Solution manual fundamentals of electric circuits
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Chapter 1, Solution 1
(a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C
(b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C
(c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C
(d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C
Chapter 1, Solution 2
(a)
(b)
(c)
(d)
(e)
i = dq/dt = 3 mA
i = dq/dt = (16t + 4) A
i = dq/dt = (-3e-t + 10e-2t) nA
i=dq/dt = 1200π cos 120π t pA
i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) µ A
Chapter 1, Solution 3
(a) q(t) = ∫ i(t)dt + q(0) = (3t + 1) C
(b) q(t) = ∫ (2t + s) dt + q(v) = (t 2 + 5t) mC
(c) q(t) = ∫ 20 cos (10t + π / 6 ) + q(0) = (2sin(10t + π / 6) + 1) µ C
(d)
10e -30t
( −30 sin 40t - 40 cos t)
900 + 1600
= − e - 30t (0.16cos40 t + 0.12 sin 40t) C
q(t) = ∫ 10e -30t sin 40t + q(0) =
Chapter 1, Solution 4
q = ∫ idt = ∫
=
10
−5
5sin 6 π t dt =
cos 6π t
6π
0
5
(1 − cos 0.06π ) = 4.698 mC
6π
Chapter 1, Solution 5
q = ∫ idt = ∫
=
1
e dt mC = - e -2t
2
1
(1 − e 4 ) mC = 490 µC
2
Chapter 1, Solution 6
(a) At t = 1ms, i =
dq 80
=
= 40 mA
dt
2
(b) At t = 6ms, i =
dq
= 0 mA
dt
(c) At t = 10ms, i =
dq 80
=
= - 20 mA
4
dt
Chapter 1, Solution 7
25A,
dq
i=
= - 25A,
dt
25A,
2
-2t
0
which is sketched below:
0
Chapter 1, Solution 8
q = ∫ idt =
10 × 1
+ 10 × 1 = 15 µC
2
Chapter 1, Solution 9
1
(a) q = ∫ idt = ∫ 10 dt = 10 C
0
3
5 ×1
q = ∫ idt = 10 × 1 + 10 −
+ 5 ×1
0
(b)
2
= 15 + 10 − 25 = 22.5 C
5
(c) q = ∫ idt = 10 + 10 + 10 = 30 C
0
Chapter 1, Solution 10
q = ixt = 8 x10 3 x15 x10 − 6 = 120 µ C
Chapter 1, Solution 11
q = it = 85 x10-3 x 12 x 60 x 60 = 3,672 C
E = pt = ivt = qv = 3672 x1.2 = 4406.4 J
Chapter 1, Solution 12
For 0 < t < 6s, assuming q(0) = 0,
t
∫
t
∫
q (t ) = idt + q (0 ) = 3tdt + 0 = 1.5t 2
0
0
At t=6, q(6) = 1.5(6)2 = 54
For 6 < t < 10s,
t
t
∫
∫
q (t ) = idt + q (6 ) = 18 dt + 54 = 18 t − 54
6
6
At t=10, q(10) = 180 – 54 = 126
For 10
∫
t
∫
q (t ) = idt + q (10 ) = ( −12)dt + 126 = −12t + 246
10
10
At t=15, q(15) = -12x15 + 246 = 66
For 15
∫
q (t ) = 0 dt + q (15) =66
15
Thus,
1.5t 2 C, 0 < t < 6s
18 t − 54 C, 6 < t < 10s
q (t ) =
−12t + 246 C, 10 < t < 15s
66 C, 15 < t < 20s
The plot of the charge is shown below.
