Electric Circuits, 9th Edition P80 pdf

766
Answers to Selected Problems
2.34 a) i = 385 mA,so a warning sign should be
posted and precautions taken.
b) Use the following resistors: 390 ft, 47
(1,
and 220 ft.
2.35
2.36 a)
P
arm
= 59.17 W;/>
leg
= 29.59 W;
P
tmnk
= 7.40 W
b) ^m = 1414.23 s;
r
leg
= 7071.13 s;
'trunk = 70.677.37 S
c) All values are much greater than a few minutes.
2.37 a) 40 V
b) No, 12 V/800 ft = 15 mA will cause a shock.
2.38 3000 V
Chapter 3
3.1 a) 6 kfl and 12 kft, 9 kfl and 7 kft; simplified
circuit is
10 V
b) 3 kft, 5 kft, and 7 kft; simplified circuit is

3 mA
c) 300 ft, 400 ft and 500 ft; simplified circuit is
1200 Q
200 mV
3.2 a) 10 ft and 40 ft, 10011 and 25 ft; simplified
circuit is
60 V
b) 9 kfl and 18 kft, and 6 kfl; simplified circuit is
50 mA
c) 600 ft, 200 ft, and 300 ft; simplified circuit is
250 fl
0.2 A
3.3
3.4
3.15
3.16
3.18
3.23
3.24
a)
b)
c)
a)
b)
c)
a)
b)
c)
a)
b)

21.2 ft
10 kft
1600 ft
30 ft
5 kft
80
fl
66 V
1.88
W,
1.32 W
17,672 ft, 12,408 ft
1200
fl,
300 ft
1 W
1875 ft, 3750ft, 7.5 kf
a)
b)
c)
d)
e)
a)
b
)
c)
d)
R
1
)

36
V
2A
0.96 A
24 V
6.4 V
25 mA
250 V
50 V
25 mA
4.167 mA
Answers to Selected Problems 767
3.31 a) 49,98012
b) 498012
c) 23012
d) 5ft
3.35 a) R,„ = 50fl;
(25/12)
3.53
3.56
3.58
3.72
3.73
50 + (25/12)
(k
25
b)
2500
mcas
c) Yes

3.51 a)
1500
(1
b) 28.8 mA
c) 750 a, 276.48 mW
d) 100012,92.16 mW
23.2
V,
21
V
a) A-connected R2—R3—R4 becomes Y-connected
512—2012—4ft; equivalent resistance is 3312.
b) Y-connected R
2
—R4—R>, becomes A-connected
10012 —8012—20 ft; equivalent resistance
is 3312.
c) Convert the delta connection
R
4
—R
s
—Rf,
to
its equivalent wye. Convert the wye connection
Ri—Rj
— Rh
to its equivalent delta.
9012
R

{
=
1.0372ft,
R
2
=
1.1435
ft,/?
3
=
1.212,
R
4
=
1.1435
ft, R
5
=
1.0372
ft, R
a
= 0.0259 ft,
R
b
= 0.006812, R
c
= 0.006812 R
L{
= 0.025912
P

diss
= 624W = P
dd
3.74 a) R
]
= 0.426912, R
2
= 0.4617 ft, R
3
= 0.48 ft,
R
4
= 0.4617ft, R
5
= 0.4269 ft, R
a
= 0.0085 ft,
R
h
= 0.0022 ft, rt
c
= 0.002212, R
d
= 0.008512
b) i, = 26.51 A,
/?/?,
= 300 W or 200 W/m;
i
2
= 25.49 A, i\R

