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: Voltage divider
Voltages share inversely with capacitance: $V_1 : V_2 = C_2 : C_1$.

Step 2: Sum
$V_1 + V_2 = V$ so $V_2 = V\dfrac{C_1}{C_1 + C_2}$.

Step 3: Answer
Option (B).

Final Answer:
Option (B). \[ \boxed{\frac{C_1V}{C_1 + C_2}} \]