Question

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

• A. $\frac{1}{3}$ A and 3 V
• B. $\frac{1}{6}$ A and 4 V
• C. $\frac{1}{9}$ A and 9 V
• D. $\frac{1}{12}$ A and 12 V

Answer 1

Answer: (A)

$\frac{1}{3}$ A and 3 V

Answer 2

Since the batteries are connected in reverse polarities, the net potential applied to the circuit $=8 V -4 V =4 V$
The net resistance in the circuit $= R + r _{1}+ r _{2}=9 \Omega+1 \Omega+2 \Omega=12 \Omega$
Hence, the current in the circuit $=\frac{4 V }{12 \Omega}=\frac{1}{3} A$
Potential difference across $P$ and $Q = IR =\frac{1}{3} A \times 9 \Omega=3 V$