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

A series combination of $n_1$, capacitors each of capacitance $C_1$ are connected to $4\, V$ source as shown in the figure.

Total capacitance of the series combination of the capacitors is
$ \frac{1}{ C_s} = \frac{ 1}{ C_1 } + \frac{ 1}{ C_1 } + \frac{ 1}{ C_1 } + ....$ upto $\, n_1 \, $ terms $ = \frac{ n_1}{ C_1} $
or $ C_s = \frac{ C_1}{ n_1} \,$ ...(i)
Total energy stored in a series combination of the capacitors is
$ U_s = \frac{1}{2} C_s ( 4 V)^2 = \frac{1}{2} \bigg( \frac{ C_1}{ n_1} \bigg) (4V)^2 $ (Using (i)) ..... (ii)
A parallel combination of $ n_2 $ capacitors each of capacitance $ C_2$ are connected to V source as shown in the figure.

Total capacitance of the parallel combination of capacitors is
$ C_p = C_2 + C_2 + .......... +$ upto $\, n_2$ trems $= n_2 \, C_2 $
or $ C_p = n_2 C_2 $ ...(iii)
Total energy stored in a parallel combination of capacitors is
$ U_p = \frac{1}{2} C_p \, V^2 $
= $ \frac{1}{2} ( n_2 C_2) (V)^2 $ (Using (iii))...(iv)
According to the given problem,
$ U_s = U_p$
Substituting the values of $U_s$ and $U_p$ from equations (ii) and (iv), we get
$ \frac{1}{2} \frac{ C_1}{ n_1} ( 4 V)^2 = \frac{1}{ 2} (n_2 C_2) (V)^2 $
or $ \frac{ C_1 16} { n_1} = n_2 C_2 $ or $ C_2 = \frac{ 16 C_1 }{ n_1 n_2} $