2
= 300 W or 200 W/m;
i
b
= 52 A, ilR
h
= 6 W or 200 W/m;
P
dd
= 1548 W =P
{Sss
Chapter
4.1
a)
b)
c)
d)
e)
f)
g)
4
11
10
9
s
6
4
6
4.2 a) 8
b) 3

c) 4
d) Avoid the topmost mesh and the leftmost
mesh, which both contain current sources.
4.3 a) 2
b) 5
c) 7
d) 1,4,7
4.4
4.8
4.9
4.13
4.17
4.19
4.24
4.26
4.27
4.33
4.34
4.38
4.39
4.42
4.44
4.48
a) 5
b) 3
c) -i
s
+ ij + U = 0; —/1 + /4 + /3 = 0;
/5
—

/
2
—
/3 = 0
d) 2
e) /?,/! +
tf
3
/
3
- R
2
i
2
= 0;
/?3*
3
+ R
5
i
5
- R4I4 = 0
120
V,
96 V
4V
a) -6.8 A, 2.7 A, -9.5 A, 2.5 A, -12 A
b) 3840W
a) 8800W
b) 8800W

750 W
3.2 V
a) -37.5
V,
75 W
b) -37.5
V,
75 W
c) Part (b), because there are fewer equations to
write and solve.
-20 V
a) 5.6 A, 3.2 A,-2.4 A
b) -8.8 A,-1.6 A, 7.2 A
a) 17,940 W
b) 17,940 W
259.2 W
2700 W
a) 162.92 W
b) 518.52 W
c) Power delivered equals power absorbed.
a) 2 raA
b) 304 mW
c) 0.9 mW
740 W
768
Answers to Selected Problems
4.51
a)
5.7 A, 4.6 A, 0.97 A,-1.1 A, 3.63 A
b) 5>dev= £/^

=
1319.685W
4.52
a)
The constraint equations are easier to formu-
late in the node voltage method, making it the
preferred method.
b) 480 mW
4.53
a)
Minimize the number of equations to write and
solve by using the mesh current method.
b) 4mW
c) No, since the mesh current method still minimizes
the number of equations to write and solve.
d) 200 mW
4.59
a)
-1mA
b) -1 mA
4.60
a)
-0.85 A
b) -0.85 A
4.63 60 V source, positive at the top, in series with a
1011 resistor
4.64 1 mA current source, with current flowing from top
to bottom, in parallel with a 3.75 O resistor
4.71
a)

51.3 V
b) -5%
4.74 160 V source, positive at the bottom, in series with
a 56.4 kft resistor
4.77
8
ft
(The voltage source is zero because there are
no independent sources in the circuit.)
4.83 2.5 H and 22.5 O
4.87
a) 6ft
b) 24 W
4.91
a)
50 V
b) 250W
4.96 30 V
4.105
v
{
= 39.583 V,v
2
= 102.5 V
4.106 t»i
=
37.5 V,v
2
=
105

V
4.107 V] = 52.0833
V,i>
2
=
117.5 V
Chapter
5
5.1
a)
inverting
input
non-inverting
input
positive
power supply
output
negative
power supply
b) The input resistance; i
n
= 0
c) The open-loop voltage gain; (v
p
-
v
n
)
= 0
d)

^ =
9V
5.2 -1 mA
5.3
a)
-15 V (saturates)
b) -10 V
c)
-4V
d)
7V
e) 15 V (saturates)
f) -1.08 V
<
v
a
<
4.92 V
5.8
a)
Many possible designs; one uses a single 3.3 kft
input resistor and three series-connected
3.3 kft resistors in the feedback path.
b) ±15 V
5.9
a)
0
<
or
<

0.40
b) 556.25 At A
5.11
0<i?
f
<60kft
5.12
a)
Inverting summing amplifier
b) -6V
c) -0.5V<v„<2V
5.14
a)
14 V
b) 3.818 V
<
v
a
<
9.273 V
5.17
a)
Non-inverting amplifier
b)
2v
s
c) -6V
< v
s
<

4 V
5.18
a)
10.54 V
5.25
5.26
5.28
5.33
5.34
5.43
5.45
b) -4.55 V
<
v
g
<
4.55 V
c) 181.76 kft
a) -15.1V
b) 34.3 kft
c) 250 kft
20 kft
a)
16
V
b) -4.2 V
< v
b
<3.8V
19.93 kft

