Question

A $5\,\mu {\text{F}}$ capacitor is fully charged by a $12\,{\text{V}}$ battery and then disconnected. If it is connected now parallel to an uncharged capacitor, the voltage across it is $3\,{\text{V}}$. Then, the capacity of the uncharged capacitor is
A. $5\,\mu {\text{F}}$
B. $25\,\mu {\text{F}}$
C. $50\,\mu {\text{F}}$
D. $10\,\mu {\text{F}}$
E. $15\,\mu {\text{F}}$

Answer 1

Use the formula for charge on the plates of the capacitor. This formula gives the relation between the charge on the plates of capacitor, capacitance and potential difference across the plates of capacitor. Then use the formula for net capacitance of two capacitors connected in parallel and determine the capacity of the uncharged capacitor.

Formula used:
The charge $Q$ stored on the plates of capacitor is given by
$Q = CV$ …… (1)
Here, $C$ is the capacitance of the capacitor and $V$ is the potential difference across the plates of the capacitor.

Complete step by step answer:
We have given that a capacitor of capacitance $5\,\mu {\text{F}}$ is fully charged by a battery of potential difference $12\,{\text{V}}$.
${Q_1} = 5\,\mu {\text{F}}$
${V_1} = 12\,{\text{V}}$
Let us first determine the charge on this capacitor.
Rewrite equation (1) for the charge on the capacitor.
${Q_1} = {C_1}{V_1}$
Substitute $5\,\mu {\text{F}}$ for ${C_1}$ and $12\,{\text{V}}$ for ${V_1}$ in the above equation.
${Q_1} = \left( {5\,\mu {\text{F}}} \right)\left( {12\,{\text{V}}} \right)$
$\Rightarrow {Q_1} = 60\,\mu {\text{C}}$
Hence, the total charge on the fully charged capacitor is $60\,\mu {\text{C}}$.
We have given that this fully charged capacitor is disconnected from the battery of $12\,{\text{V}}$ and connected in parallel to an uncharged capacitor.
The common potential of these two capacitors connected in parallel is $3\,{\text{V}}$.
$V = 3\,{\text{V}}$
Since the fully charged capacitor and uncharged capacitor are connected in parallel with each other, the potential difference across both of them will be the same.
Hence, the net charge $Q$ on both of these capacitors is equal to the charge ${Q_1}$.
$Q = {Q_1}$
Hence, the net capacitance $C$ of the fully charged capacitor and uncharged capacitor are connected in parallel with each other is
$C = {C_1} + {C_2}$
Here, ${C_2}$ is the capacitance of the uncharged capacitor.
Rearrange the above equation for ${C_2}$.
$\Rightarrow {C_2} = C - {C_1}$
Substitute $\dfrac{Q}{V}$ for $C$ in the above equation.
$\Rightarrow {C_2} = \dfrac{Q}{V} - {C_1}$
Substitute $60\,\mu {\text{C}}$ for $Q$, $3\,{\text{V}}$ for $V$ and $5\,\mu F$ for ${C_1}$ in the above equation.
$\Rightarrow {C_2} = \dfrac{{60\,\mu {\text{C}}}}{{3\,{\text{V}}}} - \left( {5\,\mu F} \right)$
$\Rightarrow {C_2} = \left( {20\,\mu {\text{F}}} \right) - \left( {5\,\mu F} \right)$
$\Rightarrow {C_2} = 15\,\mu F$

Therefore, the capacity of the uncharged capacitor is $15\,\mu F$.

So, the correct answer is “Option E”.

Note:
The students may think that the net charge on the two capacitors connected in parallel is equal to the charge on a fully charged capacitor. As there is no other component providing charge to the uncharged capacitor. Hence, the total charge on the fully charged capacitor gets divided between these two capacitors and the net charge is equal to the charge on the fully charged capacitor.