# Circuits Solutions

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Problem 1.1 Use appropriate multiple and submultiple prefixes to express thefollowing quantities:

(a) 3,620 watts (W)(b) 0.000004 amps (A)(c) 5.2106 ohms ()(d) 3.91011 volts (V)(e) 0.02 meters (m)(f) 32105 volts (V)

Solution:(a) 3,620 W = 3.62 kW.(b) 0.000004 A = 4 A.(c) 5.2106 = 5.2 .(d) 3.91011 V = 390 GV.(e) 0.02 m = 20 mm.(f) 32105 V = 3.2 MV.

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Problem 1.2 Use appropriate multiple and submultiple prefixes to express thefollowing quantities:

(a) 4.71108 seconds (s)(b) 10.3108 watts (W)(c) 0.00000000321 amps (A)(d) 0.1 meters (m)(e) 8,760,000 volts (V)(f) 3.161016 hertz (Hz)

Solution:(a) 4.71108 s = 47.1 ns.(b) 10.3108 W = 1.03 GW.(c) 0.00000000321 A = 3.21 nA.(d) 0.1 = 10 cm.(e) 8,760,000 V = 8.76 MV.(f) 3.161016 Hz = 316 aHz.

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Problem 1.3 Convert:(a) 16.3 m to mm(b) 16.3 m to km(c) 4106 F (microfarad) to pF (picofarad)(d) 2.3 ns to s(e) 3.6107 V to MV(f) 0.03 mA (milliamp) to A

Solution:(a) 16.3 m = 16,300 mm.(b) 16.3 m = 0.0163 km.(c) 4106 F = 4 pF.(d) 2.3 ns = 2.3103 s.(e) 3.6107 V = 36 MV.(f) 0.03 mA = 30 A.

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Problem 1.4 Convert:(a) 4.2 m to m(b) 3 hours to seconds(c) 4.2 m to km(d) 173 nm to m(e) 173 nm to m(f) 12 pF (picofarad) to F (farad)

Solution:(a) 4.2 m = 4.2106 m.(b) 3 hours = 1.081010 s.(c) 4.2 m = 4.2103 km.(d) 173 nm = 1.73107 m.(e) 173 nm = 0.173 m.(f) 12 pF = 1.21011 F.

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Problem 1.5 For the circuit in Fig. P1.5:(a) Identify and label all distinct nodes.(b) Which of those nodes are extraordinary nodes?(c) Identify all combinations of 2 or more circuit elements that are connected in

series.(d) Identify pairs of circuit elements that are connected in parallel.

4 5 2

1 3

16 V+

_

Figure P1.5: Circuit for Problem 1.5.

Solution:

4 5 2

1 3

16 V+

_

a b

d

c

Fig. P1.5 (a)

(a) Nodes identified in Fig. P1.5(a).(b) Nodes b, c, and d are extraordinary.(c) Series connections: 16 V and 1 .(d) Parallel connections: 4 and 5 .

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Problem 1.6 For the circuit in Fig. P1.6:(a) Identify and label all distinct nodes.(b) Which of those nodes are extraordinary nodes?(c) Identify all combinations of 2 or more circuit elements that are connected in

series.(d) Identify pairs of circuit elements that are connected in parallel.

2

4 4

8 V12 V+

_

+

_

+

_

+

_

Figure P1.6: Circuit for Problem 1.6.

Solution:

Fig. P1.6 (a)

2

4 4

8 V12 V+

_

+

_

+

_

+

_

a b

d

c

(a) Nodes identified in Fig. P1.6(a).(b) Nodes b and d are extraordinary.(c) Series connections: 12 V and 4

4 and 8 V.(d) Parallel connections: none.

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Problem 1.7 For the circuit in Fig. P1.7:(a) Identify and label all distinct nodes.(b) Which of those nodes are extraordinary nodes?(c) Identify all combinations of 2 or more circuit elements that are connected in

series.(d) Identify pairs of circuit elements that are connected in parallel.

1 0.1

0.2

0.3

0.4

1 4 V

+

_

Figure P1.7: Circuit for Problem 1.7.

Solution:

Fig. P1.7 (a)

1 0.1

0.2

0.3

0.4

1 4 V

+

_

ba

e

c d

(a) Nodes identified in Fig. P1.7(a).(b) Nodes b, c, d, and e are extraordinary.(c) Series connections: 4 V and 1 .(d) Parallel connections: None.