< R
x
<
20.07 kft
a) 24.98
b) -0.04
c) 624.5
a) -19.9844
b) 736.1 jaV
c) 5003.68
ft
d) -20,0 V, 5000
ft
a) 13.49
b) 999.446 mV, 999.834 mV
c)
387.78/AV
d) 692.47 pA
e) 13.5,1V,0V,0A
Answers to Selected Problems 769
5.49 a) 2kfl
b) 12 mfl
Chapter 6
6.1 »
f
(mV)
e) v(V)
6.4
4
3

2
1
0
(
-1
-2
-3
-4
—
)
1
2
l
/*""~3
l
4
1.4
1.2
1 F
0.8
0.6
0.4 -
0.2 -
0
0 1
a) / = 0
i
=
50/ A
/ = 0.5 - 50/A

/ = 0
b) v = 0
v = lV
v = -\V
v = 0
p = 0
p = 50/ W
/? = 50/ - 0.5 W
/7 = 0
w = 0
w = 25/
2
J
/(s)
/(s)
4
/ < 0
0 < / < 5 ms
5 < / < 10 ms
10 ms < /
/ < 0
0 < / < 5 ms
5 < / < 10 ms
10 ms < /
/ < 0
0 < / < 5 ms
5 < / < 10 ms
10 ms < /
/ < 0
0 < / < 5 ms

u>
= 25/
2
- 0.5/ +
0.0025
J 5 < / < 10 ms
to = 0 10 ms < /
6.16 a) -50 X 10
4
/ +
15
V
b) 10
6
/V
c) 1.6 X 10
6
/ -
12
V
d)
52
V
6.17
/(jxs)
t (ms)
6.21 8 H
6.25 a) -5e~
Al
A

b) -4e~
4?
- 6 A
c) -e
_4i
+ 6 A
d) 40 J
e) 400 J
f) 360 J
9) 1(4)(-6)
2
+ j(16)(6)
2
= 360 J (checks)
6.26
2
^cF,
initial voltage is 25 V
6.31 a) -20e"
25
'V
b) -16<T
25
' + 21 V
c) -4<T
25/
- 21 V
d) 320/AJ
e) 2525/AJ
f) 2205/xJ

g) \(2 X 10"
6
)(21)
2
+ |(8 X 1()"
6
)(-21)
2
=
2205 /xJ
6.39 a) 16 ^ + 32/
2
= 2-±
at at
b) -16<T
f
+ 32e
_2f
+ 32e
-
' - 32e~
2
' = 16e"
c) 34<T' - 4e~
2t
V
d) 30
V,
which is consistent with the circuit's
behavior

6.44 a) 2721.6 mJ
b) 2721.6 mJ
c) 518.4 mJ
d) 518.4 mJ
770 Answers to Selected Problems
6.45 a) -4.5 A
b) No
6.48 a) 50mH,2.4
b) 0.2 X 10~
6
Wb/A,0.2 X 10
_6
Wb/A
6.49 0.8 nWb/A, 1.2 nWb/A
6.51 v(t) = 0333v
s
(t)
6.53 There is no difference in the output voltage for
these two circuits.
Chapter
7.4
7.5
7.7
7.23
7.26
7.35
7.36
a)
b)
c)

d)
e)
a)
b)
c)
d)
e)
f)
9)
h)
i)
i)
k)
I)
a)
b)
c)
d)
a)
b)
c)
a)
b)
c)
a)
b)
a)
b)
c)
d)

7
5 mA, 15 mA
5 mA, -5 mA
5,,-20.000/
mA
-5e-
20
-
,)00
'mA
The current in a resistor can change
instantaneously.
0A
160mA
65 mA
160 mA
225 mA
0A
160e"
2n
"' mA
OV
-3.2 V
OV
-32e~
2m
V
225 - 160e"
200
' mA

