Solution manual fundamentals of electric circuits 3rd edition chapter06

Chapter 6, Problem 1.
If the voltage across a 5-F capacitor is 2te-3t V, find the current and the power.
Chapter 6, Solution 1.
i=C

(

)

dv
= 5 2e −3t − 6 te −3t = 10(1 - 3t)e-3t A
dt

p = vi = 10(1-3t)e-3t ⋅ 2t e-3t = 20t(1 - 3t)e-6t W
Chapter 6, Problem 2.
A 20-μF capacitor has energy w(t) = 10 cos2 377t J. Determine the current through
the capacitor.
Chapter 6, Solution 2.
1
w = Cv2
2

2W 20 cos2 377t
⎯⎯
→ v =
=
= 106 cos2 377t
−6
C
20 x10
2

v = ±103cos(377t) V, let us assume the v = +cos(377t) mV, this then leads to,
i = C(dv/dt) = 20x10–6(–377sin(377t)10–3) = –7.54sin(377t) A.
Please note that if we had chosen the negative value for v,
then i would have been positive.
Chapter 6, Problem 3.
In 5 s, the voltage across a 40-mF capacitor changes from 160 V to
220 V. Calculate the average current through the capacitor.
Chapter 6, Solution 3.
i=C

dv
220 − 160
= 40 x10 −3
= 480 mA
dt
5

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Chapter 6, Problem 4.
A current of 6 sin 4t A flows through a 2-F capacitor. Find the
voltage v(t) across the capacitor given that v(0) = 1 V.
Chapter 6, Solution 4.
1 t

v = ∫ idt + v(0)
C o
t

1 t
⎛ 3
⎞
= ∫ 6 sin 4 tdt + 1 = ⎜ − cos 4t ⎟ + 1 = −0.75 cos 4t + 0.75 + 1
0
2
⎝ 4
⎠0
= 1.75 – 0.75 cos 4t V
Chapter 6, Problem 5.
The voltage across a 4-μF capacitor is shown in Fig. 6.45. Find the current waveform.
v (V)
10
t (ms)
0

2

–10
Figure 6.45

4

6

8

For Prob. 6.5.

Chapter 6, Solution 5.

⎧ 5000t , 0 < t < 2ms
⎪
v = ⎨ 20 − 5000t , 2 < t < 6ms
⎪− 40 + 5000t , 6 < t < 8ms
⎩
⎧ 5,
dv 4 x10 −6 ⎪
i=C
=
⎨−5,
dt
10 −3 ⎪
⎩ 5,

0 < t < 2ms ⎧ 20 mA,
⎪
2 < t < 6ms = ⎨−20 mA,
6 < t < 8ms ⎪⎩ 20 mA,

0 < t < 2ms
2 < t < 6ms
6 < t < 8 ms

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Chapter 6, Problem 6.

The voltage waveform in Fig. 6.46 is applied across a 30-μF capacitor. Draw the current
waveform through it.

Figure 6.46
Chapter 6, Solution 6.
dv
i=C
= 30 x10 −6 x slope of the waveform.
dt
For example, for 0 < t < 2,
dv
10
=
dt 2 x10 −3
dv
10
i= C
= 30 x10 −6 x
= 150mA
dt
2 x10 −3
Thus the current i is sketched below.

i(t)

t

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Chapter 6, Problem 7.

At t=0, the voltage across a 50-mF capacitor is 10 V. Calculate the voltage across the
capacitor for t > 0 when current 4t mA flows through it.
Chapter 6, Solution 7.
v=

1
1
idt + v( t o ) =
∫
C
50 x10 −3

=

2t 2
+ 10 = 0.04t2 + 10 V
50

t

∫ 4tx10
o

−3

dt + 10

Chapter 6, Problem 8.

A 4-mF capacitor has the terminal voltage
t≤0
50 V,
⎧
v = ⎨ -100t
-600 t
+ Be
V,
t≥0
⎩Ae

If the capacitor has initial current of 2A, find:
(a) the constants A and B,
(b) the energy stored in the capacitor at t = 0,
(c) the capacitor current for t > 0.
Chapter 6, Solution 8.

