The following system of equations was generated by applying the mesh current law to the circuit in the figure. Solve for the currents. 60I1 – 4012 = 200 -40I1 + 150I2 – 100I3 = 0 -100I2 + 13013 = 230 20 2 10 Ω 80 V ww 40 n 100 n ) 3 30 n 200 V 10 A |

Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter7: Parallel Circuits
Section: Chapter Questions
Problem 3PP: Using the rules for parallel circuits and Ohmslaw, solve for the missing values....
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The following system of equations was generated by applying the mesh current law to the
circuit in the figure. Solve for the currents.
60I1 – 4012 = 200
-40I1 + 150I2 – 100I3 = 0
-100I2 + 13013 = 230
20 2
10 Ω
80 V
ww
40 n
100 n
) 3 30 n
200 V
10 A
|
Transcribed Image Text:The following system of equations was generated by applying the mesh current law to the circuit in the figure. Solve for the currents. 60I1 – 4012 = 200 -40I1 + 150I2 – 100I3 = 0 -100I2 + 13013 = 230 20 2 10 Ω 80 V ww 40 n 100 n ) 3 30 n 200 V 10 A |
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