2A
20 ms
2e~
5i)t
A, -160e~
5lM
V, -144<T
50
' V
69.92%
1.6e~
5()f
mA,32<r
5()
'
+ 8V, -8<T
50
' + 8V
800/d
160/xJ,640/J
24«-«"
raA
-
Se
-5om
+ 80V
2880/d
-2-3e-
500,,
'A,48-48e-

5,K,(,
'V
60V,0V
-5 mA
0.333
mA
5/AS
0.333
-5.333<r
2(,a(,0,)
'mA
7.37
7.51
7.53
7.55
7.56
7.68
7.69
7.71
7.78
7.85
7.87
7.95
7.96
7.103
7.104
7.105
a)
b)
a)

b)
c)
d)
e)
5 + 15e-
1000
'A
50-450<T
1(loa,
V
60 - 60e~
im
V
1 -
0.6e-
m)l
mA
1 + 2Ae~
m)t
mA
4 - 2Ae~
m
' mA
3.4 mA
-60 + 90e~
20,)0
' V
a)
b)
c)

d)
e)
f)
a)
b)
c)
d)
a)
b)
c)
d)
e)
a)
b)
c)
d)
e)
50 V
-24 V
0.1
fis
-18.5 A
-24 + 74<T
lt,7
'V
-18.5e-
1()7
'A
90 V
-60 V

1000/AS
916.3/AS
4-4<T
20
'A
80e"
20f
V
2.4 - 2.4e~
20
' A
1.6 -
1.6<T
20
'mA
Yes
40 - 40e"
500(
" mA
10e
-50oo,
v
16 - \6e'
5m)t
mA
24 - 24e"
5000
' mA
Yes
-5.013 V

-5V, 0 < t < 5 s; -5£>-
(U(
'"
5)
V, 5 s < t <
83.09 ms
a)
b)
2.25
272.1
/AS
-1600/
+ 8V,-15 + lle-
200
'V,
23
80
a)
b)
a)
b)
a)
b)
c)
-
1600* - lle"
200
'V
ms
1.091 MO

0.29 s
8.55 flashes/min
559.3 kH
24.32 flashes/min
99.06 mA
$43.39 per year
CO
Answers
to Selected Problems 771
Chapter
8
8.1 a) -10,000 rad/s,-40,000 rad/s
b) overdamped
c) 3125 ft
d) -16,000 +/12,000 rad/s,
-16,000 - /12,000 rad/s
e) 2500ft
8.7 a) 25 nF
b) 2500ft
c) 75 V
d) 30 mA
8.8
8.9
8.18
8.19
8.20
8.29
8.30
8.31
8.50

e)
-8000,
(30 cos 6000? + 71.25 sin 6000?) mA
a) 8 kft, 40 H, 625 rad/s, 500 rad/s
b)
-U-
25()l
+ 4e~
imt
mA,
O.&T
250
* -0.8<T
10()
"
f
mA,
02e-
25{)[
- 32e-
xmt
mA
a) lkft,l/xF,6000V/s,8V
b) (-3000? + 2)^
5(lll,
mA
8.11 a) 500 rad/s, 400 rad/s,
1.5625
H,4/xF,
-15 mA, 60 mA

b) 18.75e"
200
' - 18.75<r
8()0
'V
c) 75<T
200
' -
75^
800
'
mA
800/
d) -60e-
200
' + 15(T
ftUUf
mA
5e~
2im
+ \Qe-*
m)
<V
\5e~
25m
cos 3122.5? + 721e~
2mi
sin 3122.5? V
15e-
4m]

'V
60 - 120c-™ + 15e-
2iumi
mA
60 - 105e"
W)00
' cos 6000? - 90e"
800<)r
sin 6000? mA
60 - 750,000?e"
104
' - 105eT
104
' mA
60-80ff-
8n0r
+ 2Off-
32Wf
V
8.51 60 - 120,000?e
-2000(
-
60e
2000(
V
8.52 60 - 60e>-
200
(" cos 1500? - 80<r
20(K)/
sin 1500? V