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Problem 1.8 For the circuit in Fig. P1.8:(a) Identify and label all distinct nodes.(b) Which of those nodes are extraordinary nodes?(c) Identify all combinations of 2 or more circuit elements that are connected in

series.(d) Identify pairs of circuit elements that are connected in parallel.

25 40

30

10

20

60 5

12 V

15

+

_

Figure P1.8: Circuit for Problem 1.8.

Solution:

Fig. P1.8 (a)

25 40

30

10

20

60 5

12 V

15

+

_

a b

e

f c d

(a) Nodes identified in Fig. P1.8(a).(b) Nodes b, c, e, and f are extraordinary.(c) Series connections: 12 V and 25

10 and 20 (d) Parallel connections: None.

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Problem 1.9 For the circuit in Fig. P1.9:(a) Identify and label all distinct nodes.(b) Which of those nodes are extraordinary nodes?(c) Identify all combinations of 2 or more circuit elements that are connected in

series.(d) Identify pairs of circuit elements that are connected in parallel.

3 6

2 2 4

4 A

48 V+

_

Figure P1.9: Circuit for Problem 1.9.

Solution:

3 6

2

2 4

4 A

48 V+

_

b

e

a

d

c

Fig. P1.9 (a)

(a) Nodes identified in Fig. P1.9(a).(b) Nodes a, b, c, and e are extraordinary.(c) Series connections: 2 and 48 V.(d) Parallel connections: None.

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Problem 1.10 For the circuit in Fig. P1.10:(a) Identify and label all distinct nodes.(b) Which of those nodes are extraordinary nodes?(c) Identify all combinations of 2 or more circuit elements that are connected in

series.(d) Identify pairs of circuit elements that are connected in parallel.

6 16

32 8

4 12 V+

_

Figure P1.10: Circuit for Problem 1.10.

Solution:

6 16

32

10

8

4 12 V+

_

a b c d

b

Fig. P1.10 (a)

(a) Nodes identified in Fig. P1.10(a).(b) Nodes b and c are extraordinary.(c) Series connections: 8 and 6 .(d) Parallel connections: 12 V and 4

4 and short circuit.

Problem 1.11 For the circuit in Fig. P1.11:(a) Identify and label all distinct nodes.(b) Which of those nodes are extraordinary nodes?(c) Identify all combinations of 2 or more circuit elements that are connected in

series.(d) Identify pairs of circuit elements that are connected in parallel.

6

2

4

3

1

5 20 V

+

_

+

_

Figure P1.11: Circuit for Problem 1.11.

Solution:

6

2

4

3

1

5 20 V

+

_

e

c d

b

a

Fig. P1.11

(a) Nodes identified in Fig. P1.11(a).(b) Nodes a, b, and d are extraordinary.(c) Series connections: 5 and 20 V

4 and 6 .(d) Parallel connections: 2 and 1 .

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Problem 1.12 The total charge contained in a certain region of space is 1 C. Ifthat region contains only electrons, how many does it contain?

Solution:ne =

Qqe

=

11.61019 = 6.2510

18 electrons.

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Problem 1.13 A certain cross section lies in the xy plane. If 3 1020 electronsgo through the cross section in the z-direction in 4 seconds, and simultaneously,1.51020 protons go through the same cross section in the negative z-direction, whatis the magnitude and direction of the current flowing through the cross section?

Solution: Negatively charged electrons moving along +z-direction constitute acurrent in the z-direction:

Ie =Qt =

310201.610194

= 12 A, along z-direction.

Positively charged protons moving along z-direction constitute a current in thez-direction:

Ip =Qt =

1.510201.610194

= 6 A, along z-direction.

Total net current is:

I = Ie + Ip = 12+6 = 18 A, along z-direction.

