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H.W. I.C. (1) Let R=100, C=1₁V₁ (0) =0² V(t) = 100 for Oct 23 0 50 for 34 ²10 D.E. (2) for 100 D.E. (1) 50 -3 t210 D.E. (3) -10 Find Vc (12) IC (2)= end condition of DE (1) Solution IC (3) = end cond. of DE (2) sol. A 1 to 07 O XE 10 T १ 0 2 123 t original timescale 2nd timescale 3rd timescale

H.W. I.C. (1) Let R=100, C=1₁V₁ (0) =0² V(t) = 100 for Oct 23 0 50 for 34 ²10 D.E. (2) for 100 D.E. (1) 50 -3 t210 D.E. (3) -10 Find Vc (12) IC (2)= end condition of DE (1) Solution IC (3) = end cond. of DE (2) sol. A 1 to 07 O XE 10 T १ 0 2 123 t original timescale 2nd timescale 3rd timescale

Introductory Circuit Analysis (13th Edition)
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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H.W. I.C. (1) Let R=100, C=1₁V₁ (0) = 0° V (t) = 100 for 0≤t <3 50 for 32 ²10 D.E. (2) 6210 D.E. (3) for Vc 100 D.E. (1) 507 -3 -10 Find Vc (12) IC (2)= end condition of DE (1) Solution IC (3) = end cord. of DE (2) sol. w O -7 10 7 9 O 2 t 0+3 original timescale 2nd timescale 3rd timescale
Transcribed Image Text:H.W. I.C. (1) Let R=100, C=1₁V₁ (0) = 0° V (t) = 100 for 0≤t <3 50 for 32 ²10 D.E. (2) 6210 D.E. (3) for Vc 100 D.E. (1) 507 -3 -10 Find Vc (12) IC (2)= end condition of DE (1) Solution IC (3) = end cord. of DE (2) sol. w O -7 10 7 9 O 2 t 0+3 original timescale 2nd timescale 3rd timescale
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