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Liver IR (P-1) = 8MA = 8mL0 A. (0) IL (P-P) = 12-6m1 = 12-6m -90 A E.(P-P) = 8 Lo volts. From the Urcuit Is (P-P) = IR (P-P) + IL (P-P) Taking E(P-P) as reference phasor the phasor diagram is J drawn as follows: IR(P-P) त्रे →ē (1-p) 3LIP-P) \' Is (P-P) Is (P-P) = 8L0 mA +12-6L-90 mA =(8-j12.6)mA (() Is (P-P) = 14.925 L-63-986 MA

Liver IR (P-1) = 8MA = 8mL0 A. (0) IL (P-P) = 12-6m1 = 12-6m -90 A E.(P-P) = 8 Lo volts. From the Urcuit Is (P-P) = IR (P-P) + IL (P-P) Taking E(P-P) as reference phasor the phasor diagram is J drawn as follows: IR(P-P) त्रे →ē (1-p) 3LIP-P) \' Is (P-P) Is (P-P) = 8L0 mA +12-6L-90 mA =(8-j12.6)mA (() Is (P-P) = 14.925 L-63-986 MA

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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How did you get Is(p-p) in polar form?

Liver IR (P-1) = 8MA = 8mL0 A. (0) IL (P-P) = 12-6m1 = 12-6m -90 A E.(P-P) = 8 Lo volts. From the Urcuit Is (P-P) = IR (P-P) + IL (P-P) Taking E(P-P) as reference phasor the phasor diagram is J drawn as follows: IR(P-P) त्रे →ē (1-p) 3LIP-P) \' Is (P-P) Is (P-P) = 8L0 mA +12-6L-90 mA =(8-j12.6)mA (() Is (P-P) = 14.925 L-63-986 MA
Transcribed Image Text:Liver IR (P-1) = 8MA = 8mL0 A. (0) IL (P-P) = 12-6m1 = 12-6m -90 A E.(P-P) = 8 Lo volts. From the Urcuit Is (P-P) = IR (P-P) + IL (P-P) Taking E(P-P) as reference phasor the phasor diagram is J drawn as follows: IR(P-P) त्रे →ē (1-p) 3LIP-P) \' Is (P-P) Is (P-P) = 8L0 mA +12-6L-90 mA =(8-j12.6)mA (() Is (P-P) = 14.925 L-63-986 MA
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