Using Kirchoff's junction rule at b
I1 = I2 + I3
Using Kirchoff's loop rule in abefa
-V1 + I1R1 + I3R3 + I1R5 = 0
I1(R1 + R5) + I3R3 = V1
Using Kirchoff's lop rule in bcdeb
V2 + I2R2 + I2R4 - I3R3 = 0
I3R3 - I2(R2+ R4) = V2
Answer :
Required equations
I1 = I2 + I3
I1(R1 + R5) + I3R3 = V1
I3R3 - I2(R2+ R4) = V2
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5) (15 pts) Set up the equations that would be necessary to solve for the currents...
5) (15 pts) Set up the equations that would be necessary to solve for the currents labeled 11, 12 & Iz in the following circuit diagram 1 12 R2 Ri AAA 13 Vi AM R3 R4 $ AAA RS V2
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