(a) i = C

dv
= −100 ACe −100 t − 600 BCe −600 t
dt

i(0) = 2 = −100 AC − 600BC

⎯
⎯→

(1)

5 = − A − 6B

v ( 0 + ) = v (0 − )
⎯
⎯→ 50 = A + B
Solving (2) and (3) leads to
A=61, B=-11

(b) Energy =

(2)
(3)

1 2
1
Cv (0) = x 4 x10 −3 x 2500 = 5 J
2
2

(c ) From (1),
i = −100 x61x 4 x10 −3 e −100t − 600 x11x 4 x10 −3 e −600t = − 24.4e −100t − 26.4e −600t A
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Chapter 6, Problem 9.
The current through a 0.5-F capacitor is 6(1-e-t)A.
Determine the voltage and power at t=2 s. Assume v(0) = 0.
Chapter 6, Solution 9.
1 t
−t
−t t
6
1
−
e
dt
+
0
=
12
t
+
e
V = 12(t + e-t) – 12

v(t) =
∫
0
12 o
-2
v(2) = 12(2 + e ) – 12 = 13.624 V

(

)

(

)

p = iv = [12 (t + e-t) – 12]6(1-e-t)
p(2) = [12 (2 + e-2) – 12]6(1-e-2) = 70.66 W
Chapter 6, Problem 10.
The voltage across a 2-mF capacitor is shown in Fig. 6.47. Determine the current
through the capacitor.

Figure 6.47
Chapter 6, Solution 10
dv
dv
i=C
= 2 x10 −3
dt
dt
⎧ 16t , 0 < t < 1μs

⎪
v = ⎨ 16, 1 < t < 3 μs
⎪64 - 16t, 3 < t < 4μs
⎩
⎧ 16 x10 6 , 0 < t < 1μs
dv ⎪
= ⎨ 0, 1 < t < 3 μs
dt ⎪
6
⎩- 16x10 , 3 < t < 4 μs
0 < t < 1μs
⎧ 32 kA,
⎪
i (t ) = ⎨ 0, 1 < t < 3 μs
⎪- 32 kA, 3 < t < 4 μs
⎩
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Chapter 6, Problem 11.
3. A 4-mF capacitor has the current waveform shown in Fig. 6.48. Assuming that
v(0)=10V, sketch the voltage waveform v(t).

i (mA)
15
10

5
0
–5

00

2

4

6

88 t(s)

–10
Figure 6.48

For Prob. 6.11.

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Chapter 6, Solution 11.
t

v=

t

1
1
idt + v(0) = 10 +
i(t)dt
∫
C0
4 x10 −3 ∫0
t

103
15dt = 10 + 3.76t
v = 10 +
For 04 x10 −3 ∫0

v(2) = 10+7.5 =17.5
For 2 < t <4, i(t) = –10 mA
t
t
1
10 x10 −3
+
=
−
v(t) =
i
(

t
)
dt
v
(2)
dt + 17.5 = 22.5 + 2.5t
4 x10 −3 ∫2
4 x10 −3 ∫2
v(4)=22.5-2.5x4 =12.5
t

1
v(t) =
0dt + v(4) =12.5
4 x10 −3 ∫2

For 4For 6t

v(t) =

10 x103
dt + v(6) =2.5(t − 6) + 12.5 = 2.5t − 2.5
4 x10 −3 ∫4

Hence,
⎧ 10 + 3.75t V,
⎪22.5 − 2.5t V,
⎪

v(t) = ⎨
⎪ 12.5 V,
⎪⎩ 2.5t − 2.5 V,
which is sketched below.
v(t)

0 < t < 2s
2 < t < 4s
4 < t < 6s
6 < t < 8s

20
15
10
5
t (s)
0

2

4

6

8

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Chapter 6, Problem 12.

A voltage of 6e−2000 t V appears across a parallel combination of a 100-mF capacitor
and a 12-Ω resistor. Calculate the power absorbed by the parallel combination.

Chapter 6, Solution 12.

v 6 −2000t
e
=
= 0.5e−2000t
R 12
dv
= 100 x10 −3 x6(−2000)e−2000t = −1200e−2000t
ic = C
dt

iR =

i = iR + iC = −1199.5e−2000 t

p = vi = −7197e−4000t W

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Chapter 6, Problem 13.

Find the voltage across the capacitors in the circuit of Fig. 6.49 under dc
conditions.

30 Ω

Figure 6.49
Chapter 6, Solution 13.

Under dc conditions, the circuit becomes that shown below:
1

5

+
v1

+
+
−

v2

i2 = 0, i1 = 60/(30+10+20) = 1A
v1 = 30i1 = 30V, v2 = 60–20i1 = 40V
Thus, v1 = 30V, v2 = 40V

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Chapter 6, Problem 14.

Series-connected 20-pF and 60-pF capacitors are placed in parallel with seriesconnected 30-pF and 70-pF capacitors. Determine the equivalent capacitance.

Chapter 6, Solution 14.