140
120
100
q(t)
80
60
40
20
0
0
5
10
t
15
20
Chapter 1, Solution 13
2
2
w = ∫ vidt = ∫ 1200 cos 2 4 t dt
0
0
2
= 1200 ∫ ( 2 cos 8t - 1)dt (since, cos 2 x = 2 cos 2x - 1)
0
2
2
1
= 1200 sin 8t − t = 1200 sin 16 − 2
8
0
4
= - 2.486 kJ
Chapter 1, Solution 14
q = ∫ idt = ∫ 10(1 - e -0.5t )dt = 10(t + 2e -0.5t )
1
(a)
(b)
0
= 10(1 + 2e
-0.5
− 2 ) = 2.131 C
1
0
p(t) = v(t)i(t)
p(1) = 5cos2 ⋅ 10(1- e-0.5) = (-2.081)(3.935)
= -8.188 W
Chapter 1, Solution 15
(a)
q = ∫ idt = ∫
2
0
− 3 2t
3e dt =
e
2
2
-2t
= −1.5(e − 1) = 1.297 C
0
-2
(b)
5di
= −6e 2t ( 5) = −30e -2t
dt
p = vi = − 90 e − 4 t W
v=
3
(c) w = ∫ pdt = -90∫ e -4t dt =
0
3
− 90 -4t
e
= − 22.5 J
−4
0
Chapter 1, Solution 16
0
i(t) =
,
100 - 25t mA 2 < t < 4
1
0< t <1
10t V
v(t) = 10 V
1< t < 3
40 - 10t V 3 < t < 4
2
3
4
2
3
w = ∫ v(t)i(t)dt = ∫ 10 + (25t)dt + ∫ 10( 25t)dt + ∫ 10(100 − 25t)dt + ∫ ( 40 − 10t)(100 - 25t)mJ
0
1
=
1
3
2
4
250 3
250
t2
t +
+ 250 4 t - + ∫ 250( 4 − t) 2 dt
3
2 1
2 2 3
0
4
250 250
9
t2
2
( 3) + 25012 − − 8 + 2 + 25016 t - 4t +
=
+
3
2
2
3 3
= 916.7 mJ
Chapter 1, Solution 17
Σ p = 0 → -205 + 60 + 45 + 30 + p3 = 0
p3 = 205 – 135 = 70 W
Thus element 3 receives 70 W.
Chapter 1, Solution 18
p1 = 30(-10) = -300 W
p2 = 10(10) = 100 W
p3 = 20(14) = 280 W
p4 = 8(-4) = -32 W
p5 = 12(-4) = -48 W
Chapter 1, Solution 19
∑p=0
→
−4 I s − 2 x6 − 13 x 2 + 5 x10 = 0
→
Is = 3 A
Chapter 1, Solution 20
Since Σ p = 0
-30×6 + 6×12 + 3V0 + 28 + 28×2 - 3×10 = 0
72 + 84 + 3V0 = 210 or 3V0 = 54
V0 = 18 V
Chapter 1, Solution 21
i=
=
∆q
photon 1 electron
= 4 × 1011
⋅ 1. 6 × 1019 ( C / electron)
⋅
∆t
sec 8 photon
4
× 1011 × 1. 6 × 10 −19 C/s = 0.8 × 10 -8 C/s = 8 nA
8
Chapter 1, Solution 22
It should be noted that these are only typical answers.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Light bulb
Radio set
TV set
Refrigerator
PC
PC printer
Microwave oven
Blender
60 W, 100 W
4W
110 W
700 W
120 W
18 W
1000 W
350 W
Chapter 1, Solution 23
(a) i =
p 1500
=
= 12.5 W
v 120
(b) w = pt = 1. 5 × 103 × 45 × 60 ⋅ J = 1.5 ×
(c) Cost = 1.125 × 10 = 11.25 cents
45
kWh = 1.125 kWh
60
Chapter 1, Solution 24
p = vi = 110 x 8 = 880 W
Chapter 1, Solution 25
4
Cost = 1.2 kW × hr × 30 × 9 cents/kWh = 21.6 cents
6
Chapter 1, Solution 26
0. 8A ⋅ h
= 80 mA
10h
(b) p = vi = 6 × 0.08 = 0.48 W
(c) w = pt = 0.48 × 10 Wh = 0.0048 kWh
(a) i =
Chapter 1, Solution 27
(a) Let T = 4h = 4 × 36005
T
q = ∫ idt = ∫ 3dt = 3T = 3 × 4 × 3600 = 43.2 kC
0
T
T
0 . 5t
( b) W = ∫ pdt = ∫ vidt = ∫ ( 3) 10 +
dt
