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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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 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 4

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

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part

of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior

written permission of the publisher, or used beyond the limited distribution to teachers and educators

permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,

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