vd +
26.2vd +
660
660
vd − 12va − 10vb − 3vc =
vo
Rf
Rf
660
660
vd =
vo (Rf in kΩ);
Rf
Rf
Problems
262 +
6600
660
− 192 − 120 + 18 =
vo ;
Rf
Rf
6600 − 32Rf = 660vo
vo = ±12
.·.
but
32Rf = 6600 − 660vo ;
so
Rf > 0;
32Rf = 6600 − 660(−12)
[b] vo = −12 V;
io +
so
Rf = 453.75 kΩ.
A KCL equation at the output node gives
−12 − 10
−12
+
= 0;
33,000
453,750
.·. io = 412.12 µA.
"
P 5.15
#
Rf
Rf
Rf
vo = −
(0.2) +
(0.15) +
(0.4) ;
4000
5000
20,000
−6 = −0.1 × 10−3 Rf ;
P 5.16
Rf = 60 kΩ;
.·. 0 ≤ Rf ≤ 60 kΩ.
[a]
150,000
=5
Ra
so
Ra =
150,000
= 30 kΩ;
5
150,000
= 10
Rb
so
Rb =
150,000
=3
Rc
so
Rc =
150,000
= 15 kΩ;
10
150,000
= 50 kΩ.
3
[b] −5va − 10(−3) − 3(4) = −5va + 18;
−5va + 18 = −12
−5va + 18 = 12
so
so
5va = −30
thus
va =
30
= 6 V;
5
− 5va = −6
thus
va =
6
= 1.2 V.
5
5–11
5–12
CHAPTER 5. The Operational Amplifier
Thus,
1.2 V ≤ vb ≤ 6 V.
P 5.17
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:
vo = −
60,000
60,000
60,000
60,000
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 5.18
[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.
P 5.19
[a] This circuit is an example of the non-inverting amplifier.
Problems
5–13
[b] Use voltage division to calculate vp :
vp =
75,000
3vs
vs =
.
25,000 + 75,000
4
Write a KCL equation at vn = vp = 3vs /4:
3vs /4 3vs /4 − vo
+
= 0.
8000
32,000
Solving,
vo = 12vs /4 + 3vs /4 = 3.75vs .
[c] 3.75vs = 15
3.75vs = −9
so
vs = 4 V;
vs = −2.4 V;
so
Thus, −2.4 V ≤ vs ≤ 4 V.
P 5.20
[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
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:
60,000
(0.400) = 0.32 V.
vp =
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.
5–14
P 5.21
CHAPTER 5. The Operational Amplifier
[a] From Eq. 5.7,
vo =
Rs + Rf
vg
Rs
So,
Rf
= 4.
Rs
Thus,
Rs =
so
vo
Rf
=1+
= 5.
vg
Rs
Rf
100,000
=
= 25 kΩ.
4
4
[b] vo = 5vg .
When vg = −3 V, vo = 5(−3) = −15 V.
When vg = 2 V, vo = 5(2) = 10 V.
The power supplies can be set at 10 V and −15 V.
P 5.22
[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:
[b] vo = 4vg = 12
so
vo = 4vg = −12
vg = 3 V;
so
vg = −3 V.
Therefore,
−3 V ≤ vg ≤ 3 V.
P 5.23
[a] This circuit is an example of a non-inverting summing amplifier.
Problems
[b] Write a KCL equation at vp and solve for vp in terms of vs :
vp − vs
vp + 4
+
= 0;
12,000
48,000
4vp − 4vs + vp + 4 = 0
so
vp = 4vs /5 − 4/5.
Now write a KCL equation at vn and solve for vo :
vn − vo
vn
+
=0
so
vo = 5vn .
10,000
40,000
Since we assume the op amp is ideal, vn = vp . Thus,
vo = 5(4vs /5 − 4/5) = 4vs − 4.
[c] 4vs − 4 = 10
so
4vs − 4 = −10
vs = 3.5 V;
so
vs = −1.5 V;
Thus, −1.5 V ≤ vs ≤ 3.5 V.
P 5.24
[a]
vp − va vp − vb vp − vc
+
+
= 0;
Ra
Rb
Rc
Rb Rc
Ra Rc
Ra Rb
va +
vb +
vc ,
.·. vp =
D
D
D
where D = Rb Rc + Ra Rc + Ra Rb .
vn
vn − vo
+
= 0;
20,000
Rf
!
Rf
+ 1 vn = vo .
20,000
Let
Rf
+ 1 = k.
20,000
vo = kvn = kvp ;
kRb Rc
kRa Rc
kRa Rb
va +
vb +
vc ;
.·. vo =
D
D
D
kRb Rc
= 4;
D
kRa Rc
= 1;
D
Thus,
so Rc =
D
;
1000k
kRb (D/1000k)
= 4 and Rb = 4000 = 4 kΩ;
D
kRa Rb
= 2;
D
so Rb =
2D
;
1000k
5–15
5–16
CHAPTER 5. The Operational Amplifier
kRc (2D/1000k)
= 4 and Rc = 2000 = 2 kΩ;
D
Thus,
.·. D = (4000)(2000) + (1000)(2000) + (1000)(4000) = 14 × 106 ;
4D
(4)(14) × 106
.·. k =
=
= 7;
Rb Rc
(4)(2) × 106
.·.
Rf
+ 1 = 7,
20,000
Rf = 120 kΩ.
[b] vo = 4(0.75) + 1.0 + 2(1.5) = 7 V;
vn = vo /7 = 1 V = vp ;
P 5.25
[a]
ia =
va − vp
0.75 − 1
=
= −250 µA;
1000
1000
ib =
1−1
vb − vp
=
= 0 µA;
4000
4000
ic =
vc − vp
1.5 − 1
=
= 250 µA.
2000
2000
vp − va vp − vb vp − vc
vp
+
+
+
= 0;
Ra
Rb
Rc
Rg
Rb Rc Rg
Ra Rc Rg
Ra Rb Rg
.·. vp =
va +
vb +
vc ,
D
D
D
where D = Rb Rc Rg + Ra Rc Rg + Ra Rb Rg + Ra Rb Rc .
vn − vo
vn
+
= 0;
Rs
Rf
vn
1
1
+
Rs Rf
!
=
vo
;
Rf
Rf
.·. vo = 1 +
vn = kvn ,
Rs
Rf
where k = 1 +
.
Rs
vp = vn ;
.·. vo = kvp ;
or
vo =
kRg Rb Rc
kRg Ra Rc
kRg Ra Rb
va +
vb +
vc .
D
D
D
Problems
kRg Rb Rc
=6
D
kRg Ra Rc
=3
D
6
Rb
= =2
.·.
Ra
3
Since
kRg Ra Rb
= 4;
D
Rc
3
= = 0.75
Rb
4
Ra = 1 kΩ,
5–17
Rb = 2 kΩ
Rc
6
= = 1.5.
Ra
4
and
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
135 × 109
= 15
9 × 109
k=
Rf
;
.·. 15 = 1 +
Rs
Rf
= 14;
Rs
Rf = (14)(15,000) = 210 kΩ.
[b] vo = 6(0.5) + 3(2.5) + 4(1) = 14.5 V;
vn = vp =
P 5.26
vo
14.5
=
= 0.967 V;
k
15
is =
vn
0.967
=
= 64.44 µA.
15,000
15,000
io =
vn − vo
vo
−
= −4.46 mA.
210,000 3300
[a] This is a difference amplifier circuit.
[b] Use Eq. 5.8 with Ra = 10 kΩ, Rb = 20 kΩ, Rc = 15 kΩ, Rd = 25 kΩ, and
vb = 8 V:
vo =
[c]
Rd (Ra + Rb )
Rb
25(10 + 20)
20
vb −
(8) − va = 15 − 2va .
va =
Ra (Rc + Rd )
Ra
10(15 + 25)
10
Rf
25,000(10,000 + Rf )
(8) −
(1.625)
10,000(15,000 + 25,000)
10,000
=
5(10,000 + Rf ) 1.625Rf
3.375Rf
−
=5+
;
10,000
10,000
10,000
5+
3.375Rf
=9
10,000
5+
3.375Rf
= −9
10,000
so
so
Rf =
(9 − 5)10,000
= 11,852 Ω.
3.375
Rf < 0 which is not a possible solution.
5–18
P 5.27
CHAPTER 5. The Operational Amplifier
[a] vo =
33(100)
Rd (Ra + Rb )
Rb
va =
vb −
(0.90) − 4(0.45);
Ra (Rc + Rd )
Ra
20(80)
vo = 1.8563 − 1.8 = 56.25 mV.
(0.90)(33)
= 371.25 mV;
80
[b] vn = vp =
ia =
(450 − 371.25)10−3
= 3.9375 µA;
20 × 103
Ra =
va
450 × 10−3
=
= 114.3 kΩ.
ia
3.9375 × 10−6
[c] Rin b = Rc + Rd = 80 kΩ.
P 5.28
vp =
20,000
(−4) = −0.8 V = vn ;
100,000
−0.8 + 4 −0.8 − vo
+
= 0;
2000
Rf
.·. vo = 0.0016Rf − 0.8.
vo = 20 V;
Rf = 13 kΩ;
vo = −20 V;
.·. Rf = 13 kΩ.
Rf ≥ 0,
But
P 5.29
Rf = −12 kΩ.
vp = 1000ib
1000ib 1000ib − vo
+
− ia = 0;
Ra
Rf
.·.
.·.
1
1
+
Ra Rf
1000ib
1000ib
Rf
1+
Ra
!
− ia =
vo
;
Rf
− R f ia = v o .
By hypopthesis, vo = 5000(ib − ia ). Therefore,
Rf = 5 kΩ
(use two 10 kΩ resistors in parallel).
Rf
1000 1 +
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.
Problems
P 5.30
vo =
5–19
Rd (Ra + Rb )
Rb
va ;
vb −
Ra (Rc + Rd )
Ra
By hypothesis: Rb /Ra = 5;
Rc + Rd = 600 kΩ;
Rd (Ra + 5Ra )
=2
.·.
Ra 600,000
Rd = 200 kΩ;
so
Rd (Ra + Rb )
= 2.
Ra (Rc + Rd )
Rc = 400 kΩ.
Combine a 180 kΩ resistor and a 220 kΩ resistor in series for Rc ; combine two
100 kΩ resistors in series for Rd . Also, when vo = 0 we have
vn − va
vn
+
= 0;
Ra
Rb
Ra
.·. vn 1 +
Rb
ia =
= va ;
vn = (5/6)va .
va − (5/6)va
1 va
=
;
Ra
6 Ra
.·. Ra = 3 kΩ;
Rin =
va
= 6Ra = 18 kΩ;
ia
Rb = 15 kΩ.
Combine a 1.2 kΩ resistor and a 1.8 kΩ resistor in series for Ra ; use a 15 kΩ
resistor for Rb .
P 5.31
[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,
in = 0.
Ra
Rb
Therefore
Rb
vo0
va
=− ,
vo0 = − va .
Rb
Ra
Ra
Assume vb is acting alone. Replace va with a short circuit. Now
vb Rd
vp = vn =
;
Rc + Rd
vn − vo00
vn
+
+ in = 0,
Ra
Rb
1
1
+
Ra Rb
vo00 =
in = 0;
Rd
vo00
vb −
= 0;
Rc + Rd
Rb
Rb
+1
Ra
vo = vo0 + vo00 =
Rd
Rd
vb =
Rc + Rd
Ra
Rd
Ra
Ra + Rb
vb ;
Rc + Rd
Ra + Rb
Rb
vb −
va .
Rc + Rd
Ra
5–20
CHAPTER 5. The Operational Amplifier
Rd
[b]
Ra
Ra + Rb
Rc + Rd
=
Rb
,
Ra
Rd Ra = Rb Rc ,
When
Rd
Ra
therefore Rd (Ra + Rb ) = Rb (Rc + Rd );
therefore
Ra + Rb
Rc + Rd
=
Eq. 5.8 reduces to vo =
P 5.32
Ra
Rc
=
.
Rb
Rd
Rb
.
Ra
Rb
Rb
Rb
vb −
va =
(vb − va ).
Ra
Ra
Ra
[a]
vp
vp − vc vp − vd
+
+
= 0;
60,000
20,000
30,000
.·. 6vp = 3vc + 2vd = 6vn .
vn − va vn − vb
vn − vo
+
+
= 0;
10,000
15,000
375,000
.·. vo
=
63.5vn − 37.5va − 25vb
=
63.5[(1/2)vc + (1/3)vd ] − 37.5va − 25vb
=
63.5(0.1 + 0.2) − 37.5(0.4) − 25(0.8) = −15.95 V.
[b] vo = 31.75vc + 21.167(0.6) − 37.5(0.4) − 25(0.8);
±20 = 31.75vc − 22.3;
.·. vc = 1332.28 mV
and
vc = 72.44 mV;
.·. 72.44 ≤ vc ≤ 1332.28 mV.
P 5.33
Ra = Rc = 1.5 kΩ;
Then, Acm =
Rb = 12 kΩ.
1.5Rd − 1.5(12)
;
1.5(1.5 + Rd )
Adm =
Rd (13.5) + 12(1.5 + Rd )
.
2(1.5)(1.5 + Rd )
Problems
Set the CMRR equal to 100 and solve for Rd :
Adm
CMRR =
=
Acm
Rd (13.5) + 12(1.5 + Rd )
25.5Rd + 18
3(1.5 + Rd )
= 100.
=
1.5Rd − 18
3Rd − 36
1.5(1.5 + Rd )
Therefore, Rd = 13.18 kΩ.
Set the CMRR equal to −100 and solve for Rd :
CMRR =
25.5Rd + 18
= −100.
3Rd − 36
Therefore, Rd = 11 kΩ.
Thus, 11 kΩ ≤ Rd ≤ 13.18 kΩ.
(24)(26) + (25)(25)
= 24.98.
(2)(1)(25)
(1)(24) − 25(1)
[b] Acm =
= −0.04.
1(25)
24.98
[c] CMRR =
= 624.50.
0.04
P 5.34
[a] Adm =
P 5.35
αRg
vg
[a] vp =
αRg + (Rg − αRg )
vo
vn = vp = αvg
vn − vg vn − vo
+
=0
R1
Rf
(vn − vg )
Rf
+ vn − vo = 0
R1
α
vo
0.0
−15 V
α
vo
0.4 −7.8 V
=
Rf
Rf
1+
αvg −
vg ;
R1
R1
=
(αvg − vg )5 + αvg ;
=
[(α − 1)5 + α]vg ;
=
(6α − 5)vg
=
(6α − 5)(3) = 18α − 15.
α
vo
0.8 −0.6 V
0.1 −13.2 V
0.5
−6 V
0.9
1.2 V
0.2 −11.4 V
0.6 −4.2 V
1.0
3V
−9.6 V
0.7 −2.4 V
0.3
5–21
5–22
CHAPTER 5. The Operational Amplifier
[b] Rearranging the equation for vo from (a) gives
Rf
Rf
+ 1 vg α + −
vg .
vo =
R1
R1
Therefore,
slope =
Rf
+ 1 vg ;
R1
intercept = −
Rf
vg .
R1
[c] Using the equations from (b),
−8 =
Rf
+ 1 vg ;
R1
5=−
Rf
vg .
R1
Solving,
vg = −3 V;
P 5.36
Rf
= 5/3.
R1
(400 × 10−3 )2
= 16 µW.
104
10
[b] v10 kΩ =
(400) = 80 mV;
50
[a] p10 kΩ =
p10 kΩ =
(80 × 10−3 )2
= 0.64 µW.
(10 × 103 )
pa
16
=
= 25.
pb
0.64
[d] Yes, the operational amplifier serves several useful purposes:
[c]
• First, it enables the source to control 25 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.
Problems
5–23
• 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 5.37
[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 5.38
[a]
vn − va vn − vo
+
= 0;
R
R
2vn − va = vo ;
va − vn va − vo
va
+
+
= 0;
Ra
R
R
va
1
2
vn
vo
+
−
= ;
Ra R
R
R
va
R
2+
− vn = vo
Ra
vn = vp = va + vg ;
.·. 2vn − va = 2va + 2vg − va = va + 2vg ;
.·. va − vo = −2vg
(1).
5–24
CHAPTER 5. The Operational Amplifier
2va + va
R
− va − vg = vo ;
Ra
R
.·. va 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.
[b] At saturation vo = ± Vcc ,
.·. va = ± Vcc − 2vg
(3)
and
R
.·. va 1 +
Ra
= ± Vcc + vg .
(4)
Dividing Eq (4) by Eq (3) gives
1+
R
± Vcc + vg
=
;
Ra
± Vcc − 2vg
.·.
± Vcc + vg
3vg
R
=
−1=
,
Ra
± Vcc − 2vg
± Vcc − 2vg
or Ra =
P 5.39
(± Vcc − 2vg )
R
3vg
Q.E.D.
[a] Assume the op-amp is operating within its linear range, then
iL =
3
= 2 mA.
1.5
For RL = 2.5 kΩ
vo = (2.5 + 1.5)(2) = 8 V.
Now since vo < 9 V our assumption of linear operation is correct,
therefore
iL = 2 mA.
[b] 9 = 2(1.5 + RL );
RL = 3 kΩ.
Problems
5–25
[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 ≤ 3 kΩ. Therefore when RL = 6.5 kΩ the op-amp is saturated.
We can estimate the value of iL by assuming ip = in iL . Then
iL = 9/(1.5 + 6.5) = 1.125 mA. To justify neglecting the current into the
op-amp assume the drop across the 47 kΩ resistor is negligible, and the
input resistance to the op-amp is at least 500 kΩ. Then
ip = in = (3 − 1.5)/500,000 = 3 µA. But 3 µA 1.125 mA, hence our
assumption is reasonable.
[d]
P 5.40
vp =
5.6
vg = 0.7vg = 2.8 sin(5π/3)t V;
8.0
vn
vn − vo
+
= 0;
2000
18,000
10vn = vo ;
vn = vp ;
.·. vo = 28 sin(5π/3)t V
vo = 0
0 ≤ t ≤ ∞.
t ≤ 0.
At saturation
5π
28 sin
t = ±14;
3
π
5π
.·.
t= ,
3
6
t = 0.10 s,
sin
5π
t = ±0.5;
3
5π
,
6
7π
,
6
11π
,
6
0.50 s,
0.70 s,
etc.
etc.;
5–26
P 5.41
CHAPTER 5. The Operational Amplifier
It follows directly from the circuit that vo = −5vg
From the plot of vg we have vg = 0, t < 0.
vg
=
4t
0 ≤ t ≤ 0.5;
vg
=
4 − 4t
0.5 ≤ t ≤ 1.5;
vg
=
4t − 8
1.5 ≤ t ≤ 2.5;
vg
=
12 − 4t 2.5 ≤ t ≤ 3.5;
vg = 4t − 16 3.5 ≤ t ≤ 4.5, etc.
Therefore
vo = −20t
0 ≤ t ≤ 0.5;
vo
=
20t − 20
0.5 ≤ t ≤ 1.5;
vo
=
40 − 20t
1.5 ≤ t ≤ 2.5;
vo
=
20t − 60
2.5 ≤ t ≤ 3.5;
vo = 80 − 20t
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 ±6, the output is clipped at ±6. The plot is
shown below.
Problems
5–27
P 5.42
i1 =
14.7 − 10
= 1 mA;
4700
i2 + i1 + 0 = 5 mA;
i2 = 4 mA;
vo2 = 10 + (1500)(0.004) = 16 V;
i3 =
14.7 − 16
= −0.1 mA;
13,000
i4 = i3 + i1 = 0.9 mA;
vo1 = 14.7 + 3000(0.0009) = 17.4 V.
P 5.43
[a] Let vo1 = output voltage of the amplifier on the left. Let vo2 = output
voltage of the amplifier on the right. Then
−90
(−0.5) = 3 V;
15
vo2 − vo1
ia =
= −4.6 mA.
1000
vo1 =
vo2 =
−120
(0.4) = −1.6 V;
30
[b] ia = 0 when vo1 = v02 so from (a) vo2 = 3 V.
Thus
−120
(vright ) = 3;
30
90
vright = −
= −750 mV.
120
5–28
P 5.44
CHAPTER 5. The Operational Amplifier
From Eq. 5.28,
1
vref
1
1
= vn
+
+
R + ∆R
R + ∆R R − ∆R Rf
!
−
vo
.
Rf
Substituting Eq. 5.30 for vp = vn :
1
1
+ R−∆R
+ R1f
vref R+∆R
vref
vo
−
=
.
1
R + ∆R
Rf
+ 1 + 1
(R − ∆R)
R+∆R
R−∆R
Rf
Rearranging,
1
vo
1
= vref
−
.
Rf
R − ∆R R + ∆R
Thus,
!
vo = vref
P 5.45
2∆R
Rf .
2
R − (∆R)2
[a] Replace the op amp with the model from Fig. 5.18:
Write two node voltage equations, one at the left node, the other at the
right node:
vn − vg vn − vo
vn
+
+
= 0;
16,000
24,000
480,000
vo + 105 vn vo − vn
vo
+
+
= 0.
2000
24,000
400
Simplify and place in standard form:
51vn − 20vo = 30vg ;
(12 × 105 − 1)vn + 73vo = 0.
Let vg = 1 V and solve the two simultaneous equations:
vo = −1.499767 V;
vn = 91.236 µV.
Thus the voltage gain is vo /vg = −1.499767.
Problems
5–29
[b] From the solution in part (a), vn = 91.236 µV.
[c] ig =
vg − vn
1 − 91.236 × 10−6
=
;
16,000
16,000
Rg =
vg
16,000
=
= 16,001.46 Ω.
ig
1 − 91.236 × 10−6
[d] For an ideal op amp, the voltage gain is the ratio between the feedback
resistor and the input resistor:
vo
24,000
=−
= −1.5.
vg
16,000
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 5.46
Rg = 16 kΩ.
[a]
vn − vg
vn − vo
+
= 0;
25,000
150,000
.·. vo = 7vn − 6vg .
Also
vo = A(vp − vn ) = −Avn ;
−vo
.·. vn =
;
A
7
.·. vo 1 +
= −6vg ;
A
vo =
−6A
vg .
(7 + A)
−6(150)(0.5)
= −2.866 V.
(7 + 150)
[c] vo = −6(0.5) = −3 V.
[b] vo =
5–30
CHAPTER 5. The Operational Amplifier
[d] −2.94 =
−6(0.5)A
7+A
.·. A = 343.
P 5.47
[a]
vn
vn − vg
vn − vo
+
+
=0
8000 600,000 240,000
or
78.5vn − 2.5vo = vg ;
vo
vo − vn
vo − 100,000(vp − vn )
+
+
= 0;
30,000 240,000
5000
57vo − vn − 48 × 105 (vp − vn ) = 0;
vp = vg +
(vn − vg )(160)
= (11/15)vg + (4/15)vn ;
600
57vo − vn − 48 × 105 [(11/15)vg − (11/15)vn ] = 0;
57vo + 3,520,000vn = 3,520,000vg ;
∆=
No =
vo =
[b] N1 =
vn =
78.5
−2.5
3.52 × 106 57
78.5
= 8,804,474.5;
vg
3.52 × 106 3.52 × 106 vg
No
= 30.98vg ;
∆
vg
= 272.8 × 106 vg ;
vo
= 30.98.
vg
−2.5
3.52 × 106 vg 57
N1
= 0.9995vg ;
∆
= 8,800,057vg ;
vn = 999.5 mV.
vp = (11/15)(1000) + (4/15)(999.5) = 999.87 mV.
[c] vp − vn = 367.94 µV.
(1000 − 999.87)10−3
= 836.22 pA.
160 × 103
vg vg − vo
+
= 0,
since vn = vp = vg ;
[e]
8
240
vo
.·. vo = 31vg ,
= 31.
vg
[d] ig =
vn = vp = 1 V;
vp − vn = 0 V;
ig = 0 A.
Problems
P 5.48
[a]
vn
vn − vTh
vn − 0.4
+
+
= 0;
5000
500,000
100,000
vTh + 3 × 105 vn vTh − vn
+
= 0.
5000
100,000
Solving, vTh = −7.9994 V.
Short-circuit current calculation:
vn − 0.4
vn − 0
vn
+
+
= 0;
500,000
5000
100,000
.·. vn = 0.37736 V.
isc =
vn
3 × 105
−
vn = −22.64 A;
100,000
5000
RTh =
vTh
= 353.33 mΩ.
isc
5–31
5–32
CHAPTER 5. The Operational Amplifier
[b] The output resistance of the inverting amplifier is the same as the
Thévenin resistance, i.e.,
Ro = RTh = 353.33 mΩ.
[c]
vo =
500
(−7.9994) = −7.99375 V.
500.35333
vn
vn + 7.99375
vn − 0.4
+
+
= 0;
5000
500,000
100,000
.·. vn = 294.811 µV.
ig =
0.4 − 294.811 × 10−6
= 79.94 µA;
5000
Rg =
P 5.49
0.4
= 5003.7 Ω.
ig
[a] vTh = −
100,000
(0.4) = −8 V;
5000
RTh = 0, since op-amp is ideal.
Problems
5–33
[b] Ro = RTh = 0 Ω.
[c] Rg = 5 kΩ
P 5.50
since
vn = 0.
[a] Use the approximation for Eq. 5.31 to solve for Rf ; note that since we are
using 2% strain gages, δ = ∆R/R = 0.02:
Rf =
vo R
(4)(150)
=
= 1250 Ω.
2δvref
(2)(0.02)(12)
[b] Now solve for δ given vo = 30 mV:
δ=
(0.03)(150)
vo R
=
= 150 × 10−6 .
2Rf vref
2(1250)(12)
The change in strain gage resistance that corresponds to a 30 mV change
in output voltage is thus
(δ)R = (150 × 10−6 )(150) = 22.5 mΩ.
P 5.51
[a]
Let R1 = R + ∆R
vp vp − vin
vp
+
+
= 0;
Rf
R
R1
.·. vp
"
#
1
1
1
vin
+ +
=
;
Rf
R R1
R1
.·. vp =
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
"
#
R + 2Rf
vin
vo
.·. vn
−
=
RRf
R
Rf ;
.·.
"
vo
R + 2Rf
=
Rf
RRf
#"
#
RRf vin
vin
−
;
RR1 + Rf R1 + Rf R
R
5–34
CHAPTER 5. The Operational Amplifier
.·.
"
#
R + 2Rf
1
vo
=
−
vin ;
Rf
RR1 + Rf R1 + Rf R R
2
[R + 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
Rf (R + Rf )(−∆R)vin
vo ≈
.
R2 (R + 2Rf )
[b] vo ≈
350 × 103 (365 × 103 )(−150)12
≈ −1.42937 V.
(15 × 103 )2 (715 × 103 )
[c] vo =
−∆R(R + Rf )Rf vin
.
R[(R + ∆R)(R + Rf ) + RRf ]
=
−(150)(365,000)(350,000)(12)
15,000[(15,150)(365,000) + (15,000)(350,000)]
= −1.422111 V.
P 5.52
[a] vo ≈
(R + Rf )Rf (−∆R)vin
;
R2 (R + 2Rf )
vo =
(R + Rf )(−∆R)Rf vin
;
R[(R + ∆R)(R + Rf ) + RRf ]
R[(R + ∆R)(R + Rf ) + RRf ]
approx value
.·.
=
;
true value
R2 (R + 2Rf )
2
R[(R + ∆R)(R + Rf ) + RRf ] − R (R + 2Rf )
.·. Error =
R2 (R + 2Rf )
=
∆R (R + Rf )
.
R (R + 2Rf )
∆R(R + Rf )
.·. % error =
× 100.
R(R + 2Rf )
[b] % error =
150(365,000) × 100
= 0.51%.
15,000(715,000)
Problems
P 5.53
1=
∆R(365,000)
× 100;
15,000(715,000)
.·. ∆R = 293.8356 Ω;
293.8356
× 100 ≈ 1.96%.
.·. % change in R =
15,000
P 5.54
[a] It follows directly from the solution to Problem 5.51 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 5.51 that
R[(R − ∆R)(R + Rf ) + RRf ]
approx value
=
.
.·.
true value
R2 (R + 2Rf )
(R − ∆R)(R + Rf ) + RRf − R(R + 2Rf )
.·. Error =
R(R + 2Rf )
=
−∆R(R + Rf )
.
R(R + 2Rf )
−∆R(R + Rf )
.·. % error =
× 100.
R(R + 2Rf )
[c] R − ∆R = 14,820 Ω
.·. ∆R = 15,000 − 14,820 = 180 Ω;
(365,000)(350,000)(180)(12)
.·. vo ≈
≈ 1.715 V;
15,0002 (715,000)
[d] % error =
−180(365,000)(100)
= −0.6126%.
15,000(715,000)
5–35
Inductance, Capacitance, and
Mutual Inductance
Assessment Problems
AP 6.1 [a] v = L
dig
;
dt
dig
= 18[t(−10e−10t ) + e−10t ] = 18e−10t (1 − 10t);
dt
v = (50 × 10−6 )(18)e−10t (1 − 10t)
= 0.9e−10t (1 − 10t) mV,
t ≥ 0.
[b] p = vig .
v(200 ms) = (0.9 × 10−3 )e−2 (1 − 2) = −121.8 µV.
ig (200 ms) = 18(0.2)e−2 = 487.2 mA.
p(200 ms) = (−121.8 × 10−6 )(487.2 × 10−3 ) = −59.34 µW.
[c] delivering.
1
1
[d] w = Li2g = (50 × 10−6 )(487.2 × 10−3 )2 = 5.93 µJ.
2
2
[e] The energy is a maximum where the current is a maximum:
dig
= 18e−10t (1 − 10t);
dt
dig
= 0 when t = 0.1 s.
dt
ig(max) = 18(0.1)e−1 = 662.2 mA;
1
wmax = (50 × 10−6 )(662.2 × 10−3 )2 = 10.96 µJ.
2
6–1
6–2
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
AP 6.2 [a] i = C
dv
= (5 × 10−6 )[500t(−2500)e−2500t + 500e−2500t ]
dt
= 2.5 × 10−3 e−2500t (1 − 2500t) A.
[b] v(100 × 10−6 ) = 500(100 × 10−6 )e−0.25 = 38.94 mV;
i(100 × 10−6 ) = (2.5 × 10−3 )e−0.25 (1 − 0.25) = 1.46 mA;
p(100 × 10−6 ) = vi = (38.94 × 10−3 )(1.46 × 10−3 ) = 56.86 µW.
[c] p > 0, so the capacitor is absorbing power.
AP 6.3 [a] v(100 × 10−6 ) = 38.94 mV;
1
1
w = Cv 2 = (5 × 10−6 )(38.94 × 10−3 )2 = 3.79 nJ.
2
2
[b] The energy is maximum when the voltage is maximum:
dv
= 0 when (1 − 2500t) = 0 or t = 0.4 ms;
dt
vmax = 500(0.4 × 10−3 )e−1 = 73.58 mV;
1 2
pmax = Cvmax
= 13.53 nJ.
2
(24)(48)
= 16 mH.
72
[b] i(0+ ) = −0.6 + 0.2 = −0.4 A.
1 Zt
[c] i =
(0.1e−25x ) dx − 0.4 = (−250e−25t − 150) mA.
0.016 0+
1 Zt
[d] i1 =
(0.1e−25x ) dx − 0.6 = (−83.33e−25t − 516.67) mA;
+
0.048 0
AP 6.4 [a] Leq =
i2 =
1 Zt
(0.1e−25x ) dx + 0.2 = (−166.67e−25t + 366.67) mA;
+
0.024 0
i1 + i2 = i.
(100)(25)
= 20 µF.
125
[b] v(0+ ) = 15 − 10 = 5 V.
Z t
1
[c] v(t) =
(0.05e−200x ) dx + 5 = (−12.5e−200t + 17.5) V.
20 × 10−6 0+
AP 6.5 [a] Ceq =
Problems
Z t
1
[d] v1 =
0.05e−200x dx + 15 = (−2.5e−200t + 17.5) V;
−6
+
100 × 10
0
v2 =
Z t
1
0.05e−200x dx − 10 = −10e−200t V;
25 × 10−6 0+
v1 (∞) = 17.5 V,
v2 (∞) = 0 V;
1
w = (100 × 10−6 )(17.5)2 = 15.3125 mJ.
2
AP 6.6 [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
= v g 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
pabs = 720 + 180 + 60 = 960 W;
.·.
X
pdev =
X
pabs = 960 W.
22.8
M
=√
= 0.95.
L1 L2
576
√
[b] Mmax = 576 = 24 mH.
AP 6.7 [a] k = √
L1
N 2 P1
N1
[c]
= 12
=
L2
N2 P2
N2
2
;
N1 2
60
=
= 6.25.
N2
9.6
N1 √
= 6.25 = 2.5.
N2
.·.
6–3
6–4
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
Problems
P 6.1
[a] 0 ≤ t ≤ 1 ms :
i=
106 Z t
1Zt
vs dx + i(0) =
6 × 10−3 dx + 0
L 0
300 0
t
= 20x = 20t A.
0
1 ms ≤ t ≤ 2 ms :
106 Z t
i=
(12 × 10−3 − 6x) dx + 20 × 10−3 ;
−3
300 10
.·. i = 40t − 10,000t2 − 10 × 10−3 A.
2 ms ≤ t ≤ ∞ :
i=
106 Z t
(0) dx + 30 × 10−3 = 30 mA.
300 2×10−3
[b]
P 6.2
[a] i = 8e−300t − 8e−1200t A;
v=L
di
= −9.6e−300t + 38.4e−1200t V,
dt
t > 0+ ;
v(0+ ) = −9.6 + 38.4 = 28.8 V.
[b] v = 0 when 38.4e−1200t = 9.6e−300t
−1500t
or t = (ln 4)/900 = 1.54 ms.
[c] p = vi = 384e
− 76.8e
− 307.2e−2400t W.
dp
[d]
= 0 when e1800t − 12.5e900t + 16 = 0.
dt
Let x = e900t
x = 1.44766,
−600t
and solve the quadratic x2 − 12.5x + 16 = 0.
t=
ln 1.45
= 411.05 µs;
900
ln 11.05
= 2.67 ms;
900
p is maximum at t = 411.05 µs.
x = 11.0523,
t=
Problems
6–5
[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.
P 6.3
[a] i(0) = A1 + A2 = 1;
di
= −2000A1 e−2000t − 8000A2 e−8000t ;
dt
v = −30A1 e−2000t − 120A2 e−8000t V;
v(0) = −30A1 − 120A2 = 60.
Solving, A1 = 2
and A2 = −1.
Thus,
i = (2e−2000t − e−8000t ) A
v = 0.015
t ≥ 0;
di
= −60e−2000t + 120e−8000t V,
dt
t ≥ 0.
[b] p = vi = 300e−10,000t − 120e−4000t − 120e−16,000t ;
p=0
when
Let x = e6000t ;
300e6000t − 120e12,000t − 120 = 0.
then
300x − 120x2 − 120 = 0.
Thus x2 − 2.5x + 1 = 0
so
x = 0.5 and x = 2.
If x = e6000t = 0.5, t will be negative. Hence, the solution for t > 0 must
be x = 2:
e6000t = 2
Thus, t =
P 6.4
so
6000t = ln 2.
ln 2
= 115.52 µs.
6000
[a] From Problem 6.3 we have
i = A1 e−2000t + A2 e−8000t A;
v = −30A1 e−2000t − 120A2 e−8000t V;
i(0) = A1 + A2 = 1;
v(0) = −30A1 − 120A2 = −300.
6–6
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
A1 = −2;
Solving,
Thus,
A2 = 3.
i = −2e−2000t + 3e−8000t A t ≥ 0;
v = 0.015
di
= 60e−2000t − 360e−8000t V t ≥ 0.
dt
[b] i = 0 when 3e−8000t = 2e−2000t ;
.·. e6000t = 1.5 so t = (ln 1.5)/6000 = 67.58 µs.
Thus,
i > 0 for 0 ≤ t ≤ 67.58 µs
i < 0 for 67.58 µs ≤ t < ∞;
and
v = 0 when 60e−2000t = 360e−8000t ;
.·. t = (ln 6)/6000 = 298.63 µs.
Thus,
v < 0 for 0 ≤ t ≤ 298.63 µs
and
v > 0 for 298.63 µs ≤ t < ∞.
and
298.63 µs ≤ t < ∞
Therefore,
p < 0 for 0 ≤ t ≤ 67.58 µs
(inductor delivers energy);
p > 0 for 67.58 µs ≤ t ≤ 298.63 µs (inductor stores energy).
[c] p = vi = 900e−10,000t − 120e−4000t − 1080e−16,000t W;
.·. wstored =
Z t2
p dx + w(0).
t1
"
wstored = 10−3 −90e−10,000x
t2
+30e−4000x
t1
t2
+67.5e−16,000x
t1
t2
#
+ 7.5 × 10−3
t1
= 30e−4000t2 + 67.5e−16,000t2 − 90e−10,000t2 + 90e−10,000t1 − 30e−4000t1
− 67.5e−16,000t1 + 7.5 mJ
where t1 = 67.58 µs
.·. wstored = 5.11 + 7.5 = 12.61 mJ.
and
t2 = 298.63 µs;
Problems
Z t1
wextracted =
=
Z t1
p dt +
0
Z ∞
p dt
t2
[900e−10,000x − 120e−4000x − 1080e−16,000x ] dx
0
Z ∞
+
[900e−10,000x − 120e−4000x − 1080e−16,000x ] dx
t2
= 10−3 −90e−10,000x
t1
+30e−4000x
0
−3
−10
−10,000x
t1
+67.5e−16,000x
0
∞
90e
−4000x
!
0
∞
+30e
t2
t1
−16,000x
∞
+67.5e
t2
t2
= 90e−10,000t2 − 30e−4000t2 − 67.5e−16,000t2 + 30e−4000t1
+ 67.5e−16,000t1 − 90e−10,000t1 − 7.5 mJ
where t1 = 67.58 µs
and
t2 = 298.63 µs;
.·. wextracted = −12.61 mJ.
Thus, the energy stored equals the energy extracted.
P 6.5
i = (B1 cos 5t + B2 sin 5t)e−t ;
i(0) = B1 = 25 A;
di
= (B1 cos 5t + B2 sin 5t)(−e−t ) + e−t (−5B1 sin 5t + 5B2 cos 5t)
dt
= [(5B2 − B1 ) cos 5t − (5B1 + B2 ) sin 5t]e−t ;
v=2
di
= [(10B2 − 2B1 ) cos 5t − (10B1 + 2B2 ) sin 5t]e−t ;
dt
v(0) = 100 = 10B2 − 2B1 = 10B2 − 50
.·. B2 = 150/10 = 15 A.
Thus,
i = (25 cos 5t + 15 sin 5t)e−t A,
v = (100 cos 5t − 280 sin 5t)e−t V,
i(0.5) = −6.70 A;
t ≥ 0;
t ≥ 0;
v(0.5) = −150.23 V;
p(0.5) = (−6.70)(−150.23) = 1007.00 W absorbing.
!
6–7
6–8
P 6.6
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[a] vL = L
di
= [56 cos 140t + 92 sin 140t]e−20t mV;
dt
dvL
.·.
= [11,760 cos 140t − 9680 sin 140t]e−20t mV/s.
dt
dvL
= 0 when
dt
.·. t = 6.30 ms.
tan 140t =
Also 140t = 0.8821 + π
11,760
= 1.21;
9680
etc.
Because of the decaying exponential vL will be maximum the first time
the derivative is zero.
[b] vL (max) = [56 cos 0.8821 + 92 sin 0.8821]e−0.12602 = 93.997 mV;
vL max ≈ 94 mV.
Note: When
P 6.7
t=
0.8821 + π
;
140
vL = −60 mV.
[a] i
=
0
t < 0;
i
=
16t A
0 ≤ t ≤ 25 ms;
i
=
0.8 − 16t A
25 ≤ t ≤ 50 ms;
i
=
0
50 ms < t.
[b] v = L
di
= 375 × 10−3 (16) = 6 V
dt
v = 375 × 10−3 (−16) = −6 V
0 ≤ t ≤ 25 ms;
25 ≤ t ≤ 50 ms.
v
=
0
t < 0;
v
=
6V
0 < t < 25 ms;
v
= −6 V
25 < t < 50 ms;
v
=
50 ms < t.
0
p = vi.
p
=
0
t < 0;
p
=
96t W
0 < t < 25 ms;
p
=
96t − 4.8 W
25 < t < 50 ms;
p
=
0
50 ms < t.
Problems
1
w = Li2 .
2
P 6.8
w
=
0
t < 0;
w
=
48t2 J
0 < t < 25 ms;
w
=
48t2 − 4.8t + 0.12 J
25 < t < 50 ms;
w
=
0
50 ms < t.
[a] i
=
1000 Z t
250 sin 1000x dx − 5
50 0
=
5000
Z t
sin 1000x dx − 5
0
[b]
− cos 1000x t
−5
1000
0
=
5000
=
5(1 − cos 1000t) − 5;
i
= −5 cos 1000t A.
p
= vi = (250 sin 1000t)(−5 cos 1000t)
= −1250 sin 1000t cos 1000t;
p
w
w
= −625 sin 2000t W.
1 2
Li
=
2
=
1
(50 × 10−3 )25 cos2 1000t
2
=
625 cos2 1000t mJ;
=
[312.5 + 312.5 cos 2000t] mJ.
6–9
6–10
P 6.9
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[c] Absorbing power:
Delivering power:
0.5π ≤ t ≤ π ms;
0 ≤ t ≤ 0.5π ms;
1.5π ≤ t ≤ 2π ms;
π ≤ t ≤ 1.5π ms.
0 ≤ t < 0.5 s :
iL =
t
−2.8
1 Zt
2.8e−20x dx + 2 =
e−20x +2
0.05 0
(0.05)(20)
0
= −2.8e−20t + 4.8 A.
iL (0.5) = 4.8 A.
0.5 s ≤ t < ∞ :
iL =
t
1 Zt
2.8
−2.8e−20(x−0.5) dx + 4.8 =
e−20(x−0.5)
+4.8
0.05 0.5
(0.05)(20)
0.5
Problems
= 2.8e−20(t−0.5) + 2 A.
P 6.10
[a] 0 ≤ t ≤ 2 s :
v = −25t;
i=
1 Zt
x2 t
−25x dx + 0 = −10
= −5t2 A.
2.5 0
2 0
2s ≤ t ≤ 6s :
v = −100 + 25t;
i(2) = −20 A;
.·. i
=
1 Zt
(25x − 100) dx − 20
2.5 2
=
10
Z t
2
=
Z t
x dx − 40
dx − 20
2
5(t2 − 4) − 40(t − 2) − 20
= 5t2 − 40t + 40 A.
6 s ≤ t ≤ 10 s :
v = 200 − 25t;
6–11
6–12
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
i(6) = 5(36) − 240 + 40 = −20 A;
i
=
1 Zt
(200 − 25x) dx − 20
2.5 6
=
80
Z t
Z t
dx − 10
6
x dx − 20
6
= 80(t − 6) − 10(t2 − 36)/2 − 20 = 80t − 5t2 − 320 A.
10 s ≤ t ≤ 12 s :
v = 25t − 300;
i(10) = 800 − 500 − 320 = −20 A;
i
=
=
1 Zt
(25x − 300) dx − 20
2.5 10
Z t
Z t
x dx − 120
10
10
dx − 20
10
=
5(t2 − 100) − 120(t − 10) − 20
=
5t2 − 120t + 680 A.
t ≥ 12 s :
v = 0;
i(12) = 5(12)2 − 120(12) + 680 = −40 A;
i
=
1 Zt
0 dx − 40
2.5 12
= −40 A.
[b] For 0 ≤ t ≤ 2 s,
v=0
v = −25t V;
when
t=0
so
i = −5t2 A.
i = 0 A.
For 2 ≤ t ≤ 6 s,
v = −100 + 25t V;
v=0
t = 4s
when
For 6 ≤ t ≤ 10 s,
v=0
when
when
For t ≥ 12 s,
i = 5(4)2 − 40(4) + 40 = −40 A.
v = 200 − 25t V;
t = 8s
For 10 ≤ t ≤ 12 s,
v=0
so
so
v = 0;
so
i = −5t2 + 80t − 320 A.
i = −5(8)2 + 80(8) − 320 = 0 A.
v = 25t − 300 V;
t = 12 s
i = 5t2 − 40t + 40 A.
i = 5t2 − 120t + 680 A.
i = 5(12)2 − 120(12) + 680 = −40 A.
i = −40 A.
Problems
6–13
[c]
P 6.11
For 0 ≤ t ≤ 1.2 s:
1Zt
iL =
14 × 10−3 dx + 0 = 0.7 × 10−3 t;
20 0
iL (1.2 s) = (0.7 × 10−3 )(1.2) = 0.84 mA;
Rm = (25)(1000) = 25 kΩ;
vm (1.2 s) = (0.84 × 10−3 )(25 × 103 ) = 21 V.
P 6.12
p = vi = 40t[e−10t − 10te−20t − e−20t ].
w=
Z ∞
p dx =
0
Z ∞
40x[e−10x − 10xe−20x − e−20x ] dx = 0.2 J.
0
This is energy stored in the inductor at t = ∞. We can verify this using the
energy equation for inductors. Remember that as t → ∞, i = 2 A:
1
1
w = Li2 = (0.1)(2)2 = 0.2 J. (checks)
2
2
P 6.13
[a] v(20 µs)
=
12.5 × 109 (20 × 10−6 )2 = 5 V (end of first interval);
v(20 µs)
=
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).
=
10 V (start of third interval).
v(40 µs)
v(40 µs)
[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,
p(30 µs) = vi = (8.75)(0.05) = 62.5 mW. (checks)
i(30 µs) = 0.05 A;
6–14
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[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.
P 6.14
[a] i = C
d
dv
= 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,
[b] i
π
ms = −31.66 mA,
80
v
i(0+ ) = 0.72 A.
π
ms = 20.505 V,
80
p = vi = −649.23 mW.
[c] w
π
1
ms = Cv 2 = (0.3 × 10−6 )(20.505)2 = 126.13 µJ.
80
2
P 6.15
[a] v =
1 Zt
i dx + v(0+ )
C 0+
=
Z t
1
3 cos 50,000x dx = 100 sin 50,000t V.
0.6 × 10−6 0+
[b] p(t) = vi = [300 cos 50,000t] sin 50,000t
= 150 sin 100,000t W,
p(max) = 150 W.
1 2
[c] w(max) = Cvmax
= (0.3 × 10−6 )(100)2 = 3000 µJ = 3 mJ.
2
P 6.16
[a] v
=
0
t < 0;
v
= −2t V
v
=
2t − 20 V
5 ≤ t ≤ 15 s;
v
=
40 − 2t V
15 ≤ t ≤ 20 s;
v
=
0
20 s < t.
[b] i = C
0 ≤ t ≤ 5 s;
dv
:
dt
i
=
0
i
= −1 mA
0 < t < 5 s;
i
=
1 mA
5 < t < 15 s;
i
= −1 mA
15 < t < 20 s;
i
=
20 s < t.
0
t < 0;
Problems
6–15
p = vi :
p
=
0
t < 0;
p
=
(−2t)(−0.001) = 2t mW
0 < t < 5 s;
p
=
(2t − 20)(0.001) = 2t − 20 mW
5 < t < 15 s;
p
=
(40 − 2t)(−0.001) = 2t − 40 mW
15 < t < 20 s;
p
=
0
20 s < t.
w=
w
w
Z
=
=
p dx :
0
Z t
t < 0;
t
(0.002x) dx = 0.001x2 = t2 mJ
0
w
=
Z t
0 < t < 5 s;
0
(0.002x − 0.02) dx + 0.025
5
t
=
(0.001x2 − 0.02x)
+ 0.025
5
= t2 − 20t + 100 mJ
w
=
Z t
5 < t < 15 s;
(0.002x − 0.04) dx + 0.025
15
t
=
(0.001x2 − 0.04x)
+ 0.025
15
w
= t2 − 40t + 400 mJ
15 < t < 20 s;
=
20 s < t.
0
[c]
From the plot of power above, it is clear that power is being absorbed for
6–16
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
0 < t < 5 s and for 10 s < t < 15 s, because p > 0. Likewise, power is
being delivered for 5 s < t < 10 s and for 15 s < t < 20 s, because p < 0.
P 6.17
1
1
[a] w(0) = C[v(0)]2 = (400 × 10−9 )(25)2 = 125 µJ.
2
2
−1500t
[b] v = (A1 t + A2 )e
;
v(0) = A2 = 25 V.
dv
dt
= −1500e−1500t (A1 t + A2 ) + e−1500t (A1 )
(−1500A1 t − 1500A2 + A1 )e−1500t ;
=
dv
(0) = A1 − 1500A2 ;
dt
i=C
dv
,
dt
i(0) = C
dv(0)
;
dt
i(0)
90 × 10−3
dv(0)
.·.
=
=
= 225 × 103 ;
dt
C
0.40 × 10−6
.·. 225 × 103 = A1 − 1500(25).
Thus, A1 = 2.25 × 105 + 3.75 × 104 = 262,500
V
.
s
[c] v = (262,500t + 25)e−1500t ;
dv
d
= 0.40 × 10−6 (262,500t + 25)e−1500t ;
dt
dt
d
=
[(0.105t + 10 × 10−6 )e−1500t ]
dt
i=C
i
P 6.18
[a]
v
=
(0.105t + 10 × 10−6 )(−1500)e−1500t + e−1500t (0.105)
=
(−157.5t − 15 × 10−3 + 0.105)e−1500t
=
(0.09 − 157.5t)e−1500t A,
=
(90 − 157,500t)e−1500t mA,
=
5 × 106
Z 250×10−6
t≥0
t ≥ 0.
100 × 10−3 e−1000t dt − 60.6
0
w
−1000t
3e
250×10−6
=
500 × 10
=
500(1 − e−0.25 ) − 60.6 = 50 V;
=
1
Cv 2 = 12 (0.2)(10−6 )(50)2 = 250 µJ.
2
−1000 0
−60.6
Problems
[b] v = 500 − 60.6 = 439.40 V;
1
w = (0.2) × 10−6 (439.40)2 = 19.31 mJ = 19,307.24 µJ.
2
P 6.19
[a] 0 ≤ t ≤ 100 µs :
1
= 5 × 106 ;
C
C = 0.2 µF
6
v = 5 × 10
Z t
− 0.04 dx + 40;
0
v = −200 × 103 t + 40 V.
v(100 µs) = −20 + 40 = 20 V.
[b] 100 µs ≤ t ≤ 300 µs :
6
v = 5 × 10
Z t
100×10−6
0.08 dx + 20 = 4 × 105 t − 40 + 20;
v = 4 × 105 t − 20V.
v(300 µs) = 4 × 105 (300 × 10−6 ) − 20 = 100 V.
[c] 300 µs ≤ t < ∞ :
6
v = 5 × 10
Z t
300×10−6
0 dx + 100 = 100 V.
[d]
P 6.20
dv
= 0, t < 0.
dt
dv
= 5e−1000t [cos 3000t + 13 sin 3000t] mA,
[b] i = C
dt
[c] no,
v(0− ) = −30 V;
v(0+ ) = 10 − 40 = −30 V.
[a] i = C
[d] yes,
i(0− ) = 0 A;
i(0+ ) = 5 mA.
t ≥ 0.
6–17
6–18
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[e] v(∞) = 10 V;
1
1
w = Cv 2 = (0.5 × 10−6 )(10)2 = 25 µJ.
2
2
P 6.21
P 6.22
iC = C(dv/dt).
0<t<1:
iC = 0.5 × 10−6 (120)t2 = 60t2 µA;
1<t<2:
iC = 0.5 × 10−6 (120)(2 − t)2 (−1) = −60(2 − t)2 µA;
[a] i =
50 × 10−3
t = 5 × 103 t
10 × 10−6
i = 50 × 10−3
q
=
0 ≤ t ≤ 10 µs;
10 ≤ t ≤ 30 µs;
Z 10×10−6
3
5 × 10 t dt +
Z 30×10−6
10×10−6
0
=
10×10−6
2
3t
5 × 10
2
+50 × 10−3 (20 × 10−6 )
0
=
5 × 103 ( 12 )(100 × 10−12 ) + 1000 × 10−3 × 10−6
=
1.25 µC.
[b] i = 200 × 10−3 − 5 × 10−3 t
q
50 × 10−3 dt
30 µs ≤ t ≤ 50 µs;
Z 50×10−6
=
1.25 × 10−6 +
=
1.25 × 10−6 + 200 × 10−3 (20 × 10−6 ) − 5 × 103
30×10−6
[200 × 10−3 − 5 × 103 t] dt
−6
=
−6
1.25 × 10
−9
+ 4000 × 10
3
− 5 × 10
= 1.25 µC.
Since q = vC,
.·. v = 1.25/0.25 = 5 V.
t2 50×10
2 30×10−6
2500 − 900
10−12
2
Problems
[c] i = −300 × 10−3 + 5 × 10−3 t
q
50 µs ≤ t ≤ 60 µs;
=
1.25 × 10−6 +
=
1.25 × 10−6 − 300 × 10−3 (10 × 10−6 )
3
+5 × 10
=
P 6.23
Z 60×10−6
50×10−6
[−300 × 10−3 + 5 × 103 t] dt
3600 − 2500
10−12
2
1 µC.
v=
1 × 10−6
= 4 V;
0.25 × 10−6
w=
C 2 1
v = (0.25) × 10−6 (16) = 2 µJ.
2
2
[a] 6k14 = 4.2 H;
15.8 + 4.2 = 20 H;
20k60 = 15 H;
15 + 5 = 20 H;
20k80 = 16 H;
16 + 24 = 40 H;
40k10 = 8 H;
Lab = 12 + 8 = 20 H.
[b] 10k(15 + 25) = 8 H;
8k12 = 4.8 H;
44k(1.2 + 4.8) = 5.28 H;
21k4 = 3.36 H;
5.28 + 3.36 = 8.64 H.
P 6.24
[a] Combine two 10 mH inductors in parallel to get a 5 mH equivalent
inductor. Then combine this parallel pair in series with two 10 mH
inductors:
10 mk10 m + 10 m + 10 m = 25 mH.
6–19
6–20
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[b] Combine two 100 µH inductors in parallel to get a 50 µH inductor. Then
combine this parallel pair in series with one more 100 µH inductor:
100 µk100 µ + 100 µ = 150 µH.
[c] Combine two 100 µH inductors in parallel to get a 50 µH inductor. Then
combine this parallel pair with three 10 µH inductors in series:
100 µk100 µ + 10 µ + 10 µ + 10 µ = 80 µH.
P 6.25
[a]
i(t)
= −
1 Zt
− 1800e−20x dx − 12
7.5 0
=
e−20x t
−12
−20 0
240
= −12(e−20t − 1) − 12;
i(t)
[b] i1 (t)
= −12e−20t A.
= −
1Zt
− 1800e−20x dx + 4
10 0
=
e−20x t
+4
−20 0
180
= −9(e−20t − 1) + 4;
i1 (t)
[c] i2 (t)
= −9e−20t + 13 A .
1Zt
= −
− 1800e−20x dx − 16
30 0
=
60
e−20x t
−16
−20 0
= −3(e−20t − 1) − 16;
i2 (t)
= −3e−20t − 13 A.
[d] p = vi = (−1800e−20t )(−12e−20t ) = 21,600e−40t W;
Problems
w
=
Z ∞
p dt =
0
Z ∞
21,600e−40t dt
0
=
21,600
=
540 J.
e−40t ∞
−40 0
1
1
[e] w = (10)(16) + (30)(256) = 3920 J.
2
2
[f ] wtrapped = winitial − wdelivered = 3920 − 540 = 3380 J.
1
1
checks.
[g] wtrapped = (10)(13)2 + (30)(13)2 = 3380 J
2
2
P 6.26
[a] io (0) = i1 (0) + i2 (0) = 5 A.
[b]
"
io
=
5(e−25t − 1) + 5 = 5e−25t A,
va
=
3.6
vc
= va + vb = −450e−25t + 1250e−25t
t ≥ 0.
[c]
d
(5e−25t ) = −450e−25t V;
dt
800e−25t V;
1Z t
800e−25x dx + 10
= −
8 0
=
i1
i1
t
e−25x
1Zt
1250e−25x dx + 5 = −125
= −
+5
10 0
−25 0
=
4e−25t − 4 + 10;
=
4e−25t + 6 A
t ≥ 0.
6–21
6–22
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[d] i2
1Zt
= −
800e−25x dx − 5
32 0
= e−25t − 1 − 5;
i2
= e−25t − 6 A,
t ≥ 0.
1
1
1
[e] w(0) = (8)(100) + (32)(25) + (3.6)(25) = 845 J.
2
2
2
1
[f ] wdel = (10)(25) = 125 J.
2
[g] wtrapped = 845 − 125 = 720 J.
P 6.27
Referring to the figure in the solution for Problem 6.26, vb = 1250e−25t V;
io = 5e−25t A;
p = 6250e−50t W;
w=
Z t
6250e−50x dx = 6250
0
e−50x t
= 125(1 − e−50t ) W;
−50 0
wtotal = 125 J.
80%wtotal = 100 J.
Thus,
125 − 125e−50t = 100;
P 6.28
e50t = 5;
.·. t = 32.19 ms.
From Figure 6.17(a) we have
1 Zt
1 Zt
i dx + v1 (0) +
i dx + v2 (0) + · · · ;
v=
C1 0
C2 0
Z t
1
1
+
+ ···
i dx + v1 (0) + v2 (0) + · · · ;
v=
C1 C2
0
1
1
1
=
+
+ ··· ,
Ceq
C1 C2
Therefore
P 6.29
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 .
Problems
P 6.30
[a]
1
1
1
5
= +
= ;
C1
8 32
32
6–23
C1 = 6.4 nF;
C2 = 5.6 + 6.4 = 12 nF.
1
1
1
10
=
+
=
C3
18 12
72
C3 = 7.2 nF;
C4 = 12.8 + 7.2 = 20 nF.
1
1
1
1
1
= +
+
= ;
C5
8 20 40
5
C5 = 5 nF.
Equivalent capacitance is 5 nF with an initial voltage drop of −10 V.
1
7
1
+
=
.·. Ceq = 12 µF;
[b]
21 28
84
−10 V − 5 V = −15 V;
24 + 12 = 36 µF.
1
1
2
+
=
36 36
36
.·. Ceq = 18 µF;
−15 V + 2 V = −13 V;
12 + 20 = 32 µF.
6–24
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
18 + 14 = 32 µF;
1
1
2
+
=
32 32
32
.·. Ceq = 16 µF;
8 V − 13 V = −5 V.
P 6.31
[a] Combine a 470 µF capacitor and three 10 µF capacitors in parallel to get a
500 µF capacitor:
470k10k10k10 = 500 µF.
[b] Create a 15 nF capacitor by combining two 10 nF (0.01 µF) capacitors in
series to get 5 nF. Then connect another 10 nF capacitor in parallel:
(0.01 µ)k(0.01 µ + 0.01 µ) = 10 n + 5 n = 15 nF.
[c] Combine two 100 pF capacitors in series to get a 50 pF capacitor. Then
combine the series pair in parallel with a 220 pF capacitor, a 470 pF
capacitor and a 10 pF capacitor to get 750 pF:
220k470k10k(100 + 100) = 220k470k10k50 = 750 pF.
P 6.32
[a]
vo
109 Z t
= −
900 × 10−6 e−2500x dx + 30
12 0
= −75,000
=
e−2500x t
+30
−2500 0
30e−2500t V,
t ≥ 0.
Problems
[b] v1
= −
=
[c] v2
[d] p
t ≥ 0.
109
e−2500x t
(900 × 10−6 )
−15
30
−2500 0
12e−2500t − 27 V,
t ≥ 0.
= vi = (30e−2500t )(900 × 10−6 )e−2500t
=
w
18e−2500t + 27 V,
= −
=
109
e−2500x t
(900 × 10−6 )
+45
20
−2500 0
=
27 × 10−3 e−5000t .
Z ∞
27 × 10−3 e−5000t dt
0
=
27 × 10−3
e−5000t ∞
−5000 0
= −5.4 × 10−6 (0 − 1) = 5.4 µJ.
[e] w
=
1
(20 × 10−9 )(45)2 + 12 (30 × 10−9 )(15)2
2
=
20.25 × 10−6 + 3.375 × 10−6
=
23.625 µJ.
[f ] wtrapped = winitial − wdelivered = 23.625 − 5.4 = 18.225 µJ.
[g] wtrapped
=
1
(20 × 10−9 )(27)2 + 12 (30 × 10−9 )(27)2
2
=
(10 + 15)(27)2 × 10−9
= 18.225 µJ.
CHECK: 18.225 + 5.4 = 23.625 µJ.
P 6.33
C1 = 1 + 1.5 = 2.5 nF;
1
1
1
1
1
=
+
+
= ;
C2
2.5 12.5 50
2
.·. C2 = 2 nF.
vd (0) + va (0) − vc (0) = 40 + 15 + 45 = 100 V.
[a]
6–25
6–26
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
vb
= −
109 Z t
50 × 10−6 e−250x dx + 100
2 0
= −25,000
[b] va
=
100(e−250t − 1) + 100
=
100e−250t V
109 Z t
50 × 10−6 e−250x dx + 15
= −
12.5 0
= −4000
[c] vc
e−250x t
+100
−250 0
e−250x t
+15
−250 0
=
16(e−250t − 1) + 15
=
16e−250t − 1 V.
=
109 Z t
50 × 10−6 e−250x dx − 45
50 0
=
1000
e−250x t
−45
−250 0
= −4(e−250t − 1) − 45
= −4e−250t − 41 V.
[d] vd
109 Z t
50 × 10−6 e−250x dx + 40
= −
2.5 0
= −20,000
=
80(e−250t − 1) + 40
=
80e−250t − 40 V.
CHECK: vb
[e] i1
e−250x t
+40
−250 0
= −10−9
= vd + va − vc
=
80e−250t − 40 + 16e−250t − 1 + 4e−250t + 41
=
100e−250t V (checks).
d
[80e−250t − 40]
dt
= −10−9 (−20,000e−250t )
=
20e−250t µA.
Problems
= −1.5 × 10−9
[f ] i2
d
[80e−250t − 40]
dt
= −1.5 × 10−9 (−20,000e−250t )
= 30e−250t µA.
CHECK: i1 + i2 = 50e−250t µA = ib .
P 6.34
[a] w(0)
[b]
=
[ 21 (2.5)(40)2 + 12 (12.5)(15)2 + 12 (50)(45)2 ] × 10−9
=
54,031.25 nJ.
va (∞)
= −1 V;
vc (∞)
= −41 V;
vd (∞)
= −40 V.
w(∞)
= [ 12 (2.5)(40)2 + 12 (12.5)(1)2 + 21 (50)(41)2 ] × 10−9
=
[c] w =
Z ∞
44,031.25 nJ.
(100e−250t )(50e−250t ) × 10−6 dt = 10,000 nJ.
0
CHECK: 54,031.25 − 44,031.25 = 10,000.
10,000
× 100 = 18.51%.
[d] % delivered =
54,031.25
[e] w
=
5 × 10−3
Z t
e−500x dx
0
=
104 (1 − e−500t ) nJ;
.·. 104 (1 − e−500t ) = 5000;
Thus, t =
P 6.35
e−500t = 0.5.
ln 2
= 1.39 ms.
500
v2 (t) = 20 × 10−3
dio
dt
= (20 × 10−3 )(50 × 10−3 ){e−8000t [−6000 sin 6000t + 12,000 cos 6000t]
+ (−8000e−8000t )[cos 6000t + 2 sin 6000t]}
= e−8000t {4 cos 6000t − 22 sin 6000t} V;
.·. v2 (0+ ) = 4 V.
i0 (0+ ) = 50 mA;
6–27
6–28
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
vR (0+ ) = 320(50 × 10−3 ) = 16 V;
v1 (0+ ) = 16 + 4 = 20 V.
P 6.36
−106 Z t −80x
vC =
e
sin 60x dx − 300
20
0
= 5e−80t [80 sin 60t + 60 cos 60t] − 300 − 300
= 400e−80t sin 60t + 300e−80t cos 60t − 600 V;
vL = 5
dio
dt
= −5[−80e−80t sin 60t + 60e−80t cos 60t]
= 400e−80t sin 60t − 300e−80t cos 60t V;
vo = −(vC + vL )
= −(400e−80t sin 60t + 300e−80t cos 60t − 600 + 400e−80t sin 60t − 300e−80t cos 60t)
= 600 − 800e−80t cos 60t V.
P 6.37
[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
di2
dig
di1
− 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
Thus,
4
di1
= −1280e−5t + 1088e−4t ;
dt
25i1 = 100 + 1600e−5t − 1700e−4t ;
Problems
8
6–29
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 . (OK)
8
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 . (OK)
P 6.38
[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 ).
6–30
P 6.39
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[a] vab = L1
0 = L1
di2
d(i1 − i2 )
+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 )
L1 + L2 − 2M
di1
=
vab
dt
L1 L2 − M 2
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 6.40
L1 L2 − M 2
.
L1 + L2 + 2M
[a] Yes, using KVL around the lower right loop
vo = v20Ω + v60Ω = 20(i2 − i1 ) + 60i2 .
[b] vo
=
20(1 − 52e−5t + 51e−4t − 4 − 64e−5t + 68e−4t )+
60(1 − 52e−5t + 51e−4t )
=
vo
[c] vo
20(−3 − 116e−5t + 119e−4t ) + 60 − 3120e−5t + 3060e−4t ;
= −5440e−5t + 5440e−4t V.
= L2
d
di1
(ig − i2 ) + M
dt
dt
d
d
(15 + 36e−5t − 51e−4t ) + 8 (4 + 64e−5t − 68e−4t )
dt
dt
−5t
−4t
−5t
= −2880e + 3264e − 2560e + 2176e−4t ;
=
vo
16
= −5440e−5t + 5440e−4t V.
Problems
P 6.41
[a] Summing the voltages around mesh 1 yields
d(i2 + ig )
di1
+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
di2
dig
di1
− 20i1 + 16
+ 800i2 = −16 .
8
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
= −0.4A;
.·. i1 (∞) = −
+
20
780
From the solutions for i1 and i2 we have
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 ;
i2 (∞) = −
7.8
= −0.01A.
780
6–31
6–32
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)
Problems
6–33
P 6.42
When the switch is closed, the induced voltage in the coil connected to the
source is negative at the dotted terminal. Since the dc voltmeter kicks
up-scale, the induced voltage in the coil connected to the voltmeter is positive
at the upper terminal. Therefore, dot the lower terminal of the coil connected
to the voltmeter.
P 6.43
[a] 5
dig
di2
+ 40
+ 90i2 = 0;
dt
dt
40
di2
dig
+ 90i2 = −5 .
dt
dt
[b] i2 = e−t − 5e−2.25t A;
di2
= −e−t + 11.25e−2.25t A/s;
dt
ig = 10e−t − 10 A
dig
= −10e−t A/s;
dt
40[−e−t + 11.25e−2.25t ] + 90[e−t − 5e−2.25t ] = 50e−t ;
−5[−10e−t ] = 50e−t .
[c] v1
di2
dig
+5
dt
dt
=
10
=
10(−10e−t ) + 5(−e−t + 11.25e−2.25t )
= −105e−t + 56.26e−2.25t V,
t > 0.
[d] v1 (0) = −105 + 56.25 = −48.75 V;
Å
dig
di2
v1 (0) = 10 (0) + 5 (0)
dt
dt
= 10(−10) + 5(−1 + 11.25) = −100 + 56.26 = −48.75 V.
Yes, the initial value of v1 is consistent with known circuit behavior.
P 6.44
[a] Dot terminal 2; with current entering terminal 2, the flux is right-to-left
coil 1-2. Assign the current into terminal 4; the flux is left-to-right in coil
3-4. The flux is in the same direction, due to the topology of the core, so
dot terminal 4. Hence, 2 and 4 or 1 and 3.
[b] Dot terminal 1; with current entering terminal 1 the flux is up in coil 1-2.
Assign the current into terminal 4; the flux is left-to-right in coil 3-4.
Therefore the flux is in the same direction, due to the topology of the
core, so dot terminal 4. Hence, 1 and 4 or 2 and 3.
6–34
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
1
P11
P22
P11
[a] 2 = 1 +
1+
= 1+
k
P12
P12
P21
Therefore
P12 P21
k2 =
.
(P21 + P11 )(P12 + P22 )
P 6.45
P22
1+
.
P12
Now note that
φ1 = φ11 + φ21 = P11 N1 i1 + P21 N1 i1 = N1 i1 (P11 + P21 ),
and similarly
φ2 = N2 i2 (P22 + P12 ).
It follows that
(P11 + P21 ) =
φ1
N 1 i1
and
!
φ2
.
N 2 i2
(P22 + P12 ) =
Therefore
φ12 φ21
(φ12 /N2 i2 )(φ21 /N1 i1 )
=
k2 =
(φ1 /N1 i1 )(φ2 /N2 i2 )
φ1 φ2
or
v
u
u
k=t
φ21
φ1
!
!
φ12
.
φ2
[b] The fractions (φ21 /φ1 ) and (φ12 /φ2 ) are by definition less than 1.0,
therefore k < 1.
P 6.46
M2
k 2 L1
[a] L2 =
N1
=
N2
[b] P1 =
=
L1
=
L2
(0.1)2
= 160 mH;
(0.5)2 (0.250)
s
250
= 1.25.
160
L1
0.250
=
= 0.25 × 10−6 Wb/A;
2
2
N1
(1000)
P2 =
P 6.47
s
!
L2
0.16
=
= 0.25 × 10−6 Wb/A.
2
N2
(800)2
[a] L1 = N12 P1 ;
P11
= 0.5;
P21
P1 =
288 × 10−3
= 288 nWb/A;
106
P21 = 2P11 ;
Problems
6–35
.·. 288 × 10−9 = P11 + P21 = 3P11 .
P11 = 96 nWb/A;
q
P21 = 192 nWb/A;
q
M = k L1 L2 = (1/3) (0.288)(0.162) = 72 mH;
N2 =
[b] P2 =
M
72 × 10−3
=
= 375 turns.
N1 P21
(1000)(192 × 10−9 )
L2
162 × 10−3
= 1152 nWb/A.
=
N22
(375)2
[c] P11 = 96 nWb/A [see part (a)].
φ22
P22
P2 − P12
P2
[d]
=
=
=
− 1;
φ12
P12
P12
P12
P21 = P21 = 192 nWb/A;
P2 = 1152 nWb/A;
φ22
1152
=
− 1 = 5.
φ12
192
P 6.48
[a] w = (0.5)L1 i21 + (0.5)L2 i22 + M i1 i2 ;
q
M = 0.8 (0.025)(0.1) = 40 mH;
w = (0.5)(0.025)(10)2 + (0.5)(0.1)(15)2 + (0.04)(10)(15) = 18.5 J.
[b] w = (0.5)(0.025)(−10)2 + (0.5)(0.1)(−15)2 + (0.04)(−10)(−15) = 18.5 J.
[c] w = (0.5)(0.025)(−10)2 + (0.5)(0.1)(15)2 + (0.04)(−10)(15) = 6.5 J.
[d] w = (0.5)(0.025)(10)2 + (0.5)(0.1)(−15)2 + (0.04)(10)(−15) = 6.5 J.
P 6.49
q
[a] M = 1.0 (0.025)(0.1) = 50 mH,
i1 = 10 A.
Therefore w = 0.5(0.025)(10)2 + 0.5(0.1)i22 + 0.05(10)i2 = 0.
Place the equation in the standard quadratic form to give
0.05i22 + 0.5i2 + 1.25 = 0.
Solving,
i2 =
−0.5 ±
q
0.52 − 4(0.05)(1.25)
2(0.05)
= −5 A.
[b] No, setting w equal to a negative value will make the quantity under the
square root sign negative.
6–36
P 6.50
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
P1 =
L1
400 × 10−6
=
= 6.4 nWb/A;
N12
2502
P2 =
L2
900 × 10−6
=
= 3.6 nWb/A;
N22
5002
q
M = k L1 L2 = 450 µH;
M
450 × 10−6
P12 = P21 =
=
= 3.6 nWb/A;
N1 N2
(250)(500)
P11 = P1 − P21 = 6.4 − 3.6 = 2.8 nWb/A.
P 6.51
Since P1 = P2 ,
L1
0.025
N12
=
=
= 0.25.
2
N2
L2
0.1
Therefore,
N1 √
= 0.25 = 0.5,
N2
so
N2 =
P 6.52
N1
500
=
= 1000 turns.
0.5
0.5
When the touchscreen in the mutual-capacitance design is touched at the point
x, y, the touch capacitance Ct is present in series with the mutual capacitance
at the touch point, Cmxy . Remember that capacitances combine in series the
way that resistances combine in parallel. The resulting mutual capacitance is
0
Cmxy
=
P 6.53
Cmxy Ct
.
Cmxy + Ct
[a] The self-capacitance and the touch capacitance are effectively connected in
parallel. Therefore, the capacitance at the x-grid electrode closest to the
touch point with respect to ground is
Cx = Cp + Ct = 80 pF + 20 pF = 100 pF.
The same capacitance exists at the y-grid electrode closest to the touch
point with respect to ground.
[b] The mutual-capacitance and the touch capacitance are effectively
connected in series. Therefore, the mutual capacitance between the
x-grid and y-grid electrodes closest to the touch point is
0
Cmxy
=
Cmxy Ct
(80)(20)
=
= 16 pF.
Cmxy + Ct
80 + 20
Problems
6–37
[c] In the self-capacitance design, touching the screen increases the
capacitance being measured at the point of touch. For example, in part
(a) the measured capacitance before the touch is 80 pF and after the
touch is 100 pF. In the mutual-capacitance design, touching the screen
decreases the capacitance being measured at the point of touch. For
example, in part (b) the measured capacitance before the touch is 80 pF
and after the touch is 16 pF.
P 6.54
[a] The four touch points identified are the two actual touch points and two
ghost touch points. Their coordinates, in inches from the upper left
corner of the screen, are
(1.8, 2.4);
(3.0, 4.6);
(1.8, 4.6);
and
(3.0, 2.4).
These four coordinates identify a rectangle within the screen, shown
below.
[b] The touch points identified at time t1 are those listed in part (a). The
touch points recognized at time t2 are
(2.2, 2.8);
(2.5, 3.4);
(2.2, 3.4);
and
(2.5, 2.8).
The first two coordinates are the actual touch points and the last two
coordinates are the associated ghost points. Again, the four coordinates
identify a rectangle at time t2 , as shown here:
6–38
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
Here, the rectangle at time t2 is smaller than the rectangle at time t1 , so
the software would recognize the two fingers are moving toward the
middle of the screen. This pinch gesture thus specifies a zoom-out for the
screen.
[c] The touch points identified at time t1 are those listed in part (a). The
touch points recognized at time t2 are
(1.2, 1.8);
(3.6, 5.0);
(1.2, 5.0);
and
(3.6, 1.8).
The first two coordinates are the actual touch points and the last two
coordinates are the associated ghost points. Again, the four coordinates
identify a rectangle at time t2 , as shown here:
Note that the rectangle at time t2 is larger than the rectangle at time t1 ,
so the software would recognize the two fingers are moving toward the
edges of the screen. This pinch gesture thus specifies a zoom-in for the
screen.
Response of First-Order RL and
RC Circuits
Assessment Problems
60
= 0.5 A.
120
0.32
L
=
= 2 ms.
[b] τ =
R
160
[c] i = 0.5e−500t A,
t ≥ 0;
AP 7.1 [a] i(0) =
v1 = L
d
(0.5e−500t ) = −80e−500t V
dt
v2 = −70i = −35e−500t V
t ≥ 0+ ;
t ≥ 0+ .
1
[d] w(0) = (0.32)(0.5)2 = 40 mJ;
2
w90Ω =
Z t
90(0.25e−1000x ) dx = 22.5
0
e−1000x t
= 22.5(1 − e−1000t ) mJ;
−1000 0
w90Ω (1 ms) = 0.0225(1 − e−1 ) = 14.22 mJ;
% dissipated =
14.22
(100) = 35.6%.
40
AP 7.2 [a] t < 0:
iL (0+ ) = 8 A.
7–1
7–2
CHAPTER 7. Response of First-Order RL and RC Circuits
t > 0:
Req =
τ=
(10)(40)
+ 10 = 18 Ω;
50
0.072
L
=
= 4 ms;
Req
18
1
= 250;
τ
.·. iL = 8e−250t A.
diL
.·. vo = −10iL − 0.072
= −80e−250t + 144e−250t
dt
= 64e−250t A t ≥ 0+ .
1
[b] w(0) = (72 × 10−3 )(8)2 = 2304 mJ;
2
vo2
642 −500t
p40Ω =
=
e
= 102.4e−500t W;
40
40
w40Ω =
Z ∞
%diss =
102.4e−500t dt = 204.8 mJ;
0
204.8
(100) = 8.89%.
2304
AP 7.3 [a] For t < 0:
v(0) = 150 V.
1
1
[b] w(0) = Cv(0)2 = (40 × 10−9 )(150)2 = 450 µJ.
2
2
Problems
[c] For t > 0:
Req = 5000 + 30,000k60,000 = 25 kΩ;
τ = Req C = (25,000)(40 × 10−9 ) = 1 ms.
[d] v(t) = v(0)e−t/τ = 150e−1000t V,
t ≥ 0.
AP 7.4 [a] Calculate the initial voltage drop across the capacitor:
v(0) = (2700k3300)(40 mA) = (1485)(40 × 10−3 ) = 59.4 V.
The equivalent resistance seen by the capacitor is
Req = 3000k(2400 + 3600) = 3000k6000 = 2 kΩ;
τ = Req C = (2000)(0.5 × 10−6 ) = 1000 µs;
v = v(0)e−t/τ = 59.4e−1000t V
1
= 1000;
τ
t ≥ 0.
v
= 9.9e−1000t mA, t ≥ 0+ .
2400 + 3600
1
[c] w(0) = (0.5 × 10−6 )(59.4)2 = 882.09 µJ;
2
[b] i =
i3k =
59.4e−1000t
= 19.8e−1000t mA;
3000
p3k = [(19.8 × 10−3 )e−1000t ]2 (3000) = 1.176e−2000t ;
−6
1.176 −1
e−2000x 500×10
=
(e − 1) = 371.72 µJ;
w3k (500 µs) = 1.176
−2000 0
−2000
%=
371.72
× 100 = 42.14%.
882.09
7–3
7–4
CHAPTER 7. Response of First-Order RL and RC Circuits
AP 7.5 [a] For t < 0, source-transform the circuit to the right of the 200 mH
inductor. We get
i(0− ) =
30 − 250
= −4 mA;
25,000 + 30,000
i(0− ) = i(0+ ) = −4 mA.
[b] For t > 0, the circuit reduces to
Therefore i(∞) = 30/30,000 = 1 mA.
[c] τ =
L
200 × 10−3
=
= 6.67 µs.
R
30,000
[d] i(t) = i(∞) + [i(0+ ) − i(∞)]e−t/τ
= 0.001 + [−0.004 − 0.001]e−150,000t = 1 − 5e−150,000t mA,
AP 7.6 [a] vc (0+ ) = −50 V.
[b] Find the Thévenin equivalent with respect to the terminals of the
capacitor:
VTh = v20Ω =
20
(−30) = −24 V,
25
RTh = 20k5 = 4 Ω.
Therefore τ = Req C = (4)(25 × 10−3 ) = 0.1 s.
[c] Use voltage division to find the final value of voltage:
vc (∞) =
20
(−30) = −24 V.
25
t ≥ 0.
Problems
The simplified circuit for t > 0 is:
−24 − (−50)
= 6.5 A.
4
[e] vc = vc (∞) + [vc (0+ ) − vc (∞)]e−t/τ
[d] i(0+ ) =
= −24 + (−50 + 24)e−t/τ = −24 − 26e−10t V,
[f ] i = C
dvc
= (25 × 10−3 )(−10)(−26e−10t ) = 6.5e−10t A,
dt
AP 7.7 For t < 0:
Io = i(0) =
20,000k60,000k10,000
(20 × 10−3 ) = 12 mA.
10,000
For t ≥ 0:
Req = 10,000 + (10,000k20,000k60,000) = 16 kΩ;
τ=
L
0.05
=
= 3.125 µs;
Req
16,000
i(t) = Io e−t/τ = 12e−320,000t mA
t ≥ 0.
t ≥ 0.
t ≥ 0+ .
7–5
7–6
CHAPTER 7. Response of First-Order RL and RC Circuits
AP 7.8 [a] For t < 0:
vo (0) =
10,000
(75) = 50 V.
15,000
For t ≥ 0:
vo (∞) =
40,000
(−100) = −80 V;
50,000
Req = 40,000k10,000 = 8 kΩ;
τ = Req C = (8000)(40 × 10−9 ) = 0.32 ms;
vo (t) = vo (∞) + (vo (0) − vo (∞))e−t/τ = −80 + (50 + 80)e−3125t
= −80 + 130e−3125t V.
[b] For t ≥ 0:
130e−3125t − 80 + 100
io =
= (13e−3125t + 2) mA.
10,000
Problems
7–7
AP 7.9 [a] Use the circuit shown below, for t < 0, use voltage division to calculate the
initial voltage drop across the capacitor:
vC (0− ) =
20,000k60,000
(80) = 40 V.
(20,000k60,000) + 15,000
Now use the next circuit, valid for 0 ≤ t ≤ 20 ms, to calculate vC (t) for
that interval:
For 0 ≤ t ≤ 20 ms:
τ = RC = (20,000)(0.5 × 10−6 ) = 10 ms;
vC (t) = vC (0− )et/τ = 40e−100t V 0 ≤ t ≤ 20 ms.
[b] Calculate the starting capacitor voltage in the interval t ≥ 20 ms, using
the capacitor voltage from the previous interval:
vC (0.02) = 40e−100(0.02) = 5.4134 V.
Now use the next circuit, valid for t ≥ 20 ms, to calculate vC (t) for that
interval:
For t ≥ 20 ms :
Req = 80 kΩk20 kΩ = 16 kΩ;
τ = Req C = (16 × 103 )(0.5 × 10−6 ) = 8 ms.
Therefore vC (t) = vC (0.02)e−(t−0.02)/τ = 5.4134e−125(t−0.02) V,
t ≥ 0.02 s.
[c] To calculate the energy dissipated in the 20 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.
w20 k =
Z 0.02
0
Z ∞
[40e−100t ]2
[5.4134e−125(t−0.02) ]2
dt +
dt
20,000
20,000
0.02
7–8
CHAPTER 7. Response of First-Order RL and RC Circuits
=
1600 e−200t 0.02 5.41342 e−250(t−0.02) ∞
+
20,000 (−200) 0
20,000 (−250)
0.02
= − 400 × 10−6 (e−4 − 1) − 5.861 × 10−6 (0 − 1) = 398.5 µJ.
[d] Repeat the process in part (c), but recognize that the voltage across this
resistor is non-zero only for the second interval:
w80 kΩ =
∞
[5.4134e−125(t−0.02) ]2
5.41342
dt =
e−250(t−0.02)
= 1.5 µJ.
80,000
80,000(−250)
0.02
0.02
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 20 kΩ
resistor and the 80 kΩ resistor.
Check: wstored = (1/2)(0.5 × 10−6 )(40)2 = 400 µJ;
wdiss = 398.5 + 1.5 = 400 µJ.
AP 7.10 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.
Problems
5 ≤ t < ∞:
τ=
5
= 5 s;
1
1/τ = 0.2;
iL (t) = 2e−0.2(t −5) A,
t ≥ 5 s.
AP 7.11 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:
d
−36 − 14
+ 2.5 × 10−6 (−36 − vo ) = 0;
80,000
dt
.·.
2.5 × 10−6
−50
dvo
=
.
dt
80,000
Separate the variables and integrate:
dvo
= −250
dt
Z vo (t)
.·.
dx = −250
vo (0)
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
.·.
t=
20
= 80 ms.
250
7–9
7–10
CHAPTER 7. Response of First-Order RL and RC Circuits
Z t
1
−4t
AP 7.12 [a] vo = −
4 dx + 0 =
;
−6
R(0.5 × 10 ) 0
R(0.5 × 10−6 )
−4(15 × 10−3 )
= −10;
R(0.5 × 10−6 )
.·.
[b] vo =
.·.
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 )
Problems
7–11
Problems
P 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:
75
v
i(0− ) = − = − = −12.5 A.
6
6
1
1 2
[b] w(0) = Li (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− )e−t/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) = 21 Li2 (5 ms) = 12 (8) × 10−3 (3.58)2 = 51.3 mJ.
w (dis) = 625 − 51.3 = 573.7 mJ;
573.7
100 = 91.8%.
% dissipated =
625
P 7.2
[a] t < 0
15 kΩk15 kΩ = 7.5 kΩ;
ig (0− ) =
9
= 0.4 mA;
(15 + 7.5) × 103
7–12
CHAPTER 7. Response of First-Order RL and RC Circuits
i1 (0− ) = i2 (0− ) = (0.4) × 10−3
(15)
= 0.2 mA.
(30)
[b] i1 (0+ ) = i1 (0− ) = 0.2 mA;
i2 (0+ ) = −i1 (0+ ) = −0.2 mA
[c] τ =
(when switch is open).
L
30 × 10−3
=
= 10−6 ;
R
15 × 103 + 15 × 103
1
= 106 ;
τ
i1 (t) = i1 (0+ )e−t/τ .
6
i1 (t) = 0.2e−10 t mA,
t≥0
when t ≥ 0+
[d] i2 (t) = −i1 (t)
6
.·. i2 (t) = −0.2e−10 t mA,
t ≥ 0+
[e] The current in a resistor can change instantaneously. The switching
operation forces i2 (0− ) to equal 0.2 mA and i2 (0+ ) = −0.2 mA.
P 7.3
[a] For t = 0− the circuit is:
io (0− ) = 0
since the switch is open for t < 0. vL (0− ) = 0 since the
inductor behaves like a short circuit for t < 0.
300 Ωk300 Ω = 150 Ω;
.·. ig =
25
= 125 mA;
50 + 150
300
ig = 62.5 mA.
iL (0 ) =
600
−
[b] For t = 0+ the circuit is:
300 Ωk100 Ω = 75 Ω;
Problems
.·. ig =
7–13
25
= 200 mA.
50 + 75
300
ia =
200 = 150 mA;
400
.·. io (0+ ) = 150 − 62.5 = 87.5 mA.
iL (0+ ) = iL (0− ) = 62.5 mA.
.·. vL (0+ ) = −12.5 V
200(0.0625) + vL (0+ ) = 0;
[c] io (∞) = ia = 150 mA.
iL (∞) = 0,
since the switch short circuits the branch containing the
200 Ω resistor and the 50 mH inductor.
vL (∞) = 0,
since the current in the inductor is a constant at t = ∞.
[d] τ =
L
50 × 10−3
=
= 0.25 ms;
R
200
1
= 4000;
τ
.·. iL = 0 + (62.5 − 0)e−4000t = 62.5e−4000t mA,
[e] io = ia − iL = 150 − 62.5e−4000t mA,
t ≥ 0+ .
[f ] vL (t) = 0 + (−12.5 − 0)e−4000t = −12.5e−4000t V,
P 7.4
[a] For t < 0
ig =
80
= 2 A;
15 + (50k50)
iL (0− ) =
2(50)
= 1 A = iL (0+ ).
(100)
For t > 0
iL (t) = iL (0+ )e−t/τ A,
τ=
t ≥ 0;
L
0.20
=
= 0.01 s;
R
3 + (60k20) + 2
t ≥ 0.
t ≥ 0+ .
7–14
CHAPTER 7. Response of First-Order RL and RC Circuits
iL (0+ ) = 1 A;
iL (t) = e−100t A,
t ≥ 0.
vo (t) = −(60k20)iL (t) = −15e−100t V,
[b] p20Ω =
t ≥ 0+ .
vo2
= 11.25e−200t W;
20
wdiss =
Z 0.01
11.25e−200t dt
0
11.25 −200t 0.01
e
−200
0
= 56.25 × 10−3 (1 − e−2 ) = 48.64 mJ.
=
1
wstored = (0.2)(1)2 = 100 mJ.
2
48.64
× 100 = 48.64%.
% diss =
100
P 7.5
[a] iL (0) =
80
= 2 A;
16 + 12 + 4 + 8
io (0+ ) =
80
− 2 = 4 − 2 = 2 A;
16 + 4
io (∞) =
80
= 4 A.
16 + 4
[b] iL = 2e−t/τ ;
τ=
20
L
=
× 10−3 = 1 ms;
R
12 + 8
iL = 2e−1000t A;
io = 4 − iL = 4 − 2e−1000t A,
t ≥ 0+ .
[c] 4 − 2e−1000t = 3.8;
0.2 = 2e−1000t .
e1000t = 10
P 7.6
.·. t = 2.30 ms.
First, use the circuit for t < 0 to find the initial current in the inductor:
Problems
7–15
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+ .
P 7.7
From Problem 7.6, 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:
v4Ω (t) = L
di
= 0.32(−10)(4e−10t ) = −12.8e−10t V,
dt
2
v4Ω
p4Ω (t) =
= 40.96e−20t W,
4
w4Ω (t) =
Z ∞
t ≥ 0+ ;
t ≥ 0+ ;
40.96e−20t dt = 2.048 J.
0
Find the percentage of the initial energy in the inductor dissipated in the 4 Ω
resistor:
% dissipated =
2.048
100 = 80%.
2.56
7–16
P 7.8
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] vo (t) = vo (0+ )e−t/τ ;
−3
.·. vo (0+ )e−5×10 /τ = 0.25vo (0+ );
−3
.·. e5×10 /τ = 4;
−3
5 × 10
L
.·. τ =
=
R
ln 4
;
−3
250 × 10
.·. L =
ln 4
= 180.34 mH.
1
= 10 mA = iL (0+ );
6
−
[b] iL (0 ) = 60
1
1
wstored = LiL (0+ )2 = (Rτ )(100 × 10−6 ) = 2500τ µJ.
2
2
iL (t) = 10e−t/τ mA.
p50Ω = i2L (50) = 5000 × 10−6 e−2t/τ ;
wdiss =
Z 5×10−3
5000 × 10−6 e−2t/τ dt
0
−6
= 5000 × 10
e−2t/τ 5×10
(−2/τ ) 0
= 2500 × 10−6 τ 1 − e
e
−10×10−3
τ
−3
−10×10−3
τ
.
= e−2 ln 4 = 0.0625;
wdiss = 2500 × 10−6 τ (0.9375);
% diss =
P 7.9
2500 × 10−6 τ (0.9375)
× 100 = 93.75%.
2500 × 10−6 τ
1
w(0) = (20 × 10−3 )(102 ) = 1 J;
2
0.5w(0) = 0.5 J;
iR = 10e−t/τ ;
pdiss = i2R R = 100Re−2t/τ ;
wdiss =
Z t
0
R(100)e−2x/τ dx;
Problems
wdiss = 100R
e−2x/τ
−2/τ
t
7–17
= −50τ R(e−2t/τ − 1) = 50L(1 − e−2t/τ );
0
50L = (50)(20) × 10−3 = 1;
to = 10 µs;
1 − e−2to /τ = 0.5;
2to
2to R
=
= ln 2;
τ
L
e2to /τ = 2;
R=
P 7.10
20 × 10−3 ln 2
L ln 2
=
= 693.15 Ω.
2to
20 × 10−6
1
[a] w(0) = LIg2 ;
2
e−2t/τ to
(−2/τ ) 0
0
1 2
1
= Ig Rτ (1 − e−2to /τ ) = Ig2 L(1 − e−2to /τ ).
2
2
wdiss =
Z to
Ig2 Re−2t/τ dt = Ig2 R
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
(20 × 10−3 ) ln[1/0.5]
;
20 × 10−6
R = 693.15 Ω.
P 7.11
[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,
R=
R=
0.01
= 50 Ω.
0.2 × 10−3
7–18
CHAPTER 7. Response of First-Order RL and RC Circuits
We can construct a 50 Ω resistor using two parallel 100 Ω resistors. The
resulting circuit is
[b] i(t) = Io e−t/τ = 40e−5000t mA,
t ≥ 0.
1
1
[c] w(0) = LIo2 = (0.01)(0.04)2 = 8 µJ;
2
2
1
w(t) = (0.01)(0.04e−5000t )2 = 8 × 10−6 e−10,000t .
2
3
So 8 × 10−6 e−10,000t = w(0) = 6 × 10−6;
4
e−10,000t = 0.75
then
ln(4/3)
.·. t =
= 28.77 µs.
10,000
P 7.12
e10,000t = 4/3;
(for a 10 mH inductor)
v
100e−80t
=R=
= 25 Ω.
i
4e−80t
1
[b] τ =
= 12.5 ms.
80
L
[c] τ =
= 12.5 × 10−3 ;
R
[a]
L = (12.5)(25) × 10−3 = 312.5 mH.
1
1
[d] w(0) = L[i(0)]2 = (0.3125)(16) = 2.5 J.
2
2
[e] wdiss =
Z t
400e−160x dx = 2.5 − 2.5e−160t ;
0
0.8w(0) = (0.8)(2.5) = 2 J;
2.5 − 2.5e−160t = 2
.·. e160t = 5;
Solving, t = 10.06 ms.
P 7.13
[a] t < 0
40 kΩk20 kΩ = 13.33 kΩ;
Problems
60 kΩk30 kΩ = 20 kΩ;
(120 × 10−3 )(13.33 × 103 ) = 1600 V.
iL (0− ) =
1600
= 48 mA.
33,333.33
t>0
τ=
L
0.2
=
= 2.5 µs;
R
80,000
iL (t) = 48e−400,000t mA,
1
= 400,000;
τ
t ≥ 0.
p60k = (0.048e−400,000t )2 (60,000) = 138.24e−800,000t W.
w60k =
Z t
138.24e−800,000x dx = 172.8 × 10−6 [1 − e−800,000t ] J;
0
1
wL (0) = (.2)(48 × 10−3 )2 = 230.4 µJ;
2
0.25wL (0) = 57.6 µJ;
172.8(1 − e−800,000t ) = 57.6;
t=
.·. e800,000t = 1.5;
ln 1.5
= 0.507 µs.
800,000
[b] wdiss (t) = 230.4(1 − e−800,000t ) µJ;
wdiss (0.507 µs) = 76.82 µJ;
% = (76.82/230.4)(100) = 33.3%.
7–19
7–20
P 7.14
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] t < 0 :
ig (0) =
54
= 9 A;
3 + (9k4.5)
iL (0) =
9
ig (0) = 6 A.
9 + 4.5
t > 0:
i∆ =
iT (200)
2
= iT ;
300
3
vT = 50i∆ + iT
(100)(200)
2 200
= 50iT +
iT ;
300
3
3
100 200
vT
= RTh =
+
= 100 Ω.
iT
3
3
τ=
200
L
=
× 10−3
R
100
iL = 6e−500t A,
1
= 500;
τ
t ≥ 0.
[b] vL = 200 × 10−3 (−3000e−500t ) = −600e−500t V,
[c]
vL = 50i∆ + 100i∆ = 150i∆ ;
i∆ =
vL
= −4e−500t A
150
t ≥ 0+ .
t ≥ 0+ .
Problems
P 7.15
1
w(0) = (200 × 10−3 )(36) = 3.6 J;
2
p50i∆ = −50i∆ iL = −50(−4e−500t )(6e−500t ) = 1200e−1000t W;
w50i∆ =
Z ∞
−1000t
1200e
0
% dissipated =
P 7.16
e−1000t ∞
dt = 1200
= 1.2 J;
−1000 0
1.2
(100) = 33.33%.
3.6
t<0
iL (0− ) = iL (0+ ) = 25 A.
t > 0:
Find Thévenin resistance seen by inductor:
iT = 5vT ;
τ=
vT
1
= RTh = = 0.2 Ω;
iT
5
L
50 × 10−3
=
= 250 ms;
R
0.2
io = 25e−4t A,
t ≥ 0;
1/τ = 4.
7–21
7–22
CHAPTER 7. Response of First-Order RL and RC Circuits
vo = L
P 7.17
dio
= (50 × 10−3 )(−100e−4t ) = −5e−4t V,
dt
t ≥ 0+ .
[a] t < 0:
t = 0+ :
t > 0:
iR = 5e−t/τ mA;
τ=
L
= 20 × 10−6 ;
R
iR = 5e−50,000t mA.
vR = (2.5 × 103 )(5 × 10−3 )e−50,000t = 12.5e−50,000t V;
v1 = 20 × 10−3 [5 × 10−3 (−50,000)e−50,000t ] = −5e−50,000t V;
vo = −v1 − vR = −7.5e−50,000t V.
Problems
[b] io =
P 7.18
7–23
103 Z t
−7.5e−50,000x dx + 0 = 3.125e−50,000t − 3.125 mA.
48 0
[a] From the solution to Problem 7.17,
iR = 5 × 10−3 e−50,000t A;
pR = (25 × 10−6 e−100,000t )(2.5 × 103 ) = 62.5 × 10−3 e−100,000t W;
wdiss =
Z ∞
62.5 × 10−3 e−100,000t dt
0
−100,000t
−3 e
= 62.5 × 10
−105
∞
= 625 nJ.
0
1
1
[b] wtrapped = Leq i2R (0) = (50 × 10−3 )(5 × 10−3 )2 = 625 nJ.
2
2
CHECK:
w(0) = 12 (20)(25 × 10−6 ) × 10−3 + 12 (80)(25 × 10−6 ) × 10−3 = 1250 nJ;
.·. w(0) = wdiss + wtrapped .
P 7.19
[a] t < 0 :
t = 0+ :
220 = iab + (50/3) + (10/3),
iab = 200 A,
t = 0+ .
7–24
CHAPTER 7. Response of First-Order RL and RC Circuits
[b] At t = ∞:
t → ∞.
iab = 220/1 = 220 A,
2
× 10−3 = 0.167 ms;
12
[c] i1 (0) = 50/3,
τ1 =
i2 (0) = 10/3,
τ2 =
15
× 10−3 = 0.25 ms;
60
i1 (t) = (50/3)e−6000t A,
t ≥ 0;
i2 (t) = (10/3)e−4000t A,
t ≥ 0;
iab = 220 − (50/3)e−6000t − (10/3)e−4000t A,
220 − (50/3)e−6000t − (10/3)e−4000t = 210;
30 = 50e−6000t + 10e−4000t ;
3 = 5e−6000t + e−4000t ;
By trial and error
t = 123.1 µs.
P 7.20
[a] t < 0:
iL (0− ) =
−2.5(16)
= −2 mA.
(20)
t ≥ 0;
Problems
7–25
t ≥ 0:
40 × 10−3
= 40 × 10−6 ;
1/τ = 25,000;
103
vo = −1000(−2 × 10−3 )e−25,000t = 2e−25,000t V,
τ=
t ≥ 0+ .
1
[b] wdel = (40 × 10−3 )(4 × 10−6 ) = 80 nJ.
2
[c] 0.95wdel = 76 nJ;
.·. 76 × 10−9 =
Z to
0
4e−50,000t
dt;
1000
.·. 76 × 10−9 = −80 × 10−9 e−50,000t
to
= 80 × 10−9 (1 − e−50,000to );
0
.·. e−50,000to = 0.05.
50,000to = ln 20 so to = 59.9 µs;
59.9
to
=
= 1.498 so to ≈ 1.5τ.
.·.
τ
40
P 7.21
[a] t < 0 (after source-transforming the parallel resistor and current source):
8
× 10−3 = 0.2 mA;
40
10,000
vC (0− ) = v10k (0− ) =
(10 − 2) = 2 V.
20,000 + 5000 + 10,000 + 5000
i1 (0− ) = i2 (0− ) =
[b] t = 0+ :
i1 (0+ ) =
2
× 10−3 = 0.2 mA;
10
7–26
CHAPTER 7. Response of First-Order RL and RC Circuits
i2 (0+ ) =
−2
× 10−3 = −0.2 mA.
10
[c] Capacitor voltage cannot change instantaneously, therefore,
i1 (0+ ) =
2
= 0.2 mA.
10,000
[d] Switching can cause an instantaneous change in the current in a resistive
branch. In this circuit
i2 (0− ) = 0.2 mA and i2 (0+ ) = −0.2 mA.
[e] vc = 2e−t/τ V,
t ≥ 0;
τ = Re C = 5000(0.4) × 10−6 = 2 × 10−3 ;
vc = 2e−500t V,
i1 =
[f ] i2 =
P 7.22
t ≥ 0;
vc
= 0.2e−500t mA,
10,000
t ≥ 0.
−vc
= −0.2e−500t mA,
5000 + 5000
t ≥ 0+ .
[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
(7.5 × 10−3 ) = 4 mA.
80 × 103 + 20 × 103 + 50 × 103
Use Ohm’s law to find the voltage drop:
v(0− ) = (50 × 103 )i50k = (50 × 103 )(0.004) = 200 V.
i50k =
[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,
1
1
[d] w(0) = Cv 2 = (0.4 × 10−6 )(200)2 = 8 mJ.
2
2
t ≥ 0.
Problems
7–27
1
1
[e] w(t) = Cv 2 (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 ,
P 7.23
e100t = 4,
t = (ln 4)/100 = 13.86 ms.
For t < 0,
V0 =
25,000
(60) = 15 V.
75,000 + 25,000
For t ≥ 0,
Req = 25 kΩ,
τ = Req C = (25,000)(5 × 10−9 ) = 125 µs.
so
There is no source in the circuit as t → ∞ so Vf = 0. Thus,
vo (t) = Vf + (V0 − Vf )e−t/τ
= 0 + (15 − 0)e−t/125µ
= 15e−8000t V,
P 7.24
t ≥ 0.
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:
Find the current in the loop and use it to find the initial voltage drops across
the two RC circuits:
i=
15
= 0.2 mA,
75,000
v5 (0− ) = 4 V,
v1 (0− ) = 8 V.
7–28
CHAPTER 7. Response of First-Order RL and RC Circuits
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.
P 7.25
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;
1
1
2
−6
2 ∼
w1 (60 ms) = 2 Cv1 (60 ms) = 2 (1 × 10 )(1.79) = 1.59 µJ;
w5 (60 ms) = 12 Cv52 (60 ms) = 21 (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 + 21 (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 %.
P 7.26
[a] Note that there are many different possible correct solutions to this
problem.
τ
R= .
C
Choose a 10 µF capacitor from Appendix H. Then,
0.004
= 400 Ω.
10 × 10−6
Construct a 400 Ω resistor by combining a 180 Ω resistor and a 220 Ω
resistor in series:
R=
[b] v(t) = Vo e−t/τ = 120e−250t V,
−250t
[c] 120e
= 30
so
ln 4
.·. t =
= 5.545 ms.
250
250t
e
t ≥ 0.
= 4;
Problems
P 7.27
[a] t < 0:
vo (0) =
12,000k68,000
(60) = 51 V.
1800 + (12,000k68,000)
t > 0:
1
τ = (12) × 10−3 = 2 ms;
6
vo = 51e−500t V,
p=
1
= 500;
τ
t ≥ 0.
vo2
× 10−3 = 216.75 × 10−3 e−1000t W.
12
wdiss =
Z 2×10−3
216.75 × 10−3 e−1000t dt
0
= 216.75 × 10−6 (1 − e−2 ) = 187.42 µJ.
1
2
1
(51)2 × 10−6 = 216.75 µJ;
6
[b] w(0) =
0.95w(0) = 205.9125 µJ;
Z to
216.75 × 10−3 e−1000x dx = 205.9125 × 10−6 ;
0
Z to
e−1000x dx = 0.95 × 10−3 ;
0
.·. 1 − e−1000to = 0.95;
P 7.28
e1000to = 20;
so to = 3 ms.
v
= 20 kΩ.
i
1
1
1
[b]
=
= 1000;
C=
= 0.05 µF.
3
τ
RC
(10 )(20 × 103 )
1
[c] τ =
= 1 ms.
1000
1
[d] w(0) = (0.05 × 10−6 )(104 ) = 250 µJ.
2
[a] R =
7–29
7–30
CHAPTER 7. Response of First-Order RL and RC Circuits
[e] wdiss =
Z to
0
= 0.5
Z to
v2
(104 )e−2000t
dt =
dt
R
(20 × 103 )
0
e−2000t to
= 250(1 − e−2000to ) µJ.
−2000 0
0.8(250) = 200 = 250(1 − e−2000to );
.·. e−2000to = 0.2;
e2000to = 5;
1
ln 5;
2000
to ∼
= 804.72 µs.
to =
P 7.29
[a] τ = RC = RTh (0.2) × 10−6 = 10−3 ;
1000
.·. RTh =
= 5 kΩ.
0.2
vT = 20 × 103 (iT − αv∆ ) + 10 × 103 iT ;
v∆ = 10 × 103 iT ;
vT = 30 × 103 iT − 20 × 103 α10 × 103 iT ;
vT
= 30 × 103 − 200 × 106 α = 5 × 103 ;
iT
.·. 30 − 200,000α = 5;
α = 125 × 10−6 A/V.
[b] vo (0) = (0.018)(5000) = 90 V
t > 0:
vo = 90e−1000t V,
t ≥ 0.
t < 0.
Problems
v∆ − vo
v∆
+
− 125 × 10−6 v∆ = 0;
3
10 × 10
20,000
2v∆ + v∆ − vo − 2500 × 10−3 v∆ = 0;
.·. v∆ = 2vo = 180e−1000t V.
P 7.30
[a]
pds = (−90e−1000t )(22.5 × 10−3 e−1000t ) = −2025 × 10−3 e−2000t W;
Z ∞
wds =
0
pds dt = −1012.5 µJ;
.·. dependent source is delivering 1012.5 µJ.
[b] p10k =
(180)2 e−2000t
;
10 × 103
w10k =
p20k =
w20k =
Z ∞
0
p10k dt = 1620 µJ;
(90)2 e−2000t
;
20 × 103
Z ∞
0
p20k dt = 202.5 µJ;
1
wc (0) = (0.2) × 10−6 (90)2 = 810 µJ;
2
X
wdev = 810 + 1012.5 = 1822.5 µJ;
X
wdiss = 202.5 + 1620 = 1822.5 µJ.
7–31
7–32
P 7.31
CHAPTER 7. Response of First-Order RL and RC Circuits
t < 0:
t > 0:
vT
=
2 × 104 io + 60,000iT
=
20,000(−iT ) + 60,000iT = 40,000iT ;
vT
.·.
= RTh = 40 kΩ.
iT
τ = RC = 1 ms;
vo = 25e−1000t V,
io = 25 × 10−9
P 7.32
1
= 1000;
τ
t ≥ 0;
d
[25e−1000t ] = −625e−1000t µA,
dt
t ≥ 0+ .
[a] At t = 0− the voltage on each capacitor will be 25 × 10−3 × 200 = 5 V,
positive at the upper terminal. Hence at t = 0+ we have
5 5
.·. isd (0+ ) = 0.025 + + = 1.65 A.
5 8
Problems
7–33
At t = ∞, both capacitors will have completely discharged.
.·. isd (∞) = 25 mA.
[b] isd (t) = 0.025 + i1 (t) + i2 (t);
5
τ1 = (5)(2 × 10−6 ) = 10 µs so v2µ = 5e−10 t V;
τ2 = (8)(50 × 10−6 ) = 400 µs so v50µ = 5e−2500t V.
d
5
5
.·. i1 (t) = −(2 × 10−6 ) (5e−10 t ) = e−10 t A,
dt
i2 (t) = −(50 × 10−6 )
d
(5e−2500t ) = 0.625e−2500t A,
dt
.·. isd = 25 + 1000e−100,000t + 625e−2500t mA,
P 7.33
t ≥ 0+ ;
t ≥ 0+ ;
t ≥ 0+ .
[a] The equivalent circuit for t > 0:
τ = 0.4 ms;
1/τ = 2500;
vo = 20e−2500t V,
t ≥ 0;
io = 2e−2500t mA,
t ≥ 0+ ;
−2500t
i25kΩ = 0.002e
15
= 0.75e−2500t mA,
40
t ≥ 0+ ;
p25kΩ = (0.5625 × 10−6 e−5000t )(25,000) = 14,062.5 × 10−6 e−5000t W;
w25kΩ =
Z ∞
14,062.5 × 10−6 e−5000t dt = −2.8125 × 10−6 (0 − 1) = 2.8125 µJ;
0
1
1
w(0) = (0.2 × 10−6 )(100) + (0.05 × 10−6 )(900) = 32.5 µJ;
2
2
% diss (25 kΩ) =
2.8125
× 100 = 8.65%.
32.5
[b] p625Ω = 625(2 × 10−3 e−2500t )2 = 2.5 × 10−3 e−5000t W;
w625Ω =
Z ∞
0
p625 dt = 0.50 µJ;
% diss (625Ω) =
0.5
× 100 = 1.54%.
32.5
7–34
CHAPTER 7. Response of First-Order RL and RC Circuits
−2500t
i15kΩ = 0.002e
25
= 1.25e−2500t mA,
40
t ≥ 0+ ;
p15kΩ = (1.25 × 10−3 e−2500t )2 (15,000) = 23.4375 × 10−3 e−5000t W;
w15kΩ =
Z ∞
23.4375 × 10−3 e−5000t dt = 4.6875 µJ;
0
% diss (15kΩ) = 14.42%.
[c]
X
wdiss = 2.8125 + 0.50 + 4.6875 = 8 µJ;
wtrapped = w(0) −
% trapped =
X
wdiss = 32.5 − 8 = 24.5 µJ;
24.5
× 100 = 75.38%.
32.5
Check: 8.65 + 1.54 + 14.42 + 75.38 = 99.99 ≈ 100%.
P 7.34
[a] v1 (0− ) = v1 (0+ ) = 75 V
v2 (0+ ) = 0;
Ceq = 2µk8µ = 1.6 µF.
τ = (5)(1.6) × 10−3 = 8ms;
1
= 125;
τ
75e−125t
= 15e−125t mA,
5000
t ≥ 0+ .
i=
vC = 75e−125t V.
−106 Z t
15 × 10−3 e−125x dx + 75 = 60e−125t + 15 V,
2
0
t ≥ 0;
106 Z t
v2 =
15 × 10−3 e−125x dx + 0 = −15e−125t + 15 V,
8 0
t ≥ 0.
v1 =
1
[b] w(0) = (2 × 10−6 )(5625) = 5625 µJ.
2
Problems
7–35
1
1
[c] wtrapped = (2 × 10−6 )(225) + (8 × 10−6 )225 = 1125 µJ.
2
2
1
wdiss = (1.6 × 10−6 )(5625) = 4500 µJ.
2
Check: wtrapped + wdiss = 1125 + 4500 = 5625 µJ;
w(0) = 5625 µJ.
P 7.35
t < 0:
iL (0− ) = 6 A.
t > 0:
iL (∞) =
τ=
32 + 48
= 4 A;
20
L
5 × 10−3
=
= 250 µs;
R
20
1
= 4000;
τ
iL = 4 + (6 − 4)e−4000t = 4 + 2e−4000t A,
t ≥ 0;
vo = −8iL + 48 = −8(4 + 2e−4000t ) + 48 = 16 − 16e−4000t V,
t ≥ 0+ .
diL
= 5 × 10−3 [−8000e−4000t ] = −40e−4000t V,
dt
t ≥ 0+ ;
[b] vL = 5 × 10−3
vL (0+ ) = −40 V;
vo (0+ ) = 0 V.
Check: at t = 0+ the circuit is:
vL (0+ ) = 32 − 72 + 0 = −40 V,
vo (0+ ) = 48 − 48 = 0 V.
7–36
P 7.36
CHAPTER 7. Response of First-Order RL and RC Circuits
[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.
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
Problems
P 7.37
7–37
[a] t < 0:
KVL equation at the top node:
vo (0− ) vo (0− ) vo (0− )
+
+
.
4
20
5
Multiply by 20 and solve:
−40 =
−800 = (5 + 1 + 4)vo ;
vo = −80 V;
vo
.·. io (0− ) =
= −80/5 = −16 A.
5
t > 0:
Use voltage division to find the Thévenin voltage:
20
VTh = vo =
(240) = 60 V.
20 + 60
Remove the voltage source and make series and parallel combinations of
resistors to find the equivalent resistance:
RTh = 5 + 20k60 = 5 + 15 = 20 Ω.
The simplified circuit is:
L
10 × 10−3
1
=
= 0.5 ms;
= 2000.
R
20
τ
60
io (∞) =
= 3 A;
20
.·. io = io (∞) + [io (0+ ) − io (∞)]e−t/τ
τ=
= 3 + (−16 − 3)e−2000t = 3 − 19e−2000t A,
t ≥ 0.
7–38
CHAPTER 7. Response of First-Order RL and RC Circuits
[b] vo
vo
P 7.38
=
dio
;
dt
15 − 95e−2000t + 0.01(38,000)(e−2000t );
=
15 − 95e−2000t + 380e−2000t ;
=
15 + 285e−2000t V,
=
5io + (0.01)
t ≥ 0+ .
[a] t < 0:
ig =
240
= 3.75 A;
60 + 20k5
20k5
.·. io (0− ) =
(3.75) = 3 A.
5
t > 0:
io (∞) =
4k20k5
(−40) = −16 A;
5
Req = 5 + 4k20 = 8.33 Ω;
τ=
10 × 10−3
L
=
= 1.2 ms;
Req
8.33
1
= 833.33;
τ
.·. io = io (∞) + [io (0+ ) − io (∞)]e−t/τ
= −16 + (3 + 16)e−833.33t = 19e−833.33t − 16 A,
[b] vo
dio
dt
=
5io + L
=
5(19e−833.33t − 16) + 0.01(−833.33)(19e−833.33t )
= −63.33e−833.33t − 80 V,
P 7.39
t ≥ 0.
t ≥ 0+ .
[a] Note that there are many different possible solutions to this problem.
R=
L
.
τ
Problems
7–39
Choose a 10 mH inductor from Appendix H. Then,
0.01
= 80 Ω.
125 × 10−6
Construct the resistance needed by combining 33 Ω, and 47 Ω resistors in
series:
R=
[b] i(t) = If + (Io − If )e−t/τ ;
Io = 0 A;
If =
Vf
16
=
= 200 mA;
R
80
.·. i(t) = 200 + (0 − 200)e−8000t mA = 200 − 200e−8000t mA,
t ≥ 0.
[c] i(t) = 0.2 − 0.2e−8000t = (0.5)(0.2) = 0.1;
e−8000t = 0.5
e8000t = 2;
so
ln 2
= 86.64 µs.
.·. t =
8000
P 7.40
[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
R1
.
R1 + R2
io (0+ ) = Ig ;
io (∞) = Ig
Therefore
io (t) =
Ig R1
+
R1 +R2
io (t) =
R1 Ig
2 Ig
+ (RR1 +R
e−[(R1 +R2 )/L]t .
(R1 +R2 )
2)