8.63 a) 0 <
?
< 0.5" s:
v
n
(t) = l0t
2
V,v
lA
(t) = -1.6? V;
0.5
+
< ? < ?
sal
:
v
0
(t) = -5?
2
+ 15? - 3.75 V,
v
ol
(t) = 0.8? - 1.2 V
b) 3.5 s
8.64 0 < ? < 0.5
~
s:
v
a
{t) = 10 - 20e"' + Hfe~*V,

v
oi
(t) = -0.8 + 0.8e~
a
V,
0.5
+
< ? < ?
sal
:
Vo
{t) = -5 +
19.42ff"
(t-a5)
- 12.87e
_2(f
-
a5)
V,
v
ol
(0 = 0.4 - 0.91e-
2(
'
_05)
V
8.68 a) 55.23fis
b) 262.42 V
c) ?
max

=
53.63
/xs,
v(t
max
) = 262.15 V
8.69 a) 40 mJ
b) -27,808.04 V
c) 568.15 V
Chapter
9
9.1 a)
80
V
b) 500 Hz
c) 3141.59 rad/s
d) -0.5236 rad
e) -30°
f) 2 ms
g) 166.67 ^s
h) -80sinl0007r?V
i)
333.33/AS
j) 166.67jiis
9.4 a) 600 Hz
b) 1.67 ms
c) 10V
d) 6 V
e) -53.13°,-0.9273 rad
f) 662.64/AS

g) 245.97^s
2
a) -195.72<r
l066
-
67f
+ 200cos(800? - 11.87°) mA
b) -195.72^-
1066
-
67/
mA,
200 cos(800? - 11.87°) mA
c) 28.39 mA
d) 200 mA, 800 rad/s, -11.87°
e) 36.87°
9.11 a) 111.8 cos(500? - 3.43°)
b) 102.99 cos(377? + 40.29°)
c) 161.59 cos(100? - 29.96°)
d) 0
9.13 a) 502,654.82 rad/s
b) 90°
c) -39.79 ft
d) 0.05/AF
e) -/39.79 ft
9.14 a) 400 Hz
b) -90°
9.8
9.9
772 Answers to Selected Problems

9.85
9.15
9.24
9.28
9.29
9.32
9.37
9.45
9.46
9.49
9.55
9.59
9.60
9.64
9.76
9.77
9.83
9.84
c)
5fi
d) 1.99 mH
e)
/5 ft
a) 40
a
/40
a
AAA
/~W~lT\
j

VVv
600/20° vf J 7-
b) 8.32 /76.31° A
^
-/100 ft
c) 8.32cos(8000f + 76.31°) A
a) 200 /36.87° mS
b) 160 mS
c) 120 mS
d) 10 A
500 rad/s
42.43 cos(50,000f + 45°) V
42.43 cos(2000? + 45°) V
2/3 ft
227.68 / -18.43° V, (3.6 + ./10.8)(1
2 /-36.87° A, (100 - /50) ft
10/-45° A, (1.6 + /3.2) ft
188.43/-42.88° V
/80 = 80 /90° V
36 cos
2000
fV
56.57 cos(10,000f -45°)V
a) 0.3536
b) 2 A
a) 5 cos(5000f - 36.87°) A,
1
cos(5000f - 180°) A
b) 0.5
c) 9mJ,12mJ

