Question

Two batteries of emf 4V and 8V with internal resistance of 1 omega and 2 omega are connected in a circuit with resistance of 9 omega as shown in the fig. The current and potential difference between the points P and Q are

Answer 1

Answer:

Since the batteries are connected in reverse polarities, the net potential applied to the circuit =8V−4V=4V

The net resistance in the circuit =R+r1+r2 =9Ω+1Ω+2Ω=12Ω

Hence, the current in the circuit =12Ω4V =31A

Potential difference across P and Q =IR =31A×9Ω =3V