Making use of the passive sign convention, determine the current flowing through a 100 pF capacitor for t ≥ 0 if its voltage v C ( t ) is given by ( a ) 5 V; ( b ) 10 e − t V; ( c ) 2 sin 0.01 t V; ( d ) −5 + 2 sin 0.01 t V.

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Answer 1

Textbook Question

Chapter 7, Problem 1E

Making use of the passive sign convention, determine the current flowing through a 100 pF capacitor for t ≥ 0 if its voltage v_C(t) is given by (a) 5 V; (b) 10e⁻^t V; (c) 2 sin 0.01t V; (d) −5 + 2 sin 0.01t V.

(a)

Expert Solution

To determine

Find the current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 0 A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 5 V.

The capacitance of the capacitor is 100 pF.

Formula used:

The expression for the current in the capacitor is given as follows,

iC=CdvC(t)dt (1)

Here,

iC is the current passing through the capacitor.

C is the capacitance of the capacitor.

vC(t) is the voltage across the capacitor.

Calculation:

Substitute 100 pF for C and 5 V for vC(t) in equation (1).

iC=(100 pF)d(5 V)dt=0 A

Conclusion:

Thus, the current flowing through the capacitor is 0 A.

(b)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is (−10−9)e−t A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 10e−t V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 10e−t V for vC(t) in equation (1).

iC=(100 pF)d(10e−t V)dt=(100×10−12 F)(−10e−t Vs) {∵1 pF=1×10−12 F}=(−10−9)e−t A

Conclusion:

Thus, the current flowing through the capacitor is (−10−9)e−t A.

(c)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Explanation of Solution

Given Data:

The voltage across the capacitor is 2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵ 1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵ 1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

(d)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Explanation of Solution

Given Data:

The voltage across the capacitor is −5+2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and −5+2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(−5+2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

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Answer 2

Textbook Question

Chapter 7, Problem 1E

Making use of the passive sign convention, determine the current flowing through a 100 pF capacitor for t ≥ 0 if its voltage v_C(t) is given by (a) 5 V; (b) 10e⁻^t V; (c) 2 sin 0.01t V; (d) −5 + 2 sin 0.01t V.

(a)

Expert Solution

To determine

Find the current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 0 A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 5 V.

The capacitance of the capacitor is 100 pF.

Formula used:

The expression for the current in the capacitor is given as follows,

iC=CdvC(t)dt (1)

Here,

iC is the current passing through the capacitor.

C is the capacitance of the capacitor.

vC(t) is the voltage across the capacitor.

Calculation:

Substitute 100 pF for C and 5 V for vC(t) in equation (1).

iC=(100 pF)d(5 V)dt=0 A

Conclusion:

Thus, the current flowing through the capacitor is 0 A.

(b)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is (−10−9)e−t A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 10e−t V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 10e−t V for vC(t) in equation (1).

iC=(100 pF)d(10e−t V)dt=(100×10−12 F)(−10e−t Vs) {∵1 pF=1×10−12 F}=(−10−9)e−t A

Conclusion:

Thus, the current flowing through the capacitor is (−10−9)e−t A.

(c)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Explanation of Solution

Given Data:

The voltage across the capacitor is 2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵ 1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵ 1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

(d)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Explanation of Solution

Given Data:

The voltage across the capacitor is −5+2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and −5+2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(−5+2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

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Answer 3

Textbook Question

Chapter 7, Problem 1E

Making use of the passive sign convention, determine the current flowing through a 100 pF capacitor for t ≥ 0 if its voltage v_C(t) is given by (a) 5 V; (b) 10e⁻^t V; (c) 2 sin 0.01t V; (d) −5 + 2 sin 0.01t V.

(a)

Expert Solution

To determine

Find the current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 0 A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 5 V.

The capacitance of the capacitor is 100 pF.

Formula used:

The expression for the current in the capacitor is given as follows,

iC=CdvC(t)dt (1)

Here,

iC is the current passing through the capacitor.

C is the capacitance of the capacitor.

vC(t) is the voltage across the capacitor.

