Question

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}

• A. $\frac{QV_{1}}{Q-Q_{1}}$
• B. $\frac{Q_{1}V_{1}}{Q+Q_{1}}$
• C. $\frac{Q_1}{Q}$
• D. $\frac{Q}{Q_{1}}$

Hint:

In a given capacitor, $Q \propto V$ as long as $C$ is constant.

Answer 1

The Correct Option is A

Step 1: Formula
$Q = CV \implies C = Q/V$.

Step 2: Analysis
For the second case: $Q_1 = C(V - V_1)$.
Since capacitance $C$ remains constant, $Q/V = Q_1/(V - V_1)$.

Step 3: Calculation
$Q(V - V_1) = Q_1 V \implies QV - QV_1 = Q_1 V$
$QV - Q_1 V = QV_1 \implies V(Q - Q_1) = QV_1$
$V = \frac{QV_1}{Q - Q_1}$.

Step 4: Conclusion
Hence, the initial potential $V$ is $\frac{QV_1}{Q - Q_1}$. Final Answer: (A)