0
0
3600
4×3600
0. 25t 2
= 310t +
3600 0
= 475.2 kJ
( c)
= 3[40 × 3600 + 0. 25 × 16 × 3600]
W = 475.2 kWs, (J = Ws)
475.2
Cost =
kWh × 9 cent = 1.188 cents
3600
Chapter 1, Solution 28
(a) i =
P 30
=
= 0.25 A
V 120
( b) W = pt = 30 × 365 × 24 Wh = 262.8 kWh
Cost = $0.12 × 262.8 = $31.54
Chapter 1, Solution 29
(20 + 40 + 15 + 45)
30
hr + 1.8 kW hr
60
60
= 2.4 + 0.9 = 3.3 kWh
Cost = 12 cents × 3.3 = 39.6 cents
w = pt = 1. 2kW
Chapter 1, Solution 30
Energy = (52.75 – 5.23)/0.11 = 432 kWh
Chapter 1, Solution 31
Total energy consumed = 365(4 +8) W
Cost = $0.12 x 365 x 12 = $526.60
Chapter 1, Solution 32
(20 + 40 + 15 + 45)
30
hr + 1.8 kW hr
60
60
= 2.4 + 0.9 = 3.3 kWh
Cost = 12 cents × 3.3 = 39.6 cents
w = pt = 1. 2kW
Chapter 1, Solution 33
i=
dq
→ q = ∫ idt = 2000 × 3 × 10 3 = 6 C
dt
Chapter 1, Solution 34
(b) Energy =
∑ pt
= 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2
= 10,000 kWh
(c) Average power = 10,000/24 = 416.67 W
Chapter 1, Solution 35
( a) W = ∫ p( t ) dt = 400 × 6 + 1000 × 2 + 200 × 12 × 1200 × 2 + 400 × 2
= 7200 + 2800 = 10.4 kWh
( b)
10.4 kW
= 433.3 W/h
24 h
Chapter 1, Solution 36
160A ⋅ h
=4A
40
160Ah 160, 000h
( b) t =
=
= 6,667 days
0.001A 24h / day
(a)
i=
Chapter 1, Solution 37
q = 5 × 10 20 (− 1. 602 × 10 −19 ) = −80. 1 C
W = qv = −80. 1 × 12 = − 901.2 J
Chapter 1, Solution 38
P = 10 hp = 7460 W
W = pt = 7460 × 30 × 60 J = 13.43 × 106 J
Chapter 1, Solution 39
p = vi → i =
p 2 × 10 3
=
= 16.667 A
v
120
Chapter 2, Solution 1
v = iR
i = v/R = (16/5) mA = 3.2 mA
Chapter 2, Solution 2
p = v2/R →
R = v2/p = 14400/60 = 240 ohms
Chapter 2, Solution 3
R = v/i = 120/(2.5x10-3) = 48k ohms
Chapter 2, Solution 4
(a)
(b)
i = 3/100 = 30 mA
i = 3/150 = 20 mA
Chapter 2, Solution 5
n = 9; l = 7; b = n + l – 1 = 15
Chapter 2, Solution 6
n = 12;
l = 8;
b = n + l –1 = 19
Chapter 2, Solution 7
7 elements or 7 branches and 4 nodes, as indicated.
30 V
1
20 Ω
2
3
++++ -
2A
30 Ω
60 Ω
4
40 Ω
10 Ω
Chapter 2, Solution 8
12 A
a
i1
b
8A
i3
i2
12 A
c
At node a,
At node c,
At node d,
9A d
8 = 12 + i1
9 = 8 + i2
9 = 12 + i3
i1 = - 4A
i2 = 1A
i3 = -3A
Chapter 2, Solution 9
Applying KCL,
i1 + 1 = 10 + 2
1 + i2 = 2 + 3
i2 = i3 + 3
i1 = 11A
i2 = 4A
i3 = 1A
Chapter 2, Solution 10
2
4A
1
-2A
i2
i1
3
3A
At node 1,
At node 3,
4 + 3 = i1
3 + i2 = -2
i1 = 7A
i2 = -5A
Chapter 2, Solution 11
Applying KVL to each loop gives
-8 + v1 + 12 = 0
-12 - v2 + 6 = 0
10 - 6 - v3 = 0
-v4 + 8 - 10 = 0
v1 = 4v
v2 = -6v
v3 = 4v
v4 = -2v
Chapter 2, Solution 12
