Exercise 5-8 Determine the current i in the circuit of Fig. E5-8, under dc conditions.

Figure E5-8

Solution: Under dc conditions, capacitors act like open circuits. Hence, the circuit becomes:

Voltage division gives

i = 1.5× 40k

40k+15k+5k= 1 A.

Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c© 2013 National Technology Press

Problem 5.18 Reduce the circuit in Fig. P5.18 into a single equivalent capacitor atterminals (a,b). Assume that all initial voltages are zero at t = 0.

Solution:

C C

C C

CC C

a b

C C

C

2C 2C

2C 2C

a b

Ca b

C

C

2C

Ca b

a b

Ceq = 2C/3

Figure P5.18

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Exercise 5-10 Suppose the circuit of Fig. E5-9 is connected to a dc voltage source V0 = 12 V. Assuming

that the capacitors had no charge before they were connected to the voltage source, determine υ1 and υ2, given

that C1 = 6 μF, C2 = 4 μF, and C3 = 8 μF.

Solution:

According to Eq. (5.46),

C1υ1 = (C2 ‖C3)υ2,

or

υ2 =C1υ1

C2 +C3=

6×10−6

4×10−6 +8×10−6υ1 =

υ1

2.

But

υ1 +υ2 = 12 V.

Hence,

υ1 = 8 V and υ2 = 4 V.

Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c© 2013 National Technology Press

Problem 5.15 Determine voltages υ1 to υ3 in the circuit of Fig. P5.15 under dcconditions.

(a) Circuit

(b) Under dc conditions

3 Ω

20 μF10 Ω 18 Ω

4 Ω

6 Ω

+_

υ2+ _

υ1+ _

60 μF

10 μF

υ3+ _

40 V

3 Ω

20 μF

10 Ω

18 Ω

4 Ω

6 Ω

+_

+ _

+ _

60 μF

10 μF

υ3υ2

υ1

+ _

40 V

Va

Vb

Loop 1

Loop 2

Figure P5.15: Circuit for Problem 5.15.

Solution: KCL at nodes Va and Vb:

Va −4010

+Va −Vb

18+

Va −Vb

6+3= 0

Vb

4+

Vb −Va

18+

Vb −Va

6+3= 0

Solution givesVa = 20 V, Vb = 8 V.

For Loop 1,

−υ1 −Vb +(Vb −Va)39

= 0,

which givesυ1 = −12 V.

For Loop 2,

−υ3 −υ2 +(Va −Vb)69

= 0

which gives

υ3 +υ2 = (20−8)69

= 8 V. (1)

For two capacitors in series, Eq. (5.47) gives

υ2C2 = υ3C3

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or20υ2 = 60υ3 (2)

Simultaneous solution of Eqs. (1) and (2) leads to

υ2 = 6 V, υ3 = 2 V.

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Problem 5.16 Determine the voltages across the two capacitors in the circuit ofFig. P5.16 under dc conditions.

(a) Circuit

(b) Under dc conditions

40 kΩ

40 kΩ

3 kΩ

20 kΩ

20 μF

10 kΩ 3 kΩ

+_

υ2

υ1

+_+

_

10 V

2 V

40 μF

+_

40 kΩ

40 kΩ

3 kΩ

20 kΩ

20 μF

10 kΩ 3 kΩ

+_

υ1

υ2+_+

_

10 V

2 V

40 μF

+_

i2

i1

Figure P5.16: Circuit for Problem 5.16.

Solution: Mesh current equations are given by

40ki1 −10+20ki1 +40k(i1 − i2) = 0

3ki2 +40k(i2 − i1)+3ki2 −2 = 0

Simultaneous solution gives:

i1 = 0.18 mA,

i2 = 0.2 mA.

υ1 = −3ki2 = −3×103 ×0.2×10−3 = −0.6 V

υ2 = 3ki2 −2+20ki1 = 0.6−2+3.6 = 2.2 V.

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Problem 5.22 Determine (a) the amount of energy stored in each of the threecapacitors shown in Fig. P5.22, (b) the equivalent capacitance at terminals (a,b),and (c) the amount of energy stored in the equivalent capacitor.

5 μF6 μF

10 kΩ

20 μFa

b15 V

+_

υ1 υ3

υ2 ++

_

_+

_

Figure P5.22

Solution: (a) Under dc conditions, the capacitors behave like open circuits. Hence,no currents flow anywhere in the circuit, and no voltage drop exists across the resistor.Hence,υab = 15 V.

For 6-µF capacitor,

w1 =12

Cυ2ab =

12×6×10−6×152 = 0.675 (mJ).

For the 20-µF and 5-µF capacitors, from Eq. (5.47) for two capacitors in series,

20µυ2 = 5µυ3

orυ2 =

υ3

4. (1)

Also,υ2 +υ3 = 15. (2)

Solution of Eqs. (1) and (2) gives:

υ2 = 3 V, υ3 = 12 V.

w2 =12×20×10−6×32 = 0.09 (mJ)

w3 =12×5×10−6×122 = 0.36 (mJ)

(b)

Ceq = 6 µF+

(

20µ ×5µ20µ +5µ

)

F = 10 µF

(c)

weq =12

Ceq152 =12×10×10−6×152 = 1.125 (mJ)

w1 +w2 +w3 = (0.675+0.09+0.36) mJ= 1.125 (mJ),

as expected.

