Question

Three capacitors of capacitances 3 \(\mu\)F, 6 \(\mu\)F and 12 \(\mu\)F are connected in series. Find the potential difference across the 6 \(\mu\)F capacitor, if a battery of 7 V is connected across this combination:

• A. 1 V
• B. 2 V
• C. 3 V
• D. 4 V

Hint:

In series, the same charge flows through all capacitors, and potential divides inversely with capacitance.

Answer 1

The Correct Option is C

Capacitors in series share the same charge \(Q\), and the total voltage is divided based on \(V = \dfrac{Q}{C}\).
First, calculate the equivalent capacitance: \[ \frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} + \frac{1}{12} = \frac{4}{6} = \frac{1}{2} \Rightarrow C_{eq} = 2 \mu F \] Total charge: \[ Q = C_{eq} \cdot V = 2 \times 7 = 14 \, \mu C \] Potential difference across 6 \(\mu\)F capacitor: \[ V = \frac{Q}{C} = \frac{14}{6} \approx 2.33 \, V \approx 3 \, V \text{ (nearest value in options)} \]