Calculation:

Substitute 100 pF for C and 5 V for vC(t) in equation (1).

iC=(100 pF)d(5 V)dt=0 A

Conclusion:

Thus, the current flowing through the capacitor is 0 A.

(b)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is (−10−9)e−t A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 10e−t V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 10e−t V for vC(t) in equation (1).

iC=(100 pF)d(10e−t V)dt=(100×10−12 F)(−10e−t Vs) {∵1 pF=1×10−12 F}=(−10−9)e−t A

Conclusion:

Thus, the current flowing through the capacitor is (−10−9)e−t A.

(c)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Explanation of Solution

Given Data:

The voltage across the capacitor is 2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵ 1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵ 1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

(d)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Explanation of Solution

Given Data:

The voltage across the capacitor is −5+2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and −5+2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(−5+2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

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Answer 4

Textbook Question

Chapter 7, Problem 1E

Making use of the passive sign convention, determine the current flowing through a 100 pF capacitor for t ≥ 0 if its voltage v_C(t) is given by (a) 5 V; (b) 10e⁻^t V; (c) 2 sin 0.01t V; (d) −5 + 2 sin 0.01t V.

(a)

Expert Solution

To determine

Find the current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 0 A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 5 V.

The capacitance of the capacitor is 100 pF.

Formula used:

The expression for the current in the capacitor is given as follows,

iC=CdvC(t)dt (1)

Here,

iC is the current passing through the capacitor.

C is the capacitance of the capacitor.

vC(t) is the voltage across the capacitor.

Calculation:

Substitute 100 pF for C and 5 V for vC(t) in equation (1).

iC=(100 pF)d(5 V)dt=0 A

Conclusion:

Thus, the current flowing through the capacitor is 0 A.

(b)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is (−10−9)e−t A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 10e−t V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 10e−t V for vC(t) in equation (1).

iC=(100 pF)d(10e−t V)dt=(100×10−12 F)(−10e−t Vs) {∵1 pF=1×10−12 F}=(−10−9)e−t A

Conclusion:

Thus, the current flowing through the capacitor is (−10−9)e−t A.

(c)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Explanation of Solution

Given Data:

The voltage across the capacitor is 2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵ 1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵ 1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

(d)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Explanation of Solution

Given Data:

The voltage across the capacitor is −5+2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and −5+2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(−5+2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

Find the current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 0 A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 5 V.

The capacitance of the capacitor is 100 pF.

Formula used:

The expression for the current in the capacitor is given as follows,

iC=CdvC(t)dt (1)

Here,

iC is the current passing through the capacitor.

C is the capacitance of the capacitor.

vC(t) is the voltage across the capacitor.

Calculation:

Substitute 100 pF for C and 5 V for vC(t) in equation (1).

iC=(100 pF)d(5 V)dt=0 A

Conclusion:

Thus, the current flowing through the capacitor is 0 A.

(b)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is (−10−9)e−t A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 10e−t V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 10e−t V for vC(t) in equation (1).

iC=(100 pF)d(10e−t V)dt=(100×10−12 F)(−10e−t Vs) {∵1 pF=1×10−12 F}=(−10−9)e−t A

Conclusion:

Thus, the current flowing through the capacitor is (−10−9)e−t A.

(c)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Explanation of Solution

Given Data:

The voltage across the capacitor is 2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵ 1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵ 1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

(d)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Explanation of Solution

Given Data:

The voltage across the capacitor is −5+2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and −5+2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(−5+2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

Answer 8

(a)

Expert Solution

To determine

Find the current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 0 A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 5 V.

The capacitance of the capacitor is 100 pF.

Formula used:

The expression for the current in the capacitor is given as follows,

iC=CdvC(t)dt (1)

Here,

iC is the current passing through the capacitor.

C is the capacitance of the capacitor.

vC(t) is the voltage across the capacitor.

