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Question

A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 Ω as shown in the figure. Find the value of current in the circuit.

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Solution

Given: E1=10V;E2=200 V, r = 38 Ω,I=?

Net emf, E=E2E1 = 200 V - 10 V = 190 V

Current,I=Net emfResistance=19038=5 A.

Thus, a current of 5A flows in the circuit.

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