Problem 61 (Phasor transform) Given the circuit in Figure 6, find the value of R and L Given: ● ● H(₂51-+(1) 002 10 = (1) ³ R=2 L=? w m W M Ⓒu(t)=10¹√cos (wt + 30") V[ vs(t) = 10 √2 cos(wt +30°) V is(t) = 0.1 cos (wt -15°)A Ro=800 C=625nF=625 10°F w = 10^rad Figure 6 [R₁=80₁ = C = 625m
Problem 61 (Phasor transform) Given the circuit in Figure 6, find the value of R and L Given: ● ● H(₂51-+(1) 002 10 = (1) ³ R=2 L=? w m W M Ⓒu(t)=10¹√cos (wt + 30") V[ vs(t) = 10 √2 cos(wt +30°) V is(t) = 0.1 cos (wt -15°)A Ro=800 C=625nF=625 10°F w = 10^rad Figure 6 [R₁=80₁ = C = 625m
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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