Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
11th Edition
ISBN: 9780134746968
Author: James W. Nilsson, Susan Riedel
Publisher: PEARSON
Question
Book Icon
Chapter 7, Problem 1P

a.

To determine

Find the current values i1(0) and i2(0).

a.

Expert Solution
Check Mark

Answer to Problem 1P

The current values i1(0) and i2(0) are 5 mA and 15 mA respectively.

Explanation of Solution

PSPICE Circuit:

Refer to the Figure P7.1 in the textbook.

Draw the given circuit diagram in PSPICE as shown in Figure 1.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip  1

Simulation settings:

Provide the simulation settings as shown in Figure 2.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip  2

PSPICE output:

After run the PSPICE circuit a black output screen will be displayed. Right click on the mouse by keeping cursor on the output screen, click the option “Add Trace” and type the expression “-I(R3)” in trace expression box.

The current plot i1(t) is shown in Figure 3.

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip  3

From PSPICE output, the initial values of current are,

i1(0)=5mAi1(0)=i1(0+)i1(0+)=5mA

Similarly, type the expression “-I(R2)” in trace expression box to obtain the current i2(t).

Electric Circuits. (11th Edition), Chapter 7, Problem 1P , additional homework tip  4

From PSPICE output, the initial values of current are,

i2(0)=15mAi2(0+)=5mA

Conclusion:

Therefore, the values of i1(0) and i2(0) are 5 mA and 15 mA respectively.

b.

To determine

Find the current values i1(0+) and i2(0+).

b.

Expert Solution
Check Mark

Answer to Problem 1P

The current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

Explanation of Solution

Calculation:

From Figure 2 and Figure 3, the current values are,

i1(0+)=5mAi2(0+)=5mA

Conclusion:

Therefore, the current values i1(0+) and i2(0+) are 5 mA and –5 mA respectively.

c.

To determine

Find the expression i1(t) for t0.

c.

Expert Solution
Check Mark

Answer to Problem 1P

The expression i1(t) for t0 is 5e20,000tmA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2+6=8

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 for Req and 400 mH for L.

τ=400mH8=400×1038×103=50μs

The expression i1(t) is,

i1(t)=i1(0+)etτ

Substitute 5 mA for i1(0+) and 50μs for τ.

i1(t)=5et50μsmA=5e20,000tmA

Conclusion:

Therefore, the expression i1(t) for t0 is 5e20,000tmA.

d.

To determine

Find the expression i2(t) for t0.

d.

Expert Solution
Check Mark

Answer to Problem 1P

The expression i2(t) for t0 is 5e20,000tmA.

Explanation of Solution

Calculation:

Find the equivalent resistance after the switch is opened at t=0.

Req=2+6=8

Find time constant from the circuit diagram.

τ=LReq

Here,

L is the inductance.

Req is the equivalent resistance across the load.

Substitute 8 for Req and 400 mH for L.

τ=400mH8=400×1038×103=50μs

The expression i2(t) is,

i2(t)=i2(0+)etτ

Substitute –5 mA for i1(0+) and 50μs for τ.

i2(t)=5et50μsmA=5e20,000tmA

Conclusion:

Therefore, the expression i2(t) for t0 is 5e20,000tmA.

e.

To determine

Explain the reason for why i2(0)i2(0+).

e.

Expert Solution
Check Mark

Explanation of Solution

Calculation:

The current in the resistor changes instantaneously. The switching operation makes the current i2(0) is equal to 15 mA and i2(0+) is equal to –5 mA.

Conclusion:

Therefore, the reason for why i2(0)i2(0+) is stated.