20 pF is in series with 60pF = 20*60/80=15 pF
30-pF is in series with 70pF = 30x70/100=21pF
15pF is in parallel with 21pF = 15+21 = 36 pF

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Chapter 6, Problem 15.

Two capacitors (20 μF and 30 μF) are connected to a 100-V source. Find the energy
stored in each capacitor if they are connected in:
(a) parallel

(b) series

Chapter 6, Solution 15.

In parallel, as in Fig. (a),
v1 = v2 = 100

+
+
−

C+

+
v1

C

+
−

v2

−

C

+

C

v2

1 2 1
Cv = x 20 x10 −6 x100 2 = 100 mJ
2
2
1
w30 = x 30 x10 −6 x100 2 = 150 mJ
2

w20 =

(b)

When they are connected in series as in Fig. (b):

v1 =

C2
30
V=
x100 = 60, v2 = 40
C1 + C 2
50

w20 =

1

x 30 x10 −6 x 60 2 = 36 mJ
2

w30 =

1
x 30 x10 −6 x 40 2 = 24 mJ
2

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Chapter 6, Problem 16.

The equivalent capacitance at terminals a-b in the circuit in Fig. 6.50 is 30 μF.
Calculate the value of C.

Figure 6.50

Chapter 6, Solution 16
C eq = 14 +

Cx80
= 30
C + 80

⎯
⎯→

C = 20 μF

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Chapter 6, Problem 17.

Determine the equivalent capacitance for each of the circuits in
Fig. 6.51.

Figure 6.51

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Chapter 6, Solution 17.

(a)

4F in series with 12F = 4 x 12/(16) = 3F
3F in parallel with 6F and 3F = 3+6+3 = 12F
4F in series with 12F = 3F
i.e. Ceq = 3F

(b)

Ceq = 5 + [6x(4 + 2)/(6+4+2)] = 5 + (36/12) = 5 + 3 = 8F

(c)

3F in series with 6F = (3 x 6)/9 = 2F
1
1 1 1
= + + =1
C eq 2 6 3
Ceq = 1F

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Chapter 6, Problem 18.

Find Ceq in the circuit of Fig. 6.52 if all capacitors are 4 μF

Ceq

Figure 6.52

For Prob. 6.18.

Chapter 6, Solution 18.

4 μF in parallel with 4 μF = 8μF
4 μF in series with 4 μF = 2 μF
2 μF in parallel with 4 μF = 6 μF
Hence, the circuit is reduced to that shown below.
8μF

6 μF

6 μF

Ceq
1
1 1 1
= + + = 0.4583
Ceq 6 6 8

⎯⎯
→ Ceq = 2.1818 μF

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Chapter 6, Problem 19.

Find the equivalent capacitance between terminals a and b in the circuit of Fig.
6.53. All capacitances are in μF.

Figure 6.53
Chapter 6, Solution 19.

We combine 10-, 20-, and 30- μ F capacitors in parallel to get 60 μ F. The 60 - μ F
capacitor in series with another 60- μ F capacitor gives 30 μ F.
30 + 50 = 80 μ F, 80 + 40 = 120 μ F
The circuit is reduced to that shown below.
12

120

12

80

120- μ F capacitor in series with 80 μ F gives (80x120)/200 = 48
48 + 12 = 60
60- μ F capacitor in series with 12 μ F gives (60x12)/72 = 10 μ F

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Chapter 6, Problem 20.

Find the equivalent capacitance at terminals a-b of the circuit in Fig. 6.54.
a

1μF

2μF

3μF

1μF

2μF

3μF

2μF

3μF

3μF

b

Figure 6.54

For Prob. 6.20.

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Chapter 6, Solution 20.

Consider the circuit shown below.
C1
C2

C3

C1 = 1+ 1 = 2μ F
C2 = 2 + 2 + 2 = 6μ F
C3 = 4 x3 = 12 μ F

1/Ceq = (1/C1) + (1/C2) + (1/C3) = 0.5 + 0.16667 + 0.08333 = 0.75x106
Ceq = 1.3333 µF.

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Chapter 6, Problem 21.

Determine the equivalent capacitance at terminals a - b of the circuit in Fig. 6.55.

12 µF

Figure 6.55

Chapter 6, Solution 21.

4μF in series with 12μF = (4x12)/16 = 3μF
3μF in parallel with 3μF = 6μF
6μF in series with 6μF = 3μF
3μF in parallel with 2μF = 5μF
5μF in series with 5μF = 2.5μF
Hence Ceq = 2.5μF

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Chapter 6, Problem 22.

Obtain the equivalent capacitance of the circuit in Fig. 6.56.

Figure 6.56

Chapter 6, Solution 22.

