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Question

The circuit shown in Figure contains three resistors R1=100Ω, R2=50Ω, and R3=20Ω and cells of emfs E1=2V and E2. The ammeter indicates a current of 50 mA. Determine the currents in the resistors and the emf of the second cell. The internal resistance of the ammeter and of the cells should be neglected.
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Solution

Given : E1=2 V R1=100Ω R2=50Ω R3=20Ω
Kirchhoff'l loop law in ABCDA : E1+(I+0.05)R1+IR2=0
2+(I+0.05)(100)+I(50)=0 I=0.02 A

Kirchhoff'l loop law in BGEFB : E2+(0.05)R3+(I+0.05)R1=0
E2+(0.05)(20)+(0.02+0.05)(100)=0 E2=4 V
Hence current flowing through R1, I1=0.050.02=0.03 A =30 mA towards right

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