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Chapter 7 Solutions

Electric Circuits. (11th Edition)

Ch. 7.7 - There is no energy stored in the capacitor at the...Ch. 7.7 - Prob. 12APCh. 7 - Prob. 1PCh. 7 - In the circuit shown in Fig. P 7.2, the switch...Ch. 7 - Prob. 3PCh. 7 - The switch shown in Fig. P 7.4 has been open for a...Ch. 7 - Prob. 5PCh. 7 - For the circuit of Fig. P 7.5, what percentage of...Ch. 7 - Prob. 7PCh. 7 - In the circuit in Fig. P 7.8, the voltage and...Ch. 7 - Prob. 9PCh. 7 - Prob. 10PCh. 7 - The switch in the circuit seen in Fig. P 7.11 has...Ch. 7 - In the circuit in Fig. P 7.11, let Ig represent...Ch. 7 - The two switches in the circuit seen in Fig. P...Ch. 7 - Prob. 14PCh. 7 - Prob. 15PCh. 7 - Prob. 16PCh. 7 - Prob. 17PCh. 7 - Prob. 18PCh. 7 - Prob. 19PCh. 7 - For the circuit seen in Fig. P 7.19, find the...Ch. 7 - Prob. 21PCh. 7 - Prob. 22PCh. 7 - Prob. 23PCh. 7 - Prob. 24PCh. 7 - The switch in the circuit in Fig. P 7.25 is closed...Ch. 7 - In the circuit shown in Fig. P 7.26, both switches...Ch. 7 - In the circuit in Fig. P 7.27 the voltage and...Ch. 7 - Prob. 28PCh. 7 - Prob. 29PCh. 7 - The switch in the circuit seen in Fig. P 7.30 has...Ch. 7 - In Problem 7.30 how many microjoules of energy are...Ch. 7 - Prob. 33PCh. 7 - Prob. 34PCh. 7 - Prob. 35PCh. 7 - Prob. 36PCh. 7 - Prob. 37PCh. 7 - The switch in the circuit shown in Fig. P 7.38 has...Ch. 7 - Prob. 39PCh. 7 - Prob. 40PCh. 7 - Prob. 41PCh. 7 - Prob. 42PCh. 7 - Prob. 43PCh. 7 - Prob. 44PCh. 7 - Prob. 45PCh. 7 - Prob. 46PCh. 7 - For the circuit in Fig. P 7.4, find (in...Ch. 7 - Prob. 48PCh. 7 - Prob. 49PCh. 7 - Prob. 50PCh. 7 - Prob. 51PCh. 7 - Prob. 52PCh. 7 - Prob. 53PCh. 7 - Prob. 54PCh. 7 - The switch in the circuit of Fig. P 7.55 has been...Ch. 7 - The switch in the circuit seen in Fig. P 7.56 has...Ch. 7 - Prob. 57PCh. 7 - Prob. 58PCh. 7 - Prob. 59PCh. 7 - The switch in the circuit shown in Fig. P 7.61 has...Ch. 7 - Prob. 62PCh. 7 - Prob. 63PCh. 7 - Prob. 64PCh. 7 - Prob. 65PCh. 7 - Prob. 66PCh. 7 - Prob. 67PCh. 7 - Prob. 68PCh. 7 - Prob. 69PCh. 7 - Prob. 70PCh. 7 - Prob. 71PCh. 7 - Prob. 72PCh. 7 - Prob. 73PCh. 7 - For the circuit in Fig. P 7.73, how many...Ch. 7 - Prob. 75PCh. 7 - Prob. 76PCh. 7 - Prob. 77PCh. 7 - Prob. 78PCh. 7 - Prob. 79PCh. 7 - Prob. 80PCh. 7 - Prob. 81PCh. 7 - Prob. 82PCh. 7 - Prob. 84PCh. 7 - Prob. 85PCh. 7 - Prob. 86PCh. 7 - Prob. 87PCh. 7 - Prob. 88PCh. 7 - Prob. 90PCh. 7 - Prob. 91PCh. 7 - Prob. 92PCh. 7 - Prob. 93PCh. 7 - Prob. 94PCh. 7 - Prob. 95PCh. 7 - Prob. 100PCh. 7 - Prob. 101PCh. 7 - Prob. 102PCh. 7 - Prob. 103PCh. 7 - Prob. 104PCh. 7 - Prob. 105PCh. 7 - Prob. 106PCh. 7 - Prob. 107P
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