+ 15v -
loop 2
– 25v +
+
20v
-
+ 10v +
v1
-
loop 1
For loop 1,
For loop 2,
For loop 3,
+ v2 -
loop 3
-20 -25 +10 + v1 = 0
-10 +15 -v2 = 0
-v1 +v2 +v3 = 0
+
v3
-
v1 = 35v
v2 = 5v
v3 = 30v
Chapter 2, Solution 13
2A
1
I2
7A
2
3
I4
4
4A
I1
3A
I3
At node 2,
3 + 7 + I2 = 0
→
I 2 = −10 A
At node 1,
I1 + I 2 = 2
→
I 1 = 2 − I 2 = 12 A
At node 4,
2 = I4 + 4
→
I 4 = 2 − 4 = −2 A
At node 3,
7 + I4 = I3
→
I3 = 7 − 2 = 5 A
Hence,
I 1 = 12 A,
I 2 = −10 A,
I 3 = 5 A,
I 4 = −2 A
Chapter 2, Solution 14
+
3V
-
+
I3
4V
+
V3 -
→
V4 = 7V
For mesh 2,
+4 + V3 + V4 = 0
→
V3 = −4 − 7 = −11V
→
V1 = V3 + 3 = −8V
→
V2 = −V1 − 2 = 6V
For mesh 3,
−3 + V1 − V3 = 0
For mesh 4,
−V1 − V2 − 2 = 0
Thus,
V1 = −8V ,
V2 = 6V ,
+
- V4
For mesh 1,
−V4 + 2 + 5 = 0
V1
V3 = −11V ,
I4
2V -
+
I2
+
-
V4 = 7V
+
V2
+
I1
5V
-
Chapter 2, Solution 15
+
+
+
12V
-
1
- 8V +
v2
-
v1
-
3
+
2
v3
10V
+
-
For loop 1,
8 − 12 + v2 = 0
→
v2 = 4V
For loop 2,
− v3 − 8 − 10 = 0
→
v3 = −18V
→
v1 = −6V
For loop 3,
− v1 + 12 + v3 = 0
Thus,
v1 = −6V ,
v2 = 4V ,
v3 = −18V
Chapter 2, Solution 16
+ v1 -
6V
loop 1
+
-
+-
12V
10V
+-
+
v1
-
loop 2
+ v2 -
Applying KVL around loop 1,
–6 + v1 + v1 – 10 – 12 = 0
v1 = 14V
Applying KVL around loop 2,
12 + 10 – v2 = 0
v2 = 22V
Chapter 2, Solution 17
+ v1 -
24V
+
loop 1
-
+
v3
-
-
v2
+
loop 2
-+
12V
It is evident that v3 = 10V
Applying KVL to loop 2,
v2 + v3 + 12 = 0
v2 = -22V
Applying KVL to loop 1,
-24 + v1 - v2 = 0
v1 = 2V
Thus,
v1 = 2V, v2 = -22V, v3 = 10V
Chapter 2, Solution 18
Applying KVL,
-30 -10 +8 + I(3+5) = 0
8I = 32
I = 4A
-Vab + 5I + 8 = 0
Vab = 28V
+
-
10V
Chapter 2, Solution 19
Applying KVL around the loop, we obtain
-12 + 10 - (-8) + 3i = 0
i = -2A
Power dissipated by the resistor:
p 3Ω = i2R = 4(3) = 12W
Power supplied by the sources:
p12V = 12 (- -2) = 24W
p10V = 10 (-2) = -20W
p8V = (- -2) = -16W
Chapter 2, Solution 20
Applying KVL around the loop,
-36 + 4i0 + 5i0 = 0
i0 = 4A
Chapter 2, Solution 21
Apply KVL to obtain
10 Ω
-45 + 10i - 3V0 + 5i = 0
+ v0 -
But v0 = 10i,
-45 + 15i - 30i = 0
P3 = i2R = 9 x 5 = 45W
i = -3A
45V
+
+
-
5Ω
3v0
Chapter 2, Solution 22
4Ω
+ v0 6Ω
10A
2v0
At the node, KCL requires that
v0
+ 10 + 2 v 0 = 0
4
v0 = –4.444V
The current through the controlled source is
i = 2V0 = -8.888A
and the voltage across it is
v = (6 + 4) i0 = 10
v0
= −11.111
4
Hence,
p2 vi = (-8.888)(-11.111) = 98.75 W
Chapter 2, Solution 23
8//12 = 4.8, 3//6 = 2, (4 + 2)//(1.2 + 4.8) = 6//6 = 3
The circuit is reduced to that shown below.
