Two batteries of emf 4 V and 8 V with internal resistance 1Ω and 2Ω respectively are connected to an external resistance R = 9Ω as shown in figure. The current in circuit and the potential difference between P and Q respectively will be

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 = 4V/12Ω = 1/3A
Potential difference across P and Q = IR = (1/3)A × 9Ω = 3V