\( n_1 \) identical capacitors, each of capacitance \( C_1 \) are connected in series and charged by a source of potential difference 3 V. \( n_2 \) identical capacitors, each of capacitance \( C_2 \) are connected in parallel and charged by a source of potential difference 2 V. Total energy stored in both the combinations is the same. What is the value of capacitance \( C_2 \) in terms of \( C_1 \)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{9C_1}{4n_1n_2} \)
The energy stored in a combination is \( U = \frac{1}{2}C_{\mathrm{eq}}V^{2} \). So find the equivalent capacitance of each combination, then equate the two energies.
Series combination (\( n_1 \) capacitors of \( C_1 \), 3 V): the reciprocals add, so \( C_s = \frac{C_1}{n_1} \).
\( \displaystyle U_1 = \frac{1}{2}\,\frac{C_1}{n_1}\,(3)^{2} = \frac{9C_1}{2n_1} \)
Parallel combination (\( n_2 \) capacitors of \( C_2 \), 2 V): the capacitances add, so \( C_p = n_2C_2 \).
\( \displaystyle U_2 = \frac{1}{2}\,n_2C_2\,(2)^{2} = 2n_2C_2 \)
The energies are equal, \( U_1 = U_2 \):
\( \displaystyle \frac{9C_1}{2n_1} = 2n_2C_2 \)
\( \displaystyle C_2 = \frac{9C_1}{4n_1n_2} \)