ix
1Ω
+
6A
2Ω
vx
3Ω
Applying current division,
ix =
2
(6 A) = 2 A,
2 + 1+ 3
v x = 1i x = 2V
The current through the 1.2- Ω resistor is 0.5ix = 1A. The voltage across the 12- Ω
resistor is 1 x 4.8 = 4.8 V. Hence the power is
p=
v 2 4.8 2
=
= 1.92W
12
R
Chapter 2, Solution 24
(a)
I0 =
Vs
R1 + R2
V0 = −α I0 (R3 R4 ) = −
αV0
R1 + R 2
⋅
R3 R4
R3 + R4
V0
− αR3 R4
=
Vs (R1 + R2 )(R3 + R4 )
(b)
If R1 = R2 = R3 = R4 = R,
V0
α R α
=
⋅ = = 10
VS
2R 2 4
Chapter 2, Solution 25
V0 = 5 x 10-3 x 10 x 103 = 50V
Using current division,
I20 =
5
(0.01x50) = 0.1 A
5 + 20
V20 = 20 x 0.1 kV = 2 kV
p20 = I20 V20 = 0.2 kW
α = 40
Chapter 2, Solution 26
V0 = 5 x 10-3 x 10 x 103 = 50V
Using current division,
I20 =
5
(0.01x50) = 0.1 A
5 + 20
V20 = 20 x 0.1 kV = 2 kV
p20 = I20 V20 = 0.2 kW
Chapter 2, Solution 27
Using current division,
i1 =
4
(20) = 8 A
4+6
i2 =
6
(20) = 12 A
4+6
Chapter 2, Solution 28
We first combine the two resistors in parallel
15 10 = 6 Ω
We now apply voltage division,
v1 =
14
(40) = 20 V
14 + 6
v2 = v3 =
Hence,
6
(40) = 12 V
14 + 6
v1 = 28 V, v2 = 12 V, vs = 12 V
Chapter 2, Solution 29
The series combination of 6 Ω and 3 Ω resistors is shorted. Hence
i2 = 0 = v2
v1 = 12, i1 =
12
= 3A
4
Hence v1 = 12 V, i1 = 3 A, i2 = 0 = v2
Chapter 2, Solution 30
8Ω
i1
i
9A
By current division, i =
6Ω
+
v
-
4Ω
12
(9) = 6 A
6 + 12
i1 = 9 − 6 = 3A, v = 4i1 = 4 x 3 = 12 V
p6 = 12R = 36 x 6 = 216 W
Chapter 2, Solution 31
The 5 Ω resistor is in series with the combination of 10 (4 + 6) = 5Ω .
Hence by the voltage division principle,
v=
5
(20V) = 10 V
5+5
by ohm's law,
i=
v
10
=
= 1A
4 + 6 4+ 6
pp = i2R = (1)2(4) = 4 W
Chapter 2, Solution 32
We first combine resistors in parallel.
20 30 =
20 x30
= 12 Ω
50
10 40 =
10x 40
= 8Ω
50
Using current division principle,
8
12
i1 + i 2 =
(20) = 8A, i 3 + i 4 =
(20) = 12A
8 + 12
20
i1 =
20
(8) = 3.2 A
50
i2 =
30
(8) = 4.8 A
50
i3 =
10
(12) = 2.4A
50
i4 =
40
(12) = 9.6 A
50
Chapter 2, Solution 33
Combining the conductance leads to the equivalent circuit below
i
+
v
-
9A
1S
i
4S
4S
6x3
= 25 and 25 + 25 = 4 S
9
Using current division,
6 S 3S =
i=
1
1
1+
2
(9) = 6 A, v = 3(1) = 3 V
9A
+
v
-
1S
2S
Chapter 2, Solution 34
By parallel and series combinations, the circuit is reduced to the one below:
Thus i1 =
8Ω
i1
10 x15
= 6Ω
10 ( 2 + 13 ) =
25
15 x15
15 (4 + 6) =
= 6Ω
25
12 (6 + 6) = 6Ω
28V
+
v1
-
+
-
6Ω
28
= 2 A and v1 = 6i1 = 12 V
8+6
We now work backward to get i2 and v2.
i1 = 2A
8Ω
6Ω
1A
1A
28V
+
12V
-
+
-
8Ω
i1 = 2A
6Ω
+
6V
-
12 Ω
6Ω
4Ω
1A
0.6A
1A
28V
Thus, v2 =
+
12V
-
+
-
12 Ω
+
6V
-
+
15 Ω
3.6V
v
13
(3 ⋅ 6) = 3 ⋅ 12, i2 = 2 = 0.24
13
15
p2 = i2R = (0.24)2 (2) = 0.1152 W
i1 = 2 A, i2 = 0.24 A, v1 = 12 V, v2 = 3.12 V, p2 = 0.1152 W
Chapter 2, Solution 35
i
70 Ω
50V
+
-
a
+
V1
i1 -
30 Ω
I0
+
20 Ω
i2
b
V0 5 Ω
-
-
6Ω
