DI3kD2128Vo4D3-41k-8-12Positive peak =negative peak =12D28KtV;Vo43D3D2kD1-41k-8-12Positive peak =negative peak =%3D129.ViD1D2Vo6KHY33D1.3k ?-33D-6-9-12Positive peak =negative peak =3kD1129.63kVo33DtowwKD21.-3to3k 2 V D3-6-9-12Positive peak =negative peak = For each circuit shown below, the input v, is a 60HZ, 12 volt peak sine wave. Sketch thewaveform resulting at vo. Assume ideal diodes. What are its positive and negative peak values?

Note: We are authorized to answer three subparts at a time, since you have not mentioned which part…

D1 3k D2 12 8 Vo 4 D3 -4 1k -8 -12 Positive peak = negative peak = 12 D2 8 Vo 4 3D w- 2k D1 -4 1k -8 -12 Positive peak negative peak = 12 9 Vi D1 D2 6 Vo KHY 3 3D 3k -3 1. -6 -9 -12 Positive peak = negative peak = ww ww

D1 3k D2 12 8 Vo 4 D3 -4 1k -8 -12 Positive peak = negative peak = 12 D2 8 Vo 4 3D w- 2k D1 -4 1k -8 -12 Positive peak negative peak = 12 9 Vi D1 D2 6 Vo KHY 3 3D 3k -3 1. -6 -9 -12 Positive peak = negative peak = ww ww

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

DI
3k
D2
12
8
Vo
4
D3
-4
1k
-8
-12
Positive peak =
negative peak =
12
D2
8
Kt
V;
Vo
4
3D
3D
2k
D1
-4
1k
-8
-12
Positive peak =
negative peak =
%3D
12
9.
Vi
D1
D2
Vo
6
KHY
3
3D
1.
3k ?
-3
3D
-6
-9
-12
Positive peak =
negative peak =
3k
D1
12
9.
6
3k
Vo
3
3D
to
wwK
D2
1.
-3
to
3k 2 V D3
-6
-9
-12
Positive peak =
negative peak =

For each circuit shown below, the input v, is a 60HZ, 12 volt peak sine wave. Sketch the
waveform resulting at vo. Assume ideal diodes. What are its positive and negative peak values?

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