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Problem 5.32 Determine Leq at terminals (a,b) in the circuit of Fig. P5.32. Allinductor values are in millihenrys.

Solution:

Leq

b

a

3 33

33

3

3

1

1

4

4

( )13

13 + 3

+ = 2

−1

Leq

b

a

3 2

1

1

4

4

( )14

12 + 2

+ + 4 = 6

−1Leq

b

a 1

1

3

1 + 1 + = 4 mHLeq

b

a

3 6

3 + 6

Figure P5.32

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Exercise 5-12 Determine currents i1 and i2 in the circuit of Fig. E5-12, under dc conditions.

Solution: Under dc conditions, inductors act like short circuits.

The 6-A current will flow entirely through the short circuit representing L3. Hence,

i1 = 0, i2 = 6 A.

Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c© 2013 National Technology Press

Problem 5.29 For the circuit in Fig. P5.29, determine the voltages across C1 and C2

and the currents through L1 and L2 under dc conditions.

Solution:

5 Ω

4 Ω

6 Ω

10 Ω

30 V

+_

L1 = 2 H

L2 = 6 H

C1 = 1 μF

C2 = 2 μF

5 Ω

4 Ω

6 Ω

10 Ω

30 V

+_

L1

i L2

C1

C2

Figure P5.29

i =30

5+6+4= 2 A.

υc2 = 6i = 12 V,

υc1 = 30−5i = 30−10 = 20 V,

iL1 = 0,

iL2 = i = 2 A.

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Exercise 5-14 If in the circuit of Fig. E5-14, υ(0−) = 24 V, determine υ(t) for t ≥ 0.

Figure E5-14

Solution:

υ(t) = υ(0) e−t/τ

= υ(0) e−t/RC

= 24e−10t V, for t ≥ 0.

Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c© 2013 National Technology Press

Problem 5.34 Repeat Problem 5.33, but with the switch having been in position 2for a long time, and then moved to position 1 at t = 0.

R1

R2

υC

iC

C

i1

V0 R31

2

+_

Figure P5.34: Circuit for Problem 5.34.

Solution: (a) At t = 0−

At t = 0−

R1

υC(0−)

iC(0−) = 0

C

i1

V0 R3

2

+_

iC(0−) = 0

υC(0−) = i1R3

=VoR3

R1 +R3

=12×6030+60

= 8 V.

(b) At t = 0

At t = 0

υC(0) = υC(0−) = 8 V

R1

R2

iR2

iR3

iC(0)

C

i1

V0 R31

2

+_

120 kΩ60 kΩ

υC(0+) = 8 V.

iR2 =8

R2=

8120k

=230

mA,

iR3 =8

R3=

860k

=430

mA,

iC(0) = −iR2 − iR3 = −0.2 mA.

(c)

iC(∞) = 0

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υC(∞) = 0

(all energy in capacitor already dissipated into R2 and R3).(d)

υC(t) = υC(∞)+ [υC(0)−υC(∞)]e−t/τ

= 0+(8−0)e−t/τ

= 8e−t/τ, for t ≥ 0,

with

τ =

(

R2R3

R2 +R3

)

C =

(

120×60120+60

)

×103 ×10−4 = 4 s.

(e)

iC(t) = CdυC

dt

= 10−4 ddt

(8e−0.25t)

= −0.2e−0.25t (mA).

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Problem 5.45 Determine υC(t) in the circuit of Fig. P5.45 for t ≥ 0, given that theswitch had been closed for a long time prior to t = 0.

20 V

10 μF

υC+_

+ _

2 kΩ

1 kΩ

1 kΩ

2 kΩ

1 kΩ

t = 0

Figure P5.45: Circuit for Problem 5.45.

Solution: Before opening the switch at t = 0, the state of the circuit is as shown inFig. P5.45(a).

Fig. P5.45 (a) At t = 0−

20 V

υC (0−)

υ1 (0−)

υ2 (0−) υ3 (0

−)+_

+ _

2 kΩ

1 kΩ

1 kΩ

2 kΩ

1 kΩ

Node voltage analysis gives:

υ1(0−)−201k

+υ1(0−)

(2+1)k+

υ1(0−)

(1+2)k= 0,

which leads toυ1(0

−) = 12 V.

By voltage division,

υ2(0−) = υ1(0

−)

(

2k1k+2k

)

= 12×23

= 8 V,

υ3(0−) = υ1(0

−)

(

1k1k+2k

)

= 12×13

= 4 V.

Hence,

υC(0−) = υ2(0−)−υ3(0

−)

= 8−4 = 4 V.

The state of the circuit at t = 0+ (after opening the switch) is

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Fig. P5.45 (b) At t > 0

10 μF

υC+ _

2 kΩ

1 kΩ 2 kΩ

1 kΩ

Req = 3k ‖ 3k = 1.5 kΩ,

τ = ReqC = 1.5×103 ×10−5 = 1.5×10−2 s,

υC(∞) = 0 (no sources after t = 0).

