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I = 0.3 AZ 60° XL 802 XC50 + E = 10 V Z0° 000 Figure 1:

I = 0.3 AZ 60° XL 802 XC50 + E = 10 V Z0° 000 Figure 1:

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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1) Using superposition, determine the current through the inductance XL

(a) Find the current(I1) flowing through XL by current source

I1 = 30mA∠150° 
(b) Find the current(I2) flowing through XL by voltage source

I2 = 2.5A∠-90° 
(c) Find the current IL = I1 + I2

IL = 30mA∠150° +  2.5A∠-90° 

IL = 2.52A∠-89.41° 

I = 0.3 AZ 60° XL 802 XC50 + E = 10 V Z0° 000 Figure 1:
Transcribed Image Text:I = 0.3 AZ 60° XL 802 XC50 + E = 10 V Z0° 000 Figure 1:
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