Question

Capacitors of capacities \(C_1\) and \(C_2\) are connected in series. If the combination is connected to a supply of V volt then the potential difference across capacitor \(C_2\) is

• A. \(\frac{C_1+C_2}{C_1V}\)
• B. \(\frac{C_1V}{C_1+C_2}\)
• C. \(\frac{C_1+C_2}{C_2V}\)
• D. \(\frac{C_2V}{C_1+C_2}\)

Hint:

In series each capacitor carries the same charge, so voltage is inversely proportional to capacitance.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In series, the charge Q is the same on both capacitors, and the supply voltage splits as \(V = V_1 + V_2\), where \(V_1 = Q/C_1\) and \(V_2 = Q/C_2\).

Step 2: Compute
The equivalent capacitance is \(C_s = \dfrac{C_1C_2}{C_1 + C_2}\), so \(Q = C_sV = \dfrac{C_1C_2V}{C_1 + C_2}\).
\[ V_2 = \frac{Q}{C_2} = \frac{C_1V}{C_1 + C_2} \]
So the voltage across \(C_2\) contains \(C_1\) in the numerator. Option (D) would be the voltage across \(C_1\).

Final Answer:
The potential difference across \(C_2\) is \(\dfrac{C_1V}{C_1 + C_2}\), option (B). \[ \boxed{\frac{C_1V}{C_1 + C_2}} \]