Q132 / 150 · MHT CET 2025 (25 Apr Shift 2) · Physics

Two batteries of e.m.f 4 V and 8 V with internal resistance $1 \Omega$ and $2 \Omega$ respectively are connected in a circuit with a resistance of $9 \Omega$ as shown in the figure. The current and potential difference between the points ' P ' and ' Q ' is Figure

  • A
    $\frac{1}{3} \mathrm{~A}$ and 4 V
  • Correct answer
    B
    $\frac{1}{3} \mathrm{~A}$ and 3 V
  • C
    $\frac{1}{2} \mathrm{~A}$ and 5 V
  • D
    $\frac{1}{6} \mathrm{~A}$ and 3 V

Solution

Net circuit EMF: Batteries in series with opposing polarities yield $E_{\text{net}} = 8 - 4 = 4\ \text{V}$, with direction determined by the larger EMF source.

Total resistance: All resistances including internals are in series: $R_{\text{total}} = 1 + 2 + 9 = 12\ \Omega$.

Current calculation: Applying Ohm's law, $I = \frac{4}{12} = \frac{1}{3}\ \text{A}$.

Potential difference $V_{PQ}$: This equals the voltage drop across the external resistor: $V_{PQ} = \frac{1}{3} \cdot 9 = 3\ \text{V}$.

The calculated values correspond to option B.

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

Relevant Questions