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20. In the circuit shown, initially the switch is open and all the capacitors are uncharged. Consider the first 100 s interval after the switch is closed. Find the total energy stored in the capacitors and the total energy dissipated in the resistance in this 100 s interval. Internal resistance of the battery does not exceed a few ohms. Ans: 9x10-2 J -2 T

20. In the circuit shown, initially the switch is open and all the capacitors are uncharged. Consider the first 100 s interval after the switch is closed. Find the total energy stored in the capacitors and the total energy dissipated in the resistance in this 100 s interval. Internal resistance of the battery does not exceed a few ohms. Ans: 9x10-2 J -2 T

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Kindly check the answer in picture containing the question before submitting the solution .

Kindly solve using kirchoffs law and integration I.e  detailed solution from scratch.

 

Answer is .09 and .01 Joules.

20. In the circuit shown, initially the switch is open and all the capacitors are uncharged. Consider the first 100 s interval after the switch is closed. Find the total energy stored in the capacitors and the total energy dissipated in the resistance in this 100 s interval. Internal resistance of the battery does not exceed a few ohms. Ans: 9x10-2 J 1x10-2 C, = 3 µF C3 6 µF R=100 k2 C2 = 6 µF Vo 200 V C= 3 µF
Transcribed Image Text:20. In the circuit shown, initially the switch is open and all the capacitors are uncharged. Consider the first 100 s interval after the switch is closed. Find the total energy stored in the capacitors and the total energy dissipated in the resistance in this 100 s interval. Internal resistance of the battery does not exceed a few ohms. Ans: 9x10-2 J 1x10-2 C, = 3 µF C3 6 µF R=100 k2 C2 = 6 µF Vo 200 V C= 3 µF
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