Question

Two batteries of emf 6V and 3V with internal resistances $1\Omega$ and $2\Omega$ respectively are connected as shown in figure. the potential difference across A and B is

• A. 9V
• B. 5V
• C. 3V
• D. 1V

Answer 1

Answer: (B)

According to the question, we draw the circuit as below

Let the current in the circuit is $I$,
Now applying $KVL$ rule,
$6-1I-2I-3=0$
$\Rightarrow 6-3=3I$
$\Rightarrow I=\frac{3}{3}=IA$
As $V_{AB}$ is the potential across battery $6V$ and resistance of $1\Omega$.
So, $V_{AB}=6-1I$
$\Rightarrow V_{AB}=6-1=2V$
Hence, the potential difference across $A$ and $B$ is $5V$.