vsw =
R1 Ig
R2 Ig
+ (1+R
e−[(R1 +R2 )/L]t ,
(1+R1 /R2 )
1 /R2 )
Therefore
[d] |vsw (0+ )| → ∞;
h
duration → 0.
i
R1
Ig − RI1g+R
e−[(R1 +R2 )/L]t
2
t ≥ 0+ .
7–40
CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.41
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 7.40), causing the switch to arc over. At the same time, the large
voltage across L damages the meter movement.
P 7.42
[a] From Eqs. 7.1 and 7.15
Vs
Vs −(R/L)t
+ Io −
e
;
R
R
i=
v = (Vs − Io R)e−(R/L)t ;
Vs
= 10;
.·.
R
Io −
Vs − Io R = 200;
Vs
= −10;
R
R
= 500;
L
Vs
.·. Io = −10 +
= 0 A.
R
Therefore,
Vs = 200V.
i(∞) = 10 =
200
R
L=
so
R = 20 Ω;
R
= 40 mH.
500
[b] i = 10 − 10e−500t ;
i2 = 100 − 200e−500t + 100e−1000t ;
1
1
w = Li2 = (0.04)[100 − 200e−500t + 100e−1000t ] = 2 − 4e−500t + 2e−1000t ;
2
2
w(∞) = 2 J;
w(to ) = 2 − 4e−500to + 2e−1000to = 0.25(2);
.·. 1 − 2x + x2 = 0.25
Solving,
x = 1.5 and x = 0.5
.·. 0.5 = e−500to
P 7.43
and thus
so
to =
but only the second solution is possible;
ln 2
= 1.386 ms.
500
[a]
−
x2 − 2x + 0.75 = 0.
Vs
v
1Z t
+ +
v dt + Io = 0.
R R L 0
Problems
Differentiating both sides,
1
1 dv
+ 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
R
dv
= − dt;
v
L
Z v(t)
RZ t
dx
=−
dy;
L 0
Vo x
ln
v(t)
R
= − t;
Vo
L
.·.
P 7.44
v(t) = Vo e−(R/L)t = (Vs − RIo )e−(R/L)t .
For t < 0:
vx − 250
vx
10
= 0;
+ 0.9
(vx − 250) +
20
10 + 40
10 + 40
vx
(vx − 250)
+ 10
= 0;
20
50
5vx − 5000 + 20vx = 0;
io (0− ) = 200/20 = 10 A.
t > 0:
vx = 200 V;
7–41
7–42
CHAPTER 7. Response of First-Order RL and RC Circuits
After source-transforming to simplify the right-hand side of the circuit we get
Find Thévenin equivalent with respect to a, b:
VTh − 80
(VTh − 80)
+9
=0
18
18
vT = (iT − 0.9vφ )18 = iT − 0.9
vT = 18iT − 9vT
VTh = 80 V.
10vT
18
.·. 10vT = 18iT ;
vT
= RTh = 1.8 Ω.
iT
io (∞) = 80/21.8 = 3.67 A;
18;
Problems
τ=
87.2
× 10−3 = 4 ms;
21.8
1/τ = 250;
io = 3.67 + (10 − 3.67)e−250t = 3.67 + 6.33e−250t A,
P 7.45
t < 0;
calculate io (0− ):
io (0− ) =
5
(0.002) = 0.5 mA;
5 + 15
t ≥ 0.
io (0+ ) = io (0− ) = 0.5 mA.
t > 0;
calculate vo (0+ ) using the node voltage method:
−0.002 +
va va − vo
+
= 0;
5
15
vo − va
+ 5 × 10−4 + i∆ = 0;
15
i∆ =
vo
− 4i∆ − 0.001.
4
Solving,
vo (0+ ) = 2 mV.
We also know that
vo (∞) = 0.
7–43
7–44
CHAPTER 7. Response of First-Order RL and RC Circuits
Find the Thévenin resistance seen by the 2 mH inductor:
iT =
vT
vT
+
− 4i∆ ;
20
4
i∆ =
vT
− 4i∆
4
iT =
vT
4vT
vT
+
−
;
20
4
20
vT
;
.·. 5i∆ =
4
i∆ =
vT
;
20
1
1 1
2
iT
=
+ − =
= 0.1 S;
vT
20 4 5
20
.·. RTh = 10Ω;
τ=
2 × 10−3
= 0.2 ms;
10
1/τ = 5000;
.·. vo = 0 + (2 − 0)e−5000t = 2e−5000t mV,
P 7.46
t ≥ 0+ .
[a] Let v be the voltage drop across the parallel branches, positive at the top
node, then
v
1 Zt
1 Zt
+
v dx +
v dx = 0;
−Ig +
R g L1 0
L2 0
v
1
1 Zt
v dx = Ig ;
+
+
Rg
L1 L2 0
v
1 Zt
+
v dx = Ig ;
R g Le 0
1 dv
v
+
= 0;
Rg dt Le
dv Rg
+
v = 0.
dt
Le
Therefore v = Ig Rg e−t/τ ;
τ = Le /Rg .
Problems
Thus
i1 =
1 Zt
Ig Rg e−x/τ t Ig Le
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 7.47
L2
Ig ;
L1 + L2
i2 (∞) =
L1
Ig .
L1 + L2
t > 0:
τ=
4
1
= ;
80
20
io = −5e−20t A,
t ≥ 0;
vo = 80io = −400e−20t V,
−400e−20t = −80;
t ≥ 0+ ;
e20t = 5;
1
.·. t =
ln 5 = 80.47 ms.
20
P 7.48
1
1
[a] wdiss = Le i2 (0) = (4)(25) = 50 J.
2
2
1Zt
[b] i12H =
(−400)e−20x dx + 5
12 0
=
−100 e−20x t
5
10
+5 = e−20t +
A
3 −20 0
3
3
1Z t
(−400)e−20x dx + 0
i6H =
6 0
=
−200 e−20x t
10
10
A.
+ 0 = e−20t −
3 −20 0
3
3
1
wtrapped = (18)(100/9) = 100 J.
2
7–45
7–46
CHAPTER 7. Response of First-Order RL and RC Circuits
1
[c] w(0) = (12)(25) = 150 J.
2
P 7.49
For t < 0,
i40mH (0) = 75/5 = 15 A.
For t > 0, after making a Thévenin equivalent we have
Vs
Vs −t/τ
i=
+ Io −
e
;
R
R
R
4
1
=
=
= 40;
τ
Leq
100 × 10−3
Io = 15 A;
Vs
100
=
= 25 A;
R
4
i = 25 + (15 − 25)e−40t = 25 − 10e−40t A,
vo = 0.04
P 7.50
di
= 0.04(400e−40t ) = 16e−40t V,
dt
t ≥ 0;
t > 0+ .
[a] t < 0:
t > 0:
iL (0− ) = iL (0+ ) = 20 A;
τ=
4.8
= 0.1 s;
48
1
= 10;
τ
iL (∞) = 10 A;
iL = 10 + [20 − 10]e−10t = 10 + 10e−10t A,
vo = 4.8[−100e−10t ] = −480e−10t V,
t ≥ 0;
t ≥ 0+ .
Problems
1 Zt
[b] i1 =
−480e−10x dx + 8 = 4e−10t + 4 A,
12 0
1Z t
[c] i2 =
−480e−10x dx + 12 = 6e−10t + 6 A,
8 0
P 7.51
7–47
t ≥ 0.
t ≥ 0.
[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+ .
[b] t ≥ 0+ , since there is no requirement that the voltage be continuous in a
resistor.
P 7.52
[a] vc (0+ ) = 120 V.
Use voltage division to find the final value of voltage:
150
(−200) = −150 V.
150 + 50
Find the Thévenin equivalent with respect to the terminals of the
capacitor:
vc (∞) =
VTh = −150 V,
RTh = 12.5 k + 150 kk50 k = 50 kΩ,
Therefore τ = Req C = (50,000)(40 × 10−9 ) = 2 ms.
7–48
CHAPTER 7. Response of First-Order RL and RC Circuits
The simplified circuit for t > 0 is:
vc = vc (∞) + [vc (0+ ) − vc (∞)]e−t/τ ;
= −150 + [120 − (−150)]e−t/τ = −150 + 270e−500t V,
t ≥ 0.
[b] For t < 0:
150
(−200) = −150 V.
150 + 50
For t ≥ 0:
vc (0) =
vc (∞) = 120 V;
τ = Req C = 10 × 103 (40 × 10−9 ) = 0.4 ms;
vc (t) = vc (∞) + (vc (0) − vc (∞))e−t/τ = 120 + (−150 − 120)e−2500t = 120 − 270e−2500t V.
P 7.53
For t < 0:
.·. vo (0− ) = vo (0+ ) = −90 V.
Problems
7–49
t > 0:
vo (∞) = 15 V;
τ = RC = (5 k)(0.05 µ) = 0.25 ms;
1
= 4000;
τ
vo = vo (∞) + [vo (0+ ) − vo (∞)]e−t/τ = 15 + [−90 − 15]e−4000t
= 15 − 105e−4000t V
P 7.54
t ≥ 0.
[a] Find the voltage across the capacitor in the direction of the capacitor
current. Call this voltage vc . The initial capacitor charge is zero, so
V0 = 0. For t ≥ 0,
Req = 20,000 + 30,000 = 50,000 Ω;
τ = Req C = (50,000)(0.1 × 10−6 ) = 5 ms.
As t → ∞,
Vf = (20,000)(0.0075) = 150 V.
Therefore,
vC = Vf + (V0 − Vf )e−t/τ = 150 + (0 − 150)e−t/0.005 = 150 − 150e−200t V,
Thus,
i=C
dvc
= (0.1 × 10−6 )[−200(−150e−200t )]
dt
= 3e−200t mA, t ≥ 0+ .
t ≥ 0.
7–50
CHAPTER 7. Response of First-Order RL and RC Circuits
[b] i20k = (7.5 − 3e−200t ) mA;
v = 20,000i20k = 150 − 60e−200t V,
P 7.55
vo (0) =
t ≥ 0+ .
120
(80) = 80 V;
120
vo (∞) = −6(25) = −150 V;
1
= 1000;
τ
τ = (25 × 103 )(40 × 10−9 ) = 10−3 s;
vo = −150 + (80 + 150)e−1000t = −150 + 230e−1000t V,
P 7.56
t ≥ 0.
[a] Use voltage division to find the initial value of the voltage:
10 k
(−75) = −30 V.
10 k + 15 k
[b] Use Ohm’s law to find the final value of voltage:
vc (0+ ) = v10k =
vc (∞) = v5k = (5 × 10−3 )(5000) = 25 V.
[c] Find the Thévenin equivalent with respect to the terminals of the
capacitor:
VTh = 25 V,
RTh = 5 k + 20 k = 25 kΩ;
τ = RTh C = 2.5 ms.
[d] vc = vc (∞) + [vc (0+ ) − vc (∞)]e−t/τ
= 25 + (−30 − 25)e−400t = 25 − 55e−400t V,
t ≥ 0.
We want vc = 25 − 55e−400t = 0:
Therefore t =
P 7.57
ln(55/25)
= 1.97 ms.
400
t < 0;
io (0− ) = (0.015)
20
= 6 mA;
50
vo (0− ) = (6)(10) = 60 V.
t = ∞:
20
io (∞) = −0.010
= −4 mA;
50
vo (∞) = io (∞)(10) = −40 V;
Problems
RTh = 10 kΩk40 kΩ = 8 kΩ;
τ = (8000)(125 × 10−9 ) = 1 ms;
1
= 1000;
τ
.·. vo (t) = −40 + 100e−1000t V,
t ≥ 0+ .
dvo
= −12.5e−1000t mA,
dt
t ≥ 0+ ;
ic = C
i10 =
P 7.58
C = 125 nF;
vo
= −4 + 10e−1000t mA,
0.010
t ≥ 0+ ;
io = ic + i10 = −(4 + 2.5e−1000t ) mA,
t ≥ 0+ .
[a] v = Is R + (Vo − Is R)e−t/RC
.·. Is R = 80,
i = Is −
Vo −t/RC
e
;
R
Vo − Is R = −80;
.·. Vo = 0 V.
Is −
Vo
= 50 × 10−3 ;
R
.·. Is = 50 mA.
Vo − Is R = 0 − 0.05R = −80 so R =
1
= 2500;
RC
τ = RC =
C=
80
= 1600 Ω.
0.05
1
1
=
= 250 nF;
2500R
(2500)(1600)
1
= 400 µs.
2500
[b] v(∞) = 80 V;
1
w(∞) = (250 × 10−9 )(6400) = 800 µJ;
2
0.64w(∞) = 512 µJ;
1 2
512 × 10−6
v (to ) =
= 2048;
2
250 × 10−9
80 − 80e−2500to = 64;
v(to ) = 64 V;
e2500to = 5;
.·. to = 643.775 µs.
7–51
7–52
P 7.59
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,
0.002
= 200 Ω.
10 × 10−6
Construct the resistance needed by combining two 100 Ω resistors in
series:
R=
[b] v(t) = Vf + (V0 − Vf )e−t/τ ;
Vf = (If )(R) = (100 × 10−3 )(200) = 20 V;
V0 = −40 V;
.·. v(t) = 20 + (−40 − 20)e−500t V = 20 − 60e−500t V,
[c] v(t) = 20 − 60e−500t = 0
so
ln 3
.·. t =
= 2.2 ms.
500
P 7.60
[a]
1 Zt
Is R = Ri +
i dx + Vo ;
C 0+
0=R
di
i
+ + 0;
dt C
di
i
.·.
+
= 0.
dt RC
[b]
di
i
=−
;
dt
RC
Z i(t)
di
dt
=−
;
i
RC
dy
1 Zt
=−
dx;
RC 0+
i(0+ ) y
e−500t =
1
3
t ≥ 0.
Problems
ln
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 7.61
For t > 0,
VTh = (−15,000)(30)ib = −450 × 103 ib ;
ib =
400(12)
= 100 µA;
48
VTh = −450 × 103 (100 × 10−6 ) = −45 V;
RTh = 15 kΩ.
vo (0+ ) = 0;
vo (∞) = −45 V;
τ = (15, 000)(8)10−6 = 120 ms;
vo = −45 + 45e−8.33t V,
1/τ = 8.33;
t ≥ 0;
1
w(t) = (8 × 10−6 )vo2 = 8100(1 − 2e−8.33t + e−16.67t ) µJ;
2
w(∞) = 8100 µJ;
.·. 8100(1 − 2e−8.33to + e−16.67to ) = 0.90(8100);
.·. 1 − 2x + x2 = 0.90;
x = e−8.33to ;
.·. x2 − 2x + 0.10 = 0;
x1 = 1.9487,
x2 = 0.0513;
e−(25/3)to = 0.0513;
(25/3)to = ln 19.4868;
to = 356.4 ms.
7–53
7–54
P 7.62
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
vo = 50 + (80 − 50)e−t/τ ;
τ = RC = (40 × 103 )(5 × 10−9 ) = 200 × 10−6 s;
vo = 50 + 30e−5000t V.
P 7.63
vo (0) = 50 V;
vo (∞) = 80 V;
RTh = 16 kΩ;
τ = (16)(5 × 10−6 ) = 80 × 10−6 s;
1
= 12,500;
τ
v = 80 + (50 − 80)e−12,500t = 80 − 30e−12,500t V.
1
= 5000;
τ
Problems
P 7.64
[a] Let i be the current Zin the clockwise
direction around the circuit. Then
1 Zt
1 t
i dx +
i dx
Vg = iRg +
C1 0
C2 0
1
1 Zt
1 Zt
= iRg +
+
i dx.
i dx = iRg +
C1 C2
Ceq 0
0
Now differentiate the equation
0 = Rg
i
di
+
dt Ceq
Therefore i =
or
di
1
+
i = 0.
dt Rg Ceq
Vg −t/Rg Ceq
Vg −t/τ
e
=
e
;
Rg
Rg
Vg e−x/τ
1 Z t Vg −x/τ
e
dx =
v1 (t) =
C1 0 Rg
Rg C1 −1/τ
τ = Rg Ceq .
t
=−
0
v1 (t) =
Vg C2
(1 − e−t/τ );
C1 + C2
τ = Rg Ceq ;
v2 (t) =
Vg C1
(1 − e−t/τ );
C1 + C2
τ = Rg Ceq .
[b] v1 (∞) =
P 7.65
7–55
C2
Vg ;
C1 + C2
v2 (∞) =
C1
Vg .
C1 + C2
[a] t < 0:
t > 0:
vo (0− ) = vo (0+ ) = 60 V;
vo (∞) = 100 V;
τ = (0.2)(5) × 10−3 = 1 ms;
vo = 100 − 40e−1000t V,
1/τ = 1000;
t ≥ 0.
Vg Ceq −t/τ
(e
− 1);
C1
7–56
CHAPTER 7. Response of First-Order RL and RC Circuits
[b] io = −C
dvo
= −0.2 × 10−6 [40,000e−1000t ]
dt
= −8e−1000t mA;
[c] v1 =
t ≥ 0+ .
−106 Z t
−8 × 10−3 e−1000x dx + 40
0.3 0
= 66.67 − 26.67e−1000t V,
[d] v2 =
t ≥ 0.
−106 Z t
−8 × 10−3 e−1000x dx + 20
0.6 0
= 33.33 − 13.33e−1000t V,
t ≥ 0.
1
1
[e] wtrapped = (0.3)10−6 (66.67)2 + (0.6)10−6 (33.33)2
2
2
= 666.67 + 333.33 = 1000 µJ.
P 7.66
[a] For t > 0:
τ = RC = (250 × 103 )(8 × 10−9 ) = 2 ms;
vo = 50e−500t V,
[b] io =
1
= 500;
τ
t ≥ 0+ .
50e−500t
vo
=
= 200e−500t µA;
250,000
250,000
Z t
−1
200 × 10−6 e−500x dx + 50 = 10e−500t + 40 V,
v1 =
40 × 10−9 0
P 7.67
[a] Leq =
(3)(15)
= 2.5 H;
3 + 15
τ=
Leq
2.5
1
=
= s;
R
7.5
3
io (0) = 0;
io (∞) =
.·. io = 16 − 16e−3t A,
120
= 16 A;
7.5
t ≥ 0.
vo = 120 − 7.5io = 120e−3t V,
t ≥ 0+ ;
t ≥ 0.
Problems
1Z t
40 40 −3t
i1 =
120e−3x dx =
− e A,
3 0
3
3
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 7.68
[a] Leq = 4 + 8 − 2(5) = 2 H;
τ=
2
1
L
=
=
s;
R
50
25
i = 4 − 4e−25t A,
1
= 25;
τ
t ≥ 0.
di
di
di
− 5 = − = −(100e−25t ) = −100e−25t V, t ≥ 0+ .
dt
dt
dt
di
di
di
[c] v2 (t) = 8 − 5 = 3 = 3(100e−25t ) = 300e−25t V, t ≥ 0+ .
dt
dt
dt
[d] i(0) = 4 − 4 = 0, which agrees with initial conditions.
[b] v1 (t) = 4
200 = 50i1 + v1 + v2 = 50(4 − 4e−25t ) − 100e−25t + 300e−25t = 200 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.
7–57
7–58
P 7.69
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] Leq = 4 + 8 + 2(5) = 22 H;
L
22
=
s;
R
50
1
= 2.273;
τ
i = 4 − 4e−2.273t A,
t ≥ 0.
τ=
di
di
di
+ 5 = 9 = 9(9.09e−2.273t ) = 81.81e−2.273t V, t ≥ 0+ .
dt
dt
dt
di
di
di
[c] v2 (t) = 8 + 5 = 13 = 13(9.09e−2.273t ) = 118.18e−2.273t V, t ≥ 0+ .
dt
dt
dt
[d] i(0) = 0, which agrees with initial conditions.
[b] v1 (t) = 4
200 = 50i1 + v1 + v2 = 50(4 − 4e−2.273t ) + 81.81e−2.273t + 118.18e−2.273t = 200 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 7.70
[a] From Example 7.10,
Leq =
τ=
0.125 − 0.0625
L1 L2 − M 2
=
= 50 mH;
L1 + L2 + 2M
0.75 + 0.5
Leq
1
=
s;
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,
di1
di2
[c] vo = 0.5
− 0.25
= 10e−5000t V;
dt
dt
io = i1 + i2 ;
di1 di2
dio
=
+
= 200e−5000t A/s;
dt
dt
dt
di2
di1
.·.
= 200e−5000t −
;
dt
dt
di1
di1
.·. 10e−5000t = 0.5
− 50e−5000t + 0.25 ;
dt
dt
di1
.·. 0.75
= 60e−5000t ;
dt
Z i1
0
i1 =
dx =
Z t
di1 = 80e−5000t dt;
80e−5000y dy;
0
80 −5000y t
e
= 16 − 16e−5000t mA,
−5000
0
t ≥ 0.
t ≥ 0+ .
Problems
7–59
[d] i2 = io − i1 = 0.040 − 0.040e−5000t − 0.016 + 0.016e−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
t ≥ 0+ (checks).
di2
di1
− 0.25
= 10e−5000t V,
dt
dt
t ≥ 0+ (checks).
vo = 0.25
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 × 10−6 )(500) − (24 × 10−6 )(250) = 2 mWb-turns (ok);
i2 (∞)L2 − i1 (∞)M = (24 × 10−6 )(250) − (16 × 10−6 )(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.
P 7.71
[a] From Example 7.10,
Leq =
(0.008)(0.020) − (0.010)2
L1 L2 − M 2
=
= 7.5 mH;
L1 + L2 − 2M
0.008 + 0.020 − 2(0.010)
τ=
Leq
7.5 × 10−3
1
=
=
;
R
75
10,000
io =
15 15 −10,000t
− e
= 0.2 − 0.2e−10,000t A t ≥ 0.
75 75
[b] vo = 15 − 75io = 15 − 75(0.2 − 0.2e−10,000t ) = 15e−10,000t V t ≥ 0+ .
di2
di1
[c] vo = 0.008
+ 0.01 ;
dt
dt
io = i1 + i2 ;
dio
di1 di2
=
+
dt
dt
dt;
di2
dio di1
di1
=
−
= 2000e−10,000t −
;
dt
dt
dt
dt
!
di1
di1
.·. 15e−10,000t = 0.008
+ 0.01 2000e−10,000t −
;
dt
dt
7–60
CHAPTER 7. Response of First-Order RL and RC Circuits
di1
.·.
= 2500e−10,000t .
dt
di1 = 2500e−10,000t dt;
Z i1
dx = 2500
0
Z t
e−10,000y dy;
0
−10,000y
t
e
.·. i1 = 2500
= 0.25 − 0.25e−10,000t A,
−10,000 0
[d] i2
= io − i1
=
0.2 − 0.2e−10,000t − 0.25 + 0.25e−10,000t
= −50 + 50e−10,000t mA,
[e] vo
t ≥ 0.
= L2
=
t ≥ 0.
di1
di2
+M
dt
dt
0.02(−500e−10,000t ) + 0.01(2500e−10,000t )
= 15e−10,000t V, t ≥ 0+
(checks).
i1 (0) = 0.25 − 0.25 = 0; agrees with initial conditions;
i2 (0) = −0.05 + 0.05 = 0; agrees with initial conditions;
The final values of io , i1 , and i2 can be checked via the conservation of
Wb-turns:
io (∞)Leq = 0.2 × (7.5 × 10−3 ) = 1.5 mWb-turns;
i1 (∞)L1 + i2 (∞)M = 0.25(0.008) − 0.05(0.01) = 1.5 mWb-turns;
i2 (∞)L2 + i1 (∞)M = −0.05(0.02) + 0.25(0.01) = 1.5 mWb-turns.
Thus our solutions make sense in terms of known circuit behavior.
P 7.72
For t < 0:
i(0) =
0.08(150)
= 60 mA.
200
Problems
0 ≤ t ≤ 250 µs:
i = 60e−2000t mA;
i(250µs) = 60e−0.5 = 36.39 mA.
250 µs ≤ t ≤ 650 µs:
Req =
(50)(75)
= 30 Ω;
125
R
30
1
=
=
× 103 = 1200;
τ
L
25
i = 36.39e−1200(t−250×10
−6 )
mA.
650 µs ≤ t < ∞:
i(650µs) = 36.39e−0.48 = 22.52 mA;
i = 22.52e−2000(t−650×10
v=L
di
;
dt
−6 )
mA;
L = 25 mH;
di
−6
−6
= (22.52 × 10−3 )(−2000)e−2000(t−650×10 ) = −45.035e−2000(t−650×10 ) ;
dt
v = (25 × 10−3 )(−45.035)e−2000(t−650×10
= −1.126e−2000(t−650×10
−6 )
V,
v(1ms) = −1.126e−2000(350)×10
−6
−6 )
t > 650+ µs.
= −559.1 mV.
7–61
7–62
P 7.73
CHAPTER 7. Response of First-Order RL and RC Circuits
From the solution to Problem 7.72, the initial energy is
1
w(0) = (25 mH)(60 mA)2 = 45 µJ.
2
For 650 µs ≤ t < ∞:
1
−6
w(t) = (25 mH)(22.52e−2000(t−650×10 ) mA)2 = (0.04)(45 µJ).
2
Solving,
t = 964.72 µs.
P 7.74
[a] Prior to switch a closing at t = 0, there are no sources connected to the
inductor; thus, i(0− ) = 0.
At the instant switch 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=
= = 2 s.
R
1
Therefore,
i = if + [i(0+ ) − if ]e−t/τ = 3 − 3e−0.5t A,
0 ≤ t ≤ 1 s.
For part (b) we need the value of i(t) at t = 1 s:
i(1) = 3 − 3e−0.5 = 1.18 A.
.
Problems
7–63
[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:
2
L
=
= 0.8 s.
3k(9 + 6) = 2.5 Ω
τ=
R
2.5
Therefore,
i = if + [i(1+ ) − if ]e−(t−1)/τ
= −4.8 + 5.98e−1.25(t−1) A,
P 7.75
t ≥ 1 s.
[a] t < 0:
ig =
500
= 5 A;
4 + (480k120)
i(0− ) =
5(480)
= 4 A = i(0+ ).
600
[b] 0 ≤ t ≤ 100 µs:
i = 4e−t/τ ;
1
R
120 + 96k480
=
=
= 10,000;
τ
L
20 × 10−3
i = 4e−10,000t ;
4
i(25µs) = 4e−10 (25×10
−6 )
= 4e−0.25 = 3.12 A.
7–64
CHAPTER 7. Response of First-Order RL and RC Circuits
[c] i(100µs) = 4e−1 = 1.47 A.
100 µs ≤ t < ∞:
R
120
1
=
=
= 6000;
τ
L
0.02
i = 1.47e−6000(t−100×10
−6 )
A;
i(200µs) = 1.47e−6000(100)×10
−6
= 1.47e−0.6 = 807.59 mA.
[d] 0 ≤ t ≤ 100 µs:
i = 4e−10,000t ;
v=L
di
4
4
= (20 × 10−3 )(4)(−104 )e−10 t = −800e−10 t V;
dt
4
v(100− µs) = −800e−10 (100×10
−6 )
= −800e−1 = −294.30 V.
[e] 100 µs ≤ t < ∞:
i = 1.47e−6000(t−100×10
−6 )
;
v = (20 × 10−3 )(1.47)(−6000)e−6000(t−100×10
= −176.58e−6000(t−100×10
−6 )
−6 )
V;
v(100+ µs) = −176.58 V.
P 7.76
0 ≤ t ≤ 50 µs:
5
τ=
× 10−9 (180,000) = 100 µs;
9
Re = 720,000k240,000 = 180 kΩ;
vc = 494.6e−10,000t mV;
vc (50 µs) = 494.6e−0.5 = 300 mV.
Problems
50 µs ≤ t < ∞:
Re = 120,000k40,000 + 600,000k200,000 = 30,000 + 150,000 = 180 kΩ;
5
τ=
× 10−9 (180,000) = 100 µs;
9
1
= 10,000;
τ
vc = 300e−10,000(t − 50 µs) mV;
v1 =
30
vc = 50e−10,000(t − 50 µs) mV;
180
v2 =
150
vc = 250e−10,000(t − 50 µs) mV;
180
i1 =
v1
= 416.7e−10,000(t − 50 µs) nA;
120 × 103
i2 =
v2
= 416.7e−10,000(t − 50 µs) nA;
600 × 103
isw = i1 − i2 = 0 A;
isw (100 µs) = 0 A.
P 7.77
0 ≤ t ≤ 3 µs:
τ = RC = (2 × 103 )(5 × 10−9 ) = 10 µs;
vo (0) = 0 V;
vo (∞) = 6 V;
vo = 6 − 6e−100,000t V
0 ≤ t ≤ 3 µs.
1/τ = 100,000;
7–65
7–66
CHAPTER 7. Response of First-Order RL and RC Circuits
vo (3 µs) = 6 − 6e−0.3 = 1.555 V;
3 µs ≤ t < ∞:
t = ∞:
i=
6 − (−12)
= 1.8 mA;
10,000
vo (∞) = 6 − 2000i = 2.4 V;
vo = 2.4 + (1.555 − 2.4)e−(t − 3 µs)/τ ;
RTh = 2 kΩk8 kΩ = 1.6 kΩ;
P 7.78
τ = (1600)(5 × 10−9 ) = 8 µs;
1/τ = 125,000;
vo = 2.4 − 0.845e−125,000(t − 3 µs)
3 µs ≤ t < ∞.
Note that for t > 0, vo = (35/40)vc , where vc is the voltage across the 50 nF
capacitor. Thus we will find vc first.
t<0
vc (0) =
10
(280) = 56 V.
50
Problems
0 ≤ t ≤ 400 µs:
τ = Req C,
Req =
(10,000)(40,000)
= 8 kΩ;
50,000
τ = (8 × 103 )(50 × 10−9 ) = 400 µs,
vc = 56e−2500t V,
1
= 2500;
τ
t ≥ 0.
vc (400 µs) = 56e−1 = 20.60 V.
400 µs ≤ t ≤ 1.4 ms:
τ = (40 × 103 )(50 × 10−9 ) = 2 ms,
vc = 20.60e−500(t−400×10
−6 )
1
= 500;
τ
V.
−6
= 20.60e−0.5 = 12.50 V;
−3
= 12.50e−0.5 = 7.58 V.
vc (1.4ms) = 20.60e−500(1400−400)10
1.4 ms ≤ t < ∞:
τ = 400 µs,
1
= 2500;
τ
vc = 12.50e−2500(t−1.4×10
−3 )
V.
vc (1.6ms) = 12.50e−2500(1.6−1.4)10
vo = (35/40)(7.58) = 6.63 V.
7–67
7–68
P 7.79
CHAPTER 7. Response of First-Order RL and RC Circuits
1
w(0) = (50 × 10−9 )(56)2 = 78.4 µJ.
2
0 ≤ t ≤ 400 µs:
vc = 56e−2500t :
vc2 = 3136e−5000t ;
p10k = 3136 × 10−4 e−5000t .
w10k
=
Z 400×10−6
0
=
3136 × 10−4 e−5000t dt
400×10−6
−5000t
−4 e
3136 × 10
−5000 0
= −6272 × 10 (e−2 − 1) = 54.23 µJ.
1.4 ms ≤ t < ∞:
−8
vc = 12.50e−2500(t−1.4×10
−3 )
vc2 = 156.13e−5000(t−1.4×10
V;
p10k = 156.13 × 10−4 e−5000(t−1.4×10
w10k
=
=
Z ∞
1.4×10−3
−3 )
;
.
156.13 × 10−4 e−5000(t−1.4×10
−5000(t−1.4×10−3 )
156.13 × 10−4
−3 )
e
−3 )
dt
∞
−5000
1.4×10−3
= −312.26 × 10 (0 − 1) = 3.12 µJ.
−8
w10k = 54.23 × 10−6 + 3.12 × 10−6 = 57.35 µJ;
%=
57.35
(100) = 73.15%.
78.4
To check, find the energy dissipated in the 35 + 5 = 40 kΩ resistance:
0 ≤ t ≤ 400 µs:
vc = 56e−2500t ;
w40k
=
vc2 = 3136e−5000t ;
−4
784 × 10
Z 400×10−6
0
−5000t
=
−4 e
p40k =
3136 −5000t
e
;
40,000
e−5000t dt
400×10−6
784 × 10
−5000 0
= −156.8(10 )(e−2 − 1) = 13.56 µJ.
400 µs ≤ t ≤ 1.4 ms:
−7
vc = 20.60e−500(t−400×10
−6 )
;
vc2 = 424.41e−1000(t−400×10
−6 )
;
Problems
w40k
=
106.10 × 10−4
=
106.10 × 10−4
Z 1.4×10−3
e−1000(t−400×10
400×10−6
−1000(t−400×10−6 )
e
−6 )
7–69
dt
1.4×10−3
−1000
400×10−6
= −106.10(10−7 )(e−1 − 1) = 6.71 µJ.
1.4 ms ≤ t < ∞:
vc = 12.49e−2500(t−1.4×10
−3 )
;
vc2 = 156.13e−5000(t−1.4×10
−3 )
.
Note in this interval the energy dissipated in the 40k resistor will be 1/4th
that dissipated in the 10k resistor, so
1
w40k = (3.12) = 0.78 µJ.
4
w40k(total) = 13.56 × 10−6 + 6.71 × 10−6 + 0.78 × 10−6 = 21.05 µJ;
w40k + w10k = 57.35 × 10−6 + 21.05 × 10−6 = 78.40 µJ.
P 7.80
t < 0:
vc (0− ) = (50)(2000) × 10−3 = 100 V = vc (0+ ).
0 ≤ t ≤ 250 ms:
τ = ∞;
1/τ = 0;
vo = 100e−0 = 100 V.
250 ms ≤ t < ∞:
τ = (6.25)(0.16)10−3 = 1 ms;
1/τ = 1000;
vo = 100e−1000(t − 0.25) V.
7–70
CHAPTER 7. Response of First-Order RL and RC Circuits
Summary:
0 ≤ t ≤ 250 ms;
vo = 100 V,
vo = 100e−1000(t − 0.25) V,
P 7.81
250 ms ≤ t < ∞.
[a] 0 ≤ t ≤ 2 µs :
iL (0) = 0;
τ=
iL (∞) = 5 mA;
0.04
L
=
= 2 µs;
R
20,000
iL = 5 − 5e−500,000t mA,
vo = L
0 ≤ t ≤ 2 µs;
diL
= (0.04)[(500,000)(0.005)e−500,000t ] = 100e−500,000t V,
dt
2 µs ≤ t < ∞ :
iL (2 µs) = 5 − 5e−1 ≈ 3.16 mA;
iL (∞) = 0;
τ = 2 µs;
1/τ = 500,000;
iL = 0 + (3.16 − 0)e−500,000(t−2 µs) mA
= 3.16e−500,000(t−2 µs) mA,
2 µs ≤ t < ∞;
diL
= (0.04)(3.16 × 10−3 )[−500,000e−500,000(t−2 µs) ]
dt
= (−5)(4)(3.16)e−500,000(t−2 µs)
vo = L
= −63.21e−500,000(t−2 µs) V,
2 µs ≤ t < ∞.
0+ ≤ t < 2 µs.
Problems
[b]
[c] vo (4 µs) = −63.21e−1 = −23.25 V.
P 7.82
[a] io (0) = 0;
io (∞) = 25 mA;
1
R
8000
=
=
= 32,000;
τ
L
0.25
io = (25 − 25e−32,000t ) mA,
vo = 0.25
0 ≤ t ≤ 50 µs;
dio
= 200e−32,000t V,
dt
0 ≤ t ≤ 50 µs.
50 µs ≤ t < ∞:
io (50µs) = 25 − 25e−1.6 = 19.95 mA;
io = 19.95e−32,000(t−50×10
vo = (0.25)
−6 )
io (∞) = 0;
mA;
dio
= −159.62e−32,000(t−50µs) .
dt
.·. t < 0 :
vo
=
0;
0 < t < 50 µs :
vo
=
200e−32,000t V;
50 µs < t < ∞ :
vo
= −159.62e−32,000(t−50µs) .
[b] vo (50− µs) = 200e−1.6 = 40.38 V;
vo (50+ µs) = −159.62 V.
[c] io (50− µs) = io (50+ µs) = 19.95 mA.
P 7.83
[a] 0 ≤ t ≤ 6 ms:
vc (0+ ) = 0;
vc (∞) = 40 V;
RC = 500 × 103 (0.02 × 10−6 ) = 10 ms;
1/RC = 100;
vc = 40 − 40e−100t ;
vo = 40 − 40 + 40e−100t = 40e−100t V,
0 ≤ t ≤ 6 ms.
7–71
7–72
CHAPTER 7. Response of First-Order RL and RC Circuits
6 ms ≤ t < ∞:
vc (6 ms) = 40 − 40e−0.6 = 18.05 V;
vc (∞) = 0 V;
τ = 10 ms;
1/τ = 100;
vc = 18.05e−100(t − 0.006) V;
vo = −vc = −18.05e−100(t − 0.006) V,
t ≥ 6 ms.
[b]
P 7.84
[a] t < 0 :
vo = 0.
0 ≤ t ≤ 10 ms:
vo (0) = 0;
vo (∞) = 40 V;
τ = (50,000)(0.4 × 10−6 ) = 20 ms;
vo = 40 − 40e−50t V,
1/τ = 50;
0 ≤ t ≤ 10 ms;
vo (10 ms) = 40(1 − e−0.5 ) = 15.74 V.
10 ms ≤ t ≤ 20 ms:
vo (∞) = −40 V;
vo = −40 + 55.74e−50(t−0.01) V;
vo (20 ms) = −40 + 55.74e−0.5 = −6.19 V.
20 ms ≤ t ≤ ∞:
vo (∞) = 0;
vo = −6.19e−50(t−0.02) V.
Problems
[b]
[c] t ≤ 0 :
vo = 0.
0 ≤ t ≤ 10 ms:
vo (0) = 0;
vo (∞) = 40 V;
τ = 10,000(0.4 × 10−6 ) = 4 ms;
vo = 40 − 40e−250t V,
0 ≤ t ≤ 10 ms;
vo (10 ms) = 40 − 40e−2.5 = 36.72 V.
10 ms ≤ t ≤ 20 ms:
vo (∞) = −40 V;
vo = −40 + 76.72e−250(t−0.01) V,
10 ms ≤ t ≤ 20 ms;
vo (20 ms) = −40 + 76.72e−2.5 = −33.7 V.
20 ms ≤ t ≤ ∞:
vo (∞) = 0;
vo = −33.7e−250(t−0.02) V,
20 ms ≤ t ≤ ∞.
7–73
7–74
P 7.85
CHAPTER 7. Response of First-Order RL and RC Circuits
[a]
τ = (250 × 10−9 )(40 × 103 ) = 10 ms;
vc (0+ ) = 120 V;
1/τ = 100;
vc (∞) = 0;
vc = 120e−100t V.
.·. 120e−100t = 12;
e100t = 10;
t=
ln 10
= 23.026 ms.
100
[b] 0+ < t < 23.026− ms:
dvC
= (250 × 10−9 )(−100)(120e−100t ) = −3e−100t mA.
dt
t ≥ 23.026+ ms:
i=C
τ = (250 × 10−9 )(20 × 103 ) = 5 ms;
vc (23.026+ ms) = 12 V;
1/τ = 200;
vc (∞) = 120 V
vc = 120 − 108e−200(t−0.023026) V,
t ≥ 23.026 ms;
i = 250 × 10−9 (21,600)e−200(t−0.023026)
= 5.4e−200(t−0.023026) mA,
t ≥ 23.026+ ms.
[c] 120 − 108e−200∆t = 0.75(120) = 90;
.·.
108e−200∆t = 30;
e200∆t = 3.6;
P 7.86
∆t =
ln 3.6 ∼
= 6.4 ms.
200
Find the Thévenin equivalent with respect to the terminals of the capacitor.
RTh calculation:
iT =
vT
vT
vT
+
− 11.5
;
8000 20,000
20,000
Problems
iT
2.5 + 1 − 11.5
−8
=
=
;
vT
20,000
20,000
−20,000
vT
=
.·.
= −2500 Ω.
iT
8
Open circuit voltage calculation:
voc
voc − v1
+
− 11.5i∆ = 0;
8000
4000
v1 − voc
v1
+
− 1.6 × 10−3 = 0;
4000
16,000
i∆ =
v1
;
16,000
Solving, voc = −12.4 V.
vc (0) = 0;
vc (∞) = −12.4 V;
τ = RC = (−2500)(16 × 10−9 ) = −40 × 10−6 ;
1
= −25,000;
τ
vc = −12.4 + 12.4e25,000t = 930;
e25,000t = 76;
25,000t = ln 76;
t = 173.23 µs.
7–75
7–76
CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.87
vT = 2000iT + 6000(iT + 15 × 10−4 vφ ) = 8000iT + 9vφ
= 8000iT + 9(−2000iT );
vT
= −10,000.
iT
τ=
8
= −0.8 ms;
−10,000
1/τ = −1250;
i = 25e1250t mA;
.·. 25e1250t × 10−3 = 12;
P 7.88
t=
ln 480
= 4.94 ms.
1250
t > 0:
vT = −75 × 103 i∆ + 20 × 103 iT ;
i∆ =
100
iT = 0.8iT ;
125
.·. vT = −60 × 103 iT + 20 × 103 iT .
RTh =
vT
= −40 kΩ;
iT
Problems
τ = RC = −40 × 103 (0.025) × 10−6 = −10−3; s
vc = 25e1000t V;
1000t = ln 2000
P 7.89
25e1000t = 50,000;
.·.
t = 7.6 ms.
[a]
vT = 2000iσ ;
6
iσ = (iT + βiσ ) = 0.75iT + 0.75βiσ ;
8
iσ (1 − 0.75β) = 0.75iT ;
iσ =
0.75iT
;
1 − 0.75β
RTh =
1500iT
= vT ;
(1 − 0.75β)
1500
vT
=
= −3000;
iT
1 − 0.75β
1 − 0.75β = −0.5
[b] Find
2000iσ =
.·. β = 2.
VTh :
VTh − 120
VTh
(VTh − 120)
+
−2
= 0;
2000
6000
2000
VTh = 180 V.
i(0) = 0;
i(∞) = −60 mA;
τ=
L
0.3
=
= −0.1 ms.
R
−3000
7–77
7–78
CHAPTER 7. Response of First-Order RL and RC Circuits
.·. i = −60 + 60e10,000t mA.
v=L
di
= (0.3)(10,000)(0.06e10,000t ) = 180e10,000t = 36,000;
dt
ln 200
.·. t =
= 529.83 µs.
10,000
P 7.90
0 ≤ t ≤ 32 ms:
vo = −
t
1 Zt
1
1
−10 dx + 0 = −
(−10x) =
(10t);
RCf 0
RCf
RCf
0
RCf = (200 × 103 )(0.2 × 10−6 ) = 40 × 10−3
.·.
so
1
= 25;
RCf
vo (t) = 250t so vo (0.032) = (250)(0.032) = 8 V.
t ≥ 32 ms:
t
1
1 Zt
1
vo = −
5 dy + 8 = −
(5y)
+8 = −
5(t − 32 × 10−3 ) + 8;
RCf 32×10−3
RCf
RCf
32×10−3
RCf = (250 × 103 )(0.2 × 10−6 ) = 50 × 10−3
.·.
so
1
= 20;
RCf
vo (t) = −20(5)(t − 32 × 10−3 ) + 8 = −100t + 11.2 V.
The output will saturate at the negative power supply value:
−15 = −100t + 11.2
.·.
t = 262 ms.
Problems
P 7.91
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 ) = 1.6 × 10−3 s;
vp = −2 + 2e−625t V;
1/τ = 625;
vn = vp .
Write a KCL equation at the inverting input, and use it to determine vo :
vn − vo
vn
+
= 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;
P 7.92
e−625t = 1/2;
t = (ln 2)/625 = 1.11 ms.
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 Problem 7.91:
vo = 5vp = −10 + 15e−625t V.
The output will saturate at the negative power supply value:
−10 + 15e−625t = −5;
P 7.93
e−625t = 1/3;
t = (ln 3)/625 = 1.76 ms.
[a] RC = (200 × 103 )(25 × 10−9 ) = 5 × 10−3 ;
0 ≤ t ≤ 5 µs:
1
= 200.
RC
vg = 0.6 × 106 t V;
vo = −200
Z t
0.6 × 106 x dx + 0
0
= −12 × 107
x2 t
= −6 × 107 t2 V;
2 0
vo (5 µs) = −6 × 107 (5 × 10−6 )2 = −1.5 × 10−3 V.
7–79
7–80
CHAPTER 7. Response of First-Order RL and RC Circuits
5 µs ≤ t ≤ 15 µs:
vg = 6 − 0.6 × 106 t V;
vo = −200
Z t
5×10−6
(6 − 0.6 × 106 x) dx − 1.5 × 10−3
x2 t
−1.5 × 10−3
−6
−6
2
5×10
5×10
= −1200t + 6 × 10−3 + 6 × 107 t2 − 1.5 × 10−3 − 1.5 × 10−3
= 6 × 107 t2 − 1200t + 3 × 10−3 V.
t
+12 × 107
= −1200x
vo (15 µs) = 6 × 107 (15 × 10−6 )2 − 1200(15 × 10−6 ) + 3 × 10−3
= −1.5 × 10−3 V.
15 µs ≤ t ≤ 20 µs:
vg = −12 + 0.6 × 106 t V;
vo = −200
Z t
15×10−6
t
(−12 + 0.6 × 106 x) dx − 1.5 × 10−3
7x
2
t
−12 × 10
= 2400x
−1.5 × 10−3
2
= 2400t − 36 × 10 − 6 × 107 t2 + 13.5 × 10−3 − 1.5 × 10−3
= −6 × 107 t2 + 2400t − 24 × 10−3 V.
15×10−6
15×10−6
−3
vo (20 µs) = −6 × 107 (20 × 10−6 )2 + 2400(20 × 10−6 ) − 24 × 10−3 = 0.
[b]
[c] The output voltage will also repeat. This follows from the observation that
at t = 20 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 = 20 µs as
it was at t = 0, thus as vg repeats itself, so will vo .
P 7.94
[a] RC = 40(50) × 10−6 = 2 ms;
1
= 500;
RC
[b] 0 ≤ t ≤ 50 ms :
vo = −500
Z t
0
−0.50 dx = 250t V.
vo = 0,
t < 0.
Problems
7–81
[c] 50 ms ≤ t ≤ 100 ms;
vo (0.05) = 250(0.05) = 12.5 V;
vo (t) = −500
Z t
0.50 dx + 12.5 = −250(t − 0.05) + 12.5 = −250t + 25 V.
0.05
[d] 100 ms ≤ t ≤ ∞ :
vo (0.1) = −25 + 25 = 0 V;
vo (t) = 0 V.
P 7.95
Write a KCL equation at the inverting input to the op amp, where the voltage
is 0:
0 − vg 0 − vo
d
+
+ Cf (0 − vo ) = 0;
Ri
Rf
dt
.·.
dvo
1
vg
+
vo = −
.
dt
Rf Cf
Ri Cf
Note that this first-order differential equation is in the same form as Eq. 7.18
if Is = −vg /Ri . Therefore, its solution is the same as Eq. 7.19:
−vg Rf
−vg Rf −t/Rf Cf
vo =
+ Vo −
e
.
Ri
Ri
[a] vo = 0,
t < 0.
[b] Rf Cf = (4 × 106 )(50 × 10−9 ) = 0.2;
1
= 5;
Rf Cf
−vg Rf
−(−0.5)(4 × 106 )
=
= 50;
Ri
40,000
Vo = vo (0) = 0;
.·.
vo = 50 + (0 − 50)e−5t = 50(1 − e−5t ) V,
0 ≤ t ≤ 50 ms.
7–82
CHAPTER 7. Response of First-Order RL and RC Circuits
[c]
1
= 5;
Rf Cf
−vg Rf
−(0.5)(4 × 106 )
=
= −50;
Ri
40,000
Vo = vo (0.05) = 50(1 − e−0.25 ) ∼
= 11.06 V;
.·.
vo = −50 + [11.06 − (−50)]e−5(t−0.05) ;
= 61.06e−5(t−0.05) − 50 V,
[d]
50 ms ≤ t ≤ 100 ms.
1
= 5;
Rf Cf
−vg Rf
= 0;
Ri
Vo = vo (0.10) = 61.06e−0.25 − 50 ∼
= −2.45 V;
vo = 0 + (−2.45 − 0)e−5(t−0.1) = −2.45e−5(t−0.1) V,
P 7.96
[a]
Cdvp vp − vb
+
= 0;
dt
R
therefore
100 ms ≤ t ≤ ∞.
dvp
1
vb
+
vp =
;
dt
RC
RC
vn − va
d(vn − vo )
+C
= 0;
R
dt
therefore
dvo
dvn
vn
va
=
+
−
.
dt
dt
RC RC
But vn = vp .
So
dvn
vn
dvp
vp
vb
+
=
+
=
.
dt
RC
dt
RC
RC
Therefore
dvo
1
=
(vb − va );
dt
RC
vo =
1 Zt
(vb − va ) dy.
RC 0
Problems
7–83
[b] The output is the integral of the difference between vb and va and then
scaled by a factor of 1/RC.
1 Zt
[c] vo =
(vb − va ) dx;
RC 0
RC = (40 × 103 )(25 × 10−9 ) = 1 ms;
vb − va = 50 mV;
vo = 1000
P 7.97
Z t
0
0.05 dx = 50t;
50tsat = 12;
tsat = 240 ms.
[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
It follows that vce2 = VCC − VCC e−t/RL C1 .
[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/R2 C2 = 0,
or when t = R2 C2 ln 2.
7–84
CHAPTER 7. Response of First-Order RL and RC Circuits
[d] When t = R2 C2 ln 2,
we have
vce2 = VCC − VCC e−[(R2 C2 ln 2)/(RL C1 )] = 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/RL C1
+
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]
P 7.98
[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
3000 18,000
9
9 −25/6 ∼
+
e
[f ] ib1 =
= 0.5465 mA
18,000 3000
[e] ib1 =
Problems
7–85
[g] vce2 = 9 − 9e−25/6 ∼
= 8.86 V
P 7.99
[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 =
× 10−3 + 3 × 10−3 e−35/9 ∼
= 0.561 mA
18
[g] vce2 = 9 − 9e−35/9 ∼
= 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 7.100 If R1 = R2 = 20RL = 100 kΩ,
then
50 × 10−6
C1 =
= 721.35 pF;
100 × 103 ln 2
If R1 = R2 = 2RL = 10 kΩ,
C1 =
50 × 10−6
= 7.21 nF;
10 × 103 ln 2
75 × 10−6
C2 =
= 1.08 nF.
100 × 103 ln 2
then
C2 =
75 × 10−6
= 10.82 nF.
10 × 103 ln 2
Therefore 721.35 pF ≤ C1 ≤ 7.21 nF and 1.08 nF ≤ C2 ≤ 10.82 nF.
P 7.101 [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 . So vbe2 = 0 when t = RC ln 2,
therefore T2 stays OFF for RC ln 2 seconds.
P 7.102 [a] For t < 0, vce2 = 0. When the switch is momentarily closed, vce2 jumps to
VCC
12
R1 =
(8000) = 2 V.
R1 + RL
8000 + 40, 000
T2 remains open for (36,067)(100 × 10−12 ) ln 2 ∼
= 2.5 µs.
vce2 =
7–86
CHAPTER 7. Response of First-Order RL and RC Circuits
[b] ib2 =
VCC
= 332.71 µA,
R
ib2 = 0,
ib2
−2 ≤ t ≤ 0 µs;
0 < t < RC ln 2(= 2.5 µs);
=
VCC VCC −(t−RC ln 2)/RL C
+
e
R
RL
=
332.71 + 300e−250,000(t−2.5×10
−6 )
µA,
2.5 µs < t;
P 7.103 [a] 0 ≤ t ≤ 0.5:
i = If + (I0 − If )e−t/τ =
9
12
9
+
−
e−t/τ
30
30 30
i = 0.3 + 0.1e−30t/L ;
i(0.1) = 0.3 + 0.1e−3/L = 0.35;
.·. e3/L = 2;
L=
3
= 4.33 H.
ln 2
[b] 0 ≤ t ≤ tr , where tr is the time the relay releases:
12
i=0+
− 0 e−30t/L = 0.4e−30t/L ;
30
.·. 0.35 = 0.4e−30tr /L ;
e30tr /L = 1.143;
where τ = L/R.
Problems
tr =
7–87
4.33 ln 1.143 ∼
= 19.3 ms.
30
P 7.104 From the Practical Perspective,
vC (t) = 0.75VS = VS (1 − e−t/RC ).
0.25 = e−t/RC
t = −RC ln 0.25.
so
In the above equation, t is the number of seconds it takes to charge the
capacitor to 0.75VS , so it is a period. We want to calculate the heart rate,
which is a frequency in beats per minute, so H = 60/t. Thus,
H=
60
.
−RC ln 0.25
P 7.105 In this problem, Vmax = 0.8VS , so the equation for heart rate in beats per
minute is
H=
60
.
−RC ln 0.2
Given R = 100 kΩ and C = 20 µF,
H=
60
= 74.56.
−(100,000)(5 × 10−6 ) ln 0.2
Therefore, the heart rate is about 75 beats per minute.
P 7.106 From the Practical Perspective,
vC (t) = Vmax = VS (1 − e−t/RC ).
Solve this equation for the resistance R:
Vmax
= 1 − e−t/RC
VS
so
−t
Vmax
= ln 1 −
;
RC
VS
Then,
.·.
e−t/RC = 1 −
R=
−t
Vmax
C ln 1 −
VS
.
Vmax
.
VS
7–88
CHAPTER 7. Response of First-Order RL and RC Circuits
In the above equation, t is the time it takes to charge the capacitor to a
voltage of Vmax . But t and the heart rate H are related as follows:
H=
60
.
t
Therefore,
R=
−60
Vmax
HC ln 1 −
VS
.
P 7.107 From Problem 7.106,
R=
−60
Vmax
HC ln 1 −
VS
.
Note that from the problem statement,
Vmax
= 0.7.
VS
Therefore,
R=
−60
= 69.215 kΩ.
(72)(10 × 10−6 ) ln (1 − 0.7)
Natural and Step Responses of
RLC Circuits
Assessment Problems
AP 8.1 [a] α =
1
1
=
= 250;
2RC
2(1000)(2 × 10−6 )
ω02 =
1
1
=
= 40,000;
LC
(12.5)(2 × 10−6 )
s1,2 = −250 ±
q
2502 − 40,000 = −250 ± 150;
s1 = −100 rad/s;
s2 = −400 rad/s.
[b] Overdamped.
1
1
=
= 160;
[c] α =
2RC
2(1562.5)(2 × 10−6 )
ω02 =
1
1
=
= 40,000;
LC
(12.5)(2 × 10−6 )
s1,2 = −160 ±
q
1602 − 40,000 = −160 ± j120;
s1 = −160 + j120 rad/s;
[d] α =
q
40,000 =
1
;
2RC
s2 = −160 − j120 rad/s.
.·. R =
1
= 1250 Ω.
2(200)(2 × 10−6 )
AP 8.2 From the given values of R, L, and C,
α=
1
1
=
= 2500 rad/s;
2RC
2(5000)(40 × 10−9 )
1
1
=
= 20002 rad2 /s2 ;
LC
6.25(40 × 10−9 )
√
s1 , s2 = −2500 ± 25002 − 20002 so s1 = −1000 rad/s and s2 = −4000 rad/s.
ω02 =
8–1
8–2
CHAPTER 8. Natural and Step Responses of RLC Circuits
−10
= −2 mA.
5000
[b] iC (0+ ) = −(iL (0+ ) + iR (0+ )) = −(8 − 2) = −6 mA.
dvc (0+ )
−0.006
dvc (0+ )
= ic (0+ ) = −0.006,
therefore
=
= −150,000 V/s.
[c] C
dt
dt
C
[d] v = [A1 e−1000t + A2 e−4000t ] V,
t ≥ 0+ ;
[a] v(0− ) = v(0+ ) = −10,
v(0+ ) = A1 + A2 ,
therefore iR (0+ ) =
dv(0+ )
= −1000A1 − 4000A2 .
dt
Therefore A1 + A2 = −10,
−A1 − 4A2 = −150;
A1 = −63.33 V.
[e] A2 = 53.33 V.
[f ] v = [53.33e−4000t − 63.33e−1000t ] V,
t ≥ 0.
v(t)
v(t)
=
= 10.67e−4000t − 12.67e−1000t mA.
R
5000
dv(t)
[b] iC (t) = C
= 40 × 10−9 [(−4000)53.33e−4000t + (1000)63.33e−1000t ]
dt
AP 8.3 [a] iR (t) =
= 2.533e−1000t − 8.533e−4000t mA.
[c] iL (t) = −[iR (t) + iC (t)] = 2.133e−4000t − 10.133e−1000t mA.
AP 8.4 [a] α = 400;
ωd =
ωd = 300;
q
ω02 − α2 ;
.·. ω02 = ωd2 + α2 = 9 × 104 + 16 × 104 = 25 × 104 .
1
= 25 × 104 ;
LC
1
L=
= 16 mH.
4
(25 × 10 )(250 × 10−6 )
ω02 =
[b] α =
1
;
2RC
.·. R =
1
1
=
= 5 Ω.
2αC
(800)(250 × 10−6 )
[c] V0 = v(0) = 120 V.
[d] I0 = iL (0) = −iR (0) − iC (0);
iR (0) =
120
= 24 A;
5
iC (0) = C
dv
(0) = 250 × 10−6 [−400(120) + 300(80)] = −6 A;
dt
.·. I0 = −24 + 6 = −18 A.
Problems
[e] iC (t) = 250 × 10−6
dv(t)
= e−400t (−17 sin 300t − 6 cos 300t) A;
dt
v(t)
= e−400t (24 cos 300t + 16 sin 300t) A;
5
iR (t) =
iL (t) = −iR (t) − iC (t)
= e−400t (−18 cos 300t + sin 300t) A,
t ≥ 0.
Check:
diL
= 16 × 10−3 e−400t [7500 cos 300t + 5000 sin 300t];
L
dt
v(t) = e−400t [120 cos 300t + 80 sin 300t] V.
AP 8.5 [a]
1
2RC
2
=
.·. C =
[b]
1
= (500)2 ;
LC
1
= 1 µF.
(500)2 (4)
1
= 500;
2RC
.·. R =
1
= 1 kΩ.
2(500)(10−6 )
[c] v(0) = D2 = 8 V;
8
= 8 mA;
[d] iR (0) =
1000
iC (0) = −8 + 10 = 2 mA;
dv
2 × 10−3
= 2000 V/s;
(0) = D1 − 500D2 =
dt
10−6
.·. D1 = 2000 + 500(8) = 6000 V/s.
[e] v = 6000te−500t + 8e−500t V,
t ≥ 0;
dv
= [−3 × 106 t + 2000]e−500t ;
dt
iC = C
dv
= (−3000t + 2)e−500t mA,
dt
t ≥ 0+ .
V0
50
=
= 4 A.
R
12.5
[b] iC (0+ ) = I − iR (0+ ) − iL (0+ ) = 2 − 4 − 1 = −3 A.
Vo
50
diL (0+ )
[c]
=
=
= 2000 A/s.
dt
L
0.025
AP 8.6 [a] iR (0+ ) =
8–3
8–4
CHAPTER 8. Natural and Step Responses of RLC Circuits
1
1
= 640;
= 640,000;
s1,2 = −640 ± j480 rad/s.
2RC
LC
[e] iL = If + B10 e−αt cos ωd t + B20 e−αt sin ωd t,
If = I = 2 A;
[d] α =
iL (0+ ) = 1 = If + B10 ,
therefore B10 = −1 A;
diL (0+ )
= 2000 = −αB10 + ωd B20 ,
dt
therefore B20 = 2.833 A;
Therefore iL (t) = 2 + e−640t [− cos 480t + 2.833 sin 480t] A,
[f ] v(t) =
AP 8.7 α =
LdiL
= e−640t [50 cos 480t − 33.33 sin 480t] V
dt
t ≥ 0.
R
200
=
= 400 rad/s;
2L
2(0.25)
s
ω0 =
1
=
LC
α2 < ω02 :
ωd =
√
s
1
(250 × 10−3 )(16 × 10−6 )
= 500 rad/s;
underdamped.
5002 − 4002 = 300 rad/s;
vo = Vf + B10 e−400t cos 300t + B20 e−400t sin 300t;
vo (∞) = 200(0.08) = 16 V;
vo (0) = 0 = Vf + B10 = 0
so
dvo
(0) = 0 = −400B10 + 300B20
dt
B10 = −16 V;
so
B20 = −21.33 V;
.·. vo (t) = 16 − 16e−400t cos 300t − 21.33e−400t sin 300t V,
AP 8.8 α =
R
250
=
= 500 rad/s;
2L
2(0.25)
s
ω0 =
1
=
LC
α2 = ω02 :
s
1
(250 × 10−3 )(16 × 10−6 )
critically damped.
vo = Vf + D10 te−500t + D20 e−500t ;
= 500 rad/s;
t ≥ 0.
t ≥ 0.
Problems
vo (0) = 0 = Vf + D20 ;
.·. D20 = −20 V;
vo (∞) = (250)(0.08) = 20 V;
dvo
(0) = 0 = D10 − αD20
dt
.·.
AP 8.9 α =
D10 = (500)(−20) = −10,000 V/s;
vo (t) = 20 − 10,000te−500t − 20e−500t V,
t ≥ 0.
R
312.5
=
= 625 rad/s;
2L
2(0.25)
s
ω0 =
so
1
=
LC
α2 > ω02 :
s
1
(250 × 10−3 )(16 × 10−6 )
= 500 rad/s;
overdamped.
s1,2 = −625 ±
√
6252 − 5002 = −250, −1000 rad/s;
vo = Vf + A01 e−250t + A02 e−1000t ;
vo (0) = 0 = Vf + A01 + A02 ;
vo (∞) = (312.5)(08) = 25 V;
.·. A01 + A02 = −25 V;
dvo
(0) = 0 = −250A01 − 1000A02 ;
dt
Solving,
A01 = −33.33 V;
A02 = 8.33 V;
vo (t) = 25 − 33.33e−250t + 8.33e−1000t V,
t ≥ 0.
8–5
8–6
CHAPTER 8. Natural and Step Responses of RLC Circuits
Problems
P 8.1
1
1
=
,
therefore C = 500 nF.
2
(2RC)
LC
1
[b] α = 5000 =
,
therefore C = 1 µF;
2RC
[a]
s
25 × 106 −
s1,2 = −5000 ±
[c] √
1
= 20,000,
LC
s1,2 = −40 ±
therefore C = 125 nF;
q
(40)2 − 202
s1 = −5.36 krad/s,
P 8.2
s1 = −α +
q
α2 − ω02 ;
s1 + s2 = −2α
R=
→
(103 )(106 )
= (−5000 ± j5000) rad/s.
20
103 ,
s2 = −74.64 krad/s.
s2 = −α +
q
α2 − ω02 ;
α = −(s1 + s2 )/2 = −(−200 − 1800)/2 = 1000 rad/s.
1
106
=
= 27.78 Ω;
2αC
(2000)(18)
v(0+ ) = −24 V;
iR (0+ ) =
−24
= −864 mA;
27.78
dv
= 2400e−200t + 21,600e−1800t ;
dt
dv(0+ )
= 2400 + 21,600 = 24,000 V/s;
dt
iC (0+ ) = 18 × 10−6 (24,000) = 432 mA;
iL (0+ ) = −[iR (0+ ) + iC (0+ )] = −[−864 + 432] = 432 mA.
Problems
P 8.3
[a] −α +
q
α2 − ω02 = −5000;
−α −
q
α2 − ω02 = −20,000;
.·. −2α = −25,000;
α = 12,500 rad/s;
106
1
=
= 12,500;
2RC
2R(0.05)
R = 800 Ω;
q
2 α2 − ω02 = 15,000;
4(α2 − ω02 ) = 225 × 106 ;
.·. ω0 = 10,000 rad/s.
ωo2 = 108 =
.·. L =
[b] iR =
1
= 200 mH.
108 C
v(t)
= −6.25e−5000t + 25e−20,000t mA,
R
t ≥ 0+ ;
dv(t)
= 1.25e−5000t − 20e−20,000t mA,
dt
t ≥ 0+ ;
iL = −(iR + iC ) = 5e−5000t − 5e−20,000t mA,
t ≥ 0+ .
iC = C
P 8.4
1
;
LC
[a] ω02 =
109
1
=
= 25 × 106 ;
LC
40
ω0 = 5000 rad/s;
1
= 5000;
2RC
R=
R=
1
;
10,000C
109
= 12.5 kΩ.
8 × 104
[b] v(t) = D1 te−5000t + D2 e−5000t ;
v(0) = −25 V = D2 ;
dv
= (D1 t − 25)(−5000e−5000t ) + D1 e−5000t ;
dt
iC (0)
dv
(0) = 125 × 103 + D1 =
;
dt
C
8–7
8–8
CHAPTER 8. Natural and Step Responses of RLC Circuits
iC (0) = −iR (0) − iL (0);
iR (0) =
−25
= −2 mA;
12,500
.·. iC (0) = 2 − (−1) = 3 mA;
3 × 10−3
dv
(0) =
= 0.375 × 106 = 3.75 × 105 ;
.·.
dt
8 × 10−9
.·. 1.25 × 105 + D1 = 3.75 × 105 .
D1 = 2.5 × 105 = 25 × 104 V/s;
.·. v(t) = (25 × 104 t − 25)e−5000t V,
[c] iC (t) = 0 when
t ≥ 0.
dv
(t) = 0;
dt
dv
= (25 × 104 t − 25)(−5000)e−5000t + e−5000t (25 × 104 )
dt
= (375,000 − 125 × 107 t)e−5000t ;
dv
= 0 when 125 × 107 t1 = 375,000;
dt
.·. t1 = 300 µs;
v(300µs) = 50e−1.5 = 11.1565 V.
[d] iL (300µs) = −iR (300µs) =
11.1565
= 0.89252 mA;
12.5
wC (300µs) = 4 × 10−9 (11.1565)2 = 497.87 nJ;
wL (300µs) = (2.5)(0.89252)2 × 10−6 = 1991.48 nJ;
w(300µs) = ωC + ωL = 2489.35 nJ;
w(0) = 4 × 10−9 (625) + 2.5(10−6 ) = 5000 nJ.
% remaining =
P 8.5
[a] iR (0) =
2489.35
(100) = 49.79%.
5000
90
= 45 mA;
2000
iL (0) = −30 mA;
iC (0) = −iL (0) − iR (0) = 30 − 45 = −15 mA.
Problems
[b] α =
1
1
=
= 25,000;
2RC
2(2000)(10 × 10−9 )
ω02 =
1
1
=
= 4 × 108 ;
LC
(0.25)(10 × 10−9 )
s1,2 = −25,000 ±
q
6.25 × 108 − 108 (4) = −25,000 ± 15,000;
s1 = −10,000 rad/s;
s2 = −40,000 rad/s;
v = A1 e−10,000t + A2 e−40,000t .
v(0) = A1 + A2 = 90;
dv
−15 × 10−3
(0) = −104 A1 − 4A2 × 104 =
= −1.5 × 106 V/s;
−9
dt
10 × 10
−A1 − 4A2 = −150;
.·. −3A2 = −60;
A2 = 20;
v = 70e−10,000t + 20e−40,000t V,
[c] iC
= C
=
A1 = 70.
t ≥ 0.
dv
dt
10 × 10−9 [−70 × 104 e−10,000t − 80 × 104 e−40,000t ]
= −7e−10,000t − 8e−40,000t mA;
iR = 35e−10,000t + 10e−40,000t mA;
iL = −iC − iR = −28e−10,000t − 2e−40,000t mA,
P 8.6
α=
t ≥ 0.
1
1
=
= 20,000;
2RC
2(2500)(10 × 10−9 )
α2 = 4 × 108 ;
.·. α2 = ωo2 .
Critical damping:
v = D1 te−αt + D2 e−αt ;
iR (0+ ) =
90
= 36 mA;
2500
iC (0+ ) = −[iL (0+ ) + iR (0+ )] = −[−30 + 36] = −6 mA;
v(0) = D2 = 90;
8–9
8–10
CHAPTER 8. Natural and Step Responses of RLC Circuits
dv
= D1 [t(−αe−αt ) + e−αt ] − αD2 e−αt ;
dt
iC (0)
−6 × 10−3
dv
(0) = D1 − αD2 =
=
= −6 × 105 ;
−9
dt
C
10 × 10
D1 = αD2 − 6 × 105 = (2 × 104 )(90) − 6 × 105 = 120 × 104 ;
v = (120 × 104 t + 90)e−20,000t V,
P 8.7
t ≥ 0.
1
1
=
= 12,000;
2RC
2(12,500/3)(10 × 10−9 )
1
= 4 × 108 ;
LC
s1,2 = −12,000 ± j16,000 rad/s;
.·. response is underdamped.
v(t) = B1 e−12,000t cos 16,000t + B2 e−12,000t sin 16,000t;
v(0+ ) = 90 V = B1 ;
iR (0+ ) =
90
= 21.6 mA;
(12,500/3)
iC (0+ ) = −[iL (0+ ) + iR (0+ )] = −[−30 + 21.6] = 8.4 mA;
dv(0+ )
8.4 × 10−3
=
= 840,000 V/s;
dt
10 × 10−9
dv(0+ )
= −12,000B1 + 16,000B2 = 840,000,
dt
or − 3B1 + 4B2 = 210;
.·. B2 = 120 V;
v(t) = 90e−12,000t cos 16,000t + 120e−12,000t sin 16,000t V,
P 8.8
√
1
= 1000 2,
ω0 = 103 ,
2RC
s1 = −414.21,
s2 = −2414.21;
[a] α =
t ≥ 0.
therefore overdamped.
therefore v = A1 e−414.21t + A2 e−2414.21t ;
Therefore
dv(0+ )
iC (0+ )
=
= 98,000 V/s;
dt
C
− 414.21A1 − 2414.21A2 = 98,000;
A1 = 49,
A2 = −49;
v(0+ ) = 0 = A1 + A2 ;
v(t) = 49[e−414.21t − e−2414.21t ] V,
t ≥ 0.
Problems
[b]
Example 8.4: vmax ∼
= 74.1 V at 1.4 ms;
Example 8.5: vmax ∼
= 36.1 V at 1.0 ms;
Problem 8.8: vmax ∼
= 28.2 V at 0.9 ms.
P 8.9
1
= 8000,
therefore R = 62.5 Ω.
2RC
10 V
[b] iR (0+ ) =
= 160 mA;
62.5 Ω
[a]
iC (0+ ) = −(iL (0+ ) + iR (0+ )) = −80 − 160 = −240 mA = C
Therefore
dv(0+ )
−0.240
=
= −240 kV/s.
dt
C
[c] B1 = v(0+ ) = 10 V,
dvc (0+ )
= ωd B2 − αB1 .
dt
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 +
[a]
1
2RC
2
=
B2 = (−80/3) V.
dv
;
dt
80
sin 6000t] V.
3
Therefore iR = e−8000t [160 cos 6000t −
P 8.10
dv(0+ )
.
dt
1280
sin 6000t] mA.
3
460
sin 6000t] mA;
3
82
sin 6000t] mA,
3
t ≥ 0.
1
1
1
=
,
→
= 500,
LC
(0.4)(10 × 10−6 )
2RC
R = 100 Ω.
8–11
8–12
CHAPTER 8. Natural and Step Responses of RLC Circuits
[b] 0.5CV02 = 12.5 × 10−3 ,
[c]
0.5LI02 = 12.5 × 10−3 ,
I0 = 250 mA.
dv(0+ )
= D1 − αD2 ;
dt
[d] D2 = v(0+ ) = 50,
iR (0+ ) =
therefore V0 = 50 V.
50
= 500 mA.
100
Therefore iC (0+ ) = −(500 + 250) = −750 mA;
−0.75
dv(0+ )
=
= −75,000 V/s;
dt
C
Therefore
Therefore D1 − αD2 = −75,000;
α=
1
= 500,
2RC
D1 = −50,000 V/s.
[e] v = [50e−500t − 50,000te−500t ] V;
iR =
P 8.11
v
= [0.5e−500t − 500te−500t ] A,
R
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
K 1 = 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 )
= −(α + jωd )K1 − K2
Problems
1
and N2 =
K1
8–13
= K2 + (α − jωd )K1 .
(−α + jωd ) K2
It follows that A1 =
and A2 =
ωd K1 − j(αK1 + K2 )
N1
=
;
∆
2ωd
N2
ωd K1 + j(αK1 + K2 )
.
=
∆
2ωd
We see from these expressions that
P 8.12
A1 = A∗2 .
By definition, B1 = A1 + A2 . From the solution to Problem 8.11 we have
A1 + A2 =
2ωd K1
= K1 .
2ωd
But K1 is v(0), therefore, B1 = v(0), which is identical to Eq. 8.16.
By definition, B2 = j(A1 − A2 ). From Problem 8.11 we have
B2 = j(A1 − A2 ) =
αK1 + K2
j[−2j(αK1 + K2 )]
=
.
2ωd
ωd
It follows that
K2 = −αK1 + ωd B2 ,
dv(0+ )
but K2 =
dt
and K1 = B1 .
Thus we have
dv +
(0 ) = −αB1 + ωd B2 ,
dt
which is identical to Eq. 8.17.
P 8.13
1
= 1 rad/s;
2RC
1
= 10;
ω02 =
LC
√
ωd = 10 − 1 = 3 rad/s;
[a] α =
.·. v = B1 e−t cos 3t + B2 e−t sin 3t.
v(0) = B1 = 0;
v = B2 e−t sin 3t;
iR (0+ ) = 0 A;
iC (0+ ) = −iL (0+ ) = 3 A;
12 = −αB1 + ωd B2 = −1(0) + 3B2 ;
.·. B2 = 4;
.·. v = 4e−t sin 3t V,
t ≥ 0.
dv +
3
(0 ) =
= 12 V/s;
dt
0.25
8–14
CHAPTER 8. Natural and Step Responses of RLC Circuits
[b]
dv
= 4e−t (3 cos 3t − sin 3t);
dt
dv
= 0 when 3 cos 3t = sin 3t or
dt
.·. 3t1 = 1.25,
t1 = 416.35 ms;
3t2 = 1.25 + π,
t2 = 1463.55 ms;
3t3 = 1.25 + 2π,
t3 = 2510.74 ms.
tan 3t = 3;
2π
2π
=
= 2094.40 ms.
ωd
3
Td
2094.40
[d] t2 − t1 = 1047.20 ms;
=
= 1047.20 ms.
2
2
[e] v(t1 ) = 4e−(0.41635) sin 3(0.41635) = 2.50 V;
[c] t3 − t1 = 2094.40 ms;
Td =
v(t2 ) = 4e−(1.46355) sin 3(1.46355) = −0.88 V;
v(t3 ) = 4e−(2.51074) sin 3(2.51074) = 0.31 V.
[f ]
P 8.14
[a] α = 0;
ωd = ω0 =
√
10 = 3.16 rad/s;
v = B1 cos ω0 t + B2 sin ω0 t;
C
v(0) = B1 = 0;
dv
(0) = −iL (0) = 3;
dt
12 = −αB1 + ωd B2 = −0 +
√
.·. B2 = 12/ 10 = 3.79 V;
v = 3.79 sin 3.16t V,
[b] ωd = 3.16 rad/s.
[c] 3.79 V.
√
10B2 ;
t ≥ 0.
v = B2 sin ω0 t;
Problems
P 8.15
[a] s1 = −α +
q
α2 − ω02 = −40;
s2 = −α −
.·. s1 + s2 = −2α = −200;
q
α2 − ω02 = −160;
α = 100 rad/s;
q
s1 − s2 = 2 α2 − ω02 = 120;
ω0 = 80 rad/s;
C=
1
1
=
= 25 µF;
2αR
200(200)
L=
1
106
=
= 6.25 H;
ω02 C
(80)2 (25)
iC (0+ ) = A1 + A2 = 15 mA;
diL (0) diR (0)
diC (0)
=−
−
;
dt
dt
dt
diL (0)
0
=
= 0 A/s;
dt
6.25
1 dv(0)
1 iC (0)
15 × 10−3
diR (0)
=
=
=
= 3 A/s;
dt
R dt
R C
(200)(25 × 10−6 )
diC (0)
.·.
= −3 A/s;
dt
.·. 160A1 + 40A2 = 3.
4A1 + A2 = 75 × 10−3 ;
.·. A1 = 20 mA;
.·. iC = 20e−160t − 5e−40t mA,
A2 = −5 mA;
t ≥ 0.
[b] By hypothesis
v = A3 e−160t + A4 e−40t ,
t ≥ 0;
v(0) = A3 + A4 = 0;
15 × 10−3
dv(0)
=
= 600 V/s
dt
25 × 10−6
−160A3 − 40A4 = 600;
.·. A3 = −5 V;
v = −5e−160t + 5e−40t V,
t ≥ 0.
v
= −25e−160t + 25e−40t mA,
200
[d] iL = −iR − iC ;
[c] iR (t) =
iL = 5e−160t − 20e−40t mA,
t ≥ 0.
A4 = 5 V
t ≥ 0+ .
8–15
8–16
P 8.16
CHAPTER 8. Natural and Step Responses of RLC Circuits
t<0:
Vo = 60 V,
Io = 45 mA.
t > 0:
iR (0) =
60
= 37.5 mA;
1600
iL (0) = 45 mA;
iC (0) = −37.5 − 45 = −82.5 mA;
α=
1
109
=
= 5000 rad/s;
2RC
3200(62.5)
ω02 =
109
1
=
= 16 × 106 ;
LC
62.5
s1,2 = −5000 ±
√
25 × 106 − 16 × 106 = −5000 ± 3000;
s1 = −2000 rad/s;
s2 = −8000 rad/s;
.·. vo = A1 e−2000t + A2 e−8000t .
A1 + A2 = vo (0) = 60;
dvo
−82.5 × 10−3
(0) = −2000A1 − 8000A2 =
= −1320 × 103 ;
dt
62.5 × 10−9
Solving,
A1 = −140 V,
A2 = 200 V;
.·. vo = −140e−2000t + 200e−8000t V,
t ≥ 0.
Problems
P 8.17
ω02 =
α=
1
1
=
= 25 × 106 ;
−9
LC
0.64(62.5 × 10 )
1
1
=
= 4000 rad/s;
2RC
2(2000)(62.5 × 10−9 )
ωd =
q
α2 = 16 × 103 ;
(25 − 16) × 106 = 3000 rad/s;
s1,2 = −4000 ± j3000rad/s;
vo (t) = B1 e−4000t cos 3000t + B2 e−4000t sin 3000t.
vo (0) = B1 = 60 V;
iR (0) =
60
= 30 mA;
2000
iL (0) = 45 mA;
iC (0) = −iR (0) − iL (0) = −75 mA;
iC (0)
−75 × 10−3
=
= −12 × 105 ;
C
62.5 × 10−9
dvo
(0) = −4000B1 + 3000B2 = −12 × 105 ;
dt
.·. 3B2 = 4B1 − 1200 = 240 − 1200 = −960;
.·. B2 = −320 V;
vo (t) = 60e−4000t cos 3000t − 320e−4000t sin 3000t V,
P 8.18
ω02 =
α=
1
1
=
= 108 ;
LC
(0.16)(62.5 × 10−9 )
1
1
=
= 104 ;
−9
2RC
2(800)(62.5 × 10 )
.·. α2 = ω02 (critical damping).
vo (t) = D1 te−10,000t + D2 e−10,000t ;
vo (0) = D2 = 60 V;
ω0 = 104 ;
t ≥ 0.
8–17
8–18
CHAPTER 8. Natural and Step Responses of RLC Circuits
iR (0) =
60
= 75 mA;
800
iL (0) = 45 mA;
iC (0) = −(75 + 45) = −120 mA;
dvo
iC (0)
−120 × 10−3
(0) = −10,000D2 + D1 =
=
= −1920 × 103 ;
dt
C
62.5 × 10−9
.·.
D1 = −1320 × 103 V/s;
vo (t) = (60 − 132 × 104 t)e−10,000t V,
t > 0.
P 8.19
vT = 104
iT (150 × 103 ) (150)(60)106
+
iT ;
210 × 103
210 × 103
1500 × 103 9000 × 103
10,500
vT
=
+
=
× 103 = 50 kΩ;
iT
210
210
210
75
(6) = 45 V;
10
Vo =
Io = 0;
iC (0) = −iR (0) − iL (0) = −
45
= −0.9 mA;
50,000
iC (0)
−0.9 × 10−3
=
= −720 × 103 ;
C
1.25 × 10−9
ω02 =
α=
1
1
=
= 108 ;
LC
(8)(1.25 × 10−9 )
ω0 = 104 rad/s;
1
1
=
= 8000 rad/s;
2RC
(2)(50,000)(1.25 × 10−9 )
ωd =
q
(100 − 64) × 106 = 6000 rad/s;
Problems
vo = B1 e−8000t cos 6000t + B2 e−8000t sin 6000t.
vo (0) = B1 = 45 V;
dvo
(0) = 6000B2 − 8000B1 = −720 × 103 ;
dt
.·. 6000B2 = 8000(45) − 720 × 103 ;
.·. B2 = −60 V;
vo = 45e−8000t cos 6000t − 60e−8000t sin 6000t V,
t ≥ 0.
P 8.20
3
(150)(60)106
4 iT (150 × 10 )
vT = 10
+
iT ;
210 × 103
210 × 103
vT
1500 × 103 9000 × 103
10,500
=
+
=
× 103 = 50 kΩ;
iT
210
210
210
75
(6) = 45 V;
10
Vo =
Io = 0;
iC (0) = −iR (0) − iL (0) = −
45
= −0.9 mA;
50,000
iC (0)
−0.9 × 10−3
=
= −900 × 103 ;
−9
C
10
ω02 =
α=
1
1
=
= 108 ;
LC
(10)(10−9 )
ω0 = 10,000 rad/s;
1
1
=
= 10,000 rad/s;
2RC
(2)(50,000)(10−9 )
α2 = ω02
so the response is critically damped.
vo = D1 te−10,000t + D2 e−10,000t ;
8–19
8–20
CHAPTER 8. Natural and Step Responses of RLC Circuits
vo (0) = D2 = 45 V;
dvo
(0) = D1 − αD2 = −900 × 103 ;
dt
.·. D1 = −900 × 103 + (10,000)(45);
.·. D1 = −450,000 V/s;
vo = −450,000te−10,000t + 45e−10,000t V,
t ≥ 0.
P 8.21
vT = 104
iT (150 × 103 ) (150)(60)106
+
iT ;
210 × 103
210 × 103
1500 × 103 9000 × 103
10,500
vT
=
+
=
× 103 = 50 kΩ;
iT
210
210
210
75
(6) = 45 V;
10
Vo =
Io = 0;
iC (0) = −iR (0) − iL (0) = −
45
= −0.9 mA;
50,000
−0.9 × 10−3
iC (0)
=
= −1125 × 103 ;
C
0.8 × 10−9
ω02 =
α=
1
1
=
= 108 ;
LC
(12.5)(0.8 × 10−9 )
ω0 = 10,000 rad/s;
1
1
=
= 12,500 rad/s;
2RC
(2)(50,000)(0.8 × 10−9 )
α2 > ω02
so the response is overdamped.
vo = A1 es1 t + A2 es2 t ;
s1,2 = −α ±
q
α2 − ω02 = −12,500 ±
q
(12,500)2 − 108 = −12,500 ± 7500;
Problems
8–21
.·. s1,2 = −5000 rad/s, −20,000 rad/s;
A1 + A2 = Vo = 45
.·. A1 = −15,
− 5000A1 − 20,000A2 = −1125 × 103 ;
and
A2 = 60;
vo = −15e−5000t + 60e−20,000t V,
P 8.22
[a]
t ≥ 0.
1
= 20,0002 .
LC
There are many possible solutions. This one begins by choosing
L = 1 mH. Then,
C=
1
= 2.5 µF.
(10−3 )(20,000)2
We can achieve this capacitor value using components from Appendix H
by combining a 2.2 µF and three 0.1 µF capacitors in parallel.
Critically damped:
.·. R =
α = ω0 = 20,000
so
1
= 20,000;
2RC
1
= 10 Ω.
2(2.5 × 10−6 )(20,000)
Appendix H has a 10 Ω resistor. 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 8.23
[a] Underdamped response:
α < ω0
so
α < 20,000.
Therefore we choose a larger resistor value than the one used in Problem
8.22. Choose R = 15 Ω:
1
α=
= 13,333.33;
2(15)(2.5 × 10−6 )
s1,2 = −13,333.33 ±
q
13,333.332 − 20,0002 = −13,333.33 ± j14,907.12 rad/s.
8–22
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 8.22. Choose R = 5 Ω, which we can construct using two
parallel-connected 10 Ω resistors:
α=
1
= 40,000;
2(5)(2.5 × 10−6 )
s1,2 = −40,000 ±
q
40,0002 − 20,0002 = −40,000 ± 34,641;
= −5359 rad/s;
P 8.24
− 74,641 rad/s.
and
For t < 0 :
iL (0− ) = iL (0+ ) =
12
= 30 mA;
400
vo (0− ) = vo (0+ ) = 0.
For t > 0
α=
1
= 1000 rad/s;
2RC
s1 = −400 rad/s
ω02 =
1
= 64 × 104 ;
LC
s2 = −1600 rad/s;
vo (∞) = 0 = Vf ;
vo = A01 e−400t + A02 e−1600t ;
iC (0+ ) = −30 + 30 + 0 = 0;
dvo
iC (0+ )
·
..
=
= 0 = −400A01 − 1600A02 ;
dt
C
.·. A01 + 400A02 = 0;
.·. A01 = 0;
A02 = 0.
.·. vo = 0 for t ≥ 0.
vo (0) = A01 + A02 = 0.
Problems
Note:
vo (0) = 0;
vo (∞) = 0;
8–23
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 12 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.
P 8.25
For
t>0
α=
1
= 100;
2RC
1
= 6400;
LC
s1,2 = −100 ± 60;
s1 = −40 rad/s;
s2 = −160 rad/s;
vo = Vf + A01 e−40t + A02 e−160t ;
Vf = 0;
vo (0+ ) = 0;
iC (0+ ) = 37.5 mA;
.·. A01 + A02 = 0;
iC (0+ )
dvo (0+ )
=
= 6000 V/s;
dt
6.25 × 10−6
dvo (0+ )
= −40A01 − 160A02 ;
dt
−40A01 − 160A02 = 6000;
A01 + 4A02 = −150;
A01 + A02 = 0;
.·. A01 = 50 V;
A02 = −50 V.
vo = 50e−40t − 50e−160t V,
t ≥ 0.
8–24
P 8.26
CHAPTER 8. Natural and Step Responses of RLC Circuits
[a] From the solution to Prob. 8.25 s1 = −40 rad/s and s2 = −160 rad/s,
therefore
io = If + A01 e−40t + A02 e−160t .
If = 37.5 mA;
dio (0+ )
= 0;
dt
io (0+ ) = 0;
.·. 0 = 37.5 + A01 + A02 ;
−40A01 − 160A02 = 0.
It follows that
A01 = −50 mA;
A02 = 12.5 mA;
.·. io = 37.5 − 50e−40t + 12.5e−160t mA,
[b]
dio
= 2e−40t − 2e−160t ;
dt
vo = L
dio
= 25[2e−40t − 2e−160t ];
dt
vo = 50e−40t − 50e−160t V,
P 8.27
α=
t ≥ 0.
t ≥ 0.
106
1
=
= 100;
2RC
(1600)(6.25)
ω02 =
α2 = 104 ;
1
106
=
= 6400;
LC
(25)(6.25)
s1,2 = −100 ±
√
104 − 6400 = −100 ± 60;
s1 = −40 rad/s;
s2 = −160 rad/s;
io = If + A01 e−40t + A02 e−160t ;
If =
30
= 37.5 mA;
800
io (0) = 0;
0 = 37.5 × 10−3 + A01 + A02 ,
.·. A01 + A02 = −37.5 × 10−3 ;
dio
30
(0) =
= −40A01 − 160A02 .
dt
25
Solving,
A01 = −40 mA;
A02 = 2.5 mA.
io = 37.5 − 40e−40t + 2.5e−160t mA,
t ≥ 0.
Problems
P 8.28
[a] α =
1
106
=
= 100;
2RC
(1600)(6.25)
8–25
α2 = 104 ;
1
106
=
= 6400;
LC
(25)(6.25)
√
s1,2 = −100 ± 104 − 6400 = −100 ± 60;
ω02 =
s1 = −40 rad/s;
s2 = −160 rad/s;
vo (∞) = 0 = Vf ;
.·. vo = A01 e−40t + A02 e−160t .
vo (0) = 30 = A01 + A02 ;
iC (0+ ) = 0;
Note:
dvo
(0) = 0 = −40A01 − 160A02 .
.·.
dt
A01 = 40 V,
Solving,
A02 = −10 V.
vo (t) = 40e−40t − 10e−160t V,
[b] io (t) =
t > 0+ .
1Zt
(40e−40x − 10e−160x ) dx + I0
25 0
40e−40t 10e−160x
−
−40
−160
1
=
25
!
t
0
= (−0.04e−40x + 0.0025e−160x )
t
0
= −40e−40t + 2.5e−160t + 37.5 mA,
t ≥ 0.
!
P 8.29
P 8.30
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,
t ≥ 0+ .
diL
[a] v = L
dt
!
= 40e−32,000t sin 24,000t V,
[b] iC (t) = I − iR − iL = 24 × 10−3 −
t ≥ 0.
v
− iL
625
= [24e−32,000t cos 24,000t − 32e−32,000t sin 24,000t] mA,
P 8.31
diL
v=L
dt
!
= 960,000te−40,000t kV,
t ≥ 0.
t ≥ 0+ .
8–26
P 8.32
CHAPTER 8. Natural and Step Responses of RLC Circuits
1
vC (0+ ) = (240) = 120 V;
2
iL (0+ ) = 60 mA;
α=
iL (∞) =
240
= 48 mA;
5000
1
1
=
= 40;
2RC
2(5000k5000)(5 × 10−6 )
ω02 =
1
1
=
= 2500;
LC
80(5 × 10−6 )
α2 = 1600;
α2 < ω02 ;
.·.
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
=
0.048 + B10 e−40t cos 30t + B20 e−40t sin 30t.
iL (0) = 0.048 + B10 ;
B10 = 0.060 − 0.048 = 12 mA;
120
diL
(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,
P 8.33
t<0:
t > 0:
iL (0− ) = I0 = 3/150 = 20 mA;
300k150 = 100 Ω;
iL (0) = 20 mA,
120 − 20 = 100 mA.
iL (∞) = −100 mA;
t ≥ 0.
vC (0− ) = V0 = 0.
Problems
ω02 =
α=
1
1
=
= 64 × 106 ;
−9
LC
(0.03125)(500 × 10 )
1
1
=
= 104 ;
2RC
2(100)(500 × 10−9 )
ω0 = 8000 rad/s;
α2 = 100 × 106 ;
α2 − ω02 = (100 − 64)106 = 36 × 106 ;
s1,2 = −10,000 ± 6000;
s1 = −4000 rad/s;
s2 = −16,000 rad/s;
iL = If + A01 e−4000t + A02 e−16,000t ;
iL (∞) = If = −100 mA;
iL (0) = A01 + A02 + If = 20 mA;
.·. A01 + A02 − 100 = 20 so A01 + A02 = 120 mA;
diL
(0) = 0 = −4000A01 − 16,000A02 ;
dt
A01 = 160 mA,
Solving,
A02 = −40 mA.
iL = −100 + 160e−4000t − 40e−16,000t mA,
s
P 8.34
ω0 =
1
=
LC
s
t ≥ 0.
1
= 1250 rad/s;
(640 × 10−3 )(10−6 )
1
1
=
= 1000 rad/s
2RC
2(500)(10−6 )
√
ωd = 12502 − 10002 = 750;
α=
.·. underdamped.
If = −1 A;
iL = −1 + B10 e−1000t cos 750t + B20 e−1000t sin 750t;
iL (0) = −1 + B10 = 0.5
so
B10 = 1.5;
diL
V0
(0) = −αB10 + ωd B20 = ;
dt
L
.·.
−1000(1.5) + 750B20 =
40
,
640 × 10−3
so
iL (t) = −1 + e−1000t [1.5 cos 750t + (25/12) sin 750t] A,
B20 = 25/12;
t ≥ 0.
8–27
8–28
P 8.35
CHAPTER 8. Natural and Step Responses of RLC Circuits
α=
1
1
=
= 1000 rad/s;
2RC
2(500)(10−6 )
s
ω0 =
1
=
LC
s
1
= 800 rad/s;
(1.5625)(10−6 )
s1,2 = −1000 ±
Overdamped:
√
10002 − 8002 = −400, −1600 rad/s;
If = −1 A;
iL = −1 + A01 e−400t + A02 e−1600t;
iL (0) = −1 + A01 + A02 = 0.5
so
A01 + A02 = 1.5;
40
V0
diL
(0) = −400A01 − 1600A02 =
=
= 25.6;
dt
L
1.5625
A01 = 2.02133,
Solving,
A02 = −0.52133;
iL (t) = −1 + 2.02133e−400t − 0.52133e−1600t A,
P 8.36
α=
1
1
=
= 1000;
2RC
2(500)(10−6 )
s
ω0 =
t ≥ 0.
1
=
LC
α2 = ω02
s
1
= 1000 rad/s;
(1)(10−6 )
Critically damped.
If = −1 A;
iL = −1 + D10 te−1000t + D20 e−1000t ;
iL (0) = −1 + D20 = 0.5;
.·. D20 = 1.5 A;
V0
diL
(0) = D10 − αD20 = ;
dt
L
.·.
D10 − 1000(1.5) =
40
1
so
iL = −1 + 1540te−1000t + 1.5e−1000t A,
D10 = 1540;
t ≥ 0.
Problems
P 8.37
t < 0:
vo (0− ) = vo (0+ ) =
1000
(25) = 20 V;
1250
iL (0− ) = iL (0+ ) = 0.
t>0
−0.1 +
20
+ iC (0+ ) + 0 = 0;
200
.·. iC (0+ ) = 0;
106
1
=
= 250 rad/s;
2RC
(400)(10)
ω02 =
1
106
=
= 62,500;
LC
10(1.6)
.·. α2 = ω02
critically damped.
[a] iL = If + D10 te−250t + D20 e−250t ;
+
iL (0 ) = 0;
If = 100 mA;
.·. 0 = 0.1 + D20 ;
20
diL (0+ )
=
= 12.5 A/s;
dt
1.6
D20 = −100 mA;
−250D20 + D10 = 12.5;
D10 = −12.5 A/s;
.·. iL = 100 − 12,500te−250t − 100e−250t mA
[b] vo = L
dio
= 1.6(−12.5e−250t + 3125te−250t + 25e−250t )
dt
= 5000te−250t + 20e−250t V,
P 8.38
[a] wL =
t ≥ 0.
Z ∞
0
pdt =
Z ∞
0
t ≥ 0+ .
vo iL dt;
vo = 5000te−250t + 20e−250t V;
iL = 0.1 − 12.5te−250t − 0.1e−250t A;
p = 2e−250t + 500te−250t − 750te−500t − 62,500t2 e−500t − 2e−500t W.
8–29
8–30
CHAPTER 8. Natural and Step Responses of RLC Circuits
Z ∞
Z ∞
Z ∞
wL
−250t
−250t
=
e
dt + 250
te
dt − 375
te−500t −
2
0
0
0
Z ∞
31,250
2 −500t
te
dt −
Z ∞
0
e−500t dt
0
∞
=
∞
e−250t
250 −250t
e
(−250t − 1) −
+
−250 0
(250)2
0
∞
375 −500t
e
(−500t − 1) −
(500)2
0
∞
31,250 −500t
e
(5002 t2 + 1000t + 2) −
(−500)3
0
∞
e−500t
.
(−500) 0
All the upper limits evaluate to zero hence
1
250
375
(31,250)(2)
1
wL
=
+
−
−
−
;
4
3
6
2
250 62,500 25 × 10
(5) 10
500
wL = 8 + 8 − 3 − 1 − 4 = 8 mJ.
Note this value corresponds to the final energy stored in the inductor, i.e.
1
wL (∞) = (1.6)(0.1)2 = 8 mJ.
2
−250t
[b] v = 5000te
+ 20e−250t V;
iR =
v
= 25te−250t + 0.1e−250t A;
200
pR = viR = 2e−500t [62,500t2 + 500t + 1];
wR =
Z ∞
0
pR dt;
Z ∞
Z ∞
Z ∞
wR
e−500t dt
= 62,500
t2 e−500t dt + 500
te−500t dt +
2
0
0
0
=
∞
62,500e−500t
4 2
[25
×
10
t
+
1000t
+
2]
+
−125 × 106
0
∞
e−500t ∞
500e−500t
(−500t
−
1)
+
.
25 × 104
(−500) 0
0
Since all the upper limits evaluate to zero we have
wR
62,500(2)
500
1
=
+
+
.
2
125 × 106 25 × 104 500
wR = 2 + 4 + 4 = 10 mJ.
Problems
[c] 100 = iR + iC + iL
8–31
(mA);
iR + iL = 25,000te−250t + 100e−250t + 100 − 12,500te−250t − 100e−250t mA
= 100 + 12,500te−250t mA;
.·. iC = 100 − (iR + iL ) = −12,500te−250t mA = −12.5te−250t A;
pC = viC = [5000te−250t + 20e−250t ][−12.5te−250t ]
= −250[250t2 e−500t + te−500t ].
Z ∞
Z ∞
wC
2 −500t
= 250
te
dt +
te−500t dt;
−250
0
0
∞
∞
250e−500t
e−500t
wC
4 2
=
[25
×
10
t
+
1000t
+
2]
(−500t
−
1)
+
.
−250
−125 × 106
25 × 104
0
0
Since all upper limits evaluate to zero we have
250(1)
−250(250)(2)
−
= −1000 × 10−6 − 10 × 10−4 = −2 mJ.
125 × 106
25 × 104
Note this 2 mJ corresponds to the initial energy stored in the capacitor,
i.e.,
1
wC (0) = (10 × 10−6 )(20)2 = 2 mJ.
2
Thus wC (∞) = 0 mJ which agrees with the final value of v = 0.
wC =
[d] The energy delivered by the voltage source is the same as the energy
delivered by the equivalent current source is = 100 mA.
ps (del) = 0.1vo W
= 0.1[5000te−250t + 20e−250t ]
= 2e−250t + 500te−250t W;
Z ∞
Z ∞
ws
−250t
=
e
dt +
250te−250t dt
2
0
0
=
∞
e−250t ∞ 250e−250t
+
(−250t − 1)
−250 0
62,500
0
=
1
1
+
;
250 250
2(2)
4
=
= 16 mJ.
250
250
[e] wL = 8 mJ (absorbed);
wR = 10 mJ (absorbed);
ws =
wC = 2 mJ (delivered);
X
wdel = wabs = 18 mJ.
ws = 16 mJ (delivered);
8–32
P 8.39
CHAPTER 8. Natural and Step Responses of RLC Circuits
α = 800 rad/s;
ωd = 600 rad/s;
ωo2 − α2 = 36 × 104 ;
ωo2 = 100 × 104 ;
R
= 800;
2L
R = 1600L;
1
= 100 × 104 ;
LC
L=
α=
wo = 1000 rad/s;
106
= 2 mH;
(100 × 104 )(500)
.·. R = 3.2 Ω;
i(0+ ) = B1 = 0 A;
at t = 0+ .
12 + 0 + vL (0+ ) = 0;
vL (0+ ) = −12 V;
di(0+ )
−12
=
= −6000 A/s;
dt
0.002
di(0+ )
.·.
= 600B2 − 800B1 = −6000;
dt
.·. 600B2 = 800B1 − 6000;
.·. i = −10e−800t sin 600t A,
P 8.40
.·. B2 = −10 A;
t ≥ 0.
From Prob. 8.39 we know vc will be of the form
vc = B3 e−800t cos 600t + B4 e−800t sin 600t.
From Prob. 8.39 we have
vc (0) = 12 V = B3 ,
and
iC (0)
dvc (0)
=
= 0;
dt
C
dvc (0)
= 600B4 − 800B3 ;
dt
.·. 600B4 = 800B3 + 0;
B4 = 16 V;
vc (t) = 12e−800t cos 600t + 16e−800t sin 600t V
t ≥ 0.
Problems
P 8.41
q
α2 − ω02 = −4000;
[a] −α +
.·. α = 10,000 rad/s,
α=
R
= 10,000;
2L
ω02 =
−α −
q
α2 − ω02 = −16,000;
ω02 = 64 × 106 ;
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
.·. vc2 (0) = 576;
1
(31.25) × 10−9 vc2 (0) = 9 × 10−6;
2
vc (0) = 24 V;
24
di(0)
=
= 48 A/s.
dt
0.5
[c] i(t) = A1 e−4000t + A2 e−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
[e] imax = 4e−0.4621 − 4e−1.8484 = 1.89 mA.
di
[f ] vL (t) = 0.5 = [−8e−4000t + 32e−16,000t ] V,
dt
t ≥ 0+ .
8–33
8–34
P 8.42
CHAPTER 8. Natural and Step Responses of RLC Circuits
[a] ω02 =
α=
1
109
108
=
=
= 25 × 106 ;
LC
(0.25)(160)
4
R
= ω0 = 5000 rad/s;
2L
.·. R = (5000)(2)L = 2500 Ω.
[b] i(0) = iL (0) = 24 mA;
vL (0) = 90 − (0.024)(2500) = 30 V;
30
di
(0) =
= 120 A/s.
dt
0.25
[c] vC = D1 te−5000t + D2 e−5000t ;
vC (0) = D2 = 90 V;
dvC
iC (0)
−iL (0)
(0) = D1 − 5000D2 =
=
;
dt
C
C
D1 − 450,000 = −
24 × 10−3
= −150,000;
160 × 10−9
.·. D1 = 300,000 V/s;
vC = 300,000te−5000t + 90e−5000t V,
P 8.43
t ≥ 0+ .
[a] For t > 0:
Since i(0− ) = i(0+ ) = 0,
va (0+ ) = 300 V.
4
[b] va = 200i + 5 × 10
Z t
i dx + 300;
0
dva
di
= 200 + 5 × 104 i
dt
dt
dva (0+ )
di(0+ )
di(0+ )
= 200
+ 5 × 104 i(0+ ) = 200
;
dt
dt
dt
−L
di(0+ )
= 300;
dt
Problems
di(0+ )
= −0.2(300) = −60 A/s;
dt
dva (0+ )
.·.
= −12,000 V/s.
dt
R
800
[c] α =
=
= 80 rad/s;
2L
10
1
106
=
= 104 ;
LC
(5)(20)
√
s1,2 = −80 ± 6400 − 104 = −80 ± j60 rad/s.
ω02 =
Underdamped:
va = B1 e−80t cos 60t + B2 e−80t sin 60t;
va (0) = B1 = 300 V;
dva (0)
= −80B1 + 60B2 = −12,000;
dt
.·. B2 = 200 V;
va = 300e−80t cos 60t + 200e−80t sin 60t V,
P 8.44
iL (0− ) = iL (0+ ) =
t ≥ 0+ .
70
= 280 mA;
50 + 200
vc (0− ) = vc (0+ ) = 200(0.280) = 56 V;
ω02 =
α=
1
1
=
= 50 × 106 ;
LC
(0.100)(200 × 10−9 )
R
200
=
= 1000;
2L
2(0.100)
α2 < ω02
.·.
α2 = 106 ;
underdamped.
s1,2 = −1000 ± j7000 rad/s;
i = B1 e−1000t cos 7000t + B2 e−1000t sin 7000t;
i(0) = B1 = 280 mA;
di
1
(0) = 7000B2 − 1000B1 =
[−200(0.28) − (−56)] = 0;
dt
0.1
1
.·. B2 = B1 = 40 mA;
7
i = 280e−1000t cos 7000t + 40e−1000t sin 7000t mA,
t ≥ 0+ .
8–35
8–36
P 8.45
CHAPTER 8. Natural and Step Responses of RLC Circuits
[a] t < 0:
120
= 15 mA;
vo = (5000)(0.015) = 75 V.
8000
t > 0:
R
5000
α=
=
= 2500 rad/s;
2L
2(1)
io =
ω02 =
109
1
=
= 4 × 106 = 400 × 104 ;
LC
(1)(250)
α2 − ω02 = 625 × 104 − 400 × 104 = 225 × 104 ;
.·. s1,2 = −2500 ± 1500.
s1 = −1000 rad/s
s2 = −4000 rad/s;
.·. io (t) = A1 e−1000t + A2 e−4000t .
io (0) = A1 + A2 = 15 × 10−3 ;
dio
1
(0) = −1000A1 − 4000A2 = [−5000(0.015) − (−75)] = 0;
dt
1
Solving,
A2 = −5 mA.
A1 = 20 mA;
io (t) = 20e−1000t − 5e−4000t mA,
t ≥ 0+ .
[b] vo (t) = A1 e−1000t + A2 e−4000t ;
vo (0) = A1 + A2 = 75;
−15 × 10−3
dvo
(0) = −1000A1 − 4000A2 =
;
dt
250 × 10−9
Solving,
A1 = 80 V;
A2 = −5 V.
vo (t) = 80e−1000t − 5e−4000t V,
t ≥ 0+ .
Check:
5000io + 1
dio
= vo ;
dt
5000io = 100e−1000t − 25e−4000t ;
dio
= −20e−1000t + 20e−4000t ;
dt
dio
.·. 5000io +
= 80e−1000t − 5e−4000t V
dt
(checks).
Problems
P 8.46
t < 0:
i(0) =
100
100
=
= 5 A;
4 + [10k(10 + 30)] + 8
20
10
= 70 V.
vo (0) = 100 − 5(4) − 10(5)
50
t > 0:
α=
20
R
=
= 5,
2L
4
ω02 =
α2 = 25;
1
100
=
= 50;
LC
2
ω02 > α2
underdamped.
vo = B1 e−αt cos ωd t + B2 e−αt sin ωd t;
ωd =
√
50 − 25 = 5;
vo = B1 e−5t cos 5t + B2 e−5t sin 5t;
vo (0) = B1 = 70 V;
C
dvo
(0) = −5,
dt
dvo
−5
=
× 103 = −500 V/s;
dt
10
dvo
(0) = −5B1 + 5B2 = −500;
dt
5B2 = −500 + 5B1 = −500 + 350;
.·. vo = 70e−5t cos 5t − 30e−5t sin 5t V,
B2 = −150/5 = −30 V;
t ≥ 0.
8–37
8–38
P 8.47
CHAPTER 8. Natural and Step Responses of RLC Circuits
[a]
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:
α = ω0 = 5000
so
R
= 5000;
2L
.·. R = 2(0.01)(5000) = 100 Ω.
Appendix H has a 100 Ω resistor. The final circuit:
q
[b] s1,2 = −α ± α2 − ω02 = −5000 ± 0.
Therefore there are two repeated real roots at −5000 rad/s.
P 8.48
[a] Underdamped response:
α < ω0
so
α < 5000.
Therefore we choose a smaller resistor value than the one used in Problem
8.47 to give a smaller value of α. For convenience, pick α = 4000 rad/s:
R
= 4000
so
R = 2(4000)(0.01) = 80 Ω.
2L
We can create an 80 Ω resistance by combining a 33 Ω resistor and a 47 Ω
resistor in series.
√
s1,2 = −4000 ± 40002 − 50002 = −4000 ± j3000 rad/s.
α=
[b] Overdamped response:
α > ω0
so
α > 5000.
Therefore we choose a larger resistor value than the one used in Problem
8.47. Choose R = 125 Ω, which can be created by combining a 100 Ω
resistor, a 15 Ω resistor and a 10 Ω resistor in series:
R
= 6250;
2L
√
s1,2 = −6250 ± 62502 − 50002 = −6250 ± 3750
α=
= −2500 rad/s;
and
− 10,000 rad/s.
Problems
P 8.49
8–39
i(0+ ) = 0, since there is no source connected to L for t < 0.
!
15,000
(80) = 50 V;
15,000 + 9000
−
+
vC (0 ) = vC (0 ) =
s
R
α=
= 8000;
2L
q
ωd =
ω0 =
1
= 10,000;
LC
Vf = 100 V.
the response is underdamped.
10,0002 − 80002 = 6000;
vC = Vf + B10 e−8000t cos 6000t + B20 e−8000t sin 6000t;
Vf + B10 = V0
so
B10 = V0 − Vf = 50 − 100 = −50;
−αB10 + ωd B20 =
I0
,
C
so
B20 =
1
ωd
1
I0
+ αB10 =
[0 + 8000(−50)] = −66.67;
C
6000
Therefore,
vC = 100 − e−8000t (50 cos 6000t + 66.67 sin 6000t) V,
P 8.50
i=C
t ≥ 0.
dvC
dt
= 2 × 10−6 [8000e−8000t (50 cos 6000t + 66.67 sin 6000t)
− e−8000t (6000)(−50 sin 6000t + 66.67 cos 6000t)]
= 1.67e−8000t sin 6000t A,
P 8.51
t > 0.
[a] From Problem 8.49, I0 = 0, V0 = 50 V, and Vf = 100 V. Then,
R
α=
= 10,000;
2L
s
ω0 =
1
= 10,000;
LC
the response is critically damped.
vC = Vf + D10 te−10,000t + D20 e−10,000t ;
Vf + D20 = V0
D10 − αD20 =
so
D20 = V0 − Vf = 50 − 100 = −50;
I0
,
C
so
I0
+ αD20 = 0 + 10,000(−50)] = −5 × 105 ;
C
Therefore,
D10 =
vC = 100 − 500,000te−10,000t − 50e−10,000t V,
t ≥ 0.
8–40
CHAPTER 8. Natural and Step Responses of RLC Circuits
[b] i = C
dvC
dt
= 2 × 10−6 [−500,000e−10,000t + 5 × 109 te−10,000t + 500,000e−10,000t ]
= 10,000te−10,000t A,
P 8.52
t > 0.
t < 0:
io (0− ) =
48
= 3 mA;
16,000
vC (0− ) = 20 − (12,000)(0.003) = −16 V.
t = 0+ :
12 kΩk24 kΩ = 8 kΩ;
.·. vo (0+ ) = (0.003)(8000) − 16 = 24 − 16 = 8 V,
and vL (0+ ) = 20 − 8 = 12 V.
vo (t) = 8000io + vC ;
dvo
dio dvC
(t) = 8000
+
;
dt
dt
dt
dvo +
dio
dvC +
(0 ) = 8000 (0+ ) +
(0 );
dt
dt
dt
vL (0+ ) = L
dio +
(0 );
dt
Problems
8–41
vL (0+ )
12
dio +
(0 ) =
=
= 60 A/s;
dt
L
0.2
C
dvc +
(0 ) = io (0+ );
dt
3 × 10−3
dvc +
.·.
(0 ) =
= 375,000;
dt
8 × 10−9
dvo +
.·.
(0 ) = 8000(60) + 375,000 = 855,000 V/s.
dt
1
109
=
= 625 × 106 ;
LC
1.6
ω0 = 25,000 rad/s;
R
8000
=
= 20,000 rad/s;
2L
0.4
α2 = 400 × 106 ;
ω02 =
α=
α2 < ω02
underdamped.
s1,2 = −20,000 ± j15,000 rad/s;
vo (t) = Vf + B10 e−20,000t cos 15,000t + B20 e−20,000t sin 15,000t;
Vf = vo (∞) = 20 V;
8 = 20 + B10 ;
B10 = −12 V;
−20,000B10 + 15,000B20 = 855,000.
Solving,
B20 = 41 V.
.·. vo (t) = 20 − 12e−20,000t cos 15,000t + 41e−20,000t sin 15,000t V,
P 8.53
iC (0) = 0;
α=
vo (0) = 200 V;
R
4
=
= 50 rad/s;
2L
2(0.04)
ω02 =
1
103
=
= 2500;
LC
0.4
.·. α2 = ωo2 ;
critical damping.
t ≥ 0+ .
8–42
CHAPTER 8. Natural and Step Responses of RLC Circuits
vo (t) = Vf + D10 te−50t + D20 e−50t ;
Vf = 100 V;
vo (0) = 100 + D20 = 200;
D20 = 100 V;
dvo
(0) = −50D20 + D10 = 0;
dt
D10 = 50D20 = 5000 V/s;
vo = 100 + 5000te−50t + 100e−50t V,
P 8.54
t ≥ 0.
t < 0:
iL (0) =
−160
= −100 mA;
1600
vC (0) = 1000iL (0) = −100 V.
t > 0:
α=
R
1000
=
× 103 = 5000 rad/s;
2L
200
ω02 =
1
(109 )(103 )
108
=
=
= 25 × 106 ;
LC
(100)(400)
4
ω0 = 5000 rad/s
.·.
critical damping.
vC (t) = Vf + D10 te−5000t + D20 e−5000t ;
vC (0) = −100 V;
Vf = −60 V;
Problems
.·. −100 = −60 + D20 ;
C
8–43
D20 = −40 V;
dvC
(0) = iL (0) = −100 × 10−3 ;
dt
dvC
−100 × 10−3
(0) =
= −250,000 V/s;
dt
400 × 10−9
.·. D10 = 5000(−40) − 250,000 = −450,000.
vC (t) = −60 − 450,000te−5000t − 40e−5000t V,
P 8.55
t ≥ 0.
[a] vC = V + [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 = −V
and B20 =
αB10
.
ωd
When these values are substituted into the expression for [dvC /dt], we get
dvC
=
dt
But
α2
+ ωd V e−αt sin ωd t.
ωd
!
α2 + ωd2
ω2
α2
+ ωd =
= 0.
ωd
ωd
ωd
Therefore
[b]
ω02
V e−αt sin ωd t.
ωd
!
dvC
=
dt
dvC
= 0 when
dt
sin ωd t = 0,
or ωd t = nπ,
where n = 0, 1, 2, 3, . . . .
Therefore t =
[c] When tn =
and
nπ
,
ωd
nπ
.
ωd
cos ωd tn = cos nπ = (−1)n ,
sin ωd tn = sin nπ = 0.
Therefore vc (tn ) = V [1 − (−1)n e−αnπ/ωd ].
8–44
CHAPTER 8. Natural and Step Responses of RLC Circuits
[d] It follows from [c] that
v(t1 ) = V + V e−(απ/ωd )
Therefore
But
vC (t1 ) − V
e−(απ/ωd )
= −(3απ/ω ) = e(2απ/ωd ) .
d
vC (t3 ) − V
e
2π
= t3 − t1 = Td ,
ωd
(
P 8.56
and vc (t3 ) = V + V e−(3απ/ωd ) .
vc (t1 ) − V
1
ln
α=
Td
vc (t3 ) − V
thus α =
)
;
Td = t3 − t1 =
13.505
12,000
ln
= 5000;
α=
2π
0.985
1
[vC (t1 ) − V ]
ln
.
Td
[vC (t3 ) − V ]
ωd =
3π
π
2π
−
=
ms.
12 12
12
2π
= 12,000 rad/s.
Td
ω02 = ωd2 + α2 = 144 × 106 + 25 × 106 = 169 × 106 ;
L=
P 8.57
1
= 29.6 mH;
(169)(0.2)
R = 2αL = 295.86 Ω.
[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.
L
di(0)
= 0,
dt
therefore
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;
Therefore
B20 =
α 0
α Vg
B1 =
.
ωd
ωd R
Problems
α2 Vg ωd Vg
di
+
sin ωd t e−αt
vo = L = − L
dt
ωd R
R
(
!
LVg
=−
R
)
Vg L
=−
R
α2 + ωd2 −αt
e sin ωd t
ωd
Vg L
=−
R
ω02 −αt
e sin ωd t
ωd
Vg L
=−
Rωd
!
!
1
e−αt sin ωd t;
LC
Vg −αt
e sin ωd t V,
RCωd
vo = −
)
α2
+ ωd sin ωd t e−αt
ωd
(
[b]
!
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).
1
ωd
tan−1
.
ωd
α
t=
P 8.58
[a] From Problem 8.57 we have
vo =
−Vg −αt
e sin ωd t;
RCωd
α=
R
120
=
= 12,000 rad/s;
2L
0.01
ω02 =
ωd =
1012
1
=
= 400 × 106 ;
LC
2500
q
ω02 − α2 = 16 krad/s;
−Vg
−(−600)109
=
= 625;
RCωd
(120)(500)(16) × 103
.·. vo = 625e−12,000t sin 16,000t V.
[b] From Problem 8.57
1
ωd
td =
tan−1
ωd
α
td = 57.96 µs.
1
16,000
=
tan−1
16,000
12,000
!
8–45
8–46
CHAPTER 8. Natural and Step Responses of RLC Circuits
[c] vmax = 625e−0.012(57.96) sin[(0.016)(57.96)] = 249.42 V.
[d] R = 12 Ω;
α = 1200 rad/s;
ωd = 19,963.97 rad/s;
vo = 5009.02e−1200t sin 19,963.97t V,
t ≥ 0;
td = 75.67 µs;
vmax = 4565.96 V.
P 8.59
[a]
d2 vo
1
=
vg ;
2
dt
R1 C1 R2 C2
1
1
=
= 125;
R1 C1 R2 C2
(50,000)(20,000)(2 × 10−6 )(4 × 10−6 )
d2 vo
.·.
= 125vg .
dt2
0 ≤ t ≤ 0.2− :
vg = 400 mV;
d2 vo
= 50.
dt2
Let g(t) =
dvo
,
dt
Z g(t)
Z t
dx = 50
g(0)
then
dg
= 50 or dg = 50 dt.
dt
g(0) =
dvo
(0) = 0;
dt
dy;
0
g(t) − g(0) = 50t,
g(t) =
dvo
= 50t;
dt
dvo = 50t dt;
Z vo (t)
dx = 50
Z t
vo (0)
x dx;
0
vo (t) = 25t2 V,
vo (t) − vo (0) = 25t2 ,
0 ≤ t ≤ 0.2− ;
dvo1
1
=−
vg = −10vg = −4;
dt
R1 C1
dvo1 = −4 dt;
Z vo1 (t)
vo1 (0)
dx = −4
Z t
0
dy;
vo (0) = 0;
Problems
vo1 (t) − vo1 (0) = −4t,
vo1 (0) = 0;
0 ≤ t ≤ 0.2− .
vo1 (t) = −4t V,
0.2+ ≤ t ≤ tsat :
d2 vo
= −12.5,
dt2
let g(t) =
dg(t)
= −12.5;
dt
dg(t) = −12.5 dt;
Z g(t)
Z t
g(0.2+ )
dx = −12.5
dvo
;
dt
dy;
0.2
g(t) − g(0.2+ ) = −12.5(t − 0.2) = −12.5t + 2.5;
g(0.2+ ) =
C
dvo (0.2+ )
;
dt
0 − vo1 (0.2+ )
dvo
;
(0.2+ ) =
dt
20 × 103
vo1 (0.2+ ) = vo1 (0.2− ) = −4(0.2) = −0.80 V;
dvo (0.2+ )
0.80
.·. C
= 40 µA.
=
dt
20 × 103
40 × 10−6
dvo
(0.2+ ) =
= 10 V/s;
dt
4 × 10−6
dvo
.·. g(t) = −12.5t + 2.5 + 10 = −12.5t + 12.5 =
;
dt
.·. dvo = −12.5t dt + 12.5 dt.
Z vo (t)
vo (0.2+ )
dx =
Z t
0.2+
−12.5y dy +
vo (t) − vo (0.2+ ) = −6.25y 2
Z t
0.2+
12.5 dy;
t
t
+ 12.5y
0.2
;
0.2
vo (t) = vo (0.2+ ) − 6.25t2 + 0.25 + 12.5t − 2.5;
vo (0.2+ ) = vo (0.2− ) = 1 V;
.·. vo (t) = −6.25t2 + 12.5t − 1.25 V,
dvo1
= −10(−0.1) = 1,
dt
dvo1 = dt;
Z vo1 (t)
vo1 (0.2+ )
0.2+ ≤ t ≤ tsat .
0.2+ ≤ t ≤ tsat ;
dx =
Z t
0.2+
dy;
8–47
8–48
CHAPTER 8. Natural and Step Responses of RLC Circuits
vo1 (0.2+ ) = vo1 (0.2− ) = −0.8 V
vo1 (t) − vo1 (0.2+ ) = t − 0.2;
.·. vo1 (t) = t − 1 V,
0.2+ ≤ t ≤ tsat .
Summary:
0 ≤ t ≤ 0.2− s :
vo1 = −4t V,
0.2+ s ≤ t ≤ tsat :
vo = 25t2 V;
vo1 = t − 1 V,
vo = −6.25t2 + 12.5t − 1.25 V.
[b] −10 = −6.25t2sat + 12.5tsat − 1.25;
.·. 6.25t2sat − 12.5tsat − 8.75 = 0.
t2sat − 2tsat − 1.4 = 0;
√
tsat = 1 ± 2 + 1.4 = 1 ± 1.844;
.·. tsat = 2.844 sec.
vo1 (tsat ) = 1.844 − 1 = 0.844 V.
P 8.60
τ1 = (0.25 × 106 )(2 × 10−6 ) = 0.50 s;
1
= 2;
τ1
τ2 = (0.25 × 106 )(4 × 10−6 ) = 1 s;
.·.
1
= 1;
τ2
dvo
d2 vo
+3
+ 2vo = 50.
.·.
2
dt
dt
s2 + 3s + 2 = 0;
(s + 1)(s + 2) = 0;
s1 = −1,
vo = Vf + A01 e−t + A02 e−2t ;
s2 = −2;
Vf = (250/50)(250/20)(0.4) = 25 V;
vo = 25 + A01 e−t + A02 e−2t ;
vo (0) = 0 = 25 + A01 + A02 ;
.·. A01 = −50,
A02 = 25 V.
vo (t) = 25 − 50e−t + 25e−2t V,
dvo1
+ 2vo1 = −4;
dt
dvo
(0) = 0 = −A01 − 2A02 ;
dt
0 ≤ t ≤ 0.2 s;
.·. vo1 = −2 + 2e−2t V,
