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,
$6 - 1I - 2I -3 = 0$
$\Rightarrow 6- 3 = 3I$
$\Rightarrow I = \frac{3}{3} = I A$
As $V_{AB}$ is the potential across battery $6V$ and resistance of $1 \Omega$.
So, $V_{AB} = 6 - 1 I$
$\Rightarrow V_{AB} = 6- 1 =2 V $
Hence, the potential difference across $A$ and $B$ is $5\, V$.