Problem 2: Assume the capacitor is initially discharged with vc(0) = 0V. Sketch the capacitor voltage vc(t) and capacitor current ic(t) of the full-wave rectifier circuit shown below with C = 1µF assuming identical germanium diodes. Hint: The slope of the line in the triangle waveform may be used to determine the point in time that v;(t) = 2V. v,(t) D1 D2 6V v;(t) 0.2 0.6/ [s] ict + -6V vc(t)
Problem 2: Assume the capacitor is initially discharged with vc(0) = 0V. Sketch the capacitor voltage vc(t) and capacitor current ic(t) of the full-wave rectifier circuit shown below with C = 1µF assuming identical germanium diodes. Hint: The slope of the line in the triangle waveform may be used to determine the point in time that v;(t) = 2V. v,(t) D1 D2 6V v;(t) 0.2 0.6/ [s] ict + -6V vc(t)
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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![Problem 2: Assume the capacitor is initially discharged with vc(0) = 0V. Sketch the capacitor voltage
vc(t) and capacitor current ic(t) of the full-wave rectifier circuit shown below with C = 1µF assuming
identical germanium diodes. Hint: The slope of the line in the triangle waveform may be used to determine
the point in time that v¡(t) = 2Vķ-
%3D
v,(t)
D1
D2
6V
t [s] -
v;(t)
0.2
ict
0.6
1
-6V
D3
Da
C
vc(t)
OV
for t < Os
0.02s, Vpeak = 5.4V, vc(t) = {30t - 0.6V for 0.02s <t < 0.2s
for t 2 0.2s
Answers: tstart
5.4V
ic(t) = {30HA 0.02s < t < 0.2s
otherwise
S30µA 0.02s <t< 0.2s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1148f2e-12e2-4d7a-aa19-e54736499ad8%2Fff036d51-acdc-4840-9e99-83878227cad8%2Feh4dcet_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem 2: Assume the capacitor is initially discharged with vc(0) = 0V. Sketch the capacitor voltage
vc(t) and capacitor current ic(t) of the full-wave rectifier circuit shown below with C = 1µF assuming
identical germanium diodes. Hint: The slope of the line in the triangle waveform may be used to determine
the point in time that v¡(t) = 2Vķ-
%3D
v,(t)
D1
D2
6V
t [s] -
v;(t)
0.2
ict
0.6
1
-6V
D3
Da
C
vc(t)
OV
for t < Os
0.02s, Vpeak = 5.4V, vc(t) = {30t - 0.6V for 0.02s <t < 0.2s
for t 2 0.2s
Answers: tstart
5.4V
ic(t) = {30HA 0.02s < t < 0.2s
otherwise
S30µA 0.02s <t< 0.2s
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