0 ≤ t ≤ 0.2 s.
Problems
8–49
vo (0.2) = 25 − 50e−0.2 + 25e−0.4 = 0.8215 V;
vo1 (0.2) = −2 + 2e−0.4 = −0.6594 V.
At t = 0.2 s
iC =
C
0 + 0.6594
= 32.97 µA;
20 × 103
dvo
= 32.97 µA;
dt
dvo
32.97
=
= 8.24 V/s.
dt
4
0.2 s ≤ t < ∞:
d2 vo
dvo
+3
+ 2 = −12.5;
2
dt
dt
vo (∞) = −6.25;
.·. vo = −6.25 + A01 e−(t−0.2) + A02 e−2(t−0.2) .
0.8215 = −6.25 + A01 + A02 ;
dvo
(0.2) = 8.24 = −A01 − 2A02 ;
dt
.·. A01 + A02 = 7.07;
A01 = 22.38;
−A01 − 2A02 = 8.24.
A02 = −15.31;
.·. vo = −6.25 + 22.38e−(t−0.2) − 15.31e−2(t−0.2) V,
0.2 ≤ t < ∞.
dvo1
+ 2vo1 = 1;
dt
.·. vo1 = 0.5 + (−0.6594 − 1)e−2(t−0.2) = 0.5 − 1.66e−2(t−0.2) V,
0.2 ≤ t < ∞.
8–50
P 8.61
CHAPTER 8. Natural and Step Responses of RLC Circuits
[a]
2C
dva va − vg va
+
+
= 0.
dt
R
R
(1) Therefore
va
vg
dva
+
=
.
dt
RC
2RC
d(0 − vb )
0 − va
+C
=0
R
dt
(2) Therefore
dvb
va
+
= 0,
dt
RC
va = −RC
dvb
.
dt
dvb
d(vb − vo )
2vb
+C
+C
= 0.
R
dt
dt
(3) Therefore
dvb
vb
1 dvo
+
=
.
dt
RC
2 dt
From (2) we have
d2 vb
dva
= −RC 2
dt
dt
and va = −RC
When these are substituted into (1) we get
(4) − RC
d2 vb dvb
vg
−
=
.
2
dt
dt
2RC
Now differentiate (3) to get
(5)
1 dvb
1 d2 vo
d2 vb
+
=
.
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
R C
dvb
.
dt
Problems
8–51
d2 vo
vg
= 2 2.
2
dt
R C
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 8.62
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 8.63
Problem 8.63(a) — Example 8.14, with R1 and R2 removed:
[a] Ra = 100 kΩ;
d2 vo
1
=
2
dt
Ra C1
C1 = 0.1 µF;
Rb = 25 kΩ;
1
vg ;
Rb C2
1
= 100
Ra C1
vg = 250 × 10−3 ;
therefore
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
Ra C1
[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.
Problem 8.63(b) — Example 8.14 with vo1 (0) = 5 V and vo (0) = −10 V:
[a] Initial conditions will not change the differential equation; hence the
equation is the same as Example 8.14.
[b] vo = 5 + A01 e−10t + A02 e−20t
(from Example 8.14);
vo (0) = −10 = 5 + A01 + A02 .
8–52
CHAPTER 8. Natural and Step Responses of RLC Circuits
iC (0+ ) = 0.1 m − 0.2 m = −0.1 mA;
iC (0+ ) = C
dvo (0+ )
= −0.1 mA;
dt
−0.1 × 10−3
dvo (0+ )
=
= −100 V/s;
dt
10−6
dvo
= −10A01 e−10t − 20A02 e−20t ;
dt
dvo +
(0 ) = −10A01 − 20A02 = −100.
dt
Therefore −A01 − 2A02 = −10 and A01 + A02 = −15.
Thus, A01 = −40 and A02 = 25.
vo = 5 − 40e−10t + 25e−20t V.
[c] Same as Example 8.14:
dvo1
+ 20vo1 = −25.
dt
[d] From Example 8.14:
vo1 (∞) = −1.25 V;
v1 (0) = 5 V (given).
Therefore
vo1 = −1.25 + (5 + 1.25)e−20t = −1.25 + 6.25e−20t V.
P 8.64
[a] From Example 8.13
therefore
d2 vo
= 2;
dt2
dg(t)
= 2,
dt
g(t) − g(0) = 2t;
g(t) =
dvo
;
dt
g(t) = 2t + g(0);
g(0) =
dvo (0)
.
dt
Problems
iR =
8–53
dvo (0)
12
× 10−3 = 24 µA = iC = −C
;
500
dt
dvo (0)
−24 × 10−6
=
= −24 = g(0);
dt
1 × 10−6
dvo
= 2t − 24;
dt
dvo = 2t dt − 24 dt;
vo − vo (0) = t2 − 24t;
vo = t2 − 24t − 6,
vo (0) = −6 V;
0 ≤ t ≤ tsat .
[b] t2 − 24t − 6 = −9;
t2 − 24t + 3 = 0;
t∼
= 125.66 ms.
Check the output from the first stage to make sure it doesn’t saturate at
an earlier time. From Eq. 8.38
dvo1
= −40(25 × 10−3 ) = −1;
dt
dvo1 = −dt,
so
If saturation occurs,
vo1 − vo1 (0) = −t;
vo1 = −t + 12 = 5.
Therefore, the first stage would saturate at 7 s, which is later than the
time the second stage saturates.
s
P 8.65
[a] ω0 =
.·.
[b] I0 =
1
=
LC
f0 =
s
1
(8 × 10−9 )(5 × 10−12 )
= 5 × 109 rad/sec;
ω0
5 × 109
=
= 0.796 × 109 Hz = 0.796 GHz.
2π
2π
12
= 0.24 A;
50
1
1
w(0) = LI02 = (8 × 10−9 )(0.24)2 = 2.304 × 10−10 J = 0.23 nJ.
2
2
8–54
CHAPTER 8. Natural and Step Responses of RLC Circuits
[c] Because the inductor and capacitor are assumed to be ideal, none of the
initial energy will ever be dissipated, so for all t ≥ 0 the 0.23 nJ will be
continually exchanged between the inductor and capacitor.
P 8.66
[a] ω0 = 2πf0 = 2π(5 × 109 ) = 10π × 109 rad/s;
ω02 =
1
;
LC
.·.
C=
1
1
=
= 2.03 × 10−12 = 2.03 pF.
2
−9
9
2
Lω0
(0.5 × 10 )(10π × 10 )
V
12
sin ω0 t =
sin 10π × 109 t
9
ω0 RC
(10π × 10 )(25)(2.03 × 10−12 )
[b] vo (t) =
= 7.5 sin 10π × 109 t V,
P 8.67
[a] α =
0.024
R
=
= 1.5 × 106 rad/s;
−9
2L
2(8 × 10 )
s
ω0 =
1
=
LC
[b] ω02 > α2
[c] ωd =
t ≥ 0.
q
f0 =
s
1
(8 × 10−9 )(5 × 10−12 )
= 5 × 109 rad/s.
so the response is underdamped.
ω02 − α2 =
q
(5 × 109 )2 − (1.5 × 106 )2 ≈ 5 × 109 rad/s;
ω0
5 × 109
=
= 0.796 × 109 Hz = 0.796 GHz.
2π
2π
Therefore the addition of the 24 mΩ resistance does not change the
frequency of oscillation.
[d] Because of the added resistance, the oscillation now occurs within a
decaying exponential envelope (see Fig. 8.9). The form of the
6
exponential envelope is e−αt = e−1.5×10 t . Let’s assume that the
6
oscillation is described by the function Ke−1.5×10 t cos 5 × 109 t, where the
maximum magnitude, K, exists at t = 0. How long does it take before
the magnitude of the oscillation has decayed to 0.01K?
6
Ke−1.5×10 t = 0.01K
.·.
t=
so
6
e−1.5×10 t = 0.01;
ln 0.01
= 3.07 × 10−6 s = 3.07 µs.
−1.5 × 106
Therefore, the oscillations will persist for only 3.07 µs, due to the
presence of a small amount of resistance in the circuit. This is why an
LC oscillator is not used in the clock generator circuit.
Sinusoidal Steady State Analysis
Assessment Problems
AP 9.1 [a] I = 25/60◦ mA.
[b] 45 sin(50t − 30◦ ) = 45 cos(50t − 120◦ );
.·.
V = 45/−120◦ V.
[c] V = 10/53.13◦ + 4.47/−116.565◦
= 5.66/45◦ V.
[d] sin(10πt − 45◦ ) = cos(10πt − 135◦ ).
Thus,
I = −150/−135◦ + 150/45◦ = 300/45◦ A.
AP 9.2 [a] i = 400 cos(ωt + 38◦ ) mA.
[b] V = 50/−50◦ − 80/60◦ = 107.87/−94.18◦ .
Therefore v = 107.87 cos(ωt − 94.18◦ ) V.
[c] V = 80 − j40 + 25/−75◦ = 107.67/−36.57◦ .
v = 107.67 cos(ωt − 36.57◦ ) V.
AP 9.3 [a] 400 Hz.
[b] θv = 0◦ ;
100
100/0◦
/−90◦ ;
=
I=
jωL
ωL
θi = −90◦ .
100
= 20;
ωL = 5 Ω.
ωL
5
[d] L =
= 1.99 mH.
800π
[c]
9–1
9–2
CHAPTER 9. Sinusoidal Steady State Analysis
[e] ZL = jωL = j5 Ω.
AP 9.4 [a] ω = 2πf = 314,159.27 rad/s.
10 × 10−3 /0◦
V
=
[b] I =
= jωC(10 × 10−3 )/0◦ = 10 × 10−3 ωC /90◦ ;
ZC
1/jωC
.·. θi = 90◦ .
[c] 628.32 × 10−6 = 10 × 10−3 ωC;
1
10 × 10−3
=
= 15.92 Ω,
ωC
628.32 × 10−6
[d] C =
.·. XC = −15.92 Ω.
1
1
=
;
15.92(ω)
(15.92)(100π × 103 )
C = 0.2 µF.
−1
[e] Zc = j
ωC
= −j15.92 Ω.
AP 9.5 Convert the three currents to phasors and add the phasors:
I4 = −(80/30◦ − 100/−225◦ + 50/−90◦ ) = −161.59/−29.965◦ = 161.59/150.035◦ .
i4 = 161.59 cos(ωt + 150.035◦ ) A.
AP 9.6 [a] V =
(40/−90◦ )(100/45◦ )
;
|Z|/θZ
But −45 − θZ = −90◦
.·. θZ = 45◦ ;
Z = 90 − j40 + jXL .
To make the phase angle of Z equal 45◦ , the real part of Z must equal
the imaginary part of Z, or −40 + XL = 90.
.·. XL = 130 Ω;
XL = jωL = 130;
130
.·. L =
= 26 mH.
5000
[b] V =
−j40(100/45◦ )
= 31.43/−90◦ V;
(90 + j130 − j40)
.·. |V| = 31.43 V.
Problems
AP 9.7 [a]
ω = 8000 rad/s;
ωL = 64 Ω,
−1
= −100 Ω;
ωC
Zxy = 100k − j100 + 30 + j64 = 50 − j50 + 30 + j64 = (80 + j14) Ω.
[b] ω = 4000 rad/s so ωL = 32 Ω,
−1
= −200 Ω;
ωC
Zxy = 100k − j200 + 30 + j32 = 80 − j40 + 30 + j32 = (110 − j8) Ω.
[c] Zxy = 100k(−j/ωC) + 30 + ωL =
−100(j/ωC)
+ 30 + jωL
100 − (j/ωC)
100 − j1002 ωC
+ 30 + jωL.
=
(100ωC)2 + 1
The impedance will be purely resistive when the j terms cancel, i.e.,
(1002 C/L) − 1
1002 ωC
2
=
ωL
so
ω
=
= 36 × 106 .
(100ωC)2 + 1
1002 C 2
Solving for ω yields ω = 6000 rad/s.
[d] Zxy = 100k − j133.33 + 30 + j48 = 64 − j48 + 30 + j48 = 94 Ω.
AP 9.8 The frequency 6000 rad/s was found to give Zxy = 94 Ω in Assessment
Problem 9.7. Thus,
V = 470/0◦ ,
Is =
V
470/0◦
=
= 5/0◦ A.
Zxy
94
Using current division,
IC =
100k − j133.33
(5) = 1.8 + j2.4 = 3/53.13◦ A;
−j133.33
iC = 3 cos(6000t + 53.13◦ ) A,
Im = 3 A.
9–3
9–4
CHAPTER 9. Sinusoidal Steady State Analysis
AP 9.9
Convert the Y connection consisting of the 60 Ω, 75 Ω and j30 Ω impedances
into the equivalent Delta connection.
Za =
60(75) + 75(j30) + 60(j30)
= 75 + j67.5 Ω,
60
Zb =
60(75) + 75(j30) + 60(j30)
= 60 + j54 Ω,
75
Zc =
60(75) + 75(j30) + 60(j30)
= 135 − j150 Ω.
j30
Now make series and parallel combination of impedances to find the
equivalent impedance seen by the voltage source.
Zeq = 25 + [(15kZc − j45kZb )kZa
"
= 25 +
!
#
15(135 − j150) −j45(60 + j54)
+
k(75 + j67.5)
150 − j150
60 + j9
= 25 + [14.25 − j0.75 + 33.007 − j49.95]k(75 + j67.5)
= 25 + 55.25 − j12.6 = 80.25 − j12.6 Ω.
Therefore,
I=
−j50
= 0.62/−81.07◦ A.
80.25 − j12.6
AP 9.10 Calculate impedances:
ZL = j(500)(.12) = j60 Ω;
ZC =
−j
= −j100 Ω.
(500)(20 × 10−6 )
Step 1 to Step 2: source transform to get two voltage sources,
(4)(50) = 200 V;
(−j2)(−j100) = −200 V.
Problems
Now use Ohm’s law to calculate the phasor current:
I=
400
200 − (−200
=
= 4 + j2 = 4.47/26.57◦ A.
30 + 50 + j60 − j100
80 − j40
Therefore,
i(t) = 4.47 cos(500t + 26.57◦ ) A
AP 9.11 Short circuit current
Isc =
5Vx
−40 + j40 + 5Vx
+
;
−j10
10
Vx = −40 + j40 + 5Vx
.·.
Isc =
so
Vx = 10 − j10;
50 − j50 −40 + j40 + 50 − j50
+
= 6 + j4 A.
−j10
10
9–5
9–6
CHAPTER 9. Sinusoidal Steady State Analysis
Open circuit voltage
I=
−40 + j40
= −4 A;
10 − j10
Vx = 10I = −40 V;
Voc = 5Vx − j10I = −200 + j40 V;
ZrmN =
−200 + j40
= −20 + j20 Ω.
6 + j4
AP 9.12 Calculate impedances:
jωL = j(2500)(1.6 × 10−3 ) = j4 Ω;
1
−j
=
= −j4 Ω;
jωCtop
(2500)(100 × 10−6 )
1
−j
=
= −j8 Ω;
jωCmid
(2500)(50 × 10−6 )
Ig = 5/0◦ A;
−5 +
Vg = 20/90◦ V.
V1 − V2 V1 − j20
+
= 0;
−j8
−j4
Problems
V2 − V1 V2 V2 − j20
+
+
= 0.
−j8
j4
12
Solving,
V2 = −8 + j4 V;
Io =
V2
= 1 + j2 = 2.24/63.43◦ A;
j4
io = 2.24 cos(2500t + 63.43◦ ) A.
AP 9.13 Calculate impedances:
jωL = j5000(60 × 10−3 ) = j300 Ω;
1
−j
=
= −j100 Ω.
jωC
(5000)(2 × 10−6 )
−400/0◦ + (50 + j300)Ia − 50Ib − 150(Ia − Ib ) = 0;
(150 − j100)Ib − 50Ia + 150(Ia − Ib ) = 0.
Solving,
Ia = −0.8 − 1.6 A;
Ib = −1.6 + j0.8 A;
Vo = 100Ib = −160 + j80 = 178.89/153.43◦ ;
vo = 178.89 cos(5000t + 153.43◦ ) V.
9–7
9–8
CHAPTER 9. Sinusoidal Steady State Analysis
AP 9.14 [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.
2
M
= √ = 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;
1
1
1
w = L1 i21 + L2 i22 + M i1 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
Problems
AP 9.15
Mesh current equations:
−200 + (10 + j5)I1 + V1 = 0;
(4000 − j8000)I2 − V2 = 0.
Ideal transformer equations:
V1
V2
=− ;
1
20
I1 = −20I2 .
Place in standard form and solve using a calculator:
V2 = 3577.71/153.43◦ V
I2 = 0.4/−143.13◦ A
9–9
9–10
CHAPTER 9. Sinusoidal Steady State Analysis
Problems
P 9.1
[a] Right as φ becomes more negative.
[b] Left.
P 9.2
[a]
T
1250 250
=
+
= 250 µs;
2
6
6
f=
T = 500 µs;
106
1
=
= 2000Hz.
T
500
[b] v = Vm sin(ωt + θ; )
ω = 2πf = 4000π rad/s;
−250
4000π
× 10−6 + θ = 0;
6
π
.·. θ = rad = 30◦ ;
6
v = Vm sin[4000πt + 30◦ ];
75 = Vm sin 30◦ ;
Vm = 150 V;
v = 150 sin[4000πt + 30◦ ] = 150 cos[4000πt − 60◦ ] V.
P 9.3
[a] By hypothesis
i = 10 cos(ωt + θ);
di
= −10ω sin(ωt + θ);
dt
.·. 10ω = 20,000π;
ω = 2000π rad/s.
Problems
[b] f =
ω
= 1000 Hz;
2π
T =
1
= 1 ms = 1000 µs;
f
150
3
3
.·. θ = −90 − (360) = −144◦ ;
= ,
1000
20
20
.·. i = 10 cos(2000πt − 144◦ ) A.
√
P 9.4
Vm =
P 9.5
[a] 170 V.
2Vrms =
√
2(230) = 325.27 V.
[b] 2πf = 120π;
f = 60Hz.
[c] ω = 120π = 376.99 rad/s.
−π
−π
(60) =
= −1.05 rad.
[d] θ(rad) =
180
3
[e] θ = −60◦ .
1
1
= 16.67 ms.
[f ] T = =
f
60
π
1
[g] 120πt − = 0;
.·. t =
= 2.78 ms.
3
360
0.125
π
[h] v = 170 cos 120π t +
−
18
3
= 170 cos[120πt + (15π/18) − (π/3)]
= 170 cos[120πt + (π/2)]
= −170 sin 120πt V.
[i] 120π(t − to ) − (π/3) = 120πt − (π/2);
π
.·. 120πto = ;
6
P 9.6
to =
[a] ω = 2πf = 240π rad/s,
25
ms.
18
f=
ω
= 120 Hz.
2π
[b] T = 1/f = 8.33 ms.
[c] Vm = 100 V.
[d] v(0) = 100 cos(45◦ ) = 70.71 V.
45◦ (2π)
π
[e] φ = 45◦ ;
φ=
=
= 0.7854 rad.
360◦
4
[f ] V = 0 when 240πt + 45◦ = 90◦ . Now resolve the units:
(240π rad/s)t =
45◦
π
= rad,
◦
57.3 /rad
4
t = 1.042 ms.
9–11
9–12
CHAPTER 9. Sinusoidal Steady State Analysis
[g] (dv/dt) = (−100)240π sin(240πt + 45◦ );
(dv/dt) = 0 when 240πt + 45◦ = 180◦
135◦
3π
240πt =
=
rad.
◦
57.3 /rad
4
or
Therefore t = 3.125 ms.
s
P 9.7
Vrms =
Z T /2
0
1 Z T /2 2 2 2π
t dt;
Vm sin
T 0
T
Vm2 sin2
2π
V 2 Z T /2
V 2T
4π
t dt = m
1 − cos t dt = m ;
T
2 0
T
4
s
Therefore Vrms =
P 9.8
Z to +T
to
1 Vm2 T
Vm
=
.
T 4
2
Vm2 cos2 (ωt + φ) dt = Vm2
=
Z to +T
Vm2
t
1 1
+ cos(2ωt + 2φ) dt
2 2
(o
R to +T
)
Z to +T
dt +
cos(2ωt + 2φ) dt
to
2
to
i
1 h
Vm2
to +T
sin(2ωt + 2φ) |to
=
T+
2
2ω
V2
1
= m T+
[sin(2ωto + 4π + 2φ) − sin(2ωto + 2φ)]
2
2ω
1
2 T
2 T
+
(0) = Vm
.
= Vm
2
2ω
2
P 9.9
[a] The numerical values of the terms in Eq. 9.7 are
Vm = 100,
R/L = 533.33,
√
R2 + ω 2 L2 = 50;
ωL = 30;
φ = 60◦ ,
θ = 36.87◦ ;
θ = tan−1 30/40,
h
i
i = −1.84e−533.33t + 2 cos(400t + 23.13◦ ) A,
t ≥ 0.
[b] Transient component = −1.84e−533.33t A;
Steady-state component = 2 cos(400t + 23.13◦ ) A.
[c] By direct substitution into the equation from part (a),
i(1.875 ms) = 133.61 mA.
[d] 2 A,
400 rad/s,
23.13◦ .
[e] The current lags the voltage by 36.87◦ .
Problems
P 9.10
[a] From Eq. 9.7 we have
L
di
Vm R cos(φ − θ) −(R/L)t ωLVm sin(ωt + φ − θ)
√
e
−
;
= √ 2
dt
R + ω 2 L2
R 2 + ω 2 L2
Ri =
−Vm R cos(φ − θ)e−(R/L)t Vm R cos(ωt + φ − θ)
√
√
+
;
R 2 + ω 2 L2
R 2 + ω 2 L2
#
"
di
R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ)
√
L + Ri = Vm
.
dt
R 2 + ω 2 L2
But
√
R
= cos θ
R 2 + ω 2 L2
and
√
ωL
= sin θ.
R 2 + ω 2 L2
Therefore the right-hand side reduces to
Vm cos(ωt + φ).
At t = 0, Eq. 9.7 reduces to
i(0) =
[b] iss = √
−Vm cos(φ − θ) Vm cos(φ − θ)
√
+ √ 2
= 0.
R 2 + ω 2 L2
R + ω 2 L2
Vm
R 2 + ω 2 L2
cos(ωt + φ − θ).
Therefore
diss
−ωLVm
L
sin(ωt + φ − θ)
=√ 2
dt
R + ω 2 L2
and
Riss = √
Vm R
R 2 + ω 2 L2
cos(ωt + φ − θ).
"
diss
R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ)
√
L
+ Riss = Vm
dt
R 2 + ω 2 L2
= Vm cos(ωt + φ).
P 9.11
[a] Y = 100/45◦ + 500/ − 60◦ = 483.86/ − 48.48◦ ;
y = 483.86 cos(300t − 48.48◦ ).
[b] Y = 250/30◦ − 150/50◦ = 120.51/4.8◦ ;
y = 120.51 cos(377t + 4.8◦ ).
[c] Y = 60/60◦ − 120/ − 215◦ + 100/90◦ = 152.88/32.94◦ ;
y = 152.88 cos(100t + 32.94◦ ).
#
9–13
9–14
CHAPTER 9. Sinusoidal Steady State Analysis
[d] Y = 100/40◦ + 100/160◦ + 100/ − 80◦ = 0;
y = 0.
P 9.12
[a] ωL = (104 )(20 × 10−3 ) = 200 Ω.
[b] ZL = jωL = j200 Ω.
[c] VL = IZL = (10/30◦ )(200/90◦ ) = 2000/120◦ V.
[d] vL = 2 cos(10,000t + 120◦ ) kV.
P 9.13
[a] XC =
−1
−1
=
= −50 Ω.
ωC
4000(5 × 10−6 )
[b] ZC = jXC = −j50 Ω.
30/25◦
V
=
= 0.6/115◦ A.
[c] I =
ZC
50/−90◦
[d] i = 0.6 cos(4000t + 115◦ ) A.
P 9.14
[a] Vg = 150/20◦ ;
Ig = 30/ − 52◦ ;
Vg
= 5/72◦ Ω.
.·. Z =
Ig
[b] ig lags vg by 72◦ :
2πf = 8000π;
f = 4000 Hz;
T = 1/f = 250 µs;
72
.·. ig lags vg by
(250) = 50 µs.
360
P 9.15
[a] Z1 = R1 + jωL1 ;
Z2 =
R2 (jωL2 )
ω 2 L22 R2 + jωL2 R22
=
;
R2 + jωL2
R22 + ω 2 L22
Z1 = Z2
[b] R1 =
when R1 =
ω 2 L22 R2
R22 + ω 2 L22
(4 × 108 )(6.25)(5 × 104 )
= 2.5 × 104 ;
25 × 108 + (4 × 108 )(6.25)
.·. R1 = 25 kΩ;
L1 =
and L1 =
(25 × 108 )2.5
= 1.25 H.
50 × 108
R22 L2
.
R22 + ω 2 L22
Problems
P 9.16
[a] Y2 =
1
j
−
;
R2 ωL2
R1 − jωL1
1
= 2
.
R1 + jωL1
R1 + ω 2 L21
Y1 =
Therefore
R2 =
Y2 = Y1
R12 + ω 2 L21
R1
when
and L2 =
R12 + ω 2 L21
.
ω 2 L1
25 × 106 + 108 (0.25)
= 10 × 103 ;
5 × 103
[b] R2 =
.·. R2 = 10 kΩ.
50 × 106
= 1 H.
108 (0.5)
L2 =
P 9.17
[a] Z1 = R1 − j
1
;
ωC1
R2
R2 − jωR22 C2
R2 /jωC2
=
=
;
R2 + (1/jωC2 )
1 + jωR2 C2
1 + ω 2 R22 C22
Z2 =
Z1 = Z2
when R1 =
ωR22 C2
1
=
ωC1
1 + ω 2 R22 C22
1 + ω 2 R22 C22
.
ω 2 R22 C2
1 + (64 × 108 )(25 × 104 )(625 × 10−18 )
2
C1 =
[a] Y2 =
or C1 =
and
500
[b] R1 =
P 9.18
R2
1 + ω 2 R22 C22
(64 × 108 )(25 × 104 )(25 × 10−9 )
= 250 Ω;
= 50 nF.
1
+ jωC2 ;
R2
1
jωC1
ω 2 R1 C12 + jωC1
=
=
.
R1 + (1/jωC1 )
1 + jωR1 C1
1 + ω 2 R12 C12
Y1 =
Therefore
R2 =
[b] R2 =
C2 =
Y1 = Y2
1 + ω 2 R12 C12
ω 2 R1 C12
when
and C2 =
C1
.
1 + ω 2 R12 C12
1 + (4 × 108 )(4 × 106 )(2500 × 10−18 )
= 2500 = 2.5kΩ;
(4 × 108 )(2 × 103 )(2500 × 10−18 )
50 × 10−9
= 10 nF.
5
9–15
9–16
P 9.19
CHAPTER 9. Sinusoidal Steady State Analysis
[a] Y =
1
1
1
+
+
4 − j3 16 + j12 −j100
= 0.16 + j0.12 + 0.04 − j0.03 + j0.01
= 0.2 + j0.1 = 223.6/26.57◦ mS.
[b] G = 200 mS.
[c] B = 100 mS.
[d] I = 50/0◦ A,
V=
I
50
= 223.61/−26.57◦ V;
=
Y
0.223/26.57◦
V
223.6/−26.57◦
IC =
= 2.24/63.43◦ A;
=
◦
/
ZC
100 −90
iC = 2.24 cos(ωt + 63.43◦ ) A,
P 9.20
Im = 2.24 A.
[a]
500/60◦
= 1/23.13◦ A.
400 + j700 − j400
[c] i = 1 cos(8000t + 23.13◦ ) A.
[b] I =
P 9.21
[a] jωL = j(5 × 104 )(40 × 10−6 ) = j2 Ω;
106
1
= −j
= −j20 Ω;
jωC
5 × 104
Ig = 20/−20◦ A.
[b] Vo = 20/−20◦ Ze ;
Ze =
1
;
Ye
Ye =
1
1
1
+j +
;
20
20 1 + j2
Ye = 0.05 + j0.05 + 0.20 − j0.40 = 0.25 − j0.35 S;
Ze =
1
= 2.32/54.46◦ Ω;
0.25 − j0.35
Vo = (20/−20◦ )(2.32/54.46◦ ) = 46.4/34.46◦ V.
Problems
9–17
[c] vo = 46.4 cos(5 × 104 t + 34.46◦ ) V.
P 9.22
[a] Using the notation and results from Problem 9.16:
RkL = 20 + j40 so R1 = 20,
L1 =
R2 =
202 + 10002 (0.04)2
= 100 Ω;
20
L2 =
202 + 10002 (0.04)2
= 50 mH;
10002 (0.04)
40
= 40 mH;
1000
R2 kjωL2 = 100kj50 = 20 + j40 Ω. (checks)
The circuit, using combinations of components from Appendix H, is
shown here:
[b] Using the notation and results from Problem 9.18:
RkC = 20 − j40 so R1 = 20,
R2 =
C2 =
C1 = 25 µF;
1 + 10002 (20)2 (25 µ)2
= 100 Ω;
10002 (20)(25 µ)2
25 µ
1 + 10002 (20)2 (25 µ)2
= 20 µF;
R2 k(−j/ωC2 ) = 100k(−j50) = 20 − j40 Ω. (checks)
The circuit, using combinations of components from Appendix H, is
shown here:
P 9.23
[a] (20 + j40)k(−j/ωC) = 100kj50k(−j/ωC).
To cancel out the j50 Ω impedance, the capacitive impedance must be
−j50 Ω:
−j
1
= −j50 so C =
= 20 µF.
1000C
(50)(1000)
9–18
CHAPTER 9. Sinusoidal Steady State Analysis
Check:
RkjωLk(−j/ωC) = 100kj50k(−j50) = 100 Ω.
Create the equivalent of a 20 µF capacitor from components in Appendix
H by combining two 10 µF capacitors in parallel.
[b] (20 − j40)k(jωL) = 100k(−j50)k(jωL).
To cancel out the −j50 Ω impedance, the inductive impedance must be
j50 Ω:
j1000L = j50 so L =
50
= 50 mH.
1000
Check:
RkjωLk(−j/ωC) = 100kj50k(−j50) = 100 Ω.
Create the equivalent of a 50 mH inductor from components in Appendix
H by combining five 10 mH inductors in series.
P 9.24
[a] R = 800 Ω = 330 Ω + 470 Ω;
ωL −
1
1
= −600 so 5000L −
= −600.
ωC
5000C
Choose L = 40 mH by combining four 10 mH inductors in series. Then,
1
1
= 200 + 600 so C =
= 0.25 µF.
5000C
5000(800)
We can achieve the desired capacitance by combining a 0.22 µF capacitor
in parallel with three 0.01 µF capacitors.
1
1
2
so
ω
=
= 108 ;
[b] 0.04ω =
ω(0.25 × 10−6 )
0.04(0.25 × 10−6 )
.·. ω = 10,000 rad/s.
P 9.25
Z = 400 + j(5)(40) − j
1000
= 500/ − 36.87◦ Ω;
(5)(0.4)
750/0◦ × 10−3
= 1.5/36.87◦ mA;
Io =
◦
/
500 − 36.87
io (t) = 1.5 cos(5000t + 36.87◦ ) mA.
P 9.26
ZL = j(5000)(48 × 10−3 ) = j240 Ω;
ZC =
−j
= −j80 Ω.
(5000)(2.5 × 10−6 )
Problems
Construct the phasor domain equivalent circuit:
Using current division:
I=
(80 + j240)
(0.2) = 0.1 + j0.1 A;
240 − j80 + 80 + j240
Vo = 240I = 24 + j24 = 33.94/45◦ ;
vo = 33.94 cos(5000t + 45◦ ) V.
P 9.27
1
109
=
= −j4000 Ω;
jωC
(31.25)(8000)
jωL = j8000(500)10−3 = j4000 Ω;
Vg = 64/0◦ V.
Ze =
(2000)(j4000)
= 1600 + j800 Ω;
2000 + j4000
ZT = 1600 + j800 − j4000 = 1600 − j3200 Ω;
Ig =
64/0◦
= 8 + j16 mA;
1600 − j3200
Vo = Ze Ig = (1600 + j800)(0.008 + j0.016) = j32 = 32/90◦ V;
vo = 32 cos(8000t + 90◦ ) V.
9–19
9–20
P 9.28
CHAPTER 9. Sinusoidal Steady State Analysis
Vs = 250/ − 90◦ V;
1
= −j400 Ω;
jωC
jωL = j125 Ω.
Zeq = 62.5 + j125k500k(300 − j400) = 125 + j125 Ω;
Vo
250/ − 90◦
Io =
=
= −1 − j1 = 1.414/ − 135◦ A;
Zeq
125 + j125
io = 1.414 cos(2500t − 135◦ ) A.
P 9.29
Z1 = 10 − j40 Ω;
Z2 =
(5 − j10)(10 + j30)
= 10 − j10 Ω;
15 + j20
Z3 =
20(j20)
= 10 + j10 Ω;
20 + j20
.·. Zab = Z1 + Z2 + Z3 = 30 − j40 Ω = 50/ − 53.13◦ Ω.
P 9.30
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.
Problems
P 9.31
[a] Zab = j5ω +
(4000)(109 /jω625)
4000 + (109 /j625ω)
= j5ω +
4 × 1012
25 × 105 jω + 109
= j5ω +
4 × 107
104 + j25ω
= j5ω +
4 × 1011
100 × 107 ω
−
j
;
108 + 625ω 2
108 + 625ω 2
.·. 5 =
109
;
108 + 625ω 2
5 × 108 + 3125ω 2 = 109 ;
ω = 4 × 102 = 400 rad/s.
[b] Zab (400) = j2000 +
P 9.32
[a]
(4000)(−j4000)
= 2 kΩ.
4000 − j4000
109
1
=
= −j10 Ω;
jωC
j8 × 105 (125)
jωL = j8 × 105 (25 × 10−6 ) = j20 Ω;
Ig = 5/0◦ ;
−5 +
V
V V − Vo
+
+
= 0;
−j10 20
12
Vo − V
Vo
+
= 0.
12
j20
Solving,
Vo = 44.72/ − 10.30◦ V;
vo = 44.72 cos(8 × 105 t − 10.30◦ ) V.
[b] ω = 2πf = 8 × 105 ;
T =
f=
1
π
=
= 2.5π µs;
f
4 × 105
4 × 105
;
π
9–21
9–22
CHAPTER 9. Sinusoidal Steady State Analysis
10.30
.·.
(2.5π) = 224.82 ns;
360
.·. vo lags ig by 224.82 ns.
P 9.33
[a]
Va = (120 + j40)(0.04/0◦ ) = 4.8 + j1.6 V;
Ib =
4.8 + j1.6
= 20 + j20 mA;
160 − j80
Ic = 40/0◦ + (20 + j20) + (40 + j80) mA = 100 + j100 mA;
Vg = 25Ic + Va = 25(0.100 + j0.100) + 4.8 + j1.6 = 7.3 + j4.1 V.
[b] ib = 28.28 cos(800t + 45◦ ) mA;
ic = 141.42 cos(800t + 45◦ ) mA;
vg = 8.37 cos(800t + 29.32◦ ) V.
P 9.34
[a] Yp =
1
+ j4 × 10−3 ω
10 + j2ω
=
10 − j2ω
+ j4 × 10−3 ω
100 + 4ω 2
10
j2ω
−
+ j4 × 10−3 ω.
2
100 + 4ω
100 + 4ω 2
Yp is real when
=
4 × 10−3 ω =
or
2ω
,
100 + 4ω 2
ω 2 = 100;
[b] Yp (10 rad/s) =
ω = 10 rad/s;
10
= 20 mS;
500
Zp (10 rad/s) =
103
= 50 Ω;
20
Z(10 rad/s) = 50 + 150 = 200 Ω;
f = 5/π = 1.59Hz.
Problems
9–23
Vg
10/0◦
A=
= 50/0◦ mA;
200
200
Io =
io = 50 cos 10t mA.
P 9.35
[a] Y1 =
1
= 0.2 × 10−3 S;
5000
Y2 =
=
1
1200 + j0.2ω
1200
0.2ω
−j
;
6
2
1.44 × 10 + 0.04ω
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
1.44 × 106 + 0.04(64) × 106
.·. ZT = 2000 Ω.
Vo = (2.5 × 10−3 /0◦ )(2000) = 5/0◦ ;
vo = 5 cos 8000t V.
P 9.36
109
104 (j2ω)
+
[a] Zg = 4000 − j
25ω 104 + j2ω
.·.
= 4000 − j
109
2 × 104 jω(104 − j2ω)
+
25ω
108 + 4ω 2
= 4000 − j
109
4 × 104 ω 2
2 × 108 ω
+ 8
+
j
;
25ω 10 + 4ω 2
108 + 4ω 2
0.2 × 109 ω
109
=
.
25ω
108 + 4ω 2
108 + 4ω 2 = 5ω 2 ;
ω 2 = 108 ;
ω = 10,000 rad/s.
= 0.5 × 10−3 S;
9–24
CHAPTER 9. Sinusoidal Steady State Analysis
[b] When ω = 10,000 rad/s
Zg = 4000 +
4 × 104 (104 )2
= 12,000 Ω;
108 + 4(104 )2
◦
45/0
.·. Ig =
= 3.75/0◦ mA.
12,000
Vo = Vg − Ig Z1 ;
Z1 = 4000 − j
109
= 4000 − j4000 Ω;
25 × 104
Vo = 45/0◦ − (3.75 × 10−3 )(4000 − j4000) = 45 − (15 − j15)
= 30 + j15 = 33.54/26.57◦ V;
vo = 33.54 cos(10,000t + 26.57◦ ) V.
P 9.37
Va = j2Ia = j2(−j5) = 10/0◦ V;
Vc = 60/0◦ − Va = 50/0◦ V;
Ic =
50/0◦
Vc
=
= 5/53.13◦ = 3 + j4 A;
6 − j8
10/ − 53.13◦
Ib = Ic − Ia = 3 + j4 − (−j5) = 3 + j9 A = 9.49/71.57◦ A;
Vb = Ib (j5) = (3 + j9)(j5) = −45 + j15 V;
Vz = Vb + Vc = −45 + j15 + 50 + j0 = 5 + j15 V;
Vd + Vz = 60/0◦ ;
.·. Vd = 60 − 5 − j15 = 55 − j15 V.
Problems
Id =
9–25
Vd
= 3 + j11 A;
−j5
Iz = Id − Ib = 3 + j11 − 3 − j9 = j2 A;
Z=
P 9.38
5 + j15
Vz
=
= 7.5 − j2.5 Ω.
Iz
j2
V2 is the voltage across the −j10 Ω impedance.
V1 − Vg V1 V1 − V2
+
+
= 0;
20
j5
Z
(40 + j30) − (100 − j50) 40 + j30 (40 + j30) − V2
+
+
= 0;
20
j5
Z
.·. V2 = 40 + j30 + (3 − j4)Z.
V2
V2 − Vg
V2 − V1
+
− Ig +
= 0;
Z
−j10
3 + j1
V2 − (40 + j30)
V2
V2 − (100 − j50)
+
− (20 + j30) +
= 0.
Z
−j10
3 + j1
Substituting the expression for V2 found at the start and simplifying yields
Z = 12 + j16 Ω.
P 9.39
Ig = 100/ − 8.13◦ mA;
Vg = 50/ − 45◦ V;
Z=
Vg
= 500/ − 36.87◦ Ω = 400 − j300 Ω;
Ig
2.5 × 106
Z = 400 + j 0.04ω −
;
ω
!
6
2.5 × 10
.·. 0.04ω −
= −300;
ω
.·. ω 2 + 7500ω − 62.5 × 106 = 0;
.·. ω = −3750 ±
ω > 0,
q
(3750)2 + 62.5 × 106 = −3750 ± 8750.
.·. ω = 5000 rad/s.
9–26
P 9.40
CHAPTER 9. Sinusoidal Steady State Analysis
[a] Z1 = 1600 − j
Z2 =
109
= 1600 − j1600 Ω;
104 (62.5)
4000(j104 L)
4 × 105 L2 + j16 × 104 L
=
;
4000 + j104 L
16 + 100L2
ZT = Z1 + Z2 = 1600 +
16 × 104 L
4 × 105 L2
−
j1600
+
j
.
16 + 100L2
16 + 100L2
ZT is resistive when
16 × 104 L
= 1600,
16 + 100L2
or
L2 − L + 0.16 = 0.
Solving, L1 = 0.8 H and L2 = 0.2 H.
[b] When L = 0.8 H:
ZT = 1600 +
Ig =
4 × 105 (0.64)
= 4800 Ω;
16 + 64
96/0◦
× 10−3 = 20/0◦ mA;
4.8
ig = 20 cos 10,000t mA.
When L = 0.2 H:
ZT = 1600 +
4 × 105 (0.04)
= 2400 Ω;
16 + 4
ig = 40 cos 10,000t mA.
P 9.41
R
R
jωC
[a] Zp =
=
R + (1/jωC)
1 + jωRC
=
12,500
12,500
=
1 + j(1000)(12,500)C
1 + j12.5 × 106 C
=
12,500(1 − j12.5 × 106 C)
1 + 156.25 × 1012 C 2
=
12,500
156.25 × 109 C
−
j
;
1 + 156.25 × 1012 C 2
1 + 156.25 × 1012 C 2
jωL = j1000(5) = j5000;
.·. 5000 =
156.25 × 109 C
;
1 + 156.25 × 1012 C 2
Problems
.·. 781.25 × 1015 C 2 − 156.25 × 109 C + 5000 = 0;
.·. C 2 − 20 × 10−8 C + 64 × 10−16 = 0;
√
.·. C1,2 = 10 × 10−8 ± 100 × 10−16 − 64 × 10−16 .
C1 = 10 × 10−8 + 6 × 10−8 = 16 × 10−8 = 0.16 µF;
C2 = 10 × 10−8 − 6 × 10−8 = 4 × 10−8 = 0.04 µF.
12,500
.
1 + 156.25 × 1012 C 2
When C = 160 nF
Re = 2500 Ω;
[b] Re =
250/0◦
= 0.1/0◦ A;
ig = 100 cos 1000t mA.
2500
When C = 40 nF
Re = 10,000 Ω;
Ig =
Ig =
P 9.42
250/0◦
= 0.025/0◦ A;
10,000
ig = 25 cos 1000t mA.
Convert the Y on the left side to a delta.
The numerator for the three delta-connected impedances is the same:
N = −j4(20 + j60) − j40 + 10(20 + j60) = 440 + j480.
Then,
Za =
440 + j480
= 44 + j48 Ω;
10
Zb =
440 + j480
= −120 + j110 Ω;
−j4
Zc =
440 + j480
= 9.4 − j4.2 Ω.
20 + j60
9–27
9–28
CHAPTER 9. Sinusoidal Steady State Analysis
Substitute the delta-connected impedances for the Y-connected impedances to
give
Now combine the impedances in series and in parallel to give the equivalent
impedance seen by the voltage source:
Zeq = (44 + j48)k{[(9.4 − j4.2)k(63.2 + j2.4)] + [(−120 + j110)k(−j20)]} = 18 − j24 Ω.
Therefore,
I0 =
P 9.43
120/0◦
= 2.4 + j3.2 = 4/53.13◦ A.
18 − j24
[a] jωL = j(5000)(50) × 10−3 = j250 Ω;
1
1
= −j
= −j500 Ω.
jωC
(5000)(400 × 10−9 )
Using voltage division,
Vab =
(250 + j250)k(−j500)
(23.36/26.565◦ ) = 20/0◦ ;
j250 + (250 + j250)k(−j500)
VTh = Vab = 20/0◦ V.
[b] Remove the voltage source and combine impedances in parallel to find
ZTh = Zab :
1
1
1
Yab =
+
+
= 2 − j4 mS;
j250 250 + j250 −j500
ZTh = Zab =
1
= 100 + j200 Ω.
Yab
Problems
[c]
P 9.44
Step 1 to Step 2:
75/0◦
= −j4.167 = 4.167/ − 90◦ A.
j18
Step 2 to Step 3:
(j18)k24 =
(j18)(24)
= 8.64 + j11.52 Ω.
24 + j18
Step 3 to Step 4:
(4.167/ − 90◦ )(8.64 + j11.52) = 60/ − 36.87◦ V.
P 9.45
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
9–29
9–30
CHAPTER 9. Sinusoidal Steady State Analysis
V2
96
= −j
= −j4.8 A.
20
20
Combine the parallel impedances:
Y =
1
1
1
j5
1
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.
P 9.46
IN =
0.1/120◦
+ 1/30◦ mA,
ZN
IN =
0.3/ − 60◦
+ (−3/210◦ ) mA,
ZN
ZN in kΩ.
ZN in kΩ;
Problems
0.3/ − 60◦
0.1/120◦
+ 1/30◦ =
+ (−3/210◦ ; )
ZN
ZN
0.3/ − 60◦ − 0.1/120◦
= 1/30◦ + 3/210◦ ;
ZN
ZN =
0.3/ − 60◦ − 0.1/120◦
= 0.2/90◦ = j0.2 kΩ;
◦
◦
/
/
1 30 + 3 210
IN =
0.1/120◦
+ 1/30◦ = 1.5/30◦ mA.
0.2/90◦
P 9.47
(27 + j12)Ia − 3Ib = −87/0◦ ;
−3Ia + (27 − j12)Ib = 87/0◦ .
Solving,
Ia = −2.4167 + j1.21;
Ib = 2.4167 + j1.21;
VTh = 12Ia + (12 − j12)Ib = 14.5/0◦ V.
9–31
9–32
CHAPTER 9. Sinusoidal Steady State Analysis
Short Circuit Test:
(27 + j12)Ia − 3Ib − 12Isc = −87;
−3Ia + (27 − j12)Ib − (12 − j12)Isc = 87;
−12Ia − (12 − j12)Ib + (24 − j12)Isc = 0.
Solving,
Isc = 1/0◦ ;
.·.
IN = Isc = 1/0◦ ;
ZN =
VTh
14.5/0◦
= 14.5 Ω.
=
Isc
1/0◦
Alternate calculation for ZN :
X
Z = 12 + 3 + 12 − j12 = 27 − j12;
Z1 =
36
12
=
27 − j12
9 − j4;
Z2 =
36 − j36
12 − j12
=
;
27 − j12
9 − j4
Problems
Z3 =
12(12 − j12)
48 − j48
=
.
27 − j12
9 − j4
Za = 12 + j12 +
Zb = 12 +
Za kZb =
12
12(14 + j5)
=
;
9 − j4
9 − j4
12 − j12
12(10 − j5)
=
;
9 − j4
9 − j4
165 − j20
18 − j8;
ZN = Z3 + Za kZb =
48 − j48 165 − j20
+
= 14.5 Ω.
9 − j4
18 − j8
P 9.48
V1 − 75
0.02V1 (40)
V1
−
+
= 0;
150(4 + j1)
40 − j150
40 − j150
75(4 − j15)
.·. V1 =
.
16 − j12
VTh =
=
40V1
4
75(4 − j15)
=
·
40 − j150
4 − j15
16 − j12
75
= 15/36.87◦ V.
4 − j3
9–33
9–34
CHAPTER 9. Sinusoidal Steady State Analysis
Isc =
75
1
= A;
600
8
ZTh =
P 9.49
VTh
= 120/36.87◦ = 96 + j72 Ω.
Isc
The Thévenin voltage equals the open circuit voltage between the terminals a
and b, which is the voltage across the capacitor. Identify and label the node
voltages.
Write the two node voltage equations and the dependent source constraint
equation:
−2/45◦ +
V1 V1 − 10Ix
+
= 0;
20
j10
V2
V2 − 10Ix
+
= 0;
10
−j10
Ix =
V1
.
20
Solving these equations for V2 = 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
Problems
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.
P 9.50
V2 25 − 4V2 V2 − 4V2
+
+
= 0.
50
1000
−j250
Solving,
V2 = −1 − j0.75 V = 1.25/216.87◦ V;
Isc = −Iφ =
ZTh =
−25/0◦
= −25/0◦ mA;
1000
1.25/216.87◦
= 50/36.87◦ Ω = 40 + j30 Ω;
−25 × 10−3 /0◦
IN = Isc = −25/0◦ mA.
ZN = ZTh = 50/36.87◦ = 40 + j30 Ω.
9–35
9–36
P 9.51
CHAPTER 9. Sinusoidal Steady State Analysis
ωL = j1.6 × 106 (25 × 10−6 ) = j40 Ω;
1
10−6 × 109
=
= −j25 Ω.
jωC
j1.6(25)
VT = j40IT + 15I∆ + 25I∆ ;
I∆ =
IT (−j25)
−jIT
=
;
25 − j25
1 − j1
VT = j40IT + 40
Zab =
P 9.52
(−jIT )
;
1 − j1
VT
= Zab = j40 + 20(−j)(1 + j) = 20 + j20 Ω = 28.28/45◦ Ω.
IT
1
(10−3 )(109 )
=
= 12 kΩ;
ωC1
25(10/3)
1
(10−3 )(109 )
=
= 24 kΩ.
ωC2
25(5/3)
VT = (1 − j12)IT + 20IT (0.125);
ZTh =
VT
= 3.5 − j12 kΩ.
IT
Problems
P 9.53
9–37
[a]
IT =
VT
VT − αVT /10
+
;
10
j10
1
IT
(1 − α/10)
(10 − α) + j10
=
+
=
;
VT
10
j10
j100
VT
1000 + j100(10 − α)
.·. ZTh =
=
.
IT
(10 − α)2 + 100
ZTh is real when α = 10.
1000
[b] ZTh =
= 10 Ω.
100
[c] ZTh = 5 + j5;
1000
= 5;
(10 − α)2 + 100
.·. 10 − α = ±10;
α = 0;
(10 − α)2 = 100;
α = 10 ∓ 10;
α = 20.
But the j term can only equal the real term with α = 0. Thus, α = 0.
[d] ZTh will be inductive when α < 10.
P 9.54
V1 − 100 V1
V1
+
+
= 0.
j40
40
60 + j20
Solving for V1 yields
V1 = 30 − j40 V;
V1
Vo =
(j20) =
60 + j20
!
j
V1 ;
3+j
Vo = 15 + j5 V = 15.81/18.43◦ V.
9–38
P 9.55
CHAPTER 9. Sinusoidal Steady State Analysis
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
Solving, V = 10 − j30 = 31.62/−71.57◦ .
Therefore v = 31.62 cos(50,000t − 71.57◦ ) V.
P 9.56
jωL = j(5000)(0.4 × 10−3 ) = j2 Ω;
106
1
= −j
= −j4 Ω;
jωC
(5000)(50)
Vg1 = 10/53.13◦ = 6 + j8 V;
Vg2 = 8/−90◦ = −j8 V.
Vo − 6 − j8 Vo Vo + (−j8)
+
+
= 0.
j2
6
−j4
Solving,
Vo = 12/0◦ ;
vo (t) = 12 cos 5000t V.
Problems
P 9.57
−15/0◦ +
I∆ =
Vo Vo − 2.5I∆
Vo
+
+
= 0;
8
j5
−j10
Vo
.
−j10
Solving,
Vo = 72 + j96 = 120/53.13◦ V.
P 9.58
Vo
Vo
+
+ 32Io = 0;
25
−j50
(2 + j)Vo = −1600Io ;
Vo = (−640 + j320)Io ;
Io =
V1 − (Vo /4)
;
j40
.·. V1 = (−160 + j120)Io .
17 =
V1
+ Io = (−8 + j6)Io + Io = (−7 + j6)Io ;
20
.·. Io =
17
= −1.4 − j1.2 A = 1.84/ − 139.40◦ A.
(−7 + j6)
Vo = (−640 + j320)Io = 1280 + j320 = 1319.39/14.04◦ V.
P 9.59
The phasor domain circuit is as shown in the following diagram:
The mesh current equations are
(5 − j2.22)I1 + (j2.22)I2 + (0)I3 = 50;
9–39
9–40
CHAPTER 9. Sinusoidal Steady State Analysis
(j2.22)I1 + (j5 − j2.22)I2 + (−j5)I3 = 0;
(0)I1 + (−j5)I2 + (20 + j5)I3 = j100.
Solving,
I1 = 8 + j6 A;
I2 = −5.5 + j1.5 A;
I3 = 0.5 + j3.5 A.
Therefore,
V = 5(10 − I1 ) = 5(2 − j6) = 10 − j30 = 31.623/ − 71.565◦ V.
v(t) = 31.623 cos(50,000t − 71.565◦ ) V.
P 9.60
jωL = j(5000)(0.4 × 10−3 ) = j2 Ω;
106
1
= −j
= −j4 Ω;
jωC
(5000)(50)
Vg1 = 10/53.13◦ = 6 + j8 V;
Vg2 = 8/−90◦ = −j8 V.
10/53.13◦ = (6 + j2)I1 − 6I2 ;
8/−90◦ = −6I1 + (6 − j4)I2 ;
Vo = (I1 − I2 )6.
Solving,
Vo = 12/0◦ V;
vo (t) = 12 cos 5000t V.
Problems
P 9.61
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.
P 9.62
9–41
Va = 60/0◦ V;
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.
9–42
P 9.63
CHAPTER 9. Sinusoidal Steady State Analysis
jωL = j104 (1.2 × 10−3 ) = j12 Ω;
1
−j106
=
= −j20 Ω;
jωC
5 × 104
Va = 100/ − 90◦ = −j100 V;
Vb = 500/0◦ = 500 V.
The mesh current equations:
(80 + j12)I1 + (−j12)I2 + (−80)I3 = −j100;
(−j12)I1 + (20 − j8)I2 + (−20)I3 = 0;
(−80)I1 + (−20)I2 + (100)I3 = −500;
Solving,
I1 = −14 − j17 A;
I2 = 5 − j25 A;
Thus,
Ia = I1 = −14 − j17 = 22.02/ − 129.47◦ A;
ia = 22.02 cos(10,000t − 129.47◦ ) A.
Ib = −I3 = 15.2 + j18.6 = 24.02/50.74◦ A;
ib = 24.02 cos(10,000t + 50.74◦ ) A.
I1 = −15.2 − j18.6 A.
Problems
P 9.64
100/0◦ = (5 + j5)I1 − 5I2 − j5I3 ;
50/0◦ = −5I1 + (5 − j5)I2 + j5I3 ;
−10/0◦ = −j5I1 + j5I2 + 5I3 .
Solving,
I1 = 58 − j20 A;
I2 = 58 + j10 A;
I3 = 28 + j0 A;
Ia = I3 + 2 = 30 + j0 A;
Ib = I1 − I3 = 58 − j20 − 28 = 30 − j20 A;
Ic = I2 − I3 = 58 + j10 − 28 = 30 + j10 A;
Id = I1 − I2 = 58 − j20 − 58 − j10 = −j30 A.
P 9.65
The impedances are
ZL = j(5000)(0.4) = j2000 Ω;
Vo =
ZC =
−j
= −j800 Ω.
(5000)(250 × 10−9 )
600 + j2000
(75/0◦ ) = 98 + j36 = 104.4/20.17◦ V;
300 + j2000 + 600 − j800
vo = 104.4 cos(5000t + 20.17◦ ) V.
9–43
9–44
P 9.66
CHAPTER 9. Sinusoidal Steady State Analysis
1
106
= −j 4 = −j100 Ω;
jωC
10
jωL = j(500)(1) = j500 Ω;
Let Z1 = 50 − j100 Ω;
Z2 = 250 + j500 Ω
Ig = 125/0◦ mA;
Io =
Ig Z1
125/0◦ (50 − j100)
=
Z1 + Z2
(300 + j400)
= −12.5 − j25 mA = 27.95/ − 116.57◦ mA;
io = 27.95 cos(500t − 116.57◦ ) mA.
P 9.67
[a] Superposition must be used because the frequencies of the two sources are
different.
[b] For ω = 200 rad/s:
Using voltage division,
V1 =
j50k − j25
−j50
(100/135◦ ) =
(100/135◦ ) = 70.71/90◦ V.
(j50k − j25) + 50
50 − j50
Thus v1 = 70.71 cos(200t + 90◦ ) V.
For ω = 100 rad/s:
Using voltage division,
V2 =
50kj25
10 + j20
(50/45◦ ) =
(50/45◦ ) = −35.355/0◦ V.
(50kj25) − j50
10 − j30
Thus v2 = −35.355 cos 100t V.
Therefore,
vo (t) = v1 (t) + v2 (t) = [70.71 cos(200t + 90◦ ) − 35.355 cos 100t] V.
Problems
P 9.68
9–45
[a] Superposition must be used because the frequencies of the two sources are
different.
[b] For ω = 500 rad/s:
Using voltage division,
V1 =
100kj100
50 + j50
(−j40) =
(−j40) = 40/0◦ V.
(100kj100) − j100
50 − j50
Thus v1 = 40 cos 500t V.
For ω = 250 rad/s:
Using voltage division,
V2 =
100k − j200
80 − j40
(60/7.125◦ ) =
(60/7.125◦ ) = 66.564/153.43◦ V.
(100k − j200) + j50
80 + j10
Thus v2 = 66.564 cos(250t + 153.43◦ ) V.
Therefore,
vo (t) = v1 (t) + v2 (t) = [40 cos 500t + 66.564 cos(250t + 153.43◦ )] V.
P 9.69
1
106
=
= −j10 kΩ.
jωC
j100
Let Va = voltage across 1 µF capacitor, positive at upper terminal.
Then:
Vg = 1.2/0◦ V;
Va − 1.2/0◦
Va
Va
+
+
= 0;
10
−j10
10
0 − Va 0 − Vo
+
= 0;
10
200
.·. Va = (0.48 − j0.24) V;
Vo = −20Va ;
.·. Vo = −9.6 + j4.8 = 10.73/153.43◦ V.
vo = 10.73 cos(100t + 153.43◦ ) V.
9–46
P 9.70
CHAPTER 9. Sinusoidal Steady State Analysis
[a]
Va
Va − 1.2/0◦
+ jωCo Va +
= 0;
10,000
10,000
Va =
1.2
;
2 + j104 ωCo
Vo = −20Va
Vo =
(see solution to Prob. 9.69);
−24
24/180◦
=
;
2 + j106 Co
2 + j106 Co
.·. denominator angle = 60◦ ;
√
tan 60◦ = 3;
106 Co √
= 3,
2
√
√
2 3
=
2
3 µF = 3.46 µF.
or Co =
106
[b] Vo =
24/180◦
√ = 6/120◦ V.
2 + j2 3
vo = 6 cos(100t + 120◦ ) V.
P 9.71
1
1012
=
= −j10 kΩ;
jωC1
j106 (100)
1012
1
=
= −j20 kΩ.
jωC2
j(106 )(50)
Problems
Va − 20 Va − Vo Va
Va
+
+
+
= 0;
−j20
20
40
10
.·. (−2 + j7)Va − jVo = j40.
0 − Va 0 − Vo
+
= 0;
10
−j10
.·. Va = −jVo ;
.·. (7 + j)Vo = j40.
Vo =
j40
= 0.8 + j5.6 = 5.657/81.87◦ V;
7+j
vo (t) = 5.657 cos(106 t + 81.87◦ ) V.
P 9.72
[a] Vg = 2/0◦ V;
Vp =
80
Vg = 1.6/0◦ ;
100
Vn = Vp = 1.6/0◦ V;
1.6
1.6 − Vo
+
= 0;
160
Zp
Zp =
(200)(1/jωC)
;
200 + (1/jωC)
1
109
=
= −j105 = −j100 kΩ;
jωC
j105 (0.1)
Zp =
200(−j100)
= 40 − j80 kΩ;
200 − j100
Vo = 1.6 +
Zp
= 2 − j0.8 = 2.15/ − 21.80◦ .
100
vo = 2.15 cos(105 t − 21.80◦ ) V.
9–47
9–48
CHAPTER 9. Sinusoidal Steady State Analysis
[b] Vp = 0.8Vm /0◦ ;
Vn = Vp = 0.8Vm /0◦ ;
0.8Vm 0.8Vm − Vo
+
= 0;
160
40 − j80
40 − j80
.·. Vo = 0.8Vm +
Vm (0.8) = 0.8Vm (1.25 − j0.5);
160
.·. |0.8Vm (1.25 − j0.5)| ≤ 5;
.·. Vm ≤ 4.64 V.
P 9.73
[a]
1
−j109
=
= −j400 Ω;
jωC
(2 × 105 )(12.5)
Vn
Vn − Vo
+
= 0;
200
−j400
Vn
Vn
Vo
=
+
;
−j400
200 −j400
Vo = Vn − j2Vn = (1 − j2)Vn ;
Vp =
Vg (1/jωCo )
Vg
=
;
500 + (1/jωCo )
1 + j(500)(2 × 105 )Co
Vg = 10/0◦ V;
Vp =
10/0◦
= Vn ;
1 + j108 Co
◦
(1 − j2)10/0
.·. Vo =
;
1 + j108 Co
√
5(10)
|Vo | = q
= 10.
1 + 1016 Co2
Solving,
Co = 20 nF.
[b] Vo =
10(1 − j2)
= 10/ − 126.87◦ ;
1 + j2
vo = 10 cos(2 × 105 t − 126.87◦ ) V.
P 9.74
[a] jωL1 = j(50)(5) = j250 Ω;
jωL2 = j(50)(20) = j1000 Ω;
1
109
=
= −j500 Ω
jωC
j(50 × 103 )(40)
Problems
9–49
.·. Z22 = 75 + 300 + j1000 − j500 = 375 + j500 Ω;
∗
.·. Z22
= 375 − j500 Ω;
q
M = k L1 L2 = 10k × 10−3 ;
ωM = (50)(10k) = 500k;
"
500k
Zr =
625
#2
(375 − j500) = k 2 (240 − j320) Ω;
Zin = 120 + j250 + 240k 2 − j320k 2 ;
1
|Zin | = [(120 + 240k 2 )2 + (250 − 320k 2 )2 ] 2 ;
d|Zin |
1
1
= [(120 + 240k 2 )2 + (250 − 320k 2 )2 ]− 2 ×
dk
2
[2(120 + 240k 2 )480k + 2(250 − 320k 2 )(−640k)];
d|Zin |
= 0 when
dk
960k(120 + 240k 2 ) − 1280k(250 − 320k 2 ) = 0;
√
.·. k 2 = 0.32;
.·. k = 0.32 = 0.5657.
[b] Zin (min) = 120 + 240(0.32) + j[250 − 0.32(320)]
= 196.8 + j147.6 = 246/36.87◦ Ω;
I1 (max) =
369/0◦
= 1.5/ − 36.87◦ A;
246/36.87◦
.·. i1 (peak) = 1.5 A.
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 = 360 − j70 = 366.74/ − 11◦ Ω.
Thus,
|Zin |k=1 > |Zin |k=√0.32 .
If k = 0,
Zin = 120 + j250 = 277.31/64.36◦ Ω.
Thus,
|Zin |k=0 > |Zin |k=√0.32 .
9–50
P 9.75
CHAPTER 9. Sinusoidal Steady State Analysis
jωL1 = j(25 × 103 )(3.2 × 10−3 ) = j80 Ω;
jωL2 = j(25 × 103 )(12.8 × 10−3 ) = j320 Ω;
109
1
=
= −j160 Ω;
jωC
j(25 × 103 )(250)
q
jωM = j(25 × 10 )k (3.2)(12.8) × 10−3 = j160k Ω;
3
Z22 = 40 + j320 − j160 = 40 + j160 Ω;
∗
Z22
= 40 − j160 Ω;
"
160k
Zr =
|40 + j160|
#2
(40 − j160) = 37.647k 2 − j150.588k 2 ;
Zab = 10 + j80 + 37.647k 2 − j150.588k 2 = (10 + 37.647k 2 ) + j(80 − 150.588k 2 ).
Zab is resistive when
80 − 150.588k 2 = 0
or
k 2 = 0.53125.
.·. Zab = 10 + (37.647)(0.53125) = 30 Ω.
P 9.76
√
[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
Zr =
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.
jωM
j80
[c] I2 =
I1 =
(0.5/ − 53.13◦ ) = 0.08/0◦ A;
Z22
500/36.87◦
i2 = 80 cos 800t mA.
Problems
P 9.77
ZTh = 30 + j200 + (50/25)2 (15 − j20) = 90 + j120 Ω;
VTh =
P 9.78
225/0◦
(j50) = 450/36.87◦ V.
15 + j20
[a] jωL2 = j(500)103 (500)10−6 = j250 Ω;
109
1
=
= −j100 Ω;
jωC
j(500 × 103 )(20)
Z22 = 150 + 50 + j250 − j100 = 200 + j150 Ω;
∗
Z22
= 200 − j150 Ω;
ωM = (500 × 103 )(100 × 10−6 ) = 50 Ω;
Zr =
50
250
2
[200 − j150] = 8 − j6 Ω.
[b] Zab = R1 + jωL1 + 8 − j6;
jωL1 = j(500 × 103 )(80 × 10−6 ) = j40 Ω;
Zab = 20 + j34 Ω.
P 9.79
In Eq. 9.34 replace ω 2 M 2 with k 2 ω 2 L1 L2 and then write Xab as
Xab = ωL1 −
k 2 ω 2 L1 L2 (ωL2 + ωLL )
2
R22
+ (ωL2 + ωLL )2
(
= ωL1 1 −
k 2 ωL2 (ωL2 + ωLL )
}
2
R22
+ (ωL2 + ωLL )2
For Xab to be negative requires
2
R22
+ (ωL2 + ωLL )2 < k 2 ωL2 (ωL2 + ωLL ),
or
2
+ (ωL2 + ωLL )2 − k 2 ωL2 (ωL2 + ωLL ) < 0,
R22
which reduces to
2
R22
+ 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.
9–51
9–52
CHAPTER 9. Sinusoidal Steady State Analysis
P 9.80
Zab =
V1
;
I1
V1
V2
=− ;
50
1
50I1 = −I2 ;
V2
−50V2
= 2500 .
.·. Zab =
−I2 /50
I2
V2
V3
=
;
1
25
I2 = 25I3
2500 V3
V3 /25
=
.·. Zab = 2500
25I2
625 I3
= 4ZL = 4(200 + j150) = (800 + j600) Ω.
P 9.81
[a]
N1 I1 = N2 I2 ,
Zab =
I2 =
N1
I1 ;
N2
V2
V2
Vab
=
=
;
I1 + I2
I1 + I2
(1 + N1 /N2 )I1
V1
N1
=
,
V2
N2
V1 =
N1
V2 ;
N2
N1
V1 + V2 = ZL I1 =
+ 1 V2 ;
N2
Problems
Zab =
I1 ZL
;
(N1 /N2 + 1)(1 + N1 /N2 )I1
.·. Zab =
ZL
.
[1 + (N1 /N2 )]2
[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 =
P 9.82
ZL I1
V2
=
.
(1 − N1 /N2 )I1
[1 − (N1 /N2 )]2 I1
Zab =
ZL
.
[1 − (N1 /N2 )]2
Zab =
Vab
V1 + V2
=
;
I1
I1
[a]
V1
V2
=
,
N1
N2
V2 =
N1 I1 = N2 I2 ,
N2
V1 ;
N1
I2 =
N1
I1 ;
N2
V2 = (I1 + I2 )ZL = I1
N1
ZL ;
1+
N2
N1
N1
V1 + V2 =
+ 1 V2 = 1 +
N2
N2
2
(1 + N1 /N2 ) ZL I1
.·. Zab =
.
I1
Zab = 1 +
N1
N2
2
ZL .
2
ZL I1 ;
9–53
9–54
CHAPTER 9. Sinusoidal Steady State Analysis
[b] Assume dot on N2 is moved to the lower terminal, then
−V2
V1
=
,
N1
N2
V1 =
N1 I1 = −N2 I2 ,
−N1
V2 ;
N2
I2 =
−N1
I1 .
N2
As in part [a]
V2 = (I2 + I1 )ZL
Zab =
and Zab =
V1 + V2
;
I1
(1 − N1 /N2 )V2
(1 − N1 /N2 )(1 − N1 /N2 )ZL I1
=
;
I1
I1
Zab = [1 − (N1 /N2 )]2 ZL .
P 9.83
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 9.84
[a] I =
200 200
+
= (8 − j4) A;
25
j50
Vs = 200/0◦ + (1 + j2)(8 − j4) = 216 + j12 = 216.33/3.2◦ V.
Problems
9–55
[b] Use the capacitor to eliminate the j component of I, therefore
Ic = j4 A,
Zc =
200
= −j50 Ω.
j4
The capacitive reactance is −50 Ω.
Vs = 200 + (1 + j2)8 = 208 + j16 = 208.6/4.4◦ V.
[c] Let Ic denote the magnitude of the current in the capacitor branch. Then
I = (8 − j4 + jIc ) = 8 + j(Ic − 4) A.
Vs = 200/α = 200 + (1 + j2)[8 + j(Ic − 4)]
= (216 − 2Ic ) + j(12 + Ic ).
It follows that
200 cos α = (216 − 2Ic ) and 200 sin α = (12 + Ic ).
Now square each term and then add to generate the quadratic equation
5Ic2 − 840Ic + 46,000 = 2002 ;
Ic = 84 ± 75.472 A.
Therefore
Ic = 8.53 A (smallest value) and Zc = 200/j8.53 = −j23.45 Ω.
Therefore, the capacitive reactance is −23.45 Ω.
P 9.85
[a]
I` =
440 440
+
= 20 − j20 A;
22
j22
V` = (0.2 + j1.6)(20 − j20) = 36 + j28 = 45.61/37.87◦ V;
Vs = 440/0◦ + V` = 476 + j28 = 476.82/3.37◦ V.
[b]
9–56
CHAPTER 9. Sinusoidal Steady State Analysis
[c] I` =
440 440
440
+
+
= 20 + j0 A;
22
j22 −j22
V` = (0.2 + j1.6)(20 + j0) = 4 + j32 = 32.25/82.87◦ ;
Vs = 440 + V` = 444 + j32 = 445.15/4.12◦ .
P 9.86
[a]
120 = (R + 0.3 + j0.3)I1 − (0.2 + j0.2)I2 − RI3 ;
120 = −(0.2 + j0.2)I1 + (R + 0.3 + j0.3)I2 − RI3 .
Subtracting the above two equations gives
0 = (R + 0.5 + j0.5)I1 − (R + 0.5 + j0.5)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]
240 = (275.2 + j0.2)Ia − 275Ib ;
Problems
9–57
0 = −275Ia + 285Ib .
Solving,
Ia = 24.3576 − j0.4946 A;
Ib = 23.503 − j0.47726 A;
I = Ia − Ib = 0.8546 − j0.01734 A;
V1 = 250I = 213.65 − j4.335 = 213.694/ − 1.16◦ V;
V2 = 25I = 21.365 − j0.4335 = 21.3694/ − 1.16◦ V.
[d]
120 = (250.3 + j0.3)I1 − (0.2 + j0.2)I2 − 250I3 ;
120 = −(0.2 + j0.2)I1 + (25.3 + j0.3)I2 − 25I3 ;
0 = −250I1 − 25I2 + 285I3 .
Solving,
I1 = 23.94 − j0.515 A;
I2 = 28.118 − j0.651 A;
I3 = 23.468 − j0.509 A;
V1 = 250(I1 − I3 ) = 118.4/ − 0.74◦ V;
V2 = 25(I2 − I3 ) = 116.3/ − 1.75◦ V.
[e] Because an open neutral can result in unbalanced voltages across the
120 V loads.
P 9.87
[a] 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. Then
20,000
240
=
;
N1
2N2
N2
3
.·.
=
= a.
N1
500
9–58
CHAPTER 9. Sinusoidal Steady State Analysis
In part c),
IP = 2aIa ;
3
2N2 Ia
=
.·. IP =
Ia
N1
250
=
3
(24.3576 − j0.4946);
250
IP = 292.29 − j5.93 mA.
In part d),
IP N1 = I1 N2 + I2 N2 ;
N2
(I1 + I2 )
.·. IP =
N1
=
3
(23.94 − j0.515 + 28.12 − j0.651)
500
=
3
(52.06 − j1.166);
500
IP = 312.36 − j6.996 mA.
[b] Yes, because the neutral conductor carries non-zero current whenever the
load is not balanced.
P 9.88
[a] The circuit is redrawn, with mesh currents identified:
The mesh current equations are:
125/0◦ = 86Ia − 4Ib − 80Ic ;
125/0◦ = −4Ia + 66Ib − 60Ic ;
0 = −80Ia − 60Ib + 240Ic .
Problems
9–59
Solving,
Ia = 3.82653/0◦ A;
Ib = 4.2517/0◦ A;
Ic = 2.3384/0◦ A.
The branch currents are:
I1 = Ia = 3.82653/0◦ A;
I2 = Ib − Ia = 0.4252/180◦ A;
I3 = −Ib = 4.2517/180◦ A;
I4 = Ic = 2.3384/0◦ A;
I5 = Ia − Ic = 1.4881/0◦ A;
I6 = Ib − Ic = 1.9133/0◦ 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. Therefore,
250
15,000
=
N1
2N2
or
N2
1
=
.
N1
120
The ampere turn balance requires
N1 IP = N2 I1 + N2 I3 .
Therefore,
IP =
P 9.89
N2
1
(I1 + I3 ) =
(3.8265 + 4.2517) = 67.32/0◦ mA.
N1
120
[a]
The three mesh current equations are
125/0◦ = 86Ia − 4Ib − 80Ic ;
125/0◦ = −4Ia + 86Ib − 80Ic ;
0 = −80Ia − 80Ib + 260Ic .
9–60
CHAPTER 9. Sinusoidal Steady State Analysis
Solving,
Ia = 3.815/0◦ A;
Ib = 3.815/0◦ A;
Ic = 2.347/0◦ A;
.·. I2 = Ib − Ia = 0 A.
[b] IP =
N2
N2
(I1 + I3 ) =
(Ia + Ib )
N1
N1
1
(3.815 + 3.815) = 63.58/0◦ mA.
120
[c] Yes; when the two 125 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
No, the motor current drops to 5 A, well below its normal running value of
22.86 A.
P 9.91
After fuse A opens, the current in fuse B is only 15 A.
Sinusoidal Steady State Power
Calculations
Assessment Problems
1
AP 10.1 [a] P = (100)(10) cos(50 − 15) = 500 cos 35◦ = 409.58 W (abs);
2
Q = 500 sin 35◦ = 286.79 VAR (abs).
1
[b] P = (40)(5) cos(−15 − 60) = 100 cos(−75◦ ) = 25.88 W (abs);
2
Q = 100 sin(−75◦ ) = −96.59 VAR (del).
1
[c] P = (400)(10) cos(30 − 150) = 2000 cos(−120◦ ) = −1000 W (del);
2
Q = 2000 sin(−120◦ ) = −1732.05 VAR (del).
1
[d] P = (200)(5) cos(160 − 40) = 500 cos(120◦ ) = −250 W (del);
2
Q = 500 sin(120◦ ) = 433.01 VAR (abs).
AP 10.2 [a] pf = cos(θv − θi ) = cos(50 − 15) = cos(35◦ ) = 0.82 lagging;
rf = sin(θv − θi ) = sin(35◦ ) = 0.57.
[b] pf = cos(−15 − 60) = cos(−75◦ ) = 0.26 leading;
rf = sin(−75◦ ) = −0.97.
[c] pf = cos(30 − 150) = cos(−120◦ ) = −0.5 leading;
rf = sin(−120◦ ) = −0.87.
10–1
10–2
CHAPTER 10. Sinusoidal Steady State Power Calculations
[d] pf = cos(160 − 40) = cos(120◦ ) = −0.5 lagging;
rf = sin(120◦ ) = 0.87.
0.24
Ip
AP 10.3 From Ex. 9.4 Irms = √ = √ A;
3
3
2
P = Irms
R=
AP 10.4 jωL = j25 Ω;
Io =
0.0576
(1000) = 19.2 W.
3
1
= −j50 Ω.
jωC
√
150
= 2.4 + j1.2 = 7.2/26.565◦ A;
50 − j25
√
V50 = 50Io = 50 7.2/26.565◦ V;
√
Vj25 = (25/90◦ )Io = 25 7.2/116.565◦ V.
√
1
1 √
[a] P = Vm Im cos(θv − θi ) = (50 7.2)( 7.2) cos(0) = 180 W.
2
2
√
1
1 √
[b] Q = Vm Im sin(θv − θi ) = (25 7.2)( 7.2) sin(90) = 90 VAR.
2
2
[c] S = P + jQ = 180 + j90 VA;
|S| = 201.25 VA.
[d] pf = cos[tan−1 (Q/P )] = cos[tan−1 (0.5)] = cos 26.565◦ = 0.89 lagging.
AP 10.5 [a] Z = (120 + j90)k(−j125) = 120 − j90 = 150/ − 36.87◦ Ω;
Therefore, the line current is I` =
465/0◦
= 3.07/35.054◦ A (rms).
120 − j90 + 4 + j3
VL = ZI` = (150/ − 36.87◦ )(3.07/35.054◦ ) = 460.47/ − 1.82◦ V (rms);
IL =
VL
= 3.07/ − 38.69◦ A (rms).
120 + j90
Problems
[b] SL = VL I∗L = (460.47/ − 1.82◦ )(3.07/38.69◦ ) = 1413.64/36.87◦
= (1130.91 + j848.19) VA;
PL = 1130.91 W;
QL = 848.19 VAR;
[c] P` = |I` |2 (4) = (3.07)2 (4) = 37.7 W;
Q` = |I` |2 (3) = 28.27 VAR.
[d] Sg (delivering) = 465I∗` = (1168.61 − j819.91) VA.
Therefore the source is delivering 1168.61 W and absorbing 819.91
magnetizing VAR.
|VL |2
(460.47)2
[e] Qcap =
=
= −1696.26 VAR.
−125
−125
Therefore the capacitor is delivering =1696.26 magnetizing VAR.
Check:
28.27 + 848.19 + 819.91 = 1696.37 VAR (rounding) and
1130.91 + 37.7 = 1168.61 W.
AP 10.6 Series circuit derivation:
S = 400I∗ = (6000 + j8000);
Therefore I∗ = 15 + j20 = 25/53.13◦ A (rms).
I = 25/ − 53.13◦ A (rms);
Z=
V
400
=
= 16/53.13◦ = (9.6 + j12.8) Ω;
I
25/ − 53.13◦
Therefore R = 9.6 Ω,
XL = 12.8 Ω.
Parallel circuit derivation
(400)2
= 26.67 Ω;
6000
P =
(400)2
;
R
therefore R =
Q=
(400)2
;
XL
therefore XL =
AP 10.7 [a] S1 = 16 + j18 kVA;
(400)2
= 20 Ω.
8000
S2 = 6 − j8 kVA;
S3 = 8 + j0 kVA;
ST = S1 + S2 + S3 = 30 + j10 kVA;
250I∗ = (30 + j10) × 103 ;
Z=
.·. I = 120 − j40 A (rms);
250
= 1.875 + j0.625 Ω = 1.98/18.43◦ Ω.
120 − j40
10–3
10–4
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] pf = cos(18.43◦ ) = 0.9487 lagging.
AP 10.8 [a] The phasor domain equivalent circuit and the Thévenin equivalent are
shown below:
Phasor domain equivalent circuit after source transforming the voltage
source and series-connected resistor:
Thévenin equivalent:
ZTh = 64 + (320kj160) − j640 = 128 − j512 = 527.76/ − 75.96◦ Ω.
VTh = (320kj160)(0.25) = 16 + j32 = 35.777/63.435◦ V;
For maximum power transfer, ZL = (128 + j512) Ω.
35.777/63.435◦
[b] I =
= 0.14/63.435◦ A;
256
Therefore P =
0.14
√
2
!2
128 = 1.25 W.
[c] RL = |ZTh | = 527.76 Ω.
35.777/63.435◦
[d] I =
= 0.043/101.417◦ A;
655.76 − j512
Therefore P =
0.043
√
2
!2
(527.76) = 487.9 mW.
Problems
AP 10.9
Mesh current equations:
180 = (3 + j4)I1 + j9(I1 − I2 ) − j3I1 + j3(I2 − I1 );
0 = j9(I2 − I1 ) + j3I1 + 9I2 .
In standard form:
180 = (3 + j7)I1 − j6I2 ;
0 = −j6I1 + (9 + j9)I2 .
Solving,
I2 = 12/0◦ A;
.·. P = (12)2 (9) = 1296 W.
AP 10.10 [a]
j80I1 − j100I2 + 75(I1 − I2 ) = 496;
0 = 75(I2 − I1 ) + j200I2 − j100I1 + 80I2 .
In standard form:
(75 + j80)I1 + (−75 − j100)I2 = 496;
(−75 − j100)I1 + (155 + j200)I2 = 0.
Solving,
I1 = 8 − j6.2 A;
I2 = 4 − j3 = 5/ − 36.87◦ A;
1
.·. P = (25)(80) = 1000 W.
2
10–5
10–6
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] I1 − I2 = 4 − j3.2 A;
1
P75 = |I1 − I2 |2 (75) = 984 W.
2
1
[c] Pg = (496)(8) = 1984 W;
2
X
Pabs = 1000 + 984 = 1984 W. (checks)
AP 10.11 [a] Open circuit voltage:
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 .
Also,
40/0◦ = 4(I1 + I3 ) + 12I3 .
Solving,
I1 = 6 A;
I3 = 1 A;
Isc = I1 /4 + I3 = 2.5 A;
Problems
RTh =
[b] I =
VTh
40
=
= 16 Ω.
Isc
2.5
40/0◦
= 1.25/0◦ A(rms);
32
P = (1.25)2 (16) = 25 W.
[c]
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;
25
.·. % delivered =
(100) = 10.87%.
230
10–7
10–8
CHAPTER 10. Sinusoidal Steady State Power Calculations
Problems
P 10.1
[a] V = 100/ − 45◦ V,
I = 20/15◦ A.
Therefore
1
A → B;
P = (100)(20) cos[−45 − (15)] = 500 W,
2
Q = 1000 sin −60◦ = −866.03 VAR,
B → A.
[b] V = 100/ − 45◦ ,
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(−150◦ ) = −866.03 W,
B → A;
Q = 1000 sin(−150◦ ) = −500 VAR,
B → A.
[d] V = 100/0◦ ,
I = 20/120;◦
P = 1000 cos(−120◦ ) = −500 W,
B → A;
Q = 1000 sin(−120◦ ) = −866.03 VAR,
P 10.2
p = P + P cos 2ωt − Q sin 2ωt;
dp
= 0 when
dt
cos 2ωt = √
B → A.
dp
= −2ωP sin 2ωt − 2ωQ cos 2ωt;
dt
− 2ωP sin 2ωt = 2ωQ cos 2ωt or
P
P 2 + Q2
;
sin 2ωt = − √
Q
P 2 + Q2
Q
tan 2ωt = − .
P
.
Let θ = tan−1 (−Q/P ), then p is maximum when 2ωt = θ and p is minimum
when 2ωt = (θ + π).
Therefore pmax = P + P · √
and pmin = P − P · √
q
P
Q(−Q)
√
−
=
P
+
P 2 + Q2 ,
P 2 + Q2
P 2 + Q2
q
P
Q
√
−
Q
·
=
P
−
P 2 + Q2 .
P 2 + Q2
P 2 + Q2
Problems
P 10.3
10–9
[a] To find the power used by an appliance, divide the yearly kW-hours used
by the product of the number of hours per month used and the number
of months:
1080 k
central AC =
= 3000 W;
(120)(3)
dishwasher =
293 k
= 813.89 W;
(30)(12)
refrigerator =
533 k
= 61.7 W;
(720)(12)
microwave =
101 k
= 935.2 W.
(9)(12)
The total power consumed by these four appliances is
P4 = 3000 + 813.89 + 61.7 + 035.2 = 4810.76 W.
When 120 V(rms) is used, the amount of current required is
4810.8
= 40.1 A(rms).
120
Therefore, the 60 A current breaker will not interrupt the current.
I4 =
[b] the power used by the dryer is
dryer =
901 k
= 3128.47 W.
(24)(12)
With this fifth appliance added, the total power consumed is
P5 = 4810.76 + 3128.47 = 7939.2 W.
Now the current required is
7939.2
= 66.2 A(rms).
120
Now the 60 A current breaker will interrupt the current. So before
turning on the dryer, the dishwasher must be turned off:
I5 =
7939.2 − 813.89 = 7125.31 W;
the resulting current required is
7125.31
= 59.4 A(rms).
120
The 60 A circuit breaker will not interrupt the current if the dishwasher
is off.
10–10
P 10.4
CHAPTER 10. Sinusoidal Steady State Power Calculations
[a] From the solution to Problem 9.57 we have:
Vo = 72 + j96 = 120/53.13◦ V;
1
1
Pg = − Vm Im cos(θv − θi ) = − (120)(15) cos(53.13◦ ) = −540 W;
2
2
1
1
Qg = − Vm Im sin(θv − θi ) = − (120)(15) sin(53.13◦ ) = −720 W;
2
2
Therefore, the independent current source is delivering 540 W and 720
magnetizing vars.
I1 =
Vo
= 15/53.13◦ A;
8
1
1
P8Ω = Vm Im cos(θv − θi ) = (120)(15) cos(0◦ ) = 900 W.
2
2
Therefore, the 8 Ω resistor is absorbing 900 W.
I∆ =
Vo
= −9.6 + j7.2 = 12/143.13◦ A;
−j10
1
1
Qcap = Vm Im sin(θv − θi ) = (120)(12) sin(53.13◦ − 143.13◦ ) = −720 VAR.
2
2
Therefore, the −j10 Ω capacitor is delivering 720 magnetizing vars.
2.5I∆ = −24 + j18 V;
I2 =
Vj5 = 72 + j96 + 24 − j18 = 123.69/39.1◦ V;
Vj5
123.69/39.1◦
=
= 24.74/ − 50.91◦ A.
j5
j5
1
1
Qj5 = Vm Im sin(θv − θi ) = (123.69)(24.74) sin(39.1◦ + 50.9◦ ) = 1530 VAR.
2
2
Therefore, the j5 Ω inductor is absorbing 1530 magnetizing vars.
1
1
P2.5I∆ = Vm Im cos(θv − θi ) = [(2.5)12](24.74) cos(143.13◦ + 50.9◦ ) = −360 W;
2
2
1
1
Q2.5I∆ = Vm Im sin(θv − θi ) = [(2.5)12](24.74) sin(143.13◦ + 50.9◦ ) = −90 W;
2
2
Thus the dependent source is delivering 360 W and 90 magnetizing vars.
[b]
X
Pgen = 360 + 540 = 900 W =
X
Pabs .
Problems
[c]
P 10.5
X
Qgen = 720 + 90 + 720 = 1530 VAR =
X
10–11
Qabs .
[a] From the solution to Problem 9.62 we have
Ia = 2.25 − j2.25 = 3.182/ − 45◦ A;
Ib = −6.75 + j0.75 = 6.79152/173.66◦ A;
Io = 9 − j3 = 9.4868/ − 18.435◦ A;
1
1
P60V = − Vm Im cos(θv − θi ) = − (60)(3.182) cos(45◦ ) = −67.5 W;
2
2
1
1
Q60V = − Vm Im sin(θv − θi ) = − (60)(3.182) sin(45◦ ) = −67.5 VAR;
2
2
Thus, the 60 V source is developing 67.5 W and 67.5 magnetizing vars.
1
1
P90V = − Vm Im cos(θv − θi ) = − (90)(6.79152) cos(90◦ − 173.66◦ ) = −33.75 W;
2
2
1
1
Q90V = − Vm Im sin(θv − θi ) = − (90)(6.79152) sin(90◦ − 173.66◦ ) = 303.75 W;
2
2
Thus, the 90 V source is delivering 33.75 W and absorbing 303.75
magnetizing vars.
V20Ω = 20Ia = 63.64/ − 45◦ V;
1
P20Ω = (63.64)(3.182) cos(0◦ ) = 101.25 W.
2
Thus the 20 Ω resistor is absorbing 101.25 W.
V−j20Ω = −j20Ib = 135.83/83.66◦ V;
1
Q−j20Ω = (135.83)(6.79152) sin(−90◦ ) = −461.25 VAR.
2
Thus the −j20 Ω capacitor is developing 461.25 magnetizing vars.
Vj5Ω = j5Io = 47.434/71.565◦ V;
1
Qj5Ω = (47.434)(9.4868) sin(90◦ ) = 225 VAR.
2
Thus the j5 Ω inductor is absorbing 225 magnetizing vars.
[b]
X
Pdev = 67.5 + 33.75 = 101.25 W =
X
Pabs .
10–12
[c]
CHAPTER 10. Sinusoidal Steady State Power Calculations
X
Qdev = 67.5 + 461.25 = 528.75 VAR;
X
P 10.6
Qabs = 225 + 303.75 = 528.75 VAR =
Ig = 30/0◦ mA;
X
Qdev .
1
106
=
= −j1 Ω;
jωC
j(25 × 103 )(40)
jωL = j(25 × 103 )(40) × 10−6 = j1 Ω.
Z1 = −j1k(5 + j1) = 0.2 − j1 Ω;
Zeq = 2 + Z1 = 2.2 − j1 Ω;
Vg = Zeq Ig = 66 − j30 mA = 72.5/ − 24.444◦ mA;
1
1
Pg = Vg Ig cos(θv − θi ) = (0.0725)(0.03) cos(−24.444◦ ) = 990 µW.
2
2
P 10.7
[a] P =
1 (90)2
= 3 W;
2 1350
Q=
1 (90)2
= 4 VAR;
2 (1012.5)
From Problem 10.2,
pmax = P +
q
P 2 + Q2 = 3 +
q
(3)2 + (4)2 = 8 W.
[b] pmin = 3 − 5 = −2 W.
[c] P = 3 W from (a).
[d] Q = 4 VAR from (a).
[e] absorb, because Q > 0.
[f ] pf = cos(θv − θi ; )
I=
90
90
+
= 0.0667 − j0.08889 = 111.11/ − 53.13◦ mA;
1350 j1012.5
.·. pf = cos(0 + 53.13◦ ) = 0.6 lagging.
[g] rf = sin(53.13◦ ) = 0.8.
Problems
P 10.8
jωL = j10,000(10−3 ) = j10 Ω;
−15 +
.·. Vo
1
106
=
= −j40 Ω.
jωC
j10,000(2.5)
Vo
Vo + 10(Vo / − j40)
+
= 0;
−j40
20 + j10
"
#
1
1 + j0.25
+
= 15;
−j40
20 + j10
.·. Vo = 300 − j100 V;
Vo
= 2.5 + j7.5 A.
.·. I∆ =
−j40
Io = 15/0◦ − I∆ = 15 − 2.5 − j7.5 = 12.5 − j7.5 = 14.5774/ − 30.96◦ A;
V20Ω = 20Io = 291.5476/ − 30.96◦
1
1
P20Ω = Vm Im cos(θv − θi ) = (291.5476)(14.5774) cos(0◦ ) = 2125 W.
2
2
P 10.9
109
1
=
= 2500 Ω;
ωC
(5000)(80)
Zf =
−j2500(7500)
= 750 − j2250 Ω;
7500 − j2500
Zi = 1500 Ω;
Zf
750 − j2250
.·.
=
= 0.5 − j1.5.
Zi
1500
V1kΩ = −
Zf
Vg ;
Zi
Vg = 4/0◦ V;
V1kΩ = (−0.5 + j1.5)(4) = −2 + j6 = 6.325/108.43◦ V;
I1kΩ =
V1kΩ
= 6.325/108.43◦ mA;
1000
1
1
P = Vm Im cos(θv − θi ) = (6.325)(6.325 × 10−3 ) cos(0◦ ) = 20 mW.
2
2
10–13
10–14
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.10 [a] line loss = 50,000 − 40,000 = 10 kW;
Pline = V20Ω Iline cos(0◦ ) = (20|Ig |)(|Ig |) = 10,000;
√
.·. |Ig | = 500 A (rms).
Now find the load impedance:
PL = (RL |Ig |)|Ig | cos(0◦ ) = 40,000
40,000
.·. RL =
= 80 Ω;
500
QL = (XL |Ig |)|Ig | sin(90◦ ) = 30,000
30,000
.·. XL =
= 60 Ω;
500
|Ig |2 XL = 30,000
.·. XL = 60 Ω.
Thus,
|Z| =
q
(100)2 + (60 − X` )2
2500
|Ig | = q
10,000 + (60 − X` )2
;
4
625 × 10
.·. 10,000 + (60 − X` )2 =
= 12,500.
500
Solving,
(60 − X` ) = ±50.
Thus, X` = 10 Ω
or
X` = 110 Ω.
[b] If X` = 10 Ω:
2500
= 20 − j10 = 22.36/ − 26.565◦ A (rms);
100 + j50
√
Pg = −(2500)( 500) cos(26.565◦ ) = −50,000 W;
√
Qg = −(2500)( 500) sin(26.565◦ ) = −25,000 VAR;
Ig =
Thus, the voltage source is delivering 50 kW and 25 magnetizing kvars.
√
√
Q−j10 = (10 500)( 500) sin(−90◦ ) = −5000 VAR.
Problems
10–15
Therefore the line reactance is generating 5 magnetizing kvars.
√
√
Qj60 = (60 500)( 500) sin(90◦ ) = 30,000 VAR.
Therefore the load reactance is absorbing 30 magnetizing kvars.
X
Qdev = 25,000 kVAR =
X
Qabs.
If X` = 110 Ω:
2500
= 20 + j10 = 22.36/26.565◦ A (rms);
Ig =
100 − j50
√
Pg = −(2500)( 500) cos(−26.565◦ ) = −50,000 W;
√
Qg = −(2500)( 500) sin(−26.565◦ ) = 25,000 VAR;
Thus, the voltage source is delivering 50 kW and absorbing 25
magnetizing kvars.
√
√
Q−j110 = (110 500)( 500) sin(−90◦ ) = −55 kVAR.
Therefore the line reactance is generating 55 magnetizing kvars. The load
continues to absorb 30 magnetizing kvars.
X
X
Qabs .
Z to +T
vs2
dt;
R
Qdev = 55 kVAR =
Vdc2
T;
P 10.11 Wdc =
R
Ws =
to
Z to +T 2
vs
Vdc2
·
..
T =
dt.
R
R
to
Vdc2 =
1 Z to +T 2
vs dt;
T to
s
Vdc =
1 Z to +T 2
vs dt = Vrms .
T to
P 10.12 [a] Ieff = 60/110 ∼
= 0.545 A;
P 10.13 i(t) =
30
× 103 t = 750t
40
i(t) = M −
30
× 103 t
10
[b] Ieff = (60 + 80)/110 ∼
= 1.273 A.
0 ≤ t ≤ 40 ms;
40 ms ≤ t ≤ 50 ms;
i(t) = 0 when t = 50 ms;
.·. M = 3000(50 × 10−3 ) = 150.
10–16
CHAPTER 10. Sinusoidal Steady State Power Calculations
i(t) = 150 − 3000t
s
.·. Irms =
Z 0.04
1
0.05
40 ms ≤ t ≤ 50 ms;
Z 0.04
(750)2 t2 dt +
0
(750)2 t2 dt = (750)2
0
Z 0.05
(150 − 3000t)2 dt
.
0.04
t3 0.04
= 12;
3 0
(150 − 3000t)2 = 22,500 − 9 × 105 t + 9 × 106 t2 ;
Z 0.05
22,500 dt = 225;
0.04
Z 0.05
9 × 105 t dt = 45 × 104 t2
0.04
0.05
= 405;
0.04
9 × 106
Z 0.05
t2 dt = 3 × 106 t3
0.04
.·. Irms =
0.05
= 183;
0.04
q
20{12 + (225 − 405 + 183)} =
√
300 = 17.32 A.
3
24 × 10
= 80 Ω.
.·. R =
300
2
P 10.14 P = Irms
R
P 10.15 [a] Area under one cycle of vg2 : A = (−25)2 (2) + (25)2 (2) = 2500;
mean value =
Vrms =
√
A
= 312.5;
8
312.5 = 17.68 V(rms).
2
Vrms
312.5
=
= 125 mW.
R
2500
312.5
V2
= 500 Ω.
[c] R = rms =
P
0.625
[b] P =
P 10.16 [a] Area under one cycle of v 2 :
A = 4(20)2 (20 × 10−6 ) + 2(100)2 (20 × 10−6 )
= 21,600(20 × 10−6 ).
Mean value of vg2 :
A
21,600(20 × 10−6 )
=
= 3600;
120 × 10−6
120 × 10−6
√
.·. Vrms = 3600 = 60 V(rms).
M.V. =
Problems
[b] P =
10–17
2
Vrms
3600
=
= 300 W.
R
12
P 10.17 [a] ZL = j(2500)(0.405) = j1012.5 Ω.
The phasor domain circuit is
P =
1 |V|2
1 902
=
= 3 W.
2 R
2 1350
1 |V|2
1 902
=
= 4 VAR.
2 X
2 1012.5
[c] S = P + jQ = 3 + j4 = 5/53.13◦ VA;
[b] Q =
so
|S| = 5 VA.
[d] pf = cos 53.13◦ = 0.6 lagging.
P 10.18 [a] Let VL = VL /0◦ :
SL = 250(0.6 + j0.8) = 150 + j200 VA;
I∗` =
200
150
+j
;
VL
VL
I` =
150
200
−j
;
VL
VL
200
150
240/θ = VL +
−j
(1 + j8);
VL
VL
240VL /θ = VL2 + (150 − j200)(1 + j8) = VL2 + 1750 + j1000;
240VL cos θ = VL2 + 1750;
240VL sin θ = 1000;
(240)2 VL2 = (VL2 + 1750)2 + 10002 ;
57,600VL2 = VL4 + 3500VL2 + (3.0625 + 1) × 106 ,
or
VL4 − 54,100VL2 + 4,062,500 = 0.
Solving,
VL2 = 27,050 ± 26,974.8;
VL = 232.43 V(rms) and VL = 8.67 V(rms).
10–18
CHAPTER 10. Sinusoidal Steady State Power Calculations
If VL = 232.43 V(rms):
1000
sin θ =
= 0.0179;
(232.43)(240)
If VL = 8.67 V(rms):
1000
= 0.4805;
sin θ =
(8.67)(240)
.·. θ = 1.03◦ .
.·. θ = 28.72◦ .
[b]
1
106
= −j10 Ω;
P 10.19 [a]
=
jωC
j105
jωL = j105 (50 × 10−6 ) = j5 Ω.
Z = −j10 +
Ig =
(5)(j5)
+ 7.5 = 10 − j7.5 Ω;
5 + j5
50/0◦
= 3.2 + j2.4 A;
10 − j7.5
1
Sg = − Vg I∗g = −25(3.2 − j2.4) = −80 + j60 VA;
2
P = 80 W(dev);
Q = 60 VAR(abs);
|S| = |Sg | = 100 VA.
Problems
Ig (j5)
1
= (3.2 + j2.4)(1 + j1) = 0.4 + j2.8 A;
5 + j5
2
[b] I1 =
1
P5Ω = |I1 |2 (5) = 20 W;
2
1
P7.5Ω = |Ig |2 (7.5) = 60 W;
2
X
Pabs = 20 + 60 = 80 W =
[c] Ij5 =
X
Pdev .
Ig 5
1
= (3.2 + j2.4)(1 − j1) = 2.8 − j0.4 A;
5 + j5
2
1
Qj5Ω = |Ij5 |2 (5) = 20 VAR(abs);
2
1
Q−j10Ω = |Ig |2 (−10) = −80 VAR(dev);
2
X
Qabs = 20 + 60 = 80 VAR =
X
Qdev .
P 10.20 [a]
Vo − 340
Vo
Vo
+
+
= 0;
−j100
50
80 + j60
.·. Vo = 238 − j34 V(rms).
Ig =
340 − 238 + j34
= 2.04 + j0.68 A(rms);
50
Sg = −Vg I∗g = −(340)(2.04 − j0.68)
= −693.6 + j231.2 VA.
[b] Source is delivering 693.6 W.
[c] Source is absorbing 231.2 magnetizing VAR.
Vo
[d] I1 =
= 0.34 + j2.38 A(rms);
−j100
S1 = Vo I∗1 = (238 − j34)(0.34 − j2.38)
= 0 − j578 VA;
10–19
10–20
CHAPTER 10. Sinusoidal Steady State Power Calculations
I2 =
Vo
238 − j34
=
= 1.7 − j1.7 A(rms);
80 + j60
80 + j60
S2 = Vo I∗2 = (238 − j34)(1.7 + j1.7)
= 462.4 + j346.8 VA;
S50Ω = |Ig |2 (50) + j0 = (2.15)2 (50) = 231.2 W.
[e]
[f ]
X
Pdel = 693.6 W;
X
Pdiss = 462.4 + 231.2 = 693.6 W;
.·.
X
X
Pdel =
X
Pdiss = 693.6 W.
Qabs = 231.2 + 346.8 = 578 VAR;
X
Qdel = 578 VAR;
.·.
X
mag VAR del =
P 10.21 ST = 52,800 − j
X
mag VAR abs = 578.
52,800
(0.6) = 52,800 − j39,600 VA;
0.8
S1 = 40,000(0.96 + j0.28) = 38,400 + j11,200 VA;
S2 = ST − S1 = 14,400 − j50,800 = 52,801.52/ − 74.17◦ VA;
rf = sin(−74.17◦ ) = −0.9621;
pf = cos(−74.17◦ ) = 0.2727 leading.
P 10.22 [a] S1 = 24,960 + j47,040 VA;
S2 =
|VL |2
(480)2
=
= 23,040 − j23,040 VA;
Z2∗
5 + j5
S1 + S2 = 48,000 + j24,000 VA;
480I∗L = 48,000 + j24,000;
.·. IL = 100 − j50 A(rms);
Vg = VL + IL (0.02 + j0.20) = 480 + (100 − j50)(0.02 + j0.20)
= 492 + j19 = 492.37/2.21◦ Vrms;
|Vg | = 492.37 Vrms.
Problems
[b] T =
1
1
=
= 16.67 ms;
f
60
2.21◦
t
=
;
◦
360
16.67 ms
.·. t = 102.39 µs;
[c] VL lags Vg by 2.21◦ or 102.39 µs.
P 10.23 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(rms);
Vs = 200 + jI = 200 + j69 + 42 = 242 + j69 = 251.645/15.91◦ V (rms).
P 10.24 [a] Z1 = 12 + j(2π)(60)(15 × 10−3 ) = 13.27/25.23◦ Ω;
pf = cos(25.23◦ ) = 0.9 lagging;
rf = sin(25.23◦ ) = 0.43;
Z2 = 80 −
j
= 184.08/ − 64.24◦ Ω;
−6
2π(60)(16 × 10 )
pf = cos(−64.24◦ ) = 0.43 leading;
rf = sin(−64.24◦ ) = −0.9;
Z3 = 400 + Zp ;
Zp =
jωL(1/jωC)
jωL
=
jωL + 1/jωC
1 − ω 2 LC
=
j(120π)(20)
= −j570.67 Ω;
1 − (120π)2 (20)(5 × 10−6 )
.·. Z3 = 400 − j570.67 = 696.90/ − 54.97◦ Ω.
pf = cos(−54.97◦ ) = 0.57 leading;
rf = sin(−54.97◦ ) = −0.82.
10–21
10–22
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] Y = Y1 + Y2 + Y3 ;
Y1 =
1
;
13.27/25.23◦
Y2 =
1
;
184.08/ − 64.24◦
Y3 =
1
;
696.90/ − 54.97◦
Y = 71.35 − j26.05 mS;
Z=
1
= 13.16/20.06◦ Ω;
Y
pf = cos(20.06◦ ) = 0.94 lagging;
rf = sin(20.06◦ ) = 0.343.
P 10.25
2400I∗1 = 24,000 + j18,000;
I∗1 = 10 + j7.5;
.·. I1 = 10 − j7.5 A(rms);
2400I∗2 = 48,000 − j30,000;
I∗2 = 20 − j12.5;
I3 =
.·. I2 = 20 + j12.5 A(rms);
2400/0◦
= 40 + j0 A(rms);
60
I4 =
2400/0◦
= 0 − j5 A(rms);
j480
Ig = I1 + I2 + I3 + I4 = 70 A(rms);
Vg = 2400 + (70)(j10) = 2400 + j700 = 2500/16.26◦ V(rms).
P 10.26 [a]
120I∗1 = 1800 + j600;
.·. I1 = 15 − j5 A(rms);
Problems
.·. I2 = 10 + j7.5 A(rms);
120I∗2 = 1200 − j900;
I3 =
10–23
240 240
+
= 20 − j5 A(rms);
12
j48
Ig1 = I1 + I3 = 35 − j10 A(rms);
Sg1 = −120(35 + j10) = −4200 − j1200 VA.
Thus the Vg1 source is delivering 4200 W and 1200 magnetizing vars.
Ig2 = I2 + I3 = 30 + j2.5 A(rms; )
Sg2 = −120(30 − j2.5) = −3600 + j300 VA.
Thus the Vg2 source is delivering 3600 W and absorbing 300 magnetizing
vars.
[b]
X
P 10.27 S1 =
Pdev = 4200 + 3600 = 7800 W;
X
(240)2
= 7800 W =
Pdev ;
12
X
Pabs = 1800 + 1200 +
X
Qdev = 1200 + 900 = 2100 VAR;
X
Qabs = 300 + 600 +
X
(240)2
= 2100 VAR =
Qdev .
48
336 k
101 k
284 k
+
+
= 3603.7 + j0 VA;
(720)(12) (9)(12) (9)(12)
3603.7
.·. I1 =
= 30.03 A(rms).
120
S2 =
360 k
72 k
54 k
+
+4
= 505 + j0 VA;
(240)(12) (150)(6)
(60)(12)
505
.·. I2 =
= 4.21 A(rms).
120
S3 =
108 k
901 k
+
= 3449.9 + j0 VA;
(28)(12) (24)(12)
3449.9
.·. I3 =
= 14.37 A(rms).
240
Ig1 = I1 + I3 = 44.4 A(rms);
Ig2 = I2 + I3 = 18.58 A(rms).
Neither current exceeds the 50 A limit, so no interruptions occur.
10–24
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.28 [a] From the solution to Assessment Problem 10.7 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
Q` = (16,000)(0.08) = 1280 VAR.
Qs = 10,000 + 1280 = 11.28 kVAR.
P 10.29 [a]
I1 =
5000 − j2000
= 40 − j16 A (rms);
125
I2 =
3750 − j1500
= 30 − j12 A (rms);
125
I3 =
8000 + j0
= 32 + j0 A (rms);
250
.·. Ig1 = 72 − j16 A (rms.)
In = I1 − I2 = 10 − j4 A (rms);
Ig2 = 62 − j12 A;
Vg1 = 0.05Ig1 + 125 + j0 + 0.15In = 130.1 − j1.4 V(rms);
Vg2 = −0.15In + 125 + j0 + 0.05Ig2 = 126.6 + j0 V(rms);
Sg1 = [(130.1 − j1.4)(72 + j16)] = [9389.6 + j1980.8] VA;
Sg2 = [(126.6)(62 + j12)] = [7849.2 + j1519.2] VA.
Note: Both sources are delivering average power and magnetizing VAR to
the circuit.
Problems
[b] P0.05 = |Ig1 |2 (0.05) = 272 W;
P0.15 = |In |2 (0.15) = 17.4 W;
P0.05 = |Ig2 |2 (0.05) = 199.4 W;
X
Pabs = 272 + 17.4 + 199.4 + 5000 + 3750 + 8000 = 17,238.8 W;
X
Pdel = 9389.6 + 7849.2 = 17,238.8 W =
X
Qabs = 2000 + 1500 = 3500 VAR;
X
Qdel = 1980.8 + 1519.2 = 3500 VAR =
P 10.30 [a] So = original load = 1800 + j
Sf = final load = 2400 + j
X
X
Pabs ;
Qabs .
1800
(0.8) = 1800 + j2400 kVA;
0.6
2400
(0.28) = 2400 + j700 kVA;
0.96
.·. Qadded = 700 − 2400 = −1700 kVAR.
[b] Deliver.
[c] Sa = added load = 600 − j1700 = 1802.78/ − 70.56◦ kVA;
pf = cos(−70.56) = 0.3328 leading.
[d] I∗L =
(1800 + j2400) × 103
= 375 + j500 A(rms);
4800
IL = 375 − j500 = 625/ − 53.13◦ A(rms);
|IL | = 625 A(rms).
[e] I∗L =
(2400 + j700) × 103
= 500 + j145.83;
4800
IL = 500 − j145.83 = 520.83/ − 16.26◦ A(rms);
|IL | = 520.83 A(rms).
P 10.31 [a] Pbefore = (625)2 (0.02) = 7812.50 W;
Pafter = (520.83)2 (0.02) = 5425.35 W.
10–25
10–26
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] Vs (before) = 4800 + (375 − j500)(0.02 + j0.16) = 4887.5 + j50
= 4887.5/0.59◦ V(rms);
|Vs (before)| = 4887.76 V(rms);
Vs (after) = 4800 + (500 − j145.83)(0.02 + j0.16)
= 4833.33 + j77.08 = 4833.95/0.91◦ V(rms);
|Vs (after)| = 4833.95 V(rms).
P 10.32 [a] SL = 24 + j7 kVA;
I∗L = 192 + j56 A(rms);
125I∗L = (24 + j7) × 103 ;
.·. IL = 192 − j56 A(rms).
Vs = 125 + (192 − j56)(0.006 + j0.048) = 128.84 + j8.88
= 129.15/3.94◦ V(rms);
|Vs | = 129.15 V(rms).
[b] P` = |I` |2 (0.006) = |IL |2 (0.006) = (200)2 (0.006) = 240 W.
[c]
(125)2
= −7000;
XC
−
1
= −2.23;
ωC
XC = −2.23 Ω;
C=
1
= 1188.36 µF.
(2.23)(120π)
[d] I` = 192 + j0 A(rms);
Vs = 125 + 192(0.006 + j0.048) = 126.152 + j9.216
= 126.49/4.18◦ V(rms);
|Vs | = 126.49 V(rms).
[e] P` = (192)2 (0.006) = 221.184 W.
P 10.33 [a] I =
250/0◦
= 4 − j3 = 5/ − 36.87◦ A(rms);
40 + j30
P = (5)2 (1) = 25 W.
[b] YL =
1
= 17.75 − j11.83 mS;
39 + j26
.·. XC =
1
= −84.5 Ω.
−11.83 × 10−3
Problems
10–27
1
= 56.34 Ω.
17.75 × 10−3
250/0◦
[d] I =
= 4.35/ − 3.99◦ A;
57.34 + j4
[c] ZL =
P = (4.35)2 (1) = 18.92 W.
18.92
(100) = 75.7%.
25
Thus the power loss after the capacitor is added is 75.7% of the power
loss before the capacitor is added.
[e] % =
P 10.34
IL =
240,000 − j70,000
= 96 − j28 A(rms);
2500
IC =
2500
2500
=j
= jIC ;
−jXC
XC
I` = 96 − j28 + jIC = 96 + j(IC − 28);
Vs = 2500 + (1 + j8)[96 + j(IC − 28)]
= (2820 − 8IC ) + j(740 + IC );
|Vs |2 = (2820 − 8IC )2 + (740 + IC )2 = (2500)2 ;
.·. 65IC2 − 43,640IC + 2,250,000 = 0.
IC =
43,640 ±
q
(43,640)2 − 4(65)(2,250,000)
2(65)
= 335.69 ± 279.42 = 56.27 A(rms)∗ .
*Select the smaller value of IC to minimize the magnitude of I` .
2500
.·. XC =
= 44.43;
56.27
.·. C =
1
= 59.7 µF.
(44.43)(120π)
10–28
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.35 [a] Using the results of Problem 9.78,
I1 =
125/0◦
125
=
= 1 − j2 A(rms);
20 + j34 + 5 + j16
25 + j50
jωM
j50
I1 =
(1 − j2)
Z22
200 + j150
= 0.44 − j0.08 = 0.45/ − 10.30◦ A(rms);
I2 =
VL = (150 − j100)(0.44 − j0.08) = 58 − j56
= 80.62/ − 43.99◦ V(rms);
|VL | = 80.62 V(rms).
[b] Pg (ideal) = 125(1) = 125 W;
Pg (practical) = 125 − |I1 |2 (5) = 125 − 25 = 100 W;
PL = |I2 |2 (150) = 30 W;
% delivered =
30
(100) = 30%.
100
P 10.36 [a]
340/0◦ = 10Ig + j50Ig + j70(Ig − I2 ) − j30I2
+j70Ig − j40I2 + j100(Ig − I2 );
0 = j100(I2 − Ig ) − j70Ig + j40I2 + j20I2
+j40(I2 − Ig ) − j30Ig + 40I2 ;
Solving,
Ig = 5 − j1 A(rms);
I2 = 6/0◦ A(rms);
P40Ω = (6)2 (40) = 1440 W.
[b] Pg (developed) = (340)(5) = 1700 W.
Vg
340
[c] Zab =
− 10 =
− 10 = 55.38 + j13.08 = 56.91/13.28◦ Ω.
Ig
5−j
Problems
[d] P10Ω = |Ig |2 (10) = 260 W;
X
Pdiss = 1440 + 260 = 1700 W =
X
Pdev .
P 10.37 [a]
20 = j2(I1 − I2 ) + j1(I2 − I3 ) − j1(I1 − I3 );
0 = 1I2 + j1(I2 − I3 ) + j1(I1 − I2 ) + j2(I2 − I1 ) − j1(I2 − I3 );
0 = −j1(I3 − I1 ) + j1(I3 − I2 ) − j1(I1 − I2 ) + 1I3 .
Solving,
I1 = 20 − j20 A(rms);
I2 = 20 + j0 A(rms);
I3 = 0 A(rms).
Ia = I1 = 20 − j20 A
Ib = I1 − I2 = −j20 A;
Ic = I2 = 20 A
Id = I3 − I2 = −20 A;
Ie = I1 − I3 = 20 − j20 A
If = I3 = 0 A.
[b]
Va = 20 + j0 V
Vb = j2Ib − j1Id = 40 + j20 V;
Vc = 1Ic = 20 + j0 V
Vd = j1Id − j1Ib = −20 − j20 V;
Ve = −j1Ie = −20 − j20 V
Vf = 1If = 0 V;
Sa = −20I∗a = −400 − j400 VA;
Sb = Vb I∗b = −400 + j800 VA;
10–29
10–30
CHAPTER 10. Sinusoidal Steady State Power Calculations
Sc = Vc I∗c = 400 + j0 VA;
Sd = Vd I∗d = 400 + j400 VA;
Se = Ve I∗e = 0 − j800 VA;
Sf = Vf I∗f = 0 + j0 VA.
[c]
X
Pdev = 400 W;
X
Pdiss = −400 + 400 + 400 = 400 W.
Note that the total power dissipated by the coupled coils is zero:
−400 + 400 = 0 = Pb + Pd
[d]
X
Qgen = 400 + 800 = 1200 VAR;
Both the source and the capacitor are generating magnetizing vars.
X
Qdiss = 400 + 800 = 1200 VAR;
X
Q dissipated by the coupled coils is Qb + Qd .
P 10.38 [a]
30 = 3000I1 + V1 + 1000(I1 − I2 );
0 = 1000(I2 − I1 ) − V2 + 2000I2 ;
1
V2 = V1 ;
3
I2 = 3I1 .
Solving,
V1 = 28.8 V(rms);
V2 = 9.6 V(rms);
I1 = 1.2 mA(rms);
I2 = 3.6 mA(rms);
V10mA = V1 + 1000(I1 − I2 ) = 26.4 V(rms);
.·. P = −(26.4)(10 × 10−3 ) = −264 mW.
Thus 264 mW is delivered by the current source to the circuit.
Problems
10–31
[b] I1kΩ = I1 − I2 = −2.4 mA(rms);
.·. P1kΩ = (−0.0024)2 (1000) = 5.76 mW.
P 10.39 [a]
a1 Ig = I1 ;
a2 Ig = I2 ;
P320 = |I1 |2 (320);
.·.
a1
I1
= ;
a2
I2
so
P80 = |I2 |2 (80);
|I2 |2 (80) = 16[|I1 |2 (320)]
P80 = 16P320 ;
thus
a1
I1
1
=
= .
a2
I2
8
The load impedances are matched to the source impedance:
a21 (320) + a22 (80) = 136,000
.·.
[b] Ig =
a21 = 25
so
a21 (320) + (8a1 )2 (80) = 136,000;
so
a1 = 5
and
680/0◦
= 2.5/0◦ mA(rms);
(136 + 136)103
I2 = 40Ig = 100 mA(rms);
.·.
P80Ω = (0.1)2 (80) = 800 mW.
[c] I1 = 5Ig = 12.5/0◦ mA(rms);
V320 = (12.5 × 10−3 )(320) = 4 V(rms).
N1
P 10.40 [a] Zab = 1 +
N2
.·. I1 =
2
(4 − j8) = 36 − j72 Ω;
250/0◦
= 5/36.87◦ A.
4 + j42 + 36 − j72
P4(left) = |I1 |2 (4) = (5)2 (4) = 100 W;
I2 =
N1
I1 = 10/36.87◦ A;
N2
.·. IL = 15/36.87◦ A(rms);
P4(right) = (225)(4) = 900 W.
a2 = 8a1 = 40.
10–32
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] Pg = (250)(5) cos(36.87◦ ) = 1000 W(developed);
X
P 10.41 [a]
Pabs = 100 + 900 = 1000 W =
X
Pdev .
240 − j80 − 480 240 − j80
+
= 0;
ZTh
100
−100(240 + j80)
= 80 + j60 Ω;
.·. ZTh =
−(240 − j80)
.·. ZL = 80 − j60 Ω.
[b] I =
480/0◦
= 3/0◦ A(rms);
160/0◦
P = (9)(80) = 720 W.
[c] Construct the 80 Ω resistor by connecting a 33 Ω resistor and a 47 Ω
resistor in series.
1
C=
= 66.667 µF.
(250)(60)
Construct this capacitance (approximately) by combining three 22 µF
and three 0.22 µF all in parallel to give 66.66 µF.
P 10.42 [a] From Problem 9.77, ZTh = 90 + j120 Ω and VTh = 450/36.87◦ V. Thus, for
∗
maximum power transfer, ZL = ZTh
= 90 − j120 Ω:
I2 =
450/36.87◦
= 2.5/36.87◦ = 2 + j1.5 A.
180
225/0◦ = (15 + j20)I1 − j50(2 + j1.5);
150 + j100
.·. I1 =
= 6.8 − j2.4 A.
15 + j20
Sg (del) = 225(6.8 + j2.4) = 1530 + j540 VA;
Pg = 1530 W.
Problems
10–33
[b] Ploss = |I1 |2 (15) + |I2 |2 (30) = 780 + 187.5 = 967.5 W;
% loss =
967.50
(100) = 63.24%.
1530
P 10.43 ZL = |ZL |/θ◦ = |ZL | cos θ◦ + j|ZL | sin θ◦ .
|VTh |
Thus |I| = q
.
(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
2
the relationship RTh
+ XTh
= |ZL |2 or |ZTh | = |ZL |.
P 10.44 [a] ZTh =
(60,000)(j60,000)
1
+
= −j40,000 + 30,000 + j30,000 = 30 − j10 kΩ;
jωC
60,000 + j60,000
∗
.·. ZL = ZTh
= 30 + j10 kΩ.
[b] VTh =
√
90/0◦ (60)
= 45(1 − j1) = 45 2/ − 45◦ V(rms).
60 + j60
√
√
45 2/ − 45◦
I=
=
0.75
2/ − 45◦ mA(rms);
60 × 103
|Irms | = 0.75 mA(rms);
Pload = (0.75)2 × 10−6 (30 × 103 ) = 16.875 mW.
10–34