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Problem 1.14 Determine the current i(t) flowing through a resistor if thecumulative charge that has flowed through it up to time t is given by

(a) q(t) = 3.6t mC(b) q(t) = 5sin(377t) C(c) q(t) = 0.3[1 e0.4t ] pC(d) q(t) = 0.2t sin(120pit) nC

Solution:(a) i(t) = dqdt = ddt (3.6t103) = 3.6103 = 3.6 (mA).(b) i(t) = dqdt = ddt [(5sin377t)106] = 5377106 cos377t = 1.885cos377t(mA).(c) i(t) = dqdt = ddt [0.3(1 e0.4t)1012] = 0.31012 (0.4) (e0.4t)

= 0.12e0.4t (pA).(d) i(t) = dqdt = ddt [(0.2t sin120pit)109]

= (0.2sin120pit +0.2t120pi cos120pit)109

= 0.2sin120pit +75.4t cos120pit (nA).

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Problem 1.15 Determine the current i(t) flowing through a certain device if thecumulative charge that has flowed through it up to time t is given by

(a) q(t) =0.45t3 C(b) q(t) = 12sin2(800pit) mC(c) q(t) =3.2sin(377t)cos(377t) pC(d) q(t) = 1.7t[1 e1.2t ] nC

Solution:(a)

i(t) =dqdt =

ddt [0.45t

3106] =0.453t2106 =1.35t2 (A).

(b)

i(t) =dqdt =

ddt [(12sin

2 800pit)103] = 212800pi103 sin800pit cos800pit

= 60.32sin800pit cos800pit (A).

(c)

i(t) =dqdt =

ddt [(3.2sin377t cos377t)10

12]

= [(3.2377cos2 377t +3.2377sin2 377t)1012]

= 1.21(sin2 377t cos2 377t) (nA).

(d)

i(t) =dqdt =

ddt [1.7t(1 e

1.2t)109]

= [1.7(1 e1.2t)+1.7t(1.2)(e1.2t)]109

= 1.7(1 e1.2t +1.2te1.2t) (nA).

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Problem 1.16 Determine the net charge Q that flowed through a resistor over thespecified time interval for each of the following currents:

(a) i(t) = 0.36 A, from t = 0 to t = 3 s(b) i(t) = [40t +8] mA, from t = 1 s to t = 12 s(c) i(t) = 5sin(4pit) nA, from t = 0 to t = 0.05 s(d) i(t) = 12e0.3t mA, from t = 0 to t =

Solution:(a)

Q(0,3) = 3

0i dt =

30

0.36 dt = 0.36t|30 = 1.08 (C).

(b)

Q(1,12) = 12

1i dt =

[ 121

(40t +8)dt]103

=

(40t2

2+8t

)12

1103 = 2.948 (C).

(c)

Q(0,0.05) = 0.05

0i dt =

[ 0.050

5sin4pit dt]109

=

5cos4pit4pi

0.05

0109

= (0.32+0.40)109 = 80 (pC).

(d)

Q(0,) =

0i dt =

[

012e0.3t dt

]103 = 12e

0.3t

0.3

0103 = 40 (mC).

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Problem 1.17 Determine the net charge Q that flowed through a certain deviceover the specified time intervals for each of the following currents:

(a) i(t) = [3t +6t3] mA, from t = 0 to t = 4 s(b) i(t) = 4sin(40pit)cos(40pit) A, from t = 0 to t = 0.05 s(c) i(t) = [4et 3e2t ] A, from t = 0 to t = (d) i(t) = 12e3t cos(40pit) nA, from t = 0 to t = 0.05 s

Solution:(a)

Q(0,4) = 4

0i dt =

[ 40

(3t +6t3)dt103]

=

(3t2

2+

6t44

)4

0103 = 408 (mC).

(b)

Q(0,0.05) = 0.05

0i dt =

[ 0.050

4sin40pit cos40pit dt]106

=

4240pi sin

2 40pit|0.050 106 = 0.

(c)

Q(0,) =

0i dt =

0(4et 3e2t)dt =

(4et +

32

e2t)

0= 2.5 (C).

(d)Q(0,0.05) =

0.050

i dt =[ 0.05

012e3t cos40pit dt

]109.

From Tables of Integrals,

eax cosbx dx = eax (acosbx+bsinbx)a2 +b2 .

Hence,

Q(0,0.05) =[

12e3t(3cos40pit +40pi sin40pit)

9+(40pi)2

0.05

0

]109 = 0.32 (pC).

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Problem 1.18 If the current flowing through a wire is given by i(t) = 3e0.1t mA,how many electrons pass through the wires cross section over the time interval fromt = 0 to t = 0.3 ms?

