Q103 / 150 · MHT CET 2025 (26 Apr Shift 2) · Physics

A capacitor of unknown capacity is connected across a battery of V volt. The charge stored in it is Q coulomb. When potential across the capacitor is reduced by $V_1$ volt, the charge stored in it becomes $Q_1$ coulomb. The potential V is

  • Correct answer
    A
    $\frac{Q V_1}{Q-Q_1}$
  • B
    $\frac{Q_1 V_1}{Q+Q_1}$
  • C
    $\frac{Q_1}{Q}$
  • D
    $\frac{\mathrm{Q}}{\mathrm{Q}_1}$

Solution

The capacitance $C$ relates charge and voltage as $Q = CV$. When the potential decreases by $V_1$, the new charge becomes $Q_1 = C(V - V_1)$.

Substituting $C = \frac{Q}{V}$ gives $Q_1 = \frac{Q}{V}(V - V_1)$, leading to $Q_1 V = Q(V - V_1)$.

Rearranging yields $V(Q - Q_1) = QV_1$, so the solution is:

$V = \frac{QV_1}{Q - Q_1}$

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

Relevant Questions