CHAPTER 10. Sinusoidal Steady State Power Calculations
[c] The closest resistor value from Appendix H is either 27 kΩ or 33 kΩ. Find
the inductor value:
10,000L = 10,000
so
L = 1 H.
The closest inductor value is 10 mH, which has a reactance of 100 Ω.
Choosing the 27 kΩ resistor, the current in the load is
45/ − 45◦
= 777.83/ − 35.15◦ µA(rms);
57,000 − j9900
Irms =
Pload = (777.83 × 10−6 )2 (27,000) = 16.3 mW
(instead of 16.875 mW).
Choosing the 33 kΩ resistor, the current in the load is
45/ − 45◦
= 705.63/ − 36.07◦ µA(rms);
63,000 − j9900
Irms =
Pload = (705.63 × 10−6 )2 (33,000) = 16.43 mW
P 10.45 [a]
V1
V1 − 250
− 0.2Vσ +
= 0;
5 − j10
10 + j5
Vσ =
−j5V1
−jV1
=
;
10 + j5
2 + j1
−0.2Vσ =
.·. V1
"
j0.2V1
;
2 + j1
#
1
j0.2
1
250
+
+
.
=
5 − j10 2 + j1 10 + j5
5 − j10
Thus, V1 = 10(10 + j5).
VTh =
j5
V1 = j50 = 50/90◦ V(rms).
10 + j5
(instead of 16.875 mW).
Problems
Short circuit current:
Isc =
50
250/0◦
=
A(rms);
15 − j10
3 − j2
ZTh =
VTh
j50
=
(3 − j2) = 2 + j3 Ω;
Isc
50
.·. ZL = 2 − j3 Ω.
IL =
50/90◦
= 12.5/90◦ A(rms);
4
P = (12.5)2 (2) = 312.50 W.
[b] VL = (2 − j3)(j12.5) = 37.5 + j25 V(rms).
I1 =
VL
37.5 + j25
=
= 5 − j7.5 A(rms);
j5
j5
I2 = I1 + IL = 5 − j7.5 + j12.5 = 5 + j5 A(rms);
Vcs = VL + 10I2 = 37.5 + j25 + 50 + j50 = 87.5 + j75 V(rms);
Vσ = −VL = −37.5 − j25;
0.2Vσ = −7.5 − j5;
Scs = −Vcs I∗cs = −(87.5 + j75)(−7.5 + j5) = 1031.25 + j125 VA.
10–35
10–36
CHAPTER 10. Sinusoidal Steady State Power Calculations
Therefore, the dependent source is absorbing 1031.25 W and 125
magnetizing vars. Only the independent voltage source is developing
power.
Ig = −0.2Vσ + I2 = 7.5 + j5 + 5 + j5 = 12.5 + j10 A;
Sg = −250I∗g = −3125 + j2500 VA;
.·. Pdev = 3125 W.
% delivered =
312.5
(100) = 10%.
3125
Thus, 10% of the developed power is delivered to the load.
Checks:
√
P10Ω = (5 2)2 10 = 500 W;
P2Ω = 312.5 W;
√
P5Ω = ( 256.25)2 5 = 1281.25 W;
.·.
X
Pdev =
X
Pabs = 500 + 312.5 + 1281.25 + 1031.25 = 3125 W.
VAR Check:
The 250 V source is absorbing 2500 vars; the dependent current source is
absorbing 125 vars; the j5 Ω inductor is absorbing
|37.5 + j25|2 /5 = 406.25 vars. Thus,
X
Qabs = 2625 + 406.25 = 3031.25 VAR;
X
Qdev = (12.5)2 (3) + 256.25(10) = 3031.25 VAR =
P 10.46 [a] ZTh = 8 + j15 +
X
Qabs .
(−j24)(18 + j6)
= 24 + j7 = 25/16.26◦ Ω;
18 − j18
.·. R = |ZTh | = 25 Ω.
[b] VTh =
I=
√
−j24
(630/0◦ ) = 420 − j420 = 420 2/ − 45◦ V(rms).
18 + j6 − j24
420 − j420
= 12/ − 53.13◦ ;
49 + j7
P = (12)2 (25) = 3600 W = 3.6 kW.
|I| = 12;
Problems
[c] The resistor from Appendix H that is closest to 25 Ω is 22 Ω.
P 10.47 [a]
Vφ − 600 Vφ
+
− 0.05Vφ = 0;
10
j10
.·. Vφ = 240 + j480 V(rms).
VTh = Vφ + 0.05Vφ (−j20) = Vφ (1 − j1) = 720 + j240 V(rms).
Short circuit current:
Isc = 0.05Vφ +
Vφ
= (0.05 + j0.05)Vφ ;
−j20
Vφ − 600 Vφ
Vφ
+
+
= 0;
10
j10 −j20
.·. Vφ = 480 + j240 V(rms).
Isc = (0.05 + j0.05)(480 + j240) = 12 + j36 A(rms);
ZTh =
720 + j240
VTh
=
= 12 − j16 = 20/ − 53.13◦ Ω;
Isc
12 + j36
.·. Ro = 20 Ω.
[b]
I=
√
720 + j240
= 15 + j15 = 15 2/45◦ A(rms);
32 − j16
10–37
10–38
CHAPTER 10. Sinusoidal Steady State Power Calculations
√
P = (15 2)2 (20) = 9000 W = 9 kW.
[c]
720 + j240
= 30 + j10 A(rms);
24
√
P = ( 1000)2 (12) = 12 kW.
I=
[d]
Vφ − 200 − j600
Vφ − 600 Vφ
+
+
= 0;
10
j10
−j20
.·. Vφ = 200 + j200 V(rms).
0.05Vφ = 10 + j10 A(rms);
10 + j10 + IC = 30 + j10;
IL =
.·. IC = 20 + j0 A(rms);
Vφ
= 20 − j20 A(rms);
j10
IR = IC + IL = 40 − j20 A(rms);
Ig = IR + 0.05Vφ = 50 − j10 A(rms);
Sg = −600I∗g = −30,000 − j6000 VA;
600 = Vcs + 200 + j600;
Vcs = 400 − j600 V(rms);
Scs = (400 − j600)(10 − j10) = −2000 − j10,000 VA;
X
Pdev = 30,000 + 2000 = 32,000 W = 32 kW;
% delivered to Zo =
12
(100) = 37.50%.
32
Problems
Check:
X
Pabs = 12,000 + |IR |2 (10) = 32 kW =
X
Qdev = 6000 + 10,000 + |IC |2 (20) = 24 kVAR;
X
Qabs = |IL |2 (10) + |Io |2 (16) = 24 kVAR =
X
Pdev ;
X
Qdev .
1
1
= 240 Ω;
C=
= 11.05 µF.
ωC
(240)(120π)
4800 4800
+
= 30 − j20 A(rms);
[b] Iwo =
160
j240
P 10.48 [a]
Vswo = 4800 + (30 − j20)(1 + j8) = 4990 + j220
= 4994.85/2.52◦ V(rms);
Iw =
4800
4800 4800
+
+
= 30 + j0 A(rms);
160
j240 −j240
Vsw = 4800 + 30(1 + j8) = 4830 + j240 = 4835.96/2.84◦ V(rms);
% increase =
4994.85
− 1 (100) = 3.29%.
4835.96
[c] P`wo = |30 − j20|2 1 = 1300 W;
P`w = 302 (1) = 900 W;
1300
− 1 (100) = 44.44%.
% increase =
900
P 10.49 [a] First find the Thévenin equivalent:
106
1
=
= −j100 Ω;
jωC
j104
ZTh =
300(−j100)
= 30 − j90 Ω;
300 − j100
VTh =
150(−j100)
= 15 − j45 V(rms);
300 − j100
jωL = j104 (6 × 10−3 ) = j60 Ω.
I=
15 − j45
1.5
=
(13 − j9) A(rms);
40 − j30
25
10–39
10–40
CHAPTER 10. Sinusoidal Steady State Power Calculations
|I| =
1.5 √
250 A(rms);
25
P =
2.25
(250)(10) = 9 W.
625
[b] Since ZTh = 30 − j90 Ω and ω = 10,000 rad/s, set Lo as close as possible
to 90/10,000 = 9 mH. Therefore, set Lo = 8 mH.
Set Ro as close as possible to
q
√
Ro = (30)2 + (90 − 80)2 = 1000 = 31.62 Ω;
.·. Ro = 20 Ω.
[c] I =
15 − j45
3 − j9
=
A(rms);
50 − j10
10 − j2
s
.·. |I| =
90
.
104
P = |I|2 (20) =
90
(20) = 17.31 W.
104
Yes;
17.31 W > 9 W.
15 − j45
1 − j3
[d] I =
=
A(rms);
60
4
√ !2
10
30 = 18.75 W.
P =
4
[e] Ro = 30 Ω;
Lo = 9 mH.
[f ] Yes;
18.75 W > 17.31 W.
√
P 10.50 [a] Lo = 8 mH;
Ro = 302 + 102 = 31.62 Ω;
√
15(1 − j3)
15 10
/ − 62.35◦ A(rms);
=
I=
61.62 − j10
62.43
√ !2
15 10
(31.62) = 18.26 W.
P =
62.43
[b] Yes;
18.26 W > 17.31 W.
[c] Yes;
18.26 W < 18.75 W.
Problems
10–41
P 10.51 [a]
660
√ = 230I1 + j50I1 − j40(I2 − I1 ) + j100(I1 − I2 ) + j40I1 ;
2
0 = 100I2 + j100(I2 − I1 ) − j40I1 .
Solving,
I1 = 1.2245 − j0.4928 A(rms);
I2 = 1.2021 + j0.5122 A(rms);
.·. Vo = (1.30667)(100) = 130.667 V(rms).
[b] P = (1.30667)2 (100) = 170.74 W.
!
660
[c] Sg = − √ (1.2245 − j0.4928) = −571.46 + j229.985 VA
2
% delivered =
170.74
(100) = 29.88%.
571.46
P 10.52 [a] Open circuit voltage:
660
√ = 230I1 + j50I1 + j40I1 + j100I1 + j40I1 ;
2
√
660/ 2
·
. . I1 =
= 1.0145 − j1.0145 A(rms).
230 + j230
VTh = j100I1 + j40I1 = j140I1 = 142.04 + j142.04 V(rms).
.·. Pg = −571.46 W;
10–42
CHAPTER 10. Sinusoidal Steady State Power Calculations
Short circuit current:
660
√ = 230I1 + j50I1 − j40(Isc − I1 ) + j100(I1 − Isc ) + j40I1 ;
2
0 = j100(Isc − I1 ) − j40I1 .
Solving,
Isc = 2.78 − j0.411 A;
ZTh =
IL =
VTh
142.04 + j142.04
=
= 42.61 + j57.39 Ω.
Isc
2.78 − j0.411
142.04 + j142.04
= 1.67 + j1.67 = 2.357/45◦ A(rms);
85.22
PL = (2.357)2 (42.61) = 236.72 W.
[b] Using the equivalent circuit,
S = −(142.04 + j142.04)(1.67 − j1.67) = −474.4 + j0 VA;
Pdev = 474.4 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
57.39 =
so
C=
= 46.22 µF.
2π(60)C
2π(60)(57.39)
Choose the capacitor value closest to this capacitance from Appendix H,
which is 47 µF. Then,
XL = −
1
= −56.44 Ω.
2π(60)(47 × 10−6 )
Problems
Now set RL as close as possible to
RL =
q
10–43
q
2
RTh
+ (XL + XTh )2 :
42.612 + (−56.44 + 57.39)2 = 42.62 Ω.
The closest single resistor value from Appendix H is 47 Ω. The resulting
real power developed by the source is calculated below, using the
Thévenin equivalent circuit:
142.04 + j142.04
= 2.24/44.4◦ ;
IL =
42.61 + j57.39 + 47 − j56.44
PL = (2.24)2 (47) = 236 W
(instead of 236.72 W).
P 10.53 [a] VTh = 210 V;
V2 = − 41 V1 ;
Short circuit equations:
I1 = 41 I2 .
840 = 80I1 − 20I2 + V1 ;
0 = 20(I2 − I1 ) + V2 ;
.·. I2 = 14 A(rms);
210
Pmax =
30
RTh RL =
210
= 15 Ω.
14
2
15 = 735 W.
[b] VTh = −4(146/0◦ ) = −584/0◦ V (rms);
I1 = −4I2 .
V2 = 4V1 ;
Short circuit equations:
146/0◦ = 80I1 − 20I2 + V1 ;
0 = 20(I2 − I1 ) + V2 ;
.·. I2 = −146/365 = −0.40 A;
−584
P =
2920
2
1460 = 58.40 W.
P 10.54 [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 ;
RTh =
−584
= 1460 Ω.
−0.4
10–44
CHAPTER 10. Sinusoidal Steady State Power Calculations
0 = j10Ig + (30 + j40)IL .
Solving,
Ig = 4 − j3 A;
IL = −1 A.
Thus,
ig = 5 cos(5000t − 36.87◦ ) A;
iL = 1 cos(5000t − 180◦ ) A.
2
M
= √ = 0.5.
L1 L2
16
[c] When t = 100π µs:
[b] k = √
5000t = (5000)(100π) × 10−6 = 0.5π rad = 90◦ ;
ig (100π µs) = 5 cos(53.13◦ ) = 3 A;
iL (100π µs) = 1 cos(−90◦ ) = 0 A;
1
1
1
w = L1 i21 + L2 i22 + M i1 i2 = (2 × 10−3 )(9) + 0 + 0 = 9 mJ.
2
2
2
When t = 200π µs:
5000t = π rad = 180◦ ;
ig (200π µs) = 5 cos(180 − 36.87◦ ) = −4 A;
iL (200π µs) = 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
[d] From (a), Im = |IL | = 1 A,
1
.·. P = (1)2 (30) = 15 W.
2
√
70
[e] VTh =
(j10) = 35 2/45◦ V;
10 + j10
10
√
ZTh = j40 +
10 2
√
.·. RL = 25 2 Ω.
!2
(10 − j10) = 5 + j35 =
√
1250/81.78◦ Ω;
Problems
[f ]
√
35 2/45◦
√
= 0.93/4.07◦ A;
I=
(5 + 25 2) + j35
√
1
P = (0.93)2 (25 2) = 15.18 W.
2
∗
[g] ZL = ZTh
= 5 − j35 Ω.
√
√
35 2/45◦
[h] I =
= 3.5 2/45◦ ;
10
√
1
P = (3.5 2)2 (5) = 61.25 W.
2
P 10.55
VTh =
80
80
(j20) +
(j20) = 128 + j64 V(rms).
10 + j20
10 + j20
80 = 10I1 + j20(I1 − Isc ) − j20Isc ;
0 = j20(Isc − I1 ) + j20Isc + j40Isc − j20(I1 − Isc ).
Solving,
Isc = 2.76 − j1.10 A;
ZTh =
128 + j64
= 32 + j36 Ω;
2.76 − j1.10
10–45
10–46
CHAPTER 10. Sinusoidal Steady State Power Calculations
128 + j64
.·. IL =
= 2 + j1 A.
64
80 = 10I1 + j20(I1 − I2 ) − j20I2 ;
I2 = 2 + j1 A.
Solving,
I1 = 4/0◦ A;
Zg = 80/4 = 20 + j0 Ω.
P 10.56 [a] VTh =
760/0◦
(j50) = 380/16.26◦ V(rms);
28 + j96
ZTh = 31 + j100 +
50
100
2
(28 − j96) = 38 + j76 Ω;
.·. ZL = 38 − j76 Ω.
IL =
380/16.26◦
= 4.8 + j1.4 = 5/16.26◦ A(rms);
76
PL = |IL |2 (38) = 950 W.
[b]
760/0◦ = I1 (28 + j96) − j50(4.8 + j1.4);
690 + j240
.·. I1 =
= 7.31/ − 54.56◦ = 4.24 − j5.95 A.
100/73.74◦
Sg (delivered) = 760(4.24 + j5.95) = 3219.36 + j4523.52 VA;
Problems
Ploss = |I1 |2 (8) = 426.96 W;
Pin (transformer) = 3219.36 − 426.96 = 2792.40 W;
% delivered to ZL =
950
(100) = 34.02%.
2792.4
P 10.57 [a]
240 = 20I1 + j40(I1 − I2 ) − j60I2 ;
0 = j40(I2 − I1 ) + j60I2 + j160I2 + j60(I2 − I1 ) + 140I2 .
Solving,
I1 = 6.4 − j2.8 A(rms);
I2 = 2/0◦ A(rms);
Vo = 140I2 = 280/0◦ V(rms).
[b] P = |I2 |2 (140) = 560 W.
[c] Pg = (240)(6.4) = 1536 W;
% delivered =
560
(100) = 36.46%.
1536
P 10.58 [a]
VTh =
240/0◦
240/0◦
(j40) +
(j60) = 480 + j240 V(rms).
20 + j40
20 + j40
240 = (20 + j40)I1 − j100Isc ;
10–47
10–48
CHAPTER 10. Sinusoidal Steady State Power Calculations
0 = −j100I1 + j320Isc .
Solving,
Isc = 3.15 − j1.377;
ZTh =
480 + j240
VTh
=
= 100 + j120 = 156.20/50.19◦ Ω;
Isc
3.15 − j1.377
.·. RL = 156.20 Ω.
[b]
I=
536.66/26.57◦
= 1.90/1.47◦ ;
282.92/25.10◦
P = |I|2 (156.20) = 562.05 W.
P 10.59 [a]
240 = 20I1 + j40(I1 − I2 ) + j80kI2 ;
0 = j40(I2 − I1 ) − j80kI2 + j160I2 + j80k(I1 − I2 ) + 140I2 ,
or
12 = (1 + j2)I1 + j(4k − 2)I2 ;
0 = j(4k − 2)I1 + [7 + j(10 − 8k)]I2 ;
Solve the second equation for I1 , use it to eliminate I1 in the first
equation, and solve for I2 to get
I2 =
−j12(4k − 2)
.
(4k − 2)2 + (1 + j2)[7 + j(10 − 8k)]
I2 = 0 when 4k − 2 = 0;
.·. k = 0.5.
Vo = 0 when I2 = 0;
Problems
10–49
[b] When I2 = 0,
I1 =
12
= 2.4 − j4.8 A(rms);
1 + j2
Pg = (240)(2.4) = 576 W.
Check:
Ploss = |I1 |2 (20) = 576 W.
P 10.60 [a] Replace the circuit to the left of the primary winding with a Thévenin
equivalent:
VTh =
0.25/0◦
(j50) = 200 + j100 mV(rms);
25 + j50
ZTh = 20 +
(25)(j50)
= 40 + j10 Ω.
25 + j50
Transfer the secondary impedance to the primary side:
Zp =
1
XC
(160 − jXC ) = 10 − j
Ω.
16
16
Now match the load impedance to the source impedance by setting
(XC /16) = 10 Ω:
.·. C =
10−3
= 125 nF.
(160)(50)
0.2 + j0.1
= 4 + j2 mA(rms);
50
√
|I| = 20 mA(rms);
[b] I =
P = (20 × 10−6 )(10) = 200 µW.
Ro
= 40 Ω;
.·. Ro = 640 Ω.
16
0.2 + j0.1
[d] I =
= 2.5 + j1.25 mA(rms);
80
[c]
P = |I|2 (40) = 312.50 µW.
10–50
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.61
Vo = 4Va ;
4Io = Ia ;
Vb
−Va
=
;
1
2.5
therefore
Ib = −2.5Ia ;
Va
= 250 Ω.
Ia
therefore
Vb
250
=
= 40 Ω.
Ib
6.25
Therefore Ib = [100/(10 + 40)] = 2 A (rms); since the ideal transformers are
lossless, P4kΩ = P40Ω , and the power delivered to the 4 kΩ resistor is 22 (40) or
160 W.
P 10.62 [a]
Ia = −2.5Ib ;
10Ia + V1 = 100;
.·.
10(−2.5Ib ) − V2 /2.5 = 100.
Ib = aIc = 0;
.·.
10[−2.5(0)] − Voc /2.5a = 100;
Ia = −2.5Ib ;
V1 = −V2 /2.5;
10(−2.5Ib ) − V2 /2.5 = 100.
V2 = V3 /a = 0;
.·.
V2 = Voc /a;
Voc = −250a.
10Ia + V1 = 100;
.·.
V1 = −V2 /2.5;
Ib = aIsc ;
Isc = 100/(−2.5a) = −4/a.
10[−2.5(aIsc )] − 0 = 100;
Problems
10–51
Thus,
ZTh =
−250a
Voc
=
= 62.5a2 .
Isc
−4/a
For maximum power to the 4 kΩ load,
4000 = ZTh = 62.5a2 ;
so
a = 8.
[b] The circuit, with everything to the left of the 4 kΩ load resistor replaced
by its Thevenin equivalent:
PL =
V2L
(−1000)2
=
= 250 W.
4000
4000
P 10.63 [a]
1
V2 = V1 ;
4
I2 = 4I1 ;
25 = 5I1 + V1 .
0 = −V2 + 1.25I2 ;
.·. I1 = 1 A;
I2 = 4 A.
25 = (1)I1 + VTh + (1)I2 ;
.·. VTh = 20 V.
25 = (Isc + I1 )(1) + 4I1 + V1 ;
25 = (Isc + I1 )(1) + (Isc + 4I1 )(1);
10–52
CHAPTER 10. Sinusoidal Steady State Power Calculations
V1
= 4I1 (0.25) + (Isc + 4I1 )(1).
4
Solving,
Isc = 20 A;
RTh =
20
VTh
=
= 1 Ω.
Isc
20
P = (10)2 (1) = 100 W.
[b]
25 = (10 + I1 )(1) + 4I1 + V1 ;
V1
= 4I1 (0.25) + (4I1 + 10)(1).
4
Solving,
I1 = −1 A;
.·. Psource = (25)(10 − 1) = 225 W;
% delivered =
100
(100) = 44.44%.
225
[c] Pdev = 25(10 − 1) = 225 W;
P1Ω = (9)2 (1) = 81 W;
P1Ω = (10)2 (1) = 100 W;
P4Ω = (−1)2 (4) = 4 W;
P0.25Ω = (−4)2 (0.25) = 4 W;
P1Ω = (10 − 4)2 (1) = 36 W;
X
Pdiss = 81 + 4 + 100 + 4 + 36 = 225 W =
X
Pdev .
Problems
P 10.64 [a]
For maximum power transfer, Zab = 20 kΩ;
N1
Zab = 1 −
N2
.·.
1−
N1
1−
N2
2
ZL ;
2
=
N1
= ±20;
N2
N1
>0
N2
N2 =
20,000
= 400.
50
N1
= 1 ∓ 20;
N2
N1
.·.
= 21;
N2
2520
N1
=
= 120 turns.
21
21
[b]
P50Ω = P20kΩ =
(120)2
× 10−3 = 720 mW.
20
[c]
V1 + V2 = 120;
V2 = −
V1 −
V1
V2
=− ;
N1
N2
N2
V1
V1 = − ;
N1
21
V1
= 120;
21
.·. V1 = 126 V;
10–53
10–54
CHAPTER 10. Sinusoidal Steady State Power Calculations
.·. V2 = −6 V.
Check the power calculation:
P50Ω =
36
= 0.72 W = 720 mW.
50
[d]
Vcs = 120 + (6)(5) = 150 V;
Pcs (del) = (150)(16) = 2400 mW;
% delivered =
720
(100) = 30%.
2400
20
P 10.65 [a] ZTh = 32 + j124 +
5
2
(3 − j4) = 80 + j60 = 100/36.87◦ Ω;
.·. Zab = 100 Ω.
Zab =
ZL
;
(1 + N1 /N2 )2
(1 + N1 /N2 )2 = 3600/100 = 36;
.·. N1 /N2 = 5
or
N2 = N1 /5;
.·. N2 = 300 turns.
[b] VTh =
240/0◦
(j20) = 960/36.87◦ V.
3 + j4
√
960/36.87◦
= 1.6 10/18.43◦ A(rms);
180 + j60
√
|I| = 1.6 10 A(rms);
I=
P = |I|2 (100) = 2560 W.
Problems
[c]
240/0◦ = (3 + j4)I1 − j20(4.8 + j1.6);
.·. I1 = 40.32 − j21.76 A(rms).
Pgen = (240)(40.32) = 9676.80 W;
Pdiss = 9676.80 − 2560 = 7116.80 W;
% dissipated =
7116.80
(100) = 73.54%.
9676.80
N1
P 10.66 [a] Zab = 30,000 − j10,000 = 1 −
N2
2
ZL ;
1
.·. ZL = (30,000 − j10,000) = 7500 − j2500 Ω.
4
[b]
I1 =
480
× 10−3 = 8/0◦ mA(rms).
60
N1 I1 = −N2 I2 ;
I2 = −3I1 = −24/0◦ mA(rms);
IL = I1 + I2 = −16/0◦ mA(rms);
VL = (7500 − j2500)IL = −120 + j40 = 126.49/161.57◦ V(rms).
10–55
10–56
CHAPTER 10. Sinusoidal Steady State Power Calculations
30[6(44.28) + 18(15.77)]
= 16.49 kWh.
1000
30[6(44.28) + 18(8.9)]
[b]
= 12.78 kWh.
1000
30[6(44.28) + 18(4.42)]
[c]
= 10.36 kWh.
1000
30[6(44.28) + 18(0)]
[d]
= 7.97 kWh.
1000
Note that this is about 48 % of the amount of total power consumed in
part (a).
P 10.67 [a]
30[0.33(1433) + 23.67(3.08)]
= 16.52 kWh.
1000
[b] The standby power consumed in one month by the microwave oven when
in the ready state is
P 10.68 [a]
30[23.67(3.08)]
= 2.187 kWh.
1000
This is (2.187/16.52) ∗ 100 = 13.2% of the total power consumed by the
microwave in one month. Since it is not practical to unplug the
microwave when you are not using it, this is the cost associated with
having a microwave oven.
P 10.69 jωL1 = j(2π)(60)(0.18) = j67.86 Ω;
I=
120
= 1.75/ − 81.62◦ A(rms);
10 + j67.86
P = R1 |I|2 = 10(1.75)2 = 30.6 W.
P 10.70 jωL1 = j(2π)(60)(0.18) = j67.86 Ω;
I=
120
= 1.77/ − 89.93◦ A(rms);
0.08 + j67.86
P = R1 |I|2 = 0.08(1.77)2 = 250.6 mW.
Note that while the current supplied by the voltage source is virtually
identical to that calculated in Problem 10.69, the much smaller value of
transformer resistance results in a much smaller value of real power consumed
by the transformer.
P 10.71 An ideal transformer has no resistance, so consumes no real power. This is one
of the important characteristics of ideal transformers.
Balanced Three-Phase Circuits
Remember that all voltages and currents are rms values!
Assessment Problems
AP 11.1 Make a sketch:
We know VAN and wish to find VCA . To do this, write a KVL equation to
find VCA , and use the known phase angle relationships among the voltages.
VCA = VCN − VAN .
Since VAN , VBN , and VCN form a balanced set, and VAN = 80/50◦ V, and the
phase sequence is positive,
VCN = |VAN |//VAN + 120◦ = 80/50◦ + 120◦ = 80/170◦ V.
Then,
VCA = VCN − VAN = (80/170◦ ) − (80/50◦ ) = 138.56/ − 160◦ V.
11–1
11–2
CHAPTER 11. Balanced Three-Phase Circuits
AP 11.2 Make a sketch:
We know VBN and wish to find VBC . To do this, write a KVL equation to find
VBC , and use the known phase angle relationships among the voltages.
VBC = VBN + VNC = VBN − VCN .
Since VAN , VBN , and VCN form a balanced set, and VBN = 150/ − 30◦ V, and
the phase sequence is negative,
VCN = |VBN |//VBN + 120◦ = 150/ − 30◦ + 120◦ = 150/90◦ V.
Then,
VBC = VBN − VCN = (150/ − 30◦ ) − (150/90◦ ) = 259.81/ − 60◦ V.
AP 11.3 Make a sketch of the a-phase:
[a] Find the a-phase line current from the a-phase circuit:
IaA =
125/0◦
125/0◦
=
20 + j15
0.1 + j0.8 + 19.9 + j14.2
= 4 − j3 = 5/ − 36.87◦ A.
Find the other line currents using the acb phase sequence:
IbB = 5/ − 36.87◦ + 120◦ = 5/83.13◦ A;
IcC = 5/ − 36.87◦ − 120◦ = 5/ − 156.87◦ A.
Problems
11–3
[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.
Find the other line voltages using the acb phase sequence:
Vbc = 216.51/ − 30◦ + 120◦ = 216.51/90◦ V;
Vca = 216.51/ − 30◦ − 120◦ = 216.51/ − 150◦ V.
[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.
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;
VCN = 122.23/ − 1.36◦ − 120◦ = 122.23/ − 121.36◦ V.
[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.
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;
VCA = 211.72/ − 31.36◦ − 120◦ = 211.72/ − 151.36◦ V.
√
√
AP 11.4 IaA = ( 3/ − 30◦ )IAB = ( 3/ − 30◦ ) · 25/40◦ = 43.3/10◦ A.
AP 11.5 IbB = 10/(−30◦ − 120◦ ) = 10/ − 150◦ ;
"
!
#
!
/ − 30◦
1
◦
√ / − 30 IbB =
√
IBC =
· 10/ − 150◦
3
3
= 5.77/ − 180◦ A.
"
AP 11.6 [a] ICA =
!
#
1
√ / − 90◦ [103.92/40◦ ] = 60/70◦ A.
3
Therefore Zφ =
2400/0◦
= 40/ − 70◦ Ω.
60/70◦
11–4
CHAPTER 11. Balanced Three-Phase Circuits
"
[b] ICA =
!
#
1
√ / − 30◦ [103.92/40◦ ] = 60/ − 10◦ A.
3
Therefore Zφ = 40/ − 10◦ Ω.
240
240
+
= 80 + j60 = 100/36.87◦ A.
3
−j4
√
√
Therefore |IaA | = 3Iφ = 3(100) = 173.21 A.
AP 11.7 Iφ =
AP 11.8 [a] S1 = 10,200(0.87) + j10,200(0.493) = 8874 + j5029.13 VA;
S2 = 4200 + j1913.6 VA;
√
3VL IL sin θ3 = 7250;
Therefore
sin θ3 = √
7250
= 0.517.
3(220)(36.8)
cos θ3 = 0.856.
Therefore
7250
P3 =
× 0.856 = 12,003.9 W;
0.517
S3 = 12,003.9 + j7250 VA;
ST = S1 + S2 + S3 = 25.078 + j14.192 kVA;
Thus,
PT = 25,078 W.
1
[b] ST /φ = ST = 8359.3 + j4730.7 VA;
3
220 ∗
√ IaA = (8359.3 + j4730.7);
3
I∗aA = 65.81 + j37.24 A;
IaA = 65.81 − j37.24 = 75.62/ − 29.51◦ A,
so
[c] pf = cos(0◦ − 29.51◦ ) = 0.87 lagging.
AP 11.9 [a]
1
Ss/φ = (60)(0.96 − j0.28) × 103 = 19.2 − j5.6 kVA;
3
|IaA | = 75.62 A.
Problems
1
S1/φ = (45) = 15 + j0 kVA;
3
S2/φ = Ss/φ − S1/φ = 4.2 − j5.6 kVA;
4200 − j5600
√
.·. I∗2 =
= 11.547 − j15.396 A.
630/ 3
I2 = 11.547 + j15.396 A;
√
630/ 3
VAN
=
= 11.34 − j15.12 Ω;
Zy =
I2
11.547 + j15.396
Z∆ = 3Zy = 34.02 − j45.36 Ω.
√
(630/ 3)2
[b] R =
= 31.5 Ω;
R∆ = 3R = 94.5 Ω;
4200
√
(630/ 3)2
= −23.625 Ω;
X∆ = 3XL = −70.875 Ω.
XL =
−5600
11–5
11–6
CHAPTER 11. Balanced Three-Phase Circuits
Problems
P 11.1
[a] First, convert the cosine waveforms to phasors:
Va = 120/54◦ ;
Vb = 120/ − 66◦ ;
Vc = 120/174◦ .
Subtract the phase angle of the a-phase from all phase angles:
/V0a = 54◦ − 54◦ = 0◦ ;
/V0b = −66◦ − 54◦ = −120◦ ;
/V0c = 174◦ − 54◦ = 120◦ .
Compare the result to Eqs. 11.1 and 11.2:
Therefore abc.
[b] First, convert the cosine waveforms to phasors:
Va = 3240/ − 26◦ ;
Vb = 3240/94◦ ;
Vc = 3240/ − 146◦ .
Subtract the phase angle of the a-phase from all phase angles:
/V0a = −26◦ + 26◦ = 0◦ ;
/V0b = 94◦ + 26◦ = 120◦ ;
/V0c = −146◦ + 26◦ = −120◦ .
Compare the result to Eqs. 11.1 and 11.2:
Therefore acb.
P 11.2
[a] Va = 339/0◦ V;
Vb = 339/ − 120◦ V;
Vc = 339/120◦ V.
Balanced, positive phase sequence
[b] Va = 622/0◦ V;
Vb = 622/ − 240◦ V = 622/120◦ V;
Vc = 622/240◦ V = 622/ − 120◦ V.
Balanced, negative phase sequence
[c] Va = 933/ − 90◦ V;
Vb = 933/150◦ V;
Vc = 933/30◦ V.
Balanced, positive phase sequence
Problems
11–7
[d] Va = 170/ − 30◦ V;
Vb = 170/90◦ V;
Vc = 170/ − 150◦ V.
Balanced, negative phase sequence
[e] Unbalanced, due to unequal amplitudes.
[f ] Unbalanced, due to unequal phase angle separation.
P 11.3
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 11.4
I=
339/0◦ + 339/ − 120◦ + 339/120◦
= 0.
3(2 + j4)
P 11.5
I=
3394/70◦ + 3394/ − 140◦ + 3394/180◦
4947.07/168.22◦
=
= 322.746/156.91◦ A.
3(5 + j1)
15 + j3
P 11.6
[a] IaA =
240/0◦
= 2.4/ − 36.87◦ A;
80 + j60
IbB =
240/120◦
= 2.4/83.13◦ A;
80 + j60
IcC =
240/ − 120◦
= 2.4/ − 156.87◦ A;
80 + j60
Io = IaA + IbB + IcC = 0.
[b] VAN = (79 + j55)IaA = (79 + j55)(2.4/ − 36.87◦ ) = 231.0/ − 2.02◦ V.
[c] VBN = (79 + j52)IbB = 226.99/116.48◦ V;
.·.
VAB = VAN − VBN = 393.6/ − 32.5◦ V.
[d] Unbalanced.
11–8
P 11.7
CHAPTER 11. Balanced Three-Phase Circuits
Zga + Zla + ZLa = 80 + j60 Ω;
Zgb + Zlb + ZLb = 40 + j30Ω;
Zgc + Zlc + ZLc = 160 + j120Ω;
VN − 480 VN − 480/ − 120◦ VN − 480/120◦ VN
+
+
+
= 0.
80 + j60
40 + j30
160 + j120
20
Solving for VN yields
VN = 78.61/ − 122.69◦ V;
Io =
P 11.8
P 11.9
VN
= 3.93/ − 122.69◦ A.
20
[a] The circuit is unbalanced, because the impedance in each phase of the
load is not the same.
240/0◦
= 2.4 − j7.2 A;
[b] IaA =
10 + j30
IbB =
240/120◦
= 2.2 + j8.2 A;
20 + j20
IcC =
240/ − 120◦
= 2.96 − j4.48 A;
20 − j40
Io = IaA + IbB + IcC = 7.55 − j3.48 = 8.32/ − 24.75◦ A.
√
[a] Van = 1/ 3/ − 30◦ Vab = 120/ − 30◦ V(rms).
The a-phase circuit is
120/ − 30◦
= 2.4/ − 83.13◦ A(rms).
30 + j40
[c] VAN = (28 + j37)IaA = 111.36/ − 30.25◦ V(rms);
√
VAB = 3/30◦ VAN = 192.88/ − 0.25◦ A(rms).
[b] IaA =
Problems
11–9
P 11.10 Sketch the a-phase circuit:
[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◦
11–10
CHAPTER 11. Balanced Three-Phase Circuits
so
Vab = 4275.02/ − 28.38◦ V;
Vbc = 4275.02/91.62◦ V;
Vca = 4275.02/ − 148.38◦ V.
[c] Using KVL on the a-phase circuit
Va0 n = 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
Va0 n = 2482.05/1.93◦ V;
Vb0 n = 2482.05/121.93◦ V;
Vc0 n = 2482.05/ − 118.07◦ V.
P 11.11 VAN = 7967/0◦ V;
VBN = 7967/ + 120◦ V;
VCN = 7967/ − 120◦ V.
VAB = VAN − VBN = 13,799.25/ − 30◦ V;
VBC = VBN − VCN = 13,799.25/90◦ V;
VCA = VCN − VAN = 13,799.25/ − 150◦ V.
vAB = 13,799.25 cos(ωt − 30◦ ) V;
vBC = 13,799.25 cos(ωt + 90◦ ) V;
vCA = 13,799.25 cos(ωt − 150◦ ) V.
Problems
P 11.12 [a]
IaA = √
12,800
= 32.84/ − 16.26◦ A(rms);
3(216 + j63)
|IaA | = |IL | = 32.84 A(rms).
12,800
√
+ (32.84/ − 16.26◦ )(0.25 + j2) = 7416.61/0.47◦ ;
3
√
|Vab | = 3(7416.61) = 12,845.94 V(rms).
[b] Van =
P 11.13 [a] Van = Vbn (1/120◦ ) = 20/ − 210◦ = 20/150◦ V(rms);
Zy = Z∆ /3 = 39 − j33 Ω.
The a-phase circuit is
IaA =
20/150◦
= 0.4/ − 173.13◦ A(rms);
40 − j30
VAN = (39 − j33)IaA = 20.44/146.63◦ V(rms);
√
VAB = 3/ − 30◦ VAN = 35.39/116.63◦ A(rms).
1
[b] IAB = √ / − 30◦ IaA = 0.23/156.87◦ A(rms).
3
[c] VAB = (117 − j99)IAB = 35.39/116.63◦ V(rms).
P 11.14 [a] IAB =
33,000
= 88/ − 16.26◦ A;
360 + j105
IBC = 88/ − 136.26◦ A;
ICA = 88/103.74◦ A.
11–11
11–12
CHAPTER 11. Balanced Three-Phase Circuits
[b] IaA =
√
3/ − 30◦ IAB = 152.42/ − 46.26◦ A;
IbB = 152.42/ − 166.26◦ A;
IcC = 152.42/73.74◦ A.
[c]
Van = 19,052.56/ − 30◦ + (0.1 + j1.0)(152.42/ − 46.26◦ )
= 19,110.40/ − 29.57◦ V.
Vab =
√
3/30◦ Van = 33,100.18/0.43◦ V;
Vbc = 33,100.18/ − 119.57◦ V;
Vca = 33,100.18/120.43◦ V.
P 11.15 [a]
IAN =
7500/0◦
= 75/16.26◦ A;
96 − j28
|IAN | = 75 A.
√
7500 3/30◦
[b] IAB =
= 86.60/13.74◦ A;
144 + j42
|IAB | = 86.60 A.
[c] IaA =
7500
7500
+
= 217.02/ − 5.55◦ A;
96 − j28 48 + j14
|IaA | = 217.02 A.
Problems
[d] Van = (216 − j21)(j1.5) + 7500/0◦ = 7538.47/2.46◦ V;
√
|Vab | = 3(7538.47) = 13,057.01 V.
P 11.16 Zy = Z∆ /3 = 4 + j3 Ω.
The a-phase circuit is
IaA =
120/80◦
= 18.74/41.34◦ A(rms);
(1 + j1) + (4 + j3)
1
IAB = √ /30◦ IaA = 10.82/71.34◦ A(rms).
3
√
110
P 11.17 Van = 1/ 3/ − 30◦ Vab = √ /90◦ V;
3
Zy = Z∆ /3 = 4 − j5 Ω.
The a-phase circuit is
Zeq = (10 + j8)k(4 − j5) = 5.2 − j2.4 Ω;
5.2 − j2.4
VAN =
(1.8 + j0.4) + (5.2 − j2.4)
VAB =
√
!
110
√ /90◦ = 49.96/ − 98.83◦ V;
3
3/30◦ VAN = 86.53/ − 68.83◦ V.
P 11.18 [a] IAB =
720/0◦
= 144/ − 16.26◦ A;
4.8 + j1.4
IBC =
720/ − 120◦
= 36/ − 83.13◦ A;
16 − j12
ICA =
720/120◦
= 20.36/75◦ A.
25 + j25
11–13
11–14
CHAPTER 11. Balanced Three-Phase Circuits
[b] IaA = IAB − ICA
= 138.24 − j40.32 − 5.27 − j19.67
= 132.97 − j59.99 = 145.88/ − 24.28◦ A.
IbB = IBC − IAB
= 4.31 − j35.74 − 138.24 + j40.32
= −133.93 + j4.58 = 134.01/178.04◦ A.
IcC = ICA − IBC
= 5.27 + j19.67 − 4.31 + j35.74
= 0.96 + j55.41 = 55.42/89.01◦ A.
P 11.19 [a]
[b] IaA = √
34,500
= 19.92/16.26◦ A;
3(960 − j280)
|IaA | = 19.92 A.
[c] VAN = (959 − j288)(19.92/16.26◦ ) = 19,944.71/ − 0.46◦ V;
√
|VAB | = 3|VAN | = 34,545.25 V.
[d] Van = (959.8 − j281.6)(19.92/16.26◦ ) = 19,923.71/ − 0.09◦ V;
√
|Vab | = 3|Van | = 34,508.88 V.
|IaA |
[e] |IAB | = √ = 11.50 A.
3
[f ] |Iba | = |IAB | = 11.50 A.
P 11.20 [a] Since the phase sequence is acb (negative) we have:
Van = 7200/30◦ V;
Vbn = 7200/150◦ V;
Vcn = 7200/ − 90◦ V;
Problems
1
ZY = Z∆ = 1.8 + j9.0 Ω/φ.
3
√
[b] Vab = 7200/30◦ − 7200/150◦ = 7200 3/0◦ V.
Since the phase sequence is negative, it follows that
√
Vbc = 7200 3/120◦ V;
√
Vca = 7200 3/ − 120◦ V.
[c]
√
7200 3
Iba =
= 452.91/ − 78.69◦ A;
5.4 + j27
√
7200 3/ − 120◦
Iac =
= 452.91/ − 198.69◦ A;
5.4 + j27
IaA = Iba − Iac = 784.46/ − 48.69◦ A.
11–15
11–16
CHAPTER 11. Balanced Three-Phase Circuits
Since we have a balanced three-phase circuit and a negative phase
sequence we have:
IbB = 784.46/71.31◦ A;
IcC = 784.46/ − 168.69◦ A.
[d]
IaA =
7200/30◦
= 784.46/ − 48.69◦ A.
1.8 + j9
Since we have a balanced three-phase circuit and a negative phase
sequence we have:
IbB = 784.46/71.31◦ A;
IcC = 784.46/ − 168.69◦ A.
P 11.21 [a]
[b] IaA =
7200/30◦
= 7.2/13.74◦ A;
960 + j280
VAN = (957 + j259)(7.2/13.74◦ ) = 7138.28/28.88◦ V;
√
|VAB | = 3(7138.28) = 12,363.87 V.
Problems
11–17
7.2
[c] |Iba | = √ = 4.16 A.
3
[d] Van = (958.2 + j271)(7.20/13.74◦ ) = 7169.65/29.54◦ V;
√
|Vab | = 3(7169.65) = 12,418.20 V.
P 11.22 [a] IAB =
69,000/0◦
= 92/ − 36.87◦ A;
600 + j450
IBC = 92/ − 156.87◦ A;
ICA = 92/83.13◦ A.
√
[b] IaA = 3/ − 30◦ IAB = 159.35/ − 66.87◦ A;
IbB = 159.35/ − 186.87◦ A;
IcC = 159.35/53.13◦ A.
[c] Iba = IAB = 92/ − 36.87◦ A;
Icb = IBC = 92/ − 156.87◦ A;
Iac = ICA = 92/83.13◦ A.
P 11.23 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 = van iaA = [Vm cos ωt][Im cos(ωt − θφ )];
pb = vbn ibB = [Vm cos(ωt − 120◦ )][Im cos(ωt − θφ − 120◦ )];
pc = vcn icC = [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 .
11–18
CHAPTER 11. Balanced Three-Phase Circuits
P 11.24 The a-phase of the circuit is shown below:
I1 =
250/ − 60◦
= 10/ − 113.13◦ A (rms);
15 + j20
I∗2 =
500/45◦
= 2/105◦ A (rms);
◦
/
250 − 60
I = I1 + I2 = 10/ − 113.13◦ + 2/ − 105◦ = 11.98/ − 111.78◦ A (rms);
Sa = VI∗ = (250/ − 60◦ )(11.98/111.78◦ ) = 2995/51.78◦ VA;
ST = 3Sa = 8985/51.78◦ VA.
P 11.25 [a] S1 = 72 − j21 kVA;
S2 = 120 + j90 kVA;
S3 = 168 + j36 kVA;
ST = S1 + S2 + S3 = 360 + j105 kVA;
ST /φ = 120 + j35 kVA.
Single phase equivalent circuit:
120,000 + j35,000
.·. I∗aA =
= 48 + j14;
2500
.·. IaA = 48 − j14 A = 50/ − 16.26◦ A;
Van = 2500 + (1 + j5)(48 − j14) = 2618 + j226
= 2627.74/4.93◦ V;
√
.·. |Vab | = 3(2627.74) = 4551.4 V.
Problems
[b] PL /φ = 120 kW;
PS /φ = 120,000 + |IaA |2 (1) = 122,500 W = 122.5 kW;
120
η=
100 = 97.96%.
122.5
P 11.26 [a] ST ∆ = 24,000/53.13◦ − 12,000/36.87◦ = 12,924.37/68.2◦ VA;
S∆ = ST ∆ /3 = 4308.12/68.2◦ VA.
4000/36.87◦
= 100 V (rms);
40/60◦
√
√
|Vline | = |Vab | = 3|Van | = 100 3 = 173.205 V (rms).
[b] |Van | =
P 11.27
(48 + j36)103
∗
IaA =
= 100 + j75;
480
IaA = 100 − j75 A;
Van = 480 + j0 + (100 − j75)(0.2 + j0.4) = 530 + j25 V.
Ic =
530 + j25
= −5 + j106 A;
−j5
Ina = IaA + Ic = 95 + j31 = 99.93/18.07◦ A.
[b] Sg/φ = (530 + j25)(95 − j31) = 51,125 − j14,055 VA;
SgT = 3Sg/φ = 153,375 − j42,165 VA.
Therefore, the source is delivering 153,375 W and absorbing 42,165 vars.
[c] Pdel = 153,375 W.
Pabs = 3(48,000) + 3|IaA |2 (0.2) = 144,000 + 9375
= 153,375 W = Pdel .
11–19
11–20
CHAPTER 11. Balanced Three-Phase Circuits
[d] Qdel = 3|Ic |2 (5) = 168,915 VAR;
Qabs = 3(36,000) + 42,165 + 3|IaA |2 (0.4)
= 168,915 VAR = Qdel .
P 11.28 From the solution to Problem 11.18 we have:
SAB = (720/0◦ )(144/16.26◦ ) = 99,532.9 + j29,030.04 VA;
SBC = (720/ − 120◦ )(36/83.13◦ ) = 20,735.97 − j15,552.04 VA;
SCA = (720/120◦ )(20.36/ − 75◦ ) = 10,365.62 + j10,365.62 VA.
P 11.29 [a]
IaA =
14,000/0◦
= 8/ − 16.26◦ A;
1680 + j490
IaA
ICA = √ /150◦ = 4.62/133.74◦ A.
3
[b] Sg/φ = −14,000I∗aA = −107,520 − j31,360 VA;
.·. Pdeveloped/phase = 107.52 kW;
Pabsorbed/phase = |IaA |2 1677 = 107.328 kW;
% delivered =
P 11.30 [a] |S| =
Q=
[b] pf =
√
107.328
(100) = 99.82%.
107.52
3(480)(400) = 332,553.755 VA;
q
(332,553.755)2 − (300,000)2 = 143,499.13 VAR.
300,000
= 0.9021 lagging.
332,553.755
Problems
11–21
P 11.31 The complex power of the source per phase is
Ss = 30,000/( cos−1 0.8) = 30,000/36.87◦ = 24,000 + j18,000 kVA. This
complex power per phase must equal the sum of the per-phase power of the
two loads:
Ss = S1 + S2
so
24,000 + j18,000 = 20,000 + S2 ;
.·. S2 = 4000 + j18,000 VA.
|Vrms |2
.
Z2∗
Also,
S2 =
|Vrms | =
415.69
|Vload |
√
= √
= 240 V(rms).
3
3
Thus,
Z2∗ =
(240)2
|Vrms |2
=
= 0.68 − j3.05 Ω;
S2
4000 + j18,000
.·. Z2 = 0.68 + j3.05 Ω.
P 11.32 [a]
7200/0◦
I1 =
= 21.6 − j7.2 A;
300 + j100
I2 =
7200/0◦
= 3.2 + j1.6 A;
1800 − j900
I∗3 =
37,440 + j31,680
= 5.2 + j4.4;
7200
I3 = 5.2 − j4.4 A.
IaA = I1 + I2 + I3 = 30 − j10 A =
√
1000/ − 18.43◦ A.
Van = 7200 + j0 + (30 − j10)(1 + j10) = 7330 + j290 V;
Sφ = Van I∗aA = (7330 + j290)(30 + j10) = 217,000 + j82,000 VA;
ST = 3Sφ = 651 + j246 kVA.
11–22
CHAPTER 11. Balanced Three-Phase Circuits
[b] S1/φ = 7200(21.6 + j7.2) = 155.52 + j51.84 kVA;
S2φ = 7200(3.2 − j1.6) = 23.04 − j11.52 kVA;
S3φ = 37.44 + j31.68 kVA;
Sφ (load) = 216 + j72 kVA.
216
% delivered =
(100) = 99.54%.
217
P 11.33 [a] POUT = 746 × 200 = 149,200 W;
PIN = 149,200/(0.96) = 155,416.67 W;
√
3VL IL cos θ = 155,416.67;
155,416.67
IL = √
= 468.91 A.
3(208)(0.92)
√
√
[b] Q = 3VL IL sin φ = 3(208)(468.91)(0.39) = 66,207.79 VAR.
!
2450
/0◦ V;
P 11.34 [a] VAN = √
VAN I∗aA = Sφ = 144 + j192 kVA.
3
Therefore
(144 + j192)1000
√
I∗aA =
= (101.8 + j135.7) A;
2450/ 3
IaA = 101.8 − j135.7 = 169.67/ − 53.13◦ A;
|IaA | = 169.67 A.
[b] P =
(2450)2
;
R
therefore R =
(2450)2
;
X
therefore X =
Q=
(2450)2
= 41.68 Ω.
144,000
(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.
P 11.35 [a] From Problem 11.34, IaA = (101.8 − j135.7) A.
Therefore Icap = j135.7 A
Thus CY =
and
√
2450/ 3
ZCY =
= −j10.42 Ω.
j135.7
1
= 254.5 µF.
(10.42)(2π)(60)
ZC∆ = (−j10.42)(3) = −j31.26 Ω;
Therefore C∆ =
254.5
= 84.84 µF.
3
Problems
[b] CY = 254.5 µF.
[c] |IaA | = 101.8 A.
1
P 11.36 Sg/φ = (78)(0.8 − j0.6) × 103 = 20,800 − j15,600 VA;
3
I∗aA =
20,800 − j15,600
= 100 − j75 A;
208
IaA = 100 + j75 A.
VAN = 208 − (100 + j75)(0.04 + j0.20)
= 219 − j23 = 220.20/ − 6◦ V;
|VAB | =
√
3(220.20) = 381.41 V.
[b] SL/φ = (219 − j23)(100 − j75) = 20,175 − j18,725 VA;
SL = 3SL/φ = 60,525 − j56,175 VA.
Check:
Sg = 3(20,800 − j15,600) = 62,400 − j46,800 VA;
P` = 3|IaA |2 (0.04) = 1875 W;
Pg = PL + P` = 60,525 + 1875 = 62,400 W (checks);
Q` = 3|IaA |2 (0.20) = 9375 VAR;
Qg = QL + Q` = −56,175 + 9375 = −46,800 VAR (checks).
11–23
11–24
CHAPTER 11. Balanced Three-Phase Circuits
P 11.37 |Iline | =
1200
√ = 10 A(rms);
208/ 3
√
|V |
208/ 3
|Zy | =
=
= 12;
|I|
10
Zy = 12/25◦ Ω;
Z∆ = 3Zy = 36/25◦ = 32.63 + j15.21 Ω/φ.
P 11.38
7200I∗1 = (230.4 − j67.2)103 ;
I∗1 = 32 − j9.33 A;
I1 = 32 + j9.33 A.
1
Zy = Z∆ = 207.36 + j60.48 Ω;
3
I2 =
7200/0◦
= 32 − j9.33 A;
207.36 + j60.48
.·. IaA = I1 + I2 = 64 + j0 A.
Van = 7200 + j0 + 64(0.5 + j4) = 7236.53/2.03◦ V;
|Vab | =
√
3|Van | = 12,534.04 V.
Problems
P 11.39 Assume a ∆-connect load (series):
1
Sφ = (190.44 × 103 )(0.8 − j0.6) = 50,784 − j38,088 VA;
3
∗
Z∆φ
=
|13,800|2
= 3000/36.87◦ Ω;
50,784 − j38,088
Z∆φ = 3000/ − 36.87◦ = 2400 − j1800 Ω.
Now assume a Y-connected load (series):
1
ZY φ = Z∆φ = 800 − j600 Ω.
3
Now assume a ∆-connected load (parallel):
Pφ =
|13,800|2
;
R∆
R∆φ =
Qφ =
|13,800|2
= 3750 Ω;
50,784
|13,800|2
;
X∆
|13,800|2
X∆ φ =
= −5000 Ω.
−38,088
11–25
11–26
CHAPTER 11. Balanced Three-Phase Circuits
Now assume a Y-connected load (parallel):
1
RY φ = R∆φ = 1250 Ω;
3
1
XY φ = X∆φ = −1666.67 Ω.
3
P 11.40 [a]
1
1188
(0.8) 103 = 396,000 + j528,000 VA;
1188 + j
3
0.6
SL/φ =
I∗aA =
396,000 + j528,000
= 60 + j80 A;
6600
IaA = 60 − j80 A;
Van = 6600 + (60 − j80)(0.5 + j4)
= 6950 + j200 = 6952.88/1.65◦ V.
|Vab | =
√
3(6952.88) = 12,042.74 V.
Problems
[b]
I1 = 60 − j80 A
(from part [a]);
1
S2 = 0 − j (1920) × 103 = −j640,000 VAR;
3
I∗2 =
−j640,000
= −j96.97 A;
6600
.·. I2 = j96.97 A.
IaA = 60 − j80 + j96.97 = 60 + j16.97 A;
Van = 6600 + (60 + j16.97)(0.5 + j4)
= 6562.12 + j248.485 = 6566.82/2.17◦ V.
|Vab | =
√
3(6566.82) = 11,374.07 V.
[c] |IaA | = 100 A;
Ploss/φ = (100)2 (0.5) = 5000 W;
Pg/φ = 396,000 + 5000 = 401 kW;
%η =
396
(100) = 98.75%.
401
[d] |IaA | = 62.354 A;
P`/φ = (3887.98)(0.5) = 1943.99 W;
%η =
396,000
(100) = 99.51%.
397,944
[e] Zcap/Y = −j
6600
= −j68.062 Ω;
96.97
Zcap/∆ = 3Zcap/Y = −j204.187 Ω;
.·.
1
= 204.187;
ωC
C=
1
= 12.99 µF.
(204.187)(120π)
11–27
11–28
CHAPTER 11. Balanced Three-Phase Circuits
P 11.41 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◦ .
P 11.42 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.
Problems
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;
(checks).
Zφ = (8 − j6) Ω;
√
QT = 3[979.75 − 2476.25] = −2592 VAR;
QT = 3(12)2 (−6) = −2592 VAR;
(checks).
√
Zφ = 5(1 + j 3) Ω;
√
QT = 3[2160 − 0] = 3741.23 VAR;
√
QT = 3(12)2 (5 3) = 3741.23 VAR;
(checks).
Zφ = 10/75◦ Ω;
√
QT = 3[−645.53 − 1763.63] = −4172.80 VAR;
QT = 3(12)2 [−10 sin 75◦ ] = −4172.80 VAR;
√
3(W2 − W1 )
873,290.66
P 11.44 tan φ =
=
= 1.1918;
W1 + W2
732,777.88
(checks).
.·. φ = 50◦ ;
√
.·. 7600 3|IL | cos 80◦ = 114,291.64.
|IL | = 50 A;
|Z| =
7600
= 152 Ω
50
.·. Z = 152/50◦ Ω.
P 11.45 From the solution to Prob. 11.18 we have
IaA = 145.88/ − 24.28◦ A
and
IbB = 134.01/178.04◦ A.
11–29
11–30
CHAPTER 11. Balanced Three-Phase Circuits
[a] W1 = |Vac | |IaA | cos(θac − θaA )
= 720(145.88) cos(−60◦ + 24.28◦ ) = 85,274.70 W.
[b] W2 = |Vbc | |IbB | cos(θbc − θbB )
= 720(134.01) cos(−120◦ − 178.04◦ ) = 45,357.50 W.
[c] W1 + W2 = 130,632 W;
PAB = (144)2 (4.8) = 99,532.8 W;
PBC = (36)2 (16) = 20,736 W;
PCA = (20.36)2 (25) = 10,363.2 W;
PAB + PBC + PCA = 130,632;
therefore W1 + W2 = Ptotal .
P 11.46 [a] Zφ = 600 + j450 = 750/36.87◦ Ω;
Sφ =
(69 × 103 )2
= 5,078,400 + j3,808,800 VA;
750/ − 36.87◦
ST = 3Sφ = 15,235,200 + j11,426,400 VA.
√
[b] Wm1 = (69,000) 3(92) cos(0 + 66.87◦ ) = 4,318,082.44 W;
√
Wm2 = (69,000) 3(92) cos(60 − 53.13◦ ) = 10,916,117.56 W.
Check:
PT = 15,235,200 W = Wm1 + Wm2 .
P 11.47 [a] W1 = |VBA ||IbB | cos θ.
Positive phase sequence, using the equivalent Y-connected load
impedances:
√
VBA = 480 3/ − 150◦ V;
IaA =
480/0◦
= 24/ − 30◦ A;
20/30◦
IbB = 24/ − 150◦ A;
√
W1 = (24)(480) 3 cos 0◦ = 19,953.23 W;
W2 = |VCA ||IcC | cos θ.
√
VCA = 480 3/150◦ V;
IcC = 24/90◦ A;
√
W2 = (24)(480) 3 cos 60◦ = 9976.61 W.
Problems
√
[b] Pφ = (24)2 (20) cos 30◦ = 5760 3 W;
√
PT = 3Pφ = 17,280 3 W;
√
√
√
W1 + W2 = 11,520 3 + 5760 3 = 17,280 3 W;
.·. W1 + W2 = PT
P 11.48 IaA =
(checks).
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 11.49 [a] Z = 96 + j72 = 120/36.87◦ Ω;
.·. IaA = 6/ − 36.87◦ A;
√
VBC = VBN − VCN = 720 3/ − 90◦ V.
√
Wm = (720 3)(6) cos(−90 + 36.87◦ ) = 4489.48 W;
√
3Wm = 7776 VAR.
VAN = 720/0◦ V;
[b] Qφ = (36)(72) = 2592 VAR;
√
QT = 3Qφ = 7776 VAR = 3Wm .
P 11.50 [a] Negative phase sequence:
√
VAB = 480 3/ − 30◦ V;
√
VBC = 480 3/90◦ V;
√
VCA = 480 3/ − 150◦ V;
√
√
480 3/ − 30◦
IAB =
= 8 3/0◦ A;
◦
60/ − 30
√
√
480 3/90◦
IBC =
=
20
3/60◦ A;
◦
/
24 30
11–31
11–32
CHAPTER 11. Balanced Three-Phase Circuits
√
√
480 3/ − 150◦
ICA =
=
6
3/ − 150◦ A.
80/0◦
IaA = IAB + IAC
√
√
= 8 3/0◦ + 6 3/30◦ = 23.44/12.81◦ A;
IcC = ICB + ICA
√
√
= 20 3/ − 120◦ + 6 3/ − 150◦ = 43.95/ − 126.79◦ A.
√
Wm1 = 480 3(23.44) cos(−30 − 12.81◦ ) = 14,296.61 W;
√
Wm2 = 480 3(43.95) cos(−90 + 126.79◦ ) = 29,261.53 W.
[b] Wm1 + Wm2 = 43,558.14 W;
√
PA = (8 3)2 (60 cos 30◦ ) = 9976.61 W;
√
PB = (20 3)2 (24 cos 30◦ ) = 24,941.53 W;
√
PC = (6 3)2 (80) = 8640 W;
PA + PB + PC = 43,558.14 = Wm1 + Wm2 .
P 11.51 [a] I∗aA =
(192 + j56)103
= 41.67/16.26◦ A;
4800
IaA = 41.67/ − 16.26◦ A;
IbB = 41.67/ − 136.26◦ A.
√
VAB = 4800 3/30◦ V;
√
VBC = 4800 3/ − 90◦ V.
√
W1 = (4800 3)(41.67) cos 46.26◦ = 239,502.58 W.
[b] Current coil in line aA, measure IaA .
Voltage coil across AC, measure VAC .
[c] IaA = 41.67/ − 16.76◦ A;
√
VCA = 4800 3/150◦ V;
√
.·. VAC = 4800 3/ − 30◦ V;
√
W2 = (4800 3)(41.67) cos 13.74◦ = 336,497.42 W.
[d] W1 + W2 = 576,000 = 576kW;
PT = 600(0.96) = 576 kW = W1 + W2 .
Problems
P 11.52 [a] ZY =
Z∆
= 276 − j207 = 345/ − 36.87◦ Ω;
3
IaA =
6900/0◦
= 20/36.87◦ A;
345/ − 36.87◦
IbB = 20/ − 83.13◦ A;
√
VAC = 6900 3/ − 30◦ V;
√
VBC = 6900 3/ − 90◦ V;
√
W1 = (6900 3)(20) cos(−30 − 36.87◦ ) = 93,893.10 W;
√
W2 = (6900 3)(20) cos(−90 + 83.13◦ ) = 237,306.90 W.
[b] W1 + W2 = 331,200 W;
PT = 3(20)2 (276) = 331,200 W.
√
[c] 3(W1 − W2 ) = −248,400 VAR;
QT = 3(20)2 (−207) = −248,400 VAR.
P 11.53 [a]
[b]
[c]
11–33
11–34
CHAPTER 11. Balanced Three-Phase Circuits
[d]
P 11.54 [a] Q =
|V|2
;
XC
2
(13,800)
= 158.70 Ω;
.·. |XC | =
1.2 × 106
1
1
= 158.70;
C=
= 20.06 µF.
ωC
2π(50)(158.70)
√
(13,800/ 3)2
1
[b] |XC | =
= (158.70);
6
1.2 × 10
3
.·.
.·. C = 3(20.06) = 60.2 µF.
P 11.55 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.
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
Hence
I∗aA = −j150.61 A;
IaA = j150.61 A.
Problems
11–35
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 [a]
13,800 ∗
√ IaA = 75 × 103 + j180 × 103 − j1200 × 103 ;
3
√
√
75
3
−
j1020
3
= 9.41 − j128.02 A;
I∗aA =
13.8
.·. IaA = 9.41 + j128.02 A.
13,800 ◦
√ /0 + (0.6 + j4.8)(9.41 + j128.02)
3
= 7358.58 + j121.98 = 7359.59/0.95◦ V;
Van =
.·. |Vab | =
√
3(7359.59) = 12,747.2 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 11.58 [a]
13,800 ∗
√ IaA = (75 + j180) × 103 ;
3
√
√
75 3 + j180 3
∗
= 9.41 + j22.59 A;
IaA =
13.8
.·. IaA = 9.41 − j22.59 A.
13,800 ◦
√ /0 + (0.6 + j4.8)(9.41 − j22.59)
3
= 8081.51 + j31.61 = 8081.57/0.22◦ V;
Van =
.·. |Vab | =
[b] Yes:
√
3(8081.57) = 13,997.7 V.
13 kV < 13,997.7 < 14.6 kV.
[c] Ploss = 3|9.41 + j128.02|2 (0.6) = 29.66 kW.
[d] Ploss = 3|9.41 − j22.59|2 (0.6) = 1.078 kW.
11–36
CHAPTER 11. Balanced Three-Phase Circuits
[e] Yes, the voltage at the generating plant is at an acceptable level and the
line loss is greatly reduced.
P 11.59 [a] The capacitor from Appendix H whose value is closest to 20.06 µF is 22 µF.
|XC | =
Q=
1
1
=
= 144.686 Ω;
ωC
2π(50)(22 × 10−6 )
|V |2
(13,800)2
=
= −1,316,226.8 VAR/φ.
XC
−144.686
QT = 3Q = −3,948,680.4 VAR.
[b] I∗aA =
1,200,000 + j1,200,000 − j1,316,226.8
√
= 150.6 − j14.588 A;
13,800/ 3
13,800 ◦
√ /0 + (0.6 + j4.8)(150.6 + j14.588) = 8021.2/5.23◦ ;
3
√
|Vab | = 3(8021.2) = 13,893 V.
Van =
This voltage falls within the allowable range of 13 kV to 14.6 kV.
P 11.60 [a] The capacitor from Appendix H whose value is closest to 60.17 µF is 47 µF.
|XC | =
1
1
=
= 67.7255 Ω;
ωC
2π(50)(47 × 10−6 )
|V |2
(13,800)2
Q=
=
= −937,313.02 VAR.
3XC
3(−67.7255)
[b] I∗aA =
1,200,000 + j262,687
√
= 150.6 + j32.97 A;
13,800/ 3
13,800 ◦
√ /0 + (0.6 + j4.8)(150.6 − j32.97) = 8246.1/4.89◦ ;
3
√
|Vab | = 3(8246.1) = 14,282.7 V.
Van =
This voltage falls within the allowable range of 13 kV to 14.6 kV.
Introduction to the Laplace
Transform
Assessment Problems
AP 12.1 [a] 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
[b] cos(ωt + θ) = (cos ωt cos θ − sin ωt sin θ).
Therefore
L{cos(ωt + θ)} = cos θL{cos ωt} − sin θL{sin ωt}
s cos θ − ω sin θ
=
.
s2 + ω 2
AP 12.2 [a] Let f (t) = te−at :
F (s) = L{te−at } =
Now,
1
.
(s + a)2
L{t2 f (t)} = (−1)2
d2 F (s)
.
ds2
d2
1
6
}= 2
=
.
2
ds (s + a)
(s + a)4
"
So,
2
−at
L{t · te
[b] f1 (t) = e−at cos ωt;
#
F1 (s) =
F (s) = sF1 (s) − f1 (0− ) =
s+a
;
(s + a)2 + ω 2
−a2 − sa − ω 2
s(s + a)
−
1
=
.
(s + a)2 + ω 2
(s + a)2 + ω 2
12–1
12–2
CHAPTER 12. Introduction to the Laplace Transform
[c] f1 (t) = e−at sin ωt;
F (s) =
AP 12.3 F (s) =
F1 (s) =
ω
;
(s + a)2 + ω 2
F1 (s)
ω
=
.
s
s[(s + a)2 + ω 2 ]
K1
K2
K3
+
+
;
s+1 s+2 s+4
K1 =
8s2 + 37s + 32
= 1;
(s + 2)(s + 4) s=−1
K2 =
8s2 + 37s + 32
= 5;
(s + 1)(s + 4) s=−2
K3 =
8s2 + 37s + 32
= 2;
(s + 1)(s + 2) s=−4
f (t) = [e−t + 5e−2t + 2e−4t ]u(t).
AP 12.4 F (s) =
10s2 + 76s + 160
K1
K2
K3
=
+
+
;
(s + 2)(s + 4)(s + 6)
s+2 s+4 s+6
K1 =
40 − 152 + 160
= 6;
2(4)
K3 =
360 − 456 + 160
= 8;
(−4)(−2)
K2 =
160 − 304 + 160
= −4;
(−2)(2)
f (t) = [6e−2t − 4e−4t + 8e−6t ]u(t).
AP 12.5 [a] From Example 12.2,
V (s) =
[b] V (s) =
K1 =
Idc /C
15,000
= 2
.
2
s + (1/RC)s + (1/LC)
s + 1250s + 250,000
15,000
K1
K2
=
+
.
(s + 250)(s + 1000)
s + 250 s + 1000
15,000
= 20;
750
V (s) =
K2 =
15,000
= −20.
−750
20
20
−
.
s + 250 s + 1000
Therefore,
v(t) = (20e−250t − 20e−1000t )u(t) V.
Problems
AP 12.6 F (s) =
12–3
K1
K2
K2∗
+
+
;
s+1 s+2−j s+2+j
K1 =
22s2 + 60s + 58
= 10;
s2 + 4s + 5
s=−1
K2 =
22s2 + 60s + 58
= 6 + j8 = 10/53.13◦ ;
(s + 1)(s + 2 + j) s=−2+j
f (t) = [10e−t + 20e−2t cos(t + 53.13◦ )]u(t).
AP 12.7 [a] From Example 12.2,
V (s) =
[b] V (s) =
K=
15,000
Idc /C
= 2
.
2
s + (1/RC)s + (1/LC)
s + 800s + 250,000
K
K∗
15,000
=
+
.
(s + 400 − j300)(s + 400 + j300)
s + 400 − j300 s + 400 + j300
15,000
= −j25 = 25/ − 90◦ .
j600
V (s) =
25/ − 90◦
25/90◦
+
.
s + 400 − j300 s + 400 + j300
Therefore,
AP 12.8 F (s) =
v(t) = 2(25)e−400t cos(300t − 90◦ ) = (50e−400t sin 300t)u(t) V.
K1
K3
K2
+
+
;
2
s
(s + 2)
s+2
K1 =
80(s + 3)
= 60;
(s + 2)2 s=0
K2 =
80(s + 3)
= −40;
s
s=−2
"
#
"
#
80 80(s + 3)
d 80(s + 3)
K3 =
=
−
= −60;
ds
s
s
s2
s=−2
f (t) = [60 − 40te−2t − 60e−2t ]u(t).
AP 12.9 [a] From Example 12.2,
V (s) =
15,000
Idc /C
= 2
.
2
s + (1/RC)s + (1/LC)
s + 1000s + 250,000
12–4
CHAPTER 12. Introduction to the Laplace Transform
[b] V (s) =
15,000
K1
K2
=
+
.
2
2
(s + 500)
(s + 500)
s + 500
K2 = 0;
V (s) =
K1 = 15,000.
15,000
.
(s + 500)2
Therefore,
14,400
AP 12.10 F (s) =
(s2 + 60s + 2500)2
=
14,400
(s + 30 − j40)2 (s + 30 + j40)2
K1
K2
K1∗
+
+
(s + 30 − j40)2 (s + 30 − j40) (s + 30 + j40)2
=
+
K1 =
v(t) = 15,000te−500t u(t) V.
K2∗
;
(s + 30 + j40)
14,400
= −2.25 and K1∗ = −2.25;
(j80)2
"
#
14,400
−2(14,400)
d
=
= −j0.05625 = 0.05625/ − 90◦ ;
K2 =
2
ds (s + 30 + j40) s=−30+j40
(j80)3
K2∗ = j0.05625;
Therefore
f (t) = [−4.5te−30t cos 40t + 0.1125e−30t cos(40 − 90◦ )] u(t)
= e−30t [0.1125 sin 40t − 4.5t cos 40t] u(t).
AP 12.11
25
F (s) = s2 + 15s + 54
25s2 + 395s + 1494
25s2 + 375s + 1350
20s + 144
F (s) = 25 +
20s + 144
s2 + 15s + 54
K1 =
20s + 144
= 8;
s+9
s=−6
K2 =
20s + 144
= 12;
s+6
s=−9
= 25 +
K1
K2
+
;
s+6 s+9
f (t) = 25δ(t) + [8e−6t + 12e−9t ]u(t).
Problems
12–5
5s − 15
AP 12.12
F (s) = s2 + 7s + 10
5s3 + 20s2 − 49s − 108
5s3 + 35s2 + 50s
−15s2 − 99s − 108
−15s2 − 105s − 150
6s + 42
F (s) = 5s − 15 +
K1
K2
6s + 42
= 5s − 15 +
+
;
(s + 2)(s + 5)
s+2 s+5
K1 =
6s + 42
= 10;
s + 5 s=−2
K2 =
6s + 42
= −4;
s + 2 s=−5
f (t) = 5δ 0 (t) − 15δ(t) + [10e−2t − 4e−5t ]u(t).
AP 12.13 [a] Factoring the numerator,
10s2 + 210s + 980 = 10(s + 7)(s + 14).
There are zeros at −7 and −14. Factoring the denominator,
s3 + 14s2 + 50s = s(s + 7 + j)(s + 7 − j).
There are poles at 0, −7 − j and −7 + j.
[b] Factoring the numerator,
5s2 + 50s + 445 = 5(s + 5 + j8)(s + 5 − j8).
There are zeros at −5 − j8 and −5 + j8. Factoring the denominator,
4s3 + 40s2 + 176s + 480 = 4(s + 6)(s + 2 + j4)(s + 2 − j4).
There are poles at −6, −2 + j4 and −2 − j4.
[c] Factoring the numerator,
25s2 + 200s + 400 = 25(s + 4)2 .
There are two zeros at −4. Factoring the denominator,
s4 + 8s3 + 38s2 + 56s + 25
= (s + 3 + j4)(s + 3 − j4)(s + 1)2 .
There are poles at −3 + j4, −3 − j4, and two poles at −1.
12–6
CHAPTER 12. Introduction to the Laplace Transform
AP 12.14 lim f (t) = lim sF (s)
t→0
s→∞
"
= lim
1/s→0
10[1 + (7.6/s) + (16/(s2 ))]
= 10;
[1 + (2/s)][1 + (4/s)][1 + (6/s)]
#
.·. f (0+ ) = 10.
10s3 + 76s2 + 160s
lim f (t) = lim sF (s) = lim
= 0;
t→∞
s→0
s→0 (s + 2)(s + 4)(s + 6)
"
#
.·. f (∞) = 0.
"
lim f (t) = lim sF (s) = lim
t→0
s→∞
1/s→0
(80/s) + (240/s2 )
= 0;
[1 + (2/s)]2
#
.·. f (0+ ) = 0.
"
#
80s + 240
lim f (t) = lim sF (s) = lim
= 60;
t→∞
s→0
s→0
(s + 2)2
.·. f (∞) = 60.
"
lim f (t) = lim sF (s) = lim
t→0
s→∞
1/s→0
(14, 400/s3 )
= 0;
[1 + (60/s) + (2500/s2 )]2
#
.·. f (0+ ) = 0.
"
#
14,400s
lim f (t) = lim sF (s) = lim
= 0;
t→∞
s→0
s→0 (s2 + 60s + 2500)2
.·. f (∞) = 0.
Problems
12–7
Problems
P 12.1
[a] f (t) = 120 + 30t
f (t) = 120 − 30t
f (t) = −360 + 30t
f (t) = 0
− 4 s ≤ t ≤ 0;
0 ≤ t ≤ 8 s;
8 s ≤ t ≤ 12 s;
elsewhere.
f (t) = (120 + 30t)[(u(t + 4) − u(t)] + (120 − 30t)[u(t) − u(t − 8)]
+(−360 + 30t)[u(t − 8) − u(t − 12)].
[b] f (t) = 50 sin π2 t[u(t) − u(t − 4)]
= (50 sin π2 t)u(t) − (50 sin π2 t)u(t − 4).
[c] f (t) = (30 − 3t)[u(t) − u(t − 10)].
P 12.2
[a] (50 + 2.5t)[u(t + 20) − u(t)] + (50 − 5t)[u(t) − u(t − 10)]
= (2.5t + 50)u(t + 20) − 7.5tu(t) + (5t − 50)u(t − 10).
[b] (5t + 45)[u(t + 9) − u(t + 6)] + 15[u(t + 6) − u(t + 3)] − 5t[u(t + 3) − u(t − 3)]
−15[u(t − 3) − u(t − 6)] + (5t − 45)[u(t − 6) − u(t − 9)]
= 5(t + 9)u(t + 9) − 5(t + 6)u(t + 6) − 5(t + 3)u(t + 3) + 5(t − 3)u(t − 3)
+5(t − 6)u(t − 6) − 5(t − 9)u(t − 9).
P 12.3
12–8
P 12.4
CHAPTER 12. Introduction to the Laplace Transform
[a]
[b] f (t) = 30t[u(t) − u(t − 2)] + 60[u(t − 2) − u(t − 4)]
+60 cos( π4 t − π)[u(t − 4) − u(t − 8)]
+(30t − 300)[u(t − 8) − u(t − 10)].
P 12.5
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 ];
3
sε
L{δ 00 (t)} = lim F (s).
ε→0
After applying L’Hopital’s rule three times, we have
2s 3s
2s
s
s
sesε − esε/2 − e−sε/2 + se−sε =
.
ε→0 3
4
4
3 2
lim
Therefore L{δ 00 (t)} = s2 .
Problems
P 12.6
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 + t2
1
1
1
[a] A =
bh =
(2ε)
= 1.
2
2
ε
[b] 0.
P 12.7
[b] ∞.
P 12.8
esε − e−sε
1 −st
e dt =
2εs
−ε 2ε
The limit as ε → 0 is indeterminate so use l’Hopital’s rule:
F (s) =
Z ε
1
sesε + se−sε
1 2s
F (s) =
lim
=
·
= 1.
ε→0
2s
1
2s 1
"
P 12.9
12–9
[a] I =
Z 4
#
3
(t + 4)δ(t) dt +
−2
Z 4
4(t3 + 4)δ(t − 2) dt
−2
= 4 + 4(8 + 4) = 52.
[b] I =
Z 4
2
t δ(t) dt +
−3
Z 4
t2 δ(t + 2.5) dt + 0
−3
= 02 + (−2.5)2 + 0 = 6.25.
1
1 Z ∞ (3 + jω)
P 12.10 f (t) =
· πδ(ω) · ejtω dω =
2π −∞ (4 + jω)
2π
dn f (t)
P 12.11 L
dtn
(
!
3 + j0 −jt0
3
πe
= .
4 + j0
8
)
= sn F (s) − sn−1 f (0− ) − sn−2 f 0 (0− ) − · · · .
Therefore
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,
u = f (t),
v = δ(t − a),
du = f 0 (t) dt.
Therefore
Z ∞
f (t)δ 0 (t − a) dt = f (t)δ(t − a)
−∞
∞
−
−∞
= 0 − f 0 (a).
Z ∞
−∞
δ(t − a)f 0 (t) dt
12–10
CHAPTER 12. Introduction to the Laplace Transform
0
[b] L{δ (t)} =
P 12.13 [a] L{t} =
Z ∞
Z ∞
0−
h
i
d(e−st )
= − −se−st
= s.
dt = −
t=0
dt
t=0
#
"
0
−st
δ (t)e
∞
1
e−st
1
(−st
−
1)
(0
−
1)
=
.
=
0
−
s2
s2
s2
0
1
therefore L{te−at } =
.
(s + a)2
te−st dt =
0
1
[b] L{t} = 2 ;
s
ejωt − e−jωt
.
j2
Therefore
[c] sin ωt =
1
j2
L{sin ωt} =
=
!
ω
s2 + ω 2
1
1
−
s − jω s + jω
!
=
1
j2
!
.
eβt + e−βt
.
2
Therefore,
[d] cosh βt =
1 Z ∞ −(s−β)t
L{cosh βt} =
[e
+ e−(s+β)t ]dt
2 0−
1 e−(s−β)t ∞
e−(s+β)t ∞
=
+
2 −(s − β) 0− −(s + β) 0−
"
1
1
+
s−β s+β
1
=
2
!
=
s
s2 − β 2
#
.
eβt − e−βt
.
2
Therefore,
[e] sinh βt =
L{sinh βt} =
i
1 Z ∞ h −(s−β)t
e
− e−(s+β)t dt
2 0−
#∞
#∞
1 e−(s+β)t
1 e−(s−β)t
=
−
2 −(s − β) 0− 2 −(s + β) 0−
"
1
=
2
1
1
−
s−β s+β
(
)
(
)
"
!
d −at
s
P 12.14 [a] L
(te )u(t) =
− 0.
dt
(s + a)2
d −at
s
L
(te )u(t) =
.
dt
(s + a)2
=
β
(s2 − β 2 )
.
2jω
2
s + ω2
Problems
[b]
d −at
(te ) = −ate−at + e−at .
dt
L{−ate−at + e−at } =
−a
1
−a
s+a
+
=
+
.
2
2
(s + a)
(s + a)
(s + a)
(s + a)2
)
(
s
d −at
(te ) =
.
.·. L
dt
(s + a)2
P 12.15 L{e−at f (t)} =
Z ∞
0−
Z ∞
P 12.16 L{f (at)} =
0−
CHECKS
[e−at f (t)]e−st dt =
Z ∞
0−
f (t)e−(s+a)t dt = F (s + a).
f (at)e−st dt
Let u = at,
u = 0−
when t = 0− ,
f (u)e−(u/a)s
1
du
= F (s/a).
a
a
du = a dt,
and u = ∞ when t = ∞.
Therefore L{f (at)} =
Z t
Z ∞
0−
F (s)
1
=
.
s
s(s + a)
0−
Z t
1 e−at
−ax
.
[b]
e
dx = −
a
a
0−
P 12.17 [a] L
−ax
e
dx =
1 e−at
L
−
a
a
(
)
(
1
1 1
1
=
−
.
=
a s s+a
s(s + a)
)
sω
sω
d sin ωt
P 12.18 [a] L
u(t) = 2
− sin(0) = 2
.
2
dt
s +ω
s + ω2
s2
s2
−ω 2
d cos ωt
u(t) = 2
− cos(0) = 2
−1= 2
.
[b] L
dt
s + ω2
s + ω2
s + ω2
)
(
d3 (t2 )
2
3
u(t)
=
s
[c] L
− s2 (0) − s(0) − 2(0) = 2.
3
3
dt
s
ωs
d sin ωt
[d]
= (cos ωt) · ω,
L{ω cos ωt} = 2
;
dt
s + ω2
(
)
d cos ωt
= −ω sin ωt
dt
ω2
L{−ω sin ωt} = − 2
s + ω2
d3 (t2 u(t))
= 2δ(t);
dt3
L{2δ(t)} = 2.
12–11
12–12
CHAPTER 12. Introduction to the Laplace Transform
P 12.19 [a] f (t) = 4t[u(t) − u(t − 4)]
+(32 − 4t)[u(t − 4) − u(t − 12)]
+(4t − 64)[u(t − 12) − u(t − 16)]
= 4tu(t) − 8(t − 4)u(t − 4)
+8(t − 12)u(t − 12) − 4(t − 16)u(t − 16);
−4s
4[1 − 2e
.·. F (s) =
+ 2e−12s − e−16s ]
.
s2
[b]
f 0 (t) = 4[u(t) − u(t − 4)] − 4[u(t − 4) − u(t − 12)]
+4[u(t − 12) − u(t − 16)]
= 4u(t) − 8u(t − 4) + 8u(t − 12) − 4u(t − 16);
L{f 0 (t)} =
4[1 − 2e−4s + 2e−12s − e−16s ]
.
s
[c]
f 00 (t) = 4δ(t) − 8δ(t − 4) + 8δ(t − 12) − 4δ(t − 16);
L{f 00 (t)} = 4[1 − 2e−4s + 2e−12s − e−16s ].
P 12.20 [a] L{−20e−5(t−2) u(t − 2)} =
−20e−2s
.
(s + 5)
Problems
[b] First rewrite f (t) as
f (t) = (8t − 8)u(t − 1) + (24 − 8t − 8t + 8)u(t − 2)
+(8t − 40 − 24 + 8t)u(t − 4) − (8t − 40)u(t − 5)
= 8(t − 1)u(t − 1) − 16(t − 2)u(t − 2)
+16(t − 4)u(t − 4) − 8(t − 5)u(t − 5);
−s
8[e
.·. F (s) =
P 12.21 [a]
Z t
0−
x dx =
t2
L
2
(
− 2e−2s + 2e−4s − e−5s ]
.
s2
t2
.
2
)
1 Z ∞ 2 −st
t e dt
2 0−
=
∞#
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−
=
[b] L
x dx =
L{t}
1/s2
1
=
= 3.
s
s
s
Z t
Z t
0−
.·. L
0−
0
P 12.22 [a] L{f (t)} =
x dx =
1
.
s3
CHECKS
Z ∞
e−st
dt +
− ae−a(t−ε) e−st dt
−ε 2ε
ε
Z ε
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− ) =
P 12.23 [a] f1 (t) = e−at sinh βt;
F1 (s) =
F (s) = sF1 (s) − f1 (0− ) =
s
s
−0=
.
s+a
s+a
β
;
(s + a)2 − β 2
sβ
sβ
−0=
.
2
2
(s + a) − β
(s + a)2 − β 2
12–13
12–14
CHAPTER 12. Introduction to the Laplace Transform
[b] f1 (t) = e−at cos ωt;
F (s) =
[c]
F1 (s) =
s+a
;
(s + a)2 + ω 2
s+a
F1 (s)
=
.
s
s[(s + a)2 + ω 2 ]
d −at
[e sinh βt] = βe−at cosh βt − ae−at sinh βt;
dt
β(s + a)
aβ
sβ
−
=
.
2
2
2
2
(s + a) − β
(s + a) − β
(s + a)2 − β 2
Therefore F (s) =
Z t
0−
e−ax cos ωx dx =
−ae−at cos ωt + ωe−at sin ωt + a
.
a2 + ω 2
Therefore
−a(s + a)
ω2
a
1
+
+
F (s) = 2
a + ω 2 (s + a)2 + ω 2 (s + a)2 + ω 2 s
"
=
P 12.24 [a]
#
s+a
.
s[(s + a)2 + ω 2 ]
dF (s)
d
=
ds
ds
Z ∞
0−
f (t)e−st dt = −
Therefore L{tf (t)} = −
0−
tf (t)e−st dt
dF (s)
.
ds
d2 F (s) Z ∞ 2
[b]
=
t f (t)e−st dt;
2
−
ds
0
Therefore
Z ∞
d3 F (s) Z ∞ 3
=
−t f (t)e−st dt.
3
−
ds
0
Z ∞
dn F (s)
n
= (−1)
tn f (t)e−st dt = (−1)n L{tn f (t)}.
dsn
0−
[c] L{t5 } = L{t4 t} = (−1)4
1 d
L{t sin βt} = (−1)
ds
d4
ds4
1
s2
β
2
s + β2
=
120
;
s6
!
=
2βs
(s2 + β 2 )2
L{te−t cosh t}: .
From Problem 12.13(d),
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
Problems
12–15
Thus
L{t cosh t} =
s2 + 1
(s2 − 1)2
L{e−t t cosh t} =
P 12.25 [a]
Z ∞
F (u)du =
(s + 1)2 + 1
s2 + 2s + 2
=
.
[(s + 1)2 − 1]2
s2 (s + 2)2
Z ∞ Z ∞
s
0−
s
=
=
[b] L{t sin βt} =
Z ∞
f (t)e−ut dt du =
Z ∞
0−
f (t)
−ut
e
du dt =
Z ∞ Z ∞
0−
f (t)e−ut du dt
s
e−tu ∞
dt
f (t)
−t s
0−
s
#
"
Z ∞
f (t)
−e−st
dt = L
.
f (t)
−
−t
t
0
#
(
Z ∞"
2βu
du.
(u2 + β 2 )2
"
Z ∞
)
2βs
;
(s2 + β 2 )2
(
t sin βt
therefore L
t
)
=
s
#
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
#
dω
−1
=β
2
ω
ω
∞
=
s2 +β 2
β
s2 + β 2
P 12.26 [a] For t ≥ 0+ :
Rio + L
io = C
dio
+ vo = 0;
dt
dvo
dt
d2 vo
dio
=C 2 ;
dt
dt
2
dvo
d vo
.·. RC
+ LC 2 + vo = 0
dt
dt
or
1
d2 vo R dvo
+
+
vo = 0.
2
dt
L dt
LC
R
1
[b] s2 Vo (s) − sVdc − 0 + [sVo (s) − Vdc ] +
Vo (s) = 0;
L
LC
Vo (s) s2 +
Vo (s) =
R
1
s+
= Vdc (s + R/L);
L
LC
Vdc [s + (R/L)]
.
2
[s + (R/L)s + (1/LC)]
.
12–16
CHAPTER 12. Introduction to the Laplace Transform
P 12.27 [a] For t ≥ 0+ :
vo
dvo
+C
+ io = 0;
R
dt
vo = L
dvo
d2 io
=L 2;
dt
dt
dio
;
dt
.·.
d2 io
L dio
+ LC 2 + io = 0,
R dt
dt
or
d2 io
1 dio
1
+
+
io = 0.
2
dt
RC dt
LC
[b] s2 Io (s) − sIdc − 0 +
Io (s) s2 +
Io (s) =
P 12.28 io = C
1
1
[sIo (s) − Idc ] +
Io (s) = 0;
RC
LC
1
1
s+
= Idc (s + 1/RC);
RC
LC
Idc [s + (1/RC)]
.
2
[s + (1/RC)s + (1/LC)]
dvo
;
dt
.·. Io (s) = sCVo (s) =
sIdc
.
2
s + (1/RC)s + (1/LC)
vo − Vdc
1Zt
dvo
P 12.29 [a]
+
vo dx + C
= 0;
R
L 0
dt
.·.
[b] Vo +
dvo
RZ t
vo dx + RC
= Vdc .
vo +
L 0
dt
Vdc
R Vo
+ RCsVo =
;
L s
s
.·.
sLVo + RVo + RCLs2 Vo = LVdc ;
.·.
Vo (s) =
[c] io =
1Zt
vo dx;
L 0
Io (s) =
(1/RC)Vdc
.
2
s + (1/RC)s + (1/LC)
Vo
Vdc /RLC
=
.
2
sL
s[s + (1/RC)s + (1/LC)]
Problems
P 12.30 [a] C
dv1 v1 − v2
+
= ig ;
dt
R
v2 − v1
1Zt
=0
v2 dτ +
L 0
R
or
dv1 v1 v2
+
−
= ig ;
C
dt
R
R
v1 v2
1Zt
− +
+
v2 dτ = 0.
R
R L 0
[b] CsV1 (s) +
−
V1 (s) V2 (s)
−
= Ig (s);
R
R
V1 (s) V2 (s) V2 (s)
+
+
=0
R
R
sL
or
(RCs + 1)V1 (s) − V2 (s) = RIg (s);
−sLV1 (s) + (R + sL)V2 (s) = 0.
Solving,
V2 (s) =
sIg (s)
.
2
C[s + (R/L)s + (1/LC)]
P 12.31 [a] 180 = 100i1 + 15
0 = 20
[b]
di1
di2
+ 10 ;
dt
dt
di2
di1
+ 10
+ 200i2 .
dt
dt
180
= (15s + 100)I1 (s) + 10sI2 (s);
s
0 = 10sI1 (s) + (20s + 200)I2 (s).
[c] Solving the second equation in (b),
I2 (s) =
−10s
I1 (s);
20s + 200
Substituting into the first equation in (b),
180
(−100s2 )
= (15s + 100)
I1 (s);
s
(20s + 200)
12–17
12–18
CHAPTER 12. Introduction to the Laplace Transform
Therefore,
I1 (s) =
=
I2 (s) =
=
180(20s + 200)
3600(s + 10)
=
2
s[(15s + 100)(20s + 200) − 100s ]
200s(s2 + 25s + 100)
18(s + 10)
;
s(s + 5)(s + 20)
−10s
−180s(s + 10)
I1 (s) =
20s + 200
20s(s + 5)(s + 10)(s + 20)
−9
.
(s + 5)(s + 20)
P 12.32 ig (t) = 5 cos 10tu(t);
1
= 64;
LC
1
= 40;
RC
Therefore V =
P 12.33
R
= 5000;
L
Vo (s) =
15(s + 5000)
s2 + 5000s + 4 × 106
√
6.25 × 106 − 4 × 106
s2 = −4000 rad/s
15(s + 5000)
K1
K2
=
+
(s + 1000)(s + 4000)
s + 1000 s + 4000
15(4000)
= 20 V;
3000
Vo (s) =
1
= 40.
C
(40)(5)s2
200s2
=
.
(s2 + 40s + 64)(s2 + 100)
(s2 + 40s + 64)(s2 + 100)
s1 = −1000 rad/s;
K1 =
5s2
.
s2 + 100
1
= 4 × 106
LC
s1,2 = −2500 ±
Vo (s) =
so Ig (s) =
K2 =
20
5
−
s + 1000 s + 4000
vo (t) = [20e−1000t − 5e−4000t ]u(t) V
15(1000)
= −5 V
−3000
Problems
P 12.34
1
= 10,000;
RC
Io (s) =
0.1(s + 10,000)
s2 + 10,000s + 16 × 106
s1 = −2000;
Io (s) =
1
= 16 × 106 ;
LC
s2 = −8000;
0.1(s + 10,000)
K1
K2
=
+
;
(s + 2000)(s + 8000)
s + 2000 s + 8000
K1 =
0.1(8000)
= 0.133;
6000
K2 =
0.1(2000)
= −0.033;
−6000
Io (s) =
;
0.133
0.033
−
;
s + 2000 s + 8000
io (t) = [133.33e−2000t − 33.33e−8000t ]u(t) mA.
P 12.35 [a]
109
1
=
= 104 ;
RC
(4 × 103 )(25)
1
109
=
= 16 × 106 ;
LC
(2.5)(25)
Vo (s) =
40 × 106 Idc
s + 10,000s + 16 × 106
40 × 106 Idc
=
(s + 2000)(s + 8000)
K1 =
=
120,000
(s + 2000)(s + 8000)
=
K1
K2
+
.
s + 2000 s + 8000
120,000
= 20;
6000
Vo (s) =
K2 =
120,000
= −20;
−6000
20
20
−
;
s + 2000 s + 8000
vo (t) = [20e−2000t − 20e−8000t ]u(t) V.
12–19
12–20
CHAPTER 12. Introduction to the Laplace Transform
[b] Io (s) =
3 × 10−3 s
(s + 2000)(s + 8000)
=
K1
K2
+
.
s + 2000 s + 8000
K1 =
−(3 × 10−3 )(2000)
= −10−3 ;
6000
K2 =
(3 × 10−3 )(−8000)
= 4 × 10−3 ;
−6000
Io (s) =
−10−3
4 × 10−3
+
;
s + 2000 s + 8000
io (t) = (4e−8000t − e−2000t )u(t) mA.
[c] io (0) = 4 − 1 = 3 mA.
Yes. The initial inductor current is zero because there is no initial energy
in the circuit, the initial resistor current is zero because the initial
capacitor voltage is zero (no initial energy). Thus at t = 0 the source
current appears in the capacitor.
P 12.36 [a]
109
1
=
= 1250 × 104 ;
LC
(0.8)(100)
1
106
=
= 1000;
RC
(10)(100)
Vo (s) =
70,000
(s2 + 1000s + 1250 × 104 )
s1,2 = −500 ±
Vo (s) =
√
.
25 × 104 − 1250 × 104 = −500 ± j3500 rad/s;
70,000
(s + 500 − j3500)(s + 500 + j3500)
K
K∗
=
+
;
s + 500 − j3500 s + 500 + j3500
K=
70,000
= 10/ − 90◦ ;
(j7000)
Vo (s) =
10/ − 90◦
10/90◦
+
;
s + 500 − j3500 s + 500 + j3500
vo (t) = [20e−500t cos(3500t − 90◦ )]u(t) V = [20e−500t sin 3500t]u(t) V.
Problems
87,500
s(s + 500 − j3500)(s + 500 + j3500)
[b] Io (s) =
K1
K2
K2∗
+
+
.
s
s + 500 − j3500 s + 500 + j3500
=
87,500
= 7 mA;
1250 × 104
87,500
K2 =
= 3.5/171.87◦ mA;
(−500 + j3500)(j7000)
K1 =
io (t) = [7 + 7e−500t cos(3500t + 171.87◦ )]u(t) mA.
P 12.37
1
= 2 × 106 ;
C
V2 (s) =
1
= 4 × 106 ;
LC
30,000
s2 + 5000s + 4 × 106
s1 = −1000;
R
= 5000;
L
Ig =
.
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 12.38 [a] I1 (s) =
K2
K3
K1
+
+
;
s
s + 5 s + 20
K1 =
(18)(10)
= 1.8;
(5)(20)
K3 =
(10)(−10)
= −0.6;
(−20)(−15)
K2 =
(18)(5)
= −1.2;
(−5)(15)
1.8
1.2
0.6
I1 (s) =
−
−
;
s
s + 5 s + 20
i1 (t) = (1.8 − 1.2e−5t − 0.6e−20t )u(t) A.
K2
K1
+
;
s + 5 s + 20
−9
−9
K1 =
= −0.6;
K2 =
= 0.6;
15
−15
−0.6
0.6
I2 (s) =
+
;
s + 5 s + 20
I2 (s) =
i2 (t) = (0.6e−20t − 0.6e−5t )u(t) A.
0.015
;
s
12–21
12–22
CHAPTER 12. Introduction to the Laplace Transform
[b] i1 (∞) = 1.8 A;
i2 (∞) = 0 A.
[c] Yes, at t = ∞
180
= 1.8 A.
100
Since i1 is a dc current at t = ∞ there is no voltage induced in the 15 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 12.39 From Problem 12.32:
V (s) =
200s2
;
(s2 + 40s + 64)(s2 + 100)
s2 + 40s + 64 = (s + 38.33)(s + 1.67);
s2 + 100 = (s − j10)(s + j10).
Therefore
V (s) =
200s2
(s + 38.33)(s + 1.67)(s − j10)(s + j10)
K2
K3
K3∗
K1
+
+
+
.
=
s + 1.67 s + 38.33 s − j10 s + j10
K1 =
200s2
= 0.15;
(s + 38.33)(s2 + 100) s=−1.67
K2 =
200s2
= −5.11;
(s + 1.67)(s2 + 100) s=−38.33
K3 =
200s2
= 2.49/ − 5.14◦ .
(s + 1.67)(s + 38.33)(s + j10) s=j10
Therefore
v(t) = [4.98 cos(10t − 5.14◦ ) + 0.15e−1.67t − 5.11e−38.33t ]u(t) V.
(
−1
P 12.40 f (t) = L
K
K∗
+
s + α − jβ s + α + jβ
= Ke−αt ejβt + K ∗ e−αt e−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).
)
Problems
"
n
n
P 12.41 [a] L{t f (t)} = (−1)
dn F (s)
.
dsn
#
1
then F (s) = ,
s
Let f (t) = 1,
dn F (s)
(−1)n n!
=
.
dsn
s(n+1)
thus
Therefore L{t } = (−1)
(−1)n n!
n!
= (n+1) .
(n+1)
s
s
It follows that L{t(r−1) } =
(r − 1)!
sr
"
n
and L{t(r−1) e−at } =
n
#
(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
[b] f (t) = L
K
K∗
+
.
(s + α − jβ)r (s + α + jβ)r
)
Therefore
f (t) =
Ktr−1 −(α−jβ)t K ∗ tr−1 −(α+jβ)t
e
+
e
(r − 1)!
(r − 1)!
i
|K|tr−1 e−αt h jθ jβt
=
e e + e−jθ e−jβt
(r − 1)!
2|K|tr−1 e−αt
=
cos(βt + θ).
(r − 1)!
"
P 12.42 [a] F (s) =
#
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).
[b] F (s) =
K1
K2
K3
K4
+
+
+
s
s+2 s+4 s+6
K1 =
13s3 + 134s2 + 392s + 288
= 6;
(s + 2)(s + 4)(s + 6)
s=0
K2 =
13s3 + 134s2 + 392s + 288
= 4;
s(s + 4)(s + 6)
s=−2
)
12–23
12–24
CHAPTER 12. Introduction to the Laplace Transform
K3 =
13s3 + 134s2 + 392s + 288
= 2;
s(s + 2)(s + 6)
s=−4
K4 =
13s3 + 134s2 + 392s + 288
= 1;
s(s + 2)(s + 4)
s=−6
f (t) = [6 + 4e−2t + 2e−4t + e−6t ]u(t).
10(s2 + 119)
;
(s + 5)(s2 + 10s + 169)
√
s1,2 = −5 ± 25 − 169 = −5 ± j12;
[c] F (s) =
F (s) =
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).
√
[d] s1,2 = −7 ± 49 − 625 = −7 ± j24;
F (s) =
=
56s2 + 112s + 5000
s(s + 7 − j24)(s + 7 + j24)
K2
K2∗
K1
+
+
;
s
s + 7 − j24 s + 7 + j24
K1 =
5000
= 8;
625
K2 =
56(−7 + j24)2 + 112(−7 + j24) + 5000
(−7 + j24)j48
= 24 + j7 = 25/16.26◦ ;
.·. f (t) = [8 + 50e−7t cos(24t + 16.26◦ )]u(t).
P 12.43 [a] F (s) =
K1 =
K1
K1∗
+
;
s + 6 − j8 s + 6 + j8
480
= −j30 = 30/ − 90◦ ;
s + 6 + j8 s=−6+j8
f (t) = [60e−6t cos(8t − 90◦ )]u(t) = [60e−6t sin 8t]u(t).
Problems
[b] F (s) =
K1
K2
K2∗
+
+
;
s
s + 5 − j5 s + 5 + j5
K1 =
s2 + 15s + 30
= 0.6;
s2 + 10s + 50 s=0
K2 =
s2 + 15s + 30
= 0.2 − j0.7 = 0.728/ − 74.0546◦ ;
s(s + 5 + j5) s=−5+j5
f (t) = [0.6 + 1.456e−5t cos(5t − 74.0546◦ )]u(t).
[c] F (s) =
K2
K2∗
K1
+
+
;
s + 5 s + 10 − j15 s + 10 + j15
K1 =
10s2 + 30s + 400
= 2;
s2 + 20s + 325 s=−5
K2 =
10s2 + 30s + 400
= 4 + j4.33 = 5.9/47.3◦ ;
(s + 5)(s + 10 + j15) s=−10+j15
f (t) = [2e−5t + 11.8e−10t cos(15t + 47.3◦ )]u(t).
[d] F (s) =
K1∗
K2
K2∗
K1
+
+
+
;
s + 3 − j5 s + 3 + j5 s + 6 − j4 s + 6 + j4
10(s + 3)2
K1 =
= 0.833;
(s + 3 + j5)(s2 + 12s + 52) s=−3+j5
K2 =
10(s + 3)2
= 1.09/ − 143.13◦ ;
2
(s + 6 + j4)(s + 6s + 34) s=−6+j4
f (t) = [1.67e−3t cos 5t + 2.18e−6t cos(4t − 143.13◦ )]u(t).
P 12.44 [a] F (s) =
K1 =
K1 K2
K3
+
+
.
2
s
s
s + 10
8(s2 − 5s + 50)
400
=
= 40;
s + 10
10
s=0
d
K2 =
ds
(
8(s2 − 5s + 50)
s + 10
)
s=0
=
8(s + 10)(2s − 5) − 8(s2 − 5s + 50)(1)
(s + 10)2
s=0
=
10(−40) − 8(50)
= −8;
100
K3 =
8(s2 − 5s + 50)
8(100 + 50 + 50)
=
= 16;
2
s
100
s=−10
12–25
12–26
CHAPTER 12. Introduction to the Laplace Transform
F (s) =
40 8
16
− +
.
2
s
s s + 10
f (t) = [40t − 8 + 16e−10t ]u(t).
[b] F (s) =
K0 =
K0
K1
K2
4s2 + 7s + 1
=
+
+
;
s(s + 1)2
s
(s + 1)2 s + 1
1
= 1;
(1)2
K1 =
4−7+1
= 2;
−1
d 4s2 + 7s + 1
s(8s + 7) − (4s2 + 7s + 1)
K2 =
=
ds
s
s2
s=−1
s=−1
"
=
#
1+2
= 3;
1
Therefore f (t) = [1 + 2te−t + 3e−t ] u(t).
√
[c] s1,2 = −2 ± 4 − 5 = −2 ± j1
F (s) =
K1 =
K1 K2
K3
K3∗
+
+
+
.
s2
s
s + 2 − j1 s + 2 + j1
50
= 10;
5
d
K2 =
ds
(
s3 − 6s2 + 15s + 50
s2 + 4s + 5
)
s=0
=
(s2 + 4s + 5)(3s2 − 12s + 15) − (s3 − 6s2 + 15s + 50)(2s + 4)
(s2 + 4s + 5)2
s=0
=
5(15) − 50(4)
= −5;
25
K3 =
s3 − 6s2 + 15s + 50
s2 (s + 2 + j1)
s=−2+j1
(−2 + j1)3 = −2 + j11;
K3 =
(−2 + j1)2 = 3 − j4;
−2 + j11 − 6(3 − j4) + 15(−2 + j1) + 50
(3 − j4)(j2)
= 3 + j4 = 5/53.13◦ ;
F (s) =
10 5
5/53.13◦
5/ − 53.13◦
−
+
+
.
s2
s s + 2 − j1
s + 2 + j1
f (t) = [10t − 5 + 10e−2t cos(t + 53.13◦ )]u(t).
Problems
[d] s1,2 = −1 ±
F (s) =
K1 =
√
1 − 5 = −1 ± j2
K1
K1∗
K2
K2∗
+
+
+
.
(s + 1 − j2)2 (s + 1 + j2)2 s + 1 − j2 s + 1 + j2
16s3 + 72s2 + 216s − 128
;
(s + 1 + j2)2
s=−1+j2
(−1 + j2)3 = 11 − j2;
K1 =
(−1 + j2)2 = −3 − j4;
176 − j32 − 216 − j288 − 216 + j432 − 128
−16
= 24 − j7 = 25/ − 16.26◦ ;
d
K2 =
ds
=
(
16s3 + 72s2 + 216s − 128
(s + 1 + j2)2
s=−1+j2
)
(s + 1 + j2)2 (48s2 + 144s + 216)
(s + 1 + j2)4
s=−1+j2
−
(16s3 + 72s2 + 216s − 128)2(s + 1 + j2)
(s + 1 + j2)4
s=−1+j2
=
(j4)2 (−144 − j192 − 144 + j288 + 216) − (−384 + j112)(j8)
(j4)4
=
2048 + j1536
= 8 + j6 = 10/36.87◦ ;
256
25/16.26◦
10/36.87◦
25/ − 16.26◦
10/ − 36.87◦
+
+
F (s) =
+
.
(s + 1 − j2)2 (s + 1 + j2)2 s + 1 − j2
s + 1 + j2
f (t) = [50te−t cos(2t − 16.26◦ ) + 20e−t cos(2t + 36.87)]u(t).
P 12.45 [a] F (s) =
K1 =
K1 K2
K3
K3∗
+
+
+
.
s2
s
s + 1 − j2 s + 1 + j2
100(s + 1)
= 20;
s2 + 2s + 5 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)
s2 (s + 1 + j2)
= −6 + j8 = 10/126.87◦ ;
s=−1+j2
f (t) = [20t + 12 + 20e−t cos(2t + 126.87◦ )]u(t).
12–27
12–28
CHAPTER 12. Introduction to the Laplace Transform
[b] F (s) =
K1
K2
K3
K4
+
+
+
.
3
2
s
(s + 1)
(s + 1)
s+1
K1 =
40(s + 2)
= 80;
(s + 1)3 s=0
K2 =
40(s + 2)
= −40;
s
s=−1
"
#
"
#
40 40(s + 2)
d 40(s + 2)
=
−
= −40 − 40 = −80;
K3 =
ds
s
s
s2
s=−1
"
1 d 40 40(s + 2)
K4 =
−
2 ds s
s2
#
"
#
1
1 −40 40 80(s + 2)
= (−40 − 40 − 80) = −80;
=
− 2 +
2
3
2 s
s
s
2
s=−1
f (t) = [80 − 20t2 e−t − 80te−t − 80e−t ]u(t).
[c] F (s) =
s+8
5s2 + 29s + 32
5s2 + 29s + 32
=
;
=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 −
2
3
+
;
s+2 s+4
f (t) = 5δ(t) + [−3e−2t + 2e−4t ]u(t).
[d] F (s) =
4(s + 1)
4
2s3 + 8s2 + 2s − 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
480
;
(s + 6 + j8)(s + 6 − j8)
This function has no zeros and complex conjugate poles at −6 ± j8.
P 12.46 F (s) =
480
s2 + 12s + 100
=
Problems
F (s) =
=
12–29
10(s + 3)2
(s2 + 6s + 34)(s2 + 12s + 52)
10(s + 3)2
;
(s + 3 + j5)(s + 3 − j5)(s + 6 + j4)(s + 6 − j4)
This function has two zeros at −3, and complex conjugate poles at −3 ± j5
and −6 ± j4.
100(s + 1)
100(s + 1)
=
;
s2 (s2 + 2s + 5)
s2 (s + 1 + j2)(s + 1 − j2)
This function has a zero at −1, two poles at 0, and complex conjugate poles at
−1 ± j2.
P 12.47 F (s) =
F (s) =
40(s + 2)
;
s(s + 1)3
This function has a zero at −2, a pole at 0 and three poles at −1.
P 12.48 sIo (s) =
Idc s[s + (1/RC)]
;
2
s + (1/RC)s + (1/LC)
lim sIo (s) = 0,
s→0
lim sIo (s) = Idc ,
s→∞
P 12.49 sIo (s) =
.·. io (0+ ) = Idc .
Idc s[s + (1/RC)]
;
2
s + (1/RC)s + (1/LC)
lim sIo (s) = 0,
s→0
lim sIo (s) = Idc ,
s→∞
P 12.50 sVo (s) =
.·. io (∞) = 0;
.·. io (∞) = 0;
.·. io (0+ ) = Idc .
sVdc /RC
;
2
s + (1/RC)s + (1/LC)
lim sVo (s) = 0,
.·. vo (∞) = 0;
lim sVo (s) = 0,
.·. vo (0+ ) = 0;
s→0
s→∞
sIo (s) =
Vdc /RLC
;
2
s + (1/RC)s + (1/LC)
lim sIo (s) =
s→0
Vdc /RLC
Vdc
=
,
1/LC
R
lim sIo (s) = 0,
s→∞
Vdc
.·. io (∞) =
;
R
.·. io (0+ ) = 0.
12–30
CHAPTER 12. Introduction to the Laplace Transform
200s3
= 0;
P 12.51 [a] lim sV (s) = lim 4
s→∞
s→∞ s [1 + (40/s) + (64/s2 )][1 + (100/s2 )]
#
"
Therefore v(0+ ) = 0.
[b] Yes, all of the poles of V are in the left-half of the complex plane.
Therefore,
200s3
lim sV (s) = lim
= 0;
s→0
s→0 (s2 + 40s + 64)(s2 + 100)
"
#
Therefore v(∞) = 0.
P 12.52 [a] sF (s) =
6s3 + 26s2 + 26s
;
(s + 1)(s + 2)(s + 3)
lim sF (s) = 0,
.·. f (∞) = 0;
lim sF (s) = 6,
.·. f (0+ ) = 6.
s→0
s→∞
13s3 + 134s2 + 392s + 288
[b] sF (s) =
;
(s + 2)(s2 + 10s + 24)
lim sF (s) = 6,
s→0
lim sF (s) = 13,
s→∞
[c] sF (s) =
lim sF (s) = 0,
lim sF (s) = 10,
s→∞
.·. f (0+ ) = 10.
5000
= 8,
625
lim sF (s) = 56,
s→∞
P 12.53 [a] sF (s) =
.·. f (∞) = 0;
56s2 + 112s + 5000
;
s2 + 14s + 625
lim sF (s) =
s→0
.·. f (0+ ) = 13.
10s(s2 + 119)
;
(s + 5)(s2 + 10s + 169)
s→0
[d] sF (s) =
.·. f (∞) = 6;
.·. f (∞) = 8;
.·. f (0+ ) = 56.
480s
s2 + 12s + 100
;
lim sF (s) = 0,
.·. f (∞) = 0;
lim sF (s) = 0,
.·. f (0+ ) = 0.
s→0
s→∞
Problems
12–31
s2 + 15s + 30
;
s2 + 10s + 50
lim sF (s) = 0.6,
.·. f (∞) = 0.6;
[b] sF (s) =
s→0
lim sF (s) = 1,
s→∞
[c] sF (s) =
.·. f (0+ ) = 1.
10s3 + 30s2 + 400s
;
(s + 5)(s2 + 20s + 325)
lim sF (s) = 0,
s→0
.·. f (∞) = 0;
.·. f (0+ ) = 10.
lim sF (s) = 10,
s→∞
[d] sF (s) =
10s(s + 3)2
;
(s2 + 6s + 34)(s2 + 12s + 52)
lim sF (s) = 0,
.·. f (∞) = 0;
lim sF (s) = 0,
.·. f (0+ ) = 0.
s→0
s→∞
8(s2 − 5s + 50)
.
s(s + 10)
F (s) has a second-order pole at the origin so we cannot use the final
value theorem here.
P 12.54 [a] sF (s) =
lim sF (s) = 8,
s→∞
[b] sF (s) =
.·. f (0+ ) = 8.
4s2 + 7s + 1
;
(s + 1)2
lim sF (s) = 1,
.·. f (∞) = 1;
lim sF (s) = 4,
.·. f (0+ ) = 4.
s→0
s→∞
s3 − 6s2 + 15s + 50
;
s(s2 + 4s + 5)
F (s) has a second-order pole at the origin so we cannot use the final
value theorem here.
[c] sF (s) =
lim sF (s) = 1
s→∞
[d] sF (s) =
16s4 + 72s3 + 216s2 − 128s
.
(s2 + 2s + 5)2
lim sF (s) = 0,
s→0
lim sF (s) = 16,
s→∞
.·. f (0+ ) = 1.
.·. f (∞) = 0;
.·. f (0+ ) = 16.
12–32
CHAPTER 12. Introduction to the Laplace Transform
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 12.55 [a] sF (s) =
lim sF (s) = 0,
s→∞
[b] sF (s) =
.·. 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→∞
[c] This F (s) function is an improper rational function, and thus the
corresponding f (t) function contains impulses (δ(t)). Neither the initial
value theorem nor the final value theorem may be applied to this F (s)
function!
[d] This F (s) function is an improper rational function, and thus the
corresponding f (t) function contains impulses (δ(t)). Neither the initial
value theorem nor the final value theorem may be applied to this F (s)
function!
P 12.56 [a] ZL = j120π(0.01) = j3.77 Ω;
ZC =
−j
= −j26.526 Ω.
120π(100 × 10−6 )
The phasor-transformed circuit is
IL =
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 the expression for iL (t)
derived in the Practical Perspective. This matches the steady-state
response just derived in part (a).
P 12.57 The transient and steady-state components are both proportional to the
magnitude of the input voltage. Therefore,
K=
40
= 0.947.
42.26
Problems
12–33
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 12.58 We begin by using the Laplace transform of the describing differential
equation for the circuit in Fig. 12.19. Change the right-hand side so it is the
Laplace transform of Kte−500t to give:
15IL (s) + 0.01sIL (s) + 104
K
IL (s)
=
.
s
(s + 500)2
Solving for IL (s),
100Ks
K1
K1∗
IL (s) = 2
=
+
(s + 1500s + 106 )(s + 500)2
s + 750 − j661.44 s + 750 + j661.44
+
K1 =
K2 =
K3
K2
+
.
2
(s + 500)
s + 500
100Ks
= 0.1512K / − 172.82◦ mA;
2
(s + 750 + j661.44)(s + 500) s=−750+j661.44
100Ks
(s2 + 1500s + 106 )
"
= −100K mA;
s=−500
#
100Ks
d
= 0.3K mA.
K3 =
2
ds (s + 1500s + 106 ) s=−500
Therefore,
iL (t) = K[0.3024e−750t cos(661.44t − 172.82◦ ) − 100te−500t + 0.3e−500t ]u(t) mA.
12–34
CHAPTER 12. Introduction to the Laplace Transform
Plot the expression above with K = 1:
The maximum value of the inductor current is 0.0242K mA. Therefore,
K=
40
= 1652.9.
0.0242
So the inductor current rating will not be exceeded if the input to the RLC
circuit is 1652te−500t V.
The Laplace Transform in Circuit
Analysis
Assessment Problems
AP 13.1 [a] Y =
Z=
1
1
C[s2 + (1/RC)s + (1/LC)]
+
+ sC =
;
R sL
s
s/C
4 × 106 s
1
= 2
= 2
.
Y
s + (1/RC)s + (1/LC)
s + 2000s + 64 × 104
[b] zero at z1 = 0,
poles at −p1 = −400 rad/s and −p2 = −1600 rad/s.
AP 13.2 [a]
2.5s
−150
Vo =
5
(16 × 10 )/s + 5000 + 2.5s
s
=
=
[b] Vo =
−150s
s2 + 2000s + 64 × 104
−150s
.
(s + 400)(s + 1600)
K1
K2
+
.
s + 400 s + 1600
K1 =
−150s
= 50;
s + 1600 s=−400
13–1
13–2
CHAPTER 13. The Laplace Transform in Circuit Analysis
K2 =
−150s
= −200;
s + 400 s=−1600
Vo =
200
50
−
.
s + 400 s + 1600
vo (t) = (50e−400t − 200e−1600t )u(t) V.
AP 13.3 [a]
V
V
0.0384
+
=
;
6
1250 (s/10) + (2.5 × 10 /s)
s
Therefore,
V =
[b] I =
48(s2 + 25 × 106 )
.
s(s + 2500)(s + 10,000)
480
V
=
.
6
(s/10) + (2.5 × 10 /s)
(s + 2500)(s + 10,000)
[c] v(t) = (48 − 80e−2500t + 80e−10,000t )u(t) V.
[d] i(t) = (64e−2500t − 64e−10,000t )u(t) mA.
AP 13.4 Begin by transforming the circuit from the time domain to the s domain:
Io =
=
5
s2 + 502
400
4 + 0.24s +
s
=
20.833s
(s2 + 502 )(s2 + 16.67s + 1666.67)
K1
K1∗
K2
K2∗
+
+
+
.
s − j50 s + j50 s + 8.33 − j39.965 s + 8.33 + j39.965
Problems
K1 =
K2 =
20.833s
(s + j50)(s2 + 16.67s + 1666.67)
= 0.00884/ − 135◦ ;
s=j50
20.833s
(s2 + 502 )(s + 8.33 + j39.965)
= 0.00903/46.194◦ .
s=−8.33+j39.965
Therefore,
io (t) = [17.68 cos(50t − 135◦ ) + 18.06e−8.33t cos(39.965t + 46.194◦ )] mA.
AP 13.5 [a]
0 = 0.5s(I1 − 30/s) +
2500
(I1 − I2 ) + 100I1 ;
s
2500
−375
=
(I2 − I1 ) + 50(I2 − 30/s)
s
s
or
(s2 + 200s + 5000)I1 − 5000I2 = 30s;
−50I1 + (s + 50)I2 = 22.5.
∆=
