Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
11th Edition
ISBN: 9780134746968
Author: James W. Nilsson, Susan Riedel
Publisher: PEARSON
Question
Chapter 6, Problem 1P

(a)

To determine

Find the expression of the voltage across an inductor t>0 s.

(a)

Expert Solution
Check Mark

Answer to Problem 1P

The expression of the voltage across an inductor t>0 s is 0.9e10t(110t) mV_ for t>0.

Explanation of Solution

Given data:

The current through an inductor is iL=18te10t A for t0 s.

The inductor value is,

L=50μH

Formula used:

Write the general expression to find the voltage across an inductor as,

v=LdiLdt        (1)

Here,

L is the value of inductance of the inductor,

iL is the current through an inductor, and

dt is the change in time.

Calculation:

Substitute 50μH for L and 18te10t A for iL in equation (1) to find voltage across an inductor.

v=(5×106 H)ddt(18te10t A)v=(5×106×18)((10)te10t+e10t)              {ddt(uv)=uddt(v)+vddt(u)}v=0.0009(e10t10te10t) V

v=0.9e10t(110t) mV for t>0

Conclusion:

Thus, the expression of the voltage across an inductor t>0 s is 0.9e10t(110t) mV_ for t>0.

(b)

To determine

Calculate the power at the inductor terminals at time of t=200 μs.

(b)

Expert Solution
Check Mark

Answer to Problem 1P

The power at the inductor terminals at time of t=200 μs is 59.34μW_.

Explanation of Solution

Calculation:

Given that iL=18te10t A for t0 s. The values of current for several instant of times are,

At time t=0 s:

iL=18(0)e10(0) AiL=0 A

At time t=0.05 s:

iL=18(0.05)e10(0.05) AiL=0.545877 AiL=0.6 A

At time t=0.1 s:

iL=18(0.01)e10(0.01) AiL=0.66218299 AiL0.6622 A

At time t=0.15 s:

iL=18(0.15)e10(0.15) AiL=0.602451 AiL0.61 A

At time t=0.2 s:

iL=18(0.2)e10(0.2) AiL=0.487207019 AiL0.487207 A

Calculate the voltage at t=200ms.

v=0.9e10(200×103)(110(200×103)) mVv=0.9e2(12) mVv=121.8×103 mV

v=121.8μV

Write the formula to find the power in an inductor at t=200ms as,

p(200 ms)=v(200 ms)iL(200 ms)

Substitute 121.8μV for v(200 ms) and 0.487207 A for iL(200 ms) in above equation.

p(200 ms)=(121.8μV)(0.487207 A)p(200 ms)=(121.8×0.487207 A)μVA

p(200 ms)=59.34μW

PSpice simulation:

Use all the above calculated values to give as an input to the PSpice simulation.

From the given information, the current passing through an inductor and here a small resistor is connected in series with inductor to avoid the simulation error.

Draw the circuit as shown in below Figure 1.

Electric Circuits. (11th Edition), Chapter 6, Problem 1P , additional homework tip  1

Provide the simulation settings as shown in below Figure 2.

Electric Circuits. (11th Edition), Chapter 6, Problem 1P , additional homework tip  2

Now, save and run the simulation, then the plot for the input current and voltage across an inductor will be displays as shown in Figure 3.

Electric Circuits. (11th Edition), Chapter 6, Problem 1P , additional homework tip  3

From the Figure 3, the first function is used to read the voltage at t=200 ms, which is shown as v=127.79299 μV. The second function is used to calculate the power at t=200 ms which is showing as,

p=59.825μWp59.83μW

The power calculated at t=200 ms theoretically and with simulation, both are approximately equal.

Conclusion:

Thus, the power at the inductor terminals at time of t=200 μs is 59.34μW_.

(c)

To determine

Check whether an inductor is delivering the power or absorbing the power.

(c)

Expert Solution
Check Mark

Answer to Problem 1P

An inductor is delivering the power.

