Two batteries of emf 3 V and 6 V with internal resistances $2 \Omega$ and $4 \Omega$ are connected in a circuit with resistance of $10 \Omega$ as shown in figure. The current and potential difference between the points $P$ and $Q$ are

• A

  $\frac{3}{16} \mathrm{~A}$ and $\frac{8}{15} \mathrm{~V}$
• B

  $\frac{16}{3} \mathrm{~A}$ and $\frac{15}{8} \mathrm{~V}$
• C

  $\frac{3}{16} \mathrm{~A}$ and 8 V
• Correct answer
  D

  $\frac{3}{16} \mathrm{~A}$ and $\frac{15}{8} \mathrm{~V}$

Solution

Applying Kirchhoff's voltage law in the given loop, $\begin{aligned}-4 I+6-3-2 I-10 I & =0 \\ -16 I & =-3 \\ I & =\frac{3}{16} \mathrm{~A}\end{aligned}$ $\because$ Potential difference across $P Q=\frac{3}{16} \times 10=\frac{15}{8} \mathrm{~V}$

Hence, the correct answer is option: 'D'.