(s2 + 200s + 5000) −5000
−50
N1 =
30s −5000
= 30(s2 + 50s + 3750);
22.5 (s + 50)
I1 =
= s(s + 100)(s + 150);
(s + 50)
N1
30(s2 + 50s + 3750)
=
.
∆
s(s + 100)(s + 150)
13–3
13–4
CHAPTER 13. The Laplace Transform in Circuit Analysis
N2 =
(s2 + 200s + 5000) 30s
−50
= 22.5s2 + 6000s + 112,500;
22.5
N2
22.5s2 + 6000s + 112,500
I2 =
=
.
∆
s(s + 100)(s + 150)
[b] I1 =
30(s2 + 50s + 3750)
K1
K2
K3
=
+
+
.
s(s + 100)(s + 150)
s
s + 100 s + 150
K1 = 7.5;
K2 = −52.5;
K3 = 75;
i1 (t) = [7.5 − 52.5e−100t + 75e−150t ]u(t) A.
I2 =
22.5s2 + 6000s + 112,500
K1
K2
K3
=
+
+
.
s(s + 100)(s + 150)
s
s + 100 s + 150
K1 = 7.5;
K2 = 52.5;
K3 = −37.5;
i2 (t) = [7.5 + 52.5e−100t − 37.5e−150t ]u(t) A.
AP 13.6
VTh =
10s
40
400
40
·
=
=
;
10s + 1000 s
10s + 1000
s + 100
ZTh = 1000 + 1000k10s = 1000 +
I=
10,000s
2000(s + 50)
=
.
10s + 1000
s + 100
40/(s + 100)
40s
=
(5 × 105 )/s + 2000(s + 50)/(s + 100)
2000s2 + 600,000s + 5 × 107
0.02s
K1
K1∗
= 2
=
+
.
s + 300s + 25,000
s + 150 − j50 s + 150 + j50
Problems
K1 =
0.02s
= 31.62 × 10−3 /71.57◦ ;
s + 150 + j50 s=−150+j50
i(t) = 63.25e−150t cos(50t + 71.57◦ )u(t) mA.
AP 13.7 [a] s-domain equivalent circuit is
i2 (0+ ) = −
Note:
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 =
I1 =
[b] I1 =
s + 40
s
s
5s + 120
= 4(s + 20)(s + 60);
(8/s) − 2
s
−10
5s + 120
−200(s − 4.8)
;
s
=
N1
−50(s − 4.8)
=
.
∆
s(s + 20)(s + 60)
K1
K2
K3
+
+
.
s
s + 20 s + 60
−50(−20) + 240
= −1.55;
(−20)(40)
K1 =
240
= 0.2;
1200
K3 =
−50(−60) + 240
= 1.35;
(−60)(−40)
K2 =
i1 (t) = [0.2 − 1.55e−20t + 1.35e−60t ]u(t) A.
13–5
13–6
CHAPTER 13. The Laplace Transform in Circuit Analysis
AP 13.8 [a] The s-domain circuit with the voltage source acting alone is
V0
V 0 − 60/s V 0 s
+
+
= 0;
10
80
20 + 10s
.·. V 0 =
480(s + 2)
K1
K2
K3
=
+
+
.
s(s + 4)(s + 6)
s
s+4 s+6
K1 =
480(2)
= 40;
(4)(6)
V0 =
120
−160
40
+
+
;
s
s+4 s+6
K2 =
480(−2)
= 120;
(−4)(2)
K3 =
480(−4)
= −160;
(−6)(−2)
.·. v 0 (t) = [40 + 120e−4t − 160e−6t ]u(t) V.
[b] The s-domain circuit with the current source acting alone is
V 00 V 00 s V 00 − 30/s
+
+
= 0;
10
80
10(s + 2)
.·. V 00 =
240
K1
K2
K3
=
+
+
.
s(s + 4)(s + 6)
s
s+4 s+6
K1 =
240
= 10;
(4)(6)
V 00 =
10
−30
20
+
+
;
s
s+4 s+6
K2 =
240
= −30;
(−4)(2)
.·. v 00 (t) = [10 − 30e−4t + 20e−6t ]u(t) V.
K3 =
240
= 20;
(−6)(−2)
Problems
13–7
[c] v(t) = v 0 (t) + v 00 (t) = [50 + 90e−4t − 140e−6t ]u(t) V.
AP 13.9 [a]
Vo s
Vo
+
= Ig .
0.05s + 500 12.5 × 106
Therefore,
Vo
12.5 × 106 (s + 10,000)
= H(s) = 2
.
Ig
s + 10,000s + 250 × 106
[b] −z1 = −10,000 rad/s;
−p1 = −5000 + j15,000 rad/s;
AP 13.10 [a] Vo =
=
−p2 = −5000 − j15,000 rad/s.
12.5 × 106 (s + 10,000) 1
·
s2 + 10,000s + 250 × 106 s
Ko
K1
K1∗
+
+
;
s
s + 5000 − j15,000 s + 5000 + j15,000
Ko = 500;
K1 = 1250/3/ − 126.87◦ ;
K1∗ = 1250/3/126.87◦ ;
vo = [500 + (2500/3)e−5000t cos(15,000t − 126.87◦ )]u(t) V.
12.5 × 106 (s + 10,000)
K2
K2∗
[b] Vo = 2
·1=
+
;
s + 10,000s + 250 × 106
s + 5000 − j15,000 s + 5000 + j15,000
K2 = 6.59 × 106 / − 18.43◦ ;
K2∗ = 6.59 × 106 /18.43◦ ;
vo = [13.2e−5000t cos(15,000t − 18.43◦ )]u(t) MV.
AP 13.11 [a] H(s) = L{h(t)} = L{vo (t)}.
√
√
vo (t) = ( 1000/3) cos θe−t cos 3t − ( 1000/3) sin θe−t sin 3t
= 10e−t cos 3t + (10/3)e−t sin 3t.
Therefore H(s) =
=
[b] Vo =
10(s + 1)
(10/3)(3)
+
(s + 1)2 + (3)2 (s + 1)2 + (3)2
10(s + 2)
s2 + 2s + 10
.
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.
13–8
CHAPTER 13. The Laplace Transform in Circuit Analysis
AP 13.12 From Assessment Problem 13.9:
H(s) =
12.5 × 106 (s + 10,000)
.
s2 + 10,000s + 250 × 106
Therefore H(j20,000) =
12.5 × 106 (10,000 + j20,000)
= 1118.034/ − 63.43◦ .
250 × 106 − 400 × 106 + j200 × 106
Thus,
vo = (0.01)(1118.034) cos(20,000t − 63.43◦ ) = 11.8 cos(20,000t − 63.43◦ ) V.
AP 13.13 [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
s + 30,000
Vo =
Vn =
s + 5000
s + 5000
H(s) =
Vo
s(s + 30,000)
=
.
Vg
(s + 5000)(s + 8000)
sVg
;
s + 8000
Problems
[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.
13–9
13–10
CHAPTER 13. The Laplace Transform in Circuit Analysis
Problems
P 13.1
P 13.2
1Zt
i=
vdτ + I0 ;
L 0−
1
therefore I =
L
V
s
+
I0
V
I0
=
+ .
s
sL
s
−LI0
−I0
=
;
ZN = sL.
sL
s
Therefore, the Norton equivalent is the same as the circuit in Fig. 13.4.
IN =
P 13.3
VTh = Vab = CV0
P 13.4
[a]
Z=
1
sC
=
V0
;
s
ZTh =
1
.
sC
(R + sL)(1/sC)
(1/C)(s + R/L)
= 2
;
R + sL + (1/sC)
s + (R/L)s + (1/LC)
1000
R
=
= 2000;
L
0.5
1
106
=
= 5 × 106 ;
LC
0.2
2.5 × 106 (s + 2000)
Z= 2
.
s + 2000s + 5 × 106
√
[b] s1,2 = −1000 ± 106 − 5 × 106 = −1000 ± j2000;
Z=
2.5 × 106 (s + 2000)
;
(s + 1000 − j2000)(s + 1000 + j2000)
−z1 = −2000 rad/s;
−p1 = −1000 + j2000 rad/s;
−p2 = −1000 − j2000 rad/s.
P 13.5
[a] Z = R + sL +
1
L[s2 + (R/L)s + (1/LC)]
=
sC
s
5[s2 + 2000s + 107 ]
.
s
√
[b] s1,2 = −1000 ± 106 − 107 = −1000 ± j3000 rad/s.
Zeros at −1000 + j3000 rad/s and −1000 − j3000 rad/s;
Pole at 0.
=
Problems
P 13.6
[a] Z = 2000 +
13–11
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.
P 13.7
Z1 = 0.5s +
s2 + 25s + 100
2(50/s)
=
;
(2 + 50/s)
2s + 50
2s + 50
s2 + 75s + 1350
1
+ 2
=
;
25 s + 25s + 100
25(s2 + 25s + 100
Yab =
25(s2 + 25s + 100)
25(s + 5)(s + 20)
Zab = 2
=
.
s + 75s + 1350
(s + 30)(s + 45)
Zeros at −5 rad/s and −20 rad/s; poles at −30 rad/s and −45 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
(2s + 1)
2s + 1
= 2
;
2s + s + 1
+ 2s + 1
s
1/skZb = 1s
13–12
CHAPTER 13. The Laplace Transform in Circuit Analysis
Zab =
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.
P 13.9
[a]
Vo =
(Ig /s2 C)(sL)
Ig /C
= 2
;
R + sL + (1/sC)
s + (R/L)s + (1/LC)
9.6 × 10−3
Ig
=
= 3000 × 103 = 3 × 106 ;
−9
C
3.2 × 10
R
7000
=
= 14,000;
L
0.5
1
2
=
× 109 = 625 × 106 ;
LC
3.2
Vo =
3 × 106
.
s2 + 14,000s + 625 × 106
[b] sVo =
3 × 106 s
;
s2 + 14,000s + 625 × 106
lim sVo = 0;
.·. vo (∞) = 0;
lim sVo = 0;
.·. vo (0+ ) = 0.
s→0
s→∞
[c] s1,2 = −7000 ±
√
49 × 106 − 625 × 106 = −7000 ± j24,000 rad/s;
Vo =
3,000,000
;
(s + 7000 − j24,000)(s + 7000 + j24,000)
K1 =
3 × 106
= −j62.5 = 62.5/ − 90◦ ;
j48,000
vo = 125e−7000t cos(24,000t − 90◦ ) = [125e−7000t sin 24,000t]u(t) V.
Problems
P 13.10 IC =
Ig
Vo
−
s
sL
9.6 × 10−3 2
3 × 106
=
−
s
s (s + 7000 − j24,000)(s + 7000 + j24,000)
"
#
=
6 × 106
9.6 × 10−3
−
s
s(s + 7000 − j24,000)(s + 7000 + j24,000)
=
9.6 × 10−3 K1
K2
K2∗
−
−
−
.
s
s
s + 7000 − j24,000 s + 7000 + j24,000
K1 =
6 × 106
= 9.6 × 10−3 ;
6
625 × 10
K2 =
6 × 106
(−7000 + j24,000)(j48,000)
=
6
= 5 × 10−3 /163.74◦ ;
(−7 + j24)(j48)
−3
9.6 × 10
.·. IC =
s
−
9.6 × 10−3
− [conjugate terms]
s
−5/163.74◦
=
+ conjugate
s + 7000 − j24,000
#
5/ − 16.26◦
=
+ conjugate
s + 7000 − j24,000
#
"
× 10−3
"
× 10−3 .
iC = 10e−7000t cos(24,000t − 16.26◦ )u(t) mA.
Check:
iC (0+ ) = 10 cos(−16.26◦ ) = 9.6 mA (ok);
iC (∞) = 0 (ok).
P 13.11 [a] At t = 0− ,
0.2v1 = (0.8)v2 ;
v1 = 4v2 ;
Therefore v1 (0− ) = 80 V = v1 (0+ );
v1 + v2 = 100 V;
v2 (0− ) = 20 V = v2 (0+ ).
13–13
13–14
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] I =
(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
[c] i = 20e−1250t u(t) mA;
!
=
20 × 10−3
s + 1250
80
;
s + 1250
!
=
20
.
s + 1250
v1 = 80e−1250t u(t) V;
v2 = 20e−1250t u(t) V.
P 13.12 [a] For t < 0:
vC (0− ) vC (0− ) − 120
+
= 0;
30
50
8vC (0− ) = 2160;
.·. vC (0− ) = 270 V;
−12 +
io (0− ) =
270 − 120
= 3 A.
50
Problems
[b] Io =
(270/s) + 0.06 − (120/s)
40 + 0.02s + (106 /5s)
=
3(s + 2500)
s2 + 2000s + 107
=
3(s + 2500)
;
(s + 1000 − j3000)(s + 1000 + j3000)
K1 =
√
3(1500 + j3000)
= 0.75 5/ − 26.57◦ .
j6000
[c] io (t) = 3.35e−1000t cos(3000t − 26.57◦ )u(t) A.
P 13.13
Vo =
(30/s)(5 × 106 /s)
1500 + 0.1s + (5 × 106 /s)
=
1500 × 106
s(s2 + 15,000s + 50 × 106 )
=
1500 × 106
s(s + 5000)(s + 10,000)
=
K1
K2
K3
+
+
.
s
s + 5000 s + 10,000
K1 =
1500 × 106
= 30;
(5000)(10,000)
K2 =
1500 × 106
= −60;
(−5000)(5000)
K3 =
1500 × 106
= 30;
(−5000)(−10,000)
Vo =
30
60
30
−
+
;
s
s + 5000 s + 10,000
vo (t) = [30 − 60e−5000t + 30e−10,000t ]u(t) V.
13–15
13–16
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.14 Since we already have the solution for vo (t) when the initial voltage is zero, we
will use superposition to determine the contribution of the initial voltage of
−20 V.
Vo1 = output when vC (0) = 0;
Vo2 = output when vC (0) = −20 V.
4 × 10−6 +
.·. Vo2 =
=
Vo2 s
Vo2
+
= 0;
6
5 × 10
1500 + 0.1s
−20(s + 15,000)
s2 + 15,000s + 50 × 106
K2
K1
+
.
s + 5000 s + 10,000
K1 =
−20(10,000)
= −40;
5000
K2 =
−20(5000)
= 20;
−5000
Vo2 =
−40
20
+
.
s + 5000 s + 10,000
From the solution to Problem 13.13 we have
Vo1 =
30
60
30
−
+
;
s
s + 5000 s + 10,000
Vo = Vo1 + Vo2 ;
30
100
50
.·. Vo =
−
+
;
s
s + 5000 s + 10,000
vo (t) = [30 − 100e−5000t + 50e−10,000t ]u(t) V.
Problems
P 13.15 [a] io (0− ) =
13–17
48
× 10−3 = 12 mA = ρ.
4
Vo − 48/s Vo + ρL
+
= 0;
(1/sC)
R + sL
.·. Vo =
48(s + R/L) − ρ/C
s2 + (R/L)s + (1/LC)
.
When the numerical values are substituted we get
Vo =
48(s + 4875)
;
(s + 4000 − j3000)(s + 4000 + j3000)
K1 =
48(875 + j3000)
= 25/ − 16.26◦ ;
j6000
vo (t) = 50e−4000t cos(3000t − 16.26◦ )u(t) V.
Check: vo (0+ ) = 50 cos(−16.26◦ ) = 48 V, which agrees with the fact
that the initial capacitor voltage is zero.
ρ[s + (48/ρL)]
48/s + ρL
[b] Io =
= 2
;
R + sL + (1/sC)
s + (R/L)s + (1/LC)
12 × 10−3 (s + 8000)
Io =
;
(s + 4000 − j3000)(s + 4000 + j3000)
12 × 10−3 (4000 + j3000)
= 10 × 10−3 / − 53.13◦ ;
K1 =
j6000
io (t) = 20e−4000t cos(3000t − 53.13◦ )u(t) mA.
13–18
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.16 [a]
The two node voltage equations are
5
V2 V2 − V1 V2 − (15/s)
V1 − V2
+ 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)
2.5(s2 + 6)
.
s(s2 + 2.5s + 1)
V2 =
[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 −0.5t 5 −2t
+ e
u(t) V and
v1 (t) = 15 − e
3
3
V1 =
125 −0.5t 25 −2t
e
+ e
u(t) V.
v2 (t) = 15 −
6
3
[c] v1 (0+ ) = 15 −
50 5
+ = 0;
3
3
v2 (0+ ) = 15 −
125 25
+
= 2.5 V.
6
3
[d] v1 (∞) = 15 V;
v2 (∞) = 15 V.
P 13.17 [a]
I=
Vdc /s
Vdc /L
= 2
;
R + sL + (1/sC)
s + (R/L)s + (1/LC)
Vdc
= 40;
L
R
= 1.2;
L
1
= 1.0;
LC
Problems
I=
13–19
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.
160s
[c] V = sLI =
(s + 0.6 − j0.8)(s + 0.6 + j0.8)
K1
K1∗
=
+
;
s + 0.6 − j0.8 s + 0.6 + j0.8
K1 =
160(−0.6 + j0.8)
= 100/36.87◦ .
j1.6
[d] v(t) = [200e−0.6t cos(0.8t + 36.87◦ )]u(t) V.
P 13.18 [a]
−3 V1 − Vo V1 − (15/s)
+
+
= 0;
s
s
4
Vo Vo − V1 Vo − (15/s)
+
+
= 0.
2
s
2/s
Simplfying,
(s + 4)V1 − 4Vo = 27;
(s2 + s + 2)Vo − 2V1 = 15s;
∆=
N2 =
Vo =
s+4
−4
−2 s2 + s + 2
s + 4 27
= s(s + 2)(s + 3; )
= 15s2 + 60s + 54;
−2 15s
N2
15s2 + 60s + 54
K1
K2
K3
=
=
+
+
.
∆
s(s + 2)(s + 3)
s
s+2 s+3
13–20
CHAPTER 13. The Laplace Transform in Circuit Analysis
60 − 120 + 54
= 3;
(−2)(1)
K1 =
54
= 9;
(2)(3)
K3 =
135 − 180 + 54
= 3;
(−3)(−1)
K2 =
9
3
3
.·. Vo = +
+
.
s s+2 s+3
[b] vo (t) = (9 + 3e−2t + 3e−3t )u(t) V.
[c] At t = 0+ :
vo (0+ ) = 15 V(checks).
At t = ∞:
vo (∞)
vo (∞) − 15
−3+
= 0;
2
4
.·. 3vo (∞) = 27;
.·. vo (∞) = 9 V(checks).
P 13.19 For t < 0:
vo (0− ) + 150 vo (0− ) vo (0− )
+
+
= 0;
4000
15,000 10,000
.·. vo (0− ) = −90 V;
.·. iL (0− ) = −6 mA.
Problems
For t > 0 :
Vo
(Vo + 90/s)s
Vo − 30 × 10−3
+ 4+
= 0;
5s + 15,000
10
50 × 106
30(1000 − 3s)
Vo =
s2 + 8000s + 25 × 106
30(1000 − 3s)
;
(s + 4000 − j3000)(s + 4000 + j3000)
=
K1 =
30(1000 + 12,000 − j9000)
= 79.06/ − 124.70◦ V;
j6000
vo (t) = 158.11e−4000t cos(3000t − 124.70◦ )u(t) V.
Check: vo (0) = 158.11 cos(−124.70◦ ) = −90 V.
P 13.20 [a]
Vo − 300/s 12.5Vo
Vo s
+
+
= 0;
4000
s
21,000s + 2 × 108
.·. Vo =
12(21s + 20 × 104 )
K1
K2
=
+
.
(s + 10,000)(s + 40,000)
s + 10,000 s + 40,000
K1 = −4;
Vo =
K2 = 256;
−4
256
+
;
s + 10,000 s + 40,000
vo (t) = (256e−40,000t − 4e−10,000t )u(t) V.
13–21
13–22
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] Io =
Vo
12.5Vo
=
;
0.08s
s
Io =
150(21s + 20 × 104 )
K1
K2
K3
=
+
+
.
s(s + 10,000)(s + 40,000)
s
s + 10,000 s + 40,000
K1 = 75 × 10−3 ;
K2 = 5 × 10−3 ;
K3 = −80 × 10−3 ;
io (t) = (75 + 5e−10,000t − 80e−40,000t )u(t) mA.
[c] At t = 0+ the circuit is
300
(21) = 252 V;
.·. vo (0+ ) =
25
io (0+ ) = 0.
Both values agree with our solutions for vo and io .
At t = ∞ the circuit is
.·. vo (∞) = 0;
io (∞) = 75 mA.
Both values agree with our solutions for vo and io .
P 13.21 [a] For t < 0:
iL (0− ) =
i1 =
500
= 5 A;
30 + 100k(50 + 100) + 10
5(100)
= 2 A;
250
Problems
vC (0− ) = 500 − 5(30) − 2(100) = 500 − 350 = 150 V.
For t > 0:
[b]
Vo + 0.01
Vo − 150/s
+
= 0;
10 + 0.002s 40 + 2 × 105 /s
Vo
150
0.01
1
s
−
+
;
=
5
5
10 + 0.002s 40s + 2 × 10
40s + 2 × 10
10 + 0.002s
Vo =
Vo =
−50(s + 5000)
s2 + 25,000s + 108
=
−50(s + 5000)
;
(s + 5000)(s + 20,000)
−50
.
s + 20,000
[c] vo (t) = −50e−20,000t u(t) V.
P 13.22 [a] For t < 0:
1
1
1
1
1
=
+
+
= ;
Re
12.5 50 10
5
Re = 5 kΩ;
v1 = −20(5) = −100 V;
iL (0− ) =
−100
× 10−3 = −10 mA;
10
vC (0− ) = −v1 = 100 V.
13–23
13–24
CHAPTER 13. The Laplace Transform in Circuit Analysis
For t = 0+ :
s-domain circuit:
where
R = 10 kΩ;
L = 4 H;
[b]
C = 10 nF;
and
γ = 100 V;
ρ = 10 mA.
Vo
Vo
ρ
+ Vo sC − γC +
− = 0;
R
sL s
.·. Vo =
γ[s + (ρ/γC)]
s2 + (1/RC)s + (1/LC)
.
ρ
10 × 10−3
=
= 104 ;
γC
(100)(10)10−9
109
1
= 5 = 104 ;
RC
10
1
109
=
= 25 × 106 ;
LC
40
Vo =
[c] IL =
Vo
ρ
Vo 10 × 10−3
− =
−
;
sL s
4s
s
IL =
[d] Vo =
=
100(s + 104 )
.
s2 + 104 s + 25 × 106
25(s + 104 )
10−2
−0.01(s + 7500)
−
=
.
2
4
6
s(s + 10 s + 25 × 10 )
s
(s + 5000)2
100(s + 104 )
s2 + 104 s + 25 × 106
K1
K2
100(s + 104 )
=
+
;
2
2
(s + 5000)
(s + 5000)
s + 5000
K1 = 100(5000) = 5 × 105 ;
Problems
K2 =
d
[100(s + 10,000)]s=−5000 = 100;
ds
Vo =
5 × 105
100
+
;
2
(s + 5000)
s + 5000
vo = [5 × 105 te−5000t + 100e−5000t ]u(t) V.
−0.01(s + 7500)
(s + 5000)2
[e] IL =
K1
K2
+
;
(s + 5000)2 (s + 5000)
=
K1 = −0.01(2500) = −25;
K2 =
d
[−0.01(s + 7500)]s=−5000 = −0.01;
ds
"
#
−25,000
10
IL =
−
× 10−3 ;
2
(s + 5000)
s + 5000
iL = −[25,000t + 10]e−5000t u(t) mA.
50,000
.
(s + 30)2
P 13.23 [a] Vg =
Io =
50,000
10,000
=
;
(s + 30)2 (5s + 400)
(s + 30)2 (s + 80)
Vo = 5sIo =
[b] Io =
50,000s
.
(s + 30)2 (s + 80)
K1
K2
K3
+
+
.
2
(s + 30)
s + 30 s + 80
K1 =
10,000
= 200;
50
K2 =
d 10,000
= −4;
ds s + 80 s=−30
K3 =
10,000
= 4;
(−50)2
13–25
13–26
CHAPTER 13. The Laplace Transform in Circuit Analysis
Io =
200
4
4
−
+
;
2
(s + 30)
s + 30 s + 80
io (t) = [200te−30t − 4e−30t + 4e−80t ]u(t) A.
Vo =
K1
K2
K3
+
+
.
(s + 30)2 s + 30 s + 80
50,000(−30)
= −30,000;
50
d 50,000s
K2 =
= 1600;
ds s + 80 s=−30
K1 =
K3 =
50,000(−80)
= −1600;
(−50)2
vo (t) = [−30,000te−30t + 1600e−30t − 1600e−80t ]u(t) V.
P 13.24 Transforming the circuit into the s domain for t > 0:
1
40
s
Zeq =
+
+
40 × 106 625
s
=
−1
40 × 106 s
;
s2 + 64,000s + 16 × 108
Zeq
I=
s/40
0.024s
s2 + 40,0002
!
=
384 × 105 s
(s2 + 40,0002 )(s + 32,000 − j24,000)(s + 32,000 + j24,000)
=
K1∗
K2
K2∗
K1
+
+
+
;
s − j40,000 s + j40,000 s + 32,000 − j24,000 s + 32,000 + j24,000
K1 =
384 × 105 s
= −j0.0075 = 0.0075/ − 90◦ ;
(s + j40,000)(s2 + 64,000s + 16 × 108 ) s=j40,000
K2 =
384 × 105 s
= j0.0125 = 0.0125/90◦ .
(s2 + j40,0002 )(s + 32,000 + j24,000) s=−32,000+j24,000
Therefore,
i(t) = (15 sin 40,000t − 25e−32,000t sin 24,000t)u(t) mA.
Problems
P 13.25 [a] iL (0− ) = iL (0+ ) = 5 A, down;
vC (0− ) = vC (0+ ) = 0.
Vo
Vo
−5
+
+ Ia =
;
20 0.0075s
s
V1 (5s) V1
=
Ia =
+
106
50
Vo + 20Iφ = V1 ;
250s + 106
V1 ;
50 × 106
!
Vo + 20
V1
= V1 ;
50
.·. 0.6V1 = Vo ;
Vo
Vo
250s + 106
−5
.·.
+
+
V
=
;
o
20 0.0075s
30 × 106
s
.·. (s2 + 10,000s + 16 × 106 )Vo = −6 × 105 ;
Vo =
[b] Vo =
−6 × 105
.
s2 + 10,000s + 16 × 106
−6 × 105
K1
K2
=
+
.
(s + 2000)(s + 8000)
(s + 8000) (s + 2000)
K1 =
−6 × 105
= 100;
−6000
K2 =
−6 × 105
= −100;
6000
vo (t) = [100e−8000t − 100e−2000t ]u(t) V.
P 13.26 [a]
Vo − 5/s
Vo − 10Iφ
+ 0.02V∆ +
= 0;
40
s + (200/s)
13–27
13–28
CHAPTER 13. The Laplace Transform in Circuit Analysis
"
#
Vo − 10Iφ
s;
V∆ =
s + (200/s)
Iφ =
(5/s) − Vo
.
40
Solving for Vo yields:
Vo =
3s2 + 25s + 500
3s2 + 25s + 500
=
.
s(s2 + 25s + 100)
s(s + 5)(s + 20)
Vo =
K1
K2
K3
+
+
;
s
s + 5 s + 20
K1 =
3s2 + 25s + 500
= 5;
(s + 5)(s + 20) s=0
3s2 + 25s + 500
= −6;
K2 =
s(s + 20)
s=−5
3s2 + 25s + 500
K3 =
= 4;
s(s + 5)
s=−20
5
−6
4
.·. Vo = +
+
;
s s + 5 s + 20
.·. vo (t) = [5 − 6e−5t + 4e−20t ]u(t) V.
[b] From the equation at t = 0+ ,
From the circuit,
vo = 5 − 6 + 4 = 3 V.
vo = v∆ + 10iφ ;
iφ =
5 − vo
;
40
(5 − vo )
.·. vo = v∆ + 10
= v∆ + 1.25 − 0.25vo ;
40
.·. 1.25vo − 1.25 = v∆ ;
vo − 5
+ 0.02v∆ = 0;
40
vo − 5 + 0.8v∆ = 0;
vo − 5 + vo − 1 = 0
so
vo = 3 V(checks.)
Problems
At t = ∞, the circuit is
From the equation for vo (t), vo (∞) = 5 V. From the circuit,
v∆ = 0,
.·. vo = 5 V(checks).
iφ = 0
P 13.27 [a]
Vφ
Vφ − (600/s) Vφ − Vo
+
+
= 0;
10/s
10
20s
Vo
Vo − Vφ Vφ
+
+
= 0.
140
20s
4
Simplfying,
(2s2 + 2s + 1)Vφ − Vo = 1200;
(35s − 7)Vφ + (s + 7)Vo = 0;
∆=
2s2 + 2s + 1 −1
35s − 7
N2 =
Vo =
= 2s(s2 + 8s + 25);
s+7
2s2 + 2s + 1 1200
35s − 7
= −42,000s + 8400;
0
N2
−21,000s + 4200
−4200(5s − 1)
=
=
.
2
∆
s(s + 8s + 25)
s(s2 + 8s + 25)
[b] vo (0+ ) = s→∞
lim sVo = 0;
vo (∞) = lim sVo =
s→0
4200
= 168 V.
25
13–29
13–30
CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] At t = 0+ the circuit is
At t = ∞ the circuit is
Vφ − 600
Vφ
Vφ
+
+
= 0;
10
140
4
.·. Vφ = 168 V = Vo (∞)
[d] Vo =
(checks).
−21,000s + 4200
K1
K2
K2∗
N2
=
=
+
+
.
∆
s(s2 + 8s + 25)
s
s + 4 − j3 s + 4 + j3
K1 =
4200
= 168;
25
K2 =
−21,000(−4 + j3) + 4200
= −84 + j3612 = 3612.98/91.33◦ ;
(−4 + j3)(j6)
vo (t) = [168 + 7225.95e−4t cos(3t + 91.33◦ )]u(t) V.
Check:
vo (0+ ) = 0 V;
vo (∞) = 168 V.
P 13.28 [a]
20Iφ + 25s(Io − Iφ ) + 25(Io − I1 ) = 0;
Problems
25s(Iφ − Io ) +
Iφ − I1 =
50
Iφ + 5I1 + 25(I1 − Io ) = 0;
s
100
;
s
100
.·. I1 = Iφ −
.
s
Simplifying,
(−5s − 1)Iφ + (5s + 5)Io = −500/s;
(5s2 + 6s + 10)Iφ + (−5s2 − 5s)Io = 600;
∆=
−5s − 1
5s + 5
2
= −25(s2 + 3s + 2);
2
5s + 6s + 10 −5s − 5s
N2 =
Io =
−5s − 1
−500/s
5s2 + 6s + 10
600
=−
500 2
(s − 4.8s − 10; )
s
N2
20s2 − 96s − 200
=
.
∆
s(s + 1)(s + 2)
[b] io (0+ ) = s→∞
lim sIo = 20 A;
io (∞) = lim sIo =
s→0
−200
= −100 A.
2
[c] At t = 0+ the circuit is
20iφ + 5i1 = 0;
iφ − i1 = 100;
.·. 20iφ + 5(iφ − 100) = 0;
.·. iφ = io (0+ ) = 20 A(checks).
25iφ = 500;
13–31
13–32
CHAPTER 13. The Laplace Transform in Circuit Analysis
As t → ∞ the circuit is
io (∞) = −100 A(checks).
[d] Io =
K1
K2
K3
20s2 − 96s − 200
=
+
+
.
s(s + 1)(s + 2)
s
s+1 s+2
K1 =
−200
= −100;
(1)(2)
K3 =
80 + 192 − 200
= 36;
(−2)(−1)
Io =
K2 =
20 + 96 − 200
= 84;
(−1)(1)
−100
84
36
+
+
;
s
s+1 s+2
io (t) = (−100 + 84e−t + 36e−2t )u(t) A.
io (∞) = −100 A(checks).
io (0+ ) = −100 + 84 + 36 = 20 A(checks).
P 13.29 vC (0− ) = vC (0+ ) = 0.
Find the Thévenin equivalent with respect to the capacitor:
vTh
vTh
9vTh
+
+
= −0.005;
20,000 200,000 200,000
.·. vTh = −50 V;
isc = −5 mA;
.·. RTh = 10 kΩ.
Problems
Vo =
=
−50/s
107
·
10,000 + (107 /s) s
−50 × 103
−50
50
=
+
;
s(s + 1000)
s
s + 1000
vo (t) = [−50 + 50e−1000t ]u(t) V.
P 13.30
Vo
Vo
15
=
+ 0.4Vφ +
;
s
1.6 + 5/s
0.2s
Vφ =
5/s
5Vo
Vo =
;
1.6 + 5/s
1.6s + 5
15
Vo s
2Vo
5Vo
.·.
=
+
+
s
1.6s + 5 1.6s + 5
s
"
= Vo
s(s + 2) + 5(1.6s + 5)
;
s(1.6s + 5)
#
15(1.6s + 5) = Vo (s2 + 10s + 25);
15(1.6s + 5)
K1
K2
.·. Vo =
=
+
.
2
2
(s + 5)
(s + 5)
s+5
K1 = 15(−8 + 5) = −45;
Vo =
K2 = 24;
−45
24
+
;
2
(s + 5)
s+5
vo (t) = [−45te−5t + 24e−5t ]u(t) V.
13–33
13–34
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.31 [a]
100
100
I1 +
(I1 − I2 ) + 50(I1 − 4/s) = 0;
s
s
100
100
(I2 − 4/s) +
(I2 − I1 ) + 50I2 = 0.
s
s
Simplifying,
(s + 4)I1 − 2I2 = 4;
8
−2I1 + (s + 4)I2 = ;
s
∆=
N1 =
(s + 4)
−2
−2
(s + 4)
4
−2
= s2 + 8s + 12 = (s + 2)(s + 6; )
=
8/s (s + 4)
I1 =
N2 =
I2 =
N1
4(s + 2)2
4(s + 2)
4/3
8/3
=
=
=
+
.
∆
s(s + 2)(s + 6)
s(s + 6)
s
s+6
(s + 4) 4
−2
8/s
=
16s + 32
16(s + 2)
=
;
s
s
N2
16(s + 2)
16
8/3
8/3
=
=
=
−
.
∆
s(s + 2)(s + 6)
s(s + 6)
s
s+6
Ia = I1 =
Ib =
4s2 + 16s + 16
4(s + 2)2
=
;
s
s
4/3
8/3
+
;
s
s+6
4
8/3
8/3
− I1 =
−
.
s
s
s+6
Problems
13–35
[b] ia (t) = (4/3)(1 + 2e−6t )u(t) A;
ib (t) = (8/3)(1 − e−6t )u(t) A.
100
100
[c] Va =
Ia =
s
s
=
4/3
8/3
+
s
s+6
!
400/3
800/3
400/3 400/9 400/9
+
=
+
−
.
s2
s(s + 6)
s2
s
s+6
100
100
Vb =
(I2 − I1 ) =
s
s
=
!
400/3
1600/3
400/3 800/9 800/9
−
=
−
+
.
2
s
s(s + 6)
s2
s
s+6
100
100
Vc =
(4/s − I2 ) =
s
s
=
4/3
16/3
−
s
s+6
4/3
8/3
+
s
s+6
!
400/3
800/3
400/3 400/9 400/9
+
=
+
−
.
2
s
s(s + 6)
s2
s
s+6
[d] va (t) = (400/9)(3t + 1 − e−6t )u(t) V;
vb (t) = (400/9)(3t − 2 + 2e−6t )u(t) V;
vc (t) = (400/9)(3t + 1 − e−6t )u(t) V.
[e] Calculating the time when a capacitor’s voltage drop first reaches 1000 V:
For va (t) or vc (t) :
1000
9
= 3t + 1 − e−6t = 22.5;
400
3t − e−6t = 21.5;
t = 7.17 s.
For vb (t) :
3t − 2 + 2e−6t = 22.5;
3t + 2e−6t = 24.5;
t = 8.17 s.
Thus, the capacitors whose voltage drops are designated va and vc will
break down first, at a time of 7.17 s.
13–36
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.32 [a]
Ye =
12.5s
10s 2.5s
+ 9 =
;
9
10
10
109
Ze =
109
80 × 106
=
.
12.5s
s
[b] I1 =
4 × 10−3
80/s
=
;
20,000 + (100 × 106 /s)
s + 5000
V1 =
4 × 10−3 20 × 106
80,000
·
=
;
s + 5000
s
s(s + 5000)
V2 =
320,000
4 × 10−3 80 × 106
·
=
.
s + 5000
s
s(s + 5000)
[c] i1 (t) = 4e−5000t u(t) mA;
V1 =
16
16
−
;
s
s + 5000
v1 (t) = (16 − 16e−5000t )u(t) V;
V2 =
64
64
−
;
s
s + 5000
v2 (t) = (64 − 64e−5000t )u(t) V.
[d] i1 (0+ ) = 4 mA;
i1 (0+ ) =
80
× 10−3 = 4 mA(checks);
20
v1 (0+ ) = 0;
v1 (∞) =
v2 (0+ ) = 0(checks);
12.5
(80) = 16 V;
50 + 12.5
v2 (∞) =
50
(80) = 64 V(checks);
50 + 12.5
Problems
v1 (∞) + v2 (∞) = 80 V(checks);
(50 × 10−9 )v1 (∞) = 800 nC;
(12.5 × 10−9 )v2 (∞) = 800 nC(checks).
P 13.33 [a] For t < 0:
v1 = 75 − vo ;
50vo = 200(75 − v0 );
.·. vo = 60 V;
v1 = 15 V.
For t > 0:
[b] Io =
75/s
300
=
(25 × 106 /s) + 6250 + 0.25s
s2 + 25,000s + 108
=
300
20 × 10−3
20 × 10−3
=
−
;
(s + 5000)(s + 20,000)
s + 5000
s + 20,000
io (t) = (20e−5000t − 20e−20,000t )u(t) mA.
[c] Vo =
60 20 × 106
300
−
·
s
s
(s + 5000)(s + 20,000)
"
60
60
80
20
=
−
−
+
s
s
s + 5000 s + 20,000
=
−20
80
+
;
s + 5000 s + 20,000
vo (t) = (80e−5000t − 20e−20,000t )u(t) V.
#
13–37
13–38
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.34 [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
20(s + 2.4)
.
s(s + 2)
and 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
Problems
13–39
[b]
I=
20(s + 2.4)
VTh
=
.
ZTh + 2 + s
s(s + 3)(s + 6)
P 13.35 Begin by transforming the circuit from the time domain to the s domain:
30k0.0012s =
30s
.
s + 25,000
Use voltage division to find Vo (s):
30s
s + 25,000
Vo = 6
10
30s
+
s
s + 25,000
2 × 106
s2 + 50,0002
!
s2
= 2
(s + 33,333.33s + 0.833 × 109 )
=
!
K1
K1∗
+
s + 16,666.67 − j23,570.226 s + 16,666.67 + j23,570.226
+
K1 =
2 × 106
s2 + 50,0002
K2
K2∗
+
.
s − j50,000 s + j50,000
2 × 106 s2
(s + 16,666.67 + j23,570.226)(s2 + 50,0002 ) s=−16,666.67+j23,570.226
= 15/180◦ ;
13–40
CHAPTER 13. The Laplace Transform in Circuit Analysis
K2 =
2 × 106 s2
(s2 + 33,333.33s + 0.833 × 109 )(s + j50,000) s=j50,000
= 21.21/ − 45◦ .
Therefore,
vo (t) = [−30e−16,666.67t cos(23,570.226t) + 42.43 cos(50,000t − 45◦ )] V.
P 13.36 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
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 13.37 [a] The s-domain equivalent circuit is
I=
Vg /L
Vg
=
,
R + sL
s + (R/L)
Vg =
Vm (ω cos φ + s sin φ)
;
s2 + ω 2
K0
K1
K1∗
+
+
.
I=
s + R/L s − jω s + jω
Vm (ωL cos φ − R sin φ)
K0 =
,
R 2 + ω 2 L2
Vm /φ − 90◦ − θ(ω)
√
K1 =
2 R 2 + 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
+
R 2 + ω 2 L2
R 2 + ω 2 L2
Problems
[b] iss (t) = √
Vm
R 2 + ω 2 L2
sin[ωt + φ − θ(ω)].
Vm (ωL cos φ − R sin φ) −(R/L)t
e
.
R 2 + ω 2 L2
Vg
[d] I =
,
Vg = Vm /φ − 90◦ .
R + jωL
[c] itr =
Therefore I = √
Vm /φ − 90◦
Vm
/φ − θ(ω) − 90◦ .
√
=
2
2
2
2
2
2
/
R + ω L θ(ω)
R +ω L
Therefore iss = √
Vm
sin[ωt + φ − θ(ω)].
R 2 + ω 2 L2
[e] The transient component vanishes when
ωL cos φ = R sin φ or
tan φ =
ωL
R
or φ = θ(ω).
P 13.38 The s-domain equivalent circuit is
V1 − 12/s
V1 + 2.4
V1
+
+
= 0;
10 + (250/s)
2s
2s + 50
V1 =
−300(s + 25)
−300
= 2
;
2
(s + 25)(s + 10s + 125)
s + 10s + 125
Io =
−300
(2s + 50)(s2 + 10s + 125)
=
−150
(s + 25)(s + 5 − j10)(s + 5 + j10)
K1
K2
K2∗
=
+
+
.
s + 25 s + 5 − j10 s + 5 + j10
K1 =
−150
= −300 × 10−3 ;
625 − 250 + 125
K2 =
√
−150
= 150 5 × 10−3 /63.43◦ ;
(−5 + j10 + 25)(j20)
√
io (t) = [−300e−25t + 300 5e−5t cos(10t + 63.43◦ )]u(t) mA.
13–41
13–42
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.39 [a] i2 = 1.25e−t − 1.25e−3t ;
so
di2
= −1.25e−t + 3.75e−3t .
dt
di2
= 0 when
dt
Therefore
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] Solving the mesh current equations from Example 13.7 are
(3 + 2s)I1 + 2sI2 = 10;
and
2sI1 + (12 + 8s)I2 = 10.
Solving for I1 ,
I1 =
5(s + 2)
.
(s + 1)(s + 3)
A partial fraction expansion leads to the expression
2.5
2.5
+
.
s+1 s+3
Therefore we get
I1 =
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
−2(−2.89)
= 481.13 mA. (checks)
12
P 13.40 [a]
88.4
= 136I1 − sI1 + 1.5s(I1 − I2 );
s
0 = 1.5s(I2 − I1 ) + 11sI2 + 5000I2 .
Simplifying,
88.4
= (0.5s + 136)I1 − 1.5sI2 ;
s
!
di1
.
dt
Problems
13–43
0 = −1.5sI1 + (12.5s + 5000)I2 ;
∆=
N1 =
I1 =
[b] sI1 =
0.5s + 136
−1.5s
−1.5s
12.5s + 5000
88.4/s
−1.5s
0
12.5s + 5000
= 4(s + 200)(s + 850);
=
1105(s + 400)
;
s
N1
276.25(s + 400)
=
.
∆
s(s + 200)(s + 850)
276.25(s + 400)
;
(s + 200)(s + 850)
lim sI1 = i1 (∞) = 650 mA;
s→0
lim sI1 = i1 (0) = 0.
s→∞
[c] I1 =
K1
K2
K3
+
+
.
s
s + 200 s + 850
K1 = 650 × 10−3 ;
K2 = −425 × 10−3 ;
K3 = −225 × 10−3 ;
i1 (t) = (650 − 425e−200t − 225e−850t )u(t) mA.
P 13.41 [a] From the solution to Problem 13.40 we have
N2 =
0.5s + 136 88.4/s
−1.5s
.·. I2 =
=
= 132.6;
0
132.6
33.15
=
4(s + 200)(s + 850)
(s + 200)(s + 850)
51 × 10−3 51 × 10−3
−
;
s + 200
s + 850
i2 (t) = (51e−200t − 51e−850t )u(t) mA.
[b] Reversing the dot on the 12.5 H coil will reverse the sign of M , thus the
circuit becomes
The two simultaneous equations are
88.4
= (136 + 0.5s)I1 + 1.5sI2 ;
s
13–44
CHAPTER 13. The Laplace Transform in Circuit Analysis
0 = 1.5sI1 + (12.5s + 5000)I2 .
When these equations are compared to those derived in Problem 13.40
we see the only difference is the algebraic sign of the 1.5s term. Thus
reversing the dot will have no effect on I1 and will reverse the sign of I2 .
Hence,
i2 (t) = (−51e−200t + 51e−850t )u(t) mA.
1
1
P 13.42 [a] w = L1 i21 + L2 i22 + M i1 i2 ;
2
2
1
1
w = (40)(9) + (90)(4) + 30(6) × 10−3 = 540 mJ.
2
2
[b] The s-domain circuit:
(600 + 0.04s)I1 − 0.03sI2 = 0.18;
−0.03sI1 + (0.09s + 1350)I2 = −0.27;
∆=
0.04(s + 15,000)
−0.03s
−0.03s
0.09(s + 15,000)
= 27 × 10−4 (s + 10,000)(s + 30,000; )
N1 =
I1 =
N2 =
I2 =
0.18
−0.03s
= 81 × 10−4 (s + 30,000; )
−0.27 0.09(s + 15,000)
N1
3
=
.
∆
s + 10,000
0.04(s + 15,000) 0.18
−0.03s
= −54 × 10−4 (s + 30,000);
−0.27
N2
−2
=
.
∆
s + 10,000
[c] i1 (t) = 3e−10,000t u(t) A;
i2 (t) = −2e−10,000t u(t) A.
Problems
[d] p600Ω = (600)(9e−20,000t ) = 5400e−20,000t W;
p1350Ω = (1350)(4e−20,000t ) = 5400e−20,000t W;
w600 =
5400
× 10−3 = 270 mJ;
20
w1350 =
5400
× 10−3 = 270 mJ;
20
wT = 540 mJ.
[e] With the dot reversed,
1
1
w = L1 i21 + L2 i22 − M i1 i2 = 180 + 180 − 180 = 180 mJ.
2
2
The s-domain equivalent circuit is
Solving for I1 and I2 yields
I1 =
3
;
s + 30,000
I2 =
−2
;
s + 30,000
.·. i1 (t) = 3e−30,000t u(t) A;
w600 = 5400
Z ∞
w1350 = 5400
e−60,000t dt = 90 mJ;
0
Z ∞
wT = 180 mJ.
P 13.43 For t < 0:
i2 (t) = −2e−30,000t u(t) A.
0
e−60,000t dt = 90 mJ;
13–45
13–46
CHAPTER 13. The Laplace Transform in Circuit Analysis
For t > 0+ :
L1 + M = 3.6 + 1.2 = 4.8 H;
15 × 4.8 = 72;
L2 + M = 2.4 + 1.2 = 3.6 H;
15 × 3.6 = 54.
12Io + 4.8sIo − 72 + (Io − I2 )(6 − 1.2s) = 0;
(6 − 1.2s)(I2 − Io ) + 3.6sI2 − 54 = 0;
.·. ∆ =
3(s + 5) −(5 − s)
= 5(s + 1)(s + 10);
−(5 − s) 2(s + 2.5)
60 −(5 − s)
No =
= 75(s + 7);
45 2(s + 2.5)
Io =
No
75(s + 7)
=
∆
5(s + 1)(s + 10)
=
K1 =
K1
K2
+
.
(s + 1) (s + 10)
(15)(6)
= 10;
9
Problems
K2 =
(15)(−3)
= 5;
−9
10
5
.·. Io =
+
;
s + 1 s + 10
.·. io (t) = [10e−t + 5e−10t ]u(t) A.
P 13.44 [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
With the current source acting alone,
V 00
V 00
V 00 s
5
+
+
= ;
2
1.25s
20
s
V 00 =
50/3
50/3
100
=
−
;
(s + 2)(s + 8)
s+2 s+8
Vo = V 0 + V 00 =
50
50
−
.
s+2 s+8
[b] vo = [50e−2t − 50e−8t ]u(t) V.
13–47
13–48
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.45
Va − 0.4/s
Va s
(Va − Vo )s
+
+
= 0;
6
5000
250 × 10
250 × 106
(0 − Va )s (0 − Vo )
+
= 0;
250 × 106
25,000
−104 Vo
;
s
Va =
.·. Vo (s2 + 20,000s + 500 × 106 ) = −20,000;
Vo =
−20,000
.
(s + 10,000 − j20,000)(s + 10,000 + j20,000)
K1 =
−20,000
= j0.5 = 0.5/90;◦
j40,000
vo (t) = e−10,000t cos(20,000t + 90◦ ) = −e−10,000t sin(20,000t)u(t) V.
P 13.46 [a] Let Va be the s-domain voltage across the 0.2 µF capacitor, positive at the
upper terminal and let Vb be the s-domain voltage across the 200 kΩ
resistor, positive at the upper terminal. With Vg = 8/s V,
Va s
Va − Vg
Va
+
+
= 0;
6
5 × 10
400,000 400,000
−Va
sVb
− 7 = 0;
400,000 10
12.5
.·. Va =
Vg .
s + 25
−25
−312.5
.·. Vb =
Va =
Vg ;
s
s(s + 25)
Vb
sVb (Vb − Vo )s
+ 7 +
= 0;
200,000 10
107
"
2(s + 25)
2(s + 25)
.·. Vo =
Vb =
s
s
#"
−312.5
s(s + 25)
# 8
−5000
=
.
s
s3
Problems
[b] vo (t) = −2500t2 u(t) V.
[c] The op amp will saturate when vo = −12.5 V.
−12.5 = −2500t2 ;
P 13.47 [a] Vo = −
Zf
Vg ;
Zi
Zf =
h
t2 = 0.005;
108
s+
109
(10)(2)×104
i =
108
;
s + 5000
8000
109
Zi =
s+
s
(50)(8000)
Vg =
!
8000
(s + 2500);
s
=
20,000
;
s2
.·. Vo =
[b] Vo =
.·. t = 0.071 = 71 ms.
−250 × 106
.
s(s + 2500)(s + 5000)
K1
K2
K3
+
+
.
s
s + 2500 s + 5000
K1 =
−250 × 106
= −20;
(5000)(2500)
K2 =
−250 × 106
= 40;
(−2500)(2500)
K3 =
−250 × 106
= −20;
(−5000)(−2500)
.·. vo (t) = (−20 + 40e−2500t − 20e−5000t )u(t) V.
[c] −20 + 40e−2500ts − 20e−5000ts = −5;
.·. 40e−2500ts − 20e−5000ts = 15.
Let x = e−2500ts . Then
40x − 20x2 = 15;
or x2 − 2x + 0.75 = 0.
Solving,
x = 1 ± 0.5
so
.·. e−2500ts = 0.5;
x = 0.5;
.·. ts =
ln 2
× 10−6 = 277.26 µs.
0.0025
13–49
13–50
CHAPTER 13. The Laplace Transform in Circuit Analysis
[d] vg = m tu(t);
Vo =
=
K1 =
Vg =
m
;
s2
−108 s
m
· 2
8000(s + 2500)(s + 5000) s
−12,500m
.
s(s + 2500)(s + 5000)
−12,500m
= −m × 10−3 ;
(2500)(5000)
.·. −5 = −m × 10−3
.·. m = 5000 V/s.
P 13.48 [a]
Vp − Vg2
Vp s
+
= 0;
50 × 106
20,000
Vp =
2500
Vg2 ;
s + 2500
Vp − Vg1 Vp − Vo (Vp − Vo )s
= 0;
+
+
80,000
20,000
108
(s + 6250)Vp − (s + 5000)Vo = 1250Vg1 ;
(s + 6250)(2500)
.·. (s + 5000)Vo =
Vg2 − 1250Vg1.
(s + 2500)
Vg1 =
16
;
s
.·. Vo =
=
K1 =
8
Vg2 = ;
s
7500 × 104
s(s + 2500)(s + 5000)
K2
K3
K1
+
+
.
s
s + 2500 s + 5000
7500 × 104
750
=
= 6;
(2500)(5000)
125
Problems
K2 =
7500 × 104
= −12;
(−2500)(2500)
K3 =
7500 × 104
= 6;
(−5000)(−2500)
13–51
vo = [6 − 12e−2500t + 6e−5000t ]u(t) V.
[b] 6 − 12e−2500ts + 6e−5000ts = 5;
let x = e−2500ts ;
6 − 12x + 6x2 = 5.
1
= 0.
6
x2 − 2x +
x=1−
q
5/6 = 0.0871.
.·. e−2500ts = 0.0871;
ts = 976.15 µs.
1
1/RC
1/sC
=
=
.
R + 1/sC
RsC + 1
s + 1/RC
There are no zeros, and a single pole at −1/RC rad/s.
R
R/L
[b]
=
.
R + sL
s + R/L
There are no zeros, and a single pole at −R/L rad/s.
P 13.49 [a]
[c] There are several possible solutions. For the RL circuit choose L = 10
mH. Then R = (5000)(0.01) = 50 Ω. Use 2 100 Ω resistors in parallel to
get the equivalent 50 Ω resistance needed.
For the RC circuit choose C = 1 µF. Then R = 1/(5000 × 10−6 ) = 200 Ω.
Use 2 100 Ω resistors in series to get the 200 Ω resistance required.
R
RsC
s
=
=
.
R + 1/sC
RsC + 1
s + 1/RC
There is a single zero at 0 rad/s, and a single pole at −1/RC rad/s.
s
sL
=
.
[b]
R + sL
s + R/L
There is a single zero at 0 rad/s, and a single pole at −R/L rad/s.
P 13.50 [a]
[c] There are several possible solutions. For the RL circuit choose L = 10
mH. Then R = (250)(0.01) = 2.5 Ω. Use 4 10 Ω resistors in parallel to get
the equivalent 2.5 Ω resistance needed.
For the RC circuit choose C = 1 µF. Then R = 1/(250 × 10−6 ) = 4 kΩ.
Use a 1.8 kΩ and a 2.2 kΩ resistor in series to get the 4 kΩ resistance
required.
13–52
P 13.51 [a]
CHAPTER 13. The Laplace Transform in Circuit Analysis
1/sC + sL
s2 + 1/LC
= 2
.
1/sC + sL + R
s + (R/L)s + 1/LC
q
There are two zeros at j 1/LC rad/s, and two poles:
−p1 = −(R/2L) +
q
−p2 = −(R/2L) −
q
(R/2L)2 − (1/LC)
(R/2L)2 − (1/LC).
[b] There are several possible solutions. One is
R = 100 Ω + 150 Ω = 250 Ω;
L = 10 mH;
C = 1 µF.
These component values yield the following poles:
−p1 = −5000 rad/s
− p2 = −20,000 rad/s.
and
[c] There are several possible solutions. One is
R = 100 Ω + 100 Ω = 200 Ω;
L = 10 mH;
C = 1 µF.
These component values yield the following poles:
−p1 = −10,000 rad/s
and
− p2 = −10,000 rad/s.
[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/s
P 13.52 [a]
and
− p2 = −6000 − j8000 rad/s.
Vo
Rk1/sC
1/LC
=
= 2
.
Vi
(Rk1/sC) + sL
s + (1/RC)s + 1/LC
There are no zeros and two poles:
p1 = −(1/2RC) +
q
p2 = −(1/2RC) −
q
(1/2RC)2 − (1/LC);
(1/2RC)2 − (1/LC).
[b] There are several
qpossible solutions. Any selection of R, L, and C such
that 1/2RC > 1/LC will yield two real, distinct poles. When L = 1
mH and C = 0.1 µF, R must be less than 50 Ω. For example, let
R = 10 Ω;
L = 1 mH;
C = 0.1 µF.
These component values yield the following poles:
−p1 = −10,102.05 rad/s
and
− p2 = −989,897.95 rad/s.
Problems
[c] To get repeated real poles, 1/2RC =
R=
q
13–53
1/LC, so choose
1
= 50 Ω.
2(100,000)(0.1 × 10−6 )
We can create a resistance of 50 Ω using two parallel-connected 100 Ω
resistors. Then
R = 50 Ω;
L = 1 mH;
C = 0.1 µF,
These component values yield the following poles:
−p1 = −100,000 rad/s
and
− p2 = −100,000 rad/s,
[d] There are several possible solutions. Any resistor with a value greater
than 50 Ω will yield two complex conjugate poles. For example, let
R = 100 Ω;
L = 1 mH;
C = 0.1 µF,
These component values yield the following poles:
−p1 = −50,000 + j86,602.54 rad/s
P 13.53 [a]
− p2 = −50,000 − j86,602.54 rad/s,
1/sC
1
Vo
=
=
;
Vi
R + 1/sC
RCs + 1
H(s) =
[b]
and
(1/RC)
250
=
;
s + (1/RC)
s + 250
−p1 = −250 rad/s.
R
RCs
s
Vo
=
=
=
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 + 3 × 106
=
z1 = 0;
[d]
−p1 = −3 × 106 rad/s.
Vo
R
R/L
3 × 106
=
=
=
;
Vi
R + sL
s + (R/L)
s + 3 × 106
−p1 = −3 × 106 rad/s.
[e]
Vo s
Vo
Vo − Vi
+ 5+
= 0;
6
2.5 × 10
10
25 × 103
13–54
CHAPTER 13. The Laplace Transform in Circuit Analysis
sVo + 25Vo + 100Vo = 100Vi
H(s) =
Vo
100
=
Vi
s + 125
−p1 = −125 rad/s.
P 13.54 [a]
4
800,000Vi
Vi = Vo +
;
8
800,000 + (16 × 106 /s)
0.5Vi −
sVi
= Vo ;
s + 20
Vo
−0.5(s − 20)
.·.
= H(s) =
.
Vi
(s + 20)
[b] −z1 = 20 rad/s;
−p1 = −20 rad/s.
P 13.55 [a] Zi = 10,000 +
Zf =
109
104 (s + 5000)
=
;
20s
s
25,000
250 × 106
=
;
(25,000)(4 × 10−9 )s + 1
s + 10,000
H(s) = −
Zf
−25,000s
=
.
Zi
(s + 5000)(s + 10,000)
[b] Zero at s = 0;
Poles at −p1 = −5000 rad/s and −p2 = −10,000 rad/s.
[c] If vg (t) = u(t), Vg (s) = 1/s. Then,
Vo =
−25,000
(s + 5000)(s + 10,000)
=
5
−5
+
.
s + 5000 s + 10,000
Therefore,
vo (t) = (−5e−5000t + 5e−10,000t )u(t) V.
Note that the op amp will never saturate because the output voltage will
always stay within the two power supply voltages.
Problems
P 13.56 [a]
Va − Vg
Va s
(Va − Vo )s
+
+
= 0;
8
8000
5 × 10
1.25 × 108
Vo
−Va s
−
= 0;
8
5 × 10
62,500
Va =
−8000Vo
;
s
−8000Vo
.·.
(5s + 62,500) − 4sVo = 62,500Vg ;
s
−15,625s
Vo
.·. H(s) =
= 2
.
Vg
s + 10,000s + 125 × 106
√
s1,2 = −5000 ± 25 × 106 − 125 × 106 = −5000 ± j10,000;
H(s) =
−15,625s
.
(s + 5000 − j10,000)(s + 5000 + j10,000)
[b] −p1 = −5000 + j10,000 rad/s;
−p2 = −5000 − j10,000 rad/s;
z = 0.
[c] If vi (t) = u(t), Vi (s) = 1/s. Then,
Vo =
−15,625
(s + 5000 − j10,000)(s + 5000 + j10,000)
=
K1 =
K1
K1∗
+
.
s + 5000 − j10,000 s + 5000 + j10,000
−15,625
= 0.78125/90◦ .
s + 5000 + j10,000 s=−5000+j10,000
Therefore,
vo (t) = 1.5625e−5000t cos(10,000t + 90◦ ) V.
13–55
13–56
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.57 [a] Let R1 = 40 kΩ; R2 = 10 kΩ; C2 = 500 nF; and Cf = 250 nF. Then
(R2 + 1/sC2 )1/sCf
(s + 1/R2 C2 )
=
;
Zf = C +C
R2 + sC1 2 + sC1 f
Cf s s + C22Cf Rf2
1
= 4 × 106 ;
Cf
1
= 200 rad/s;
R2 C2
750 × 10−9
C2 + Cf
=
= 600 rad/s;
C2 Cf R2
1.25 × 10−9
6
4 × 10 (s + 200)
.·. Zf =
Ω;
s(s + 600)
Zi = R1 = 40 × 103 Ω;
H(s) =
−Zf
−100(s + 200)
Vo
=
=
.
Vg
Zi
s(s + 600)
[b] −z1 = −200 rad/s;
−p1 = 0;
−p2 = −600 rad/s.
[c] If vg (t) = u(t) then Vg (s) = 1/s. Then,
Vo =
−100(s + 200)
−33.33 −0.11
0.11
=
+
+
.
2
2
s (s + 600)
s
s
s + 600
Thus,
vo = (−33.33t − 0.11 + 0.11e−600t )u(t) V.
Note that the op amp will saturate once the output voltage exceeds the
power supply voltages.
P 13.58 [a]
Vo
Vo
+
+ Vo (50 × 10−9 )s = Ig ;
10,000 2s
.·. Vo =
20 × 106 s
· Ig
s2 + 2000s + 10 × 106
60 × 10−3 s
Vo
;
Io = 4 ;
2
6
s + 16 × 10
10
2000s
.·. H(s) = 2
.
s + 2000s + 107
Ig =
Problems
13–57
(2000s)(60 × 10−3 s)
;
(s + 1000 − j3000)(s + 1000 + j3000)(s2 + 16 × 106 )
[b] Io =
Io =
120s2
.
(s + 1000 − j3000)(s + 1000 + j3000)(s + j4000)(s − j4000)
[c] Damped sinusoid of the form
M e−1000t cos(3000t + θ1 ).
[d] Steady-state sinusoid of the form
N cos(4000t + θ2 ).
[e] Io =
K1
K1∗
K2
K2∗
+
+
+
.
s + 1000 − j3000 s + 1000 + j3000 s − j4000 s + j4000
K1 =
120(−1000 + j3000)2
= 20 × 10−3 /163.74◦ ;
(j6000)(−1000 − j1000)(−j1000 + j7000)
K2 =
120(−16 × 106 )
= 24 × 10−3 / − 36.87◦ ;
(j8000)(1000 + j1000)(j1000 + j7000)
io (t) = [40e−1000t cos(3000t + 163.74◦ ) + 48 cos(4000t − 36.87◦ )] mA.
Test:
io (0) = 0.040 cos(163.74◦ ) + 0.048 cos(−36.87◦ ) = −0.0384 + 0.0384 = 0;
Z=
1
;
Y
Y =
1
1
1
1 + j0.75
+
+
=
;
10,000 j8000 −j5000
10,000
10,000
.·. Z =
= 8000/ − 36.87◦ Ω.
1 + j0.75
Vo = Ig Z = (60 × 10−3 /0◦ )(8000/ − 36.87◦ ) = 480/ − 36.87◦ V;
Io =
Vo
= 48/ − 36.87◦ mA;
104
ioss = 48 cos(4000t − 36.87◦ ) mA(checks).
P 13.59 [a]
1000(Io − Ig ) + 500Io + µ(Ig − Io )(1000) + 10sIo = 0;
13–58
CHAPTER 13. The Laplace Transform in Circuit Analysis
.·. Io =
100(1 − µ)
Ig ;
s + 100(1.5 − µ)
.·. H(s) =
100(1 − µ)
.
s + 100(1.5 − µ)
[b] µ < 1.5.
[c]
µ
H(s)
Io
−0.5 150/(s + 200)
1500/s(s + 200)
0 100/(s + 150)
1000/s(s + 150)
1.0 0
0
1.5 −50/s
−500/s2
2.0 −100/(s − 50) −1000/s(s − 50)
µ = −0.5:
Io =
7.5
7.5
−
;
s
(s + 200)
io = [7.5 − 7.5e−200t ]u(t), A.
µ = 0:
Io =
20/3
20/3
−
;
s
s + 150
µ=1:
io =
20
[1 − e−150t ]u(t), A.
3
io = 0 A.
µ = 1.5:
−500
;
s2
µ = 2:
Io =
Io =
io = −500t u(t) A.
20
20
−
;
s
s − 50
P 13.60
Vg = 12sI1 − 8sIo ;
io = 20[1 − e50t ]u(t) A.
Problems
0 = −8sI1 + (12s + 8000 + 4 × 106 /3s)Io ;
−8s 12s + 8000 + 4 × 106 /3s
No =
Io =
−8s
12s
∆=
12s Vg
= 80(s + 200)(s + 1000);
= 8sVg ;
−8s 0
No
8sVg
=
;
∆
80(s + 200)(s + 1000)
H(s) =
Io
0.1s
=
.
Vg
(s + 200)(s + 1000)
z1 = 0;
−p1 = −200 rad/s;
−p2 = −1000 rad/s.
P 13.61 [a] 0 ≤ t ≤ 20:
y(t) =
Z t
t
(5)(1)(dλ) = 5λ = 5t.
0
0
20 ≤ t ≤ 40:
y(t) =
Z 20
t−20
20
= 5(40 − t).
(5)(1)(dλ) = 5λ
t−20
13–59
13–60
CHAPTER 13. The Laplace Transform in Circuit Analysis
t ≥ 40 :
y(t) = 0.
[b] 0 ≤ t ≤ 5:
y(t) =
Z t
t
20 dλ = 20λ = 20t.
0
0
5 ≤ t ≤ 20:
y(t) =
Z t
t
20 dλ = 20λ
t−5
= 100.
t−5
20 ≤ t ≤ 25:
y(t) =
Z 20
20
t−5
t ≥ 25 :
= 20(25 − t).
20 dλ = 20λ
t−5
y(t) = 0.
Problems
13–61
[c] The expressions are
0≤t≤1:
y(t) =
Z t
t
100 dλ = 100λ = 100t;
0
1 ≤ t ≤ 20 :
y(t) =
0
Z t
t
100 dλ = 100λ
t−1
20 ≤ t ≤ 21 :
y(t) =
Z 20
20
= 100(21 − t);
100 dλ = 100λ
t−1
21 ≤ t < ∞ :
= 100;
t−1
t−1
y(t) = 0.
[d]
[e] Yes, note that h(t) is approaching 20δ(t), therefore y(t) must approach
20x(t), i.e.
y(t) =
Z t
h(t − λ)x(λ) dλ →
0
Z t
20δ(t − λ)x(λ) dλ
0
→ 20x(t).
This can be seen in the plot, e.g., in part (c), y(t) ∼
= 20x(t).
P 13.62 [a]
y(t) = 0
t < 0;
13–62
CHAPTER 13. The Laplace Transform in Circuit Analysis
0≤t≤8:
y(t) =
Z t
400 dλ = 400t;
0
8 ≤ t ≤ 16 :
y(t) =
Z 8
400 dλ = 400(8 − t + 8) = 400(16 − t);
t−8
16 ≤ t < ∞ :
y(t) = 0.
[b]
y(t) = 0
t < 0;
0≤t≤8:
y(t) =
Z t
200 dλ = 200t;
0
8 ≤ t ≤ 10 :
y(t) =
Z t
200 dλ = 200(t − t + 8) = 1600;
t−8
10 ≤ t ≤ 18 :
y(t) =
Z 10
t−8
18 ≤ t < ∞ :
y(t) = 0.
200 dλ = 200(18 − t);
Problems
[c]
y(t) = 0
t < 0;
0 ≤ t ≤ 0.5 :
y(t) =
Z t
400 dλ = 400t;
0
0.5 ≤ t ≤ 8 :
y(t) =
Z t
400 dλ = 400(t − t + 0.5) = 200;
t−0.5
8 ≤ t ≤ 8.5 :
y(t) =
Z 8
400 dλ = 400(8.5 − t);
t−0.5
8.5 ≤ t < ∞ :
y(t) = 0.
P 13.63 [a] −1 ≤ t ≤ 4:
vo = 20
Z t+1
3λ dλ = 30λ2
0
t+1
= 30t2 + 60t + 30 V.
0
4 ≤ t ≤ 7:
Z 5
vo = 20
Z t+1
3λ dλ + 20
0
5
= 30λ2
t+1
0
t+1
−10λ2
+400λ
2
(20 − λ) dλ
5
5
= −10t + 380t − 610 V.
5
13–63
13–64
CHAPTER 13. The Laplace Transform in Circuit Analysis
7 ≤ t ≤ 12:
Z 5
vo = 20
Z t+1
3λ dλ + 20
t−7
(20 − λ) dλ
5
5
t+1
= 30λ2
t−7
t+1
−10λ2
+400λ
5
5
2
= −40t + 800t − 2080 V.
12 ≤ t ≤ 19:
Z t+1
vo = 20
t+1
t−7
t+1
−10λ2
(20 − λ) dλ = 400λ
t−7
t−7
= −160t + 3680 V.
19 ≤ t ≤ 27:
Z 20
vo = 20
20
t−7
t−7
= 10t2 − 540t + 7290 V.
[b]
P 13.64 H(s) =
Vo
1
=
;
Vi
s+1
h(t) = e−t .
20
−10λ2
(20 − λ) dλ = 400λ
t−7
Problems
For 0 ≤ t ≤ 1:
vo =
Z t
e−λ dλ = (1 − e−t ) V.
0
For 1 ≤ t ≤ ∞:
vo =
Z t
e−λ dλ = (e − 1)e−t V.
t−1
P 13.65 H(s) =
s
1
Vo
=
=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
(−e−λ ) dλ = e−λ
t−1
t
= (1 − e)e−t V.
t−1
P 13.66 [a] h(λ) =
5
λ
10
0 ≤ λ ≤ 10 s;
h(λ) = 10 −
5
λ
10
h(λ) = 0
20 ≤ λ ≤ ∞.
10 ≤ λ ≤ 20 s;
0 ≤ t ≤ 10 s:
vo =
Z t
0
(0.5λ)(4) dλ = 2
λ2 t
= t2 V.
2 0
13–65
13–66
CHAPTER 13. The Laplace Transform in Circuit Analysis
10 ≤ t ≤ 20 s:
vo =
Z 10
2λ dλ +
0
Z t
4(10 − 0.5λ) dλ;
10
vo = 100 + 40t − 400 − t2 + 100 = 40t − 200 − t2 V.
20 ≤ t ≤ ∞:
vo =
Z 10
2λ dλ +
0
Z 20
4(10 − 0.5λ) dλ;
10
vo = 100 + 400 − (400 − 100) = 200 V.
[b]
0 ≤ λ ≤ 2.5 s;
[c] h(λ) = 8λ
h(λ) = 40 − 8λ
2.5 ≤ λ ≤ 5 s;
5 ≤ λ ≤ ∞.
h(λ) = 0
0 ≤ t ≤ 2.5 s:
vo =
Z t
32λ dλ = 16t2 V.
0
2.5 ≤ t ≤ 5 s:
vo =
Z 2.5
32λ dλ +
0
Z t
4(40 − 8λ) dλ = 160t − 200 − 16t2 V.
2.5
5 ≤ t ≤ ∞:
vo =
Z 2.5
0
32λ dλ +
Z 5
2.5
4(40 − 8λ) dλ = 200 V.
Problems
13–67
[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 13.67 [a] From Problem 13.53(a)
250
H(s) =
;
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
16(250)e−250λ dλ = 16(e − 1)e−250t V.
t−0.004
[b]
2500
s + 2500
0 ≤ t ≤ 4 ms:
P 13.68 [a] H(s) =
vo =
Z t
.·. h(λ) = 2500e−2500λ .
16(2500)e−2500λ dλ = 16(1 − e−2500t ) V.
0
4 ms ≤ t ≤ ∞:
vo =
Z t
t−0.004
16(2500)e−2500λ dλ = 16(e10 − 1)e−2500t V.
13–68
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] Decrease.
[c] The circuit with R = 10 kΩ.
P 13.69 [a]
vo =
Z t
20(20e−2λ ) dλ
0
= 400
e−2λ t
= −200[e−2t − 1]
−2 0
= 200(1 − e−2t ) V,
[b]
0 ≤ t ≤ ∞.
Problems
0 ≤ t ≤ 1:
vo =
Z t
t
400(1 − λ) dλ = 400(λ − λ2 /2) = 200t(2 − t) V;
0
0
1 ≤ t ≤ ∞:
vo =
Z 1
1
400(1 − λ) dλ = 400(λ − λ2 /2) = 200 V.
0
0
[c]
P 13.70 H(s) =
Vo
8s
0.8s
=
=
Vi
50 + 10s
s+5
4
5
= 0.8 −
;
= 0.8 1 −
s+5
s+5
h(t) = 0.8δ(t) − 4e−5t .
vo =
=
Z t
75[0.8δ(λ) − 4e−5λ ] dλ
0
Z t
Z t
60δ(λ) dλ − 300
0
e−5λ dλ
0
= 60 − 300
e−5λ t
−5 0
= 60 + 60[e−5t − 1] = 60e−5t V
P 13.71 [a] H(s) =
=
0 ≤ t ≤ ∞.
Vo
1/LC
= 2
Vi
s + (R/L)s + (1/LC)
25
25
=
.
s2 + 10s + 25
(s + 5)2
h(λ) = 25λe−5λ u(λ).
13–69
13–70
CHAPTER 13. The Laplace Transform in Circuit Analysis
0 ≤ t ≤ 0.10s:
Z t
vo = 250
λe−5λ dλ
0
t
e−5λ
= 250
(−5λ − 1)
25
0
(
)
= 10[1 − e−5t (5t + 1)] V.
0.1 ≤ t ≤ ∞:
Z t
vo = 250
λe−5λ dλ
t−0.1
t
e−5λ
= 250
(−5λ − 1)
25
t−0.1
(
)
= −10e−5t [(5t + 1) − e0.5 (5t + 0.5)] V.
[b]
P 13.72 [a] Io =
800Ig
20Ig
=
;
25 + 0.025s
s + 1000
Io
800
= H(s) =
;
Ig
s + 1000
Problems
h(λ) = 800e−1000λ u(λ).
0 ≤ t ≤ 1 ms:
io =
Z t
(10 × 10−6 )(800)e−1000λ dλ = 0.008
0
e−1000λ t
−1000 0
= 8(1 − e−1000t ) µA.
1 ms ≤ t ≤ 4 ms:
io =
Z t−1×10−3
(−5 × 10−6 )(800e−1000λ dλ)
0
Z t
+
t−1×10−3
(10 × 10−6 )(800e−1000λ dλ)
e−1000λ t−1×10
= −0.004
−1000 0
−3
+0.008
e−1000λ t
−1000 t−1×10−3
13–71
13–72
CHAPTER 13. The Laplace Transform in Circuit Analysis
h
i
i
h
= 4 e−1000(t−0.001) − 1 − 8 e−1000t − e−1000(t−0.001) ;
= [12e−1000(t−0.001) − 8e−1000t − 4] µA.
4 ms ≤ t ≤ ∞:
io =
Z t−0.001
−1000λ
− 0.004e
dλ +
Z t
t−0.004
"
−1000λ
t−0.001
−1000λ
t−0.004
−1000(t−0.001)
[b] Vo = 0.025sIo =
#
t
× 10−6 ;
−8e
= 4e
= [12e
0.008e−1000λ dλ
t−0.001
t−0.001
−1000(t−0.004)
− 4e
− 8e−1000t ] µA.
20sIg
;
s + 1000
Vo
20s
20,000
= H(s) =
= 20 −
;
Ig
s + 1000
s + 1000
h(λ) = 20δ(λ) − 20,000e−1000λ .
Problems
0 < t < 0.001 s:
vo =
Z t
(10 × 10−6 )[20δ(λ) − 20,000e−1000λ ] dλ
0
= 200 × 10−6 − 0.2
e−1000λ t
−1000 0
= 200 × 10−6 + 200 × 10−6 [e−1000t − 1] = 200e−1000t µV.
0.001 s < t < 0.004 s:
vo =
Z t−0.001
(−5 × 10−6 )[20δ(λ) − 20,000e−1000λ ] dλ
0
Z t
+
(10 × 10−6 )(−20,000e−1000λ ) dλ
t−0.001
= −100 × 10−6 + 0.1
e−1000λ t−0.001
e−1000λ t
−0.2
−1000 0
−1000 t−0.001
13–73
13–74
CHAPTER 13. The Laplace Transform in Circuit Analysis
= −100 × 10−6 − 0.1 × 10−3 e−1000(t−0.001) + 0.1 × 10−3
+0.2 × 10−3 e−1000t − 0.2 × 10−3 e−1000(t−0.001)
= 200e−1000t − 300e−1000(t−0.001) µV.
0.004 s < t < ∞:
vo =
Z t−0.001
(−5 × 10−6 )(−20,000e−1000λ ) dλ
t−0.004
Z t
+
(10 × 10−6 )(−20,000e−1000λ ) dλ
t−0.001
= 200e−1000t − 300e−1000(t−0.001) + 100e−1000(t−0.004) µV.
[c] At t = 0.001− :
io = 8(1 − e−1 ) = 5.06 µA;
i20Ω = (10 − 5.06) = 4.94 µA;
.·. vo = 20(4.94 × 10−6 ) − 5(5.06 × 10−6 ) = 73.58 µV.
From the solution for vo we have
vo (0.001− ) = 200e−1 = 73.58 µV (checks).
At t = 0.001+ :
io (0.001+ ) = io (0.001− ) = 5.06 µA;
i20Ω = (−5 − 5.06) µA = −10.06 µA;
.·. vo (0.001+ ) = 20(−10.06 × 10−6 ) − 5(5.06 × 10−6 ) = −226.42 µV.
From the solution for vo we have
vo (0.001+ ) = 200e−1 − 300 = −226.42 µV (checks).
Problems
At t = 0.004− :
io = 12e−3 − 8e−4 − 4 = −3.55 µA;
i20Ω = (−5 + 3.55) = −1.45 µA;
vo = 20(−1.45 × 10−6 ) − 5(−3.55 × 10−6 ) = −11.27 µV.
From the solution for vo ,
vo (0.004− ) = 200e−4 − 300e−3 = −11.27 µV (checks).
At t = 0.004+ :
io (0.004+ ) = io (0.004− ) = −3.55 µA;
i20Ω = 3.55 µA;
vo (0.004+ ) = 20(3.55 × 10−6 ) + 5(3.55 × 10−6 ) = 88.73 µV.
From the solution for vo ,
vo (0.004+ ) = 200e−4 − 300e−3 + 100 = 88.73 µV(checks).
60
Ig ;
100
Vg
.·. Io = ;
50
P 13.73 [a] Io =
Ig =
Vg
;
30
H(s) =
Io
1
= ;
Vg
50
h(λ) = 0.02δ(λ).
[b]
0 < t < 0.5 s :
io =
Z t
100[0.02δ(λ)] dλ = 2 A.
0
0.5 s ≤ t ≤ 1.0 s:
io =
Z t−0.5
0
−100[0.02δ(λ)] dλ = −2 A.
13–75
13–76
CHAPTER 13. The Laplace Transform in Circuit Analysis
1s < t < ∞ :
vo = 0.
[c]
Yes, because the circuit has no memory.
P 13.74 [a]
Vo − Vg Vo s Vo
+
+
= 0;
40
8
10
(5s + 5)Vo = Vg ;
H(s) =
Vo
0.2
=
;
Vg
s+1
h(λ) = 0.2e−λ u(λ).
[b] 0 ≤ t ≤ 0.5 s:
vo =
Z t
100(0.2e−λ ) dλ = 20
0
vo = 20 − 20e−t V,
e−λ t
;
−1 0
0 ≤ t ≤ 0.5 s.
Problems
13–77
0.5 s ≤ t ≤ 1 s:
vo =
Z t−0.5
−λ
(−100)(0.2e
) dλ +
Z t
0
100(0.2e−λ ) dλ
t−0.5
= − 20
e−λ t
e−λ t−0.5
+20
−1 0
−1 t−0.5
= 40e−(t−0.5) − 20e−t − 20 V,
0.5 s ≤ t ≤ 1 s.
1 s ≤ t ≤ ∞:
vo =
Z t−0.5
−λ
(−100)(0.2e
) dλ +
Z t
t−1
100(0.2e−λ ) dλ
t−0.5
e−λ t
e−λ t−0.5
= −20
+20
−1 t−1
−1 t−0.5
= 40e−(t−0.5) − 20e−(t−1) − 20e−t V,
1 s ≤ t ≤ ∞.
[c]
[d] No, the circuit has memory because of the capacitive storage element.
P 13.75 [a] Y (s) =
Z ∞
y(t)e−st dt.
0
Y (s) =
Z ∞
−st
Z ∞
e
0
=
=
Z ∞
0
h(λ)x(t − λ) dλ dt
0
Z ∞Z ∞
0
e−st h(λ)x(t − λ) dλ dt
0
Z ∞
h(λ)
0
e−st x(t − λ) dt dλ.
13–78
CHAPTER 13. The Laplace Transform in Circuit Analysis
But x(t − λ) = 0 when t < λ.
Therefore Y (s) =
Z ∞
h(λ)
0
Let u = t − λ;
Y (s) =
Z ∞
Z ∞
0
=
Z ∞
Z ∞
u = 0 when t = λ;
u = ∞ when t = ∞.
e−s(u+λ) x(u) du dλ
0
h(λ)e−sλ
0
=
e−st x(t − λ) dt dλ.
λ
du = dt;
h(λ)
Z ∞
Z ∞
e−su x(u) du dλ
0
h(λ)e−sλ X(s) dλ = H(s) X(s).
0
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.
[b] F (s) =
a
1
a
= ·
= H(s)X(s);
2
s(s + a)
s (s + a)2
.·. h(t) = u(t),
x(t) = at e−at u(t).
Problems
.·. f (t) =
Z t
"
−aλ
(1)aλe
dλ = a
0
=
t
e−aλ
(−aλ − 1)
a2
0
1
1 −at
[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
K1
K2
a
=
+
+
;
2
2
s(s + a)
s
(s + a)
s+a
1
K0 = ;
a
d a
K2 =
ds s
K1 = −1;
1
1
− te−at − e−at u(t).
f (t) =
a
a
P 13.76 vi = 10 sin λ [u(λ) − u(λ − π)];
H(s) =
1
;
s+1
h(λ) = e−λ ;
h(t − λ) = e−(t − λ) = e−t eλ .
1
=− ;
a
s=−a
13–79
13–80
CHAPTER 13. The Laplace Transform in Circuit Analysis
−t
vo = 10e
Z t
"
−t
= 10e
eλ sin λ dλ
0
t
eλ
(sin λ − cos λ)
2
0
#
= 5e−t [et (sin t − cos t) + 1]
= 5(sin t − cos t + e−t ) V;
vo (2.2) = 7.539 V.
P 13.77
Vo =
40 × 103 Ig
(10 × 103 );
50 × 103 + 12.5 × 106 /s
Vo
8000s
= H(s) =
;
Ig
s + 250
250
2 × 106
H(s) = 8000 1 −
= 8000 −
;
s + 250
s + 250
h(t) = 8000δ(t) − 2 × 106 e−250t .
Problems
vo =
Z 5×10−3
(−10 × 10−3 )[8000δ(λ) − 2 × 106 e−250λ ] dλ
0
Z 7×10−3
+
5×10−3
(5 × 10−3 )[−2 × 106 e−250λ ] dλ
Z 5×10−3
= −80 + 20,000
Z 7×10−3
e−250λ dλ − 10,000
0
5×10−3
e−250λ dλ
= −80 − 80(e−1.25 − 1) + 40(e−1.75 − e−1.25 )
= −120e−1.25 + 40e−1.75 = −27.43 V.
Alternate:
Ig =
Z 2×10−3
(5 × 10−3 )e−st dt +
Z 8×10−3
0
2×10−3
(−10 × 10−3 )e−st dt
5 15 −2×10−3 s 10 −8×10−3 s
=
+ e
× 10−3 ;
− e
s
s
s
Vo = Ig H(s) =
8
−3
−3
[5 − 15e−2×10 s + 10e−8×10 s ]
s + 250
−3 s
40
120e−2×10
=
−
s + 250
s + 250
vo (t) = 40e−250t − 120e−250(t−2×10
+80e−250(t−8×10
−3 )
−3
80e−8×10 s
+
;
s + 250
−3 )
u(t − 2 × 10−3 )
u(t − 8 × 10−3 ) V;
vo (7 × 10−3 ) = 40e−1.75 − 120e−1.25 + 0 = −27.43 V (checks).
P 13.78 Vo =
75
100
25
192 × 106
−
+
=
;
s
s + 800 s + 3200
s(s + 800)(s + 3200)
240
Vo = H(s)Vg = H(s)
;
s
.·. H(s) =
800,000
.
(s + 800)(s + 3200)
H(j1600) =
8 × 105
= 0.125/ − 90◦ ;
(800 + j1600)(3200 + j1600)
.·. vo (t) = (40)(0.125) cos(1600t − 90◦ ) V = 5 sin 1600t V.
13–81
13–82
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.79 H(j8000) =
=
104 (6000 + j8000)
−64 × 106 + j7 × 106 + 88 × 106
107 (6 + j8)
= 4/36.87◦ ;
106 (24 + j7)
.·. vo (t) = 50 cos(8000t + 36.87◦ ) V.
P 13.80 H(j20) =
25(8 + j20)
= 0.44/ − 33.57◦ ;
−400 + j1200 + 150
.·. io (t) = 4.4 cos(20t − 33.57◦ ) A.
P 13.81 [a] H(s) =
−Zf
;
Zi
Zf =
(1/Cf )
4 × 109
=
;
s + (1/Rf Cf )
s + 16,000
Zi =
Ri [s + (1/Ri Ci )]
25,000(s + 8000)
=
;
s
s
H(s) =
−16 × 104 s
.
(s + 8000)(s + 16,000)
[b] H(j8000) =
√
−16 × 104 (j8000)
= 40/ − 161.57◦ ;
(8000 + j8000)(16,000 + j8000)
√
√
vo (t) = (200 10) × 10−3 ( 40) cos(8000t − 161.57◦ )
= 4 cos(8000t − 161.57◦ ) V.
P 13.82 [a] Let R1 = 10 kΩ,
R2 = 50 kΩ,
Then Vn = Vp =
Vg R2
.
R2 + (1/sC)
Also
C = 400 pF,
R2 C = 2 × 10−5 .
Vn − Vg Vn − Vo
+
= 0,
R1
R1
therefore Vo = 2Vn − 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.
Problems
[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
=
.
−6
j20 × 10 Rx + 1
Rx − j50,000
Thus,
50,000
= tan 60◦ = 1.7321,
Rx
Rx = 28,867.51 Ω.
P 13.83 [a] The s-domain circuit is
The node-voltage equation is
V
V
V
ρ
+ +
=
sL1 R sL2
s
ρR
s + (R/Le )
L1 L2
;
L1 + L2
So V =
where Le =
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
Therefore i2 =
K3 =
−ρL1
;
L1 + L2
ρL1
[1 − e−(R/Le )t ]u(t).
L1 + L2
[d] λ(t) = L1 i1 + L2 i2 = ρL1 .
13–83
13–84
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.84 [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
so V =
ρL1 L2
= ρLe ;
L1 + L2
Therefore v(t) = ρLe δ(t)
V
ρL1
=
sL2
L1 + L2
I1 = I2 =
Thus i1 = i2 =
1
;
s
ρL1
u(t) A.
L1 + L2
P 13.85 Original charge on C1 ; q1 = V0 C1 .
The charge transferred to C2 ; q2 = V0 Ce =
The charge remaining on C1 ;
V0 C1 C2
.
C1 + C2
q10 = q1 − q2 =
V0 C1
q2
=
Therefore V2 =
C2
C1 + C2
V0 C12
.
C1 + C2
q10
V0 C1
and V1 =
=
.
C1
C1 + C2
P 13.86 [a]
Vo =
0.75
· 20 × 10−3 s
100 + 25 × 10−3 s
=
0.6s
2400
= 0.6 −
;
s + 4000
s + 4000
Problems
13–85
vo (t) = 0.6δ(t) − 2400e−4000t u(t) V.
[b] At t = 0 the voltage impulse establishes a current in the inductors; thus
iL (0) =
103 Z 0+
750 × 10−3 δ(t) dt = 30 A.
25 0−
It follows that since iL (0− ) = 0 that
diL
(0) = 30δ(t);
dt
.·. vo (0) = (20 × 10−3 )(30δ(t)) = 0.6δ(t).
This agrees with our solution.
At t = 0+ our circuit is
.·. iL (t) = 30e−t/τ A,
t ≥ 0+ ;
τ = L/R = 0.25 ms;
.·. iL (t) = 30e−4000t A,
vo (t) = 20 × 10−3
t ≥ 0+ ;
diL
= −2400e−4000t V,
dt
t ≥ 0+ ;
which agrees with our solution.
P 13.87 [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,
So vo (0+ ) = 106
Z 0+ "
0−
#
δ(t)
dt = 1000 V;
1000
Therefore wC = (0.5)Cv 2 = 0.5 J.
13–86
CHAPTER 13. The Laplace Transform in Circuit Analysis
[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 .
6
10
250 + 0.05s 1000
Note that this equation is identical to that derived in part [c], therefore
the solution for Vo will be the same.
P 13.88 [a]
Vo =
=
2.7
2 × 106
·
54 × 103 + 25 × 106 /s + 2 × 106 /s
s
5.4 × 106
100
;
=
3
6
54 × 10 s + 27 × 10
s + 500
vo (t) = 100e−500t u(t) V.
Problems
13–87
[b] At t = 0 the impulsive current passes through the two capacitors. The
voltage on the 0.04 µF capacitor at t = 0+ is
v0.04 = 25 × 106
Z 0+
0−
50 × 10−6 δ(t) dt = 1250 V.
The voltage on the 0.5 µF capacitor at t = 0+ is
v0.5 = 2 × 106
Z 0+
0−
50 × 10−6 δ(t) dt = 100 V.
Note this agrees with our solution.
At t = 0+ the circuit is
The equivalent capacitance is
(0.04)(0.5) × 10−12
1
Ce =
=
µF.
−6
0.54 × 10
27
Thus, the time constant is
τ = 54 × 103 Ce = 2 ms.
Therefore, 1/τ = 500, which agrees with our solution.
It follows that
vR (t) = 1350e−500t V,
t ≥ 0+ .
Therefore
0.04
vo (t) =
vR = 100e−500t V,
t ≥ 0+ ,
0.54
which also agrees with our solution.
P 13.89 [a] For t < 0:
Req = 4 kΩk20 kΩk80 kΩ = 3.2 kΩ;
i1 (0− ) =
64,000
= 3.2 A;
20,000
i2 (0− ) =
v = 20(3200) = 64 kV;
64,000
= 0.8 A.
80,000
13–88
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] For t > 0:
i1 + i2 = 0;
400(∆i1 ) = 100(∆i2 );
i1 (0− ) + ∆i1 + i2 (0− ) + ∆i2 = 0;
therefore ∆i1 = −0.8 A;
i1 (0+ ) = 3.2 − 0.8 = 2.4 A.
∆i2 = −3.2 A;
[c] i2 (0− ) = 0.8 A.
[d] i2 (0+ ) = 0.8 − 3.2 = −2.4 A.
[e] The s-domain equivalent circuit for t > 0 is
I1 =
2.4
1.2
=
;
0.5s + 100,000
s + 2 × 105
5
i1 (t) = 2.4e−2×10 t u(t) A.
5
[f ] i2 (t) = −i1 (t) = −2.4e−2×10 t u(t) A.
−0.32(s + 650 × 103 )
[g] V = −1.28 + (0.4s + 20,000)I1 =
s + 2 × 105
= −0.32 −
144,000
;
s + 2 × 105
5
v(t) = [−0.32δ(t)] − [144,000e−2×10 t u(t)] V.
P 13.90 [a]
50 = 2sI1 − sI2 ;
Problems
13–89
1.5
0 = −sI1 + 2s +
I2 ;
s
−s
2s
∆=
= 4s2 + 3 − s2 = 3(s2 + 1);
−s (2s + 1.5/s)
−s
50
N1 =
= 100s +
0 (2s + 1.5/s)
75
100s2 + 75
100(s2 + 0.75)
=
=
;
s
s
s
N1
100(s2 + 0.75)
100 s2 + 0.75
I1 =
=
·
=
∆
s(3)(s2 + 1)
3 s(s2 + 1)
K1
K1∗
K0
+
+
.
s
s−j s+j
=
"
#
100 −1 + 0.75
12.5 ◦
/0 ;
K1 =
=
3
(j1)(j2)
3
100 0.75
= 25;
K0 =
3
1
25
.·. i1 = 25 +
cos t u(t) A.
3
2s 50
[b] N2 =
= 50s;
−s 0
N2
50s
50
K1
K1∗
s
=
=
=
+
.
∆
3(s2 + 1)
3 s2 + 1
s−j s+j
I2 =
50
K1 =
3
i2 =
j1
j2
!
=
25 ◦
/0 ;
3
50
cos t u(t) A.
3
1.5 50
K1
K1∗
1.5
s
25
I2 =
=
+
.
[c] Vo =
=
s
s
3 s2 + 1
s2 + 1
s−j s+j
K1 =
25
= −j12.5 = 12.5/ − 90◦ ;
j2
vo = 25 cos(t − 90◦ ) = 25 sin t;
vo = [25 sin t]u(t) V.
[d] Let us begin by noting i1 jumps from 0 to (100/3) A between 0− and 0+
and in this same interval i2 jumps from 0 to (50/3) A. Therefore in the
13–90
CHAPTER 13. The Laplace Transform in Circuit Analysis
derivatives of i1 and i2 there will be impulses of (100/3)δ(t) and
(50/3)δ(t), respectively. Thus
di1
100
50
=
δ(t) −
sin t A/s;
dt
3
3
di2
50
100
= δ(t) −
sin t A/s.
dt
3
3
From the circuit diagram we have
50δ(t) = 2
=
di1 di2
−
dt
dt
200
100
50δ(t) 100
δ(t) −
sin t −
+
sin t
3
3
3
3
= 50δ(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 50.
For the given polarity dots and reference directions of i1 and i2 we have
λ(0+ ) = L1 i1 (0+ ) + M i1 (0+ ) − L2 i2 (0+ ) − M i2 (0+ );
λ(0+ ) = 2
=
P 13.91 [a] Z1 =
100
50
50
100
+
−2
−
3
3
3
3
300 150
−
= 50. (checks)
3
3
1/C1
20 × 1010
=
Ω;
s + 1/R1 C1
s + 20 × 104
Z2 =
1/C2
5 × 1010
=
Ω;
s + 1/R2 C2
s + 12,500
Problems
Vo − 10/s
Vo
+
= 0;
Z2
Z1
Vo (s + 12,500) Vo (s + 20 × 104 )
10 (s + 20 × 104 )
+
=
;
5 × 1010
20 × 1010
s
20 × 1010
Vo =
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)5 × 1010
"
162,500s + 25 × 108
1+
s(s + 50,000)
"
K1
K2
1+
+
.
s
s + 50,000
−12
= 40 × 10
−12
= 40 × 10
K1 = 50,000;
#
#
K2 = 112,500;
io = 40δ(t) + [2 × 106 + 4.5 × 106 e−50,000t ]u(t) pA.
[c] When
C1 = 80 pF,
Z1 =
125 × 108
Ω.
s + 12,500
Vo (s + 12,500) Vo (s + 12,500)
10 (s + 12,500)
+
=
;
500 × 108
125 × 108
s 125 × 108
40
.·. Vo + 4V0 = ;
s
8
Vo = ;
s
vo = 8u(t) V.
V0
8 (s + 12,500)
12,500
−12
I0 =
=
=
160
×
10
1
+
;
Z2
s 5 × 1010
s
io (t) = 160δ(t) + 2 × 106 u(t) pA.
13–91
13–92
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.92 Let a =
1
1
=
.
R1 C1
R2 C2
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
P 13.93 [a] For t < 0,
125v1 = 500v2 ;
v1 + v2 = 25;
C1
.
C1 + C2
therefore v1 = 4v2 ;
therefore v1 (0− ) = 20 V.
[b] v2 (0− ) = 5 V.
[c] v3 (0− ) = 0 V.
[d] For t > 0:
I=
25/s
× 10−6 = 2 × 10−6 ;
12.5/s
i(t) = 2δ(t) µA.
109 Z 0+
2 × 10−6 δ(t) dt + 20 = −16 + 20 = 4 V.
125 0−
109 Z 0+
[f ] v2 (0+ ) = −
2 × 10−6 δ(t) dt + 5 = −4 + 5 = 1 V.
500 0−
[e] v1 (0+ ) = −
Problems
[g] V3 =
13–93
2.5 × 106
5
· 2 × 10−6 =
s
s
v3 (t) = 5u(t) V;
v3 (0+ ) = 5 V.
Check: v1 (0+ ) + v2 (0+ ) = v3 (0+ ).
P 13.94 [a] The circuit parameters are
1202
1202
1202
Ra =
= 20 Ω
Rb =
= 32 Ω
Xa =
= 40 Ω.
720
450
360
The branch currents are
120/0◦
120/0◦
= −j3 = 3/ − 90◦ A(rms);
= 6/0◦ A(rms)
I2 =
I1 =
j40
20
120/0◦
I3 =
= 3.75/0◦ A(rms);
32
.·. Io = I1 + I2 + I3 = 9.75 − j3 = 10.2/ − 17.1◦ A(rms).
Therefore,
√
i2 = 3 2 cos(ωt − 90◦ ) A
Thus,
i2 (0− ) = i2 (0+ ) = 0 A
√
io = 10.2 2 cos(ωt − 17.1◦ ) A.
and
and
√
io (0− ) = io (0+ ) = 9.75 2 A.
[b] Begin by using the s-domain circuit in Fig. 13.60 to solve for Vo
symbolically. Write a single node voltage equation:
Vo − (Vg + L` Io )
Vo
Vo
+
+
= 0;
sL`
Ra sLa
.·. Vo =
(Ra /L` )Vg + Io Ra
,
s + [Ra (La + L` )]/La L`
√
where L` = 1/60π H, La = 1/3π H, Ra = 20 Ω, and I0 Ra = 195 2 V.
Also,
Vg = 120/0◦ + (9.75 − j3)(j2) = 126 + j19.5V(rms).
Inverse phasor-transforming gives us
√
vg (t) = 2[126 cos(120πt) − 19.5 sin(120πt)]V(rms).
The Laplace transform of vg (t) is then
√
2(126s − 2340π)
Vg =
.
s2 + 14,400π 2
Thus,
√
√
√
1200π(126 2s − 2340π 2)
195 2
Vo =
+
(s + 1260π)(s2 + 14,400π 2 ) s + 1260π
13–94
CHAPTER 13. The Laplace Transform in Circuit Analysis
√
K1
K2
K2∗
195 2
=
+
+
+
.
s + 1260π s − j120π s + j120π s + 1260π
The coefficients are
√
K1 = −120.67 2 V
√
√
K2 = 60.44 2/3.357◦ V
K2∗ = 60.44 2/ − 3.357◦ .
√
√
Note that K1 + 195 2 = 74.33 2 V. Thus, the inverse transform of Vo is
√
√
vo = 74.33 2e−1260πt + 120.88 2 cos(120πt + 3.357◦ ) V.
Initially,
√
√
√
vo (0+ ) = 74.33 2 + 120.88 2 cos 3.357◦ = 195 2 V.
√
Note that at t = 0+ the initial value of i0 , which is 9.75
in the
√ 2 A, exists √
20 Ω resistor Ra . Thus, the initial value of Vo is (9.75 2)(20) = 195 2 V.
[c] The phasor domain equivalent circuit has a j2 Ω inductive impedance in
series with the parallel combination of a 20 Ω resistive impedance and a
j40 Ω inductive impedance (remember that ω = 120π rad/s). From part
(b) Vg = 126 + j19.5 = 127.5/8.8◦ V(rms). The node voltage equation in
the phasor domain circuit is
V0
V0 − 127.5/8.8◦ V0
+
+
= 0;
j2
20
j40
.·. V0 = 120.88/3.357◦ V(rms).
√
Therefore, v0 = 120.88 2 cos(120πt + 3.357◦ ) V, agreeing with the
steady-state component of the result in part (b).
[d] A plot of v0 , generated in Excel, is shown below.
Problems
P 13.95 [a] At t = 0− the phasor domain equivalent circuit is
I1 =
−j120
= −j6 = 6/ − 90◦ A (rms);
20
I2 =
−j120
= −3 = 3/180◦ A (rms);
j40
I3 =
−j120
= −j3.75 = 3.75/ − 90◦ A (rms);
32
IL = I1 + I2 + I3 = −3 − j9.75 = 10.2/ − 107.103◦ A (rms);
√
iL = 10.2 2 cos(120πt − 107.103◦ )A.
√
iL (0− ) = iL (0+ ) = −3 2A;
√
i2 = 3 2 cos(120πt + 180◦ )A.
√
i2 (0− ) = i2 (0+ ) = −3 2A;
Vg = Vo + j2IL ;
Vg = −j120 + 19.5 − j6
= 19.5 − j126 = 127.5/ − 81.2◦ V (rms);
√
vg = 127.5 2 cos(120πt − 81.2◦ )
√
= 127.5 2[cos 120πt cos 81.2◦ + sin 120πt sin 81.2◦ ]
√
√
= 19.5 2 cos 120πt + 126 2 sin 120πt V;
√
√
19.5 2s + 126 2(120π)
·
. . Vg =
.
s2 + (120π)2
s-domain circuit:
13–95
13–96
CHAPTER 13. The Laplace Transform in Circuit Analysis
where
1
1
H;
La =
H;
Ra = 20 Ω;
60π
3π
√
√
iL (0) = −3 2 A;
i2 (0) = −3 2 A.
Ll =
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 ]
Ra
= 1200π;
Ll
1
1
10( 60π
+ 3π
)
Ra (Ll + La )
=
= 1260π;
1
1
Ll La
( 3π )( 60π )
√
√
iL (0) − i2 (0) = −3 2 + 3 2 = 0;
√
√
1200π[19.5
2s
+
126
2(120π)]
.·. Vo =
2
(s + 1260π)[s + (120π)2 ]
K2
K2∗
K1
=
+
+
;
s + 1260π s − j120π s + j120π
√
√
K1 = −7.08 2
K2 = 60.44 2/ − 86.643◦ ;
√
√
.·. vo (t) = −7.08 2e−1260πt + 120.88 2 cos(120πt − 86.643◦ )V.
Check:
√
vo (0) = (−7.08 + 7.08) 2 = 0.
[b]
Problems
13–97
√
[c] In Problem 13.94, the line-to-neutral voltage spikes at 195 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.
P 13.96 [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 = 2 Ω we have
Vg = 120/θ◦ +
120/θ◦
(2)[Ra Rb + j(Ra + Rb )Xa ];
Ra Rb Xa
Ra = 20 Ω;
Rb = 32 Ω;
Xa = 40 Ω;
Vg = 120/θ◦ + 120/θ◦ (0.05 + j0.1625)
= 120/θ◦ (1.05 + j0.1625) = 127.5/(θ + 8.8)◦ ;
√
vg = 127.5 2 cos(120πt + θ + 8.8◦ )V.
Let β = θ + 8.8◦ . Then
√
vg = 127.5 2(cos 120πt cos β − sin 120πt sin β)V.
Therefore
√
127.5 2(s cos β − 120π sin β)
Vg =
.
s2 + (120π)2
The s-domain circuit becomes
where ρ1 = iL (0+ ) and ρ2 = i2 (0+ ).
13–98
CHAPTER 13. The Laplace Transform in Circuit Analysis
The s-domain node voltage equation is
Vo
Vo + ρ2 La
Vo − (Vg + ρ1 Ll )
+
+
= 0.
sLl
Ra
sLa
Solving for Vo yields
Vo =
Vg Ra /Ll + (ρ1 − ρ2 )Ra
l )Ra
]
[s + (LaL+L
a Ll
.
Substituting the numerical values
Ll =
1
H;
60π
La =
1
H;
3π
Ra = 20 Ω;
gives
Vo =
1200πVg + 20(ρ1 − ρ2 )
.
(s + 1260π)
Now determine the values of ρ1 and ρ2 .
ρ1 = iL (0+ )
IL =
=
and
ρ2 = i2 (0+ );
120/θ◦
[(Ra + Rb )Xa − jRa Rb ]
Ra Rb Xa
120/θ◦
(52(40) − j640)
640(40)
= 10.2/(θ − 17.1)◦ A(rms);
√
.·. iL = 10.2 2 cos(120πt + θ − 17.1◦ )A.
√
iL (0+ ) = ρ1 = 10.2 2 cos(θ − 17.2◦ )A;
√
√
.·. ρ1 = 9.75 2 cos θ + 3 2 sin θA.
120/θ◦
= 3/(θ − 90)◦ ;
j40
√
i2 = 3 2 cos(120πt + θ − 90◦ )A;
√
ρ2 = i2 (0+ ) = 3 2 sin θA;
√
.·. ρ1 − ρ2 = 9.75 2 cos θ.
√
(ρ1 − ρ2 )Ra = 195 2 cos θ.
I2 =
Rb = 32 Ω;
Problems
13–99
√
195
1200π
2 cos θ
· Vg +
.·. Vo =
s + 1260π
s + 1260π
√
√
"
#
1200π
127.5 2(s cos β − 120π sin β)
195 2 cos θ
=
+
s + 1260π
s2 + 14,400π 2
s + 1260π
√
K1 + 195 2 cos θ
K2
K2∗
=
+
+
.
s + 1260π
s − j120π s + j120π
Now
√
(1200π)(127.5 2)[−1260π cos β − 120π sin β]
K1 =
12602 π 2 + 14,400π 2
√
−1200(127.5 2)[1260 cos β + 120 sin β]
.
=
12602 + 14,400
Since β = θ + 8.8◦ , K1 reduces to
√
√
K1 = −120.67 2 cos θ + 7.08 2 sin θ.
From the partial fraction expansion√for Vo we see vo (t) will go directly
into steady state when K1 = −195 2 cos θ. It follows that
√
√
7.08 2 sin θ = −74.33 2 cos θ
or
tan θ = −10.5;
Therefore,
θ = −84.5597◦ .
[b] When θ = −84.5597◦ , β = −75.7623◦ .
√
2)[−120π sin(−75.7623◦ ) + j120π cos(−75.7623◦ )
1200π(127.5
·
. . K2 =
(1260π + j120π)(j240π)
√
637.5 2(116.314 + j29.5135)
=
−120 + j1260
√
= 60.44 2/ − 81.20◦ ;
√
.·. vo = 120.88 2 cos(120πt − 81.20◦ ) V t > 0
= 170.95 cos(120πt − 81.20◦ ) V t > 0.
[c] vo = 169.71 cos(120πt − 84.5597◦ )V
t < 0;
vo = 170.95 cos(120πt − 81.20◦ )V
t > 0.
13–100
CHAPTER 13. The Laplace Transform in Circuit Analysis
Introduction to
Frequency-Selective Circuits
Assessment Problems
AP 14.1 fc = 4 kHz,
ωc =
1
;
RC
ωc = 2πfc = 8π krad/s;
C = 10 nF;
1
1
=
= 3978.9 Ω.
.·. R =
3
ωc C
(8π × 10 )(10 × 10−9 )
AP 14.2 [a] ωc = 2πfc = 2π(10,000) = 20π krad/s;
R = Lωc = (10−3 )(20,000π) = 62.8 Ω.
[b] H(jω) =
ωc
20,000π
=
.
ωc + jω
20,000π + jω
When ω = 2πf = 2π(100,000) = 200,000π rad/s,
H(j200,000π) =
20,000π
1
=
= 0.0995/ − 84.3◦ ;
20,000π + j200,000π
1 + j10
.·. |H(j200,000π)| = 0.0995.
[c] .·. θ(200,000π) = −84.3◦ .
AP 14.3 ωc =
15,000
R
=
= 1.5 Mrad/s.
L
10 × 10−3
AP 14.4 [a] ωc =
1
1
=
= 425.5 rad/s.
RC
(50)(47 × 10−6 )
14–1
14–2
CHAPTER 14. Introduction to Frequency-Selective Circuits
[b] ωc =
[c] ωc =
1
(103 )(47 × 10−6 )
= 21.3 rad/s.
1
= 4.26 rad/s.
(5000)(47 × 10−6 )
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 ωo2 =
1
LC
(1/RC)
L
s + R+R
RL
1
RC
=
(1/RC)
s + K1
1
RC
,
RL
.
R + RL
so C =
ωo
ωo
=
β
R/L
Q=
Z
RL
=
R+Z
R(RL Cs + 1) + RL
1
1
=
= 40.53 nF;
2
3
ωo L
(50π × 10 )2 (1 × 10−3 )
so R =
ωo L
(50π × 103 )(1 × 10−3 )
=
= 19.63 Ω.
Q
8
AP 14.7 ωo = 2π(8000) = 16,000π rad/s;
β = 2π(500) = 1000π rad/s;
1
RC
so R =
ωo2 =
1
LC
so L =
AP 14.8 ωo2 =
1
LC
so C =
1
RC
so R =
β=
β=
C = 25 × 10−6 F;
1
1
=
= 12.73 Ω;
βC
(1000π)(25 × 10−6 )
1
ωo2 C
=
1
(16,000π)2 (25 × 10−6 )
= 15.83 µH.
1
1
=
= 1.0132 µF;
2
4
ωo L
(10 π)2 (10−3 )
1
1
=
= 785.4 Ω.
βC
400π(1.0132 × 10−6 )
Problems
AP 14.9 ωo2 =
1
LC
1
so L =
ωo2 C
=
1
14–3
= 15.83 mH.
(16,000π)2 (25 × 10−9 )
In Example 14.6(d), Q = fo /β = 5000/200 = 25. In this problem, fo = 8 kHz
but Q is unchanged, so calculate the new bandwidth:
β=
fo
8000
=
= 320 Hz.
Q
25
1
1
.·. R =
=
= 19.89 kΩ.
βC
2π(320)(25 × 10−9 )
AP 14.10 ωo = 30,000π rad/s;
L = 10 µH;
ωo2 =
1
LC
Q=
ωo
= RCωo ;
β
so C =
1
= 11.26 µF;
ωo2 L
Q
6
.·. R =
=
= 5.65 Ω.
ωo C
(30,000π)(11.26 × 10−6 )
AP 14.11 ωo = 2πfo = 2π(4000) = 8π krad/s;
Q=
ωo
ωo
=
β
(1/RC)
ωo2 =
1
LC
so L =
so C =
1
ωo2 C
=
R = 30 Ω;
Q = 6;
Q
6
=
= 7.96 µF;
Rωo
(30)(8000π)
1
(8000π)2 (7.96 × 10−6 )
= 198.9 µH.
14–4
CHAPTER 14. Introduction to Frequency-Selective Circuits
Problems
P 14.1
R
1.5 × 103
=
= 6000 rad/s;
L
0.25
6000
.·. fc =
= 954.93 Hz.
2π
R/L
6000
[b] H(s) =
=
;
s + R/L
s + 6000
[a] ωc =
H(jω) =
6000
;
6000 + jω
H(jωc ) =
6000
= 0.7071/ − 45◦ ;
6000 + j6000
H(j0.3ωc ) =
H(j3ωc ) =
6000
= 0.9578/ − 16.70◦ ;
6000 + j1800
6000
= 0.3162/ − 71.57◦ .
6000 + j18,000
[c] vo (ωc ) = 35.36 cos(6000t − 45◦ ) V;
vo (0.3ωc ) = 47.89 cos(1800t − 16.70◦ ) V;
vo (3ωc ) = 15.81 cos(18,000t − 71.57◦ ) V.
P 14.2
Vo
R
[a] H(U L) (s) =
=
=
Vi
R + sL
R
L
.
R
L
R
RL
Vo
RkRL
L R + RL
[b] H(L) (s) =
=
=
.
R
RL
Vi
RkRL + sL
s+
L R + RL
R
RL
R
ωc(L) =
[c] ωc(U L) = ;
so the cutoff frequencies are different.
L
L R + RL
H(0)(U L) = 1;
s+
H(0)(L) = 1
so the passband gains are the same.
1000
= 50,000 rad/s.
0.02
[e] From part (c), ωc(L) < ωc(U L) . Therefore,
[d] ωc(U L) =
ωc(L) = 50,000 − 0.1(50,000) = 45,000 rad/s;
1000
RL
0.02 1000 + RL
45,000 =
so
.·. 0.1RL = 900 so RL ≥ 9 kΩ.
RL
= 0.9;
1000 + RL
Problems
P 14.3
R
(R/L)
Vo
=
=
.
Vi
sL + R + Rl
s + (R + Rl )/L
(R/L)
;
[b] H(jω) = R+R l
+
jω
L
[a] H(s) =
|H(jω)| = r
(R/L)
R+Rl 2
+ ω2
L
;
|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
R + Rl 2
.·. ωc2 =
;
L
1575
= 6300 rad/s;
[e] ωc =
0.25
6000
H(jω) =
.
6300 + jω
;
+ ωc2
.·. ωc = (R + Rl )/L.
H(j0) = 0.9524;
0.9524
√ / − 45◦ = 0.6734/ − 45◦ ;
2
6000
H(j1890) =
= 0.9122/ − 16.70◦ ;
6300 + j1890
H(j6300) =
H(j18,900) =
P 14.4
R
so R = Lωc = (0.01)(2500) = 25 Ω.
L
[b] Req = 100k25 = 20 Ω;
[a] ωc =
ωc =
P 14.5
6000
= 0.3012/ − 71.57◦ .
6300 + j18,900
Req
20
=
= 2000 rad/s.
L
0.01
[a] ωc = 0.95(2500) = 2375 =
Req
0.01
Then,
23.75 = 25kRL =
Solving,
RL = 475 Ω.
25RL
.
25 + RL
so Req = 2375(0.01) = 23.75 Ω.
14–5
14–6
CHAPTER 14. Introduction to Frequency-Selective Circuits
[b] H(s) =
ωc
2375
=
;
s + ωc
s + 2375
|H(j0)| =
P 14.6
2375
= 1.
2375
R
= 5000π rad/s;
L
R = Lωc = (0.025)(5000π) = 392.7 Ω.
[a] ωo =
[b] Re = 392.7k750 = 257.745 Ω;
Re
= 10,309.8 rad/s;
L
ωloaded
.·. floaded =
= 1640.85 Hz.
2π
[c] Use a 390 Ω resistor, which gives
ωc = R/L = 390/0.025 = 15,600 rad/s = 2482.8 Hz.
ωloaded =
P 14.7
[a] ωc =
109
= 12,500 rad/s;
80 × 103
fc = 1989.44 Hz.
[b] H(jω) =
12,500
;
12,500 + jω
.·. H(jωc ) = 0.7071/ − 45◦ .
H(j0.2ωc ) =
H(j8ωc ) =
12,500
= 0.9806/ − 11.31◦ ;
12,500 + j2500
12,500
= 0.1240/ − 82.87◦ .
12,500 + j100,000
[c] vo (ωc ) = 339.41 cos(12,500t − 45◦ ) mV;
vo (0.2ωc ) = 470.68 cos(2500t − 11.31◦ ) mV;
vo (8ωc ) = 59.54 cos(100,000t − 82.87◦ ) mV.
P 14.8
[a] Let Z =
RL (1/SC)
RL
=
.
RL + 1/SC
RL Cs + 1
Then H(s) =
=
=
Z
Z +R
RL
RRL Cs + R + RL
(1/RC)
R + RL
s+
.
RRL C
Problems
(1/RC)
[b] |H(jω)| = q
ω 2 + [(R + RL )/RRL C]2
14–7
.
|H(jω)| is maximum at ω = 0.
RL
.
R + RL
RL
(1/RC)
[d] |H(jωc )| = √
=q
;
2
2(R + RL )
ωC + [(R + RL )/RRL C]2
[c] |H(jω)|max =
1
R + RL
=
(1 + (R/RL )) .
.·. ωc =
RRL C
RC
20
[e] ωc = 12,500 1 +
= 13,333.33 rad/s.
300
H(j0) =
300
= 0.9375;
320
H(jωc ) =
12,500
= 0.6629/ − 45◦ ;
13,333.33 + j13,333.33
H(j0.2ωc ) =
H(j8ωc ) =
P 14.9
12,500
= 0.9193/ − 11.31◦ ;
13,333.33 + j2666.67
12,500
= 0.1163/ − 82.87◦ .
13,333.33 + j106,666.67
[a] ωc = 2π(500) = 3141.59 rad/s.
1
1
1
so R =
=
= 6366 Ω.
[b] ωc =
RC
ωc C
(3141.59)(50 × 10−9 )
[c]
1/sC
1/RC
3141.59
Vo
=
=
=
.
Vi
R + 1/sC
s + 1/RC
s + 3141.59
Vo
(1/sC)kRL
1/RC
3141.59
[e] H(s) =
=
=
.
=
R + RL
Vi
R + (1/sC)kRL
s + 2(3141.59)
s+
1/RC
RL
[f ] ωc = 2(3141.59) = 6283.19 rad/s.
[d] H(s) =
[g] H(0) = 1/2.
P 14.10 [a] fc =
160
× 103 = 25.46 kHz.
2π
14–8
CHAPTER 14. Introduction to Frequency-Selective Circuits
[b]
1
= 160 × 103 ;
RC
R=
1
= 250 Ω.
(160 × 103 )(25 × 10−9 )
1
R
[c] ωc =
1+
;
RC
RL
.·.
R
= 0.08
RL
[d] H(j0) =
.·. RL = 12.5R = 3125 Ω.
RL
3125
=
= 0.9259;
R + RL
3375
H(j0) = 0.9259.
P 14.11 [a] R = ωc L = (160 × 103 )(25 × 10−3 ) = 4000 Ω = 4 kΩ.
R
RL
·
= 150,000;
[b]
L R + RL
.·.
RL
150,000
=
= 0.9375;
R + RL
160,000
.·. 0.0625RL = (0.9375)(4000);
P 14.12 [a] H(s) =
.·. RL = 60 kΩ.
s
s
sL
=
=
.
R + sL
s + R/L
s + 20,000
R
= 20,000 rad/s.
L
jR/L
j
1
[c] |H(jR/L)| =
=
=√ .
jR/L + R/L
j+1
2
[b] ωc =
RL
Vo
RL ksL
R + RL
P 14.13 [a] H(s) =
=
=
R
RL
Vi
R + RL ksL
s+
L R + RL
s
=
1
s
2
1
s + (20,000)
2
.
R
RL
1
= (20,000) = 10,000 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.
[b] ωc =
Problems
P 14.14 [a]
1
109
=
= 10 krad/s;
RC
(40 × 103 )(2.5)
fc =
5000
= 1591.55 Hz.
π
jω
;
10,000 + jω
[b] H(jω) =
j10,000
= 0.7071/45◦ ;
10,000 + j10,000
H(jωc ) =
H(j0.1ωc ) =
j1000
= 0.0995/84.29◦ ;
10,000 + j1000
H(j10ωc ) =
j100,000
= 0.9950/5.71◦ .
10,000 + j100,000
[c] vo (ωc ) = 565.69 cos(10,000t + 45◦ ) mV.
vo (0.1ωc ) = 79.60 cos(1000t + 84.29◦ ) mV;
vo (10ωc ) = 796.03 cos(100,000t + 5.71◦ ) mV.
P 14.15 [a] H(s) =
R
Vo
=
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)| =
ω
R
·q
2
R + Rc
ω +
1
(R+Rc )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 =
R
2(R + Rc )
1
(R + Rc )2 C 2
1
.
(R + Rc )C
when ω = ωc .
14–9
14–10
CHAPTER 14. Introduction to Frequency-Selective Circuits
[e] ωc =
1
109
=
= 8000 rad/s;
(R + RC )C
(50 × 103 )(2.5)
40
j8000
H(jωc ) =
= 0.5657/45◦ .
50 8000 + j8000
P 14.16 [a]
H(j0.1ωc ) =
(0.8)j800
= 0.0796/84.29◦ ;
8000 + j800
H(j10ωc ) =
(0.8)j80,000
= 0.7960/5.71◦ .
8000 + j80,000
1
= 2π(800) = 1600π rad/s;
RC
.·. R =
109
= 9.95 kΩ.
(1600π)(20)
[b] Re = 9.95k68 = 8.68 kΩ;
ωc =
109
= 5761.84 rad/s;
(8.68)(103 )(20)
fc =
5761.84
= 917.03 Hz.
2π
P 14.17 [a] ωc = 2π(300) = 1884.96 rad/s.
1
1
1
[b] ωc =
so R =
=
= 5305 Ω.
RC
ωc C
(1884.96)(100 × 10−9 )
[c]
Vo
s
s
R
=
=
=
.
Vi
R + 1/sC
s + 1/RC
s + 1884.96
s
Vo
RkRL
s
[e] H(s) =
=
=
=
.
R + RL
Vi
RkRL + (1/sC)
s + 2(1884.96)
s+
1/RC
RL
[f ] ωc = 2(1884.96) = 3769.91 rad/s.
[d] H(s) =
[g] H(∞) = 1.
P 14.18 ωo =
fo =
√
ωc1 ωc2 =
q
(180)(200) = 189.74 krad/s;
ωo
= 30.20 kHz.
2π
Problems
14–11
β = 200 − 180 = 20 krad/s = 3.18 kHz.
Q=
P 14.19 β =
ωo
189.74
30.20
=
= 9.49 =
.
β
20
3.18
ωo
80
=
= 10 krad/s;
Q
8