Solution:

Q(0,0.3 ms) = 0.3 ms

0i dt =

[ 0.3 ms0

3e0.1t dt]103

=

3e0.1t

0.1 103

0.3103

0

=30(e31051)103 = 9107 (C).

ne =Qe

=

91071.61019 = 5.6210

12 electrons.

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Problem 1.19 The cumulative charge in mC that entered a certain device is givenby

q(t) =

0, for t < 0,5t, for 0 t 10 s,60 t, for 10 s t 60 s

(a) Plot q(t) versus t from t = 0 to t = 60 s.(b) Plot the corresponding current i(t) entering the device.

Solution:(a)

10

50 mC

20 30 40 50 60

q(t)

t (s)

(b)

10

5 mA

1 mA

20 30 40 50 60

i(t)

t (s)

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Problem 1.20 A steady flow resulted in 3 1015 electrons entering a device in0.1 ms. What is the current?

Solution:i =

Qt =

nee

t =310151.61019

0.1103 = 4.8 A.

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Problem 1.21 Given that the current in mA flowing through a wire is given by

i(t) =

0, for t < 0,6t, for 0 t 5 s,30e0.6(t5), for t 5 s,

(a) Sketch i(t) versus t.(b) Sketch q(t) versus t.

Solution:(a)

2

30 mA

4 6 8 10 12 14

i(t)

t (s)

20 mA

10 mA

(b) q(t) = t i(t)dt.For 0 t 5 s,

q(t) =( t

06t dt

)103 = 6t

2

2

t

0103 = 3t2 (mC).

For t 5 s,

q(t) =[ 5

06t dt +

t5

30e0.6(t5) dt]103

=

[6t22

5

0+30e+3

t5

e0.6t dt]103

= [75+50(1 e0.6(t5))] (mC).

2

125

100

75

50

25

4 6 8 10 12 14

q (mC)

t (s)

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Problem 1.22 The plot in Fig. P1.22 displays the cumulative amount of charge q(t)that has entered a certain device up to time t. What is the current at

(a) t = 1 s(b) t = 3 s(c) t = 6 s

q(t)

t2 s 4 s

4 C

4 C

06 s 8 s

Figure P1.22: q(t) for Problem 1.22.

Solution:(a) a = 42 = 2 A @ t = 1 s (slope of first segment).(b) i = 0 @ t = 3 s (slope of q(t) = 0 at t = 3 s).(c) i = 84 =2 A @ t = 6 s (negative slope of third segment).

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Problem 1.23 The plot in Fig. P1.23 displays the cumulative amount of charge q(t)that has exited a certain device up to time t. What is the current at:

(a) t = 2 s(b) t = 6 s(c) t = 12 s

q(t)

t4 s 8 s

2 C

0

4 C4e0.2(t8)

Figure P1.23: q(t) for Problem 1.23.

Solution:(a) i = 0 @ t = 2 s (slope = 0 of first segment).(b) i = 4284 = 24 = 0.5 A (slope of second segment).(c)

i =dqdt =

ddt (4e

0.2(t8)) = 4e1.6

ddt e

0.2t=40.2e1.6e0.2t

=0.36 A @ t = 12 s.

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Problem 1.24 The plot in Fig. P1.24 displays the cumulative charge q(t) that hasentered a certain device up to time t. Sketch a plot of the corresponding current i(t).

q

t (s)1 2

20 C

20 C

03 54

Figure P1.24: q(t) for Problem 1.24.

Solution: Based on the slope of q(t):

i(t) =dqdt =

20 A for 0 t 1 s20 A for 1 t 3 s0 for 3 t 4 s20 A for 4 t 5 s0 for t 5 s

Fig. P1.24

i (A)

t (s)1 2

20

20

03 54

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Problem 1.25 In the circuit of Fig. P1.25, node V1 was selected as the ground node.(a) What is the voltage at node V2?(b) What is the voltage difference V32 = V3V2?(c) What are the voltages at nodes 1, 3, 4, and 5 if node 2 is selected as the ground

node instead of node 1?

R1

R3

R4

V2

V1 = 0

V3 = 32 VV5 = 20 VV4 = 10 V

48 V

R5

R2

+

_

Figure P1.25: q(t) for Problem 1.25.

