Question

Two cells having emf of 10V and 8V ,and internal resistance of 1 ohm (each) are connected as shown with an external resistance of 8 ohm .Find the current flowing through the circuit.

Answer 1

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+r

1

+r

2

=9Ω+1Ω+2Ω=12Ω

Hence, the current in the circuit =

12Ω

4V

=

3

1

A

Potential difference across P and Q =IR =

3

1

A×9Ω =3V

Hope it helps...