Apply the loop rule to loop aedcba in the figure below. E, = 18 V b I2 12 0.5 2 R2 2.5 Q 12 R, a e 13 6.0 2 R3 1.52 f 0.5 Q E2 = 45 V O (1,R,) + (I,r,) - E, + (1,R2) = 0 O (1,R2) + (I2";) - Ez + (I,R3) = 0 O (1,R3) – (12r2) + E, + (I,R3) = 0 O (1,R2) + (I2r2) – E, + (I,R,) = 0

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter21: Current And Direct Current Circuits
Section: Chapter Questions
Problem 74P
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Apply the loop rule to loop aedcba in the figure below.
Apply the loop rule to loop aedcba in the figure below.
E, = 18 V
b
I2
12
0.5 2
R2
2.5 Q
12
R,
a
e
13
6.0 2
R3
1.52
f
0.5 Q
E2 = 45 V
O (1,R,) + (I,r,) - E, + (1,R2) = 0
O (1,R2) + (I2";) - Ez + (I,R3) = 0
O (1,R3) – (12r2) + E, + (I,R3) = 0
O (1,R2) + (I2r2) – E, + (I,R,) = 0
Transcribed Image Text:Apply the loop rule to loop aedcba in the figure below. E, = 18 V b I2 12 0.5 2 R2 2.5 Q 12 R, a e 13 6.0 2 R3 1.52 f 0.5 Q E2 = 45 V O (1,R,) + (I,r,) - E, + (1,R2) = 0 O (1,R2) + (I2";) - Ez + (I,R3) = 0 O (1,R3) – (12r2) + E, + (I,R3) = 0 O (1,R2) + (I2r2) – E, + (I,R,) = 0
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