Calculation:

Substitute 100 pF for C and 5 V for vC(t) in equation (1).

iC=(100 pF)d(5 V)dt=0 A

Conclusion:

Thus, the current flowing through the capacitor is 0 A.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Find the current flowing through a 100 pF capacitor for t≥0 s.

Answer 13

Answer to Problem 1E

The current flowing through the capacitor is 0 A.

Answer 14

Explanation of Solution

Given Data:

The voltage across the capacitor is 5 V.

The capacitance of the capacitor is 100 pF.

Formula used:

The expression for the current in the capacitor is given as follows,

iC=CdvC(t)dt (1)

Here,

iC is the current passing through the capacitor.

C is the capacitance of the capacitor.

vC(t) is the voltage across the capacitor.

Calculation:

Substitute 100 pF for C and 5 V for vC(t) in equation (1).

iC=(100 pF)d(5 V)dt=0 A

Conclusion:

Thus, the current flowing through the capacitor is 0 A.

Answer 15

(b)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is (−10−9)e−t A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 10e−t V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 10e−t V for vC(t) in equation (1).

iC=(100 pF)d(10e−t V)dt=(100×10−12 F)(−10e−t Vs) {∵1 pF=1×10−12 F}=(−10−9)e−t A

Conclusion:

Thus, the current flowing through the capacitor is (−10−9)e−t A.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer 20

Answer to Problem 1E

The current flowing through the capacitor is (−10−9)e−t A.

Answer 21

Explanation of Solution

Given Data:

The voltage across the capacitor is 10e−t V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 10e−t V for vC(t) in equation (1).

iC=(100 pF)d(10e−t V)dt=(100×10−12 F)(−10e−t Vs) {∵1 pF=1×10−12 F}=(−10−9)e−t A

Conclusion:

Thus, the current flowing through the capacitor is (−10−9)e−t A.

Answer 22

(c)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Explanation of Solution

Given Data:

The voltage across the capacitor is 2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵ 1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵ 1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer 27

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Answer 28

Explanation of Solution

Given Data:

The voltage across the capacitor is 2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and 2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵ 1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵ 1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

Answer 29

(d)

Expert Solution

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Explanation of Solution

Given Data:

The voltage across the capacitor is −5+2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and −5+2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(−5+2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The current flowing through a 100 pF capacitor for t≥0 s.

Answer 34

Answer to Problem 1E

The current flowing through the capacitor is 2(cos (0.01t)) pA.

Answer 35

Explanation of Solution

Given Data:

The voltage across the capacitor is −5+2(sin (0.01t)) V.

The capacitance of the capacitor is 100 pF.

Calculation:

Substitute 100 pF for C and −5+2(sin (0.01t)) V for vC(t) in equation (1).

iC=(100 pF)d(−5+2(sin (0.01t)) V)dt=(100×10−12 F)(2×0.01(cos (0.01t)) Vs) {∵1 pF=1×10−12 F}=2×10−12(cos (0.01t)) A=2(cos (0.01t)) pA {∵1 A=1×1012 pA}

Conclusion:

Thus, the current flowing through the capacitor is 2(cos (0.01t)) pA.

Answer 36

Ch. 7.1 - Determine the current flowing through a 5 mF...Ch. 7.1 - Prob. 2P Ch. 7.1 - Prob. 3P Ch. 7.2 - 7.4 The current through a 200 mH inductor is shown...Ch. 7.2 - The current waveform of Fig. 7.14a has equal rise...Ch. 7.2 - Prob. 6P Ch. 7.2 - Let L = 25 mH for the inductor of Fig. 7.10. (a)...Ch. 7.3 - Find Ceq for the network of Fig. 7.23. FIGURE...Ch. 7.4 - If vC(t) = 4 cos 105t V in the circuit in Fig....Ch. 7.5 - Derive an expression for vout in terms of vs for...

Making use of the passive sign convention, determine the current flowing through a 100 pF capacitor for t ≥ 0 if its voltage v C ( t ) is given by ( a ) 5 V; ( b ) 10 e − t V; ( c ) 2 sin 0.01 t V; ( d ) −5 + 2 sin 0.01 t V.

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