Solution:(a) V2 = 48 V(b) V32 = V3V2 = 3248 =16 V

R1

R3

R4

V2 = 0

V1

48 V

R5

R2

+

_

V3 V5V4

Fig. P1.25 (a)

(c) V1 =48 VFrom part (b), V32 =16 V, which means V3 is below V2 by 16 V. Hence,

V3 =16 V (relative to V2).

From the earlier configuration, V4 was 22 V below V3. Hence, in the newconfiguration

V4 =2216 =38 V (relative to V2).Similarly,

V5 =1216 =28 V (relative to V2).

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Problem 1.26 In the circuit of Fig. P1.26, node V1 was selected as the ground node.(a) What is the voltage difference across R6?(b) What are the voltages at nodes 1, 3, and 4 if node 2 is selected as the ground

node instead of node 1?

R2

R3

R4R1

R6

10 V

R5

+

_

20 V

V1 = 0 V2 = 4 V

V3 = 6 V

V4 = 12 V

+

_

Figure P1.26: q(t) for Problem 1.26.

Solution:(a) V3 = 6 V, relative to V1. This includes a voltage rise of 10 V due to the voltagesource. Hence, the voltage across R6 must be 4 V.(b)

Fig. P1.26 (a)

R3

R4

R6

10 V

R5

+

_

20 V

V1 = 4 V V2 = 0

V3 = 2 V

V4 = 8 V

+

_

The new voltages are:

V1 =4 V (relative to V2),V3 = 6 V4 V = 2 V (relative to V2),V4 = 12 V4 V = 8 V (relative to V2).

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Problem 1.27 For each of the eight devices in the circuit of Fig. P1.27, determinewhether the device is a supplier or a recipient of power and how much power it issupplying or receiving.

+ _+ _

6 V

4 A 3 A

16 V

1 A

1 A

2 A

+ _12 V

+

+_

_

10 V

+

_

9 V

+

_

7 V

4 V+ _8 V

1

2

3

4 5

6

7

8

Figure P1.27: Circuit for Problem 1.27.

Solution:Device 1: p = i = 16 (4) =64 W (supplier)Device 2: p = i = 64 = 24 W (recipient)Device 3: p = i = 101 = 10 W (recipient)Device 4: p = i = 81 = 8 W (recipient)Device 5: p = i = 41 = 4 W (recipient)Device 6: p = i = 124 = 24 W (recipient)Device 7: p = i = 9 (3) =27 W (supplier)Device 8: p = i = 73 = 21 W (recipient)

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Problem 1.28 For each of the seven devices in the circuit of Fig. P1.28, determinewhether the device is a supplier or a recipient of power and how much power it issupplying or receiving.

+ _6 V

+ _4 V

5 A

4 A

2 A3 A

1 A

2 A

24 V

+

_

10 V

6 V

12 V

+_

+_

+_

+_

8 V

1

2

3 4

5

6 7

Figure P1.28: Circuit for Problem 1.28.

Solution:Device 1: p = i = 24 (5) =120 W (supplier)Device 2: p = i = 65 = 30 W (recipient)Device 3: p = i = 81 = 8 W (recipient)Device 4: p = i = 124 = 48 W (recipient)Device 5: p = i = 4 (2) =8 W (supplier)Device 6: p = i = 103 = 30 W (recipient)Device 7: p = i = 62 = 12 W (recipient)

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Problem 1.29 An electric oven operates at 120 V. If its power rating is 0.6 kW,what amount of current does it draw and how much energy does it consume in 12minutes of operation?

Solution:

i =p

=

0.6103120 = 5 (A).

w = p t = 0.61031260 = 432 (kJ).

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Problem 1.30 A 9-V flashlight battery has a rating of 1.8 kWh. If the bulb draws acurrent of 100 mA when lit, determine the following:

(a) For how long will the flashlight provide illumination?(b) How much energy in joules is contained in the battery?(c) What is the batterys rating in ampere-hours?

Solution:(a)

t = wp

=

W i

=

1.8103

9100103 hours

= 2,000 hours.

(b) W = 1.81033600 = 6.48 (MJ).(c) Ampere-hours = 1.8 kWh9 V = 0.2 (kAh).

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Problem 1.31 The voltage across and current through a certain device are given by

(t) = 5cos(4pit) V,i(t) = 0.1cos(4pit) A.

