Two batteries of e.m.f. 4 V and 8 V with internal resistances 1 Ω and 2 Ω are connected in a circuit with a resistance of 9 Ω as shown in figure. The current and potential difference between the points P and Q are

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13 A and 3 V

16 A and 4 V

19 A and 9 V

12 A and 12 V

answer is A.

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Detailed Solution

Applying Kirchhoff's voltage law in the given loop,−2i+8−4−1 × i−9i=0⇒ i=13 APotential difference across PQ=13 × 9 = 3 V

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