Explanation of Solution

Discussion:

The power calculated in an inductor has the negative sign which indicates that the power is delivering by an inductor.

Conclusion:

Thus, an inductor is delivering the power.

(d)

To determine

Calculate the energy stored in an inductor in micro joules at t=200 ms.

(d)

Expert Solution
Check Mark

Answer to Problem 1P

The energy stored in an inductor in at t=200 ms is 5.93μJ_.

Explanation of Solution

Discussion:

From the simulation results shown in Part (b), the fourth equation is written to calculate energy stored in an inductor. From the Figure 3, the energy stored in an inductor in at t=200 ms is 5.93μJ.

Conclusion:

Thus, the energy stored in an inductor in at t=200 ms is 5.93μJ_.

(e)

To determine

Find the maximum energy, and also find the instant of time that occurs.

(e)

Expert Solution
Check Mark

Answer to Problem 1P

The maximum energy is 10.96μJ_, and it occurs at the instant of time t=200 ms_.

Explanation of Solution

Discussion:

When the current is maximum energy in an inductor also maximum, since the square of current is directly proportional to the energy stored.

From the Figure 3, it is calculated that maximum current as 662.2 mA, that occurs at t=200 μs. In Figure 3, the last formula is used to get the maximum energy stored in an inductor in at t=200 ms is shown as 10.96μJ.

Conclusion:

Thus, the maximum energy is 10.96μJ_, and it occurs at the instant of time t=200 ms_.

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Chapter 6 Solutions

Electric Circuits. (11th Edition)

Ch. 6 - The current in a 200 mH inductor is The voltage...Ch. 6 - The current in a 20 mH inductor is known to...Ch. 6 - Assume in Problem 6.5 that the value of the...Ch. 6 - Evaluate the integral for Example 6.2. Comment on...Ch. 6 - Find the inductor current in the circuit in Fig....Ch. 6 - The current in and the voltage across a 5 H...Ch. 6 - The current in the 2.5 mH inductor in Fig. P6.11...Ch. 6 - Initially there was no energy stored in the 5 H...Ch. 6 - The voltage across a 5 μF capacitor is known to...Ch. 6 - The triangular voltage pulse shown in Fig. P6.15...Ch. 6 - The expressions for voltage, power, and energy...Ch. 6 - A 20µF capacitor is subjected to a voltage pulse...Ch. 6 - The initial voltage on the 0.5 μF capacitor shown...Ch. 6 - The current shown in Fig. P6.20 is applied to a...Ch. 6 - The rectangular-shaped current pulse shown in Fig....Ch. 6 - Use realistic inductor values from Appendix H to...Ch. 6 - For the circuit shown in Fig. P6.24, how many...Ch. 6 - The two parallel inductors in Fig. P6.26 are...Ch. 6 - Derive the equivalent circuit for a series...Ch. 6 - Derive the equivalent circuit for a parallel...Ch. 6 - Use realistic capacitor values from Appendix H to...Ch. 6 - Prob. 30PCh. 6 - The two series-connected capacitors in Fig. P6.31...Ch. 6 - The four capacitors in the circuit in Fig, P6.32...Ch. 6 - For the circuit in Fig. P6.32, calculate the...Ch. 6 - At t = 0. a series-connected capacitor and...Ch. 6 - The current in the circuit in Fig. P6.35 is known...Ch. 6 - Show that the differential equations derived in...Ch. 6 - Prob. 37PCh. 6 - Prob. 38PCh. 6 - Let υg represent the voltage across the current...Ch. 6 - Prob. 40PCh. 6 - Prob. 41PCh. 6 - Prob. 42PCh. 6 - Prob. 43PCh. 6 - Prob. 44PCh. 6 - Prob. 45PCh. 6 - Prob. 46PCh. 6 - Prob. 47PCh. 6 - Prob. 48PCh. 6 - The self-inductances of two magnetically coupled...Ch. 6 - Prob. 50PCh. 6 - Prob. 51PCh. 6 - Prob. 52PCh. 6 - Prob. 53P
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