MHTCET 25 Apr 2025 Shift 2 Paper

Concept:Use Kirchhoff's voltage law to find the loop current, then apply Ohm's law to find the potential difference across the $9\Omega$ resistor.Explanation:The two batteries and the $9\Omega$ resistor form a single closed loop.Let the current in the loop be $I$.The $8V$ battery has internal resistance $2\Omega$, and the $4V$ battery has internal resistance $1\Omega$.Applying Kirchhoff's voltage law around the loop:$$-2I + 8 - 4 - 1I - 9I = 0$$Combine the emf terms and the resistance terms:$$4 - (2 + 1 + 9)I = 0$$$$4 - 12I = 0$$$$I = \frac{4}{12} = \frac{1}{3}A$$So the current in the circuit is $\frac{1}{3}A$.This same current flows through the $9\Omega$ resistor between points $P$ and $Q$.Using Ohm's law:$$V_{PQ} = IR = \frac{1}{3} \times 9 = 3V$$Answer:The current is $\frac{1}{3}A$ and the potential difference between $P$ and $Q$ is $3V$.Hence, the correct option is B.