512,000/60° ft
I
V
=(V /2)-1 R
o
v in / m v
a) 247.11/1.68° V
b) -32 ft, 241.13/1.90° V
c) -26.90 ft
9.88
a) l
{
= 24 /0° A,
1
2
= 2.04 /0° A,
I
3
= 21.96/0° A, I
4
= 19.40/0° A,
I
5
= 4.6/0° A, I
6
= 2.55/0° A
b) 0.42 /0° A
9.89 a) 0A
b) 0.436/0° A
c) Yes; when the loads are equal, no power is lost

to the neutral line, so the cost of power is lower.
Chapter
10
10.1 a) 409.58 W (abs), 286.79 VAR (abs)
b) 103.53 W (abs), -386.37 VAR (del)
c) -1000W (del),-1732.05 VAR (del)
d) -250 W (del),
433.01
VAR (abs)
10.2 a) Yes
b) Yes
10.15 a) 15.81 V(rms)
b) 62.5 W
10.18 a) 6.4
W,
4.8 VAR,
8
VA
b) 6.4 W
c) 4.8 VAR
10.26 a) 0.96 lagging, 0.28; 0.8 leading,-0.6;
0.6 leading, -0.8
b) 0.74 leading,-0.67
10.27 a) 1.875 + /0.625 ft
b) 0.9487 lagging
10.44 a) 20 + /20 ft
b)
20
W
c) With 22 ft and 1 mH, the load impedance is

22 + /5 ft and the load power is 17.7 W
10.47 a) 360 mW
b)
4000ft,0.1/AF
c) 443.1
mW;
yes
d) 450 mW
e) 4000 ft, 66.67 nF
f) Yes
10.48 a) 4123.1 ft, 0.1/xF, 443.18 mW
b) Yes
c) Yes
10.64 90 W
10.65 a) 10
b) 250W
10.66 a) 28.8 ft
b) 28.8 ft
c) Yes
Answers to Selected Problems 773
10.67 a) P
L
=
PH =
V
1
R\
+ R
2
v\Ri + R

2
)
R1R2
PM =
Yl
Ri
V
2
V
2
Pi.
(Yl _ vL\(¥l\
(p
L
p
¥
)yp,)
PM - PL
b) 1125W
10.68 36ft, 24 ft
Chapter 11
11.2 a) acb
b) abc
11.3 a) Balanced, negative phase sequence
b) Balanced, positive phase sequence
c) Balanced, negative phase sequence
d) Balanced, positive phase sequence
e) Unbalanced, due to unequal amplitudes
f) Unbalanced, due to unequal phase angle
separation

11.7 v
AB
= 13,198.23 cos
a>t
V,
v
BC
= 13,198.23 cos(wr + 120°) V,
v
CA
= 13,198.23 cos(w/ - 120°) V
11.9 a) 15.24 A(rms)
b) 6583.94 V(rms)
11.11 a) I
aA
= 5/-36.87° A, I
b
B = 5/83.13° A,
I
cC
- 5/-156.87° A
b) V
ab
= 216.51 /-30°V,V
bc
= 216.51 /90° V,
V
ca
= 216.51/-150° V
c) V

AN
= 122.23/-1.36° V,
V
BN
= 122.23/118.64° V,
V
CN
= 122.23/-121.36° V
d) V
AB
= 211.72/-31.36° V,
V
BC
= 211.72/88.64° V,
V
CA
= 211.72/-151.36° V
11.12 a)
1+/3H
20/If
V(
150° /+\
rms) \-J
aA
39 - /33 n
b) 0.4 /-173.13° A(rms)
c) 35.39/176.63° V(rms)
11.13 21.64/121.34° V(rms)
11.16 159.5 /29.34° V(rms)
11.22 6120/36.61° VA

11.24 a) 1833.46 /22° VA
b) 519.62 V(rms)
11.43 a) W
2
~ W
x
= V
L
/
L
[cos (d - 30°) - cos (6 +
30°)] = 2V
L
I
L
$in0sm30
o
=
V
L
I
L
sm6
Thus,
V3(W
2
- W
x
) = Vw
L

