Question

A series combination of \[{n_1}\]capacitors, each of value ${{\text{C}}_1}$, is charged by a source of potential difference 4V. When another parallel combination of \[{n_2}\]capacitors , each of value ${{\text{C}}_2}$,is charged by a source of potential difference ${\text{V}}$,its has the same(total) energy stored in it, as the first combination has. The value of ${{\text{C}}_2}$, in terms of ${{\text{C}}_1}$, is then

(A). $\dfrac{{2{C_1}}}{{{n_1}{n_2}}}$
(B). $16\dfrac{{{n_2}}}{{{n_1}}}{C_1}$
(C). $2\dfrac{{{n_2}}}{{{n_1}}}{C_1}$
(D). $\dfrac{{16{C_1}}}{{{n_1}{n_2}}}$

Answer 1

Hint: Series combination of the capacitor of same magnitude in numbers $n$ is the gives the equivalent capacitance by dividing the magnitude of capacitance by the number of capacitor and the capacitance of the parallel combination of such capacitors gives the equivalent capacitance by multiplying both magnitude of capacitance and the numbers of the capacitance.

Complete step-by-step solution -
Given,
In series combination,
No of capacitors =\[{n_1}\]
Value of each capacitor =${{\text{C}}_1}$
Source of potential difference = applied voltage =${{\text{V}}_s}$= ${\text{4V}}$.
In parallel combination,
No of capacitors=\[{n_2}\]
Value of each capacitor=${{\text{C}}_2}$
Source of potential energy = applied voltage =${{\text{V}}_p}$=${\text{V}}$
One condition is given,
Both combinations have the same amount of stored energy.
Now in series combination equivalent capacitance =${{\text{C}}_s} = \dfrac{{{C_1}}}{{{n_1}}}$
And the energy stored in the series combination=${{\text{E}}_s} = \dfrac{1}{2}{C_s}{{\text{V}}_s}^2$
$ \Rightarrow {{\text{E}}_s} = \dfrac{1}{2}\dfrac{{{C_1}}}{{{n_1}}}{(4{\text{V)}}^2}$
$ \Rightarrow {{\text{E}}_s} = \dfrac{{8{C_1}{{\text{V}}^2}}}{{{n_1}}}$----equation (1)
Likewise for the parallel combination equivalent capacitance=${{\text{C}}_p} = {n_2}{{\text{C}}_2}$
And the energy stored in the parallel combination=${{\text{E}}_p} = \dfrac{1}{2}{C_p}{{\text{V}}_p}^2$
$ \Rightarrow {{\text{E}}_p} = \dfrac{1}{2}{n_2}{C_2}{{\text{V}}^2}$------equation (2)
Here as per the given condition that the stored energy in both the combination is same, this implies that,
Stored energy in the series combination = stored energy in the parallel combination
${{\text{E}}_s} = {{\text{E}}_p}$
$\dfrac{{8{C_1}{{\text{V}}^2}}}{{{n_1}}} = \dfrac{1}{2}{n_2}{C_2}{{\text{V}}^2}$
So now on simplification we will get the value of ${{\text{C}}_2}$in the terms of the ${{\text{C}}_1}$.
So, ${{\text{C}}_2} = \dfrac{{16{C_1}}}{{{n_1}{n_2}}}$
Thus we get this solution of the given problem, like this we can express the capacitance ${{\text{C}}_1}$in terms of ${{\text{C}}_2}$.

Hence the option (D) is the correct answer.

Note – Capacitors are the charge storage device. It is like a bucket full of charges. It stores charges and when there is a need for a high amount of charges then it just gives it’s stored charges and the electronic device gets its required voltage to run.