Hence

υC(t) = υC(∞)+ [υC(0)−υC(∞)]e−t/τ

= 4e−100t/1.5 V for t ≥ 0.

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Problem 5.46 After having been in position 1 for a long time, the switch in thecircuit of Fig. P5.46 was moved to position 2 att = 0. Given thatV0 = 12 V,R1 = 30 Ω, R2 = 120Ω, R3 = 60 Ω, andL = 0.2 H, determine:

(a) iL(0−) andυL(0−),

(b) iL(0) andυL(0),

(c) iL(∞) andυL(∞),

(d) iL(t) for t ≥ 0,

(e) υL(t) for t ≥ 0.

Solution:

(b) At t = 0−

(a)

(c) At t > 0

R3

R2

R1

L υL

iL

V0

1

2

+_

R3

R1

L υL

iL

V0

2

+_

1R3

R2

L

iL

(d) At t = 8

R3

R1

L

iL

V0

2

+_

Figure P5.46

(a) At t = 0−, the inductor part of the circuit contained no sources. Hence,

iL(0−) = 0

υL(0−) = 0

(b) After t = 0 (Fig. P5.45(c)),

iL(0) = iL(0−) = 0

υL(0) = υR3 =V0R3

R1 +R3=

12×6030+60

= 8 V.

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(c)

iL(∞) =V0

R1=

1230

= 0.4 A (no current flow throughR3)

υL(∞) = 0.

(d) For t ≥ 0:

Req =R1R3

R1 +R3=

30×6030+60

= 20 Ω

τ =L

Req=

0.220

=1

100s

iL(t) = [iL(∞)+ [iL(0)− iL(∞)]e−t/τ ]

= [0.4+[0−0.4]e−100t ]

= [0.4(1− e−100t)] (A).

(e)

υL(t) = LdiLdt

= 8e−100t (V).

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Problem 5.63 Relate iout(t) to υi(t) in the circuit of Fig. P5.63. Evaluate it forυC(0) = 3 V, R = 10 kΩ, C = 50 µF, and υi(t) = 9u(t) V.

(b) Equivalent circuit

(a)

Cυi

υC

R

+_ iout

υn

υout

υp

υi

C

υC

R

iout+_

Vcc = 12 V

Figure P5.63: Circuit for Problem 5.63.

Solution: Sinceυi = υp = υn = υout,

it follows that the circuit is equivalent to the RC circuit shown in Fig. P5.63(b), forwhich

−υi +υC + ioutR = 0.

Also,

iout = CdυC

dt.

Hence,

RCdυC

dt+υC = υi

dυC

dt+aυC = aυi (1)

where

a =1

RC.

Solution of (1) is

υC(t) eat∣

∣

t0 =

∫ t

0aυie

at dt

υC(t) = υC(0) e−at + e−at∫ t

0aυie

at dt.

For υC(0) = 3 V, υi(t) = 9u(t) (V), and a = 1/(10×103 ×50×10−6) = 2,

υC(t) = 3e−2t + e−2t [9e2t −9]

= 3e−2t +9(1− e−2t)

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= [9−6e−2t ] V, for t ≥ 0.

Since υout = υi and υi never exceeds Vcc = 12 V, the op amp will not experiencesaturation.

iout = CdυC

dt

= 50×10−6 ddt

[9−6e−2t ]

= 0.6e−2t (mA), for t ≥ 0.

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Problem 5.65 In the circuit of Fig. P5.65:

(a) Derive an expression forυout(t) for t ≥ 0 in terms ofR1, R2, R3, C, andA.

(b) Evaluate the expression forR1 = 1 kΩ, R2 = 5 kΩ, R3 = 2 kΩ, C = 0.25 mF,andA = 12 V.

R1 R2

+_

C

υs(t) = Au(t)

υout

R3

υn

υ1

in

i2

iC

υC

i1

υp

+ _

Fig. P5.65 (a)

Figure P5.65: Circuit for Problem 5.65.

Solution: (a) At nodeυ1:

in = 0 υ1 = υn = υp = 0,

i1 + in + i2 + iC = 0,

υ1−AR1

+0+υ1−υout

R3+C

dυC

dt= 0.

Also,υout(t) = −υC(t)

(becauseυ1 = 0). Hence

dυout

dt+

1R3C

υout = −A

R1C.

Setting 1/(R3C) = a, multiplying both sides byeat ′ and then integrating fromt ′ = 0to t ′ = t leads to

∫ t

0

ddt ′

(υouteat ′) dt ′ =

−AR1C

∫ t ′

0eat ′ dt ′.

Keeping in mind thatυout(0) = 0, we have

υout(t) eat = −A

R1Caeat ′

∣

∣

∣

t

0,

which simplifies to

υout(t) = −AR3

R1(1− e−at) for t ≥ 0.

Alternatively, we could have derived this expression directly from Example 5-16upon realizing thatR2 has no impact on the response.

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(b)

υout(t) = −12

(

20001000

)

(1− e−at) for t ≥ 0,

a =1

(2000)(0.25×10−3),

υout(t) = −24(1− e−2t) t ≥ 0.

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