In the given circuit, the terminal potential difference of the cell is :
- $2 \mathrm{~V}$
- $3 \mathrm{~V}$
- $4 \mathrm{~V}$
- $1.5 \mathrm{~V}$
Solution
$\begin{aligned} \mathrm{i} & =\frac{3}{1+2}=1 \mathrm{~A} \\ \mathrm{v} & =\mathrm{E}-\mathrm{ir} \\ & =3-1 \times 1=2 \mathrm{~V}\end{aligned}$
Asked in: JEE Main 2024 (08 Apr Shift 1)