Capacitors

JEE Main level, single correct, medium

141 of 180

Question 141 of 180

A series combination of n1n_1 capacitors, each of capacity C1C_1 is charged by source of potential difference 4 V4 \mathrm{~V}. When another parallel combination of n2n_2 capacitors each of capacity C2C_2 is charged by a source of potential difference VV, it has the same total energy stored in it as the first combination has. The value of C2C_2 in terms of C1C_1 is then

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Answer: (D) 16C1n1n2\frac{16 C_1}{n_1 n_2}

Equivalent capacitance of n2n_2 number of capacitors each of capacitance C2\mathrm{C}_2 in parallel =n2C2=n_2 C_2
Equivalent capacitance of n1n_1 number of capacitors each of capacitances C1\mathrm{C}_1 in series.
Capacitance of each is C1=C1n1C_1=\frac{C_1}{n_1}
According to question, total energy stored in both the combinations are same
 i.e., 12(C1n1)(4V)2=12(n2C2 b 2∴C2=16C1n1n2\begin{aligned} & \text { i.e., } \frac{1}{2}\left(\frac{C_1}{n_1}\right)(4 V)^2=\frac{1}{2}\left(n_2 C_2 \text { b }^2\right. \\ & \therefore C_2=\frac{16 C_1}{n_1 n_2} \\ & \end{aligned}

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