Determine:(a) The instantaneous power p(t) at t = 0 and t = 0.25 s.(b) The average power pav, defined as the average value of p(t) over a full time

period of the cosine function (0 to 0.5 s).

Solution:(a)

p(t) = i = (5cos4pit)(0.1cos4pit) = 0.5cos2 4pit (W).p(0) = 0.5 W @ t = 0

p(0.25 s) = 0.5cos2(4pi0.25) = 0.5 W @ t = 0.25 s.

(b)

pav =1T

T0

p(t)dt = 10.5

0.50

0.5cos2 4pit dt

=

18pi [sin4pit cos4pit +4pit]|

0.50 =

14

(W).

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Problem 1.32 The voltage across and current through a certain device are given by

(t) = 100(1 e0.2t) V

i(t) = 30e0.2t mA.

Determine:(a) The instantaneous power p(t) at t = 0 and t = 3 s.(b) The cumulative energy delivered to the device from t = 0 to t = .

Solution:(a)

p(t) = i = 100(1 e0.2t)30e0.2t103

= 3(e0.2t e0.4t) (W).

(b)

W =

0p(t)dt = 3(e0.2t e0.4t)dt

=

(3e0.2t

0.2 +3e0.4t

0.4

)

0

=

30.2

30.4 = 7.5 (J).

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Problem 1.33 The voltage across a device and the current through it are showngraphically in Fig. P1.33. Sketch the corresponding power delivered to the deviceand calculate the energy absorbed by it.

10 A

5 A

1 s 2 s

i(t)

t

5 V

0

01 s 2 s

(t)

t

Figure P1.33: i(t) and (t) of the device in Problem 1.33.

Solution: For 0 t 1 s,

p = i = 5t10 = 50t

For 1 s t 2 s,

p = i = (105t)5 = 5025t.

W = 2

0p dt =

10

50t dt + 2

1(5025t)dt

=

50t22

1

0+

(50t 25t

2

2

)2

1

= 37.5 (J).

1 s 2 s

50 W

25 W

p(t)

t (s)

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Problem 1.34 The voltage across a device and the current through it are showngraphically in Fig. P1.34. Sketch the corresponding power delivered to the deviceand calculate the energy absorbed by it.

10 A

1 s 2 s

i(t)

t

5 V

1 s 2 s

(t)

t

0

0

Figure P1.34: i(t) and (t) of the device in Problem 1.34.

Solution: For 0 t 1 s,

p(t) = i = (5t)(10t) = 50t2

For 1 s t 2 s,

= 5(2 t)i = 10(2 t)

p(t) = 50(2 t)2

1 s 2 s

50 W

p(t)

t

w =

20

p(t) dt

=

10

50t2 dt + 2

150(2 t)2 dt

= 33.3 J.

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Problem 1.35 The voltage across a device and the current through it are showngraphically in Fig. P1.35. Sketch the corresponding power delivered to the deviceand calculate the energy absorbed by it.

10 A

4 s

i(t)

t

1 s 3 s

4 s

1 s

3 s

0

2 s

(t)

t

5 V

0

5 V

Figure P1.35: i(t) and (t) of the device in Problem1.35.

Solution:For 0 t 1 s,

p(t) = i = 5(10t) = 50t.

For 1 s t 3 s, p = 0.For 3 s t 4 s,

=5 V,i = (10t +40) A,

p(t) = i = 50t200.

For t 4 s, p = 0.

41 2

p(t)

t (s)

50 W

0

50 W

3

Fig. P1.35 (a)

Energy w = 0.

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Problem 1.36 After t = 0, the current entering the positive terminal of a flashlightbulb is given by

i(t) = 2(1 e10t) (A),and the voltage across the bulb is (t) = 12e10t (V). Determine the maximum powerlevel delivered to the flashlight.

Solution:

p(t) = i

= 12e10t [2(1 e10t)]

= 24(e10t e20t).

to find pmax, we take the derivative of p(t) and equate it to zero:

d pdt =

ddt [24(e

10t e20t)]

= 24(10)e10t24(20)e20t = 0,

which simplifies toe10t2e20t = 0.

Dividing by e10t gives12e10t = 0,

or

e10t =12

.

Taking the natural log of both sides gives

ln(e10t) = ln(

12

),

10t =0.693,

or

t = 0.0693 s.