Combining the capacitors in parallel, we obtain the equivalent circuit shown below:

4

6

3

2

Combining the capacitors in series gives C1eq , where
1
1
1
1
1
=
+
+
=
1
C eq 60 20 30 10

C1eq = 10μF

Thus
Ceq = 10 + 40 = 50 μF
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Chapter 6, Problem 23.

For the circuit in Fig. 6.57, determine:
(a) the voltage across each capacitor,
(b)
the energy stored in each capacitor.

Figure 6.57

Chapter 6, Solution 23.

(a)

(b)

3μF is in series with 6μF
v4μF = 1/2 x 120 = 60V
v2μF = 60V
3
v6μF =
(60) = 20V
6+3
v3μF = 60 - 20 = 40V

3x6/(9) = 2μF

Hence w = 1/2 Cv2
w4μF = 1/2 x 4 x 10-6 x 3600 = 7.2mJ
w2μF = 1/2 x 2 x 10-6 x 3600 = 3.6mJ
w6μF = 1/2 x 6 x 10-6 x 400 = 1.2mJ
w3μF = 1/2 x 3 x 10-6 x 1600 = 2.4mJ

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Chapter 6, Problem 24.

Repeat Prob. 6.23 for the circuit in Fig. 6.58.

80 µF

Figure 6.58

Chapter 6, Solution 24.

20μF is series with 80μF = 20x80/(100) = 16μF
14μF is parallel with 16μF = 30μF
(a) v30μF = 90V
v60μF = 30V
v14μF = 60V

80
v20μF =
x 60 = 48V
20 + 80
v80μF = 60 - 48 = 12V
1 2
Cv
2
w30μF = 1/2 x 30 x 10-6 x 8100 = 121.5mJ
w60μF = 1/2 x 60 x 10-6 x 900 = 27mJ
w14μF = 1/2 x 14 x 10-6 x 3600 = 25.2mJ
w20μF = 1/2 x 20 x 10-6 x (48)2 = 23.04mJ
w80μF = 1/2 x 80 x 10-6 x 144 = 5.76mJ

(b) Since w =

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Chapter 6, Problem 25.

(a) Show that the voltage-division rule for two capacitors in series as in Fig. 6.59(a) is
v1 =

C2
vs ,

C1 + C 2

v2 =

C1
vs
C1 + C 2

assuming that the initial conditions are zero.

Figure 6.59
(b) For two capacitors in parallel as in Fig. 6.59(b), show that the
current-division rule is
i1 =

C1
is ,
C1 + C 2

i2 =

C2
is
C1 + C 2

assuming that the initial conditions are zero.

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Chapter 6, Solution 25.

(a) For the capacitors in series,
Q1 = Q2
vs = v1 + v2 =

Similarly, v1 =

v1 C 2
=
v 2 C1

C1v1 = C2v2

C + C2
C2
v2
v2 + v2 = 1
C1
C1

v2 =

C1
vs
C1 + C 2

C2
vs
C1 + C 2

(b) For capacitors in parallel

Q1 Q 2
=
C1 C 2
C
C + C2
Qs = Q1 + Q2 = 1 Q 2 + Q 2 = 1
Q2
C2
C2
v1 = v2 =

or

C2
C1 + C 2
C1
Qs
Q1 =
C1 + C 2

Q2 =

i=

dQ
dt

i1 =

C1
is ,
C1 + C 2

i2 =

C2
is
C1 + C 2

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Chapter 6, Problem 26.

Three capacitors, C1 = 5 μF, C2 = 10 μF, and C3 = 20 μF, are connected in parallel across
a 150-V source. Determine:
(a) the total capacitance,
(b) the charge on each capacitor,
(c) the total energy stored in the parallel combination.

Chapter 6, Solution 26.

(a)

Ceq = C1 + C2 + C3 = 35μF

(b)

Q1 = C1v = 5 x 150μC = 0.75mC
Q2 = C2v = 10 x 150μC = 1.5mC
Q3 = C3v = 20 x 150 = 3mC

(c)

w=

1
1
C eq v 2 = x 35x150 2 μJ = 393.8mJ
2
2

Chapter 6, Problem 27.

Given that four 4-μF capacitors can be connected in series and in parallel, find the
minimum and maximum values that can be obtained by such series/parallel
combinations.
Chapter 6, Solution 27.

If they are all connected in parallel, we get CT = 4 x4 μ F = 16μ F

If they are all connected in series, we get
1
4
=
⎯⎯
→ CT = 1μ F
CT 4 μ F
All other combinations fall within these two extreme cases. Hence,
Cmin = 1μ F, Cmax = 16μ F

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