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Two cells E(1) and E(2) in the circuit s...

Two cells `E_(1)` and `E_(2)` in the circuit shown in figure, have emfs of 5 V and 9 V and internal resistance of `0.3 Omega` and `1.2 Omega` respestivley. Calculate the value of current flowing through the resistance of `3 Omega`.

Text Solution

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Net emf `= E_(2) - E_(1) = 9-5 = 4V`.
Total resistance `= 0.3 +1.2 +4.5 + (6 xx 3)/(6+3)=8 Omega`
Current through the circuit, `I = (4V)/(8 Omega) = 0.5 A`
Current through `3 Omega` resistance `= (6 xx 0.5)/(6+3) =1/3A`
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