In the circuit shown, the current through the $4 \Omega$ resistor is $1 \mathrm{~A}$ when the points $P$ and $M$ are connected to a DC voltage source. The potential difference between the points $M$ and $N$ is

• A

  $1.5 \mathrm{~V}$
• B

  $1.0 \mathrm{~V}$
• C

  $0.5 \mathrm{~V}$
• Correct answer
  D

  $3.2 \mathrm{~V}$

Solution

Key Idea : In parallel resistances, potential difference across them is same. Potential difference across PM $V=4 \times 1=4 \text { volt (ie, across } 4 \Omega)$ Equivalent resistance of lower side arm, $R_{e q}=1+\frac{0.5 \times 0.5}{0.5+0.5}=1.25 \Omega$ Now the circuit can be shown as Let current I flows in lower branch, so $\begin{aligned} 1.25 I & =4 \mathrm{~V} \\ I & =\frac{4}{1.25}=3.2 \mathrm{~A} \end{aligned}$ Therefore, 3.2 A current flows in $1 \Omega$ resistance, hence potential difference between $M$ and $N$ is $V^{\prime}=3.2 \times 1=3.2$ volt

Hence, the correct answer is option: 'D'.