5 kN 20 k2 i(1) + VR - t = 0 5 µF v2 100 V 20 μ

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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The switch in Fig. 8.73 is moved from A to B at t = 0 after being at A for a

long time. This places the two capacitors in series, thus allowing equal and

opposite de voltages to be trapped on the capacitors

 

(d) Find vr(t), t > 0. (e) Find i(t). (f) Find vi(t) and

v2(t) from i(t) and the initial values. (g) Show that the stored energy at i = 0

plus the total energy dissipated in the 20 kS2 resistor is equal to the energy

stored in the capacitors at t = 0

5 kN
20 k2
i(1)
+ VR
-
t = 0
5 µF
v2
100 V
20 μ
Transcribed Image Text:5 kN 20 k2 i(1) + VR - t = 0 5 µF v2 100 V 20 μ
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