Question

Capacitors of capacitances \(C_1, C_2\) and \(C_3\) are connected in series. If the combination is connected to a supply of \(V\) volt, the potential difference across capacitor \(C_1\) is

• A. \( \dfrac{C_1 C_2 C_3}{V} \)
• B. \( \dfrac{C_2 C_3 + C_1 C_3 + C_1 C_2}{C_2 C_3 V} \)
• C. \( \dfrac{V}{C_1 + C_2 + C_3} \)
• D. \( \dfrac{C_2 C_3 V}{C_2 C_3 + C_1 C_3 + C_1 C_2} \)

Hint:

In series capacitors, voltage divides inversely with capacitance.

Answer 1

The Correct Option is D

Step 1: Charge in series combination.
In series, the same charge \(Q\) flows through all capacitors.
Step 2: Equivalent capacitance.
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \Rightarrow C_{\text{eq}} = \frac{C_1 C_2 C_3}{C_2 C_3 + C_1 C_3 + C_1 C_2} \]
Step 3: Charge on the combination.
\[ Q = C_{\text{eq}} V \]
Step 4: Potential difference across \(C_1\).
\[ V_1 = \frac{Q}{C_1} = \frac{C_2 C_3 V}{C_2 C_3 + C_1 C_3 + C_1 C_2} \]
Step 5: Conclusion.
The potential difference across \(C_1\) is \( \dfrac{C_2 C_3 V}{C_2 C_3 + C_1 C_3 + C_1 C_2} \).