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in Physics by (51.7k points)
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In the given figure, the emf of the battery will be

(1) 12V

(2) 13V

(3) 16V

(4) 18V

1 Answer

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(2) 13V

Current entering at point Q is 2A as it will divide equally. 

Hence current through PQ = 2A 

and current in branch PQRS = 2A 

∴ Current in branch PS = 6/4 A = 1.5 A

∴ Therefore current drawn from cell = (2.0 + 1.5)A = 3.5A

Total resistance in the circuit = 2Ω + 12/7 Ω = 26/7 Ω

Hence emf E = IR = 3.5 x 26/7 = 13V

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