Electric Circuits (10th Edition)
Electric Circuits (10th Edition)
10th Edition
ISBN: 9780133760033
Author: James W. Nilsson, Susan Riedel
Publisher: PEARSON
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Question
Chapter 6, Problem 2P

a.

To determine

Write the expressions of current i(t) in the intervals t<0,0t25ms,25mst50ms,andt>50ms.

a.

Expert Solution
Check Mark

Answer to Problem 2P

The current i(t) in the intervals t<0,0t25ms,25mst50ms,andt>50ms are 0,4tA,0.2+4tA,and0_ respectively.

Explanation of Solution

Given data:

Refer to the given figure in the respective question for the triangular current pulse.

Calculation:

PSPICE:

Design the circuit in PSPICE that contains the inductor and the given triangular pulse as shown in Figure 1.

Electric Circuits (10th Edition), Chapter 6, Problem 2P , additional homework tip  1

Provide the simulation settings as shown in Figure 2.

Electric Circuits (10th Edition), Chapter 6, Problem 2P , additional homework tip  2

Run the circuit and obtain the graph of current versus time as shown n Figure 3.

Electric Circuits (10th Edition), Chapter 6, Problem 2P , additional homework tip  3

For t<0andt>50ms, the current i(t) is zero since the input current pulse is zero.

From Figure 3, it is concluded that, for 0t25msand25mst50ms, the current i(t) are 4tAand0.2+4tA respectively.

Conclusion:

Thus, the current i(t) in the intervals t<0,0t25ms,25mst50ms,andt>50ms are 0,4tA,0.2+4tA,and0_ respectively.

b.

To determine

Derive the expressions for the inductor voltage, power, and energy.

b.

Expert Solution
Check Mark

Answer to Problem 2P

The expressions for the inductor voltage, power, and energy are v={0t<02V0t25ms2V25mst50ms0t>50ms_, p={0t<08tW0t25ms8t0.4W25mst50ms0t>50ms_, and w={0t<04t2J0t25ms4t20.4t+(10×103)J25mst50ms0t>50ms_ respectively.

Explanation of Solution

Calculation:

Write the expression for the inductor voltage.

v=Ldidt        (1)

Substitute 50 mH for L and 4t A for i in Equation (1).

v=(50mH)d(4t)dt{0t25ms}=(50×103H)(4)=2V

Substitute 50 mH for L and 0.2+4tA for i in Equation (1).

v=(50mH)ddt(0.2+4t){25mst50ms}=(50×103H)(4)=2V

For t<0andt>50ms, the inductor voltage is zero since the input current is zero.

The inductor voltage expression is,

v={0t<02V0t25ms2V25mst50ms0t>50ms

Consider the expression for the power in the inductor.

p=vi        (2)

Substitute 2V for v and 4t A for i in Equation (2).

p=(2)(4t){0t25ms}=8tW

Substitute 2V for v and (0.2+4t)A for i in Equation (2).

p=(2)(0.2+4t)A{25mst50ms}=8t0.4W

For t<0andt>50ms, the inductor power is zero since the input current is zero.

The power in the inductor is,

p={0t<08tW0t25ms8t0.4W25mst50ms0t>50ms

For t<0andt>50ms, the energy in the inductor is zero since the input current is zero.

Calculate the energy in the inductor.

w={0t<00t(8x)dx0t25ms0.025t(8x0.4)dx+2.5×10325mst50ms0t>50ms

Simplify the equation as follows.

w={0t<0[8x22]0t=4t2J0t25ms[4x20.4x]0.025t+(2.5×103)=4t20.4t+(10×103)J25mst50ms0t>50ms

Conclusion:

Thus, the expressions for the inductor voltage, power, and energy are v={0t<02V0t25ms2V25mst50ms0t>50ms_, p={0t<08tW0t25ms8t0.4W25mst50ms0t>50ms_, and w={0t<04t2J0t25ms4t20.4t+(10×103)J25mst50ms0t>50ms_ respectively.

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Chapter 6 Solutions

Electric Circuits (10th Edition)

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