1
ωc2 = 80  +
16
fc2 =
s
2
10,000
= 1.59 kHz;
2π

 = 85.16 krad/s;
85.16
= 13.55 kHz;
2π

s
1
ωc1 = 80 − +
16
fc1 =
1
1+
16
β=
1
1+
16
2

 = 75.16 krad/s;
75.16
= 11.96 kHz.
2π
P 14.20 [a] ωo =
q
1/LC
so L =
1
ωo2 C
=
1
(20,000)2 (50 × 10−9 )
Q=
ωo
β
so β =
β=
R
L
so R = Lβ = (50 × 10−3 )(4000) = 200 Ω.
β
[b] ωc1,2 = ± +
2
s
ωo
20,000
=
= 4000 rad/s.
Q
5
β2
4000
+ ωo2 = ±
+
2
2
ωc1 = 18,099.75 rad/s
P 14.21 H(jω) =
jω(4000)
Vo = (1)Vi
s
4000
2
2
+ 20,0002 = ±2000 + 20,099.75.
ωc2 = 22,099.75 rad/s.
20,0002 − ω 2 + jω(4000)
[a] H(j20,000) =
= 50 mH.
.
j20,000(4000)
20,0002 − 20,0002 + j(20,000)(4000)
.·. vo (t) = 200 cos 20,000t mV.
= 1;
14–12
CHAPTER 14. Introduction to Frequency-Selective Circuits
[b] H(j18,099.75) =
j18,099.75(4000)
20,0002 − 18,099.752 + j(18,099.75)(4000)
1
Vo = √ /45◦ Vi
2
[c] H(j22,099.75) =
.·. vo (t) = 141.42 cos(18,099.75t + 45◦ ) mV.
1
j22,099.75(4000)
/ − 45◦ ;
√
=
20,0002 − 22,099.752 + j(22,099.75)(4000)
2
1
Vo = √ / − 45◦ Vi
2
[d] H(j2000) =
.·. vo (t) = 141.42 cos(22,099.75t − 45◦ ) mV.
j2000(4000)
20,0002 − 20002 + j(2000)(4000)
j200,000(4000)
20,0002 − 200,0002 + j(200,000)(4000)
H(jω) =
= 0.02/ − 88.8◦ ;
.·. vo (t) = 4 cos(200,000t − 88.8◦ ) mV.
Vo = 0.02/ − 88.8◦ Vi
P 14.22 H(s) = 1 −
= 0.02/88.8◦ ;
.·. vo (t) = 4 cos(2000t + 88.8◦ ) mV.
Vo = 0.02/88.8◦ Vi
[e] H(j200,000) =
1
= √ /45◦ ;
2
(R/L)s
s2 + (1/LC)
=
;
s2 + (R/L)s + (1/LC)
s2 + (R/L)s + (1/LC)
20,0002 − ω 2
.
20,0002 − ω 2 + jω(4000)
[a] H(j20,000) =
Vo = (0)Vi
20,0002 − 20,0002
= 0;
20,0002 − 20,0002 + j(20,000)(4000)
.·. vo (t) = 0 mV.
[b] H(j18,099.75) =
20,0002 − 18,099.752
1
= √ / − 45◦ ;
2
2
20,000 − 18,099.75 + j(18,099.75)(4000)
2
1
Vo = √ / − 45◦ Vi
2
[c] H(j22,099.75) =
1
Vo = √ /45◦ Vi
2
[d] H(j2000) =
.·. vo (t) = 141.42 cos(18,099.75t − 45◦ ) mV.
20,0002 − 22,099.752
1
/45◦ ;
√
=
20,0002 − 22,099.752 + j(22,099.75)(4000)
2
.·. vo (t) = 141.42 cos(22,099.75t + 45◦ ) mV.
20,0002 − 20002
= 0.9998/ − 1.16◦ ;
20,0002 − 20002 + j(2000)(4000)
Vo = 0.9998/ − 1.16◦ Vi
.·. vo (t) = 199.96 cos(2000t − 1.16◦ ) mV.
Problems
[e] H(j200,000) =
20,0002 − 200,0002
= 0.9998/1.16◦ ;
20,0002 − 200,0002 + j(200,000)(4000)
.·. vo (t) = 199.96 cos(200,000t + 1.16◦ ) mV.
Vo = 0.9998/1.16◦ Vi
P 14.23 [a] β = ωc2 − ωc1
Then,
β
− +
2
v
u
u
t
β
2
ωo2 = ωc1 ωc2 .
and
!2
ωc2 − ωc1
+ ωo2 = −
+
2
s
ωc2 − ωc1
2
2
ωc1 − ωc2
=
+
2
s
2
2
ωc2
− 2ωc1 ωc2 + ωc1
+ 4ωc1 ωc2
4
ωc1 − ωc2
=
+
2
s
2
2
ωc2
+ ωc1
+ 2ωc1 ωc2
4
ωc1 − ωc2
+
=
2
s
(ωc1 + ωc2 )2
4
=
+ ωc1 ωc2
ωc1 − ωc2 ωc1 + ωc2
+
= ωc1 .
2
2
Similarly,
v
u
β u
β
+t
2
2
!2
ωc2 − ωc1
+ ωo2 =
+
2
s
ωc2 − ωc1
2
2
+ ωc1 ωc2
ωc2 − ωc1
=
+
2
s
2
2
ωc2
− 2ωc1 ωc2 + ωc1
+ 4ωc1 ωc2
4
ωc2 − ωc1
=
+
2
s
2
2
ωc2
+ ωc1
+ 2ωc1 ωc2
4
ωc2 − ωc1
=
+
2
s
(ωc1 + ωc2 )2
4
=
ωc2 − ωc1 ωc1 + ωc2
+
= ωc2 .
2
2

1
[b] ωo 
−
2Q
+
v
u
u
t
1
1+
2Q
!2

v
u
−ωo u

+ tω 2 +
=
2Q
o
ωo
2Q
!2
.
But Q = ωo /β. Thus
v
u
−ωo u
+ tω 2 +
2Q
o
14–13
ωo
2Q
!2
−ωo
=
2(ωo /β)
v
u
u
+ tω 2 +
o
ωo
2(ωo /β)
!2
14–14
CHAPTER 14. Introduction to Frequency-Selective Circuits
β
=− +
2
= ωc1
v
u
u
t
β
2
!2
+ ωo2
from part (a).
Similarly,

 1
ωo 
2Q
+
v
u
u
t
1
1+
2Q
!2

v
u
ωo u

+ tω 2 +
=
o
2Q
ωo
2Q
!2
.
But Q = ωo /β. Thus
v
u
ωo
ωo u
+ tωo2 +
2Q
2Q
β
= +
2
= ωc2
P 14.24 [a] ωo2 =
v
u
u
t
β
2
!2
v
u
u
ωo
ωo
=
+ tωo2 +
2(ωo /β)
2(ωo /β)
!2
+ ωo2
from part (a).
1
109
=
= 25 × 108 ;
LC
L(25)
Q2 =
R2 C
;
L

!2
1
[b] ωc2 = 50  +
20
.·. L =
109
= 16 mH;
625 × 108
LQ2
(0.016)102
= 64 × 106 ;
.·. R2 =
=
−9
C
25 × 10
.·. R = 8 kΩ.

s
1 
1+
= 52.56 krad/s;
400
.·. fc2 = 8.37 kHz.

1
ωc1 = 50 − +
20
s

1 
1+
= 47.56 krad/s;
400
.·. fc1 = 7.57 kHz.
ωo
[c] β =
= 5000 rad/s = 795.77 Hz.
Q
Check: β = fc2 − fc1 = 795.77 Hz.
P 14.25 [a] We need ωc = 50,000 rad/s. There are several possible approaches – this
one starts by choosing L = 1 mH. Then,
1
C=
= 0.4 µF.
2
50,000 (0.001)
Use the closest value from Appendix H, which is 0.47 µF to give
s
ωo =
1
= 46,126.56 rad/s.
(0.001)(0.47 × 10−6 )
Problems
Then, R =
14–15
Q
10
=
= 461.3 Ω.
ωo C
(46,126.56)(0.47 × 10−6 )
Use the closest value from Appendix H, which is 470 Ω to give
Q = 470(46,126.56)(0.47 × 10−6 ) = 10.2.
[b] % error in ωo =
46,126.56 − 50,000
(100) = −7.75%;
50,000
% error in Q =
P 14.26 [a] L =
10.2 − 10
(100) = 2%.
10
1
= 3.17 mH;
[2π(20,000)]2 (20 × 10−9 )
R=
ωo L
40π × 103 (3.17 × 10−3 )
=
= 79.58 Ω.
Q
5

s
1
[b] fc1 = 20 − +
10

1
[c] fc2 = 20  +
10
s

1 
1+
= 18.10 kHz.
100

1 
1+
= 22.10 kHz.
100
[d] β = fc2 − fc1 = 4 kHz
or
fo
20
=
= 4 kHz.
β=
Q
5
P 14.27 [a] We need ωc close to 2π(20,000) = 125,663.706 rad/s. There are several
possible approaches – this one starts by choosing L = 10 mH. Then,
1
C=
= 6.33 nF.
[2π(20,000)]2 (0.01)
Use the closest value from Appendix H, which is 0.0047 µF to give
s
ωo =
1
= 145,864.99 rad/s or fo = 23,215.13 Hz.
(0.01)(4.7 × 10−9 )
Then, R =
ωo L
(145,864.99)(0.01)
=
= 291.73 Ω.
Q
5
Use the closest value from Appendix H, which is 270 Ω to give
(145,864.99)(0.01)
= 5.4.
270
23,215.13 − 20,000
[b] % error in fc =
(100) = 16.1%;
20,000
Q=
% error in Q =
5.4 − 5
(100) = 8%.
5
14–16
CHAPTER 14. Introduction to Frequency-Selective Circuits
1
(103 )(1012 )
=
= 2.56 × 1012 ;
LC
(312.5)(1.25)
P 14.28 [a] ωo2 =
ωo = 1.6 × 106 rad/s;
fo =
[b] Q =
800
= 254.65 kHz.
π
ωo L
(1.6 × 106 )(312.5 × 10−3 )
=
= 8.
R + Ri
(50 + 12.5)103

800  1
[c] fc1 =
− +
π
16

800  1
[d] fc2 =
+
π 16
s
s

1 
1+
= 239.23 kHz.
256

1 
= 271.06 kHz.
1+
256
[e] β = fc2 − fc1 = 31.83 kHz
or
100
ωo
β=
= 200 krad/s =
kHz = 31.83 kHz.
Q
π
P 14.29 [a] H(s) =
(R/L)s
1
i)
s2 + (R+R
s + LC
L
.
For the numerical values in Problem 14.28 we have
16 × 104 s
H(s) = 2
;
s + 2 × 105 s + 2.56 × 1012
.·. H(jω) =
H(jωo ) =
j16 × 104 ω
.
(2.56 × 1012 − ω 2 ) + j2 × 105 ω
j16 × 104 (1.6 × 106 )
= 0.8/0◦ ;
j2 × 105 (1.6 × 106 )
.·. vo (t) = 640 cos ωt mV.

1
[b] ωc1 = 1.6 × 106 − +
16
H(jωc1 ) =
s

1 
= 1.5 × 106 rad/s.
1+
256
j16 × 104 (1.5 × 106 )
2.56 × 1012 − 1.52 × 1012 + j2 × 105 (1.5 × 106 )
= 0.57/45◦ ;
.·. vo (t) = 452.55 cos(1.5 × 106 t + 45◦ ) mV.
Problems

1
[c] ωc2 = 1.6 × 106  +
16

s
1 
1+
= 1.7 × 106 rad/s.
256
j16 × 104 (1.7 × 106 )
2.56 × 1012 − 1.72 × 1012 + j2 × 105 (1.7 × 106 )
H(jωc2 ) =
= 0.57/ − 45◦ ;
.·. vo (t) = 452.55 cos(1.7 × 106 t − 45◦ ) mV.
P 14.30 [a]
ωo2 =
1
(106 )(109 )
=
= 1012 ;
LC
(40)(25)
ωo = 106 rad/s = 1 Mrad/s.
500
kHz = 159.15 kHz.
π
[c] Q = ωo RC = (106 )(300)(25 × 10−9 ) = 7.5.
[b] fo =


s
1 
1+
= 935.55 krad/s.
225
1
[d] ωc1 = 106 − +
15
[e] .·. fc1 = 148.90 kHz.

1
[f ] ωc2 = 106  +
15

s
1 
= 1068.89 krad/s.
1+
225
[g] .·. fc2 = 170.12 kHz.
ωo
[h] β =
= 133.33 krad/s or 21.22 kHz.
Q
P 14.31 [a]
Z
1
Vo
=
where Z =
Vi
Z +R
Y
and Y = sC +
H(s) =
=
RL
1
1
LCRL s2 + sL + RL
+
=
.
sL RL
RL Ls
RL Ls
2
RLCs + (R + R
L )Ls + RRL
(1/RC)s
s2 +
h
R+RL
RL
i
1
s + LC
1
RC
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
,
K=
RL
,
R + RL
β=
1
.
(RkRL )C
14–17
14–18
CHAPTER 14. Introduction to Frequency-Selective Circuits
R + RL
[b] β =
RL
1
[c] βU =
;
RC
1
.
RC
R
R + RL
βU = 1 +
βU .
.·. βL =
RL
RL
[d] Q =
ωo
ωo RC
= R+R .
L
β
RL
[e] QU = ωo RC;
.·. QL =
[f ] H(jω) =
RL
1
QU .
QU =
R + RL
[1 + (R/RL )]
Kjωβ
;
ωo2 − ω 2 + jωβ
H(jωo ) = K.
Let ωc represent a cutoff 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
where
R
β = 1+
RL
1
1
and ωo2 =
.
RC
LC
s
β2
+ ωo2
4
Problems
1
(106 )(1012 )
=
= 625 × 1012 ;
LC
(400)(4)
P 14.32 ωo2 =
ωo = 25 Mrad/s.
Qu = ωo RC = (25 × 106 )(100 × 103 )(4 × 10−12 ) = 10;
.·.
RL
10 = 9;
R + RL
P 14.33 [a] ωo2 =
.·. RL = 9R = 900 kΩ.
(103 )(1012 )
1
=
= 25 × 1012 ;
LC
(10)(4)
ωo = 5 Mrad/s.
R + RL 1
6.25
[b] β =
·
=
RL
RC
5.0
5
ωo
[c] Q =
=
= 20.
β
0.25
RL
[d] H(jωo ) =
= 0.8/0◦ ;
R + RL
1012
(4)(1.25 × 106 )
!
= 250 krad/s.
.·. vo (t) = 600 cos(5 × 106 t) mV.
[e] β = 1 +
R
RL
1
1.25
= 1+
(200 × 103 ) rad/s;
RC
RL
ωo = 5 × 106 rad/s.
Q=
[f ]
ωo
25
=
β
1 + (1.25/RL )
where RL is in megohms.
14–19
14–20
CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.34 [a]
[b] L =
1
109
=
= 3.2 × 10−3 = 3.2 mH;
ωo2 C
(625 × 108 )5
R=
800
ωo L
=
= 80 Ω.
Q
10
[c] Re = 80k320 = 64 Ω;
Re + Ri = 64 + 36 = 100 Ω;
Qsystem =
ωo L
800
=
= 8.
Re + Ri
100
250 × 103
[d] βsystem =
=
= 31.25 krad/s;
Qsystem
8
ωo
βsystem (kHz) =
31.25
= 4.97 kHz = 4973.59 Hz.
2π
P 14.35 [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 = ∞.
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
sL(1/sC)
= 2
.
sL + 1/sC
s LC + 1
Then
Vo
R
R(s2 LC + 1)
H(s) =
=
=
Vi
Z +R
sL + R(s2 LC + 1)
=
[s2 + (1/LC)]
s2 +
1
RC
s+
1
LC
;
Problems
H(s) =
s2 + ωo2
.
s2 + βs + ωo2
[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
ωo2 − ω 2
[d] |H(jω)| = q
(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
−β
+
ωc1 =
2
v
u
u
t
v
u
β u β
ωc2 = + t
2
2
β
2
!2
+ ωo2 ;
!2
+ ωo2 .
Also note that since β = ωo /Q

v
u
u
1
 −1
ωc1 = ωo 
+ t1 +
2Q
2Q

ωc2 = ωo 

1
+
2Q
v
u
u
t
1+
1
2Q
!2


;
!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 14.36 [a] ωo2 =
q
1
(106 )(1012 )
=
= 64 × 1012 ;
LC
(625)(25)
.·. ωo = 8 Mrad/s.
R2 C/L.
14–21
14–22
CHAPTER 14. Introduction to Frequency-Selective Circuits
ωo
= 1.27 MHz.
2π
[c] Q = ωo RC = (8 × 106 )(80 × 103 )(25 × 10−12 ) = 16.
[b] fo =

[d] ωc1 = 8 × 106 −
[e] fc1 =
1
+
32

s
1+
1 
= 7.75 Mrad/s.
1024
ωc1
= 1.234 MHz.
2π


s
1
1 
[f ] ωc2 = 8 × 106  + 1 +
= 8.25 Mrad/s.
32
1024
ωc1
[g] fc2 =
= 1.31 MHz.
2π
[h] β = fc2 − fc1 = 79.58 kHz
or
β=
ωo
500 × 103
=
= 79.58 kHz.
2πQ
2π
P 14.37 [a] ωo = 2πfo = 100π krad/s;
L=
R=
1
ωo2 C
=
1
(100π × 103 )2 (100 × 10−9 )
= 101.32 µH;
Q
8
=
= 254.65 Ω.
3
ωo C
(100π × 10 )(100 × 10−9 )

1
[b] fc2 = 50k  +
16

fc1 = 50k −

s
1 
1+
= 53.22 kHz;
256
1
+
16

s
1+
1 
= 46.97 kHz.
256
[c] β = fc2 − fc1 = 6.25 kHz
or
fo
50k
β=
=
= 6.25 kHz.
Q
8
P 14.38 [a] Re = 254.65k932 = 200 Ω;
Q = ωo Re C = 100π × 103 (200)(0.1)10−6 = 2π = 6.28.
[b] β =
fo
50,000
=
= 7.96 kHz.
Q
2π

1
[c] fc2 = 50,000 
+
4π
s

1 
1+
= 54.14 kHz.
16π 2
Problems

1
[d] fc1 = 50,000 −
+
4π
14–23

s
1 
1+
= 46.18 kHz.
16π 2
P 14.39 [a] We need ωc = 2π(50,000) = 314,159.265 rad/s. There are several possible
approaches – this one starts by choosing L = 10 µH. Then,
1
= 1.01 µF.
[2π(50,000)]2 (10 × 10−6 )
C=
Use the closest value from Appendix H, which is 1 µF, to give
s
ωc =
1
(10 × 10−6 )(1 × 10−6 )
Then, R =
= 316,227.766 rad/s so fc = 50,329.2 Hz.
Q
8
=
= 25.3 Ω.
ωo C
(316,227.766)(1 × 10−6 )
Use the closest value from Appendix H, which is 22 Ω, to give
Q = 22(316,227.766)(1 × 10−6 ) = 6.957.
[b] % error in fc =
% error in Q =
P 14.40 H(s) =
50,329.2 − 50,000
(100) = 0.66%;
50,000
6.957 − 8
(100) = −13.04%.
8
1
s2 + LC
1 .
s2 + R
s + LC
L
s
[a] ωo =
1
=
LC
s
1
= 5000 rad/s.
(0.005)(8 × 10−6 )
ωo
= 795.8 Hz.
2π
s
s
L
0.005
[c] Q =
=
= 2.5.
2
2
R C
(10) (8 × 10−6 )
R
10
[d] β =
=
= 2000 rad/s.
L
0.005
[b] fo =
β (Hertz) =
2000
= 318.31 Hz.
2π
v
u
−β u
β
[e] ωc1 =
+t
2
2
!2
−2000
=
+
2
s
[f ] fc1 =
+ ωo2
2000
2
2
4099.02
= 652.38 Hz.
2π
+ 50002 = 4099.02 rad/s.
14–24
CHAPTER 14. Introduction to Frequency-Selective Circuits
β
2
!2
2000
=
+
2
s
β
[g] ωc2 = +
2
[h] fc2 =
v
u
u
t
+ ωo2
2000
2
2
+ 50002 = 6099.02 rad/s.
6099.02
= 970.69 Hz.
2π
P 14.41 [a] H(jω) =
1
− ω2
LC
1
R
− ω2 + j ω
LC
L
=
50002 − ω 2
.
50002 − ω 2 + j2000ω
ωo = 5000 rad/s :
H(jωo ) =
50002 − 50002
= 0.
50002 − 50002 + j2000(5000)
ωc1 = 4099.02 rad/s :
H(jωc1 ) =
50002 − 4099.022
= 0.7071/ − 45◦ .
50002 − 4099.022 + j2000(4099.02)
ωc2 = 6099.02 rad/s :
H(jωc2 ) =
50002 − 6099.022
= 0.7071/45◦ .
50002 − 6099.022 + j2000(6099.02)
0.1ωo = 500 rad/s :
H(j0.1ωo ) =
50002 − 5002
= 0.9992/ − 2.314◦ .
50002 − 5002 + j2000(500)
10ωo = 50,000 rad/s :
H(j10ωo ) =
50002 − 50,0002
= 0.9992/2.314◦ .
50002 − 50,0002 + j2000(50,000)
[b] ω = ωo = 5000 rad/s :
Vo = H(j5000)Vi = 0Vi ;
vo (t) = 0.
ω = ωc1 = 4099.02 rad/s :
Vo = H(j4099.02)Vi = (0.7071/ − 45◦ )(2) = 1.414/ − 45◦ ;
vo (t) = 1.414 cos(4099.02t − 45◦ ) V.
ω = ωc2 = 6099.02 rad/s :
Problems
14–25
Vo = H(j6099.02)Vi = (0.7071/45◦ )(2) = 1.414/45◦ ;
vo (t) = 1.414 cos(6099.02t + 45◦ ) V.
ω = 0.1ωo = 500 rad/s :
Vo = H(j500)Vi = (0.9992/ − 2.314◦ )(2) = 1.998/ − 2.314◦ ;
vo (t) = 1.998 cos(500t − 2.314◦ ) V.
ω = 10ωo = 50,000 rad/s :
Vo = H(j50,000)Vi = (0.9992/2.314◦ )(2) = 1.998/2.314◦ ;
vo (t) = 1.998 cos(50,000t + 2.314◦ ) V.
P 14.42 [a] ωo =
q
1/LC
so L =
1
1
=
= 781.25 mH;
ωo2 C
(4000)2 (80 × 10−9 )
Q=
ωo
β
so β =
β=
R
L
so R = Lβ = (781.25 × 10−3 )(6000) = 4687.5 Ω.
β
[b] ωc1,2 = ± +
2
s
6000
β2
+ ωo2 = ±
+
2
2
ωc1 = 2000 rad/s
P 14.43 H(jω) =
ωo
4000
=
= 6000 rad/s;
Q
2/3
s
6000
2
2
+ 40002 = ±3000 + 5000;
ωc2 = 8000 rad/s.
ωo2 − ω 2
40002 − ω 2
=
ωo2 − ω 2 + jωβ
40002 − ω 2 + jω(6000)
[a] H(j4000) =
Vo = (0)Vi
[b] H(j2000) =
40002 − 40002
=0
40002 − 40002 + j(4000)(6000)
.·. vo (t) = 0 V
40002 − 20002
1
= √ / − 45◦
2
2
4000 − 2000 + j(2000)(6000)
2
1
Vo = √ / − 45◦ Vi
2
.·. vo (t) = 88.39 cos(2000t − 45◦ ) V
14–26
CHAPTER 14. Introduction to Frequency-Selective Circuits
40002 − 80002
1
= √ /45◦
2
2
4000 − 8000 + j(8000)(6000)
2
[c] H(j8000) =
1
Vo = √ /45◦ Vi
2
[d] H(j500) =
.·. vo (t) = 88.39 cos(8000t + 45◦ ) V
40002 − 5002
= 0.982/ − 10.78◦
2
2
4000 − 500 + j(500)(6000)
Vo = 0.982/ − 10.78◦ Vi
[e] H(j32,000) =
40002 − 32,0002
= 0.982/10.78◦
40002 − 32,0002 + j(32,000)(6000)
.·. vo (t) = 122.8 cos(32,000t + 10.78◦ ) V
Vo = 0.982/10.78◦ Vi
P 14.44 H(jω) =
.·. vo (t) = 122.8 cos(500t − 10.78◦ ) V
jω(6000)
jωβ
=
ωo2 − ω 2 + jωβ
40002 − ω 2 + jω(6000)
[a] H(j4000) =
j(4000)(6000)
40002 − 40002 + j(4000)(6000)
.·. vo (t) = 125 cos 4000t V
Vo = (1)Vi
[b] H(j2000) =
j(2000)(6000)
40002 − 20002 + j(2000)(6000)
1
Vo = √ /45◦ Vi
2
[c] H(j8000) =
j(8000)(6000)
40002 − 80002 + j(8000)(6000)
1
= √ / − 45◦
2
.·. vo (t) = 88.39 cos(8000t − 45◦ ) V
j(500)(6000)
40002 − 5002 + j(500)(6000)
Vo = 0.187/79.22◦ Vi
[e] H(j32,000) =
1
= √ /45◦
2
.·. vo (t) = 88.39 cos(2000t + 45◦ ) V
1
Vo = √ / − 45◦ Vi
2
[d] H(j500) =
=1
= 0.187/79.22◦
.·. vo (t) = 23.375 cos(500t + 79.22◦ ) V
j(32,000)(6000)
40002 − 32,0002 + j(32,000)(6000)
Vo = 0.187/ − 79.22◦ Vi
= 0.187/ − 79.22◦
.·. vo (t) = 23.375 cos(32,000t − 79.22◦ ) V
Problems
P 14.45 [a] Let Z =
Z=
RL (sL + (1/sC))
;
RL + sL + (1/sC)
RL (s2 LC + 1)
.
s2 LC + RL Cs + 1
Vo
s2 RL CL + RL
Then H(s) =
=
.
Vi
(R + RL )LCs2 + RRL Cs + R + RL
Therefore
H(s) =
=
RL
[s2 + (1/LC)]
h
i
·
RRL
s
1
R + RL
s2 + R+R
+
L
LC
L
K(s2 + ωo2 )
,
s2 + βs + ωo2
RL
where K =
;
R + RL
ωo2 =
1
;
LC
RRL
β=
R + RL
1
.
LC
RRL
1
[c] β =
.
R + RL L
ωo
ωo L
[d] Q =
=
.
β
[RRL /(R + RL )]
[b] ωo = √
[e] H(jω) =
K(ωo2 − ω 2 )
;
(ωo2 − ω 2 ) + jβω
H(jωo ) = 0.
[f ] H(j0) =
Kωo2
RL
=
K
=
.
ωo2
R + RL
h
i
K (ωo /ω)2 − 1
i
o;
[g] H(jω) = nh
(ωo /ω)2 − 1 + jβ/ω
H(j∞) =
−K
RL
=K=
.
−1
R + RL
K(ωo2 − ω 2 )
;
(ωo2 − ω 2 ) + jβω
Let ωc represent a cutoff frequency. Then
[h] H(jω) =
K
|H(jωc )| = √ ;
2
K
K(ωo2 − ωc2 )
.·. √ = q
.
2
(ωo2 − ωc2 )2 + ωc2 β 2
1
.
L
14–27
14–28
CHAPTER 14. Introduction to Frequency-Selective Circuits
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
β=
P 14.46 [a] ωo2 =
RRL
1
1
· and ωo2 =
.
R + RL L
LC
1
1
=
= 1010 ;
−3
LC
(400 × 10 )(250 × 10−12 )
ωo = 105 = 100 krad/s = 15.9 kHz.
β=
1
(5000)(20,000) 1
RRL
· =
·
= 104 rad/s = 1.59 kHz;
R + RL L
25,000
0.4
Q=
105
ωo
= 4 = 10.
β
10
[b] H(j0) =
RL
20,000
=
= 0.8;
R + RL
25,000
RL
= 0.8.
R + RL
H(j∞) =