Using this value of t in the expression for p(t) gives

pmax = 24(e100.0693 e200.0693)= 24(0.50.25) = 6 W.

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Problem 1.37 Apply the law of conservation of power to determine the amount ofpower delivered to device 4 in the circuit of Fig. P1.37, given that that the amountsof power delivered to the other devices are: p1 =100 W, p2 = 30 W, p3 = 22 W,p5 = 67 W, p6 =201 W, and p7 = 120 W.

P4 = ?

2

6

4

5

1 3 7

Figure P1.37: Circuit of Problem 1.37.

Solution: Conservation of power requires that

p1 + p2 + p3 + p4 + p5 + p6 + p7 = 0.

Hence,

p4 =p1 p2 p3 p5 p6 p7=(100)302267 (201)120= 62 W.

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Problem 1.38 Determine Vy in the circuit of Fig. P1.38.

10 2

5

Vy12 V

1.2 A

+

_

Vx

I = 0.1Vx+ _

+

_

Figure P1.38: Circuit of Problem 1.38.

Solution: Give that a 1.2-A current is entering the + terminal of Vx it follows that

Vx = 1.25 = 6 V.

Hence,I = 0.1Vx = 0.6 A,

andVy = 0.62 = 1.2 V.

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Problem 1.39 Determine V , the voltage of the dependent voltage source in thecircuit of Fig. P1.39.

20

10 V

Ix

30

5

10 V = 2Ix

+

_

+

_ 15 V

+

_

Figure P1.39: Circuit of Problem 1.39.

Solution: Given that the voltage across the 5- resistor is 15 V, it follows that

Ix =155 = 3 A.

Hence,V = 2Ix = 23 = 6 V.

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Problem 1.40 For the circuit in Fig. P1.40, generate circuit diagrams that includeonly those elements that have current flowing through them for

(a) t < 0(b) 0 < t < 2 s(c) t > 2 s

+

_

V0

R1

R2 R3

R4

R6R5

t = 2 s

t = 0

Figure P1.40: Circuit for Problem 1.40.

Solution:(a) t < 0

+

_

V0

R1

R2 R3

R6R5

(b) 0 < t < 2 s

+

_

V0

R1

R2

R5

(c) t > 2 s

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+_

V0

R1

R2

R4

R6R5

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Problem 1.41 For the circuit in Fig. P1.41, generate circuit diagrams that includeonly those elements that have current flowing through them for

(a) t < 0(b) 0 < t < 2 s(c) t > 2 s

V1

V2

R1 SPST

SPST

SPDT R2R3R4

R6

R5+_

+_

t = 0

t = 0

t = 2 s

12

Figure P1.41: Circuit for Problem 1.41.

Solution:(a) t < 0

V2

R2R3R4

R6

+_

(b) 0 < t < 2 s

V1

R1

R2R3

+_

(c) t > 2 s

V1

R1

R3R4

R5

+_

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Problem 1.42 The switch in the circuit of Fig. P1.42 closes at t = 0. Whichelements are in-series and which are in-parallel at (a) t < 0 and (b) t > 0?

+

_

s R4

R6R5

R3R1

R2t = 0

2

1

3 4

Figure P1.42: Circuit for Problem 1.42.

Solution:(a) At t < 0: s and R1 are in-series

R5 and R6 are in-parallel with each other and with a short circuit

R2 is in-parallel with the series combination of s and R1

(b) At t > 0:All of the above continues to be true, but in addition:

R3 and R4 are in-series and their combination is in-parallel with R2

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prob1-1s1prob1-2s1prob1-3s1prob1-4s1prob1-5s1prob1-6s1prob1-7s1prob1-8s1prob1-9s1prob1-10s1prob1-11s1prob1-12s1prob1-13s1prob1-14s1prob1-15s1prob1-16s1prob1-17s1prob1-18s1prob1-19s1prob1-20s1prob1-21s1prob1-22s1prob1-23s1prob1-24s1prob1-25s1prob1-26s1prob1-27s1prob1-28s1prob1-29s1prob1-30s1prob1-31s1prob1-32s1prob1-33s1prob1-34s1prob1-35s1prob1-36s1prob1-37s1prob1-38s1prob1-39s1prob1-40s1prob1-41s1prob1-42s1