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why is the current equal to that?

why is the current equal to that?

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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why is the current equal to that?

5 U1 Find the steady-state expression for v(t) in the circuit shown by using the technique of source transformations. The sinusoidal voltage sources are v₁ = 240 cos (4000t + 53.13°) V, V1 V₂ = 96 sin 4000t V. V2 15 mH 30 ΩΣ + vo(t) 20 Ω m €25/6 μF V₂ +
Transcribed Image Text:5 U1 Find the steady-state expression for v(t) in the circuit shown by using the technique of source transformations. The sinusoidal voltage sources are v₁ = 240 cos (4000t + 53.13°) V, V1 V₂ = 96 sin 4000t V. V2 15 mH 30 ΩΣ + vo(t) 20 Ω m €25/6 μF V₂ +
Assessment Problem 9.10 V₁ V₂ = 96/-90° = -j96 V jwL = j(4000) (15 × 10-³) = 60 2 1 6 × 106 jwC (4000) (25) = V₂ 20 240/53.13° = 144 + j192 V Perform a source transformation: V₁ 144+j192 j60 j60 = −j- Y = = -1 Z = 96 20 Combine the parallel impedances: 1 1 1 1 + + + j60 30 -j60 20 1 Y = -j60 N = = 120 = 3.2 - j2.4 A -j4.8 A - j5 j60 - 1 12 3.2-j2.4A1 j60n 300 -j600 200 -j4.8A + 3.2+j2.4A V. 120 Vo = 12(3.2 + j2.4) = 38.4 + j28.8 V = 48/36.87° V Vo = 48 cos(4000t + 36.87°) V
Transcribed Image Text:Assessment Problem 9.10 V₁ V₂ = 96/-90° = -j96 V jwL = j(4000) (15 × 10-³) = 60 2 1 6 × 106 jwC (4000) (25) = V₂ 20 240/53.13° = 144 + j192 V Perform a source transformation: V₁ 144+j192 j60 j60 = −j- Y = = -1 Z = 96 20 Combine the parallel impedances: 1 1 1 1 + + + j60 30 -j60 20 1 Y = -j60 N = = 120 = 3.2 - j2.4 A -j4.8 A - j5 j60 - 1 12 3.2-j2.4A1 j60n 300 -j600 200 -j4.8A + 3.2+j2.4A V. 120 Vo = 12(3.2 + j2.4) = 38.4 + j28.8 V = 48/36.87° V Vo = 48 cos(4000t + 36.87°) V
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