Question

A capacitor \( C_1 = 4 \, \mu F \) is connected in series with another capacitor \( C_2 = 1 \, \mu F \). The combination is connected across a d.c. source of 200 V. The ratio of potential across \( C_2 \) to that across \( C_1 \) is

• A. 8:1
• B. 16:1
• C. 2:1
• D. 4:1

Hint:

In a series combination of capacitors, the potential divides inversely with respect to the capacitance.

Answer 1

The Correct Option is D

Step 1: Understanding the behavior of capacitors in series.
When capacitors are connected in series, the total capacitance \( C_{\text{total}} \) is given by: \[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} \] For \( C_1 = 4 \, \mu F \) and \( C_2 = 1 \, \mu F \), we have: \[ \frac{1}{C_{\text{total}}} = \frac{1}{4} + \frac{1}{1} = \frac{5}{4} \quad \Rightarrow \quad C_{\text{total}} = \frac{4}{5} \, \mu F \] Step 2: Voltage division across capacitors.
The voltage across each capacitor in a series combination is inversely proportional to its capacitance. Therefore, the ratio of the potential across \( C_2 \) to that across \( C_1 \) is given by: \[ \frac{V_{C_2}}{V_{C_1}} = \frac{C_1}{C_2} \] Substituting the values of \( C_1 \) and \( C_2 \): \[ \frac{V_{C_2}}{V_{C_1}} = \frac{4}{1} = 4 \] Step 3: Conclusion.
Thus, the ratio of the potential across \( C_2 \) to that across \( C_1 \) is 4:1, which corresponds to option (D).