Question

The resultant capacitance of parallel combination of two capacitors \( C_1 \) and \( C_2 \) is 20 µF. When these capacitors are individually connected to a voltage source of 1V, then the energy stored in \( C_2 \) is 9 times that of \( C_1 \). If these two capacitors are connected in series, the resultant capacitance value is:

• A. 1.4 µF
• B. 8 µF
• C. 18 µF
• D. 1.8 µF

Hint:

For capacitors in parallel, the total capacitance is the sum of individual capacitances, and for capacitors in series, the total capacitance is given by the reciprocal sum.

Answer 1

The Correct Option is D

We are given that the energy stored in \( C_2 \) is 9 times that stored in \( C_1 \), so: \[ C_2 = 9 C_1 \] The resultant capacitance of the two capacitors in parallel is: \[ C_{{parallel}} = C_1 + C_2 = C_1 + 9 C_1 = 10 C_1 \] Given that \( C_{{parallel}} = 20 \, \mu F \): \[ 10 C_1 = 20 \quad \Rightarrow \quad C_1 = 2 \, \mu F \] Thus: \[ C_2 = 9 C_1 = 9 \times 2 = 18 \, \mu F \] Now, the resultant capacitance for the series combination of \( C_1 \) and \( C_2 \) is: \[ \frac{1}{C_{{series}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2} + \frac{1}{18} = \frac{10}{18} \] Thus: \[ C_{{series}} = \frac{18}{10} = 1.8 \, \mu F \]