I
L
sm 0 = Q
T
b) 2592 VAR, -2592 VAR, 3741.23 VAR,
-4172.80 VAR
11.44 197.26 W, 476.64 W
11.52 a)
1.70 MVA
1.2MVAR
1.2 MW
b)
1.2
MW
11.53 a)
16.71
AIF
b) 50.14/aF
11.56 |V
ab
| = 12,548.8 V, so the voltage is below the
acceptable level of 13
kV.
Thus, when the load at
the substation drops to zero, the capacitor bank
must be switched off.
11.57 P
L(be
fore) = 81.66 kW,
P

L(after)
= 40.83 kW
Chapter 12
12.2
12.3
12.7
12.9
12.10
12.14
a) (t + \0)u(t + 10) - 2tu(t) +
(t - 10)«(f - 10)
b) -8(t + 3)u(t + 3) + 8(7 + 2)u(t + 2) +
8(r + l)«(f + 1) - 8(f - l)u(t -1)-
8(/
- 2)u{t -2) + 8(/ - 3}M(/ - 3)
a) 5t[u{t) - u{t - 2)] + 10[«(f -2)-
u(t - 6)] + (-5? + A0)[u(t - 6) - u(t - 8)]
b) 10 sin irt[u{t) - u(t - 2)]
c) 4t[u(t) - u{t - 5)]
a) 1.0
b) 0
c) oo
a) 26
b) 2.25
2/9
at
SC
°
3)
f + c,

2
M
^
2
b)
* + J
c) 2
d) check
774 Answers to Selected Problems
12.17 a)
b)
c)
d)
e)
12.22 a)
b)
12.40 a)
(.v + a)
2
CO
2
i ~>
S + CO
co
cos 0 +
s
2
+
1
->

sr
sinh 0 +
(r -
1
v(.v + a)
1
.v
sin 6
->
of
y[cosh 0]
- 1)
13.9 a)
12.41
12.42
12.50
12.55
12.56
s(s + a)
[«?"'
+ 5e'
2
' + 2e-
4
']u(t)
b) [6 + 4e~
2t
+ 2e~
4t
+ e~

(
"]u(t)
c) [4e'
1
+ 20cT'cos(2f + 36.87°)]M(?)
d) [490 + 250e"
7
'cos(r - 163.74°)]«(r)
a) [20/ - 4 + 4e-*]u(t)
b) [250 - lOOte'
1
- 250e~
f
]u(t)
c) [30r - 8 + K)e
3l
cos(t + 36.87°) ]u(t)
d) [20 - 2.5fV - I5te~' - 20e"']«(0
e) [16 + S9A4te~
2t
cos (t + 26.57°) +
I13.14e
_2r
cos0 + 98.13°) ]u(t)
c) 5'(t) - 105(f) + [3(k'
_5/
+ 20e~
10f
]«(f)
a) /(()') =

8,.f(oo)
= 0
b) /(0^) = 13,/(00) = 6
c) /(0
+
) = 20, /(oo) = 0
d) /(0
+
) = 250, /(oo) = 490
0.947
588
Chapter 13
13.4 a) —
8 X 10
7
s
13.6 a)
.v
2
+ 40,000* + 256 X 10
6
b) Zero at 0; poles at -8000 rad/s and
-32,000 rad/s
s
2
+ 8000* + 25 x 10
6
b) Zeros at -4000 + /3000 rad/s and
-4000 - /30()() rad/s; pole at 0
16 x Uf

-^vw—
5000 (2
SI
150
V-s
b)
'©
-150s
+
V„
2.5 s
a
(s + 400)(.v + 1600)
c) (SO*-*** - 200e-
l60()f
)«(r)V
13.10 a)
A/W-
+ 10012
b)
c)
13.12 a)
5 x 10
s
n
s '
V.
o.ob'
n
75.v

