For the inductive circuit in Fig. 20.10:
a. Sketch the waveforms of $\nu_L and i_L$ .
b. Find the instantaneous power level for the inductor at times $t_1 through t_5$ .
c. Plot the results of part (b) for one full period of the applied voltage.
d. Find the average value of the curve of part (c) over one full cycle of the applied voltage and compare the peak value of each pulse with the value determined by Eq. (20.13).

Question

For the inductive circuit in Fig. 20.10:
a. Sketch the waveforms of [latex] u_L and i_L[/latex].
b. Find the instantaneous power level for the inductor at times [latex]t_1 through t_5[/latex].
c. Plot the results of part (b) for one full period of the applied voltage.
d. Find the average value of the curve of part (c) over one full cycle of the applied voltage and compare the peak value of each pulse with the value determined by Eq. (20.13).

[latex]Q_L = VI\hspace{30 pt} (VAR)\hspace{30 pt} (20.13)[/latex]
e. Find the energy stored or released for any one pulse of the power curve.

Accepted Answer

a. Both waveforms have been plotted in Fig. 20.11. Note that the voltage leads the current by 90° as shown by the fact that the voltage passes through 0 V at 0° and the current passes through the horizontal axis at 90°.

The peak value of the current is simply determined by Ohm’s law:

[latex]I_m=\frac{V_m}{X_L}=\frac{12V}{4\Omega}=3A [/latex]

b. The resulting power curve appears in Fig. 20.11 with three different vertical scales for the voltage, current, and power.

At [latex]t_1, t_3, and t_5[/latex] either the voltage or current is zero resulting in

[latex]P_L = \nu_L i_L = 0 W[/latex].

At [latex]t_2, \alpha = 45° and \nu_L = 12 sin \alpha = 12 sin 45° = 12(0.7071) =8.49 V[/latex]

and [latex]i_L=\frac{V_m}{X_L}sin(\alpha-90°) =\frac{12V}{4 Ω} sin(45° -90°)=3 sin(-45°) = -2.12 A,\[0.5cm] so\; that\hspace{30 pt} P_L = \nu_Li_L = (8.49 V)(-2.12 A) = −18 W.[/latex]

The results at [latex]t_4, or \alpha = 135°[/latex] are a mirror image of that obtained at [latex]t_2 so that P_L = 18 W[/latex] as shown in Fig. 20.11.

Note the fact that the frequency of the power curve is clearly twice that of the applied voltage or current verifying the equation [latex]P_L = VI sin \omega t[/latex].

c. The average value for the curve in Fig. 20.11 is 0 W over any full cycle of the applied voltage.

The peak value of the curve is 18 W, which compares directly with that obtained from the product

[latex]VI=\frac{V_mI_m}{2}=\frac{(12 V)(3 A)}{2}=18W [/latex]

d. The energy stored or released during each pulse of the power curve is

[latex]W_L=\frac{VI}{\omega_1}=\frac{V_mI_m}{2\omega_1}=\frac{(12 V)(3 A)}{2(200 rad/s)}=90 mJ [/latex]

Question 20.2: For the inductive circuit in Fig. 20.10: a. Sketch the wavef...

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