Question

A series combination of $n_1$ capacitors, each of value $ C_1$, is charged by a source of potential difference $4V$. 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 $C_2$, in terms of $C_1$, is then

• A. $ \frac{ 2C_1}{ n_1 n_2}$
• B. 16 $ \frac{ n_2}{ n_1} C_1 $
• C. 2 $ \frac{ n_2}{ n_1} C_1 $
• D. $ \frac{ 16 \, C_1}{ n_1 n_2}$

Answer 1

The Correct Option is D

To solve this problem, we need to find the value of the capacitance $C_2$ in terms of $C_1$. The problem describes two different combinations of capacitors: one in series and another in parallel, with equal energy stored. Let's break down the steps:

1. Calculate the energy stored in the series combination:
   • The equivalent capacitance for $n_1$ capacitors in series, each of capacitance $C_1$, is: $ C_{\text{series}} = \frac{C_1}{n_1} $
   • The energy stored in a capacitor is given by the formula: $ E = \frac{1}{2} C V^2 $
   • Therefore, the energy stored in the series combination is: $ E_{\text{series}} = \frac{1}{2} \left(\frac{C_1}{n_1}\right) (4V)^2 = \frac{1}{2} \cdot \frac{C_1}{n_1} \cdot 16V^2 = \frac{8 C_1 V^2}{n_1} $
2. Calculate the energy stored in the parallel combination:
   • The equivalent capacitance for $n_2$ capacitors in parallel, each of capacitance $C_2$, is: $ C_{\text{parallel}} = n_2 C_2 $
   • The energy stored in the parallel combination is: $ E_{\text{parallel}} = \frac{1}{2} (n_2 C_2) V^2 = \frac{1}{2} n_2 C_2 V^2 $
3. Set the energies equal and solve for $C_2$:
   • Since both combinations store the same energy, set: $ \frac{8 C_1 V^2}{n_1} = \frac{1}{2} n_2 C_2 V^2 $
   • Cancel $V^2$ from both sides and solve for $C_2$: $ 8 C_1 = \frac{1}{2} n_2 C_2 \Rightarrow C_2 = \frac{16 C_1}{n_1 n_2} $

Thus, the value of $C_2$ in terms of $C_1$ is $ \frac{16 \, C_1}{ n_1 n_2}$, which matches the correct answer.