2
+ 812,500.v + 6875 X
sis
1
+ 10
4
v + 5 X 10
7
)
/ + \ 137.5
/
1.25 mV
-Hv
^_7
10
6
[137.5 +
8().04^
50()()
'cos(5000/ + 14134
0
)
]K(0
ii,
V-s
R
sC
re© 4 £0
v.
b)

c)
•48(s + 8000)
.v
2
+ 8000* + 25 X 10
6
2.4(.v + 4875)
.v
2
+ 8000s + 25 X 10
6
d) [80e"-"
)00f
cos(3000f + 126.87°) ]u(t)V
e) [2.5e-
4mt
cos(3(mt -
16.26°)]H(0A
13.21 a) [35 + 5.73e~' cos(7r + 167.91 °)]H(I)V
b) Compare solution at t = 0 and t = oo to
circuit at t = 0 and t = oo
13.22 a) [10 - Kk>""
a5
'cos0.5/]u(f)A
b) 7.07e
_OA
cos(0.5r - 45°)w(r) V
c) Compare solutions at t = 0 and t = oo to
circuit at t = 0 and t = oo
Answers to Selected Problems 775

13.29 a)
65
2
+ 6s - 18 -9r - 30.v
18
s(s + 2)(5 + 3)' s(s + 2)(5 + 3)
b) Initial values: 6 A, -9 A;
final values: -3 A, -3 A
c) [-3 + 3e~
2
' + 6e-
3,
]u(t)A,
[-3 - 3e~
2
' - 3e"
3
']u(t)A
13.34 63.25*"
ia
*cos(50> + 71.57°)w(0 mA
240(5 + 40)
13.39 a) —
' s(s + 20)(5 + 80)
b) Initial value is 0. final value is 6 A
c) (6 - 4e~
m
- 2e~
m
)u(t) A

13.40 a)
(-2e~
m
+ 2e^
)
')«(0mA
b) (2<
-2()(
2e
-hl
>(0 mA
13.42 a)
480(5 + 2.5)
5(5 + 4)(5 + 6)
b) [50 + 90e~
4
' - \4Qe~
(
"]u(t)V
13.50 a)
b)
c)
d)
e)
250
5 + 250
5
5 + 250
V
5 + 8000

8000
5 + 8000
, no zeros, pole at -250 racl/s
, zero at 0, pole at -250 rad/s
, zero at 0, pole at -8000 rad/s
, no zeros, pole at -8000 rad/s
100
, no zeros, pole at -500 rad/s
5 + 500
13.62 (e - l)e"
?
V
13.63 (1 - e)e-'V
13.77 16.97 cos (3/ +
8.13°)
V
5(5 + 30,000)
13.79 a)
1
(5 + 5000)(5 + 8000)
b) (5e-
5l)m
- 4Ae-*
m
')u(t)V
13.80 a) =1*2
;
(5 + 400)(5 + 1000)
b) 13.13 cos(400f - 156.8°) V
13.87 a) 0.8 A

b) 0.6 A
c) 0.2 A
d) -0.6 A
e) Q.6e~
zxm
'\t(t)A
f) -0.6e-
2xl
%(f)A
g) -1.6 x 1(T
3
8(0 - 72QOe'
2xlift
u(t)V
13.88 a) 80 V
b) 20 V
c) 0 V
d) 325(f)juA
e)
16
V
f) 4V
g)
20
V
13.92 a)
i
2
(0~)
= *

2
<0
+
) = 0A;
*'L(0~) = idQ
+
) = 35.36 A
1440TT( 122.92 V25 - 3000TT\/2
b) V
Q
= = =— +
J
(5 + 1475TT)(5
2
+ 14,400TT
2
)
300 V2
5 + 1475 w
v
a
= 252.89<T
l475jr/
+
172.62 cos(1207it + 6.85°) V
y„(0
+
) = 424.26 V
c) V„ = 122.06/6.85° V(rms)
d

)
»„
(V)
t (ms)
13.93 a) -20.58^
-1475
^ + 172.62 cos(
12()77-/
- 83.15°) V
b)
v„
(V)
t (ms)
c) Voltage spikes in Problem 13.92 but not here.