Question

Two batteries of emf $4V$ and $8V$ with internal resistances $1\Omega$ and $2\Omega$ are connected in a circuit with a 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 $3V$
• B. $\frac{1}{6}A$ and $4V$
• C. $\frac{1}{9}A$ and $9V$
• D. $\frac{1}{2}A$ and $12V$

Answer 1

Answer: (A)

Applying Kirchhoffs law in the given loop

$-2I+8-4-1\times I-9I=0$
$\Rightarrow I=\frac{1}{3}A$
Potential difference across $PQ$
$=\frac{1}{3}\times 9=3V$