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. $\frac{16C_1}{n_1{n}_2}$

Answer 1

Answer: (D)

A series combination of $n_1$, capacitors each of capacitance $C_1$ are connected to $4V$ 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
$U_s=\frac{1}{2}{C}_s(4V{)}^2=\frac{1}{2}(\frac{C_1}{n_1})(4V{)}^2$ (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
$\frac{1}{2}\frac{C_1}{n_1}(4V{)}^2=\frac{1}{2}(n_2{C}_2)(V{)}^2$
or $\frac{C_{1}16}{n_1}=n_2{C}_2$ or $C_2=\frac{16C_1}{n_1{n}_2}$