105  1
[c] fc2 =
+
2π 20

s

1 
= 16.73 kHz;
1+
400
105  1
fc1 =
− +
2π
20
s

1 
1+
= 15.14 kHz.
400
Check:
β = fc2 − fc1 = 1.59 kHz.
ωo
105
[d] Q =
= RRL 1
β
·L
R+R
L
40(R + RL )
5
=8 1+
RRL
RL
where RL is in kilohms.
=
Problems
14–29
[e]
P 14.47 [a] ωo2 =
1
= 1012 ;
LC
.·. L =
1
(1012 )(400 × 10−12 )
= 2.5 mH.
RL
= 0.96;
R + RL
.·. 0.04RL = 0.96R;
.·.
36,000
= 1.5 kΩ.
.·. R =
24
RL = 24R
[b] β =
Q=
1
RL
R · = 576 × 103 ;
R + RL
L
ωo
106
=
= 1.74.
β
576 × 103
4 × 106
P 14.48 [a] |H(jω)| = q
(4 × 106 − ω 2 )2 + (500ω)2
.·.
= 1;
16 × 1012 = (4 × 106 − ω 2 )2 + (500ω)2
= −8 × 106 ω 2 + ω 4 + 25 × 104 ω 2 = 0;
.·.
ω 2 = 8 × 106 − 25 × 104
so
ω = 2783.88 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
√
(4 × 106 − ω 2 )2 = 0 so ω = 4 × 106 = 2000 rad/s.
[c] |H(j2000)| = q
P 14.49 [a] H(s) =
4 × 106
(4 × 106 − 20002 )2 + [500(2000)]2
= 4.
sL
s2 LC
s2
=
=
1
1 .
RsC + s2 LC + 1
R + sL + sC
s2 + R
s + LC
L
14–30
CHAPTER 14. Introduction to Frequency-Selective Circuits
[b] When s = jω is very small (think of ω approaching 0),
H(s) ≈
s2
1
LC
= 0.
[c] When s = jω is very large (think of ω approaching ∞),
s2
= 1.
s2
[d] The magnitude of H(s) approaches 0 as the frequency approaches 0, and
approaches 1 as the frequency approaches ∞. Therefore, this circuit is
behaving like a high pass filter when the output is the voltage across the
inductor.
ωc2
1
[e] |H(jωc )| = q
=√ ;
2
(250,000 − ωc2 )2 + (8000ωc )2
H(s) ≈
2ωc4 = (250,000 − ωc2 )2 + (8000ωc )2 = 625 × 108 − 500,000ωc2 + ωc4 + 64 × 106 ωc2 .
ωc4 − 635 × 105 ωc2 − 625 × 108 = 0.
Simplifying,
Solve for ωc2 and then ωc :
ωc2 = 63,500,984.2367
so
ωc = 7968.75 rad/s.
fc = 1268.3 Hz
P 14.50 [a] H(s) =
1
sC
1
R + sL + sC
=
1
1
LC
=
1 .
RsC + s2 LC + 1
s2 + R
s + LC
L
[b] When s = jω is very small (think of ω approaching 0),
H(s) ≈
1
LC
1
LC
= 1.
[c] When s = jω is very large (think of ω approaching ∞),
1
H(s) ≈ LC2 = 0.
s
[d] The magnitude of H(s) approaches 1 as the frequency approaches 0, and
approaches 0 as the frequency approaches ∞. Therefore, this circuit is
behaving like a low pass filter when the output is the voltage across the
capacitor.
250,000
1
[e] |H(jωc )| = q
=√ ;
2
(250,000 − ωc2 )2 + (8000ωc )2
1250 × 108 = (250,000 − ωc2 )2 + (8000ωc )2
= 625 × 108 − 500,000ωc2 + ωc4 + 64 × 108 ωc2 .
Problems
14–31
ωc4 + 635 × 105 ωc2 − 625 × 108 = 0.
Simplifying,
Solve for ωc2 and then ωc :
ωc2 = 984.2367
so
ωc = 31.37 rad/s.
f = 4.99 Hz.
P 14.51 [a] Use the cutoff frequencies to calculate the bandwidth:
ωc1 = 2π(697) = 4379.38 rad/s
ωc2 = 2π(941) = 5912.48 rad/s;
β = ωc2 − ωc1 = 1533.10 rad/s.
Thus
Calculate inductance and capacitance:
C=
L=
1
1
=
= 1.087 µF;
Rβ
(600)(1533.1)
1
Cωc1 ωc2
=
1
= 35.5 mH.
(1.087 × 10−6 )(4379.38)(5912.48)
[b] At the outermost two frequencies in the low-frequency group (687 Hz and
941 Hz) the amplitudes are
|Vpeak |
√
= 0.707|Vpeak |
2
|V697Hz | = |V941Hz | =
because these are cutoff frequencies. We calculate the amplitudes at the
other two low frequencies:
ωβ
|V | = (|Vpeak |)(|H(jω)|) = |Vpeak | q
.
(ωo2 − ω 2 )2 + (ωβ)2
Therefore
(4838.05)(1533.10)
|V770Hz | = (|Vpeak |) q
(5088.522 − 4838.052 )2 + [(4838.05)(1533.10)]2
= 0.948|Vpeak |,
and
(5353.27)(1533.10)
|V852Hz | = (|Vpeak |) q
(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
14–32
CHAPTER 14. Introduction to Frequency-Selective Circuits
tones designed with this same 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.
P 14.52 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
C=
1
1
=
= 625.6 nF;
Rβ
(600)(2664)
L=
1
1
=
= 20.51 mH.
2
ωc1 ωc2 C
(2π) (1209)(1633)(625.6 × 10−9 )
At the highest of the low-band frequencies, 941 Hz, the amplitude is
|Vω | = |Vpeak | q
where
|Vω | = q
ωβ
(ωo2 − ω 2 )2 + ω 2 β 2
ωo =
√
ωc1 ωc2 . Thus,
|Vpeak |(5912)(2664)
[(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.
Problems
14–33
P 14.53 From Problem 14.51 the response to the largest of the DTMF low-band tones
is 0.948|Vpeak |. The response to the 20 Hz tone is
|V20Hz | = q
|Vring |(125.6)(1533)
[(50892 − 125.62 )2 + [(125.6)(1533)]2 ]
= 0.00744|Vring |;
.·.
|V20Hz |
|V20Hz |
0.00744|Vring |
1
=
=
= .
|V770Hz |
|V852Hz |
0.948|Vpeak |
3
.·. |V20Hz | = 42.5|V770Hz |.
Thus, the 20Hz signal can be 42.5 times as large as the DTMF tones.
Active Filter Circuits
Assessment Problems
AP 15.1 H(s) =
−8000
−(1/R1 C)
=
.
s + (1/R2 C)
s + 2000
1
= 8000;
R1 C
.·. R1 =
C = 40 nF;
1
= 3125 Ω.
(8000)(40 × 10−9 )
1
= 2000;
R2 C
.·. R2 =
AP 15.2 H(s) =
1
= 12.5 kΩ.
(2000)(40 × 10−9 )
−(R2 /R1 )s
;
s + (1/R1 C)
1
= 1 rad/s;
R1 C
R2
= 1,
R1
.·.
R1 = 1 Ω,
.·. C = 1 F.
.·. R2 = R1 = 1 Ω;
Hprototype (s) =
−s
.
s+1
15–1
15–2
CHAPTER 15. Active Filter Circuits
AP 15.3 ωc = 2πfc = 2π(4000) = 8000π rad/s;
.·. kf = 8000π = 25,132.74.
C
kf km
C0 =
.·. km =
.·.
1
;
kf km
10 × 10−6 =
1
= 3.98.
(10 × 10−6 )(25,132.74)
AP 15.4 [a] For a 2nd order Butterworth low pass filter
H(s) =
s2 +
1
√
.
2s + 1
For the circuit in Fig. 15.21
H(s) =
1
R 2 C1 C2
s2 +
2
RC1
s+
1
.
R 2 C1 C2
Equate the transfer functions. For R = 1 Ω,
√
2
2
= 2, .·. C1 = √ = 1.414 F;
RC1
2
1
R2 C1 C2
= 1,
1
.·. C2 =
= 0.707 F.
C1
[b] For a 2nd order Butterworth high pass filter
s2
√
H(s) = 2
.
s + 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 = 2 = 1.414 Ω;
R2 C
1
= 1,
R1 R2 C 2
1
.·. R1 = √ = 0.707 Ω.
2
AP 15.5 From the statement of the problem, K = 10 ( = 20 dB). Therefore for the
prototype bandpass circuit
R1 =
Q
16
=
= 1.6 Ω;
K
10
Problems
R2 =
Q
2Q2 − K
=
16
Ω;
502
R3 = 2Q = 32 Ω.
The scaling factors are
kf =
ωo0
= 2π(6400) = 12,800π;
ωo
km =
1
C
=
= 1243.40.
C 0 kf
(20 × 10−9 )(12,800π)
Therefore,
R10 = km R1 = (1.6)(1243.30) = 1.99 kΩ;
R20 = km R2 = (16/502)(1243.40) = 39.63 Ω;
R30 = km R3 = 32(1243.40) = 39.79 kΩ.
AP 15.6 ωo = 2000π rad/s;
ωo0
= 2000π;
ωo
.·.
kf =
km =
1
105
C
=
=
;
C 0 kf
(15 × 10−9 )(2000π)
3π
R 0 = km R =
σ =1−
105
(1) = 10,610 Ω;
3π
1
1
=1−
= 0.9875;
4Q
4(20)
σR0 = 10,478 Ω;
C 0 = 15 nF;
2C 0 = 30 nF.
(1 − σ)R0 = 133 Ω;
15–3
15–4
CHAPTER 15. Active Filter Circuits
Problems
P 15.1
[a] ωc =
1
R2 C
so R2 =
1
1
=
= 6366 Ω;
ωc C
2π(500)(50 × 10−9 )
R2
R1
so R1 =
R2
6366
=
= 637 Ω.
K
10
K=
[b] Both the cutoff frequency and the passband gain are changed.
P 15.2
[a] 10(0.2) = 2 V so Vcc ≥ 2 V.
−10(2π)(500)
[b] H(jω) =
;
jω + 2π(500)
H(j1000π) =
−10(1000π)
10
= −5 + j5 = √ /135◦ ;
1000π + j1000π
2
10
Vo = √ /135◦ Vi
2
[c] H(j100π) =
−10(1000π)
= 9.95/174.3◦ ;
1000π + j100π
Vo = 9.95/174.3◦ Vi
[d] H(j10,000π) =
so vo (t) = 1.99 cos(100πt + 174.3◦ ) V.
−10(1000π)
= 0.995/95.7◦ ;
1000π + j10,000π
Vo = 0.995/95.7◦ Vi
P 15.3
so vo (t) = 1.414 cos(1000πt + 135◦ ) V.
so vo (t) = 199 cos(10,000πt + 95.7◦ ) mV.
[a] K = 10(10/20) = 3.1623 =
R2
;
R1
R2 =
1
1
=
= 25,464.791 Ω;
ωc C
(2π)(2500)(2.5 × 10−9 )
R1 =
R2
25,464.791
=
= 8052.62 Ω.
K
3.1623
Problems
15–5
[b]
P 15.4
[a]
1
= 2π(2500) so Rf C = 63.662 × 10−6 .
Rf C
There are several possible approaches. Here, choose C = 22 nF. Then
63.662 × 10−6
= 2893.7 Ω.
22 × 10−9
Choose Rf = 2.7 kΩ. This gives
Rf =
ωc =
1
= 16,835.02 rad/s so fc = 2679.4 Hz.
(2700)(22 × 10−9 )
To get a passband gain of 10 dB,
Rf
2700
=
= 853.8 Ω.
3.1623
3.1623
Choose Ri = 1 kΩ to give K = 20 log10 (2700/1000) = 8.63 dB.
The resulting circuit is
Ri =
[b] % error in fc =
2679.4 − 2500
(100) = 7.2%;
2500
% error in passband gain =
8.63 − 10
(100) = −13.7%.
10
15–6
P 15.5
CHAPTER 15. Active Filter Circuits
[a] R1 =
1
1012
=
= 5.85 kΩ;
ωc C
(2π)(40)(103 )(680)
K = 100.6 = 3.98 =
R2
;
R1
.·. R2 = 3.98R1 = 23.29 kΩ.
[b]
P 15.6
[a]
1
= 2π(40,000) so R1 C = 3.98 × 10−6 .
R1 C
There are several possible approaches. Here, choose C = 220 pF. Then
3.98 × 10−6
= 18,085.8 Ω.
220 × 10−12
Choose R1 = 18 kΩ. This gives
R1 =
ωc =
1
(220 × 10−12 )(18,000)
= 252.5 krad/s so fc = 40,190.6 Hz.
To get a passband gain of 12 dB,
R2 = 3.98R1 = 3.98(18,000) = 71,640 Ω.
Choose R2 = 68 kΩ to give a passband gain of 20 log10 (68/18) = 11.54
dB. The resulting circuit is
[b] % error in fc =
40,190.6 − 40,000
(100) = 0.48%;
40,000
% error in passband gain =
11.54 − 12
(100) = −3.83%.
12
Problems
P 15.7
[a] ωc =
1
R1 C
so R1 =
R2
R1
so R2 = KR1 = (5)(5305) = 26.5 kΩ.
K=
1
1
=
= 5305 Ω;
ωc C
2π(300)(100 × 10−9 )
[b] The passband gain changes but the cutoff frequency is unchanged.
P 15.8
[a] 5(0.15) = 0.75 V so Vcc ≥ 0.75 V.
−5jω
;
[b] H(jω) =
jω + 600π
H(j600π) =
−5(j600π)
5
= √ / − 135◦ ;
600π + j600π
2
5
Vo = √ / − 135◦ Vi
2
[c] H(j60π) =
−5(j60π)
= 0.5/ − 95.7◦ ;
600π + j60π
Vo = 0.5/ − 95.7◦ Vi
[d] H(j6000π) =
so vo (t) = 74.63 cos(60πt − 95.7◦ ) mV.
−5(j6000π)
= 4.98/ − 174.3◦ ;
600π + j6000π
Vo = 4.98/ − 174.3◦ Vi
P 15.9
so vo (t) = 530.33 cos(600πt − 135◦ ) mV.
so vo (t) = 746.3 cos(6000πt − 174.3◦ ) mV.
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
15–7
15–8
CHAPTER 15. Active Filter Circuits
P 15.10 [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 )
=
−(1/C2 )[s + (1/R1 C1 )]
.·. H(s) =
[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 )
−C1
H(j0) =
C2
C1
[c] H(j∞) = −
C2
−R2
.
R1
R2 C2
R1 C1
!
−C1
.
C2
j
j
=
=
[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 .
P 15.11 [a] Zf =
(1/C2 )
;
s + (1/R2 C2 )
Zi = R 1 +
1
R1
=
[s + (1/R1 C1 )];
sC1
s
H(s) = −
(1/C2 )
s
·
[s + (1/R2 C2 )] R1 [s + (1/R1 C1 )]
=−
1
s
.
R1 C2 [s + (1/R1 C1 )][s + (1/R2 C2 )]
Problems
[b] H(jω) = −
1
R1 C2 jω +
jω
1
R1 C1
jω + R21C2
15–9
;
H(j0) = 0.
[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 15.12 [a] km =
500
= 0.125;
4000
1
(625 × 10−9 ) = 1 × 10−6
km kf
L0 =
so
kf =
625 × 10−9
= 5.
(0.125)(1 × 10−6 )
(0.125)(5)
= 125 mH.
5
[b] Zab = jωL + Rk
=
R/jωC
1
= jωL +
jωC
R + (1/jωC)
[jωL(1 + jωRC) + R](1 − jωRC)
.
(1 + jωRC)(1 − jωRC)
The denominator is purely real, so set the imaginary part of the
numerator equal to 0 and solve for ω:
ωL + ω 3 R2 LC 2 − ωR2 C = 0
.·.
ω2 =
R2 C − L
(500)2 (10−6 ) − 0.125
=
= 4 × 106 .
2
2
2
−6
2
R LC
500 (0.125)(10 )
Thus,
ω = 2 krad/s.
[c] In the original, unscaled circuit, the frequency at which the impedance Zab
is purely real is
2
ωus
=
Thus,
40002 (625 × 10−9 ) − 5
= 16 × 104 .
40002 (5)(625 × 10−9 )2
ωus = 400 rad/s.
ωscaled
2000
=
= 5 = kf .
ωus
400
15–10
CHAPTER 15. Active Filter Circuits
P 15.13 [a] kf =
4000
= 8;
500
C0 =
20 × 10−6
= 2.5 µF;
8
L0 =
1
= 125 mH.
8
Io =
(50 − j100)k(250 + j500)
(0.125) = 27.95/ − 116.565◦ mA;
250 + j500
[b]
io = 27.95 cos(4000t − 116.565◦ ) mA.
The magnitude and phase angle of the output current are the same as in
the unscaled circuit.
P 15.14 [a] The scaled circuit, transformed into the frequency domain with ω = 25
krad/s, is
The equivalent impedance seen by the source is used to find the voltage
drop across the source:
Zeq = 400 + [−j200k(1000 + j200)] = 440 − j200 Ω;
V = Zeq I = (440 − j200)(0.03) = 13.2 − j6 V;
1
1
.·. S = − VI∗ = − (13.2 − j6)(0.03) = −198 + j90 mVA.
2
2
Thus the real power delivered by the source in the scaled circuit is 198
mW.
Problems
15–11
[b] The unscaled circuit, transformed into the frequency domain with ω = 25
krad/s, is
Zeq = 2 + [−jk(5 + j)] = 2.2 − j1 Ω;
V = Zeq I = (2.2 − j1)(0.03) = 66 − j30 mV;
1
1
.·. S = − VI∗ = − (0.066 − j0.033)(0.03) = −990 + j450 µVA.
2
2
The real power delivered by the source in the unscaled circuit is 990 µ W.
Note that the ratio of the real power delivered by the source in the scaled
circuit to the real power delivered by the source in the unscaled circuit is
990 × 10−6
= 200 = km .
0.198
P 15.15 [a] The scaled circuit, transformed into the frequency domain with ω = 25
krad/s, is
The equivalent impedance seen by the source is used to find the voltage
drop across the source:
Zeq = 2 + [−j0.01k(5 + j100)] = 2 − j0.01 Ω;
V = Zeq I = (2 − j0.01)(0.03) = 60 − j0.3 mV;
1
1
.·. S = − VI∗ = − (0.06 − j0.0003)(0.03) = −900 + j4.5 µVA.
2
2
Thus the real power delivered by the source in the scaled circuit is
900 µW.
[b] The scaled circuit, transformed into the frequency domain with the source
frequency scaled by kf = 0.01 (ω = 250 rad/s), is
Zeq = 2 + [−jk(5 + j)] = 2.2 − j1 Ω;
15–12
CHAPTER 15. Active Filter Circuits
V = Zeq I = (2.2 − j1)(0.03) = 66 − j30 mV;
1
1
.·. S = − VI∗ = − (0.066 − j0.033)(0.03) = −990 + j450 mVA.
2
2
The real power delivered by the source in the scaled circuit when the
source frequency is also scaled is 990 µ W.
[c] When both the components of the circuit and the source frequency are
scaled using the frequency scale factor, the real power delivered by the
source is unchanged from the unscaled circuit.
P 15.16 [a] km =
20
= 5;
4
−9
5 × 10
.·. 100 × 10−12 =
5kf
Lscaled =
;
.·. kf = 10.
5
(80) = 40 mH;
10
R2scaled = (21)(5 × 103 ) = 105 kΩ.
[b]
Vo
Vo
Vo − (300/s)
+
+
= 0;
20,000
0.04s 105,000 + (1010 /s)
.·.
Vo =
252s + 24 × 106
−4
256
=
+
.
s2 + 500,000 + 40 × 109
s + 100,000 s + 400,000
Therefore,
vo = (256e−400,000t − 4e−100,000t )u(t) V.
P 15.17 km =
R0
1000
=
= 0.2;
R
5000
kf = 0.02.
Rf0 = km Rf = 0.2(25,000) = 5 kΩ;
C0 =
C
4 × 10−9
=
= 1 µF.
km kf
(0.2)(0.02)
Problems
15–13
Va − (0.4/s) (Va − Vo )s Va s
+
+ 6 = 0;
1000
106
10
Va s
Vo
−200
+
= 0 so Va =
Vo .
6
10
5000
s
.·.
Vo =
−400
0.5/90◦
0.5/ − 90◦
=
+
.
s2 + 400s + 200 × 103
s + 200 − j400 s + 200 + j400
vo = [e−200t cos(400t + 90◦ )]u(t) V.
P 15.18 From the solution to Problem 14.30, ωo = 106 rad/s and β = 133.33 krad/s.
Compute the two scale factors:
kf =
2π(250 × 103 )
ωo0
=
= π/2;
ωo
106
km =
1 C
2 25 × 10−9
5
=
=
.
kf C 0
π 10 × 10−9
π
Thus,
R 0 = km R =
5
(300) = 477.46 Ω;
π
L0 =
km
5/π
(40 × 10−6 ) = 40.53 µH.
L=
kf
π/2
Calculate the cutoff frequencies:
0
ωc1
= kf ωc1 = (π/2)(935.56 × 103 ) = 1469.57 krad/s;
0
ωc2
= kf ωc2 = (π/2)(1068.89 × 103 ) = 1679.01 krad/s.
To check, calculate the bandwidth:
0
0
β 0 = ωc2
− ωc1
= 209.44 krad/s = (π/2)β (checks!).
15–14
CHAPTER 15. Active Filter Circuits
P 15.19 From the solution to Problem 14.36, ωo = 8 × 106 rad/s and β = 500 krad/s.
Calculate the scale factors:
kf =
ωo0
500 × 103
=
= 0.0625;
ωo
8 × 106
km =
0.0625(50 × 10−6 )
kf L0
=
= 0.005.
L
625 × 10−6
Thus,
R0 = km R = (0.005)(80,000) = 400 Ω;
C0 =
Calculate the bandwidth:
β 0 = kf β = (0.0625)(500 × 103 ) = 31,250 rad/s.
To check, calculate the quality factor:
Q=
8 × 106
ωo
=
= 16;
β
500 × 103
ωo0
500 × 103
Q = 0 =
= 16 (checks).
β
31,250
0
P 15.20 [a] From Eq 15.1 we have
H(s) =
−Kωc
s + ωc
where K =
R2
,
R1
ωc =
0
1
;
R2 C
0
−K ωc
.·. H 0 (s) =
s + ωc0
where K 0 =
R20
R10
ωc0 =
1
R20 C 0
.
By hypothesis R10 = km R1 ;
and C 0 =
R20 = 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
C
25 × 10−12
=
= 80 nF.
km kf
(0.005)(0.0625)
Problems
[b] H(s) =
−K
.
s+1
−K
[c] H 0 (s) = s kf
=
+1
−Kkf
.
s + kf
P 15.21 [a] From Eq. 15.4
H(s) =
ωc =
R2
−Ks
where K =
and
s + ωc
R1
1
;
R1 C
0
0
R20 = km R2 ;
C0 =
−K s
R
where K 0 = 20
.·. H 0 (s) =
0
s + ωc
R1
and ωc0 =
1
R10 C 0
.
By hypothesis
R10 = km R1 ;
C
.
km kf
It follows that
K 0 = K and ωc0 = kf ωc ;
.·. H 0 (s) =
[b] H(s) =
−K(s/kf )
−Ks
= s
.
s + kf ωc
+
ω
c
kf
−Ks
.
s+1
−K(s/kf )
[c] H 0 (s) = s
kf
+1
=
−Ks
.
s + kf
P 15.22 For the RC circuit
H(s) =
s
Vo
=
;
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
15–15
15–16
CHAPTER 15. Active Filter Circuits
For the RL circuit
s
H(s) =
;
s + (R/L)
R0 = km R;
km L
;
kf
L0 =
R0
R
= kf
= kf ;
0
L
L
H 0 (s) =
s
(s/kf )
=
.
s + kf
(s/kf ) + 1
P 15.23 For the RC circuit
H(s) =
(1/RC)
Vo
=
;
Vi
s + (1/RC)
R0 = km R;
C0 =
C
;
km kf
C
1
1
.·. R0 C 0 = km R
= RC = .
km kf
kf
kf
1
= kf .
R0 C 0
(1/R0 C 0 )
kf
H (s) =
=
;
0
0
s + (1/R C )
s + kf
0
H 0 (s) =
1
.
(s/kf ) + 1
For the RL circuit
R0 = km R;
L0 =
H(s) =
km
L;
kf
R0
km R
R
= km = k f
= kf ;
0
L
L
L
kf
H 0 (s) =
(R0 /L0 )
kf
;
=
0
0
s + (R /L )
s + kf
H 0 (s) =
1
.
(s/kf ) + 1
R/L
s + R/L
so
Problems
P 15.24 H(s) =
(R/L)s
s2 + (R/L)s + (1/LC)
=
βs
s2 + βs + ωo2
15–17
.
For the prototype circuit ωo = 1 and β = ωo /Q = 1/Q.
For the scaled circuit
H 0 (s) =
(R0 /L0 )s
s2 + (R0 /L0 )s + (1/L0 C 0 )
where R0 = km R; L0 =
km
C
L; and C 0 =
;
kf
kf km
R0
km R
R
.·.
= km = k f
= kf β.
0
L
L
L
kf
kf2
1
kf km
=
= kf2 ;
=
km
L0 C 0
LC
LC
kf
Q0 =
ωo0
kf ωo
=
= Q.
0
β
kf β
Therefore the Q of the scaled circuit is the same as the Q of the unscaled
circuit. Also note β 0 = kf β.
.·.
1
Q
s
kf
+ Q1
kf
s
Q
0
;
H (s) =
k
s2 + Qf s + kf2
H 0 (s) = 2
s
kf
P 15.25 [a] L = 1 H;
R=
s
kf
.
+1
C = 1 F;
1
1
=
= 0.04 Ω.
Q
25
[b] kf = 100,000;
km =
3600
= 90,000.
0.04
Thus,
R0 = (0.04)(90,000) = 3.6 kΩ;
L0 =
C0 =
90,000
(1) = 0.9 H;
100,000
1
(105 )(9 × 104 )
=
1
nF = 0.11 nF.
9
15–18
CHAPTER 15. Active Filter Circuits
[c]
P 15.26 [a] By hypothesis, LC = 1; Thus,
C=
1
1
= F.
L
Q
[b] H(s) =
(R/L)s
s2 + (R/L)s + (1/LC)
H(s) =
(1/Q)s
s2 + (1/Q)s + 1
;
.
[c] In the prototype circuit,
R = 1 Ω;
L = 20 H;
.·. km = 5000;
C = 0.05 F;
kf = 50,000.
Thus
R0 = 5 kΩ;
L0 =
5000
(20) = 2 H;
50,000
C0 =
0.05
= 0.2 × 10−9 = 0.2 nF.
(5000)(50,000)
[d]
[e] H 0 (s) = H 0 (s) =
s
50,000
1
20
2
s
50,000
1
+ 20
s
50,000
;
+1
2500s
s2 + 2500s + 25 × 108
.
P 15.27 [a] Using the first prototype,
ωo = 1 rad/s;
C = 1 F;
L = 1 H;
R = 16 Ω;
Problems
80,000
= 5000;
16
km =
kf = 80,000.
Thus,
R0 = 80 kΩ;
C0 =
L0 =
5
(1) = 62.5 mH;
80
1
= 2.5 nF.
400 × 106
Using the second prototype
ωo = 1 rad/s;
L=
C = 16 F;
1
= 6.25 mH;
16
km = 80,000;
R = 1 Ω;
kf = 80,000.
Thus,
R0 = 80 kΩ;
C0 =
L0 =
80
(6.25) = 6.25 mH;
80
16
= 2.5 nF.
64 × 108
[b]
P 15.28 [a] For the circuit in Fig. P15.28(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. P15.28(b)
Qs + Qs
Vo
H(s) =
=
Vi
1 + Qs + Qs
=
H(s) =
Q(s2 + 1)
;
Qs2 + s + Q
s2 + 1
s2 +
1
Q
s+1
.
15–19
15–20
CHAPTER 15. Active Filter Circuits
2
s
+1
50,000
H 0 (s) = 2
s
s
1
+
+1
50,000
5 50,000
2
8
[b]
s + 25 × 10
.
s2 + 10,000s + 25 × 108
=
P 15.29 For the scaled circuit
H 0 (s) =
L0 =
.·.
s2 +
s+
s2 +
R0
L0
km
L;
kf
1
L0 C 0
C0 =
;
1
L0 C 0
C
km kf ;
kf2
1
=
;
L0 C 0
LC
R0 = km R;
R0
R
·
..
= kf
.
0
L
L
It follows then that
s2 +
H 0 (s) =
s2 +
k2
R
L
kf s + LCf
s
kf
= 2
s
kf
kf2
LC
+
2
+
s
kf
R
L
1
LC
+
1
LC
= H(s)|s=s/kf .
P 15.30 For the circuit at the bottom of Fig. 14.31
H(s) =
s2 +
1
LC
s
s2 + RC
+
1
LC
.
It follows that
H 0 (s) =
s2 + L01C 0
s2 + R0sC 0 + L01C 0
where R0 = km R;
L0 =
km
L;
kf
Problems
C0 =
.·.
C
;
km kf
kf2
1
=
.
L0 C 0
LC
1
kf
=
0
0
RC
RC.
s2 +
H 0 (s) =
s2 +
= 2
s
kf
kf2
LC
k2
kf
RC
s
kf
2
+
s + LCf
1
RC
1
+ LC
s
kf
1
+ LC
= H(s)|s=s/kf .
P 15.31 For prototype circuit (a):
H(s) =
Vo
Q
Q
=
;
1 =
Vi
Q + s2s+1
Q + s+ 1
s
H(s) =
s2 + 1
Q(s2 + 1)
=
.
Q(s2 + 1) + s
s2 + 1 s + 1
Q
For prototype circuit (b):
H(s) =
=
Vo
1
=
Vi
1 + (s(s/Q)
2 +1)
s2 + 1
s2 +
1
Q
s+1
.
P 15.32 ωo = 50,000 rad/s;
β = 300,000 rad/s;
.·. ωc2 − ωc1 = 300,000;
√
ωc1 ωc2 = ωo = 50,000.
15–21
15–22
CHAPTER 15. Active Filter Circuits
Solve for the cutoff frequencies:
ωc1 ωc2 = 25 × 108 ;
ωc2 =
25 × 108
;
ωc1
25 × 108
.·.
− ωc1 = 300,000,
ωc1
or ωc21 + 300,000ωc1 − 25 × 108 = 0;
ωc1 = 8113.88 rad/s;
.·. ωc2 = 300,000 + 8113.88 = 308,113.88 rad/s.
Thus, fc1 = 1291.4 Hz
and
fc2 = 49,037.85 Hz;
ωc2 =
1
= 308,113.88;
RL CL
RL =
1
= 21.64 Ω;
(308,113.88)(150 × 10−9 )
ωc1 =
1
= 8113.88;
RH CH
RH =
1
= 821.64 Ω.
(8113.88)(150 × 10−9 )
P 15.33 [a] Hhp =
−s
;
s+1
0
.·. Hhp
=
−s
;
s + 4000π
1
= 4000π;
RH CH
Hlp =
kf = 4000π;
−1
;
s+1
.·. RH =
kf = 16,000π;
−16,000π
0
.·. Hlp
=
;
s + 16,000π
106
= 3.98 kΩ;
(4000π)(0.02)
Problems
1
= 16,000π;
RL CL
.·. RL =
15–23
106
= 994.72 Ω;
(16,000π)(0.02)
Kωc2
= 0.8K;
ωc1 + ωc2
|H(jωo )| =
20 log10 (0.8K) = 10;
√
.·. K = 1.25 10;
√
Rf
.·.
= 1.25 10;
Ri
Ri = 10 kΩ;
√
Rf = 12.5 10 = 39.53 kΩ.
s
16,000π
−39.53
−63,248s
·
·
= 2
[b] H 0 (s) =
.
s + 4000π s + 16,000π
10
s + 20,000πs + 64 × 106 π 2
√
[c] ωo = ωc1 ωc1 = 8000π rad/s;
H 0 (jωo ) =
(16,000π)(j8000π)
39.53
·
(4000π + j8000π)(16,000π + j8000π)
10
= (0.8)(3.953) = 3.16 =
[d] 20 log10 |H 0 (jωo )| = 20 log10
√
√
10.
10 = 10 dB.
[e]
Note that because this is not a broadband filter, the cutoff frequencies,
identified where the magnitude is down 3 dB from the maximum
magnitude, are actually 1.4 kHz and 11.3 kHz.
P 15.34 ωo = 2π(5000) rad/s;
β = 2π(30,000) rad/s;
K = 4;
C = 250 nF;
15–24
CHAPTER 15. Active Filter Circuits
β = ωc2 − ωc1 = 60,000π;
ωo =
√
ωc1 ωc2 = 10,000π.
Solve for the cutoff frequencies:
.·. ωc21 + 60,000πωc1 − (10,000π)2 = 0;
ωc1 = 5098.1 rad/s;
ωc2 = 60,000π + ωc1 = 193,593.7 rad/s;
ωc1 =
1
;
RL CL
.·. RL =
1
(250 × 10−9 )(5098.1)
= 784.6 Ω;
1
= ωc2 ;
RH CH
RH =
1
(250 × 10−9 )(193,593.7)
= 20.7 Ω;
Rf
= 4;
Ri
If Ri = 1 kΩ
P 15.35 [a] ωc1 =
Rf = 4Ri = 4 kΩ.
1
= 2000π rad/s;
RL CL
RL =
109
= 31.83 kΩ;
(2000π)(5)
ωc2 =
1
= 10,000π rad/s;
RH CH
109
RH =
= 6.37 kΩ;
(10,000π)(5)
20 log10
Rf
Ri
= 10;
Choose Ri = 31.83 kΩ;
.·. Rf =
√
10Ri .
then Rf = 100.66 kΩ
Problems
[b]
[c] H(s)LP =
−2000π
−1
=
;
s/kf + 1
s + 2000π
H(s)HP =
−
−s/kf
−s
=
;
s/kf + 1
s + 10,000π
√
Rf
= − 10;
Ri
H(s) =
=
√
2000π
s
10
+
s + 2000π s + 10,000π
√
s2 + 4000πs + 20 × 106 π 2
10
.
(s + 2000π)(s + 10,000π)
"
#
#
"
√
ωc1 ωc2 = 20 × 106 π 2
√
√
= 1000π 20 = 2000π 5 rad/s;
[d] ωo =
√
√
#
j4000π(2000π 5)
√
√
H(jωo ) = 10
(2000π + j2000π 5)(10,000π + j2000π 5)
√ √
√ √
j2 5 10
j2 5 10
√
√ =
√
=
(1 + j 5)(5 + j 5)
j6 5
√
10
=
= 1.05.
3
√
"
[e] 20 log10 |H(jωo )| = 20 log10 1.05 = 0.46 dB.
15–25
15–26
CHAPTER 15. Active Filter Circuits
1−
[f ] H(jω) =
ω
√
1000 20π
2 ω
1 + j 2000π
4
2ζ = √ ;
20
w
+ j √420 · 100√
20π
ω
1 + j 10,000π
2
ζ=√ ;
20
;
ζ 2 = 0.20;
√
fo = 1000 5 = 2236.07 Hz;
q
√
fp = fo 1 − 2ζ 2 = fo 0.6 = 1732.05 Hz;
√
ωo = 2000π 5;
AdB (fp ) = 10 log10 [4ζ 2 (1 − ζ 2 )] = 10 log10 0.64 = −1.94 dB;
AdB (fo /2) = 10 log10 0.7625 = −1.18 dB;
AdB (fo ) = 20 log10 2ζ = −0.97 dB.
For the quadratic term, AdB = 0 when f =
P 15.36 H(s) =
Zf =
√
2fp = 2449.48 Hz.
Vo
−Zf
=
;
Vi
Zi
(1/C2 )
1
kR2 =
;
sC2
s + (1/R2 C2 )
Zi = R 1 +
sR1 C1 + 1
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
Problems
[a] H(s) =
15–27
−10,000s
−10,000s
−3.57(2800s)
= 2
= 2
;
5
(s + 2000)(s + 800)
s + 2800s + 16 × 10
s + 2800s + (1264.91)2
ωo = 1264.91 rad/s;
β = 2800 rad/s;
K = −3.57.
ωo
[b] Q =
= 0.45;
β
β
ωc1,2 = ± +
2
v
u
u
t
β
2
!2
ωc1 = 486.8 rad/s
P 15.37 [a] H(s) =
+ ωo2 = ±1400 +
√
14002 + 16 × 105 = ±1400 + 1886.8;
ωc2 = 3286.8 rad/s.
(1/sC)
(1/RC)
=
;
R + (1/sC)
s + (1/RC)
H(jω) =
(1/RC)
;
jω + (1/RC)
|H(jω)| = q
|H(jω)|2 =
(1/RC)
ω 2 + (1/RC)2
;
(1/RC)2
.
ω 2 + (1/RC)2
[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 =
(R2 + sL)Vin
.
R1 LCs2 + (R1 R2 C + L)s + (R1 + R2 )
But
vo =
sLVa
.
R2 + sL
Therefore
Vo =
sLVin
;
R1 LCs2 + (L + R1 R2 C)s + (R1 + R2 )
H(s) =
sL
R1 LCs2 + (L + R1 R2 C)s + (R1 + R2 )
;
15–28
CHAPTER 15. Active Filter Circuits
H(jω) =
jωL
;
[(R1 + R2 ) − R1 LCω 2 ] + jω(L + R1 R2 C)
|H(jω)| = q
|H(jω)|2 =
ωL
[R1 + R2 − R1 LCω 2 ]2 + ω 2 (L + R1 R2 C)2
;
ω 2 L2
(R1 + R2 − R1 LCω 2 )2 + ω 2 (L + R1 R2 C)2
ω 2 L2
= 2 2 2 4
.
R1 L C ω + (L2 + R12 R22 C 2 − 2R12 LC)ω 2 + (R1 + R2 )2
[c] Let Va be the voltage across R2 positive at the upper terminal. Then
Va
Va − Vin
+
+ 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 )
H(jω) =
−2R2 R3 C(jω)
;
(R1 + R2 ) + jω(2R1 R2 C)
|H(jω)| = q
|H(jω)|2 =
2R2 R3 Cω
(R1 + R2 )2 + ω 2 4R12 R22 C 2
4R22 R32 C 2 ω 2
.
(R1 + R2 )2 + 4R12 R22 C 2 ω 2
P 15.38 n = 5: 1 + (−1)5 s10 = 0;
s10 = 1/(0 + 360k)◦
so
s10 = 1;
s = 1/36k ◦ .
;
Problems
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;
15–29
15–30
CHAPTER 15. Active Filter Circuits
s12 = 1/180◦ + 360k.
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◦
Problems
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).
1
= −10 log10 (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 )
P 15.39 [a] y = 20 log10 √
x = log10 ω =
ln ω
;
ln 10
.·. ln ω = x ln 10.
dω
1 dω
= ln 10,
= ω ln 10;
ω dx
dx
dy
=
dx
dy
dω
!
dω
dx
!
=
−20nω 2n
dB/decade.
1 + ω 2n
At ω = ωc = 1 rad/s
dy
= −10n dB/decade.
dx
1
= −10n log10 (1 + ω 2 )
[b] y = 20 log10 √
2
n
[ 1+ω ]
−10n
ln(1 + ω 2 );
ln 10
dy
−10n
1
−20nω
=
.
2ω =
2
dω
ln 10 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
15–31
15–32
CHAPTER 15. Active Filter Circuits
[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 15.40 [a] n ∼
=
(−0.05)(−40) ∼
= 3.32;
log10 (4000/1000)
.·. n = 4.
[b] Gain = 20 log10 q
1
1 + (4)8
= −10 log10 (1 + 48 ) = −48.16 dB.
P 15.41 [a] From Table 15.1:
Hlp (s) =
1
(s + 1)(s2 + 0.618s + 1)(s2 + 1.618s + 1)
;
1
Hhp (s) =
[(1/s) + 1][(1/s)2 + 0.618(1/s) + 1][(1/s)2 + 1.618(1/s) + 1]
;
s5
.
(s + 1)(s2 + 0.618s + 1)(s2 + 1.618s + 1)
Hhp (s) =
P 15.42 [a] kf = 10,000;
(s/10,000)5
[(s/10,000) + 1]
0
Hhp
(s) =
·
1
[(s/10,000)2 + 0.618s/10,000 + 1][(s/10,000)2 + 1.618s/10,000 + 1]
s5
=
.
(s + 10,000)(s2 + 6180s + 108 )(s2 + 16,180s + 108 )
Problems
[b] H 0 (j10,000) =
=
j(10,000)5
[10,000(j + 1)][6180(j10,000)][16,180(j10,000)]
j(10,000)2
(1 + j)(6180)(16,180)j 2
= 0.7072/ − 45◦ ;
20 log10 |H 0 (j10,000)| = −3.01 dB.
P 15.43 [a] n ∼
=
(−0.05)(−25)
= 1.79; .·. n = 2;
log10 (100/20)
.·. H(s) =
1
√
.
2s + 1
√
C1 = 2 F;
s2 +
√
2
= 2:
C1
C2 =
√
1
1
= √ = 0.5 2 F;
C1
2
km = 3000;
kf = 40,000π;
√
C10 =
2
= 3.75 nF;
(3000)(40,000π)
1
C20 = C10 = 1.88 nF;
2
R1 = R2 = 3 kΩ.
[b]
(−0.05)(−25)
P 15.44 [a] n ∼
= 1.79;
=
log10 (5/1)
s2
√
;
s2 + 2s + 1
√
√
2
= 2;
R2 = 2 Ω;
R2
.·. n = 2;
.·. H(s) =
kf = 10,000π;
R1 =
1
1
= √ Ω;
R2
2
15–33
15–34
CHAPTER 15. Active Filter Circuits
109
4000
=
.
(10,000π)(25)
π
.·. km =
1 4000
R1 = √ ·
= 900.32 Ω;
π
2
R2 =
√ 4000 2
= 1800.63 Ω.
π
[b]
P 15.45 [a] A bandpass filter.
[b] fc1 = 5000 Hz;
fo =
Q=
q
fc2 = 20,000 Hz;
fc1 fc2 = 10,000 Hz;
ωo
fo
10,000
=
=
= 0.67.
β
fc2 − fc1
15,000
s2
√
;
s2 + 2s + 1
[c] H(s)hp =
H 0 (s)hp =
=
H(s)lp =
(s/104 π)2
√
(s/104 π)2 + 2(s/104 π) + 1
s2
;
s2 + π 2 × 104 s + 108 π 2
√
1
√
.
s2 + 2s + 1
H 0 (s)lp =
=
(s/4 × 104 π)2 +
1
√
2(s/4 × 104 π) + 1
16 × 108 π 2
;
s2 + 4π 2 × 104 s + 16 × 108 π 2
√
H(s) = H 0 (s)hp · H 0 (s)lp
=
16 × 108 π 2 s2
√
.
(s2 + π 2104 s + 108 π 2 )(s2 + 4π 2 × 104 s + 16 × 108 π 2 )
√
Problems
[d] ωo = 20,000π rad/s = 2 × 104 krad/s;
16 × 108 π 2 (−4 × 108 π 2 )
√
(−3 × 108 π 2 + jπ 2104 (2 × 104 π))
H(s) =
×
1
√
(12 × 108 π 2 + j4
2π104 (2 × 104 π))
−64
−64
√
√ =
= 0.9412.
−68
(−3 + j2 2)(12 + j8 2)
=
P 15.46 For the scaled circuit
H 0 (s) =
1/(R0 )2 C10 C20
,
s2 + R02C 0 s + (R0 )21C 0 C 0
1
2
1
where
R0 = km R;
C10 = C1 /kf km ;
C20 = C2 /kf km .
It follows that
kf2
1
=
;
(R0 )2 C10 C20
R2 C1 C2
2
R0 C10
=
2kf
.
RC1
kf2 /RC1 C2
.·. H 0 (s) =
k2
2k
s2 + RCf1 s + R2 Cf1 C2
1/RC1 C2
= 2
s
kf
P 15.47 [a] H(s) =
2
+ RC
1
s
kf
+ R2 C11 C2
.
1
(s2 + 0.765s + 1)(s2 + 1.848s + 1)
[b] fc = 1000 Hz;
H 0 (s) = =
ωc = 2000π rad/s;
.
kf = 2000π;
1
s
2000π
2
+ 0.765s
+1
2000π
2
s
+ 1.848s
+1
2000π
2000π
6 2 2
(4 × 10 π )
(s2 + 1530πs + 4 × 106 π 2 )(s2 + 3696πs + 4 × 106 π 2 )
.
15–35
15–36
CHAPTER 15. Active Filter Circuits
[c] H 0 (j8000π) =
16
;
(−60 + j12.24)(−60 + j29.568)
|H 0 (j8000π)| =
16
= 3.91 × 10−3 ;
(61.24)(66.89)
Gain = 20 log10 |H(j8000π)| = −48.16 dB (checks with Problem 15.40).
P 15.48 [a] km = 2000;
First stage:
kf = 2000π.
2
= 0.765;
C1
.·. C1 =
2
;
0.765
C10 =
2
= 208.05 nF;
(0.765)(2000)(2000π)
C2 =
1
0.765
=
;
C1
2
C20 =
0.765
= 30.44 nF.
2(2000)(2000π)
Second stage:
2
= 1.848;
C1
[b]
.·. C1 =
2
;
1.848
C10 =
2
= 86.12 nF;
(1.848)(2000)(2000π)
C2 =
1
1.848
=
;
C1
2
C20 =
1.848
= 73.53 nF.
2(2000)(2000π)
Problems
s2
P 15.49 H 0 (s) =
s2 +
1
2
s+
2 k2 )
km R2 (C/km kf )
km R1 km R2 (C 2 /km
f
.
s2
H 0 (s) =
s2 +
=
15–37
kf2
2kf
s+
R2 C
R1 R2 C 2
(s/kf )2
!
.
2
s
1
2
(s/kf ) +
+
R2 C kf
R1 R2 C 2
P 15.50 [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 1000 Hz. The order of the Butterworth is
n=
(−0.05)(−20)
= 2.51;
log10 (1000/400)
.·. n = 3.
Hhp (s) =
s3
.
(s + 1)(s2 + s + 1)
For the prototype first-order section,
R1 = R2 = 1 Ω,
C = 1 F.
For the prototype second-order section,
R1 = 0.5 Ω,
R2 = 2 Ω,
C = 1 F.
The scaling factors are,
kf = 2π(1000) = 2000π;
km =
109
104
=
.
50(2000π)
π
In the scaled first-order section,
R1 = R2 =
104
(1) = 3.183 kΩ;
π
C = 50nF.
In the scaled second-order section,
R1 = 0.5km = 1591.55 Ω;
R2 = 2km = 6.366 kΩ;
15–38
CHAPTER 15. Active Filter Circuits
C = 50 nF.
For the low-pass section the cut-off frequency is 8000 Hz. The order of
the Butterworth filter is
(−0.05)(−20)
.·. n = 3.
n=
= 2.51;
log10 (20,000/8000)
Hlp (s) =
1
.
(s + 1)(s2 + s + 1)
For the prototype first-order section,
R1 = R2 = 1 Ω,
C = 1 F.
For the prototype second-order section,
R1 = R2 = 1 Ω;
C1 = 2 F;
C2 = 0.5 F.
The low-pass scaling factors are,
km = 5 × 103 ;
kf = (8000)(2π) = 16,000π.
For the scaled first-order section,
R1 = R2 = 5 kΩ;
C=
1
= 3.98 nF.
(16,000π)(5 × 103 )
For the scaled second-order section,
R1 = R2 = 5 kΩ;
C1 =
2
= 7.96 nF;
8π × 107
C2 =
0.5
= 1.99 nF.
8π × 107
Gain amplifier:
20 log10 K = 10 dB,
.·. K = 3.16.
Since we are using 5 kΩ resistors in the low-pass stage, we will use
Rf = 15.8 kΩ and Ri = 5 kΩ in the inverting amplifier stage.
Problems
15–39
[b]
P 15.51 [a] Unscaled high-pass stage:
Hhp (s) =
s3
.
(s + 1)(s2 + s + 1)
Frequency scaling factor kf = 2000π. Therefore the scaled transfer
function is
(s/2000π)3
0
Hhp
(s) = 2
s
s
s
+1
+ 2000π + 1
2000π
2000π
=
s3
.
(s + 2000π)[s2 + 2000πs + 4 × 106 π 2 ]
Unscaled low-pass stage:
Hlp (s) =
1
.
(s + 1)(s2 + s + 1)
Frequency scaling factor kf = 16,000π. Therefore the scaled transfer
function is
1
Hlp0 (s) = 2 s
s
s
+1
+ 16,000π + 1
16,000π
16,000π
=
(16,000π)3
.
(s + 16,000π)(s2 + 16,000πs + 256 × 106 π 2 )
Thus the transfer function for the filter is
4096 × 1010 π 3 s3
0
H 0 (s) = 10Hhp
(s)Hlp0 (s) =
,
D1 D2 D3 D4
where
D1 = s + 2000π;
D2 = s + 16,000π;
15–40
CHAPTER 15. Active Filter Circuits
D3 = s2 + 2000πs + 4 × 106 π 2 ;
D4 = s2 + 16,000πs + 256 × 106 π 2 .
[b] At 400 Hz
ω = 800π rad/s:
D1 (j800π) = 800π(2.5 + j1);
D2 (j800π) = 800π(20 + j1);
D3 (j800π) = 16 × 105 π 2 (2.1 + j1.0);
D4 (j800π) = 128 × 105 π 2 (19.95 + j1).
Therefore
D1 D2 D3 D4 (j800π) = 131,072π 6 1014 (2505.11/53◦ );
H 0 (j800π) =
(4096π 3 × 1010 )(512 × 106 π 3 )/ − 90◦
131,072 × 1014 π 6 (2505.11/53◦ )
= 0.639/ − 143◦ ;
.·. 20 log10 |H 0 (j800π)| = 20 log10 (0.639) = −3.89 dB.
At f = 5000 Hz,
ω = 10,000π rad/s.
Then
D1 (j10,000π) = 2000π(1 + j5);
D2 (j10,000π) = 10,000π(1.6 + j1);
D3 (j10,000π) = 107 π 2 (−9.6 + j2);
D4 (j10,000π) = 107 π 2 (15.6 + j16).
H 0 (j10,000π) =
(4096 × π 3 × 1010 )(1012 π 3 )/ − 90◦
2 × 1021 π 6 (2108.22/ − 35.35◦ )
= 9.71/ − 54.65◦ ;
.·. 20 log10 |H 0 (j10,000π)| = 19.74 dB.
[c] From the transfer function the gain is down 19.74 + 3.89 or 23.63 dB at
400 Hz. Because the upper cut-off frequency is eight times the lower
cut-off frequency we would expect the high-pass stage of the filter to
predict the loss in gain at 400 Hz. For a 3nd order Butterworth
1
GAIN = 20 log10 q
1 + (1000/400)6
= −23.89 dB.
Problems
15–41
5000 Hz is in the passband for this bandpass filter. Hence we expect the
gain at 5000 Hz to nearly equal 20 dB as specified in Problem 15.50.
Thus our scaled transfer function confirms that the filter meets the
specifications.
P 15.52 [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
−G1 G3
2
s C1 C2 + (G1 + G2 + G3 )C1 s + G2 G3
−G1 G3 /C1 C2
h
i
(G1 +G2 +G3 )
2 G3
s+ G
C2
C1 C2
1 G2 G3
−G
G2 C1 C2
h
i
=
2 +G3 )
2 G3
s2 + (G1 +G
s+ G
C2
C1 C2
=
s2 +
−Kbo
,
2
s + b1 s + bo
where K =
and b1 =
G1
;
G2
bo =
G1 + G2 + G3
.
C2
[c] Rearranging we see that
G1 = KG2 ;
G3 =
bo C1 C2
bo C1
=
;
G2
G2 .
G2 G3
;
C1 C2
15–42
CHAPTER 15. Active Filter Circuits
Since by hypothesis C2 = 1 F
b1 =
G1 + G2 + G3
= G1 + G2 + G3 ;
C2
bo C1
.·. b1 = KG2 + G2 +
;
G2
b1 = G2 (1 + K) +
bo C1
.
G2
Solving this quadratic equation for G2 we get
b1
G2 =
±
2(1 + K)
=
b1 ±
s
b21 − bo C1 4(1 + K)
4(1 + K)2
q
b21 − 4bo (1 + K)C1
2(1 + K)
.
For G2 to be realizable
C1 <
b21
.
4bo (1 + K)
[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 15.53 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.
1
1
=
F.
4(1 + K)
36
1
1
[b] G2 =
=
S;
2(1 + K)
18
[a] C1 =
18
.·. R1 =
= 2.25 Ω;
8
1
1
1
.·. R3 = 2 Ω.
G3 =
C1 = (18)
= S;
G2
36
2
G1 = 8G2 =
8
S;
18
.·. R2 = 18 Ω;
Problems
[c] fc = 50 kHz;
kf = 105 π;
15–43
ωc = 100π krad/s;
250 × 10−12 =
1
105 πk
;
m
40
.·. km =
× 103 ;
π
R1 = 2.25(40/π)103 = 28.65 kΩ;
R2 = 18(40/π)103 = 229.18 kΩ;
R3 = 2(40/π)103 = 25.46 kΩ;
C1 =
1
5
10 π(40/π)103
[d] R1 = R2 = km =
C=
1
= 694.4 nF.
36
40
× 103 = 12.73 kΩ
π
1
= 250 pF.
kf km
[e]
P 15.54 [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 .
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) =
−C1 C3 s2
Vo
=
Vi
C2 C3 s2 + G1 (C1 + C2 + C3 )s + G1 G2 ]
= h
−C1 2
s
C2
G G
G
s2 + C2 C1 3 (C1 + C2 + C3 )s + C12 C32
i
−Ks2
.
= 2
s + b1 s + bo
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 =
G1 G2
= G1 G2 ;
(1)(1)
bo
bo
.·. G2 =
= (K + 2);
G1
b1
.·. R2 =
b1
.
bo (K + 2)
K +2
.
b1
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;
8+2
= 10 Ω;
1
R1 =
R2 =
1
= 0.1 Ω.
1(8 + 2)
P 15.55 [a] Low-pass filter with a gain of 0 dB (handle 20 dB passband gain in a
separate gain section):
n=
(−0.05)(−20)
= 3.66;
log10 (1500/800)
.·. n = 4.
In the first prototype second-order section: b1 = 0.765, bo = 1, C2 = 1 F.
(0.765)2
b21
C1 ≤
≤
≤ 0.073.
4bo (1 + K)
(4)(2)
Choose C1 = 0.05 F;
G2 =
0.765 ±
q
(0.765)2 − 4(2)(0.05)
2(1 + 1)
=
0.765 ± 0.430
.
4
Arbitrarily select the larger value for G2 , then
G2 = 0.3 S; .·. R2 = 3.33 Ω;
G1 = KG2 = 0.3 S;
G3 =
R1 = 3.33 Ω;
bo C1
(1)(0.05)
=
= 0.167;
G2
0.3
R3 = 1/G3 = 6 Ω.
Therefore in the first second-order prototype circuit
R1 = 3.33 Ω;
R2 = 3.33 Ω;
C1 = 0.05 F;
C2 = 1 F.
R3 = 6 Ω;
In the second second-order prototype circuit:
b1 = 1.848, b0 = 1, C2 = 1 F;
2
(1.848)
.·. C1 ≤
≤ 0.427.
8
Choose C1 = 0.3 F;
15–46
CHAPTER 15. Active Filter Circuits
G2 =
=
1.848 ±
q
(1.848)2 − 8(0.3)
4
;
1.848 ± 1.008
.
4
Arbitrarily select the larger value, then
G2 = 0.71 S; .·. R2 = 1.4 Ω;
G1 = KG2 = 0.71 S;
G3 =
R1 = 1.4 Ω;
(1)(0.3)
bo C1
=
= 0.42 S;
G2
0.71
R3 = 1/G3 = 2.4 Ω.
In the low-pass section of the filter
kf = 2π(800) = 1600π;
km =
C2
1
12,500
=
=
.
0
−9
C2 kf
50 × 10 kf
π
Therefore in the first scaled second-order section
R1 = 3.33km = 13.25 kΩ;
R2 = 3.33km = 13.25 kΩ;
R3 = 6km = 23.87 kΩ;
C1 =
0.05
= 2.5 nF;
(1600π)(12,500/π)
C2 = 50 nF.
In the second scaled second-order section
R1 = 1.4km = 5.57 kΩ;
R2 = 1.4km = 5.57 kΩ;
R3 = 2.4km = 9.55 kΩ;
C1 =
0.3
= 15 nF;
(1600π)(12,500/π)
C2 = 50 nF.
High-pass filter section with a gain of 0 dB (handle 20 dB passband gain
in a separate gain section):
n=
(−0.05)(−20)
= 3.66;
log10 (13,500/7200)
n = 4.
Problems
In the first prototype second-order section:
b1 = 0.765; bo = 1; C2 = C3 = 1 F.
C1 = K = 1 F;
R1 =
3
K +2
=
= 3.92 Ω;
b1
0.765
R2 =
b1
0.765
=
= 0.255 Ω.
bo (K + 2)
3
In the second prototype second-order section: b1 = 1.848; bo = 1;
C2 = C3 = 1 F.
C1 = K = 1 F
R1 =
K +2
3
=
= 1.62 Ω;
b1
1.848
R2 =
b1
1.848
=
= 0.616 Ω.
bo (K + 2)
3
In the high-pass section of the filter
kf = 2π(7200) = 14,400π;
km =
C
1
1389
=
=
.
C 0 kf
50 × 10−9 kf
π
In the first scaled second-order section
R1 = 3.92km = 1.73 kΩ;
R2 = 0.255km = 113 Ω;
C1 = C2 = C3 = 50 nF.
In the second scaled second-order section
R1 = 1.62km = 716 Ω;
R2 = 0.616km = 272 Ω;
C1 = C2 = C3 = 50 nF.
In the gain section, let Ri = 10 kΩ and Rf = 100 kΩ.
15–47
15–48
CHAPTER 15. Active Filter Circuits
[b]
P 15.56 [a] The prototype low-pass transfer function is
Hlp (s) =
1
.
(s2 + 0.765s + 1)(s2 + 1.848s + 1)
The low-pass frequency scaling factor is
kflp = 2π(800) = 1600π.
The scaled transfer function for the low-pass filter is
Hlp0 (s) = =
1
s
1600π
2
+ 0.765s
+1
1600π
s
1600π
2
65,536 × 108 π 4
.
[s2 + 1224πs + (1600π)2 ] [s2 + 2956.8πs + (1600π)2 ]
The prototype high-pass transfer function is
Hhp (s) =
+ 1.848s
+1
1600π
s4
.
(s2 + 0.765s + 1)(s2 + 1.848s + 1)
The high-pass frequency scaling factor is
kfhp = 2π(7200) = 14,400π.
Problems
15–49
The scaled transfer function for the high-pass filter is
0
Hhp
(s) = =
(s/14,400π)4
s
14,400π
2
0.765s
+ 14,400π
+1
s
14,400π
4
2
1.848s
+ 14,400π
+1
s
[s2 + 11,016πs + (14,400π)2 ][s2 + 26,611.2πs + (14,400π)2 ]
.
The transfer function for the filter is
i
h
0
H 0 (s) = Hlp0 (s) + Hhp
(s) (−10).
[b] fo =
q
fc1 fc2 =
q
(800)(7200) = 2400 Hz;
ωo = 4800π rad/s;
(jωo )2 = −2304 × 104 π 2 ;
(jωo )4 = 5,308,416 × 108 π 4 ;
Hlp0 (jωo ) =
65,536 × 108 π 4
×
[−2048 × 104 π 2 + j1224(4800π 2 )]
1
[−2048 × 104 π 2 + j2956.8(4800π 2 )]
= 0.0123/50.73◦ ;
5,308,416 × 108 π 4
0
Hhp
(jωo ) =
[18,432 × 104 π 2 + j11,016(4800π 2 )]
1
[18,432 × 104 π 2 + j26,611.2(4800π 2 )]
= 0.0123/ − 50.73◦ ;
.·. H 0 (jωo ) = 0.0123(1/50.73◦ + 1/ − 50.73◦ )(−10) = −0.1557/0◦ ;
G = 20 log10 |H 0 (jωo )| = 20 log10 (0.1557) = −16.15 dB.
P 15.57 [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.
[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
15–50
CHAPTER 15. Active Filter Circuits
or
(G1 + sC1 )Va − sC1 Vo = G1 Vi
−G2 Va + (G2 + sC2 )Vo = 0.
G2 + sC2 Vo
.·. Va =
;
G2
.·.
"
#
(G2 + sC2 )
(G1 + sC1 )
− sC1 Vo = G1 Vi ;
G2
Vo
G1 G2
.·.
=
;
Vi
(G1 + sC1 )(G2 + sC2 ) − C1 G2 s
which reduces to
Vo
bo
G1 G2 /C1 C2
= 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 =
bo
G1 G2 ·
. . 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
R2 = 1/G2 = 0.765 Ω.
In the second section bo = 1,
.·. G1 = 1.848 S.
R1 = 1/G1 = 0.541 Ω;
b1 = 1.848;
Problems
1
G2 =
(1) = 0.541 S;
1.848
R2 = 1/G2 = 1.848 Ω.
P 15.58 [a] kf = 2π(25) × 103 = 50π × 103 ;
km =
80
1012
=
× 103 .
3
50π × 10 (750)
3π
In the first section
80
1
·
(103 ) = 11.10 kΩ;
R1 =
0.765 3π
R2 = (0.765)
80
(103 ) = 6.49 kΩ.
3π
In the second section
1
80
R1 =
·
(103 ) = 4.59 kΩ;
1.848 3π
R2 = (1.848)
80
(103 ) = 15.69 kΩ.
3π
15–51
15–52
CHAPTER 15. Active Filter Circuits
[b]
P 15.59 [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 = sC1 Vi
and Va =
(G2 + sC2 )Vo
.
sC2
Thus
("
#
)
(G2 + sC2 )
(G1 + sC1 ) − G1 Vo = sC1 Vi ;
sC2
Vo {s2 C1 C2 + sC1 G2 + G1 G2 } = s2 C1 C2 Vi ;
Problems
15–53
Vo
s2
=
G1 G2
G2
Vi
2
s+
s +
C2
C1 C2
Vo
s2
=
= 2
.
Vi
s + b1 s + bo
H(s) =
[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
;
b1 =
;
[d] bo =
C1 C2
C2
.·. G2 = b1 C2 ;
G1 =
R2 =
1
.
b1 C2
b1
bo
C1 .·. R1 =
.
b1
bo 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 15.60 [a] fc = 40 kHz;
ωc = 80π krad/s;
1012
km =
= 11.05 × 103 ;
4
8π × 10 (360)
.·. R1 = R2 = km = 11 kΩ.
[b] C = 360 pF;
[c]
Ri = Rf = 11 kΩ.
.·. kf = 8π × 104 ;
15–54
CHAPTER 15. Active Filter Circuits
[d] H 0 (s) = h
=
(s/8π × 104 )3
s
8π×104
i +1
2
s
8π×104
3
s
(s + 8π × 104 )(s2 + 8π × 104 s + 64π 2 × 108 )
[e] H 0 (j8π × 104 ) =
=
−j
1
= √ /135◦ ;
j(1 + j1)
2
P 15.61 From Eq 15.19 we can write
1
−
s
R1 C
H(s) =
2
R1 + R2
s2 +
s+
R3 C
R1 R2 R3 C 2
or
R3
2
s
2R1
R3 C
.
H(s) =
2
R1 + R2
2
s +
s+
R3 C
R1 R2 R3 C 2
−
Therefore
2
ωo
=β= ;
R3 C
Q
R1 + R2
= ωo2 ;
2
R1 R2 R3 C
R3
.
2R1
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;
R1 R2 R3
Q
Q
+ R2 =
(2Q)R2 ;
K
K
.·. R2 =
Q
2Q2 − K
.
.
(j8π × 104 )3
(8π × 104 + j8π × 104 )[j(8π × 104 )(8π × 104 )]
1
GAIN = 20 log10 √ = −3.01 dB.
2
and K =
s
+ 8π×10
4 + 1
Problems
P 15.62 [a] Q = 8, K = 5, ωo = 1000 rad/s, C = 1 µF.
For the circuit in Fig 15.27
1
−
s
R1 C
!
H(s) =
R1 + R2
2
2
s+
s +
R3 C
R1 R2 R3 C 2
Kβs
= 2
.
s + βs + ωo2
R3 =
2
;
βC
β=
2
,
R3 C
β=
ωo
1000
=
= 125 rad/s;
Q
8
.·.
6
2 × 10
.·. R3 =
= 16 kΩ.
(125)(1)
Kβ =
1
;
R1 C
.·. R1 =
1
1
=
= 1.6 kΩ.
KβC
5(125)(1 × 10−6 )
ωo2 =
R1 + R2
;
R1 R2 R3 C 2
106 =
(1600 + R2 )
.
(1600)(R2 )(16,000)(10−6 )2
Solving for R2 ,
R2 =
[b]
(1600 + R2 )106
,
256 × 105
246R2 = 16,000,
R2 = 65.04 Ω.
15–55
15–56
CHAPTER 15. Active Filter Circuits
P 15.63 [a] It follows directly from Eq. 15.21 that
H(s) =
s2 + 1
.
s2 + 4(1 − σ)s + 1
Now note from Eq 15.22 that (1 − σ) equals 1/4Q, hence
H(s) =
s2 + 1
.
s2 + Q1 s + 1
[b] For Example 15.14 ωo = 5000 rad/s and Q = 5. Therefore kf = 5000 and
(s/5000)2 + 1
s2 + 25 × 106
H (s) =
.
= 2
1
s
s + 1000s + 25 × 106
(s/5000)2 +
+1
5 5000
0
P 15.64 [a] ωo = 1000 rad/s;
Q = 4;
C = 2 µF
H(s) =
s2 + ωo2
s2 + (1/R2 C 2 )
#
=
;
2 + βs + ω 2
s
4(1
−
σ)
1
o
s2 +
s+
RC
R2 C 2
"
4(1 − σ)
;
RC
ωo =
1
;
RC
R=
1
1
=
= 500 Ω;
ωo C
(1000)(2 × 10−6 )
β=
1000
ωo
=
= 250;
Q
4
.·.
β=
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.
Problems
15–57
[b]
[c] From Eq. 15.21,
H(s) =
s2 + (1/R2 C 2 )
s2 + 106
=
.
s2 + [4(1 − σ)s/RC] + (1/R2 C 2 )
s2 + 250s + 106
P 15.65 To satisfy the gain specification of 30 dB at ω = 0 and α = 1 requires
R1 + R2
= 31.6
R1
or
R2 = 30.6R1 .
Choose a standard resistor of 3.3 kΩ for R1 and a 100 kΩ potentiometer for R2 .
Since (R1 + R2 )/R1 1 the value of C1 is
C1 =
1
= 31.83 nF.
2π(50)(105 )
Use standard capacitor values of 10 nF and 22 nF in parallel to get 32 nF.
Using the selected values of R1 and R2 the maximum gain for α = 1 is
20 log10
103.3
3.3
= 29.9 dB.
α=1
When C1 = 32 nF the frequency 1/R2 C1 is
1
109
= 5
= 312.5 rad/s = 49.74 Hz.
R2 C1
10 (32)
The magnitude of the transfer function at 312.5 rad/s is
|H(j312.5)|α=1 =
|103.3 × 103 + j312.5(3.3)(100)(32)10−3 |
= 22.146.
3.3 × 103 + j312.5(3.3)(100)(32)10−3 |
Therefore the gain at 49.74 Hz is
20 log10 (22.146)α=1 = 26.9 dB.
15–58
CHAPTER 15. Active Filter Circuits
P 15.66 20 log10
R1 + R2
R1
= 15.56;
R1 + R2
.·.
= 6;
R1
Choose
.·. R2 = 5R1 .
R1 = 50 kΩ. Then
1
= 120π rad/s;
R2 C1
P 15.67 [a] |H(j0)| =
R2 = 250 kΩ;
.·. C1 =
1
= 10.61 nF.
(120π)(250 × 103 )
R1 + αR2
3.3 + α(100)
=
.
R1 + (1 − α)R2
3.3 + (1 − α)100
P 15.68 [a] Combine the impedances of the capacitors in series in Fig. P15.68(b) to
get
1−α
α
1
1
=
+
=
,
sCeq
sC1
sC1
sC1
which is identical to the impedance of the capacitor in Fig. P15.68(a).
[b]
Vx =
α/sC1
V = αV ;
(1 − α)/sC1 + α/sC1
Vy =
αR2
= αV = Vx .
(1 − α)R2 + αR2
Problems
15–59
[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. P15.68(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 :
(1−α)
· (1 − α)R2
Zi = R1 + sC1
(1 − α)R2 + (1−α)
sC1
=
R1 + (1 − α)R2 + R1 R2 C1 s
;
1 + R2 C1 s
α
· αR2
Zf = R1 + sC1
αR2 + sCα1
=
R1 + αR2 + R1 R2 C1 s
.
1 + R2 C1 s
P 15.69 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 )
When β = 1
|H(j∞)|β=1 =
Ro (R4 + R3 )
.
R3 (R4 + Ro )
If R4 Ro
Ro
|H(j∞)|β=1 ∼
> 1.
=
R3
Thus, when β = 1 we have amplification or “boost”. When β = 0
|H(j∞)|β=0 =
R3 (R4 + Ro )
.
Ro (R4 + R3 )
15–60
CHAPTER 15. Active Filter Circuits
If R4 Ro
R3
< 1.
|H(j∞)|β=0 ∼
=
R0
Thus, when β = 0 we have attenuation or “cut”.
Also note that when β = 0.5
|H(jω)|β=0.5 =
(0.5R4 + Ro )(0.5R4 + R3 )
= 1.
(0.5R4 + R3 )(0.5R4 + Ro )
Thus, the transition from amplification to attenuation occurs at β = 0.5. If
β > 0.5 we have amplification, and if β < 0.5 we have attenuation.
Also note the amplification an attenuation are symmetric about β = 0.5. i.e.
|H(jω)|β=0.6 =
1
.
|H(jω)|β=0.4
Yes, the circuit can be used as a treble volume control because
• The circuit has no effect on low frequency signals;
• Depending on β the circuit can either amplify (β > 0.5) or attenuate
(β < 0.5) signals in the treble range;
• The amplification (boost) and attenuation (cut) are symmetric around
β = 0.5. When β = 0.5 the circuit has no effect on signals in the treble
frequency range.
P 15.70 [a] |H(j∞)|β=1 =
.·.
maximum boost = 20 log10 28.3 = 29 dB.
[b] |H(j∞)|β=0 =
.·.
Ro (R4 + R3 )
(46.5)(471.5)
=
= 28.3;
R3 (R4 + Ro )
(1.5)(516.5)
1.5(271.5)
R3 (R4 + R3 )
=
= 0.0353;
Ro (R4 + Ro )
46.5(516.5)
maximum cut = 20 log10 0.0353 = −29.045 dB.
[c] R4 = 470 kΩ;
Ro = R1 + R3 + 2R2 = 45.6 kΩ;
.·. R4 = 10.1Ro .
Yes, R4 is significantly greater than Ro .
[d] |H(j/R3 C2 )|β=1 =
=
o
(2R3 + R4 ) + j R
(R4 + R3 )
R3
(2R3 + R4 ) + j(R4 + Ro )
473 + j 46.5
(471.5)
1.5
473 + j516.5
= 20.88.
20 log10 |H(j/R3 C2 )|β=1 = 20 log10 20.88 = 26.4 dB.
Problems
15–61
[e] When β = 0
|H(j/R3 C2 )|β=0 =
(2R3 + R4 ) + j(R4 + Ro )
.
Ro
(2R3 + R4 ) + j (R4 + R3 )
R3
Note this is the reciprocal of |H(j/R3 C2 )|β=1 .
.·. 20 log10 |H(j/R3 C2 )|β=0 = −26.4 dB.
[f ] The frequency 1/R3 C2 is very nearly where the gain is 3 dB off from its
maximum boost or cut. Therefore for frequencies higher than 1/R3 C2 the
circuit designer knows that gain or cut will be within 3 dB of the
maximum.
P 15.71 |H(j∞)| =
=
[(1 − β)R4 + Ro ][βR4 + R3 ]
[(1 − βR4 + R3 ][βR4 + Ro ]
[(1 − β)470 + 46.5][β470 + 1.5]
.
[(1 − β)470 + 1.5][β470 + 46.5]
Fourier Series
Assessment Problems
AP 16.1 av = 0.
ak =
2πkt
2πkt
2 Z T /2
2 Z T /4
40 cos
80 cos
dt +
dt
T 0
T
T T /4
T
2 Z 3T /4
2πkt
2ZT
2πkt
+
−40 cos
dt +
−80 cos
dt
T T /2
T
T 3T /4
T
=
2πkt T /4 160 T
2πkt T /2
80 T
sin
+
sin
T 2πk
T 0
T 2πk
T T /4
+
=−
bk =
−80 T
2πkt 3T /4 −160 T
2πkt T
sin
sin
+
T 2πk
T T /2
T 2πk
T 3T /4
πk
80
sin
,
πk
2
k odd;
ak = 0,
k even.
2 Z T /4
2πkt
2 Z T /2
2πkt
40 sin
dt +
80 sin
dt
T 0
T
T T /4
T
2πkt
2ZT
2πkt
2 Z 3T /4
−40 sin
dt +
−80 sin
dt
+
T T /2
T
T 3T /4
T
=
−80 T
2πkt T /4 160 T
2πkt T /2
cos
−
cos
T 2πk
T 0
T 2πk
T T /4
+
=
2πkt 3T /4 160 T
2πkt T
80 T
cos
+
cos
T 2πk
T T /2
T 2πk
T 3T /4
240
,
πk
k odd;
bk = 0,
16–1
k even.
16–2
CHAPTER 16. Fourier Series
AP 16.2 [a] av = 0 V.
(function is symmetric about the x-axis)
π
80
3π
80
= 8.488;
[b] a1 = − sin = −25.465; a2 = 0; a3 = − sin
π
2
3π
2
a4 = 0;
b1 =
80
5π
sin
= −5.093.
5π
2
240
= 76.394;
π
b5 =
[c] ω0 =
a5 = −
b2 = 0;
b3 =
240
= 25.465;
3π
b4 = 0;
240
= 15.279.
5π
2π
T
[d] f3 = 3f0 =
=
2π
= 31.416 rad/s.
200 × 10−3
3
= 15 Hz.
0.2
[e] v(t) = −25.465 cos 31.416t + 76.394 sin 31.416t + 8.488 cos 94.248t
+ 25.465 sin 94.248t − 5.093 cos 157.08t + 15.279 sin 157.08t + · · · V.
AP 16.3 v(t) is even and has both half- and quarter-wave symmetry, therefore av = 0,
bk = 0 for all k, ak = 0 for k-even; for k-odd we have
8 Z T /4 4Vp
ak =
t − Vp cos kω0 t dt
T 0
T
=
8Vp Z T /4
32Vp Z T /4
t
cos
kω
t
dt
−
cos kω0 t dt
0
T2 0
T 0
"
#T /4
32Vp cos kω0 t t sin kω0 t
+
=
T2
k 2 ω02
kω0
0
−
8Vp sin kω0 t T /4
T kω0
0
32Vp cos(kπ/2)
1
T 2 sin(kπ/2)
8Vp sin(kπ/2)
=
−
+
−
2
2
T2
k 2 ω0
k 2 ω0
k8π
k2π
"
#
=
−32Vp
4Vp sin(kπ/2) 4Vp sin(kπ/2)
−
+
2
T 2 k 2 ω0
kπ
kπ
=
−8Vp
.
π2k2
Therefore v(t) =
∞
−8Vp X
1
cos nω0 t V.
2
π n=1,3,5,... n2
Problems
16–3
AP 16.4 [a] From the results of Assessment Problem 16.2:
A1 = −25.465 − j76.394 = 80.53/ − 108.435◦ ;
A3 = 8.488 − j25.465 = 26.84/ − 71.57◦ ;
A2 = 0;
A4 = 0;
A5 = −5.093 − j15.279 = 16.11/ − 108.435◦ .
θ1 = −108.435◦ ;
θ2 not defined;
θ4 not defined;
θ5 = −108.435◦ .
[b] ω0
2π
T
=
θ3 = −71.57◦ ;
2π
= 31.416 rad/s.
200 × 10−3
v(t) = 80.53 cos(31.416t − 108.435◦ ) + 26.84 cos(94.284t − 71.57◦ )
+16.11 cos(157.08t − 108.435◦ ) + · · · V.
AP 16.5 vi is even and has both half- and quarter-wave symmetry, therefore av = 0,
bk = 0 for all k, ak = 0 for k-even; for odd k we have
!
kπ
4Vm
8 Z T /4
ak =
sin
.
Vm cos kω0 t dt =
T 0
πk
2
vi =
∞
4A X
nπ
1
sin
cos nω0 t;
π n=1,3,5,... n
2
ω0 =
2π
× 103 = 500 rad/s;
4π
vi = 60
∞
X
n=1,3,5,...
4A
= 60;
π
1
nπ
sin
cos 500nt V.
n
2
From the circuit
Vo =
jω
jω
Vi
· jωL =
Vi =
Vi ;
R + jωL
R/L + jω
1000 + jω
Vi1 = 60/0◦ V;
ω = 500 rad/s;
Vi3 = −20/0◦ = 20/180◦ V;
Vi5 = 12/0◦ V;
Vo1 =
3ω = 1500 rad/s;
5ω = 2500 rad/s;
j500
(60/0◦ ) = 26.83/63.43◦ V;
1000 + j500
16–4
CHAPTER 16. Fourier Series
Vo3 =
j1500
(20/180◦ ) = 16.64/ − 146.31◦ V;
1000 + j1500
Vo5 =
j2500
(12/0◦ ) = 11.14/21.80◦ V;
1000 + j2500
.·.
vo = 26.83 cos(500t + 63.43◦ ) + 16.64 cos(1500t − 146.31◦ )
+11.14 cos(2500t + 21.80◦ ) + . . . V.
AP 16.6 [a] From the solution to Assessment Problem 16.5 the Fourier series for the
input voltage is
∞
1
4A X
nπ
vg =
sin
cos nω0 t;
π n=1,3,5,... n
2
ω0 =
2π
× 103 = 2000 rad/s;
π
.·.
vg = 42
∞
X
n=1,3,5,...
1
nπ
sin
n
2
4A
= 42;
π
cos 2000nt V.
Employing the technique used in solving Assessment Problem 16.5 we
have
Vg1 = 42/0◦
ω0 = 2000 rad/s;
Vg3 = 14/180◦
3ω0 = 6000 rad/s;
Vg5 = 8.4/0◦
5ω0 = 10,000 rad/s;
Vg7 = 6/180◦
7ω0 = 14,000 rad/s.
From the circuit we have
Vo Vo − Vg
+
+ (Vo − Vg )sC = 0;
R
sL
.·.
Vo
s2 + 1/LC
= H(s) = 2
Vg
s + (s/RC) + (1/LC).
Substituting in the numerical values gives
H(s) =
s2 + 108
;
s2 + 500s + 108
H(j2000) =
96
= 0.9999/ − 0.60◦ ;
96 + j1
H(j6000) =
64
= 0.9989/ − 2.68◦ ;
64 + j3
Problems
16–5
H(j10,000) = 0;
−96
= 0.9974/4.17◦ ;
−96 + j7
H(j14,000) =
Vo1 = (42/0◦ )(0.9999/ − 0.60◦ ) = 41.998/ − 0.60◦ V;
Vo3 = (14/180◦ )(0.9989/ − 2.68◦ ) = 13.985/177.32◦ V;
Vo5 = 0 V;
Vo7 = (6/180◦ )(0.9974/4.17◦ ) = 5.984/184.17◦ V;
vo = 41.998 cos(2000t − 0.60◦ ) + 13.985 cos(6000t + 177.32◦ )
+5.984 cos(14,000t + 184.17◦ ) + . . . V.
q
[b] The 5th harmonic at the frequency 1/LC = 10,000 rad/s has been
eliminated from the output voltage by the circuit, which is a band reject
filter with a center frequency of 10,000 rad/s.
AP 16.7 [a] The voltage is even with quarter-wave symmetry.
av =
2
1 T
V
2 4 m
T
=
Vm
60
=
= 15 V;
4
4
4 Z T /4
4Vm
ak =
Vm −
t cos kωo t dt
T 0
T
#
"
kπ
4Vm
= 2 2 1 − cos
;
π k
2
bk = 0,
all k.
a1 =
240
;
π2
a2 =
120
240
(1 − cos π) = 2 ;
2
4π
π
Vrms
P =
v
"
u
2 2 #
u
1
240
120
t
= (15)2 +
+
= 24.38 V.
2
π2
(24.38)2
= 59.46 W;
10
[b] Area under v 2 ,
v 2 = 3600 −
0 ≤ t ≤ T /4 :
28,800
57,600 2
t+
t.
T
T2
π2
16–6
CHAPTER 16. Fourier Series
A=2
Z T /4 28,800
57,600 2
3600 −
t+
t dt = 600T.
T
T2
0
s
Vrms =
P =
√
√
1
600T = 600 = 24.49 V;
T
2
600 /10 = 60 W.
59.46
[c] Error =
− 1 100 = −0.9041%.
60.00
AP 16.8 Co = Av =
Vm
Vm T 1
· =
.
2
T
2
1 Z T Vm −jnωo t
Cn =
te
dt
T 0 T
#T
Vm e−jnω0 t
= 2
(−jnω0 t − 1)
T −n2 ω02
0
"
Vm e−jn2πT /T
= 2
T
−n2 ω02
"
"
#
Vm
,
2nπ
n = ±1, ±2, ±3, . . . .
Vm
1
1
= 2
(1 + jn2π) − 2 2
2
2
T n ω0
n ω0
=j
s
AP 16.9 [a] Vrms =
s
=
#
2π
1
−jn T − 1 −
(−1)
T
−n2 ω02
1ZT 2
v dt =
T 0
s
1ZT
T 0
Vm
T
2
t2 dt
Vm2 t3 T
T3 3 0
s
Vm2
Vm
=√ ;
3
3
√ 2
(120/ 3)
P =
= 480 W.
10
=
[b] From the solution to Assessment Problem 16.8
120
= 60 V;
2
C1 = j
C3 = j
120
20
=j ;
6π
π
C6 = j
10
120
=j ;
12π
π
C0 =
120
60
=j ;
2π
π
C2 = j
120
30
=j ;
4π
π
C4 = j
120
15
=j ;
8π
π
C5 = j
C7 = j
120
8.57
=j
.
14π
π
120
12
=j ;
10π
π
Problems
Vrms
16–7
v
u
∞
X
u
= tc2 + 2
|c |2
n
o
n=1
=
q
602 + (2/π 2 )(602 + 302 + 202 + 152 + 122 + 102 + 8.572 )
= 68.58 V.
[c] P =
(68.58)2
= 470.29 W;
10
% error =
470.29
− 1 (100) = −2.02%.
480
AP 16.10
1 Z T /4
Vm e−jnωo t T /4
Cn =
Vm e−jnωo t dt =
T 0
T −jnωo 0
"
#
nπ
Vm
nπ
Vm
Vm
[j(e−jnπ/2 − 1)] =
sin
+j
−1
=
cos
T nωo
2πn
2
2πn
2
Vm
nπ
nπ
sin
− j 1 − cos
2πn
2
2
=
v(t) =
∞
X
;
Cn ejnωo t .
n=−∞
1 Z T /4
Vm
Co = Av =
Vm dt =
T 0
4
or
"
Vm
sin(nπ/2)
1 − cos(nπ/2)
Co =
lim
−j
n→0
2π
n
n
"
#
(π/2) cos(nπ/2)
Vm
(π/2) sin(nπ/2)
=
lim
−j
2π n→0
1
1
#
Vm π
Vm
=
− j0 =
.
2π 2
4
Note it is much easier to use Co = Av than to use L’Hopital’s rule to find the
limit of 0/0.
16–8
CHAPTER 16. Fourier Series
Problems
P 16.1
[a] Area under v(t) for one period is zero, so av = 0.
2 Z T /2
2ZT
Vm cos kω0 t dt +
−Vm cos kω0 t dt
T 0
T T /2
ak =
=
T /2
T
2Vm
2Vm
sin kω0 t
sin kω0 t
−
kT ω0
kT ω0
0
T /2
=
Vm
Vm
(sin kπ − sin 0) −
(sin kπ − sin 2kπ) = 0.
kπ
kπ
2 Z T /2
2ZT
Vm sin kω0 t dt +
−Vm sin kω0 t dt
T 0
T T /2
bk =
=
T /2
T
2Vm
2Vm
(− cos kω0 t)
cos kω0 t
+
kT ω0
kT ω0
0
T /2
=
Vm
−2Vm
−Vm
(cos kπ − cos 0) +
(cos 2kπ − cos kπ) =
(cos kπ − 1).
kπ
kπ
kπ
bk = 0,
k even;
Thus, v(t) =
bk =
4Vm
,
kπ
k odd.
∞
4Vm X
1
sin nω0 t V.
π n=1,3,5,... n
1 Z T /2
π
−Vm
π T
[b] av =
Vm sin t dt =
cos t
T 0
T
π
T 0
=
ak =
2Vm
−Vm
(−1 − 1) =
.
π
π
2ZT
π
Vm sin t cos kω0 t dt
T 0
T
!
T
2Vm
=
T
− cos(π − 2kπ)t/T
cos(π + 2kπ)t/T
−
2(π − 2kπ)/T
2(π + 2kπ)/T
Vm
=
π
1
1
+
(− cos(π − k2π) + cos 0 − cos(π + k2π) + cos 0)
1 − 2k 1 + 2k
Vm
π
1
4Vm /π
(− cos π(1 − 2k) + cos 0 − cos π(1 + 2k) + cos 0) =
.
2
1 − 4k
1 − 4k 2
=
0
Problems
16–9
2ZT
π
bk =
Vm sin t sin kω0 t dt
T 0
T
sin(π − 2kπ)t/T
sin(π + 2kπ)t/T
−
2(π − 2kπ)/T
2(π + 2kπ)/T
2Vm
=
T
=
Vm
π
!
T
0
1
1
+
(sin(π − k2π) − sin 0 − sin(π + k2π) + sin 0) = 0.
1 − 2k 1 + 2k
∞
X
1
2Vm
Thus, v(t) =
1+2
cos nω0 t V.
2
π
n=1 1 − 4n
"
#
Vm
2π
1 Z T /2
t dt =
;
Vm sin
T 0
T
π
[c] av =
2 Z T /2
Vm
2π
ak =
Vm sin t cos kω0 t dt =
T 0
T
π
Note: ak = 0 for k-odd, ak =
!
1 + cos kπ
;
1 − k2
2Vm
π(1 − k 2 )
for k even.
2 Z T /2
2π
bk =
Vm sin t sin kω0 t dt = 0 for k = 2, 3, 4, . . . .
T 0
T
For k = 1, we have b1 =
v(t) =
P 16.2
Vm
;
2
therefore
∞
1
Vm Vm
2Vm X
cos nω0 t V.
+
sin ω0 t +
π
2
π n=2,4,6,... 1 − n2
The area under the periodic function in Fig. P16.2 for one period is zero, so
av = 0.
ak =
I1 =
I2 =
I3 =
I4 =
2
(I1 + I2 + I3 + I4 + I5 + I6 ).
T
Z T /6
0
Z T /3
T /6
Z T /2
T /3
100 cos kω0 t dt =
50 cos kω0 t dt =
T /2
T /3
50
50
sin(2πk/T )t
=
(sin(2πk/3) − sin(2πk/6));
kπ
kπ
T /6
100 cos kω0 t dt =
Z 2T /3
T /6
100
100
sin(2πk/T )t
(sin(2πk/6) − sin 0);
=
kπ
kπ
0
T /2
100
100
=
sin(2πk/T )t
(sin(kπ) − sin(2πk/3));
kπ
kπ
T /3
− 100 cos kω0 t dt =
2T /3
−100
−100
sin(2πk/T )t
=
(sin(4πk/3) − sin(kπ));
kπ
kπ
T /2
16–10
CHAPTER 16. Fourier Series
Z 5T /6
I5 =
2T /3
Z T
I6 =
5T /6
ak =
− 50 cos kω0 t dt =
5T /6
−50
−50
sin(2πk/T )t
(sin(5πk/3) − sin(4πk/3));
=
kπ
kπ
2T /3
− 100 cos kω0 t dt =
T
−100
−100
sin(2πk/T )t
(sin(2πk) − sin(5πk/3)).
=
kπ
kπ
5T /6
2
(I1 + I2 + I3 + I4 + I5 + I6 )
T
2
=
T
50
50
50
50
sin(kπ/3) −
sin(2kπ/3) −
sin(4kπ/3) +
sin(5kπ/3)
kπ
kπ
kπ
kπ
= 0.
bk =
I1 =
I2 =
I3 =
I4 =
I5 =
I6 =
bk =
2
(I1 + I2 + I3 + I4 + I5 + I6 ).
T
Z T /6
0
Z T /3
T /6
Z T /2
T /3
100 sin kω0 t dt =
50 sin kω0 t dt =
T /3
−50
−50
cos(2πk/T )t
=
(cos(2πk/3) − cos(2πk/6));
kπ
kπ
T /6
100 sin kω0 t dt =
Z 2T /3
T /2
Z 5T /6
2T /3
Z T
5T /6
T /6
−100
−100
cos(2πk/T )t
=
(cos(2πk/6) − cos 0);
kπ
kπ
0
T /2
−100
−100
cos(2πk/T )t
=
(cos(kπ) − cos(2πk/3));
kπ
kπ
T /3
− 100 sin kω0 t dt =
− 50 sin kω0 t dt =
2T /3
100
100
cos(2πk/T )t
=
(cos(4πk/3) − cos(kπ));
kπ
kπ
T /2
5T /6
50
50
cos(2πk/T )t
=
(cos(5πk/3) − cos(4πk/3));
kπ
kπ
2T /3
− 100 sin kω0 t dt =
T
100
100
cos(2πk/T )t
=
(cos(2πk) − cos(5πk/3)).
kπ
kπ
5T /6
2
(I1 + I2 + I3 + I4 + I5 + I6 )
T
2
=
T
+
100 50
50
200
−
cos(kπ/3) +
cos(2kπ/3) −
cos(kπ)
kπ
kπ
kπ
kπ
2
T
50
50
100
cos(4kπ/3) −
sin(5kπ/3) +
.
kπ
kπ
kπ
"
#
200
πk
2 − cos
V,
bk =
πk
3
bk = 0 V,
k even.
k odd;
Problems
P 16.3
16–11
In studying the periodic function in Fig. P16.3 note that it can be visualized
as the combination of two half-wave rectified sine waves, as shown in the figure
below. Hence we can use the Fourier series for a half-wave rectified sine wave
which is given as the answer to Problem 16.1(c).
In using the previously derived Fourier series for the half-wave rectified sine
wave we note v1 (t) has been shifted T /4 units to the left and v2 (t) has been
shifted T /4 units to the right. Thus,
v1 (t) =
∞
200
400 X
cos nωo (t + T /4)
+ 100 sin ωo (t + T /4) −
V.
π
π n=2,4,6,...
(n2 − 1)
Now observe the following:
sin ωo (t + T /4) = sin(ωo t + π/2) = cos ωo t;
cos nωo (t + T /4) = cos(nωo t + nπ/2) = cos
nπ
cos nωo t;
2
because n is even.
∞
X
200
400
cos(nπ/2) cos(nωo t)
.·. v1 (t) =
+ 100 cos ωo t −
V;
π
π n=2,4,6,...
(n2 − 1)
∞
X
100
200
cos nωo (t − T /4)
.·. v2 (t) =
+ 50 sin ωo (t − T /4) −
V.
π
π n=2,4,6,...
(n2 − 1)
16–12
CHAPTER 16. Fourier Series
Again, observe the following:
sin(ωo t − π/2) = − cos ωo t;
cos(nωo t − nπ/2) = cos(nπ/2) cos nωo t;
because n is even.
∞
X
100
200
cos(nπ/2) cos(nωo t)
.·. v2 (t) =
− 50 cos ωo t −
V.
π
π n=2,4,6,...
(n2 − 1)
Thus:
v = v1 + v2
∞
X
300
600
cos(nπ/2) cos(nωo t)
.·. v(t) =
+ 50 cos ωo t −
V.
π
π n=2,4,6,...
(n2 − 1)
P 16.4
2π
2π
2π
2π
=
= 31.416 rad/s;
ω0b =
=
= 785.4 rad/s.
T
0.2
T
0.008
1
1
1
1
[b] f0a = =
= 5 Hz;
f0b = =
= 125 Hz.
T
0.2
T
0.008
1 Z 2T /3
1ZT
7
Vm
[c] ava =
Vm dt +
dt = Vm = 7π V;
T 0
T 2T /3 3
9
[a] ω0a =
avb =
2(50 × 1 + 25 × 1)
= 18.75 V.
8
[d] The periodic function in Fig. P16.4(a):
"
#
Z T
2 Z 2T /3
Vm
aka =
Vm cos kω0 t dt +
cos kω0 t dt
T
3
0
2T /3
4Vm
=
3kω0 T
4kπ
sin
3
!
4kπ
6
sin
=
k
3
!
V;
"
#
Z T
2 Z 2T /3
Vm
bkb =
Vm sin kω0 t dt +
sin kω0 t dt
T
3
0
2T /3
4Vm
=
3kω0 T
"
4kπ
1 − cos
3
!#
"
6
=
k
4kπ
1 − cos
3
The periodic function in Fig. P16.4(b):
2
akb =
T
(Z
−T /8
2
+
T
−T /4
25 cos kωo t dt +
(Z
T /8
0
)
Z 0
50 cos kωo t dt +
−T /8
50 cos kωo t dt
Z T /4
T /8
)
25 cos kωo t dt
!#
V.
Problems
4
=
T
(Z
T /8
4
=
T
(
0
50 cos kωo t dt +
Z T /4
T /8
)
25 cos kωo t dt
T /8
T /4
50
25
sin kωo t
sin kωo t
+
kωo
kωo
0
T /8
(
50
kπ
kπ
kπ
=
2 sin
+ sin
− sin
kπ
4
2
4
(
50
kπ
kπ
=
sin
+ sin
kπ
4
2
2
bkb =
T
(Z
−T /8
−T /4
2
+
T
2
=
T
0
(Z
T /4
2
+
T
T /8
V.
0
)
Z 0
50 sin kωo t dt +
−T /8
50 sin kωo t dt +
50 sin kωo t dt
Z T /4
T /8
− 25 sin kωo t dt +
(Z
T /8
)
)
25 sin kωo t dt +
(Z
T /8
)
)
25 sin kωo t dt
Z T /8
0
Z T /4
T /8
)
− 50 sin kωo t dt
)
25 sin kωo t dt
= 0.
[e] For the periodic function in Fig. P16.4(a),
v(t) = 7π +
∞ X
6
1
4nπ
4nπ
6
sin
cos nωo t +
1 − cos
sin nωo t V.
n
3
n
3
For the periodic function in Fig. P16.4(b),
∞
1
nπ
nπ
50 X
sin
+ sin
cos nωo t V
v(t) = 18.75 +
π n=1 n
4
2
16–13
16–14
P 16.5
CHAPTER 16. Fourier Series
[a] I1 =
Z to +T
to
sin mω0 t dt = −
to +T
1
cos mω0 t
mω0
to
=
−1
[cos mω0 (to + T ) − cos mω0 to ]
mω0
=
−1
[cos mω0 to cos mω0 T − sin mω0 to sin mω0 T − cos mω0 to ]
mω0
=
−1
[cos mω0 to − 0 − cos mω0 to ] = 0 for all m;
mω0
I2 =
Z to +T
to
cos mω0 to dt =
to +T
1
[sin mω0 t]
mω0
to
=
1
[sin mω0 (to + T ) − sin mω0 to ]
mω0
=
1
[sin mω0 to − sin mω0 to ] = 0 for all m.
mω0
Z to +T
1 Z to +T
[b] I3 =
cos mω0 t sin nω0 t dt =
[sin(m + n)ω0 t − sin(m − n)ω0 t] dt.
2 to
to
But (m + n) and (m − n) are integers, therefore from I1 above, I3 = 0 for
all m, n.
Z to +T
1 Z to +T
[c] I4 =
sin mω0 t sin nω0 t dt =
[cos(m − n)ω0 t − cos(m + n)ω0 t] dt.
2 to
to
If m 6= n, both integrals are zero (I2 above). If m = n, we get
I4 =
[d] I5 =
1 Z to +T
T
T
1 Z to +T
dt −
cos 2mω0 t dt = − 0 = .
2 to
2 to
2
2
Z to +T
to
cos mω0 t cos nω0 t dt
1 Z to +T
=
[cos(m − n)ω0 t + cos(m + n)ω0 t] dt.
2 to
If m 6= n, both integrals are zero (I2 above). If m = n, we have
I5 =
P 16.6
1 Z to +T
1 Z to +T
T
T
dt +
cos 2mω0 t dt = + 0 = .
2 to
2 to
2
2
f (t) sin kω0 t = av sin kω0 t +
∞
X
an cos nω0 t sin kω0 t +
n=1
∞
X
bn sin nω0 t sin kω0 t.
n=1
Now integrate both sides from to to to + T. All the integrals on the right-hand
side reduce to zero except in the last summation when n = k, therefore we
have
Z to +T
to
f (t) sin kω0 t dt = 0 + 0 + bk
T
2
or bk =
2 Z to +T
f (t) sin kω0 t dt.
T to
Problems
P 16.7
(Z
0
1 Z to +T
1
av =
f (t) dt =
T to
T
Let t = −x,
f (t) dt +
−T /2
dt = −dx,
x=
Z T /2
16–15
)
f (t) dt ;
0
T
2
when t =
−T
2
and x = 0 when t = 0.
Therefore
1Z0
1Z0
1 Z T /2
f (t) dt =
f (−x)(−dx) = −
f (x) dx.
T −T /2
T T /2
T 0
Therefore av = −
1 Z T /2
1 Z T /2
f (t) dt +
f (t) dt = 0;
T 0
T 0
2Z0
2 Z T /2
ak =
f (t) cos kω0 t dt +
f (t) cos kω0 t dt.
T −T /2
T 0
Again, let t = −x in the first integral and we get
2 Z T /2
2Z0
f (t) cos kω0 t dt = −
f (x) cos kω0 x dx.
T −T /2
T 0
Therefore ak = 0
bk =
for all k.
2Z0
2 Z T /2
f (t) sin kω0 t dt +
f (t) sin kω0 t dt.
T −T /2
T 0
Using the substitution t = −x, the first integral becomes
2 Z T /2
f (x) sin kω0 x dx;
T 0
Therefore we have bk =
P 16.8
4 Z T /2
f (t) sin kω0 t dt.
T 0
2Z0
2 Z T /2
bk =
f (t) sin kω0 t dt +
f (t) sin kω0 t dt.
T −T /2
T 0
Now let t = x − T /2 in the first integral, then dt = dx, x = 0 when t = −T /2
and x = T /2 when t = 0, also
sin kω0 (x − T /2) = sin(kω0 x − kπ) = sin kω0 x cos kπ. Therefore
2Z0
2 Z T /2
f (t) sin kω0 t dt = −
f (x) sin kω0 x cos kπ dx and
T −T /2
T 0
bk =
Z T /2
2
(1 − cos kπ)
f (x) sin kω0 t dt.
T
0
Now note that 1 − cos kπ = 0 when k is even, and 1 − cos kπ = 2 when k is
odd. Therefore bk = 0 when k is even, and
4 Z T /2
bk =
f (t) sin kω0 t dt when k is odd.
T 0
16–16
P 16.9
CHAPTER 16. Fourier Series
Because the function is even and has half-wave symmetry, we have av = 0,
ak = 0 for k even, bk = 0 for all k and
ak =
4 Z T /2
f (t) cos kω0 t dt,
T 0
k odd.
The function also has quarter-wave symmetry;
therefore f (t) = −f (T /2 − t) in the interval T /4 ≤ t ≤ T /2;
thus we write
ak =
4 Z T /4
4 Z T /2
f (t) cos kω0 t dt +
f (t) cos kω0 t dt.
T 0
T T /4
Now let t = (T /2 − x) in the second integral, then dt = −dx, x = T /4 when
t = T /4 and x = 0 when t = T /2. Therefore we get
4 Z T /4
4 Z T /2
f (t) cos kω0 t dt = −
f (x) cos kπ cos kω0 x dx.
T T /4
T 0
Therefore we have
Z T /4
4
ak = (1 − cos kπ)
f (t) cos kω0 t dt.
T
0
But k is odd, hence
ak =
8 Z T /4
f (t) cos kω0 t dt,
T 0
k odd.
P 16.10 Because the function is odd and has half-wave symmetry, av = 0, ak = 0 for all
k, and bk = 0 for k even. For k odd we have
4 Z T /2
bk =
f (t) sin kω0 t dt.
T 0
The function also has quarter-wave symmetry, therefore f (t) = f (T /2 − t) in
the interval T /4 ≤ t ≤ T /2. Thus we have
bk =
4 Z T /4
4 Z T /2
f (t) sin kω0 t dt +
f (t) sin kω0 t dt.
T 0
T T /4
Now let t = (T /2 − x) in the second integral and note that dt = −dx, x = T /4
when t = T /4 and x = 0 when t = T /2, thus
Z T /4
4 Z T /2
4
f (t) sin kω0 t dt = − cos kπ
f (x)(sin kω0 x) dx.
T T /4
T
0
But k is odd, therefore the expression for bk becomes
8 Z T /4
bk =
f (t) sin kω0 t dt.
T 0
Problems
16–17
P 16.11 [a] v(t) is even and has both half- and quarter-wave symmetry, therefore
av = 0, bk = 0 for all k, ak = 0 for k-even; for odd k we have
!
kπ
8 Z T /4
4Vm
ak =
Vm cos kω0 t dt =
sin
.
T 0
πk
2
∞
4Vm X
1
nπ
v(t) =
sin
cos nω0 t V.
π n=1,3,5,... n
2
[b] Odd function with both half- and quarter-wave symmetry.
vg (t) =
6Vm
t,
T
0 ≤ t ≤ T /6;
av = 0,
ak = 0 for all k;
bk = 0 for k even;
bk =
8 Z T /4
f (t) sin kω0 t dt,
T 0
k odd
8 Z T /4
8 Z T /6 6Vm
=
t sin kω0 t dt +
Vm sin kω0 t dt
T 0
T
T T /6
=
!
kπ
12Vm
sin
;
2
2
k π
3
∞
12Vm X
1
nπ
v(t) =
sin
sin nω0 t V.
π 2 n=1,3,5,... n2
3
P 16.12 [a]
[b] Even.
[c] Yes.
[d] av = 0;
ak = 0,
bk = 0
for all k; the function is even.
k even, half-wave symmetry;
16–18
CHAPTER 16. Fourier Series
8 Z T /4
ak =
0.4t2 cos kωo t dt
T 0
=
3.2 Z T /4 2
t cos kωo t dt
T 0
3.2
=
T
(
T /4
2t
k 2 ωo2 t2 − 2
cos
kω
t
+
sin
kω
t
.
o
o
k 2 ωo2
k 3 ωo3
0
)
First term is 0 at both T /4 and 0; second term is 0 at 0, hence
3.2
ak = 3 3
k ωo T
=
(
k 2 ωo2 T 2 − 32
kπ
sin
16
2
)
kπ
T2
(k 2 4π 2 − 32) sin
.
3
3
5k (8π )
2
T 2 = 400;
40
kπ
.·. ak = 3 3 (k 2 π 2 − 8) sin
;
π k
2
∞
40 X
n2 π 2 − 8
nπ
f (t) = 3
sin
cos nωo t.
3
π n=1,3,5,...
n
2
!
[e] cos nωo (t − T /4) = cos(nωo t − nπ/2)
= sin(nπ/2) sin nωo t
since n is odd;
∞
40 X
n2 π 2 − 8
.·. f (t) = 3
sin nωo t.
π n=1,3,5,...
n3
!
P 16.13 [a]
[b] Odd.
[c] Yes.
[d] av = 0;
bk = 0
ak = 0
for all k since the function is odd.
for k even, since the function has half-wave symmetry;
Problems
8 Z T /4
bk =
f (t) sin kωo t dt,
T 0
=
16–19
k odd
3.2 Z T /4 3
t sin kωo t dt
T 0
T /4
T /4
3.2 3k 2 ωo2 t2 − 6
t(6 − k 2 ωo2 t2 )
=
sin
kω
t
cos
kω
t
+
.
o
o
T
k 4 ωo4
k 3 ωo3
0
0
"
#
Note that the first term is zero at the lower limit and the second term is
zero at both limits because
cos kωo T /4 = cos kπ/2,
k odd.
Thus
(
bk =
kπ
(3k 2 ωo2 T 2 /16) − 6
sin
4
4
k ωo
2
)
3.2
.
T
ωo T = 2π;
3.2
bk =
T
=
(
kπ
12(k 2 π 2 − 8)T 4
sin
4
4
256k π
2
)
3(k 2 π 2 − 8)T 3
kπ
sin
;
4
4
20k π
2
T = 20 s;
bk =
kπ
1200(k 2 π 2 − 8)
sin
;
k4π4
2
∞
1200 X
n2 π 2 − 8
nπ
f (t) = 4
sin
sin nωo t.
4
π n=1,3,5,...
n
2
!
[e] sin nωo (t − T /4) = sin(nωo t − nπ/2)
= − cos nωo t sin nπ/2
(n is odd);
∞
1200 X
n2 π 2 − 8
f (t) = − 4
cos nωo t.
π n=1,3,5,...
n4
!
P 16.14 [a] f =
[b] No.
[c] Yes.
[d] Yes.
[e] Yes.
1
1
=
= 100 Hz.
T
0.01
16–20
CHAPTER 16. Fourier Series
[f ] av = 0,
function is odd;
ak = 0,
for all k; the function is odd;
bk = 0,
for k even, the function has half-wave symmetry;
8 Z T /4
f (t) sin kωo t,
bk =
T 0
8
=
T
=
(Z
T /8
0
k odd
4000t sin kωo t dt +
Z T /4
T /8
)
5 sin kωo t dt
8
{Int1 + Int2};
T
Z T /8
Int1 = 4000
0
t sin kωo t dt
"
T /8
1
t
= 4000 2 2 sin kωo t −
cos kωo t
k ωo
kωo
0
=
#
4000
kπ 500T
kπ
sin
−
cos
;
2
2
k ωo
4
kωo
4
Z T /4
Int2 = 5
T /8
sin kωo t dt =
T /4
−5
5
kπ
cos kωo t
=
cos
;
kωo
kωo
4
T /8
kπ
4000
5
500T
Int1 + Int2 = 2 2 sin
+
−
k ωo
4
kωo
kωo
cos
kπ
.
4
500T = (500)(10 × 10−3 ) = 5;
.·.
Int1 + Int2 =
kπ
4000
sin
.
2
2
k ωo
4
bk =
kπ
80
kπ
8 4000
· 2 2 · T 2 sin
= 2 2 sin
,
T 4π k
4
π k
4
i(t) =
∞
80 X
sin(nπ/4)
sin nωo t A.
2
π n=1,3,5,...
n2
P 16.15 [a] ωo =
[b] No.
[c] Yes.
[d] No.
2π
π
= rad/s.
T
6
k odd;
Problems
16–21
P 16.16 [a] i(t) is even, therefore bk = 0 for all k.
1 T
1
Im
· · Im · 2 · =
A;
2 4
T
4
av =
4 Z T /4
4Im
t cos kωo t dt
ak =
Im −
T 0
T
=
4Im Z T /4
16Im Z T /4
cos kωo t dt −
t cos kωo t dt
T
T2 0
0
= Int1 − Int2 .
kπ
4Im Z T /4
2Im
sin
;
Int1 =
cos kωo t dt =
T 0
πk
2
Int2 =
16Im Z T /4
t cos kωo t dt
T2 0
16Im
=
T2
(
T /4
1
t
cos
kω
t
+
sin
kω
t
o
o
k 2 ωo2
kωo
0
)
!
4Im
kπ
2Im
kπ
= 2 2 cos
−1 +
sin
;
π k
2
kπ
2
kπ
4Im
.·. ak = 2 2 1 − cos
π k
2
!
A.
∞
X
1 − cos(nπ/2)
Im 4Im
cos nωo t A.
.·. i(t) =
+ 2
4
π n=1
n2
[b] Shifting the reference axis to the left is equivalent to shifting the periodic
function to the right:
cos nωo (t − T /2) = cos nπ cos nωo t.
Thus
i(t) =
∞
Im 4Im X
(1 − cos(nπ/2)) cos nπ
+ 2
cos nωo t A.
4
π n=1
n2
16–22
CHAPTER 16. Fourier Series
P 16.17 [a]
[b] av = 0;
ak = 0 for all k even;
For k odd,
ak =
8 Z T /4
f (t) cos kωo t dt;
T 0
8 Z T /8
8t
8 Z T /4
16t
1+
cos kωo t dt +
4−
cos kωo t dt
T 0
T
T T /8
T
ak =
bk = 0 for all k;
= Int1 + Int2 .
64 Z T /8
8 Z T /8
Int1 =
cos kωo t dt + 2
t cos kωo t dt
T 0
T 0
#T /8
"
8 sin kωo t T /8 64 cos kωo t
t
=
+
;
sin kωo t
+ 2
2
2
T kωo
T
k ωo
kωo
0
0
kωo T = 2kπ;
(kωo T )2 = 4k 2 π 2 ;
"
8
kπ
16
kπ
Int1 =
sin
+ 2 2 cos
kπ
4
k π
4
Int2 =
!
#
−1
k odd.
32 Z T /4
128 Z T /4
cos kωo t dt − 2
t cos kωo t dt
T T /8
T T /8
#T /4
"
32 sin kωo t T /4 128 cos kωo t
t
=
− 2
+
sin kωo t
;
2
2
T kωo
T
k ωo
kωo
T /8
T /8
Int2 =
kπ
32
kπ
−8
sin
+ 2 2 cos
kπ
4
k π
4
ak = Int1 + Int2
#
"
16
kπ
= 2 2 3 cos
−1 .
k π
4
k odd.
Problems
[c] a1 =
16–23
48
π 16
cos − 2 = 1.8178;
2
π
4 π
a3 =
48
3π
16
cos
− 2 = −0.5622;
2
9π
4
9π
a5 =
48
5π
16
cos
−
= −0.2024;
25π 2
4
25π 2
f (t) = 1.8178 cos ωo t − 0.5622 cos 3ωo t − 0.2024 cos 5ωo t − · · · .
[d] f (T /8) = 1.8178 cos(π/4) − 0.5622 cos(3π/4) − 0.2024 cos(5π/4) = 1.8261.
P 16.18 Let f (t) = v2 (t − T /6).
av = −(2Vm /3)(T /3)(1/T ) = −(2Vm /9) and bk = 0 since f (t) is even.
T /6
4 2Vm 1
2Vm
4 Z T /6
−
cos kωo tdt = −
sin kωo t
T 0
3
T 3 kωo
0
ak =
=−
π
8Vm
sin k
3k2π
3
=−
4Vm
π
sin k .
3kπ
3
∞
2Vm 4Vm X
1
nπ
v2 (t − T /6) = −
cos nωo t,
−
sin
9
3π n=1 n
3
Therefore,
∞
2Vm 4Vm X
1
nπ
v2 (t) = −
−
sin
cos nωo (t + T /6).
9
3π n=1 n
3
and
Then, v(t) = v1 (t) + v2 (t) = Vm + v2 (t). Simplifying,
∞
1
7Vm 4Vm X
nπ
nπ
−
sin
cos
9
3π n=1 n
3
3
v(t) =
∞
4Vm X
1
nπ
+
sin2
3π n=1 n
3
If
Vm = 9π
ak = −
bk =
then
12
nπ
sin2
n
3
cos nωo t
sin nωo t V.
av = 7π = 21.99 (Checks).
12
nπ
nπ
sin
cos
n
3
3
=
a1 = 6 sin(4π/3) = −5.2;
12
n
1
2
=−
12
n
1
2nπ
sin
2
3
1 − cos
2nπ
3
6
n
=
b1 = 6[1 − cos(4π/3)] = 9;
6
4nπ
sin
;
n
3
=
1 − cos
4nπ
3
.
16–24
CHAPTER 16. Fourier Series
a2 = 3 sin(8π/3) = 2.6;
b2 = 3[1 − cos(8π/3)] = 4.5;
a3 = 2 sin(12π/3) = 0;
b3 = 2[1 − cos(12π/3)] = 0;
a4 = 1.5 sin(16π/3) = −1.3;
b4 = 1.5[1 − cos(16π/3)] = 2.25;
a5 = 1.2 sin(20π/3) = 1.04;
b5 = 1.2[1 − cos(20π/3)] = 1.8.
All coefficients check!
P 16.19 [a] The voltage has half-wave symmetry. Therefore,
av = 0;
ak = bk = 0,
k even.
4 Z T /2
2Vm
t cos kωo t dt,
Vm −
T 0
T
k odd.
4 Z T /2
2Vm
bk =
Vm −
t sin kωo t dt,
T 0
T
k odd.
ak =
4Vm Z T /2
8Vm Z T /2
t cos kωo t dt
ak =
cos kωo t dt − 2
T
T
0
0
= Int1 − Int2 .
T /2
4Vm Z T /2
4Vm 1
Int1 =
cos kωo t dt =
·
sin kωo t
= 0;
T
T
kωo
0
0
#T /2
"
8Vm cos kωo t t sin kωo t
+
Int2 = 2
T
k 2 ωo2
kωo
0
"
#
8Vm
1
= 2
(cos kπ − 1)
2
T
k ωo2
=
−4Vm
−16Vm
=
,
k 2 (4π 2 )
π2k2
4Vm
.·. ak = 2 2 ,
π k
bk =
k odd.
k odd.
4Vm Z T /2
8Vm Z T /2
sin kωo t dt − 2
t sin kωo t dt
T
T
0
0
= Int1 − Int2 .
Problems
Int1 =
=
T /2
4Vm −1
4Vm Z T /2
cos kωo t
sin kωo t dt =
·
T
T
kωo
0
0
−4Vm
8Vm
4Vm
[cos kπ − 1] =
=
;
T kωo
kωo T
πk
8Vm Z T /2
Int2 = 2
t sin kωo t dt
T
0
#T /2
"
8Vm sin kωo t t cos kωo t
= 2
−
T
k 2 ωo2
kωo
0
2Vm
8Vm
T
cos kπ − 0 − 0 =
= 2 0−
;
T
2kωo
kπ
4Vm 2Vm
2Vm
−
=
;
.·. bk =
πk
πk
πk
2Vm
.·. Ak / − θk = ak − jbk =
πk
2
− j1 .
πk
Vm = 378π V
Ak / − θ k =
v(t) =
756 2
− j1 V
k πk
∞
X
An cos(nωo t − θn ).
n=1,3,5,...
A1 / − θ1 = 896.20/ − 57.52◦ V;
A3 / − θ3 = 257.61/ − 78.02◦ V;
A5 / − θ5 = 152.42/ − 82.74◦ V;
A7 / − θ7 = 108.45/ − 84.80◦ V;
A9 / − θ9 = 84.21/ − 85.95◦ V.
v(t) = 896.20 cos(ωo t − 57.52◦ ) + 257.61 cos(3ωo t − 78.02◦ )
+152.42 cos(5ωo t − 82.74◦ ) + 108.45 cos(7ωo t − 84.80◦ )
+84.21 cos(9ωo t − 85.95◦ ) + . . . V.
[b] v(T /8) = 896.20 cos(45 − 57.52◦ ) + 257.61 cos(135 − 78.02◦ )
+152.42 cos(225 − 82.74◦ ) + 108.45 cos(315 − 84.80◦ )
+84.21 cos(405 − 85.95◦ ) = 888.92 V;
2(378π) T
v(T /8) = 378π −
T
8
1
= 378π(1 − ) = 890.64 V.
4
16–25
16–26
CHAPTER 16. Fourier Series
The % difference based on the exact value is
888.92 − 890.64
(100) = −0.19%.
890.64
P 16.20 From Problem 16.3,
v(t) =
∞
300
600 X
cos(nπ/2) cos(nωo t)
+ 50 cos ωo t −
V.
π
π n=2,4,6,...
(n2 − 1)
Therefore,
av =
300
V;
π
an =
−600 cos(nπ/2)
π(n2 − 1)
bn = 0
a1 = 50;
for n even;
for all n.
Therefore,
A1 = 50
and
θ1 = 0◦ ;
and
An =
−600 cos(nπ/2)
π(n2 − 1)
Thus, v(t) =
and θn = 0◦
for all even n.
∞
300
600 X
cos(nπ/2) cos(nωo t)
+ 50 cos ωo t −
V.
π
π n=2,4,6,...
(n2 − 1)
P 16.21 The periodic function in Problem 16.12 is even, so bk = 0 for all k. Thus,
An / − θn = an − jbn = an = an /0◦ .
From Problem 16.12,
av = 0 = A0 ;
an =
40
nπ
.
(n2 π 2 − 8) sin
3
3
π n
2
Therefore,
An =
40
nπ
(n2 π 2 − 8) sin
,
3
3
π n
2
and
− θn = 0◦ .
∞
n2 π 2 − 8
nπ
40 X
Thus, f (t) = 3
sin
cos nωo t.
3
π n=1,3,5,...
n
2
!
Problems
16–27
P 16.22 The periodic function in Problem 16.14 is odd, so av = 0 and ak = 0 for all k.
Thus,
An / − θn = an − jbn = 0 − jbn = bn / − 90◦ .
From Problem 16.14,
bk =
80
kπ
sin
,
π2k2
4
k odd.
Therefore,
An =
nπ
80
sin
,
2
2
π n
4
n odd,
and
−θn = −90◦ ,
n odd.
∞
sin(nπ/4)
80 X
Thus, i(t) = 2
cos(nωo t − 90◦ ) A.
2
π n=1,3,5,...
n
P 16.23 The function has half-wave symmetry, thus ak = bk = 0 for k-even, av = 0; for
k-odd
ak =
4 Z T /2
8Vm Z T /2 −t/RC
Vm cos kω0 t dt −
e
cos kω0 t dt
T 0
ρT 0
h
i
where ρ = 1 + e−T /2RC .
Upon integrating we get
ak =
4Vm sin kω0 t T /2
T
kω0
0
8Vm
− cos kω0 t
e−t/RC
−
·
+ kω0 sin kω0 t
·
2
2
ρT
(1/RC) + (kω0 )
RC
(
=
#
"
T /2
0
−8Vm RC
T [1 + (kω0 RC)2 ].
4 Z T /2
8Vm Z T /2 −t/RC
bk =
Vm sin kω0 t dt −
e
sin kω0 t dt
T 0
ρT 0
=−
4Vm cos kω0 t T /2
T
kω0
0
8Vm
−e−t/RC
sin kω0 t
−
·
·
+ kω0 cos kω0 t
2
2
ρT
(1/RC) + (kω0 )
RC
(
=
4Vm
8kω0 Vm R2 C 2
−
πk
T [1 + (kω0 RC)2 ].
"
#
T /2
0
)
)
16–28
CHAPTER 16. Fourier Series
P 16.24 [a] a2k + b2k = a2k +
2
4Vm
+ kω0 RCak
πk
= a2k [1 + (kω0 RC)2 ] + 8Vπkm
(
h
2Vm
+ kω0 RCak
πk
i
.
But ak =
−8Vm RC
}
T [1 + (kω0 RC)2 ]
Therefore
a2k =
a2k + b2k =
64Vm2 R2 C 2
16Vm2
64Vm2 kω0 R2 C 2
+
−
.
T 2 [1 + (kω0 RC)2 ]
π2k2
πkT [1 + (kω0 RC)2 ]
(
64Vm2 R2 C 2
,
T 2 [1 + (kω0 RC)2 ]2
)
thus we have
Now let α = kω0 RC and note that T = 2π/ω0 , thus the expression for
a2k + b2k reduces to a2k + b2k = 16Vm2 /π 2 k 2 (1 + α2 ). It follows that
q
a2k + b2k =
4Vm
q
πk 1 + (kω0 RC)2
[b] bk = kω0 RCak +
Thus
.
4Vm
.
πk
bk
4Vm
1 + α2
1
= kω0 RC +
=α−
=− .
ak
πkak
α
α
Therefore
ak
= −α = −kω0 RC.
bk
P 16.25 Since av = 0 (half-wave symmetry), Eq. 16.20 gives us
vo (t) =
∞
X
4Vm
1
q
cos(nω0 t − θn ) where
1 + (nω0 RC)2
1,3,5,... nπ
tan θn =
bn
.
an
But tan βk = kω0 RC. It follows from Eq. 16.29 that tan βk = −ak /bk or
tan θn = − cot βn . Therefore θn = 90◦ + βn and
cos(nω0 t − θn ) = cos(nω0 t − βn − 90◦ ) = sin(nω0 t − βn ), thus our expression
for vo becomes
vo =
∞
4Vm X
sin(nω0 t − βn )
q
.
π n=1,3,5,... n 1 + (nω0 RC)2
P 16.26 [a] e−x ∼
= 1 − x for small x;
e−t/RC ∼
= 1−
t
RC
therefore
and e−T /2RC ∼
= 1−
T
.
2RC
Problems
Vm
2Vm [1 − (t/RC)]
=
vo ∼
= Vm −
2 − (T /2RC)
RC
∼
=
Vm
RC
t−
T
4
=
−8
−8
Vp =
[b] ak =
2
2
π k
π2k2
"
2t − (T /2)
2 − (T /2RC)
Vm
Vm T
t−
RC
4RC
Vm T
4RC
=
16–29
#
for 0 ≤ t ≤
T
.
2
−4Vm
.
πω0 RCk 2
P 16.27 [a] Express vg as a constant plus a symmetrical square wave. The constant is
Vm /2 and the square wave has an amplitude of Vm /2, is odd, and has
half- and quarter-wave symmetry. Therefore the Fourier series for vg is
∞
Vm 2Vm X
1
vg =
+
sin nω0 t.
2
π n=1,3,5,... n
The dc component of the current is Vm /2R and the kth harmonic phase
current is
2Vm /kπ
2Vm
q
/ − θk
Ik =
=
R + jkω0 L
kπ R2 + (kω0 L)2
!
−1
where θk = tan
kω0 L
.
R
Thus the Fourier series for the steady-state current is
i=
∞
Vm 2Vm X
sin(nω0 t − θn )
q
A.
+
2R
π n=1,3,5,... n R2 + (nω0 L)2
[b]
The steady-state current will alternate between I1 and I2 in exponential
traces as shown. Assuming t = 0 at the instant i increases toward
(Vm /R), we have
Vm
Vm −t/τ
+ I1 −
e
R
R
i=
for 0 ≤ t ≤
T
2
16–30
CHAPTER 16. Fourier Series
and i = I2 e−[t−(T /2)]/τ for T /2 ≤ t ≤ T, where τ = L/R. Now we solve for
I1 and I2 by noting that
Vm
Vm −T /2τ
I1 = I2 e
and I2 =
+ I1 −
e
.
R
R
These two equations are now solved for I1 . Letting x = T /2τ, we get
−T /2τ
(Vm /R)e−x
.
I1 =
1 + e−x
Therefore the equations for i become
"
#
Vm
Vm
i=
−
e−t/τ
R
R(1 + e−x )
for 0 ≤ t ≤
#
"
Vm
i=
e−[t−(T /2)]/τ
R(1 + e−x )
for
T
2
and
T
≤ t ≤ T.
2
A check on the validity of these expressions shows they yield an average
value of (Vm /2R):
1
Iavg =
T
1
=
T
=
(Z
T /2 0
Vm
2R
Vm
Vm T
+ τ (1 − e−x ) I1 −
+ I2
2R
R
since I1 + I2 =
Vm
.
R
P 16.28 The Fourier series for the input voltage is
∞
8Vm X
nπ
1
sin
sin nω0 (t + T /4)
π 2 n=1,3,5,... n2
2
vi =
∞
8Vm X
1
2 nπ
= 2
sin
cos nω0 t
π n=1,3,5,... n2
2
∞
8Vm X
1
= 2
cos nω0 t;
π n=1,3,5,... n2
8Vm
8(281.25π 2 )
=
= 2250 mV;
π2
π2
ω0 =
.·.
2π
2π
=
× 103 = 10;
T
200π
vi = 2250
∞
X
)
Z T
Vm −t/τ
Vm
dt +
I2 e−[t−(T /2)]/τ dt
+ I1 −
e
R
R
T /2
1
cos 10nt mV.
2
n=1,3,5,... n
Problems
From the circuit we have
Vo =
Vi
1
Vi
·
=
;
R + (1/jωC) jωC
1 + jωRC
Vo =
1/RC
100
Vi =
Vi ;
1/RC + jω
100 + jω
Vi1 = 2250/0◦ mV;
ω0 = 10 rad/s;
Vi3 =
2250 ◦
/0 = 250/0◦ mV;
9
3ω0 = 30 rad/s;
Vi5 =
2250 ◦
/0 = 90/0◦ mV;
25
5ω0 = 50 rad/s;
Vo1 =
100
(2250/0◦ ) = 2238.83/ − 5.71◦ mV;
100 + j10
Vo3 =
100
(250/0◦ ) = 239.46/ − 16.70◦ mV;
100 + j30
Vo5 =
100
(90/0◦ ) = 80.50/ − 26.57◦ mV;
100 + j50
.·.
vo = 2238.33 cos(10t − 5.71◦ ) + 239.46 cos(30t − 16.70◦ )
+80.50 cos(50t − 26.57◦ ) + . . . mV.
P 16.29 [a] ωo =
2π
= 240π rad/s.
T
∞
2(54π) 4(54π) X
cos n(240π)t
f (t) =
−
π
π n=1 4n2 − 1
= 108 − 216
∞
X
cos n(240π)t
n=1
4n2 − 1
.
vg1 =
−216
cos 240πt = −72 cos 240πt;
3
vg2 =
−216
cos 480πt = −14.4 cos 480πt;
15
vg3 =
−216
cos 720πt = −6.2 cos 720πt;
35
16–31
16–32
CHAPTER 16. Fourier Series
vg4 =
−216
cos 960πt = −3.4 cos 960πt.
63
Vg1 = 72/180◦ V;
Vg2 = 14.4/180◦ V;
Vg3 = (216/35)/180◦ V;
Vg4 = (216/63)/180◦ V.
H(s) =
Vo
(1/LC)
= 2
;
Vg
s + (1/RC)s + (1/LC)
1
106
=
= 4 × 104 ;
LC
25
H(s) =
1
106
=
= 80;
RC
(5000)(2.5)
4 × 104
.
s2 + 80s + 4 × 104
H(jω) =
4 × 104
;
4 × 104 − ω 2 + j80ω
H(j0) = 1/0◦ ;
H1 (j240π) =
4 × 104
4 × 104 − 5.76π 2 × 104 + j1.92 × 104 π
= 0.0752/ − 173.49◦ ;
H2 (j480π) =
4 × 104
4 × 104 − 23.04π 2 × 104 + j3.84 × 104 π
= 0.0179/ − 176.91◦ ;
H3 (j720π) =
4 × 104
4 × 104 − 51.84π 2 × 104 + j5.76 × 104 π
= 0.0079/ − 177.96◦ ;
H4 (j960π) =
4 × 104
4 × 104 − 92.16π 2 × 104 + j7.68 × 104 π
= 0.0044/ − 178.47◦ .
Vo1 = 72/180◦ H1 = 5.41/6.51◦ V;
Vo2 = 14.4/180◦ H2 = 0.2575/3.09◦ V;
Problems
16–33
Vo3 = (216/35)/180◦ H3 = 0.0486/2.04◦ V;
Vo4 = (216/63)/180◦ H4 = 0.0151/1.53◦ V.
Vodc = (108)(1) = 108 V;
vo = 108 + 5.41 cos(240πt + 6.51◦ ) + 0.2575 cos(480πt + 3.09◦ )
+ 0.0486 cos(720πt + 2.04◦ ) + 0.0151 cos(960πt + 1.53◦ ) + · · · V.
[b] The circuit is a low pass filter. Hence, the harmonic terms are greatly
reduced in the output voltage.
P 16.30 [a] ωo =
2π
2π
=
(103 ) = 104 rad/s;
T
0.2π
vg (t) = 840
∞
X
nπ
1
sin
cos n10,000t V
2
n=1,3,5,... n
= 840 cos 10,000t − 280 cos 30,000t + 168 cos 50,000t
− 120 cos 70,000t + · · · V;
Vg1 = 840/0◦ V;
Vg3 = 280/180◦ V;
Vg5 = 168/0◦ V;
Vg7 = 120/180◦ V;
H(s) =
β=
Vo
βs
= 2
;
Vg
s + βs + ωc2
1
109
= 4
= 5000 rad/s;
RC
10 (20)
ωc2 =
(109 )(103 )
1
=
= 25 × 108 ;
LC
400
H(s) =
5000s
s2 + 5000s + 25 × 108
H(jω) =
;
j5000ω
;
25 × 108 − ω 2 + j5000ω
H1 =
j5 × 107
= 0.02/88.81◦ ;
24 × 108 + j5 × 107
H3 =
j15 × 107
= 0.09/84.64◦ ;
16 × 108 + j15 × 107
H5 =
j25 × 107
= 1/0◦ ;
25 × 107
16–34
CHAPTER 16. Fourier Series
H7 =
j35 × 107
= 0.14/ − 81.70◦ ;
−24 × 108 + j35 × 107
Vo1 = Vg1 H1 = 17.50/88.81◦ V;
Vo3 = Vg3 H3 = 26.14/ − 95.36◦ V;
Vo5 = Vg5 H5 = 168/0◦ V;
Vo7 = Vg7 H7 = 17.32/98.30◦ V;
vo = 17.50 cos(10,000t + 88.81◦ ) + 26.14 cos(30,000t − 95.36◦ )
+ 168 cos(50,000t) + 17.32 cos(70,000t + 98.30◦ ) + · · · V.
[b] The 5th harmonic because the circuit is a passive bandpass filter with a Q
of 10 and a center frequency of 50 krad/s.
P 16.31 ωo =
2π
× 103 = 20 krad/s;
0.1π
300
= 15th harmonic.
.·. n =
20
(π 2 (15)2 − 8)
π
sin 15
3
15
2
Vg15 = 45
= −29.5 V = 29.5/180◦ V;
H(s) =
=
(1/RC)s
s2 + (1/RC)s + (1/LC)
104 s
.
s2 + 104 s + 9 × 1010
H(j300,000) = 1/0◦ ;
Vo15 = (29.5/180◦ )(1/0◦ ) = 29.5/180◦ V;
vo15 = 29.5 cos(300,000t + 180◦ ) V.
Problems
P 16.32 H(s) =
(1/LC)
Io
,
=
R
1
1
2
Ig
s2 + R1 C + L2 s + R1R+R
LC
1
where R1 = 800 Ω and R2 = 200 Ω. Thus
H(s) =
ωo =
20 × 108
.
s2 + 60,000s + 25 × 108
2π
2π
=
(103 ) = 10 krad/s;
T
0.2π
H(j50,000) =
ig (t) =
5ωo = 50 krad/s.
20 × 108
2
= −j ;
j(60,000)(50,000)
3
∞
1
8A X
sin(nπ/2) sin nωo t;
2
π n=1,3,5,... n2
2
8(30π ) 1
.·. ig5 (t) =
· (1) sin 50,000t
π2
25
= 9.6 sin 50,000t A = 9.6 cos(50,000t − 90◦ ) A;
Ig5 = 9.6/ − 90◦ ;
2
H(j50,000) = / − 90◦ ;
3
Io5 = (9.6)(2/3)/ − 180◦ = 6.4/ − 180◦ A;
.·. io5 = 6.4 cos(50,000t − 180◦ ) = −6.4 cos(50,000t) A.
P 16.33 w0 =
2π × 103
= 3 rad/s.
2094.4
jω0 k = j3k;
VR =
2
2sVg
(Vg ) = 2
;
2 + s + 1/s
s + 2s + 1
16–35
16–36
CHAPTER 16. Fourier Series
VR
Vg
H(s) =
!
=
2s
s2 + 2s + 1
H(jω0 k) = H(j3k) =
j6k
;
(1 − 9k 2 ) + j6k
vg1 = 25.98 sin ω0 t V;
H(j3) =
;
Vg1 = 25.98/0◦ V;
j6
= 0.6/ − 53.13◦ ;
−8 + j6
VR1 = 15.589/ − 53.13◦ V;
√
(15.589/ 2)2
P1 =
= 60.75 W;
2
vg3 = 0,
therefore P3 = 0 W;
vg5 = −1.04 sin 5ω0 t V;
H(j15) =
Vg5 = 1.04/180◦ ;
j30
= 0.1327/ − 82.37◦ ;
−224 + j30
VR5 = (1.04/180◦ )(0.1327/ − 82.37◦ ) = 138/97.63◦ mV;
√
(0.138/ 2)2
= 4.76 mW;
P5 =
2
P 16.34 vg (t) = 25 −
ωo =
therefore P ∼
= P1 + P3 + P5 ∼
= 60.8 W.
∞
200 X
1
cos(2n − 1)ωo t V;
2
π n=1,3,5,... (2n − 1)2
2π
2π
=
× 106 = 1 Mrad/s;
T
2π
vg (t) = 25 −
H(s) =
200
200
200
cos
ω
t
−
cos
3ω
t
−
cos 5ωo t − · · · V.
o
o
π2
9π 2
25π 2
(1/LC)
;
s2 + (1/RC)s + (1/LC)
1
(103 )(1012 )
=
= 1012 ;
LC
(20)(50)
H(s) =
1012
.
s2 + 106 s + 1012
1
1012
=
= 106 ;
RC
(20 × 103 )(50)
Problems
H(jω) =
16–37
1012
;
1012 − ω 2 + j106 ω
H(j0) = 1;
H(jωo ) = −j1;
H(j3ωo ) =
1
= 0.1170/ − 159.44◦ ;
−8 + j3
H(j5ωo ) =
1
= 0.0408/ − 168.23◦ .
−24 + j5
.·. vo = 25 − 20.26 cos(ωo t − 90◦ ) − 0.2635 cos(3ωo t − 159.44◦ )
− 0.0331 cos(5ωo t − 168.23◦ ) − · · · V.
Now note that the harmonic terms will have a negligible effect on the rms
value of vo , hence a good estimate of the power delivered to the 20 kΩ resistor
can be obtained by assuming vo ≈ 25 − 20.26 cos(ωo t − 90◦ ) V.
.·. Vo(rms) ≈
v
u
u
t
252 +
−20.26
√
2
!2
= 28.82 V;
2
(28.82)
= 41.52 mW.
.·. P ≈
20 × 103
P 16.35 [a] From Example 16.1
1
av = (270π) = 135π V;
2
ak = 0,
bk =
all k;
−270π
−270
=
πk
k
all k;
.·. v(t) = 135π − 270 sin ωo t − 135 sin 2ωo t − 90 sin 3ωo t − · · · .
Vrms =
P =
v
u
u
t
(135π)2 +
270
√
2
!2
(479.05)2
= 287.06 W.
81π 2
135
+ √
2
!2
90
+ √
2
!2
= 479.05;
16–38
CHAPTER 16. Fourier Series
270π
[b] Vrms = √ = 489.73 V;
3
2
(489.72)
= 300 W.
.·. P =
81π 2
287.06
[c] % error =
− 1 (100) = −4.31%.
300
P 16.36 [a] vg has is odd and has both half-wave and quarter-wave symmetry.
.·. av = 0, ak = 0 all k, bk = 0 k-even;
bk =
8 Z T /4
f (t) sin kωo t dt,
T 0
8
=
T
(Z
T /8
0
k-odd
Vm sin kωo t dt +
"
Z T /4
)
"
#
#
Vm
sin kωo t dt
T /8 2
8Vm
cos kωo t T /8
8Vm
cos kωo t T /4
=
−
+
−
T
kωo
2T
kωo
0
T /8
"
#
"
#
8Vm
kπ
8Vm
kπ
=
1 − cos
+
cos
−0
kωo T
4
2T kωo
4
8Vm
=
kωo T
(
kπ 1
kπ
1 − cos
+ cos
4
2
4
(
)
)
4Vm
kπ
=
1 − 0.5 cos
.
πk
4
4Vm
π
1 − 0.5 cos
π
4
b1 =
= 0.8231Vm ;
4Vm
3π
b3 =
1 − 0.5 cos
3π
4
5π
4Vm
b5 =
1 − 0.5 cos
5π
4
4Vm
7π
1 − 0.5 cos
7π
4
b7 =
= 0.5745Vm ;
= 0.3447Vm ;
= 0.1176Vm .
s
(0.8231)2 + (0.5745)2 + (0.3447)2 + (0.1176)2
;
2
Vg(rms) ≈ 0.7550Vm .
Vg(rms) ≈ Vm
"
[b] Area = 2
2Vm2
s
Vg(rms) =
T
8
Vm2
+
4
T
4
s
#
5
= Vm2 T ;
8
1 5Vm2
5
T = Vm
= 0.7906 Vm .
T 8
8
Problems
16–39
0.7550Vm
− 1 100;
[c] % Error =
0.7906Vm
Error = −4.5%.
P 16.37 [a] Use the Fourier series constructed in Problem 16.1(c):
120
80
+ 60 sin ωo t −
cos 2ωo t + · · · mA.
π
π
i(t) ≈
Irms ≈
[b] Irms =
v
u
u 120 2
t
π
602
80
+
+ √
2
2π
!2
= 59.86 mA.
120
= 60 mA;
2
% error =
59.86
− 1 (100) = −0.23%.
60
[c] Use the Fourier series constructed in Problem 16.1(b):
i(t) ≈
240 480 1
1
−
cos ωo t +
cos 2ωo t + · · · mA;
π
π 3
15
v

u
u 240 2 480 2
u

Irms ≈ t
+
π
π
1
√
3 2
!2
1
√
+
15 2
!2 
 = 84.76 mA;
120
Irms = √ = 84.85 mA.
2
% error = −0.11%.
P 16.38 [a] Half-wave symmetry av = 0, ak = bk = 0, even k:
ak =
−16Im Z T /4
4 Z T /4 −4Im
t cos kω0 t dt =
t cos kω0 t dt
T 0
T
T2
0
−16Im
=
T2
(
T /4
cos kω0 t
t
+
sin
kω
t
0
k 2 ω02
kω0
0
(
)
)
−16Im
T
kπ
1
− 2 2 ;
=
0+
sin
2
T
4kω0
2
k ω0
"
−2Im
kπ
ak =
sin
πk
2
bk =
!
#
2
−
,
πk
odd k.
4 Z T /4 −4Im
−16Im Z T /4
t sin kω0 t dt =
t sin kω0 t dt
T 0
T
T2
0
−16Im
=
T2
(
)
!
T /4
sin kω0 t
t
−4Im
kπ
−
cos
kω
t
= 2 2 sin
,
0
2
2
k ω0
kω0
π k
2
0
odd k.
16–40
CHAPTER 16. Fourier Series
−2Im
[b] ak − jbk =
πk
("
kπ
sin
2
−2Im
a1 − jb1 =
π
a3 − jb3 =
−2Im
3π
a5 − jb5 =
−2Im
5π
−2Im
a7 − jb7 =
7π
!
#
2
2
1−
−j
π
π
−1 −
1−
"
2
2
kπ
−
− j
sin
πk
πk
2
2
2
−j
5π
5π
2
2
−1 −
+j
7π
7π
;
= 0.47Im /119.72◦ ;
2
2
+j
3π
3π
!#)
= 0.26Im / − 9.93◦ ;
= 0.11Im /171.70◦ ;
= 0.10Im / − 4.77◦ .
ig = 0.47Im cos(ω0 t + 119.72◦ ) + 0.26Im cos(3ω0 t − 9.93◦ )
+0.11Im cos(5ω0 t + 171.70◦ ) + 0.10Im cos(7ω0 t − 4.77◦ ) + · · · .
[c]
v
u
u
Ig = t
∞
X
n=1,3,5,...
A2n
2
!
s
(0.47)2 + (0.26)2 + (0.11)2 + (0.10)2
∼
= 0.3927Im .
= Im
2
!
!
Z T /4 2
2
4Im 2
32Im
t3 T /4 Im
T
[d] Area = 2
t dt =
;
=
T
T2
3 0
6
0
Ig
v
u
u1
=t
T
2T
Im
6
!
Im
= √ = 0.41Im .
6
!
!
[e] % error =
0.3927Im
√ − 1 100 = −3.8%.
(Im / 6)
estimated
− 1 100 =
exact
P 16.39 [a] v = 80 + 200 cos(500t + 45◦ ) + 60 cos(1500t − 90◦ ) V;
i = 10 + 6 cos(500t − 15◦ ) + 3 cos(1500t + 30◦ ) A;
1
1
P = (80)(10) + (200)(6) cos(60◦ ) + (60)(3) cos(−120◦ ) = 1055 W.
2
2
[b] Vrms =
v
u
u
t
[c] Irms =
v
u
u
t
(80)2 +
200
√
2
6
(10)2 + √
2
!2
!2
60
+ √
2
3
+ √
2
!2
= 167.93 V.
!2
= 11.07 A.
Problems
P 16.40 [a]
Area under i2 :
A =4
"Z
T /8 0
Z T /4
16 2
8 2
4 − t dt
1 + t dt +
T
T
T /8
T
T
T
4T
T
=4
+ +
+ 2T − 4T + T +
−
8
8
24
3
6
=
44T
.
24
s
Irms =
#
1
T
44T
24
s
=
44
= 1.35.
24
2
[b] P = Irms
(54) = 99 W.
[c] From Problem 16.17:
a1 = 1.8178A;
ig ≈ 1.8178 cos ωo t A;
P =
1.8178
√
2
!2
(54) = 89.22 W.
89.22
− 1 = −9.88%.
[d] % error =
99
2
P 16.41 [a] Area under v = A = 4
Z T /4
0
=
Therefore Vrms
v
u
u1
=t
T
16Vp2 2
t dt
T2
64Vp2 t3 T /4 Vp2 T
.
=
3T 2 0
3
Vp2 T
3
!
Vp
= √ = 28.8675 V.
3
16–41
16–42
CHAPTER 16. Fourier Series
[b] From Assessment Problem 16.3,
vg = −40.53 cos ω0 t − 4.5 cos 3ω0 t − 1.62 cos 5ω0 t + · · · V.
s
Therefore Vrms ∼
=
P 16.42 [a] v(t) =
(−40.53)2 + (−4.5)2 + (−1.62)2
= 28.8571 V.
2
∞
640 X
1
sin(nπ/2) sin nωo t V;
2
π n=1,3,5,... n2
640
1
1
1
v(t) ≈ 2 sin 200πt − sin 600πt +
sin 1000πt −
sin 1400πt ;
π
9
25
49
vrms ≈
v
u
640 u
t
1
√
2
π2
640
≈ 2
π
s
!2
1
+ √
9 2
!2
1
√
+
25 2
!2
1
√
+
49 2
!2
1
1
1
1
+
+
+
≈ 46.1808 V;
2 162 1250 4802
80
[b] Vrms = √ = 46.1880 V.
3
46.1808
− 1 100 = −0.0156%.
% Error =
46.1880
[c] v(t) =
∞
1
320 X
sin nω0 t V;
π n=1,3,5,... n
v(t) ≈
320
1
1
1
sin 200πt + sin 600πt + sin 1000πt + sin 1400πt ;
π
3
5
7
v
u
320 u
t √1
vrms ≈
π
2
320
≈
π
s
!2
1
+ √
3 2
!2
1
+ √
5 2
!2
1
1
1
1
+
+
+
≈ 77.9578 V;
2 18 50 98
Vrms = 80 V.
% Error =
77.9578
− 1 100 = −2.55%.
80
P 16.43 Figure P16.43(b): ta = tb = 5 s.
v = 5t,
A1 =
Z 5
0
0 ≤ t ≤ 5;
25t2 dt =
1
+ √
7 2
3125
;
3
!2
Problems
s
Vrms =
1 3125
25
= √ V.
5
3
3
Figure P16.43(c): ta = 1 s;
tb = 3 s.
0 ≤ t ≤ 1;
v = 25t,
v = −25t + 50,
1 ≤ t ≤ 3;
v = 12.5t − 62.5,
Area 1 = A1 =
Area 2 = A2 =
Area 3 = A3 =
3 ≤ t ≤ 5.
Z 1
625t2 dt =
0
Z 3
625(t2 − 4t + 4) dt =
1
Z 5
3
3125
.
3
s
1 3125
25
= √ V.
5
3
3
Figure P16.43(d): ta = tb = 2 s;
v(t) = 12.5t,
v(t) =
A1 =
A2 =
0 ≤ t ≤ 2;
25
(t − 5),
3
Z 2
2 ≤ t ≤ 5.
156.25t2 dt =
0
1250
;
3
Z 5
625 2
625
(t − 10t + 25) dt =
;
9
3
2
A1 + A2 =
s
Vrms =
3125
.
3
1 3125
25
= √ V.
5
3
3
1250
;
3
156.25(t2 − 10t + 25) dt =
A1 + A2 + A3 =
Vrms =
625
;
3
1250
3
16–43
16–44
CHAPTER 16. Fourier Series
P 16.44 Odd function with half- and quarter-wave symmetry, therefore av = 0, ak = 0
for all k, bk = 0 for k even; for k odd we have
bk =
8 Z T /8
8 Z T /4
2 sin kω0 t dt +
8 sin kω0 t dt
T 0
T T /8
=
8
πk
"
!#
kπ
1 + 3 cos
4
,
−j4
Therefore Cn =
nπ
s
k odd.
nπ
1 + 3 cos
4
,
n odd.
2
T
3T
T
(2)2
(2) + (8)2
−
T
8
8
8
−j12.5
j1.5
j0.9
[b] C1 =
; C3 =
; C5 =
;
π
π
π
P 16.45 [a] Irms =
C7 =
−j1.8
;
π
C9 =
v
u
u
Irms = tI 2 + 2
−j1.4
;
π
∞
X
dc
C11 =
s
|Cn |2 ∼
=
n=1,3,5,...
=
√
34 = 5.831 A.
j0.4
;
π
2
(12.52 + 1.52 + 0.92 + 1.82 + 1.42 + 0.42 )
π2
∼
= 5.768 A.
5.768 − 5.831
× 100 = −1.08%.
5.831
[d] Using just the terms C1 – C9 ,
[c] % Error =
v
u
u
Irms = tI 2 + 2
∞
X
dc
s
|Cn |2 ∼
=
n=1,3,5,...
2
(12.52 + 1.52 + 0.92 + 1.82 + 1.42 )
π2
∼
= 5.765 A;
% Error =
5.765 − 5.831
× 100 = −1.13%.
5.831
Thus, the % error is still less than 1.5%.
P 16.46 T = 32 ms, therefore 8 ms requires shifting the function T /4 to the right.
∞
X
4
nπ jnω0 (t−T /4)
i=
−j
1 + 3 cos
e
nπ
4
n=−∞
n(odd)
∞
4 X
1
n
nπ jnω0 t
=−
sin π
1 + 3 cos
e
.
π n=−∞ n
2
4
n(odd)
Problems
1 Z T /2 2Vm −jnωo t
P 16.47 [a] Cn =
te
dt
T −T /2 T
T /2
2Vm e−jnωo t
= 2
(−jnω
t
−
1)
o
T
−n2 ωo2
−T /2
"
#
=
2Vm −jnπ
[e
(jnπ + 1) − ejnπ (−jnπ + 1)]
4π 2 n2
=
−jVm
[sin nπ − nπ cos nπ].
π 2 n2
sin nπ = 0
Cn =
for all n;
jVm
Vm
cos nπ.
nπ cos nπ = j
2
2
π n
nπ
[b] Co = 0;
[c]
C−1 = j72;
C1 = −j72;
C−2 = −j36;
C2 = j36;
C−3 = j24;
C3 = −j24;
C−4 = −j18;
C4 = j18.
Vo
Vo − Vg
Vo
+ Vo sC +
+
= 0;
R2
sL
R1
(1/R1 C)s
Vo
=
.·. H(s) =
R
VG
s2 + 1 +R2 s + (1/LC)
R1 R2 C
3200s
= 2
.
s + 4000s + 16 × 108
H(jnωo ) =
ωo =
j3200nωo
;
16 × 108 − n2 ωo2 + j4000nωo
2π
× 106 = 40,000 rad/s;
50π
.·. H(jnωo ) =
j1.28n
.
16(1 − n2 ) + j1.6n
H−1 = 0.8/0◦ ;
H1 = 0.8/0◦ ;
H−2 = 0.0532/86.19◦ ;
H2 = 0.0532/ − 86.19◦ ;
H−3 = 0.0300/87.85◦ ;
H3 = 0.0300/ − 87.85◦ ;
H−4 = 0.0213/88.47◦ ;
H4 = 0.0213/ − 88.47◦ .
16–45
16–46
CHAPTER 16. Fourier Series
Co = 0;
C−1 = (72/90◦ )(0.8/0◦ ) = 57.60/90◦ ;
C1 = 57.60/ − 90◦ ;
C−2 = (36/ − 90◦ )(0.0532/86.19◦ ) = 1.92/ − 3.81◦ ;
C2 = 1.92/3.81◦ ;
C−3 = (24/90◦ )(0.0300/87.85◦ ) = 0.72/177.85◦ ;
C3 = 0.72/ − 177.85◦ ;
C−4 = (18/ − 90◦ )(0.0213/88.47◦ ) = 0.38/ − 1.53◦ ;
C4 = 0.38/1.53◦ .
[d] Vo(rms)
v
u 4
u X
≈ t2
|C |2
n
n=1
=
P =
2(57.62 + 1.922 + 0.722 + 0.382 ) = 81.51 V;
(81.51)2
× 10−3 = 33.22 mW.
200
s
P 16.48 [a] Vrms =
[b] Vrms
q
2 Z T /2 4Vm2 2
Vm
72π
√
√ = 130.59 V.
t
dt
=
=
T 0
T2
3
3
v
u 4
q
u X
|c |2 = 2(722 + 362 + 242 + 182 ) = 121.49 V.
≈ t2
n
n=1
121.49
[c] % error =
− 1 (100) = −6.97%.
130.59
P 16.49 [a] From Example 16.3 we have:
40
= 10 V,
av =
4
"
!
kπ
40
ak =
sin
;
πk
2
40
kπ
bk =
1 − cos
πk
2
A1 = 18.01 V
A3 = 6 V,
A6 = 4.24 V,
!#
,
θ1 = 45◦ ,
θ3 = 135◦ ,
θ6 = 90◦ ,
Ak / − θk◦ = ak − jbk .
A2 = 12.73 V,
A4 = 0,
θ2 = 90◦ ,
A5 = 3.6 V,
A7 = 2.57 V,
θ5 = 45◦ ,
θ7 = 135◦ ;
A8 = 0.
Problems
[b] Cn =
an − jbn
,
2
C−n =
an + jbn
= Cn∗ .
2
C0 = av = 10 V;
C3 = 3/135◦ V;
C1 = 9/45◦ V;
C−3 = 3/ − 135◦ V; C−6 = 2.12/ − 90◦ V;
C−1 = 9/ − 45◦ V;
C4 = C−4 = 0
C7 = 1.29/135◦ V;
C2 = 6.37/90◦ V;
C5 = 1.8/45◦ V;
C−7 = 1.29/ − 135◦ V;
C6 = 2.12/90◦ V;
C−2 = 6.37/ − 90◦ V; C−5 = 1.8/ − 45◦ V.
P 16.50 [a] From the solution to Problem 16.35 we have
av = 135π V;
bk =
−270
k
ak = 0,
all k;
Ak / − θk = ak − jbk = j
.·. θk = −90,
all k;
270
270 ◦
/90 ;
=
k
k
all k.
A1 / − θ1 = 270/90◦ ;
A2 / − θ2 = 135/90◦ ;
16–47
16–48
CHAPTER 16. Fourier Series
A3 / − θ3 = 90/90◦ ;
A4 / − θ4 = 67.5/90◦ ;
A5 / − θ5 = 54/90◦ ;
A6 / − θ6 = 45/90◦ .
135
1
= Cn /θn .
[b] Cn = (an − jbn ) = j
2
n
135
1
;
C−n = (an + jbn ) = −j
2
n
C1 = 135/90◦ ;
C−1 = 135/ − 90◦ ;
C2 = 67.5/90◦ ;
C3 = 45/90◦ ;
C4 = 33.75/90◦ ;
C5 = 27/90◦ ;
C6 = 22.5/90◦ ;
Co = av = 424.12
C−2 = 67.5/ − 90◦ ;
C−3 = 45/ − 90◦ ;
C−4 = 33.75/ − 90◦ ;
C−5 = 27/ − 90◦ ;
C−6 = 22.5/ − 90◦ .
Problems
16–49
P 16.51 [a] v = A0 + A2 cos(2ωo t + 180◦ ) + A4 cos(4ωo t + 0◦ )
+A6 cos(6ωo t + 180◦ );
v = A0 − A2 cos 2ωo t + A4 cos 4ωo t − A6 sin 6ωo t.
[b] v(−t) = A0 − A2 cos 2ωo t + A4 cos 4ωo t − A6 cos 6ωo t;
.·. v(−t) = v(t);
even function.
[c] v(t − T /2) = A0 − A2 cos(2ωo t − 2π) + A4 cos(4ωo t − 4π)
−A6 cos(6ωo t − 6π)
= A0 − A2 cos 2ωo t + A4 cos 4ωo t − A6 cos 6ωo t;
.·. v(t − T /2) = v(t), no, the function does not have half-wave symmetry.
[d] Since the function does not have half-wave symmetry, it also does not
have quarter-wave symmetry.
P 16.52 [a] i = 441 cos(1000t − 90◦ ) + 49 cos(3000t + 90◦ ) + 17.64 cos(5000t − 90◦ )
+ 9 cos(7000t + 90◦ ) mA
= 441 sin 1000t − 49 sin 3000t + 17.64 sin 5000t − 9 sin 7000t mA.
[b] i(t) = −i(−t)
odd.
[c] Yes
An = 0 for n even.
Ao = 0,
s
4412 + 492 + 17.642 + 92
= 314.07 mA.
2
[e] C−1 = 220.50/90◦ ;
C1 = 220.50/ − 90◦ ;
[d] Irms =
C−3 = 24.50/ − 90◦ ;
C3 = 24.50/90◦ ;
C−5 = 8.82/90◦ ;
C5 = 8.82/ − 90◦ ;
C−7 = 4.50/ − 90◦ ;
C7 = 4.50/90◦ .
i = −j4.5e−j7000t + j8.82e−j5000t − j24.5e−j3000t + j220.5e−j1000t
− j220.50ej1000t + j24.5ej3000t − j8.82ej5000t + j4.5ej7000t mA.
16–50
CHAPTER 16. Fourier Series
[f ]
2(2.5π) 4(2.5π) cos 5000t
−
= 5 − (10/3) cos 5000t − · · · V;
π
π
4−1
P 16.53 vg =
H(j0) = 1;
H(j5000) =
106
√
= 0.04/ − 163.58◦ ;
6
6
6
(10 − 25 × 10 ) + j5 2 × 10
.·. vo (t) = 5 − 0.1332 cos(5000t − 163.58◦ ) − · · · V.
P 16.54 ωo =
2π
2π
=
= 0.5 rad/s;
T
4π


