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0.10 32 E, = 24 V + C d: R, = 5.0 2 ,0.5 Ω R5 = 20 2 12 ¿Ez = 48 V R, = 40 2 + a e k 13 0.20 N R3 = 78 N 13 = 0.05 N E= 36 V Ez = 6.0 V og hi 2.

0.10 32 E, = 24 V + C d: R, = 5.0 2 ,0.5 Ω R5 = 20 2 12 ¿Ez = 48 V R, = 40 2 + a e k 13 0.20 N R3 = 78 N 13 = 0.05 N E= 36 V Ez = 6.0 V og hi 2.

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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Apply the junction rule to point a shown below.

0.10 Ω E1 = 24 V + R, = 5.0 2 , = 0.5 Q 12 R5 = 20 N 12 E, = 48 V R2 = 40 N a e 13 0.20 N R = 78 Q f 13 = 0.05 N E= 36 V Ez = 6.0 V g hị
Transcribed Image Text:0.10 Ω E1 = 24 V + R, = 5.0 2 , = 0.5 Q 12 R5 = 20 N 12 E, = 48 V R2 = 40 N a e 13 0.20 N R = 78 Q f 13 = 0.05 N E= 36 V Ez = 6.0 V g hị
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