In the circuit shown in fig. $E_{1} = 3 V, E_{2} = 2 V, E_{3} = 1 V, R = r_{1} = r_{2} = r_{3} = 1 ohm$ .Find the potential difference between the points A and B and the currents through each branch.

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$2 V, I_{1} = 1 A, I_{2} = 2 A, I_{3} = 3 A$

$2 V, I_{1} = I_{2} = I_{3} = 0$

$5 V, I_{1} = I_{2} = I_{3} = 0$

$2 V, I_{1} = I_{3} = 1 A, I_{2} = 0$

answer is C.

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Detailed Solution

Effective potential difference across three batteries connected in parallel i.e V_PQ= $\frac{\frac{3}{1} + \frac{2}{1} + \frac{1}{1}}{3} = 2 V$

Since PD across PQ = 2V and E₂ = 2 V , hence current through the battery = 0 , I₃ = 0

V_PQ=2 = 3 - I₁

I₁ = 1 A which flows through E₃ also.

Hence I₁ = I₃ = 1A

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