Question

In the figure shown, battery $1$ has $\text{emf}=6\text{V}$ and internal resistance $=1\Omega$. Battery $2$ has $\text{emf}=2\text{V}$ and internal resistance $=3\Omega$. The wires have negligible resistance. What is the potential difference across the terminals of battery $2$?

• A. $4\text{V}$
• B. $1.5\text{V}$
• C. $5\text{V}$
• D. $0.5\text{V}$

Answer 1

The correct option is C $5\text{V}$

Let $i$ be the current in the circuit.


Applying KVL in loop:

$6-3i-2-i\left(1\right)=0$

$\therefore i=1\text{A}$

Now potential difference across terminals of battery $2$ is
$V_A-V_B$

Potential difference across battery $2$ can be calculated as:

$V_A-i\left(3\right)-2=V_B$

Putting $i=1$

$V_A-V_B=5\text{V}$