Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf
Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf
9th Edition
ISBN: 9781259989452
Author: Hayt
Publisher: Mcgraw Hill Publishers
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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 vC(t) is given by (a) 5 V; (b) 10et V; (c) 2 sin 0.01t V; (d) −5 + 2 sin 0.01t V.

(a)

Expert Solution
Check Mark
To determine

Find the current flowing through a 100 pF capacitor for t0 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
Check Mark
To determine

The current flowing through a 100 pF capacitor for t0 s.

Answer to Problem 1E

The current flowing through the capacitor is (109)et A.

Explanation of Solution

Given Data:

The voltage across the capacitor is 10et V.

The capacitance of the capacitor is 100 pF.

Calculation:

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

iC=(100 pF)d(10et V)dt=(100×1012 F)(10et Vs)                   {1 pF=1×1012 F}=(109)et A

Conclusion:

Thus, the current flowing through the capacitor is (109)et A.

(c)

Expert Solution
Check Mark
To determine

The current flowing through a 100 pF capacitor for t0 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=(100×1012 F)(2×0.01(cos (0.01t)) Vs)                   { 1 pF=1×1012 F}=2×1012(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
Check Mark
To determine

The current flowing through a 100 pF capacitor for t0 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=(100×1012 F)(2×0.01(cos (0.01t)) Vs)                   {1 pF=1×1012 F}=2×1012(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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Chapter 7 Solutions

Loose Leaf for Engineering Circuit Analysis Format: Loose-leaf

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