a) For Example 6.1, plot i,v , p, and w versus time. Line up the plots vertically to allow easy assessment of each variable’s behavior. b) In what time interval is energy being stored in the inductor? c) In what time interval is energy being extracted from the inductor? d) What is the maximum energy

Question

a) For Example 6.1, plot i,v , p, and w versus time. Line up the plots vertically to allow easy assessment of each variable’s behavior.
b) In what time interval is energy being stored inthe inductor?
c) In what time interval is energy being extractedfrom the inductor?
d) What is the maximum energy stored in the inductor?
e) Evaluate the integrals
[latex]\int_{0}^{0.2}{p dt }[/latex] and [latex]\int_{0.2}^{\infty}{p dt } [/latex],
and comment on their significance.
f) Repeat (a)–(c) for Example 6.2.
g) In Example 6.2, why is there a sustained current in the inductor as the voltage approaches zero?

Accepted Answer

a) The plots of i,v ,p, and w follow directly from the expressions for i and obtained in Example 6.1 and are shown in Fig. 6.8. In particular ,[latex]p=vi[/latex],and [latex]w=\left(\frac{1}{2}\right)L i^{2}[/latex] .

b) An increasing energy curve indicates that energy is being stored. Thus energy is being stored in the time interval 0 to 0.2 s. Note that this corresponds to the interval when [latex]p\gt0[/latex].

c) A decreasing energy curve indicates that energy is being extracted. Thus energy is being extracted in the time interval 0.2s to[latex] \infty[/latex] . Note that this corresponds to the interval when [latex]p\lt0[/latex].

d) From Eq. 6.12 [latex]w=\frac{1}{2}Li^{2}[/latex] we see that energy is at a maximum when current is at a maximum; glancing at the graphs confirms this. From Example 6.1, maximum current = 0.736 A. Therefore,[latex] w_{\max } = 27.07 mJ[/latex].
Thus

[latex]\int_{0}^{0.2}{p dt }=10\left[\frac{e^{-10t }}{100} \left(-10t -1\right) \right]^{0.2}_{0}[/latex]

[latex]-50\left\{\frac{t^{2}e^{-10t }}{-10}+\frac{2}{10} \left[\frac{e^{-10t }}{100}\left(-10t-1\right) \right] \right\}^{0.2}_{0} [/latex]

[latex]=0.2 e^{-2 } =27.07 mJ, [/latex],

[latex]\int_{0.2}^{\infty }{p dt }=10\left[\frac{e^{-10t }}{100} \left(-10t -1\right) \right]^{\infty}_{0.2}[/latex]

[latex]-50\left\{\frac{t^{2}e^{-10t }}{-10}+\frac{2}{10} \left[\frac{e^{-10t }}{100}\left(-10t-1\right) \right] \right\}^{\infty }_{0.2}[/latex]

[latex]=-0.2 e^{-2 } =-27.07 mJ, [/latex].

Based on the definition of p, the area under the plot of p versus t represents the energy expended over the interval of integration. Hence the integration of the power between 0 and 0.2 s represents the energy stored in the inductor during this time interval. The integral of p over the interval [latex]0.2s- \infty[/latex] is the energy extracted. Note that in this time interval, all the energy originally stored is removed; that is, after the current peak has passed, no energy is stored in the inductor.

e) From Example 6.1,

[latex]i= 10t e^{-5t } A[/latex] and [latex]v=e^{-5t }\left(1-5t\right)V [/latex].

Therefore,

[latex]p = vi = 10t e^{-10t }-50 t^{2} e^{-10t }W[/latex].

f) The plots of v , i, p, and w follow directly from the expressions for and i given in Example 6.2 and are shown in Fig. 6.9. Note that in this case the power is always positive, and hence energy is always being stored during the voltage pulse.

g) The application of the voltage pulse stores energy in the inductor. Because the inductor is ideal, this energy cannot dissipate after the voltage subsides to zero. Therefore, a sustained current circulates in the circuit. A lossless inductor obviously is an ideal circuit element. Practical inductors require a resistor in the circuit model. (More about this later.)

Question 6.3: a) For Example 6.1, plot i,v , p, and w versus time. Line up...

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