∞
8(π 2 /8)  X
1
vg =
cos nωo t
2
2
π
n
n=1,3,5,...
= cos 0.5t +
1
1
cos 1.5t +
cos 2.5t + · · · V.
9
25
The transfer function of the prototype third-order high pass Butterworth filter
is
H(s) =
s3
√
,
(s + 1)(s2 + 2s + 1)
where ωc = 1 rad/s.
H(j0.5) =
(j0.5)3
√
= 0.108/ − 159.88◦ ;
2
(1 + j0.5)[(j0.5) + 2j0.5 + 1]
H(j1.5) =
(j1.5)3
√
= 0.76/93.18◦ ;
2
(1 + j1.5)[(j1.5) + 2j1.5 + 1]
Problems
H(j2.5) =
16–51
(j2.5)3
√
= 0.92/55.76◦ .
(1 + j2.5)[(j2.5)2 + 2j2.5 + 1]
Vg1 = 1/0◦ V;
1
Vg3 = /0◦ V;
9
Vg5 =
1 ◦
/0 V.
25
Vo1 = (1)(0.108/ − 159.88◦ ) = 0.108/ − 159.88◦ V;
Vo3 = (1/9)(0.76/93.18◦ ) = 0.084/93.18◦ V;
Vo5 = (1/25)(0.92/55.76◦ ) = 0.037/55.76◦ V.
vo (t) = 0.108 cos(0.5t − 159.88◦ )
+ 0.084 cos(1.5t + 93.18◦ ) + 0.037 cos(2.5t + 55.76◦ ) + · · · V.
P 16.55 vg =
2(2.5π)
2.5π 2.5π
+
sin(5000t) −
cos(10,000t) + · · ·
π
2
3π
= 2.5 + 1.25π sin(5000t) − (5/3) cos(10,000t) − · · · V.
H(s) =
=
(s/8000)2 +
1
√
!
2s/8000 + 1
80002
√
s2 + 8000 2s + 80002
!
(s/1000)2
√
(s/1000)2 + 2s/1000 + 1
!
s2
√
;
s2 + 1000 2s + 10002
!
H(j0) = 0;
80002
√
80002 − 50002 + j(5000)8000 2
H(j5000) =
−50002
√
10002 − 50002 + j(5000)1000 2
!
= 0.931/ − 39◦ ;
H(j10,000) =
·
80002
√
80002 − 10,0002 + j(10,000)8000 2
−10,0002
√
10002 − 10,0002 + j(10,000)1000 2
!
!
= 0.539/ − 99.5◦ .
.·. vo (t) = 0 + 3.65 cos(5000t − 129◦ ) − 0.898 cos(10,000t − 99.5◦ ) − · · · V.
!
16–52
CHAPTER 16. Fourier Series
P 16.56 [a] Let Va represent the node voltage across R2 , then the node-voltage
equations are
Va − Vg
Va
+
+ Va sC2 + (Va − Vo )sC1 = 0;
R1
R2
(0 − Va )sC2 +
0 − Vo
= 0.
R3
Solving for Vo in terms of Vg yields
−1
s
Vo
R1 C1
= H(s) =
Vg
s2 + 1 1 + 1 s +
R3
C1
C2
.
R1 +R2
R1 R2 R3 C1 C2
It follows that
R1 + R2
ωo2 =
R1 R2 R3 C1 C2
1
β=
R3
1
1
+
;
C1 C2
R3
Ko =
R1
C2
.
C1 + C2
Note that
R3
−R
1
C2
1
1
+ C12 s
C1 +C2 R3 C1
.
H(s) =
2
s2 + R13 C11 + C12 s + R1 RR21R+R
3 C1 C2
[b] For the given values of R1 , R2 , R3 , C1 , and C2 we have
R3
−
R1
1
R3
1
1
+
C1 C2
C2
C1 + C2
=−
R3
400
=−
;
2R1
313
= 2000;
R1 + R2
= 0.16 × 1010 = 16 × 108 ;
R1 R2 R3 C1 C2
H(s) =
ωo =
−(400/313)(2000)s
s2 + 2000s + 16 × 108
.
2π
2π
=
× 106 = 4 × 104 rad/s;
T
50π
H(jnωo ) =
=
−(400/313)(2000)jnωo
16 × 108 − n2 ωo2 + j2000nωo
−j(20/313)n
;
(1 − n2 ) + j0.05n
Problems
H(jωo ) =
16–53
−j(20/313)
400
=−
= −1.28;
j(0.050)
313
H(j3ωo ) =
−j(20/313)(3)
= 0.0240/91.07◦ ;
−8 + j0.15
H(j5ωo ) =
−j(100/313)
= 0.0133/90.60◦ ;
−24 + j0.25
∞
1
4A X
sin(nπ/2) cos nωo t;
vg (t) =
π n=1,3,5,... n
A = 15.65π V;
vg (t) = 62.60 cos ωo t − 20.87 cos 3ωo t + 12.52 cos 5ωo t − · · · ;
vo (t) = −80 cos ωo t − 0.50 cos(3ωo t + 91.07◦ )
+ 0.17 cos(5ωo t + 90.60◦ ) − · · · V.
[c] Given how much the amplitudes of the higher harmonics have been
reduced by the filter, we expect the quality factor Q to be quite high.
[d] ωo = 4 × 104 rad/s and β = 2000 rad/s. Therefore, Q = 40,000/2000 = 20.
We expect the output voltage to be dominated by the fundamental
frequency component since the bandpass filter is tuned to this frequency!
P 16.57 [a] Let Va represent the node voltage across R2 , then the node-voltage
equations are
Va − Vg
Va
+
+ Va sC2 + (Va − Vo )sC1 = 0;
R1
R2
(0 − Va )sC2 +
0 − Vo
= 0.
R3
Solving for Vo in terms of Vg yields
−1
s
Vo
R1 C1
= H(s) =
Vg
s2 + 1 1 + 1 s +
R3
C1
It follows that
R1 + R2
ωo2 =
= 108 ;
R1 R2 R3 C1 C2
1
β=
R3
R3
Ko =
R1
1
1
+
C1 C2
C2
C1 + C2
= 400;
= 1.
C2
.
R1 +R2
R1 R2 R3 C1 C2
16–54
CHAPTER 16. Fourier Series
Note that
R3
−R
1
C2
1
1
+ C12 s
C +C
R
C1
1 2 3
.
H(s) =
2
s2 + R13 C11 + C12 s + R1 RR21R+R
3 C1 C2
[b] For the given values of R1 , R2 , R3 , C1 , and C2 we have
H(s) =
−400s
s2 + 400s + 108
;
∞
(8)(2.25π 2 ) X
1
cos nωo t
2
2
π
n=1,3,5,... n
vg =
1
1
= 18 cos ωo t + cos 3ωo t +
cos 5ωo t + · · · mV
9
25
= [18 cos ωo t + 2 cos 3ωo t + 0.72 cos 5ωo t + · · ·] mV.
ωo =
2π
× 103 = 104 rad/s;
0.2π
H(jk104 ) =
−400jk104
−jk
=
;
8
2
8
4
10 − k 10 + j400k10
25(1 − k 2 ) + jk
H1 = −1 = 1/180◦ ;
H3 =
−j3
= 0.015/90.86◦ ;
−200 + j3
H5 =
−j5
= 0.0083/90.48◦ ;
−600 + j5
vo = 18 cos(ωo t + 180◦ ) + 0.03 cos(3ωo t + 90.86◦ )
+ 0.006 cos(5ωo t + 90.48◦ ) + · · · mV.
Note – ωo = 104 rad/s and β = 400 rad/s. Therefore,
Q = 10,000/400 = 25. We expect the output voltage to be dominated by
the fundamental frequency component since the bandpass filter is tuned
to this frequency!
The Fourier Transform
Assessment Problems
AP 17.1 [a] F (ω) =
Z 2 π
t e−jωt dt
2
A sin
−2
=
2
Ae−jωt
[−jω
sin(π/2)t
−
(π/2)
cos(π/2)t]
(π 2 /4) − ω 2
−2
=
Ae−j2ω
Aej2ω
(−jω
sin
π
−
(π/2)
cos
π)
−
(jω sin π − (π/2) cos π)
(π 2 /4) − ω 2
(π 2 /4) − ω 2
2πAe−j2ω
2πAej2ω
2πA
− 2
= 2
(e−j2ω − ej2ω )
= 2
2
2
2
π − 4ω
π − 4ω
π − 4ω
=
[b] F (ω) =
−j4πA
sin 2ω.
π 2 − 4ω 2
Z 0
−τ /2
Z τ /2
−2A
2A
−jωt
t+A e
dt +
t + A e−jωt dt
τ
τ
0
=
0
τ /2
2Ae−jωt
Ae−jωt 0
2Ae−jωt
Ae−jωt τ /2
(−jωt
−
1)
−
(−jωt
−
1)
+
+
τ ω2
−jω −τ /2
τ ω2
−jω 0
−τ /2
0
=
4A
ωτ
1
−
cos
ω2τ
2
.
Z 5
Z 8
−5
1
6ejtω dω +
2ejtω dω + 6ejtω dω
2π
−8
−5
5
n
o
1
=
6e−j5t − 6e−j8t + 2ej5t − 2e−j5t + 6ej8t − 6ej5t
j2πt
Z
AP 17.2 f (t) =
1 4e−j5t − 4ej5t 6ej6t − 6e−j8t
=
+
πt
j2
j2
"
=
1
(6 sin 8t − 4 sin 5t).
πt
17–1
#
17–2
CHAPTER 17. The Fourier Transform
AP 17.3 [a] Use the sifting property of the Dirac delta function:
F (ω) =
Z ∞
−∞
δ(t − t0 )e−jωt dt = e−jωt0 .
[b] f + (t) = te−at ,
L{f + (t)} =
f − (t) = −te−at ;
1
,
(s + a)2
Therefore F (ω) =
L{f − (t)} =
−1
.
(s + a)2
1
1
−j4aω
−
= 2
.
2
2
(a + jω)
(a − jω)
(a + ω 2 )2
[c] F (s) = L{te−at } =
1
.
(s + a)2
F (ω) = F (s)
+F (s)
s=jω
;
s=−jω
1
2(a2 − ω 2 )
2(a2 − ω 2 )
1
+
=
=
.
F (ω) =
(a + jω)2
(a − jω)2
(a2 − ω 2 )2 + 4a2 ω 2
(a2 + ω 2 )2
#
"
"
#
AP 17.4 v(t) = Vm [u(t + τ ) − u(t − τ )];
"
#
"
#
1
F{u(t + τ )} = πδ(ω) +
ejωτ ;
jω
1
F{u(t − τ )} = πδ(ω) +
e−jωτ .
jω
#
"
1
[ejωτ − e−jωτ ]
Therefore V (ω) = Vm πδ(ω) +
jω
= j2Vm πδ(ω) sin(ωτ ) +
=
2Vm
sin(ωτ )
ω
(2Vm τ ) sin(ωτ )
.
ωτ
AP 17.5 f (t) = tu(t);
"
d
d
1
F(tu(t)) = j F(u(t)) = j
πδ(ω) +
dω
dω
jω
= jπδ 0 (ω) −
1
.
ω2
AP 17.6 [a] Vg (ω) = F{36sgn t} =
72
.
jω
#
Problems
Vo
.
Vg
Using voltage division,
[b] H(s) =
Vo =
60,000k(500,000/s)
41.667
Vg =
Vg ;
12,000 + 60,000k(500,000/s)
s + 50
H(s) =
41.667
,
s + 50
H(jω) =
[c] Vo (ω) = H(jω) · Vg (ω) =
=
41.667
.
50 + jω
41.667 72
3000
·
=
50 + jω jω
jω(50 + jω)
60
60
−
.
jω jω + 50
[d] vo (t) = 30sgn(t) − 60e−50t u(t) V.
[e] vo (0− ) = −30 − 0 = −30 V.
[f ] vo (0+ ) = 30 − 60 = −30 V.
[g] vo (∞) = 30 − 0 = 30 V.
AP 17.7
Vo s
Vo − Vg 100Vo
+
+
= 0;
2s
s
100s + 125 × 104
.·. Vo =
Io =
s(100s + 125 × 104 )Vg
.
125(s2 + 12,000s + 25 × 106 )
sVo
;
100s + 125 × 104
H(s) =
Io
s2
=
.
Vg
125(s2 + 12,000s + 25 × 106 )
H(jω) =
−8 × 10−3 ω 2
;
(25 × 106 − ω 2 ) + j12,000ω
Vg (ω) = 300π[δ(ω + 5000) + δ(ω − 5000)];
17–3
17–4
CHAPTER 17. The Fourier Transform
Io (ω) = H(jω)Vg (ω) =
−2.4πω 2 [δ(ω + 5000) + δ(ω − 5000)]
;
(25 × 106 − ω 2 ) + j12,000ω
−2.4π Z ∞ ω 2 [δ(ω + 5000) + δ(ω − 5000)] jtω
io (t) =
e dω
2π
−∞ (25 × 106 − ω 2 ) + j12,000ω
25 × 106 ej5000t
25 × 106 e−j5000t
+
= −1.2
−j(12,000)(5000) j(12,000)(5000)
(
6
=
12
(
e−j5000t ej5000t
+
−j
j
)
)
◦
◦
= 0.5[e−j(5000t+90 ) + ej(5000t+90 ) ];
io (t) = 1 cos(5000t + 90◦ ) A.
AP 17.8 |V (ω)| = 10 −
|V (ω)|2 = 100 −
10
ω,
5000π
0 ≤ ω ≤ 5000π;
200
100
ω2;
ω+
5000π
25π 2 × 106
1 Z 5000π
200ω
100 × 10−6 2
W1Ω =
100 −
ω dω
+
π 0
5000π
25π 2
"
#
#5000π
1
200ω 2
100 × 10−6 ω 3
=
100ω −
+
π
10,000π
75π 2
0
"
1
200
100 × 10−6 (5000π)3
=
100(5000π) −
(5000π)2 +
π
10,000π
75π 2
"
= 100(5000) −
200(5000)2 100 × 10−6 (5000)3
+
10,000
75
= 166.67 kJ.
W10kΩ =
AP 17.9 Io =
166.67 × 103
= 16.67 J.
10 × 103
0.5sIg
sIg
=
;
0.5s + 25
s + 50
H(s) =
Io
s
=
.
Ig
s + 50
#
Problems
H(jω) =
Ig (ω) =
jω
;
jω + 50
12
;
jω + 10
Io (ω) = H(jω)Ig (ω) =
|Io (ω)| = q
|Io (ω)|2 =
=
12(jω)
;
(jω + 10)(jω + 50)
12ω
(ω 2 + 100)(ω 2 + 2500)
;
144ω 2
(ω 2 + 100)(ω 2 + 2500)
−6
ω 2 + 100
+
150
ω 2 + 2500
.
1 Z ∞ 150dω
1 Z ∞ 6dω
Wo (total) =
−
π 0 ω 2 + 2500 π 0 ω 2 + 100
ω ∞
0.6
ω ∞
3
−
tan−1
= tan−1
π
50 0
π
10 0
= 1.5 − 0.3 = 1.2 J.
Wo (0 — 100 rad/s) =
0.6
3
tan−1 (2) −
tan−1 (10)
π
π
= 1.06 − 0.28 = 0.78 J.
Therefore, the percent between 0 and 100 rad/s is
0.78
(100) = 64.7%.
1.2
17–5
17–6
CHAPTER 17. The Fourier Transform
Problems
P 17.1
[a] F (ω) =
Z 0
(−Ae−jωt ) dt +
Z τ /2
−τ /2
Ae−jωt dt
0
A
[2 − ejωτ /2 − e−jωτ /2 ]
jω "
#
ejωτ /2 + e−jωτ /2
2A
1−
=
jω
2
−j2A
=
[1 − cos(ωτ /2)].
ω
=
[b] F (ω) =
Z ∞
te−at e−jωt dt =
0
P 17.2
[a] F (ω) =
Z ∞
te−(a+jω)t dt =
0
Z ∞
1
.
(a + jω)2
f (t)e−jωt dt
−∞
=
Z τ /2
−2A −jωt
2A −jωt
te
dt +
te
dt
τ
−τ /2 τ
0
Z 0
= Int1 + Int2.
−2A Z 0
te−jωt dt
Int1 =
τ
−τ /2
−2A
=
τ
(
0
e−jωt
(−jωt
−
1)
−ω 2
−τ /2
)
=
o
−2A n
jωτ /2
1
−
[e
(−jωτ
/2
+
1)]
ω2τ
=
o
2A n jωτ /2
e
(1
−
jωτ
/2)
−
1
;
ω2τ
2A Z τ /2 −jωt
Int2 =
te
dt
τ 0
2A
=
τ
=
(
τ /2
e−jωt
(−jωt
−
1)
−ω 2
0
)
2A n jωτ /2
e
(jωτ /2 + 1) − 1)] }
ω2τ
F (ω) = Int1 + Int2
2A
ωτ
ωτ
= 2 2 cos
+ ωτ sin
− 2}
ω τ
2
2
Problems
[b] After using L’Hopital’s rule we have
2Aτ cos(ωτ /2)
Aτ
=
.
ω→0
4
2
F (0) = lim
[c]
P 17.3
[a] F (ω) = j
2A
ω
ωo
−
ωo
ωo
≤ω≤ .
2
2
1 Z ωo /2 j2A jωt
f (t) =
ωe dω
2π −ωo /2 ωo
ωo /2
jA ejtω
=
(jtω
−
1)
πωo −t2
−ωo /2
"
=
#
A
[ωo t cos(ωo t/2) − 2 sin(ωo t/2)].
πωo t2
"
#
A ωo t cos(ωo t/2) − 2 sin(ωo t/2)
[b] f (t) =
;
πωo
t2
A ωo t(− ω2o sin ω2o t ) + ωo cos ω2o t − ωo cos ω2o t
f (0) = lim
t→0 πωo
2t
(
"
ωo t
A −ωo2
= lim
sin
t→0 πωo
4
2
(
"
#)
[c] When A = 2π and ωo = 2 rad/s,
f (t) =
1
[2t cos t − 2 sin t] ;
t2
f (−t) = −f (t)
odd function.
= 0.
#)
17–7
17–8
CHAPTER 17. The Fourier Transform
P 17.4
[a] F (ω) = F (s) |s=jω = L{e−at sin ω0 t}s=jω
=
ω0
ω0
.
=
2
(s + a)2 + ω0 s=jω (a + jω)2 + ω02
[b] F (s) = L{e−at sin ω0 t} =
F (ω) = F (s)
−j0.5
j0.5
ω0
=
+
.
2
2
(s + a) + ω0
(s + a) − jω0 (s + a) + jω0
+F (s)
s=jω
F (ω) =
−ja
a2 + (ω − ω0 )2
+
[c] F (s) = L{e−at cos ω0 t} =
F (ω) = F (s)
.
+F (s)
;
s=−jω
0.5
0.5
+
(a + jω) − jω0 (a + jω) + jω0
+
=
ja
a2 + (ω + ω0 )2
s+a
0.5
0.5
=
+
.
2
2
(s + a) + ω0
(s + a) − jω0 (s + a) + jω0
s=jω
F (ω) =
;
s=−jω
0.5
0.5
+
(a − jω) − jω0 (a − jω) + jω0
a
a2 + (ω − ω0 )2
+
a
a2 + (ω + ω0 )2
.
#
"
1
1
=
.
[d] F (ω) = L{f (t)}s=−jω =
2
(s + a) s=−jω (a − jω)2
−
[e] F (s) = L{t3 e−at } =
F (ω) = F (s)
+F (s)
s=jω
F (ω) =
6
.
(s + a)4
;
s=−jω
6
6
a2 − ω 2
+
=
−j48aω
.
(a + jω)4 (a − jω)4
(a2 + ω 2 )4
Problems
(
P 17.5
F{sin ω0 t} = F
=
ejω0 t
2j
)
(
−F
e−jω0 t
2j
17–9
)
1
[2πδ(ω − ω0 ) − 2πδ(ω + ω0 )]
2j
= jπ[δ(ω + ω0 ) − δ(ω − ω0 )].
P 17.6
1 Z∞
f (t) =
[A(ω) + jB(ω)][cos tω + j sin tω] dω
2π −∞
1 Z∞
=
[A(ω) cos tω − B(ω) sin tω] dω
2π −∞
j Z∞
[A(ω) sin tω + B(ω) cos tω] dω.
2π −∞
But f (t) is real, therefore the second integral in the sum is zero.
+
P 17.7
By hypothesis, f (t) = −f (−t). From Problem 17.6, we have
1 Z∞
f (−t) =
[A(ω) cos tω + B(ω) sin tω] dω.
2π −∞
For f (t) = −f (−t), the integral
f (t) is real and odd, we have
R∞
−∞ A(ω) cos tω dω must be zero. Therefore, if
−1 Z ∞
B(ω) sin tω dω.
f (t) =
2π −∞
P 17.8
F (ω) =
−j2
;
ω
therefore B(ω) =
−2
;
ω
thus we have
1 Z ∞ −2
1 Z ∞ sin tω
f (t) = −
sin tω dω =
dω.
2π −∞ ω
π −∞ ω
But
sin tω
ω
is even;
so f (t) =
2 Z ∞ sin tω
dω.
π 0
ω
Therefore,
2 π


· = 1,
t > 0; 

π 2
from a table of definite integrals

2
−π


f (t) = ·
= −1, t < 0. 
π
2

f (t) =
Thus f (t) = sgn t.
17–10
P 17.9
CHAPTER 17. The Fourier Transform
From Problem 17.4[c] we have
F (ω) =
2 + (ω − ω0 )2
+
2 + (ω + ω0 )2
.
Note that as → 0, F (ω) → 0 everywhere except at ω = ±ω0 . At ω = ±ω0 ,
F (ω) = 1/, therefore F (ω) → ∞ at ω = ±ω0 as → 0. The area under each
bell-shaped curve is independent of , that is
Z ∞
dω
dω
=
= π.
2
2
2
−∞ + (ω + ω0 )2
−∞ + (ω − ω0 )
Z ∞
Therefore as → 0,
P 17.10 A(ω) =
Z 0
F (ω) → πδ(ω − ω0 ) + πδ(ω + ω0 ).
f (t) cos ωt dt +
Z ∞
−∞
f (t) cos ωt dt = 0
0
since f (t) cos ωt is an odd function.
B(ω) = −2
Z ∞
f (t) sin ωt dt,
since f (t) sin ωt is an even function.
0
P 17.11 A(ω) =
Z ∞
f (t) cos ωt dt
−∞
=
Z 0
f (t) cos ωt dt +
−∞
Z ∞
=2
Z ∞
f (t) cos ωt dt
0
f (t) cos ωt dt,
since f (t) cos ωt is also even.
0
B(ω) = 0, since f (t) sin ωt is an odd function and
Z 0
f (t) sin ωt dt = −
−∞
Z ∞
f (t) sin ωt dt.
0
(
)
Z ∞
df (t)
df (t) −jωt
P 17.12 [a] F
=
e
dt.
dt
−∞ dt
Let u = e−jωt , then du = −jωe−jωt dt; let dv = [df (t)/dt] dt, then
v = f (t).
(
Therefore F
df (t)
dt
)
= f (t)e−jωt
∞
Z ∞
−
−∞
f (t)[−jωe−jωt dt]
−∞
= 0 + jωF (ω).
[b] Fourier transform of f (t) exists, i.e., f (∞) = f (−∞) = 0.
Problems
(
[c] To find F
(
Then F
d2 f (t)
,
dt2
)
d2 f (t)
dt2
let g(t) =
)
(
=F
(
But G(ω) = F
df (t)
dt
dg(t)
dt
df (t)
.
dt
)
= jωG(ω).
)
= jωF (ω).
(
Therefore we have F
d2 f (t)
dt2
)
= (jω)2 F (ω).
Repeated application of this thought process gives
(
F
P 17.13 [a] F
dn f (t)
dtn
)
= (jω)n F (ω).
Z t
f (x) dx =
Z ∞ Z t
−∞
−∞
Now let u =
Z t
f (x) dx e−jωt dt.
−∞
f (x) dx,
then du = f (t)dt.
−∞
Let dv = e−jωt dt,
then v =
e−jωt
.
−jω
Therefore,
F
Z ∞
∞
e−jωt Z t
e−jωt
f (t) dt
f (x) dx =
f (x) dx
−
−jω −∞
−∞ −jω
−∞
−∞
Z t
= 0+
Z ∞
[b] We require
F (ω)
.
jω
f (x) dx = 0.
−∞
Z ∞
[c] No, because
e−ax u(x) dx =
−∞
P 17.14 [a] F{f (at)} =
Z ∞
#
"
1
6= 0.
a
f (at)e−jωt dt.
−∞
Let u = at,
du = adt,
u = ±∞ when t = ±∞.
Therefore,
F{f (at)} =
Z ∞
−∞
−jωu/a
f (u)e
du
a
!
1
= F
a
ω
,
a
a > 0.
17–11
17–12
CHAPTER 17. The Fourier Transform
[b] F{e−|t| } =
1
1
2
+
=
.
1 + jω 1 − jω
1 + ω2
Therefore F{e−a|t| } =
(1/a)2
.
(ω/a)2 + 1
Therefore F{e−0.5|t| } =
4
4ω 2 + 1
,
F{e−|t| } =
2
ω2 + 1
;
F{e−2|t| } = 1/[0.25ω 2 + 1], yes as “a” increases, the sketches show that
f (t) approaches zero faster and F (ω) flattens out over the frequency
spectrum.
P 17.15 [a] F{f (t − a)} =
Z ∞
f (t − a)e−jωt dt.
−∞
Let u = t − a, then du = dt, t = u + a, and u = ±∞ when t = ±∞.
Therefore,
F{f (t − a)} =
Z ∞
f (u)e−jω(u+a) du
−∞
−jωa
=e
Z ∞
f (u)e−jωu du = e−jωa F (ω).
−∞
jω0 t
[b] F{e
f (t)} =
Z ∞
−∞
f (t)e−j(ω−ω0 )t dt = F (ω − ω0 .)
ejω0 t + e−jω0 t
[c] F{f (t) cos ω0 t} = F f (t)
2
(
"
#)
1
1
= F (ω − ω0 ) + F (ω + ω0 )
2
2
.
Problems
P 17.16 Y (ω) =
Z ∞ Z ∞
−∞
=
Z ∞
17–13
x(λ)h(t − λ) dλ e−jωt dt
−∞
Z ∞
−jωt
h(t − λ)e
x(λ)
−∞
dt dλ.
−∞
Let u = t − λ, du = dt, and u = ±∞, when t = ±∞.
Therefore Y (ω) =
Z ∞
Z ∞
x(λ)
h(u)e
−∞
=
−jω(u+λ)
du dλ
−∞
Z ∞
−jωλ
x(λ) e
Z ∞
−∞
=
−jωu
h(u)e
du dλ
−∞
Z ∞
x(λ)e−jωλ H(ω) dλ = H(ω)X(ω).
−∞
Z ∞ P 17.17 F{f1 (t)f2 (t)} =
−∞
1Z∞
F1 (u)ejtu du f2 (t)e−jωt dt
2π −∞
1Z∞
2π −∞
=
Z ∞
−∞
F1 (u)f2 (t)e−jωt ejtu du dt
Z ∞
1Z∞
F1 (u)
f2 (t)e−j(ω−u)t dt du
2π −∞
−∞
=
1Z∞
=
F1 (u)F2 (ω − u) du.
2π −∞
P 17.18 [a] f 0 (t) =
.·.
2A
,
τ
−τ
< t < 0;
2
f 0 (t) =
2A
2A
[u(t + τ /2) − u(t)] −
[u(t) − u(t − τ /2)]
τ
τ
=
f 0 (t) =
−2A
,
τ
τ
;
2
0<t<
2A
4A
2A
u(t + τ /2) −
u(t) +
u(t − τ /2);
τ
τ
τ
2A
τ
4A
2A
τ
f (t) =
δ t+
−
δ(t) +
δ t−
.
τ
2
τ
τ
2
.·.
00
2A jωτ /2 4A 2A −jωτ /2
[b] F{f (t)} =
e
−
+
e
τ
τ
τ
00
4A ejωτ /2 + e−jωτ /2
4A
ωτ
=
−1 =
cos
τ
2
τ
2
"
#
[c] F{f 00 (t)} = (jω)2 F (ω) = −ω 2 F (ω);
Thus we have F (ω) = −
1
ω2
−1 .
therefore F (ω) = −
4A
ωτ
cos
τ
2
−1
.
1
F{f 00 (t)}.
ω2
17–14
CHAPTER 17. The Fourier Transform
P 17.19 (i) F{e−at u(t)} =
dF (ω)
−j
=
.
dω
(a + jω)2
1
= F (ω);
a + jω
"
#
dF (ω)
1
Therefore j
=
.
dω
(a + jω)2
Therefore F{te−at u(t)} =
1
.
(a + jω)2
(ii) F{|t|e−a|t| } = F{te−at u(t)} − F{teat u(−t)}
1
d
=
−j
2
(a + jω)
dω
1
a − jω
!
1
1
+
.
(a + jω)2 (a − jω)2
=
(iii) F{te−a|t| } = F{te−at u(t)} + F{teat u(−t)}
1
d
=
+j
2
(a + jω)
dω
=
!
1
1
−
.
2
(a + jω)
(a − jω)2
P 17.20 [a] f1 (t) = cos ω0 t,
f2 (t) = 1,
1
a − jω
F1 (u) = π[δ(u + ω0 ) + δ(u − ω0 )];
−τ /2 < t < τ /2,
Thus F2 (u) =
and f2 (t) = 0 elsewhere;
τ sin(uτ /2)
.
uτ /2
Using convolution,
1Z∞
F (ω) =
F1 (u)F2 (ω − u) du
2π −∞
=
1Z∞
sin[(ω − u)τ /2]
π[δ(u + ω0 ) + δ(u − ω0 )]τ
du
2π −∞
(ω − u)(τ /2)
τZ ∞
sin[(ω − u)τ /2]
=
δ(u + ω0 )
du
2 −∞
(ω − u)(τ /2)
+
=
τZ ∞
sin[(ω − u)τ /2]
δ(u − ω0 )
du
2 −∞
(ω − u)(τ /2)
τ sin[(ω + ω0 )τ /2] τ sin[(ω − ω0 )τ /2]
·
+ ·
.
2 (ω + ω0 )(τ /2)
2
(ω − ω0 )τ /2
Problems
17–15
[b] As τ increases, the amplitude of F (ω) increases at ω = ±ω0 and at the
same time the width of the frequency band of F (ω) approaches zero as ω
deviates from ±ω0 .
The area under the [sin x]/x function is independent of τ, that is
Z ∞
τ Z ∞ sin[(ω − ω0 )(τ /2)]
sin[(ω − ω0 )(τ /2)]
dω =
[(τ /2) dω] = π.
2 −∞ (ω − ω0 )(τ /2)
(ω − ω0 )(τ /2)
−∞
Therefore as t → ∞,
f1 (t)f2 (t) → cos ω0 t and F (ω) → π[δ(ω − ω0 ) + δ(ω + ω0 )].
P 17.21 [a] Ig (ω) = F{10 sgn (t)} =
H(s) =
20
.
jω
Vo
.
Ig
Using current division and Ohm’s law,
4s
4
Vo = Io s =
(Ig )s =
Ig ;
4+1+s
5+s
H(s) =
4s
,
s+5
H(jω) =
Vo (ω) = H(jω) · Ig (ω) =
j4ω
.
5 + jω
j4ω
5 + jω
!
20
jω
!
=
80
.
5 + jω
vo (t) = 80e−5t u(t) V.
[b] At t = 0− the circuit is
From the circuit, vo (0− ) = 0, which agrees with our solution. Using
current division, i1 (0− ) = −2 A and io (0− ) = −8 A.
At t = 0+ the circuit is
io (0+ ) = io (0− ) = −8 A (continuity);
i1 (0+ ) = 10 − io (0− ) = 18 A;
17–16
CHAPTER 17. The Fourier Transform
.·.
vo (0+ ) = −1io (0+ ) + 4i1 (0+ ) = 8 + 72 = 80 V (KVL right mesh),
which agrees with our solution.
We also know from the circuit vo (∞) = 0, which agrees with our solution.
The time constant with respect to the terminals of the inductor is L/Req .
Thus,
1
= 5,
.·.
τ
which also agrees with our solution.
Thus our solution makes sense in terms of known circuit behavior.
τ = 1/(1 + 4) = 0.2 s;
P 17.22 [a] Ig (ω) = F{10 sgn (t)} =
H(s) =
20
.
jω
Io
.
Ig
Using current division,
Io =
4k(s + 1)
4
Ig =
Ig ;
s+1
s+5
H(s) =
4
,
s+5
H(jω) =
Io (ω) = H(jω) · Ig (ω) =
4
.
5 + jω
4
5 + jω
!
20
jω
!
=
80
16
16
=
−
.
jω(5 + jω)
jω jω + 5
io (t) = 8sgn(t) − 16e−5t u(t) A.
[b] From Problem 17.21(b),
io (0− ) = −8 A;
io (0+ ) = io (0− ) = −8 A.
This agrees with our solution.
The time constant with respect to the terminals of the inductor is L/Req .
Thus,
1
.·.
= 5,
τ
which also agrees with our solution.
Thus our solution makes sense in terms of known circuit behavior.
τ = 1/(1 + 4) = 0.2 s;
P 17.23 [a] Find the Thévenin equivalent with respect to the terminals of the
inductor. Thus,
vTh =
40
vg = 0.8vg ;
50
RTh = 10k40 = 8 Ω.
Problems
0.8Vg
0.2Vg
=
;
8 + 4s
s+2
0.2
Io
=
;
H(s) =
Vg
s+2
Io =
H(jω) =
0.2
;
jω + 2
!
1
Vg (ω) = 125 πδ(ω) +
.
jω
Io (ω) = Vg (ω)H(jω)
!
25
1
=
πδ(ω) +
jω + 2
jω
=
25πδ(ω)
25
+
jω + 2
jω(2 + jω)
= I1 (ω) + I2 (ω).
1 Z ∞ 25πδ(ω)ejωt
dt = 6.25 A.
2π −∞ 2 + jω
12.5
12.5
I2 (ω) =
−
;
jω
jω + 2
i1 (t) =
i2 (t) = 6.25sgn(t) − 12.5e−2t u(t) A.
io (t) = i1 + i2 = 6.25 + 6.25sgn(t) − 12.5e−2t u(t) A
= 12.5u(t) − 12.5e−2t u(t) A.
[b]
17–17
17–18
CHAPTER 17. The Fourier Transform
P 17.24 [a] From the solution to Problem 17.23 we have
H(s) =
0.2
Io
=
.
Vg
s+2
H(jω) =
0.2
;
jω + 2
vg = 125sgn(t) V;
Io = Vg H(jω) =
Vg (ω) =
250
;
jω
50
25
25
=
−
;
jω(jω + 2)
jω jω + 2
.·. io (t) = 12.5sgn(t) − 25e−2t u(t) A.
[b]
P 17.25 [a]
Vo 2Vo
+
= Ig ;
50
s
Vo
1
2
+
= Ig ;
50 s
50s
Vo
= H(s) =
;
Ig
s + 100
H(jω) =
Ig (ω) =
jω50
;
jω + 100
4
;
jω
Problems
Vo (ω) =
17–19
4 50(jω)
200
·
=
;
jω j + 100
jω + 100
.·. vo (t) = 200e−100t u(t) V.
[b] At t = 0− the circuit is
At t = 0+ the circuit is
From the circuit
vo (0+ ) = (4)(50) = 200 V,
which agrees with our solution.
At t = ∞
vo (∞) = 0,
since the inductor short-circuits the dc current source. This is also in
agreement with our solution.
τ = L/R = 0.5/50 = 1/100;
.·. 1/τ = 100
which agrees with our solution.
Vo
2 50sIg
P 17.26 [a] Io =
=
;
0.5s
s s + 100
100
Io
= H(s) =
;
Ig
s + 100
H(jω) =
100
;
jω + 100
Ig (ω) =
4
;
jω
Io (ω) =
400
4
4
=
−
;
jω(jω + 100)
jω jω + 100
.·. io (t) = 2sgn(t) − 4e−100t u(t) A.
[b]
• From the solution to Problem 17.25(b) we note io (0− ) = −2 A and
io (0+ ) = −2 A. Our solution agrees with these results.
17–20
CHAPTER 17. The Fourier Transform
• From the circuit, io (∞) = 2 A. Our solution agrees with this value.
• From the circuit, τ = 0.01 s which agrees with our solution.
P 17.27 [a] Vo =
Vg (1/sC)
Vg
=
;
R + (1/sC)
RCs + 1
1/RC
Vo
1
= H(s) =
=
;
Vg
s + (1/RC)
s+1
H(jω) =
1
;
jω + 1
Vg (ω) =
30
30
+
;
−jω + 5 jω + 5
Vo (ω) =
30
30
+
(−jω + 5)(jω + 1) (jω + 5)(jω + 1)
=
K1 =
K1
K2
K3
K4
+
+
+
.
−jω + 5 jω + 1 jω + 5 jω + 1
30
= 5;
6
Vo (ω) =
K2 =
30
= 5;
6
K3 =
30
= −7.5;
−4
K4 =
30
= 7.5;
4
5
12.5
7.5
+
−
−jω + 5 jω + 1 jω + 5
vo (t) = 5e5t u(−t) + (12.5e−t − 7.5e−5t )u(t) V.
[b] vo (0− ) = 5 V.
[c] vo (0+ ) = 12.5 − 7.5 = 5 V.
[d]
Vo − Vg Vo s
+
− 0.5 = 0;
10
10
Vo − Vg + Vo s − 5 = 0;
Vo (s + 1) = 5 + Vg ;
Vg =
30
;
s+5
.·. Vo =
5
30
5
7.5
7.5
12.5
7.5
+
=
+
−
=
−
;
s + 1 (s + 1)(s + 5)
s+1 s+1 s+5
s+1 s+5
vo (t) = (12.5e−t − 7.5e−5t )u(t) V.
Problems
[e] Yes, for t ≥ 0+ the solution in part (a) is also
vo (t) = (12.5e−t − 7.5e−5t )u(t) V.
P 17.28 [a] Io =
Vg
Vg s
=
;
10 + 10/s
10s + 10
H(s) =
Io
0.1
=
;
Vg
s+1
H(jω) =
0.1
;
jω + 1
Vg (ω) =
30
30
+
;
−jω + 5 jω + 5
"
30
0.1jω
30
Io (ω) = H(jω)Vg (jω) =
+
jω + 1 −jω + 5 jω + 5
=
3jω
3jω
+
(jω + 1)(−jω + 5) (jω + 1)(jω + 5)
=
K1
K2
K3
K4
+
+
+
.
jω + 1 −jω + 5 jω + 1 jω + 5
K1 =
3(−1)
= −0.5;
6
K2 =
#
3(5)
= 2.5;
6
3(−5)
3(−1)
= −0.75;
K4 =
= 3.75;
4
−4
−1.25
2.5
3.75
.·. Io (ω) =
+
+
;
jω + 1 −jω + 5 jω + 5
K3 =
io (t) = 2.5e5t u(−t) + [−1.25e−t + 3.75e−5t ]u(t) A.
[b] io (0− ) = 2.5 V.
[c] io (0+ ) = 2.5 V.
[d] Note – since io (0+ ) = 2.5 A, vo (0+ ) = 30 − 25 = 5 V.
Io =
Vg − (5/s)
sVg − 5
=
;
10 + (10/s)
10s + 10
.·. Io =
Vg =
30
;
s+5
25s − 25
2.5(s − 1)
−1.25
3.75
=
=
+
;
10(s + 1)(s + 5)
(s + 1)(s + 5)
s+1
s+5
io (t) = (−1.25e−t + 3.75e−5t )u(t) A.
17–21
17–22
CHAPTER 17. The Fourier Transform
[e] Yes, for t ≥ 0+ the solution in part (a) is also
io (t) = (−1.25e−t + 3.75e−5t )u(t) A.
P 17.29 [a] H(s) =
1/RC
100
=
;
s + 1/RC
s + 100
H(ω) =
Vo (ω) =
=
100
;
jω + 100
Vg (ω) =
!
!
30
jω
100
jω + 100
=
30
;
jω
3000
jω(jω + 100)
30
30
−
;
jω jω + 100
.·. vo (t) = 15sgn(t) − 30e−100t u(t) V.
[b] vo (0− ) = −15 V.
[c] vo (0+ ) = 15 − 30 = −15 V.
[d]
Vo − 15/s (Vo + 15/s)s
+
= 0;
50,000
5 × 106
100Vo −
1500
+ Vo s + 15 = 0;
s
K1
K2
15(100 − s)
=
+
.
.·. Vo =
s(s + 100)
s
s + 100
K1 =
15(100)
= 15;
100
K2 =
15(200)
= −30;
−100
vo (t) = (15 − 30e−100t )u(t) V.
[e] Yes, they agree. The solution from part (a) for t > 0 is
vo (t) = (15 − 30e−100t )u(t) V.
P 17.30 [a] H(s) =
Io
(1/R)s
=
;
Vg
s + 1/RC
20 × 10−6 s
H(s) =
;
s + 100
20 × 10−6 (jω)
H(ω) =
;
jω + 100
Problems
20 × 10−6 (jω) 30
600 × 10−6
·
=
;
jω + 100
jω
jω + 100
Io (ω) =
io (t) = 600e−100t u(t) µA.
[b] io (0− ) = 0.
[c] io (0+ ) = 600 µA.
[d]
Io =
=
30/s
30
=
6
50,000 + (5 × 10 /s)
50,000s + 5 × 106
600 × 10−6
;
s + 100
io (t) = 600e−100t u(t) µA.
[e] Yes they agree. Also note that at t = 0+ the circuit is
io (0+ ) =
30
= 600 µA.
50,000
which agrees with our solution.
P 17.31 [a] Vg (ω) =
H(s) =
1
1
+ πδ(ω) +
;
1 − jω
jω
1
Va
0.5k(1/s)
=
=
,
Vg
1 + 0.5k(1/s)
s+3
H(jω) =
1
;
3 + jω
Va (ω) = H(jω)Vg (jω)
1
1
πδ(ω)
=
+
+
(1 − jω)(3 + jω) jω(3 + jω) 3 + jω
1/4
1/4
1/3
1/3
πδ(ω)
=
+
+
−
+
1 − jω 3 + jω
jω
3 + jω 3 + jω
1/4
1/3
1/12
πδ(ω)
=
+
−
+
.
1 − jω
jω
3 + jω 3 + jω
Therefore va (t) =
1 t
1
1
1
e u(−t) + sgn t − e−3t u(t) +
V.
4
6
12
6
17–23
17–24
CHAPTER 17. The Fourier Transform
[b] va (0− ) =
1 1
1
1
− + 0 + = V;
4 6
6
4
va (0+ ) = 0 +
1
1
1
1
−
+ = V;
6 12 6
4
va (∞) = 0 +
1
1
1
+ 0 + = V.
6
6
3
P 17.32 [a]
vg (t) = 18e4t u(−t) − 12u(t);
.·. Vg (ω) =
18
12
− 12πδ(ω) −
.
4 − jω
jω
Using voltage division,
Vo (s) =
(103 /5s)
80
V
(s)
=
Vg ;
g
(103 /5s) + 2.5
s + 80
Vo (s)
80
=
;
.·. H(s) =
Vg (s)
s + 80
.·. H(jω) =
80
.
jω + 80
Vo (jω) = H(jω) · Vg (ω)
=
(80)(18)
(80)12πδ(ω)
(12)(80)
−
−
(jω + 80)(4 − jω)
jω + 80
jω(jω + 80)
=
(120/7) (120/7) 960πδ(ω) 12
12
+
−
−
+
;
jω + 80
4 − jω
jω + 80
jω jω + 80
vo (t) =
120 −80t
120 4t
e
u(t) +
e u(−t) − 6 − 6sgn(t) + 12e−80t u(t)V;
7
7
120
120
.·. vo (0− ) =
−6+6=
V;
7
7
vo (0+ ) =
120
120
− 6 − 6 + 12 =
V.
7
7
The voltages at 0− and 0+ must be the same since the voltage cannot
change instantaneously across a capacitor.
Vg (s)
0.4s
[b] Io (s) =
=
Vg (s);
3
(10 /5s) + 2.5
s + 80
H(s)
Io (s)
0.4s
=
;
Vg (s)
s + 80
0.4jω
.·. H(jω) =
.
jω + 80
Problems
17–25
Io (jω) = H(jω) · Vg (ω);
io (t) =
=
7.2jω
4.8πδ(ω)jω
4.8jω
−
−
(4 − jω)(jω + 80)
jω + 80
jω(jω + 80)
=
(24/70)
(48/7)
4.8πδ(ω)jω
4.8
−
−
−
4 − jω
jω + 80
jω + 80
jω + 80
=
(24/70) (816/70) 4.8πδ(ω)jω
−
−
;
4 − jω
jω + 80
jω + 80
816 −80t
24 4t
e u(−t) −
e
u(t)A;
70
70
.·. io (0− ) = 24/70A;
[c] vo (t) =
io (0+ ) = −816/70A.
120 4t
120 −80t
e
u(t) +
e u(−t) − 6 − 6sgn(t) + 12e−80t u(t)V.
7
7
P 17.33 [a]
Vo
Vo
(Vo − Vg )s
+
+
= 0;
4
5 × 10
0.3125s 50
.·. Vo =
s2 Vg
.
s2 + 1000s + 16 × 104
s2
Vo
= H(s) =
;
Vg
(s + 200)(s + 800)
H(jω) =
(jω)2
;
(jω + 200)(jω + 800)
vg = 90e−400|t| ;
Vg (ω) =
.·. Vo (ω) = H(jω)Vg (ω) =
=
K1 =
72,000
;
(jω + 400)(−jω + 400)
72,000(jω)2
(jω + 200)(jω + 400)(jω + 800)(−jω + 400)
K1
K2
K3
K4
+
+
+
.
jω + 200 jω + 400 jω + 800 −jω + 400
72,000(−200)2
= 40;
(200)(600)(600)
17–26
CHAPTER 17. The Fourier Transform
K2 =
72,000(−400)2
= −180;
(−200)(400)(800)
72,000(−800)2
K3 =
= 160;
(−600)(−400)(1200)
K4 =
72,000(400)2
= 20;
(600)(800)(1200)
.·. vo (t) = [40e−200t − 180e−400t + 160e−800t ]u(t) + 20e400t u(−t) V.
[b] vo (0− ) = 20 V;
Vo (0+ ) = 40 − 180 + 160 = 20 V;
vo (∞) = 0 V.
[c] IL =
Vo
3.2sVg
=
;
0.3125s
(s + 200)(s + 800)
H(s) =
3.2s
IL
=
;
Vo
(s + 200)(s + 800)
H(jω) =
IL (ω) =
=
K4 =
3.2(jω)
;
(jω + 200)(jω + 800)
3.2(jω)(72,000)
(jω + 200)(jω + 400)(jω + 800)(−jω + 400)
K1
K2
K3
K4
+
+
+
.
jω + 200 jω + 400 jω + 800 −jω + 400
(3.2)(400)(72,000)
= 160 mA;
(600)(800)(1200)
iL (t) = 160e400t u(−t);
.·. iL (0− ) = 160 mA.
K1 =
(3.2)(−200)(72,000)
= −640 mA;
(200)(600)(600)
K2 =
(3.2)(−400)(72,000)
= 1440 mA;
(−200)(400)(800)
K3 =
(3.2)(−800)(72,000)
= −640 mA;
(−600)(−400)(1200)
.·. iL (0+ ) = K1 + K2 + K3 = −640 + 1440 − 640 = 160 mA.
Checks, i.e.,
At t = 0− :
iL (0+ ) = iL (0− ) = 160 mA.
vC (0− ) = 90 − 20 = 70 V.
At t = 0+ :
vC (0+ ) = 90 − 20 = 70 V.
Problems
[d] We can check the correctness of out solution for t ≥ 0+ by using the
Laplace transform. Our circuit becomes
Vo
Vo
(Vo − Vg )s
160 × 10−3
−3
+
+
+ 1.4 × 10 +
= 0;
50 0.3125s
5 × 104
s
.·. (s2 + 1000s + 16 × 104 )Vo = s2 Vg − (70s + 8000).
vg (t) = 90e−400t u(t) V;
Vg =
90
;
s + 400
2
90s − (70s + 8000)(s + 400)
.
.·. (s + 200)(s + 800)Vo =
(s + 400)
2
4
20s − 36,000s − 320 × 10
.·. Vo =
(s + 200)(s + 400)(s + 800)
=
40
180
160
−
+
;
s + 200 s + 400 s + 800
.·. vo (t) = [40e−200t − 180e−400t + 160e−800t ]u(t) V.
This agrees with our solution for vo (t) for t ≥ 0+ .
P 17.34 [a] vg = 125 cos 75t;
Vg (ω) = 125π[δ(ω + 75) + δ(ω − 75)].
Va
Va − Vg
Va
+
+
= 0;
20s
5s
400
Va
1
Vg
1
1
+
+
= ;
20s 5s 400
5s
Va [20 + 80 + s] = 80Vg ;
17–27
17–28
CHAPTER 17. The Fourier Transform
80Vg
;
s + 100
Va =
Vo =
H(s) =
Vo
72
=
;
Vg
s + 100
H(ω) =
72
.
jω + 100
Vo (ω) = Vg (ω)H(ω) =
Va
(360) = 0.9Va .
400
9000π[δ(ω + 75) + δ(ω − 75)]
;
jω + 100
1 Z∞
vo (t) =
Vo (ω)ejtω dω
2π −∞
e−j75t
ej75t
+
= 4500
100 + j75 100 − j75
"
e−j75t
ej75t
= 180
+
4 + j3 4 − j3
"
#
#
◦
◦
◦
◦
= 36[ej75t e−j36.87 + e−j75t ej36.87 ]
= 36[ej(75t−j36.87 ) + e−j(75t+36.87 ) ];
vo (t) = 72 cos(75t − 36.87◦ ) V.
[b] In the phasor domain:
V − 125
V
V
+
+
= 0;
j375
j1500 400
"
#
1
1
1
125
V
+
+
=
;
j375 j1500 400
j375
V=
(144 + j192)(125)
= 64 − j48 = 80/ − 36.87◦ V;
j375
Vo =
360
(V) = 72/ − 36.87◦ V;
400
vo (t) = 72 cos(75t − 36.87◦ ) V.
Problems
P 17.35 [a]
IL =
80 × 104 /s
80 × 104
I
=
Ig ;
g
5000 + 5s + 80 × 104 /s
5s2 + 5000s + 80 × 104
16 × 104
16 × 104
IL
= H(s) = 2
=
;
Ig
s + 1000s + 16 × 104
(s + 200)(s + 800)
H(jω) =
16 × 104
;
(jω + 200)(jω + 800)
Ig (ω) =
−45
45
+
;
(−jω + 400) (jω + 400)
IL (ω) =
−45(16 × 104 )
(jω + 200)(jω + 800)(−jω + 400)
+
45(16 × 104 )
(jω + 200)(jω + 800)(jω + 400)
= IL1 + IL2 .
IL1 =
K2
K3
K1
+
+
:
jω + 200 jω + 800 −jω + 400
K1 =
−45(16 × 104 )
= −20;
(600)(600)
K2 =
−45(16 × 104 )
= 10;
(−600)(1200)
K3 =
−45(16 × 104 )
= −10;
(600)(1200)
IL1 =
−20
10
10
+
−
.
jω + 200 jω + 800 −jω + 400
IL2 =
K4
K5
K6
+
+
:
jω + 200 jω + 800 jω + 400
K4 =
45(16 × 104 )
= 60;
(200)(600)
17–29
17–30
CHAPTER 17. The Fourier Transform
K5 =
45(16 × 104 )
= 30;
(−600)(−400)
K6 =
45(16 × 104 )
= −90;
(−200)(400)
IL2 =
60
30
90
+
−
.
jω + 200 jω + 800 jω + 400
.·. IL =
40
10
90
40
+
−
−
;
jω + 200 jω + 800 −jω + 400 jω + 400
iL (t) = (40e−200t + 40e−800t − 90e−400t )u(t) − 10e400t u(−t) A.
−
[b] iL (0− ) = −10e400(0 ) u(0− ) = −10 A.
+
+
+
[c] iL (0+ ) = (40e−200(0 ) + 40e−800(0 ) − 90e−400(0 ) = −10 A.
[d] Yes, there cannot be an instantaneous change in the inductor current,
.·. iL (0− ) = iL (0+ ).
P 17.36 [a] Vg (ω) =
900
60
+
;
−jω + 5 (jω + 5)2
Vo
sVo
Vo − Vg
+
+
= 0;
12
4s + 20 300
25(s + 5)
Vo
.·. H(s) =
=
;
Vg
(s + 10)(s + 20)
H(ω) =
25(jω + 5)
.
(jω + 10)(jω + 20)
Vo (ω) = Vg (ω)H(ω)
=
1500(jω + 5)
22,500
+
(jω + 10)(jω + 20)(−jω + 5) (jω + 10)(jω + 20)(jω + 5)
= V1 (ω) + V2 (ω).
V1 (ω) =
K1
K2
K3
+
+
;
jω + 10 jω + 20 −jω + 5
K1 =
1500(−5)
= −50;
(10)(15)
K2 =
1500(−15)
= 90;
(−10)(25)
K3 =
1500(10)
= 40;
(15)(25)
Problems
V2 (ω) =
K4
K5
K6
+
+
;
jω + 10 jω + 20 (jω + 5)
K4 =
22,500
= −450;
(10)(−5)
K5 =
22,500
= 150;
(−10)(−15)
K6 =
22,500
= 300;
(5)(15)
Vo (ω) =
−50
90
40
450
+
+
−
jω + 10 jω + 20 −jω + 5 jω + 10
+
=
150
300
+
jω + 20 (jω + 5)
−500
240
40
300
+
+
+
;
jω + 10 jω + 20 −jω + 5 (jω + 5)
.·. vo (t) = [−500e−10t + 240e−20t + 300e−5t ]u(t) + 40e5t u(−t) V.
[b] vo (0− ) = 40 V.
[c] vo (0+ ) = −500 + 240 + 300 = 40 V.
P 17.37 Vo =
60
40
20
40(s2 + 20s + 150)
−
+
=
;
s
s + 5 s + 20
s(s + 5)(s + 20)
8
Vi = ;
s
H(s) =
5(s2 + 20s + 150)
;
(s + 5)(s + 20)
H(jω) =
5[(jω)2 + 20(jω) + 150]
.
(jω + 5)(jω + 20)
Vi (ω) =
16
;
(jω)
Vo (ω) =
80[(jω)2 + 20(jω) + 150]
jω(jω + 5)(jω + 20)
=
K1
K2
K3
+
+
;
jω
jω + 5 jω + 20
17–31
17–32
CHAPTER 17. The Fourier Transform
K1 =
(80)(150)
= 120;
100
K2 =
(80)(25 − 100 + 150)
= −80;
(−5)(15)
K3 =
(80)(150)
= 40;
300
.·. vo (t) = 60sgn(t) − 80e−5t u(t) + 40e−20t u(t) V.
P 17.38 v(t) = 4te−t u(t);
V (ω) =
Therefore |V (ω)| =
√
4
.
1 + ω2
"
1Z 3
4
W1Ω =
π 0
(1 + ω 2 )
(
4
.
(1 + jω)2
#2
dω
√ )
3
ω
1
−1 ω
+
tan
2 ω2 + 1
1 0
√
"
#
3 1
= 16
+
= 3.769 J;
8π
6
16
=
π
∞
8
π
8
ω
−1 ω
+
tan
=
0+
= 4 J.
W1Ω (total) =
2
π ω +1
1 0
π
2
Therefore % =
1
P 17.39 [a] f (t) =
2π
[b] W = 2
0
[c] W =
1
π
Z ∞
0
3.769
(100) = 94.23%.
4
Z 0
Z ∞
ω jtω
e e
dω +
Z ∞
−∞
−ω jtω
e
e
dω =
0
2
(1/π)
2
dt = 2
2
2
(1 + t )
π
Z ∞
1 e−2ω
π −2
∞
e−2ω dω =
0
dt
1
=
J.
2
2
(1 + t )
2π
=
0
1/π
.
1 + t2
1
J.
2π
1 Z ω1 −2ω
0.95
[d]
e
dω =
,
1 − e−2ω1 = 0.95,
π 0
2π
ω1 = (1/2) ln 20 ∼
= 1.5 rad/s.
e2ω1 = 20;
Problems
P 17.40 Io =
Ig (10)
50Ig
=
;
10 + 0.2s
s + 50
H(s) =
Io
50
=
;
Ig
s + 50
H(jω) =
50
.
jω + 50
Ig (ω) =
3
;
jω + 25
Io (ω) =
150
;
(jω + 25)(jω + 50)
|Io (ω)| = q
|Io (ω)|2 =
Wo =
150
(ω 2 + 625)(ω 2 + 2500)
22,500
(ω 2 + 625)(ω 2 + 2500)
=
12
ω 2 + 625
−
12
ω 2 + 2500
.
1 Z ∞ 12dω
1 Z ∞ 12dω
−
π 0 ω 2 + 625 π 0 ω 2 + 2500
ω
12 1
·
· tan−1
=
π 25
25
=
;
∞
12 1
ω
− ·
· tan−1
π 50
50
0
∞
0
12 π
12 π
12
6
6
· −
· =
−
=
= 120 mJ;
25π 2 50π 2
50 50
50
.·. Wo (total) = 120 mJ.
12 1
12 1
tan−1 (0.4) −
tan−1 (0.2) = 43.06 mJ;
π 25
π 50
Wo (10 rad/s) =
%=
43.06
(100) = 35.88%.
120
P 17.41 [a] Vg (ω) =
600
;
(jω + 5)(−jω + 5)
H(s) =
Vo
5
=
;
Vg
s + 25
Vo (ω) =
3000
;
(jω + 5)(jω + 25)(−jω + 5)
Vo (ω) =
15
5
10
−
+
;
jω + 5 jω + 25 −jω + 5
H(ω) =
5
.
(jω + 25)
vo (t) = [15e−5t − 5e−25t ]u(t) + 10e5t u(−t) V.
17–33
17–34
CHAPTER 17. The Fourier Transform
[b] |Vg (ω)| =
[c] |Vo (ω)| =
[d] Wi = 2
[e] Wo =
600
(ω 2 + 25)
.
3000
√
.
ω 2 + 625
(ω 2 + 25)
Z ∞
∞
−10t
3600e
0
Z 0
10t
100e
−∞
= 10 +
e−10t
= 720 J.
dt = 7200
−10 0
dt +
Z ∞
(15e−5t − 5e−25t )2 dt
0
Z ∞
[225e−10t − 150e−30t + 25e−50t ] dt
0
= 10 + 22.5 − 5 + 0.5 = 28 J.
[f ] |Vg (ω)| =
600
,
2
ω + 25
|Vg2 (ω)| =
36 × 104
;
(ω 2 + 25)2
36 × 104 Z 10
dω
Wg =
2
π
0 (ω + 25)2
36 × 104
=
π
(
1
ω
1
−1 ω
+
tan
2(25) ω 2 + 25 5
5
7200 10
1
+ tan−1 2 = 690.8 J;
π
125 5
690.8
·
.. %=
× 100 = 95.95%.
720
=
10 )
0
Problems
[g] |Vo (ω)|2 =
=
9 × 106
(ω 2 + 25)2 (ω 2 + 625)
15,000
25
25
− 2
+ 2
;
2
2
(ω + 25)
ω + 25 (ω + 625)
(
1
1
Wo =
15,000
π
2(25)
+25
!
ω
1
−1 ω
+
tan
ω 2 + 25 5
5
1
ω 10
tan−1
25
25 0
10
−25
ω 10
1
tan−1
5
5 0
0
)
300 10
1
5
1
+ tan−1 2 − tan−1 2 + tan−1 0.4
π 125 5
π
π
=
= 27.14 J;
%=
27.14
× 100 = 96.93%.
28
P 17.42 [a] Vi (ω) =
A
;
a + jω
A
;
|Vi (ω)| = √ 2
a + ω2
H(s) =
s
;
s+α
H(jω) =
jω
;
α + jω
ω
.
|H(ω)| = √ 2
α + ω2
ωA
Therefore |Vo (ω)| = q
(a2 + ω 2 )(α2 + ω 2 )
Therefore |Vo (ω)|2 =
WIN =
Z ∞
.
ω 2 A2
.
(a2 + ω 2 )(α2 + ω 2 )
A2 e−2at dt =
0
A2
;
2a
when α = a we have
(
)
Z a
A2 R a ω 2 dω
A2 Z a dω
a2 dω
WOUT =
=
−
π 0 (ω 2 + a2 )2
π
0 (a2 + ω 2 )2
0 a2 + ω 2
=
A2
4aπ
π
−1 .
2
A2 Z ∞
ω2
A2
WOUT (total) =
dω
=
.
π 0
(a2 + ω 2 )2
4a
"
Therefore
17–35
#
WOUT (a)
1
= 0.5 − = 0.1817 or 18.17%.
WOUT (total)
π
17–36
CHAPTER 17. The Fourier Transform
[b] When α 6= a we have
1Z α
ω 2 A2 dω
WOUT (α) =
π 0 (a2 + ω 2 )(α2 + ω 2 )
=
A2
π
Z α 0
K2
K1
+
dω ,
a2 + ω 2 α2 + ω 2
2
where K1 =
a
a2 − α2
and K2 =
−α2
.
a2 − α 2
Therefore
A2
α
απ
WOUT (α) =
a tan−1
−
;
2
2
π(a − α )
a
4
A2
A2
π
π
−
α
=
.
a
π(a2 − α2 ) 2
2
2(a + α)
WOUT (total) =
α
απ
2
WOUT (α)
=
· a tan−1
−
.
WOUT (total)
π(a − α)
a
4
Therefore
√
For α = a 3, this ratio is 0.2723, or√
27.23% of the output energy lies in
the frequency band between 0 and a 3.
√
of the output energy lies in
[c] For α = a/ 3, the ratio is 0.1057, or 10.57%
√
the frequency band between 0 and a/ 3.
P 17.43 [a] |Vi (ω)|2 =
[b] Vo =
104
;
ω2
|Vi (25)|2 =
104
= 16;
252
|Vi (50)|2 =
104
= 4.
502
Vi R
sRCVi
=
;
R + (1/sC)
RCs + 1
H(s) =
Vo
s
=
;
Vi
s + (1/RC)
H(jω) =
jω
;
jω + 25
1
1
=
= 25.
RC
(20,000)(2 × 10−6 )
Problems
|Vo (ω)| =
100
100
|ω|
·√ 2
=√ 2
;
|ω|
ω + 625
ω + 625
|Vo (ω)|2 =
104
,
ω 2 + 625
25 ≤ ω ≤ 50 rad/s;
104
|Vo (25)| = 2
= 8;
25 + 625
2
|Vo (ω)|2 = 0,
104
|Vo (50)| = 2
= 3.2.
50 + 625
2
104
1 50
1 Z 50 104
dω
=
[c] W1Ω =
−
π 25 ω 2
π
ω 25
1
200 ∼
104 1
=
−
=
= 63.66 J.
π 25 50
π
[d] W1Ω =
=
17–37
50
1 Z 50 104
104
−1 ω
dω
=
·
tan
π 25 ω 2 + 625
25π
25 25
400
[tan−1 2 − tan−1 1] ∼
= 40.97 J.
π
elsewhere;
Two-Port Circuits
Assessment Problems
AP 18.1
z11 =
V1
= 1 + 12 = 13 Ω;
I1 I2 =0
z21 =
V2
= 12 Ω;
I1 I2 =0
z22 =
V2
= 4 + 12 = 16 Ω;
I2 I1 =0
z12 =
V1
= 12 Ω.
I2 I1 =0
AP 18.2 With V2 = 0, attach a 1 V source at V1 and calculate the two currents:
I1 =
1
I1
0.25
= 0.25 A so y11 =
=
= 0.25 S;
1 + (4k12)
V1 V2 =0
1
18–1
18–2
CHAPTER 18. Two-Port Circuits
I2 =
−12
I2
−0.1875
I1 = −0.1875 A so y21 =
=
= −0.1875 S.
12 + 4
V1 V2 =0
1
With V1 = 0, attach a 1 V source at V2 and calculate the two currents:
I2 =
1
I2
0.203125
= 0.203125 A so y22 =
= 0.203125 S;
=
4 + (1k12)
V2 V1 =0
1
I1 =
−12
I1
−0.1875
I2 = −0.1875 A so y12 =
= −0.1875 S.
=
1 + 12
V2 V1 =0
1
AP 18.3 I1 = g11 V1 + g12 I2 ;
V2 = g21 V1 + g22 I2 ;
g11 =
I1
0.25 × 10−6
=
= 12.5 × 10−6 = 12.5 µS;
V1 I2 =0
20 × 10−3
g21 =
V2
−5
= −250;
=
V1 I2 =0 20 × 10−3
g12 =
I1
200 × 10−6
=
= 4;
I2 V1 =0 50 × 10−6
g22 =
V2
10
= 200 kΩ.
=
I2 V1 =0 50 × 10−6
AP 18.4 First calculate the a-parameters:
a11 =
V1
−10
=
= −0.67;
V2 I2 =0
15
a12 =
−V1
−(−10)
=
= 2.5 Ω;
I2 V2 =0
4
a21 =
I1
30
=
= 2 S;
V2 I2 =0 15
a22 =
−I1
−(−5)
=
= 1.25.
I2 V2 =0
4
Now the h-parameters are calculated:
∆a
−5.833
=
= −4.667;
a22
1.25
h11 =
a12
2.5
=
= 2 Ω;
a22
1.25
h12 =
h21 =
−1
−1
=
= −0.8;
a22
1.25
h22 =
a21
2
=
= 1.6 S.
a22
1.25
Problems
AP 18.5 y11 = y22 ;
y12 = y21 .
With Port 2 shorted:
Therefore,
18–3
V1 = 80 V;
4 = y11 (80) + y12 (0);
I1 = 4 A.
so y11 = y22 =
4
= 50 mS.
80
With Port 2 open:
−2.25 = 0.05V1 + y12 (100);
0 = y12 V1 + 0.05(100);
so V1 =
−5
.
y12
Substitute and solve:
!
−5
+ y12 (100);
−2.25 = 0.05
y12
.·.
2
100y12
+ 2.25y12 − 0.25 = 0;
.·.
y12 = y21 = 40 mS.
AP 18.6 Because the measurements involve setting V2 to zero (Measurement 1) and I2
to zero (Measurement 2) the easiest two-port parameters to calculate are the a
parameters. From the first measurement:
a12 = −
−4
V1
=
= 20 Ω;
I2 V2 =0 −0.2
a22 = −
I1
−0.005
=
= 0.025.
I2 V2 =0
−0.2
From the second measurement:
a11 =
V1
0.02
=
= 0.5 × 10−3 ;
V2 I2 =0
40
a21 =
I1
20 × 10−6
=
= 500 nS.
V2 I2 =0
40
From the circuit,
Zg = 250 Ω;
Vg = 5.25 mV;
18–4
CHAPTER 18. Two-Port Circuits
ZTh =
20 + (0.025)(250)
a12 + a22 Zg
=
= 42 kΩ;
a11 + a21 Zg
0.5 × 10−3 + (500 × 10−9 )(250)
VTh =
Vg
5.25 × 10−3
=
= 8.4 V.
a11 + a21 Zg
0.5 × 10−3 + (500 × 10−9 )(250)
i=
8.4
= 0.10 mA;
84,000
P = (0.10 × 10−3 )2 (42,000) = 420 µW.
AP 18.7 The Thevenin equivalent seen looking into the g-network from the right is
VTh =
g21 Vg
(800/7)(30)
=
= 1846.154 V;
1 + g11 Zg
1 + (3/35)(10)
ZTh = g22 −
=
g12 g21 Zg
1 + g11 Zg
50,000 (20/7)(800/7)10)
−
= 5384.615 Ω.
7
1 + (3/35)(10)
The simplified circuit is shown here:
Vo =
−h21 ZL Vg
(h11 + Zg )(1 + h22 ZL ) − h12 h21 ZL
=
4(15,000)(1846.154)
= 3750 V.
(5000 + 5384.6)[1 + (0.0002)(15,000)] − (0.8)(15,000)
Problems
Problems
P 18.1
With port 2 short-circuited, we have
V1 V1
+ ;
I1 =
20
5
−20
I2 =
I1 = −0.8I1 .
25
I1
= y11 = 0.25 S;
V1
When V2 = 0, we have I1 = y11 V1
and I2 = y21 V1 .
Therefore I2 = −0.8(y11 V1 ) = −0.8y21 V1 .
Thus y21 = −0.8y11 = −0.2 S.
With port 1 short-circuited, we have
V2 V2
I2 =
+ ;
15
5
I2
4
= y22 =
V2
15
S;
−15
I1 =
I2 = −0.75I2 = −0.75y22 V2 .
20
Therefore y12 =
I1
4
= (−0.75) = −0.2 S.
V2
15
Summary:
y11 = 0.25 S;
y12 = −0.2 S;
y21 = −0.2 S;
y22 =
4
S.
15
18–5
18–6
P 18.2
CHAPTER 18. Two-Port Circuits
h11 =
V1
= 20k5 = 4 Ω;
I1 V2 =0
h21 =
I2
(−20/25)I1
= −0.8;
=
I1 V2 =0
I1
h12 =
V1
(20/25)V2
= 0.8;
=
V2 I1 =0
V2
h22 =
I2
1
8
1
+
=
S.
=
V2 I1 =0 15 25
75
Summary:
h11 = 4 Ω;
P 18.3
h21 = −0.8;
h12 = 0.8;
h22 =
8
S.
75
With Port 2 open:
V1
20
4
=
= .
V2 I2 =0 15
3
V2 =
15
V1
20
I1 =
V1
V1
=
20k(5 + 15)
10
so a11 =
3
30
V2 = V1 = I1
4
4
so V1 = 10I1 ;
so a21 =
I1
2
=
S.
V2 I2 =0 15
With Port 2 shorted:
I2 = −
I1 =
V1
5
so a12 =
−V1
= 5 Ω.
I2 V2 =0
V1
V1
−5I2
=
=
20k5
4
4
so a22 =
−I1
= 1.25.
I2 V2 =0
Summary:
a11 = 4/3;
a12 = 5 Ω;
a21 = (2/15) S;
a22 = 1.25.
Problems
P 18.4
V1 = z11 I1 + z12 I2 ;
V1 = z21 I1 + z22 I2 .
z11 =
V1
= 5k20 + 16 = 20 Ω;
I1 I2 =0
z21 =
V2
= 16 + (10)(5/25) = 18 Ω;
I1 I2 =0
z12 =
V1
= 16 + (10/25)(5) = 18 Ω;
I2 I1 =0
z22 =
V2
= 10k15 + 16 = 22 Ω.
I2 I1 =0
Summary:
z11 = 20 Ω;
P 18.5
z12 = 18 Ω;
z21 = 18 Ω;
z22 = 22 Ω.
V2 = b11 V1 − b12 I1 ;
I2 = b21 V1 − b22 I1 ;
b11 =
V2
;
V1 I1 =0
b21 =
I2
.
V1 I1 =0
V1 = 2 × 10−3 V2 ;
.·. b11 =
1
= 500;
2 × 10−3
V2 = 25,000I2 ;
so
V1 = (2 × 10−3 )(25,000)I2 = 50I2 ;
18–7
18–8
CHAPTER 18. Two-Port Circuits
1
.·. b21 =
= 20 mS.
50
b12 =
−V2
;
I1 V1 =0
I1 = −
2 × 10−3 V2
;
4000
I2 = 30I1 +
b22 =
−I2
.
I1 V1 =0
.·. b12 =
4000
= 2 MΩ;
2 × 10−3
V2
4000
= 30I1 −
I1 = −50I1 ;
25,000
(2 × 10−3 )(25,000)
Summary:
b11 = 500;
P 18.6
P 18.7
b12 = 2 MΩ;
b21 = 20 mS;
b22 = 50.
g11 =
20 × 10−3
b21
=
= 0.4 mS;
b22
50
g12 =
−1
−1
=
= −0.02;
b22
50
g21 =
∆b
(500)(50) − (2 × 106 )(20 × 10−3 )
=
= −300;
b22
50
g22 =
b12
2 × 106
=
= 40 kΩ.
b22
50
h11 =
V1
;
I1 V2 =0
h21 =
I2
.
I1 V2 =0
V1
= 40k[6 + 20k5] = 40k10 = 8 Ω
I1
.·. h11 = 8 Ω;
.·. b22 = 50.
Problems
I6 =
40
I1 = 0.8I1 ;
40 + 10
I2 =
−20
I6 = −0.8I6 = −0.8(0.8)I1 = −0.64I1
20 + 5
h12 =
V1
;
V2 I1 =0
h22 =
.·. h21 = −0.64;
I2
.
V2 I1 =0
V2
= 80k[5 + 20k(40 + 6)] = 15.314 Ω
I2
.·. h22 =
Vx =
20k46
V2 ;
5 + 20k46
V1 =
40(20k46)
557.5758
40
Vx =
V2 =
V2 ;
40 + 6
46(5 + 20k46)
871.2121
.·.
h12 = 0.64.
1
= 65.3 mS;
15.314
Summary:
h11 = 8 Ω;
P 18.8
h12 = 0.64;
h21 = −0.64;
V2 = b11 V1 − b12 I1 ;
I2 = b21 V1 − b22 I1 ;
b12 =
−V2
;
I1 V1 =0
5k20 = 4 Ω;
b22 =
−I2
.
I1 V1 =0
h22 = 65.3 mS.
18–9
18–10
CHAPTER 18. Two-Port Circuits
I2 =
V2
V2
= ;
4+4
8
I1 = Ib − Ia ;
Ia =
20
I2 ;
25
5
I2 ;
25
Ib =
I1 =
5
20
−15
−3
−
I2 =
I2 =
I2 ;
25 25
25
5
b22 =
5
−I2
= ;
I1
3
−V2
−V2
b12 =
=
I1
I2
b11 =
V2
;
V1 I1 =0
I2
I1
=8
b21 =
V1 = Va − Vb ;
Va =
40
5
=
Ω;
3
3
I2
.
V1 I1 =0
20
V2 ;
25
Vb =
V1 =
20
5
15
3
V2 − V2 = V2 = V2 ;
25
25
25
5
b11 =
V2
5
= ;
V1
3
5
V2 ;
25
V2 = (20 + 5)k(20 + 5)I2 = 12.5I2 ;
I2
I2
b21 =
=
V1
V2
V2
V1
1
=
12.5
5
2
=
S.
3
15
Summary:
5
b11 = ;
3
b12 =
40
Ω;
3
b21 =
2
S;
15
5
b22 = .
3
Problems
P 18.9
a11 =
V1
;
V2 I2 =0
V2 =
V1
R3 ;
R1 + R3
R1 + R3
R1
.·. a11 =
=1+
= 1.2
R3
R3
.·. R1 = 0.2R3
a21 =
I1
;
V2 I2 =0
R1
.·.
= 0.2;
R3
(Eq 1);
V2 = I3 R3 =
R2
I1 R3 ;
R1 + R2 + R3
R1 + R2 + R3
= 20 × 10−3
.·. a21 =
R2 R3
(Eq 2).
Substitute Eq 1 into Eq 2:
0.2R3 + R2 + R3
R2 + 1.2R3
=
= 20 × 10−3
R2 R3
R2 R3
a22 = −
I1
;
I2 V2 =0
R1
= 0.4;
R2
I2 =
−R2
I1 ;
R1 + R2
(Eq 3).
R1 + R2
.·. a22 =
= 1.4;
R2
R1
0.2R3
.·. R2 =
=
= 0.5R3
0.4
0.4
(Eq 4).
Substitute Eq 4 into Eq 3:
3.4
0.5R3 + 1.2R3
=
= 20 × 10−3
(0.5R3 )R3
R3
.·. R3 = 170 Ω.
Therefore,
R1 = 0.2R3 = 0.2(170) = 34 Ω;
R2 = 0.5R3 = 0.5(170) = 85 Ω.
Summary:
R3 = 170 Ω.
R1 = 34 Ω;
R2 = 85 Ω;
18–11
18–12
CHAPTER 18. Two-Port Circuits
P 18.10 h11 =
V1
;
I1 V2 =0
h21 =
I2
.
I1 V2 =0
V1
(800)(200)
= 40 +
= 40 + 160 = 200 Ω;
I1
1000
.·. h11 = 200 Ω.
Ia = I1
200
= 0.2I1 ;
1000
16.2vσ + 160Ib = 0;
vσ = 160I1 ;
.·. 160Ib = −2592I1 ;
Ib = −16.2I1 ;
.·. Ia + Ib + I2 = 0;
0.2I1 − 16.2I1 + I2 = 0;
I2 = 16I1 ;
.·. h21 = 16.
h12 =
V1
;
V2 I1 =0
I1 = 0;
h21 =
I2
.
V2 I1 =0
vσ = V1 ;
V1
V1 − V2
+
= 0;
200
800
1
.·. h12 = = 0.2.
5
4V1 + V1 − V2 = 0;
5V1 = V2 ;
Problems
I2 =
V2 + 16.2V1 V2 − V1
+
;
160
800
800I2 = 6V2 + 80V1 ;
800I2 = 6V2 + 80(0.2V2 ) = 22V2 ;
I2
22
.·. h22 =
=
= 27.5 mS.
V2
800
Summary:
h11 = 200 Ω;
P 18.11 g11 =
I1
;
V1 I2 =0
h12 = 0.20;
g21 =
h21 = 16;
V2
.
V1 I2 =0
.·. g11 = 10−3 S;
V1 = 200I1 + 800I1 = 1000I1 ;
V− =
1000
V2 = V+ ;
1500
800
1000
V2 =
V1 ;
.·.
1500
1000
g12 =
I1
;
I2 V1 =0
I1 = 0;
g21 =
V+ =
800
V1 ;
1000
.·. g21 = 1.2;
V2
;
I2 V1 =0
.·. g12 = 0;
Also, Vo = 0;
P 18.12 g11 =
g22 =
V2
.·. g22 =
= 40 Ω.
I2
I1
5 × 10−6
=
= 0.1 mS;
V1 I2 =0 50 × 10−3
V2
200 × 10−3
=
= 4;
V1 I2 =0 50 × 10−3
I1
2 × 10−6
= 4;
g12 =
=
I2 V1 =0 0.5 × 10−6
g22 =
h22 = 27.5 mS.
V2
10 × 10−3
=
= 20 kΩ.
I2 V1 =0 0.5 × 10−6
18–13
18–14
CHAPTER 18. Two-Port Circuits
P 18.13 [a] z11 =
V1
50 × 10−3
= 10 kΩ;
=
I1 I2 =0 5 × 10−6
z21 =
V2
200 × 10−3
= 40 kΩ.
=
I1 I2 =0
5 × 10−6
V1 = z11 I1 + z12 I2
.·.
−10,000(2)
= −40 kΩ.
0.5
z12 =
V2 = z21 I1 + z22 I2
so 0.01 = 40,000(2 × 10−6 ) + z22 (0.5 × 10−6 );
0.01 − (40,000)(2 × 10−6 )
z22 =
= −140 kΩ.
0.5 × 10−6
.·.
[b] g11 = 0.1 mS;
g21 = 4;
g12 = 4;
g22 = 20 kΩ.
z11 =
1
1
=
= 10 kΩ;
g11
0.1 × 10−3
z12 =
−g12
−4
=
= −40 kΩ;
g11
0.1 × 10−3
z21 =
g21
4
= 40 kΩ;
=
g11
0.1 × 10−3
z22 =
∆g
−14
= −140 kΩ.
=
g11
0.1 × 10−3
P 18.14 h11 =
Ia =
so 0 = 10,000(2 × 10−6 ) + z12 (0.5 × 10−6 );
V1
;
I1 V2 =0
h21 =
5Io
= 5I1 = −I2 ;
1
I2
.
I1 V2 =0
.·. h21 = −5;
V1 = (25 + j10)I1 + 5I1 = (30 + j10)I1 ;
.·. h11 = 30 + j10 Ω;
h12 =
V1
;
V2 I1 =0
Io = 0
thus
h22 =
I2
.
V2 I1 =0
5Io = 0 dependent source is a short circuit;
Problems
.·. h12 = 0;
V1 = 5Io = 0;
h22 =
1 + j3
I2
=
= (1 − j/3) S.
V2
j3
Summary:
h11 = 30 + j10 Ω;
h21 = −5;
h12 = 0;
P 18.15 V1 = a11 V2 − a12 I2 ;
I1 = a21 V2 − a22 I2 ;
V1 = h11 I1 + h12 V2 ;
I2 = h21 I1 + h22 V2 ;
V1 = −a12 I2 + a11 V2 ;
I2 =
a21 V2 − I1
;
a22
a21 − I1
.·. V1 = −a12
+ a11 V2 ;
a22
a12
a11 a22 − a12 a21
V1 =
I1 +
V2 ;
a22
a22
a12
;
.·. h11 =
a22
I2 = −
h12 =
∆a
;
a22
1
a21
I1 +
V2 ;
a22
a22
1
.·. h21 = − ;
a22
h22 =
a21
.
a22
P 18.16 I1 = y11 V1 + y12 V2 ;
I2 = y21 V1 + y22 V2 ;
V2 = b11 V1 − b12 I1 ;
I2 = b21 V1 − b22 I1 ;
I1 =
b11
1
V1 −
V2 ;
b12
b12
b11
.·. y11 =
;
b12
y12 = −
"
I2 = b21 V1 − b22
I2 =
1
;
b12
#
b11
1
V1 −
V2 ;
b12
b12
b21 b12 − b11 b22
b22
V1 +
V2 ;
b12
b12
∆b
.·. y21 = − ;
b12
y22 =
b22
.
b12
h22 = (1 − j/3) S.
18–15
18–16
CHAPTER 18. Two-Port Circuits
P 18.17 I1 = g11 V1 + g12 I2 ;
V2 = g21 V1 + g22 I2 .
I1 = y11 V1 + y12 V2 ;
I2 = y21 V1 + y22 V2 .
Solving the second y-parameter equation for V2 and comparing it to the
second g-parameter equation:
V2 =
I2
y21
−
V1 ;
y22 y22
.·.
g21 = −
y21
;
y22
and g22 =
1
.
y22
Substitute the y-parameter equation for V2 from above into the first
y-parameter equation and comparing it to the first g-parameter equation:
!
I2
y21
∆y
y12
−
V1 =
V1 +
I2 ;
y22 y22
y22
y22
I1 = y11 V1 + y12
.·.
g11 =
∆y
;
y22
and g12 =
y12
.
y22
P 18.18 First, find the y parameters:
I1 = y11 V1 + y12 V2 ;
I2 = y21 V1 + y22 V2 .
Since the two-port is symmetric and reciprocal we only need to calculate two
parameters since y11 = y22 and y12 = y21 .
"
#
V1
V1
1
1
=
I1 =
+
+
V1 ;
1
s
s 1 + s+1
1+ 1
s+1
s2 + 2s + 2
I1
=
;
V1
s(s + 2)
Problems
y11 = y22 =
s2 + 2s + 2
.
s(s + 2)
Ia =
V1
;
s
Ib =
1
V1
V1
s
;
·
1
1 =
s+2
1 + s+1 1 + s
V1
V1
I2 = −(Ia + Ib ) = −
+
;
s
s+2
I2
2s + 2
=−
;
V1
s(s + 2)
y12 = y21 = −
2(s + 1)
.
s(s + 2)
Now, transform to the a parameters:
a11 =
−y22
s2 + 2s + 2
=
;
y21
2(s + 1)
a12 =
−1
s(s + 2)
=
;
y21
2(s + 1)
a21 =
−1
s(s + 2)
−∆y
=
=
;
y21
y21
2(s + 1)
a22 =
−y11
s2 + 2s + 2
=
.
y21
2(s + 1)
P 18.19 V2 = 0:
V1
s2 + s + 1
= s + [1k(1/s)] =
;
I1
s+1
s+1
V1
.·. y11 =
= 2
.
I1 V2 =0 s + s + 1
18–17
18–18
CHAPTER 18. Two-Port Circuits
−1
−s
−s
s+1
I2 =
I1 =
I1 =
V1 ;
2
1 + (1/s)
s+1
s+1 s +s+1
I2
−s
.
.·. y21 =
= 2
V1 V2 =0 s + s + 1
V1 = 0:
V2
1
s
s2 + s + 1
= (1/s) + 1ks = +
=
;
I2
s s+1
s(s + 1)
s(s + 1)
I2
= 2
.
.·. y22 =
V2 V1 =0 s + s + 1
#
"
−1
−1
s(s + 1)
I1 =
I2 =
V2 ;
s+1
s + 1 s2 + s + 1
I1
−s
.·. y12 =
.
= 2
V2 V1 =0 s + s + 1
P 18.20 y11 =
I1 =
I1
;
V1 V2 =0
y21 =
V1 V1
3V1
+
=
;
20 10
20
I2
.
V1 V2 =0
I1
3
.·. y11 =
=
= 0.15 S;
V1
20
Problems
I2 = −
2V1
V1 3V1 V1
7V1
− (I1 − V1 /20) = − −
+
=−
;
50
25
20
20
50
7
I2
= − = −0.14 S;
.·. y21 =
V1
50
y12 =
I1
;
V2 V1 =0
y22 =
I2
.
V2 V1 =0
I1 =
V2 V2
−
= 0.1V2 ;
5
10
I1
.·. y12 =
= 0.1 S;
V2
I2 =
V2
V2
6 + j2
V2
+
+
=
V2 ;
50 −j25 10
50
I2
6 + j2
.·. y22 =
=
= 0.12 + j0.04 S.
V2
50
Summary:
y11 = 0.15 S;
y12 = 0.1 S;
y21 = −0.14 S;
y22 = 0.12 + j0.04 S.
P 18.21 b11 = −
y11
−0.15
=
= −1.5;
y12
0.1
b12 = −
1
−1
=
= −10 Ω;
y12
0.1
b21 = −
∆y
−[(0.15)(0.12 + j0.04) + (0.1)(0.14)]
=
= −0.32 − j0.06 S;
y12
0.1
b22 =
y22
0.12 + j0.04
=
= 1.2 + j0.4.
y12
0.1
18–19
18–20
CHAPTER 18. Two-Port Circuits
P 18.22 [a] h11 =
V1
;
I1 V2 =0
h21 =
I2
.
I1 V2 =0
V1 = (R + sL)I1 − sM I2 ;
0 = −sM I1 + (R + sL)I2 ;
∆=
(R + sL)
−sM
−sM
(R + sL)
V1
N1 =
−sM
= (R + sL)2 − s2 M 2 ;
= (R + sL)V1 ;
0 (R + sL)
I1 =
N1
(R + sL)V1
=
;
∆
(R + sL)2 − s2 M 2
0 = −sM I1 + (R + sL)I2 ;
h12 =
V1
;
V2 I1 =0
V1 = −sM I2 ;
h22 =
I2 =
V1 =
−sM V2
;
R + sL
h22 =
I2
1
=
.
V2
R + sL
h11 =
sM
I2
.·. h21 =
=
;
I1
R + sL
I2
.
V2 I1 =0
V2
;
R + sL
h12 =
V1
(R + sL)2 − s2 M 2
=
.
I1
R + sL
V1
−sM
=
;
V2
R + sL
Problems
[b] h12 = −h21
(reciprocal);
h11 h22 − h12 h21 = 1 (symmetrical, reciprocal);
h12 =
−sM
;
R + sL
h11 h22 − h12 h21 =
=
h21 =
sM
R + sL
(checks).
(R + sL)2 − s2 M 2
1
(sM )(−sM )
·
−
R + sL
R + sL
(R + sL)2
(R + sL)2 − s2 M 2 + s2 M 2
= 1 (checks).
(R + sL)2
P 18.23 First we note that
z11 =
(Zb + Zc )(Za + Zb )
Za + 2Zb + Zc
Therefore
z12 =
and z22 =
(Za + Zb )(Zb + Zc )
Za + 2Zb + Zc
z11 = z22 .
V1
;
I2 I1 =0
Use the circuit below:
V1 = Zb Ix − Zc Iy = Zb Ix − Zc (I2 − Ix ) = (Zb + Zc )Ix − Zc I2 ;
Ix =
Zb + Zc
I2
Za + 2Zb + Zc
so
V1 =
(Zb + Zc )2
I2 − Zc I2 ;
Za + 2Zb + Zc
V1
(Zb + Zc )2
Zb2 − Za Zc
.·. Z12 =
=
− Zc =
.
I2
Za + 2Zb + Zc
Za + 2Zb + Zc
z21 =
V2
;
I1 I2 =0
Use the circuit below:
V2 = Zb Ix − Zc Iy = Zb Ix − Zc (I1 − Ix ) = (Zb + Zc )Ix − Zc I1 ;
18–21
18–22
CHAPTER 18. Two-Port Circuits
Ix =
Zb + Zc
I1
Za + 2Zb + Zc
so
V2 =
(Zb + Zc )2
I1 − Zc I1 ;
Za + 2Zb + Zc
(Zb + Zc )2
Zb2 − Za Zc
V2
.·. z21 =
=
− Zc =
= z12 .
I1
Za + 2Zb + Zc
Za + 2Zb + Zc
Thus the network is symmetrical and reciprocal.
P 18.24 V2 = b11 V1 − b12 I1 ;
V1 = Vg − I1 Zg ;
I2 = b21 V1 − b22 I1 ;
V2 = −I2 ZL .
VTh with respect to Port 2 is V2 when I2 = 0.
b21 V1 − b22 I1 = 0 so I1 =
.·.
I1 +
b21
b21
V1 =
(Vg − I1 Zg );
b22
b22
b21
b21
Zg I1 =
Vg
b22
b22
so I1 =
(b21 /b22 )Vg
.
1 + (b21 Zg /b22 )
V2 = b11 (Vg − I1 Zg ) − b12 I1 = b11 Vg − I1 (b11 Zg + b12 )
"
(b11 Zg + b12 )(b21 /b22 )Vg
b11 (b22 + b21 Zg ) − (b11 Zg + b12 )b21
= Vg
= b11 Vg −
1 + (b21 Zg /b22 )
b22 + b21 Zg
=
ZTh =
Vg ∆b
= VTh .
b22 + b21 Zg
V2
;
I2 Vg =0
If Vg = 0 then V1 = −I1 Zg ;
V2 = b11 (−I1 Zg ) − b12 I1 = −(b11 Zg + b12 )I1 ;
I2 = b21 (−I1 Zg ) − b22 I1 = −(b12 Zg + b22 )I1 ;
.·.
ZTh =
b11 Zg + b12
V2
=
.
I2 Vg =0 b12 Zg + b22
P 18.25 I1 = g11 V1 + g12 I2 ;
V1 = Vg − I1 Zg ;
V2 = g21 V1 + g22 I2 ;
V2 = −I2 ZL .
I2 =
−V2
ZL
.·.
V2 (1 + g22 /ZL ) = g21 V1 ;
so V2 = g21 V1 + g22 (−V2 /ZL );
thus
V2
g21 ZL
=
.
V1
ZL + g22
#
Problems
P 18.26 V1 = a11 V2 − a12 I2 ;
18–23
I1 = a21 V2 − a22 I2 ;
V1 = Vg − I1 Zg ;
V2 = −I2 ZL .
I1 = a21 (−I2 ZL ) − a22 I2 = −(a22 + a21 ZL )I2 ;
−1
I2
=
.
I1
a22 + a21 ZL
.·.
P 18.27 I1 = y11 V1 + y12 V2 ;
I2 = y21 V1 + y22 V2 ;
V1 = Vg − I1 Zg ;
V2 = −I2 ZL .
I1 = y11 (Vg − I1 Zg ) + y12 V2 =
y11
y12
Vg +
V2 ;
1 + y11 Zg
1 + y11 Zg
I2 = y21 (Vg − I1 Zg ) + y22 V2 = −
V2
;
ZL
Rearranging,
y21 Vg − y21 Zg I1 + (y22 + 1/ZL )V2 = 0,
or
"
y21 Vg − y21 Zg
#
y11
y12
Vg +
V2 + (y22 + 1/ZL )V2 = 0.
1 + y11 Zg
1 + y11 Zg
Collecting Vg terms on the left and V2 terms on the right of the equals sign
gives
"
#
"
#
y11 y21 Zg
y21 y12 Zg
1
y21 −
Vg =
− y22 −
V2 .
1 + y11 Zg
1 + y11 Zg
ZL
Thus,
V2
y21 − (y11 y21 Zg )/(1 + y11 Zg )
=
Vg
(y21 y12 Zg )/(1 + y11 Zg ) − (y22 ZL + 1)/ZL
=
ZL y21 (1 + y11 Zg ) − y11 y21 Zg ZL
y21 y12 Zg ZL − (1 + y11 Zg )(1 + y22 ZL )
=
y21 ZL
.
y21 y12 Zg ZL − (1 + y11 Zg )(1 + y22 ZL )
18–24
CHAPTER 18. Two-Port Circuits
P 18.28 V1 = h11 I1 + h12 V2 ;
V1 = Vg − I1 Zg ;
I2 = h21 I1 + h22 V2 ;
V2 = −I2 ZL .
V1 = h11 I1 − h12 ZL I2 ;
I2 = h21 I1 − h22 ZL I2
so I2 =
h21
I1 ;
1 + h22 ZL
!
.·.
h12 ZL h21
I1 ;
V1 = h11 −
1 + h22 ZL
.·.
Zin =
P 18.29 [a] I2 =
h12 ZL h21
V1
= h11 −
.
I1
1 + h22 ZL
−Vg
a11 ZL + a12 + a21 Zg ZL + a22 Zg
=
−50 × 10−3
(5 × 10−4 )(5 × 103 ) + 10 + (10−6 )(100)(5 × 103 ) + (−3 × 10−2 )(100)
=
−50 × 10−3
= −5 mA.
10
1
PL = (5 × 10−3 )2 (5 × 103 ) = 62.5 mW.
2
[b] ZTh =
a12 + a22 Zg
10 + (−3 × 10−2 )(100)
=
a11 + a21 Zg
5 × 10−4 + (10−6 )(100)
7
70
=
kΩ.
−4
6 × 10
6
Vg
50 × 10−3
500
[c] VTh =
=
=
V;
−4
a11 + a21 Zg
6 × 10
6
=
Therefore V2 =
P 18.30 [a]
250
V;
6
Pmax =
(1/2)(250/6)2
= 74.4 mW.
(70/6) × 103
V2
y21 ZL
;
=
Vg
y12 y21 Zg ZL − (1 + y11 Zg )(1 + y22 ZL )
y12 y21 Zg ZL = (−2 × 10−6 )(100 × 10−3 )(2500)(70,000) = −35;
1 + y11 Zg = 1 + (2 × 10−3 )(2500) = 6;
1 + y22 ZL = 1 + (−50 × 10−6 )(70 × 103 ) = −2.5;
y21 ZL = (100 × 10−3 )(70 × 103 ) = 7000;
Problems
V2
7000
7000
=
=
= −350;
Vg
−35 − (6)(−2.5)
−20
V2 = −350Vg = −350(80) × 10−3 = −28 V(rms) = 28/180◦ V(rms).
|V2 |2
= 11.2 × 10−3 = 11.20 mW.
70,000
−28/180◦
= −0.4 × 10−3 /180◦ = 400/0◦ µA;
[c] I2 =
70,000
[b] P =
y21
I2
=
;
I1
y11 + ∆yZL
∆y = (2 × 10−3 )(−50 × 10−6 ) − (−2 × 10−6 )(100 × 10−3 )
= 100 × 10−9 ;
∆yZL = (100)(70) × 103 × 10−9 = 7 × 10−3 ;
y11 + ∆yZL = 2 × 10−3 + 7 × 10−3 = 9 × 10−3 ;
I2
100
100 × 10−3
=
=
;
I1
9 × 10−3
9
.·. 100I1 = 9I2 ;
I1 =
9(400 × 10−6 )
= 36 µA(rms).
100
Pg = (80)10−3 (36) × 10−6 = 2.88 µW.
1 + y11 Zg
.
y22 + ∆yZg
From the solution to Problem 18.30
P 18.31 [a] ZTh =
1 + y11 Zg = 1 + (2 × 10−3 )(2500) = 6;
y22 + ∆yZg = −50 × 10−6 + 10−7 (2500) = 200 × 10−6 ;
ZTh =
6
× 106 = 30,000 Ω;
200
∗
ZL = ZTh
= 30,000 Ω.
[b] y21 ZL = (100 × 10−3 )(30,000) = 3000;
y12 y21 Zg ZL = (−2 × 10−6 )(100 × 10−3 )(2500)(30,000) = −15;
1 + y11 Zg = 6;
1 + y22 ZL = 1 + (−50 × 10−6 )(30 × 103 ) = −0.5;
V2
3000
3000
=
=
= −250;
Vg
−15 − 6(−0.5)
−12
18–25
18–26
CHAPTER 18. Two-Port Circuits
V2 = −250(80 × 10−3 ) = −20 = 20/180◦ V(rms).
P =
[c] I2 =
|V2 |2
400
=
× 10−3 = 13.33 mW.
30,000
30
−V2
20/0◦
2
=
= mA;
30,000
30,000
3
100 × 10−3
100 × 10−3
I2
=
=
= 20;
I1
2 × 10−3 + 10−7 (30,000)
5 × 10−3
I2
2 × 10−3
1
I1 =
=
=
mA(rms).
20
3(20)
30
Pg (developed) = (80 × 10−3 )
P 18.32 [a] ZTh = g22 −
1
8
× 10−3 = µW.
30
3
g12 g21 Zg
;
1 + g11 Zg
1
1
g12 g21 = − + j
2
2
1
1
1
−j
=j ;
2
2
2
1 + g11 Zg = 1 + 1 − j1 = 2 − j1;
.·. ZTh = 1.5 + j2.5 −
j3
= 2.1 + j1.3 Ω.
2 − j1
.·. ZL = 2.1 − j1.3 Ω.
V2
g21 ZL
=
;
Vg
(1 + g11 Zg )(g22 + ZL ) − g12 g21 Zg
1
1
g21 ZL =
(2.1 − j1.3) = 0.4 − j1.7;
−j
2
2
1 + g11 Zg = 1 + 1 − j1 = 2 − j1;
g22 + ZL = 1.5 + j2.5 + 2.1 − j1.3 = 3.6 + j1.2;
g12 g21 Zg = j3;
0.4 − j1.7
0.4 − j1.7
V2
=
=
;
Vg
(2 − j1)(3.6 + j1.2) − j3
8.4 − j4.2
V2 =
0.4 − j1.7
(42/0◦ ) = 5 − j6 V(rms) = 7.81/ − 50.19◦ V(rms).
8.4 − j4.2
The rms value of V2 is 7.81 V.
Problems
[b] I2 =
−V2
−5 + j6
=
= −3 + j1 A(rms).
ZL
2.1 − j1.3
P = |I2 |2 (2.1) = 21 W.
[c]
−g21
I2
=
;
I1
g11 ZL + ∆g
∆g =
=
1
1
−j
6
6
3
5
1
1
+j
−
−j
2
2
2
2
1
1
− +j
2
2
3
5
3
5
1
2
1
+j −j +
−j = −j .
12
12
12 12
2
3
3
1
1
0.8
3.4
g11 ZL =
−j
(2.1 − j1.3) =
−j
.
6
6
6
6
0.8
3.4 4
2
.·. g11 ZL + ∆g =
−j
+ − j = 0.8 − j0.9;
6
6
6
6
I2
−[(1/2) − j(1/2)]
=
.
I1
0.8 − j0.9
(0.8 − j0.9)I2
.·. I1 =
=
−0.5 + j0.5
!
1.6 − j1.8
I2
−1 + j1
= (−1.7 + j0.1)(−3 + j1) = 5 − j2 A(rms.)
.·. Pg (developed) = (42)(5) = 210 W.
% delivered =
21
(100) = 10%.
210
P 18.33 [a] The circuit is symmetric so y11 = y22 and y12 = y21 . Also,
y11 =
I1
V1 V2 =0
and y21 =
I2
.
V1 V2 =0
The circuit when V2 = 0 is
V1
1
s(s2 + 2)
= s + (sk1/s) = s +
= 2
;
I1
s + (1/s)
s +1
.·.
I1
s2 + 1
y11 =
=
= y22 .
V1 V2 =0 s(s2 + 2)
18–27
18–28
[b]
CHAPTER 18. Two-Port Circuits
I2 = −
s/(s2 + 1)
−1
−1 (s2 + 1)
I1 = 2
I1 = 2
V1 ;
s
s +1
(s + 1) s(s2 + 2)
.·.
y21 =
I2
−1
=
= y12 .
V1 V2 =0 s(s2 + 2)
V2
y21 ZL
=
Vg
y12 y21 Zg ZL − (1 + y11 Zg )(1 + y22 ZL )
=
−1/s(s2 + 2)
[1/s(s2 + 2)]2 − [1 + (s2 + 1)/(s(s2 + 2))]2
=
−s(s2 + 2)
1 − s2 (s2 + 2)2 − 2(s2 + 1)s(s2 + 2) − (s2 + 1)2
s(s2 + 2)
s6 + 2s5 + 5s4 + 6s3 + 6s2 + 4s
1
= 3
2
s + 2s + 3s + 2
1
;
(s + 1)(s2 + s + 2)
=
200
.
s(s + 1)(s2 + s + 2)
√
1
7
s1,2 = − ± j
;
2
2
K2
K3
K3∗
K1
√ +
√ ;
+
+
V2 =
s
s+1 s+ 1 −j 7
s+ 1 +j 7
.·. V2 =
2
K1 = 100;
K2 = −100;
2
2
2
K3 = 37.8/90◦ ;
.·. v2 (t) = [100 − 100e−t + 75.6e−0.5t cos(1.32t + 90◦ )]u(t) V.
CHECK
v2 (0) = 100 − 100 + 75.6 cos 90◦ = 0;
v2 (∞) = 100 + 0 + 0 = 100 V.
P 18.34 [a] h11 =
V1
;
I1 V2 =0
h11 =
h21 =
I2
.
I1 V2 =0
(1/sC)(sL)
(1/C)s
= 2
;
(1/sC) + sL
s + (1/LC)
Problems
I2 = −Ia ;
Ia =
I1 (1/sC)
;
sL + (1/sC)
−I1
;
2
s LC + 1
I2 =
h21 =
I2
−(1/LC)
= 2
;
I1
s + (1/LC)
h12 =
V1
;
V2 I1 =0
V1 =
V2
V2 (1/sC)
= 2
;
sL + (1/sC)
s LC + 1
h22 =
I2
.
V2 I1 =0
1/LC
V1
= h12 = 2
;
V2
s + (1/LC)
V2
(1/sC)[sL + (1/sC)]
s2 + (1/LC)
=
=
;
I2
sL + (2/sC)
sC[s2 + (2/LC)]
I2
Cs[s2 + (2/LC)]
= h22 =
.
V2
s2 + (1/LC)
[b]
1
1
=
= 25 × 106 ;
LC
(0.4)(100 × 10−9 )
h11 =
107 s
;
s2 + 25 × 106
h12 =
25 × 106
;
s2 + 25 × 106
h21 =
−25 × 106
;
s2 + 25 × 106
h22 =
10−7 s(s2 + 50 × 106 )
.
(s2 + 25 × 106 )
6
25×10
800
V2
−h21 ZL
−h21 ZL
s2 +25×106
=
=
=
;
7s
10
V1
h11 + ∆hZL
h11 + ZL
+
800
2
6
(s +25×10 )
=
25 × 106
25 × 106
=
;
s2 + 12,500s + 25 × 106
(s + 2500)(s + 10,000)
V1 =
45
;
s
V2 =
1125 × 106
45
60
15
=
−
+
;
s(s + 2500)(s + 10,000)
s
s + 2500 s + 10,000
v2 = [45 − 60e−2500t + 15e−10,000t ]u(t) V.
18–29
18–30
CHAPTER 18. Two-Port Circuits
P 18.35 I2 =
=
y21 Vg
1 + y22 ZL + y11 Zg + ∆yZg ZL
−0.25(1)
1 + (−0.04)(100) + (0.025)(10) + (−0.00125)(10)(100)
= 0.0625 A(rms).
Po = (I2 )2 ZL = (0.0625)2 (100) = 390.625 mW.
I2
y21
−0.25
=
=
= 2.5;
I1
y11 + ∆yZL
0.025 + (−0.00125)(100)
0.0625
I2
.·. I1 =
=
= 25 mA(rms).
2.5
2.5
Pg = (1)(0.025) = 25 mW (developed).
Po
390.625
=
= 15.625.
Pg
25
P 18.36 [a] a11 =
V1
;
V2 I2 =0
a21 =
V2 = −j52I1 = −j52
I1
.
V2 I2 =0
V1
;
20 + j20
a11 =
V1
20 + j20
5
=
= (−1 + j);
V2
−j52
13
a21 =
I1
1
j
=
=
S;
V2
−j52
52
a12 = −
V1
;
I2 V2 =0
a22 = −
V1 = (20 + j20)I1 − j52I2 ;
I1
.
I2 V2 =0
Problems
0 = −j52I1 + (160 + j320)I2 ;
∆=
20 + j20
−j52
−j52
160 + j320
20 + j20 V1
N2 =
I2 =
−j52
= −496 + j9600;
= j52V1 ;
0
j52V1
−496 + j9600
so
−496 + j9600
V1
1
=
= (9600 + j496);
I2
j52
52
V1
1
.·. a12 = − = (−2400 − j124) Ω.
I2
13
j52I1 = (160 + j320)I2 ;
−320 + j160
I1
.
.·. a22 = − =
I2
52
Summary:
a11 =
5
(−1 + j);
13
a12 =
a21 =
j
S;
52
−320 + j160
.
52
[b] VTh =
a22 =
1
(−2400 − j124) Ω;
13
Vg
100/0◦
=
a11 + a21 Zg
(5/13)(−1 + j) + (j/52)(10)
= −80 − j120 = 144.22/ − 123.69◦ V;
ZTh =
1
(−2400 − j124) + −320+j160
(10)
a12 + a22 Zg
52
= 13
a11 + a21 Zg
(5/13)(−1 + j) + (j/52)(10)
= 222.4 + j278.4 = 356.33/51.38◦ Ω.
[c] V2 =
144.22/ − 123.69◦
(400) = 84.607/ − 147.789◦ ;
622.4 + j278.4
v2 (t) = 84.607 cos(2000t − 147.789◦ ) V.
18–31
18–32
CHAPTER 18. Two-Port Circuits
P 18.37 z11 =
V1
100
800
=
=
Ω;
I1 I2 =0 1.125
9
z21 =
V2
104
832
=
=
Ω;
I1 I2 =0 1.125
9
z12 =
V1
20
=
= 80 Ω;
I2 I1 =0 0.25
z22 =
V2
24
=
= 96 Ω.
I2 I1 =0 0.25
ZTh = z22 −
z12 z21
(80)(832/9)
= 96 −
= 12.8 Ω;
z11 + Zg
(800/9) + 0
.·. ZL = 12.8 Ω.
z21 ZL
V2
=
;
V1
z11 ZL + ∆z
800
10,240
832
∆z =
96 − 80
=
;
9
9
9
V2
(832/9)(12.8)
10,649.60
=
=
= 0.52;
V1
(800/9)(12.8) + (10,240/9)
20,480
V2 = (0.52)(160) = 83.20 V.
P =
(83.2)2
= 540.80 W.
12.8
P 18.38 [a] Derive the a parameters of the bridged-tee circuit. When I2 = 0,
30Ia − 15Ib = 1;
−15Ia + 45Ib = 0.
Problems
Ia = I1 = 0.04;
a11 =
Ib = 0.0133;
V1
1
=
= 1.25;
V2 I2 =0 0.8
18–33
V2 = 15Ib + 15Ia = 0.8.
a21 =
I1
0.04
=
= 0.05 S.
V2 I2 =0 0.8
When V2 = 0,
30Ix − 15Iy − 15Iz = 1;
−15Ix + 45Iy − 15Iz = 0;
−15Ix − 15Iy + 30Iz = 0.
I1 = Ix = 0.111;
a12 = −
I2 = −Iz = −0.0888.
V1
1
= 11.25 Ω;
=−
I2 V2 =0
−0.0888
a22 = −
I1
0.111
= 1.25.
=−
I2 V2 =0
−0.0888
The a-parameters of the cascaded networks are
a11 = (1.25)2 + (11.25)(0.05) = 2.125;
a12 = (1.25)(11.25) + (11.25)(1.25) = 28.125 Ω;
a21 = (0.05)(1.25) + (1.25)(0.05) = 0.125 S;
a22 = a11 = 2.125,
100
= 32 V;
3.125
162
[c] P =
= 17.73 W.
14.44
[b] VTh =
RTh = (45.125/3.125) = 14.44 Ω.
therefore V2 = 16 V.
P 18.39 The a parameters of the first two port are
a011 =
z11
200
=
= −125 × 10−6 ;
z21
−1.6 × 106
a012 =
∆z
40 × 106
=
= −25 Ω;
z21
−1.6 × 106
18–34
CHAPTER 18. Two-Port Circuits
a021 =
1
1
=
= −625 × 10−9 S;
6
z21
−1.6 × 10
a022 =
z22
40,000
=
= −25 × 10−3 .
z21
−1.6 × 106
Now derive the a parameters of the second two port. When I2 = 0,
16,000Ia − 8000Ib = 1;
Ia = I1 = 75 × 10−6 ;
a11 =
−8000Ia + 24,000Ib = 0.
Ib = 25 × 10−6 ;
1
V1
=
= 1.25;
V2 I2 =0 0.8
a21 =
When V2 = 0,
16,000Ix − 8000Iy − 8000Iz = 1;
−8000Ix + 24,000Iy − 8000Iz = 0;
−8000Ix − 8000Iy + 16,000Iz = 0.
V2 = 8000Ib + 8000Ia = 0.8.
I1
75 × 10−6
=
= 93.75 µS.
V2 I2 =0
0.8
Problems
I1 = Ix = 208.33 × 10−6 ;
I2 = −Iz = −166.667 × 10−6 .
a12 = −
1
V1
= 6 kΩ;
=−
I2 V2 =0
−166.667 × 10−6
a22 = −
I1
208.33 × 10−6
= 1.25.
=−
I2 V2 =0
−166.667 × 10−6
a0011 = 1.25;
a0012 = 6 kΩ;
18–35
a0021 = 93.75 µS;
a0022 = 1.25.
The a parameters of the cascade connection are
a11 = −125 × 10−6 (1.25) + (−25)(93.75 × 10−6 ) = −2.5 × 10−3 ;
a12 = −125 × 10−6 (6000) + (−25)(1.25) = −32 Ω;
a21 = −625 × 10−9 (1.25) + (−25 × 10−3 )(93.75 × 10−6 ) = −3.125 × 10−6 S;
a22 = −625 × 10−9 (6000) + (−25 × 10−3 )(1.25) = −35 × 10−3 .
ZL
Vo
=
;
Vg
(a11 + a21 Zg )ZL + a12 + a22 Zg
a21 Zg = (−3.125 × 10−6 )(500) = −1.5625 × 10−3 ;
a11 + a21 Zg = −2.5 × 10−3 − 1.5625 × 10−3 = −4.0625 × 10−3 ;
(a11 + a21 Zg )ZL = (−4.0625 × 10−3 )(8000) = −32.5;
a22 Zg = (−35 × 10−3 )(500) = −17.5;
Vo
8000
=
= −97.56;
Vg
−32.5 − 32 − 17.5
vo = Vo = −97.56Vg = −1.46 V.
18–36
CHAPTER 18. Two-Port Circuits
P 18.40 [a] From reciprocity and symmetry
a011 = a022 ,
∆a0 = 1;
.·. 16 − 5a021 = 1,
a021 = 3 S.
For network B
a0011 =
V1
;
V2 I2 =0
V1 = (6 + j12 + 3)I1 = (9 + j12)I1 ;
V2 = 3I1 + j4I1 = (3 + j4)I1 ;
a0011 =
9 + j12
= 3;
3 + j4
a0021 =
I1
1
=
= 0.12 − j0.16 S;
V2 I2 =0 3 + j4
a0022 = a0011 = 3.
∆a00 = 1 = (3)(3) − (0.12 − j0.16)a0012 ;
.·. a0012 =
8
= 24 + j32 Ω.
0.12 − j0.16
[b] a11 = a011 a0011 + a012 a0021 = 12 + 5(0.12 − j0.16) = 12.6 − j0.8;
a12 = a011 a0012 + a012 a0022 = (4)(24 + j32) + (5)(3) = 111 + j128 Ω;
a21 = a021 a0011 + a022 a0021 = (3)(3) + (4)(0.12 − j0.16) = 9.48 − j0.64 S;
a22 = a021 a0012 + a022 a0022 = (3)(24 + j32) + (4)(3) = 84 + j96.
1
1
V2
=
=
= 0.079 + j0.005.
V1 I2 =0 a11
12.6 − j0.8
P 18.41 [a] V1 = I2 (z12 − z21 ) + I1 (z11 − z21 ) + z21 (I1 + I2 )
= I2 z12 − I2 z21 + I1 z11 − I1 z21 + z21 I1 + z21 I2 = z11 I1 + z12 I2 .
V2 = I2 (z22 − z21 ) + z21 (I1 + I2 ) = z21 I1 + z22 I2 .
Problems
18–37
[b] Short circuit Vg and apply a test current source to port 2 as shown. Note
that IT = I2 . We have
V + IT (z12 − z21 )
V
− IT +
= 0.
z21
Zg + z11 − z21
Therefore
"
#
z21 (Zg + z11 − z12 )
V =
IT
Zg + z11
and VT = V + IT (z22 − z21 ).
!
Thus
z12 z21
VT
= ZTh = z22 −
.
IT
Zg + z11
For VTh note that Voc =
z21
Vg since I2 = 0.
Zg + z11
P 18.42 [a] V1 = (z11 − z12 )I1 + z12 (I1 + I2 ) = z11 I1 + z12 I2 ;
V2 = (z21 − z12 )I1 + (z22 − z12 )I2 + z12 (I2 + I1 ) = z21 I1 + z22 I2 .
[b] With port 2 terminated in an impedance ZL , the two mesh equations are
V1 = (z11 − z12 )I1 + z12 (I1 + I2 );
0 = ZL I2 + (z21 − z12 )I1 + (z22 − z12 )I2 + z12 (I1 + I2 ).
Solving for I1 :
I1 =
V1 (z22 + ZL )
.
z11 (ZL + z22 ) − z12 z21
Therefore
V1
z12 z21
Zin =
= z11 −
.
I1
z22 + ZL
P 18.43 [a] At the input port: V1 = h11 I1 + h12 V2 ;
At the output port: I2 = h21 I1 + h22 V2 .
18–38
CHAPTER 18. Two-Port Circuits
[b]
V2
+ (100 × 10−6 V2 ) + 100I1 = 0,
104
therefore I1 = −2 × 10−6 V2 .
V20 = 1000I1 + 15 × 10−4 V2 = −5 × 10−4 V2 ;
100I10 + 10−4 V20 + (−2 × 10−6 )V2 = 0,
therefore I10 = 205 × 10−10 V2 .
Vg = 1500I10 + 15 × 10−4 V20 = 3000 × 10−8 V2 ;
105
V2
=
= 33,333.
Vg
3
P 18.44 [a] I1 = y11 V1 + y21 V2 + (y12 − y21 )V2 ;
I1 = y11 V1 + y12 V2 ;
I2 = y21 V1 + y22 V2 .
I2 = y12 V1 + y22 V2 + (y21 − y12 )V1 .
Problems
18–39
[b] Using the second circuit derived in part [a], we have
where ya = (y11 + y12 ) and yb = (y22 + y12 ).
At the input port we have
I1 = ya V1 − y12 (V1 − V2 ) = y11 V1 + y12 V2 .
At the output port we have
V2
+ (y21 − y12 )V1 + yb V2 − y12 (V2 − V1 ) = 0.
ZL
Solving for V1 gives
!
V1 =
1 + y22 ZL
V2 .
−y21 ZL
Substituting V1 into the equation for I1 and at the same time using
V2 = −ZL I2 , we get
y21
I2
=
.
I1
y11 + ∆yZL
P 18.45 [a] The g-parameter equations are I1 = g11 V1 + g12 I2 and V2 = g21 V1 + g22 I2 .
These equations are satisfied by the following circuit:
[b] Replace the two-port network described by g parameters with its
equivalent circuit from part (a) and replace the two-port network
described by h parameters with its equivalent circuit from Fig. P18.43.
Attach the source and the load to get the circuit shown here:
18–40
CHAPTER 18. Two-Port Circuits
Note from the mesh in the middle of the circuit that I10 = Ix and that
I2 = −Ix .
Write a KCL equation at the node labeled V1 :
V1 − 30
V1
20
+
+ (−Ix ) = 0.
10
35/3
7
Write a KVL equation for the mesh whose current is Ix :
50,000
800
+ 5000 Ix − 0.2V20 −
V1 = 0.
7
7
Write a KCL equation at the node labeled V20 :
−4Ix + 200 × 10−6 V20 +
V20
= 0.
15,000
Solving these three equations, we get
V1 = 20 V;
Ix = 0.25 A;
V20 = 3750 V.
Thus, the voltage across the load is 3750 V, which matches the solution
to Assessment Problem 18.7.
P 18.46 [a] To determine a11 and a21 create an open circuit at port 2. Apply a voltage
at port 1 and measure the voltage at port 2 and the current at port 1. To
determine a12 and a22 create a short circuit at port 2. Apply a voltage at
port 1 and measure the currents at ports 1 and 2.
[b] The equivalent a-parameters for the black-box amplifier can be calculated
as follows:
−∆h
−23.5
a11 =
=
= −0.01567;
h21
1500
a12 = −
h11
−500
=
= −0.333 Ω;
h21
1500
a21 = −
h22
−0.05
=
= −33.33 µS;
h21
1500
a22 =
−1
−1
=
.
h21
1500
Problems
18–41
Create an open circuit at port 2. Apply 1 V at port 1. Then,
a11 =
V1
1
= −0.01567 so V2 = −63.83 V measured;
=
V2 I2 =0 V2
a21 =
I1
I1
= −33.33 × 10−6
=
V2 I2 =0 −63.83
so I1 = 2.13 mA measured.
Create a short circuit at port 2. Apply 1 V at port 1. Then,
a12 = −
V1
−1
−1
=
=
I2 V2 =0 I2
3
a22 = −
I1
−1
−I1
=
=
I2 V2 =0
3
1500
so I2 = 3 A measured;
so I1 = 2 mA measured.
P 18.47 [a] To determine z11 and z21 create an open circuit at port 2. Apply a current
at port 1 and measure the voltages at ports 1 and 2. To determine z12
and z22 create an open circuit at port 1. Apply a current at port 2 and
measure the voltages at ports 1 and 2.
[b] The equivalent z-parameters for the black-box amplifier can be calculated
as follows:
∆h
23.5
z11 =
=
= 470 Ω;
h22
0.05
10−3
h12
=
= 20 mΩ;
z12 =
h22
0.05
z21 =
−h21
−1500
=
= −30 kΩ;
h22
0.05
z22 =
1
1
=
= 20 Ω.
h22
0.05
Create an open circuit at port 1. Apply 1 A at port 2. Then,
z11 =
V1
V1
=
= 470 so V1 = 470 V measured;
I1 I2 =0 1
z21 =
V2
V2
=
= −30,000 so V2 = −30 kV measured.
I1 I2 =0 1
Create an open circuit at port 1. Apply 1 A at port 2. Then,
z12 =
V1
V1
=
= 0.02 µS so V1 = 20 mV measured;
I2 I1 =0 1
z22 =
V2
V2
=
= 20 so V2 = 20 V measured.
I2 I1 =0 1