A series combination of n_1 capacitors, each of value C_1 , is charged by a source of potential difference 4 ~V . When another parallel combination of n_2 capacitors, each of value C_2 , is charged by a source of

Correct answer: D. (16 C_1)/(n_1 n_2) P.D. across each capacitor is (V_1= 4 V n_1 ) Energy stored across the combination is (U_1=n_1 × (1)/(2) C_1 16 V^2 n_1^2 = 16 C_1 V^2 2 n_1 ) aligned & Energy stored across parallel combination is U_2=n_2 × (1)/(2) C_2 V^2 & (16 C_1 v^2)/(2 n_1)=(n_2 C_1 v^2)/(2) & C_2=(16 C_1)/(n_1 n_2) aligned )

NEET 2010

Exam: NEET
Year: 2010
Subject: Physics
Chapter: Capacitance

NEET 2010 · Physics · Capacitance

A series combination of n_1 capacitors, each of value C_1 , is...

A series combination of

n_1

capacitors, each of value

\mathrm{C}_1

, is charged by a source of potential difference

4 \mathrm{~V}

. When another parallel combination of

n_2

capacitors, each of value

C_2

, is charged by a source of potential difference

V

, it has the same (total) energy stored in it, as the first combination has. The value of

\mathrm{C}_2

, in terms of

\mathrm{C}_1

, is then

Options

A $\frac{2 \mathrm{C}_1}{\mathrm{n}_1 \mathrm{n}_2}$
B $16 \frac{\mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$
C $2 \frac{\mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$
D $\frac{16 C_1}{n_1 n_2}$

Correct answer

D. $\frac{16 C_1}{n_1 n_2}$

Step-by-step solution

P.D. across each capacitor is (V_1=\frac{4 \mathrm{ V}}{\mathrm{n}_1}) Energy stored across the combination is (\mathrm{U}_1=\mathrm{n}_1 \times \frac{1}{2} \mathrm{C}_1 \frac{16 \mathrm{ V}^2}{\mathrm{n}_1^2}=\frac{16 \mathrm{C}_1 \mathrm{ V}^2}{2 \mathrm{n}_1})

\begin{aligned} & \text{Energy stored across parallel combination is } \mathrm{U}_2=\mathrm{n}_2 \times \frac{1}{2} \mathrm{C}_2 \mathrm{ V}^2\\ & \frac{16 C_1 v^2}{2 n_1}=\frac{n_2 C_1 v^2}{2} \\ & \Rightarrow C_2=\frac{16 C_1}{n_1 n_2} \end{aligned}

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