Problem FiveThe voltage in Fig. 5 is given by v = 400t² V for t>0 and the initial value of the shown current is givenby i(0) = 0.5 A. At time t = 0.4s, find energy: (a) stored in the capacitor, (b) stored in the inductor, and(c) dissipated by the resistor since t= 0.100210H-10 μFFig. 5

Answered: Problem Five The voltage in Fig. 5 is…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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Problem Five
The voltage in Fig. 5 is given by v = 400t² V for t>0 and the initial value of the shown current is given
by i(0) = 0.5 A. At time t = 0.4s, find energy: (a) stored in the capacitor, (b) stored in the inductor, and
(c) dissipated by the resistor since t= 0.
1002
10H
-10 μF
Fig. 5

Expert Solution

Step 1

Given data,

Step 2

The expression for the current through the resistor,

$i_{R} (t) = \frac{v (t)}{100 Ω} = \frac{400 t^{2}}{100 Ω} = 4 t^{2} A$

The expression for the current through the inductor with the initial conditions.

$I_{L} (s) = \frac{V (s) - L i_{o}}{s L} = \frac{\frac{400 \times 2}{s^{3}} - (10) (0.5)}{s (10)} = \frac{800 - 5 s^{3}}{10 s^{4}} = \frac{80}{s^{4}} - \frac{1}{2 s}$

By the Laplace transform,

$i_{L} (t) = \frac{40}{3} t^{3} - \frac{1}{2}$

(a). The expression for the energy stored in the capacitor.

$E_{C} (t) = \frac{1}{2} C v^{2}$

$= \frac{1}{2} (10 \times 10^{- 6} F) \times (400 t^{2}) = 0.8 t^{4} E_{C} (0.4 s) = 0.8 \times {(0.4)}^{4} = 20.48 mJ$

Hence the energy stored in the capacitor is $20.48 mJ$ .

b). The expression for the energy stored in the inductor.

$E_{L} (s) = \frac{1}{2} L (i_{L} {(t))}^{2} = \frac{1}{2} (10 H) \times [\frac{40}{3} t^{3} - \frac{1}{2}] = \frac{1}{2} (10 H) \times [\frac{40}{3} \times {(0.4)}^{3} - \frac{1}{2}] = 624.23 mJ$

Hence, the energy stored in an inductor is 624.23 mJ.

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