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: Understanding the Question:
We use the fundamental relationship for a capacitor to relate changes in voltage and charge.

Step 2: Key Formula or Approach:
For a capacitor of capacitance \( C \), the charge is given by \( Q = CV \).

Step 3: Detailed Explanation:
Case 1: \( Q = CV \) \dots (i)
Case 2: When the potential is reduced by \( V_1 \), the new potential is \( V - V_1 \).
The new charge is \( Q_1 = C(V - V_1) \) \dots (ii)
Dividing equation (i) by equation (ii):
\[ \frac{Q}{Q_1} = \frac{CV}{C(V - V_1)} = \frac{V}{V - V_1} \]
Cross-multiplying:
\[ Q(V - V_1) = Q_1 V \]
\[ QV - QV_1 = Q_1 V \]
\[ QV - Q_1 V = QV_1 \]
\[ V(Q - Q_1) = QV_1 \]
\[ V = \frac{QV_1}{Q - Q_1} \]

Step 4: Final Answer:
The initial potential is \( V = \frac{QV_1}{Q - Q_1} \).