Question

In the adjoining circuit diagram, $E=5V$, $r=1\Omega $, ${{R}_{2}}=4\Omega $, ${{R}_{1}}={{R}_{3}}=1\Omega $ and $C=3\mu F$. Then the numerical value of the charge on each plate of the capacitance is
(A). $20\mu C$
(B). $12\mu C$
(C). $6\mu C$
(D). $3\mu C$

Answer 1

Hint: The circuit shown contains capacitors as well as resistors connected together. As capacitors are assumed to be charged, no current flows through the arms containing the capacitors. The current only flows through the arms not containing the capacitors. We can use ohm’s law and the relation between charge, capacitance and potential to calculate charge on each capacitor.
Formulas used:
$R=\dfrac{V}{I}$
$\dfrac{1}{C}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}$
$Q=CV$

Complete answer:

We assume that, $t=\infty $ hence, the capacitors will act like open circuits and no current will flow through the arms containing the capacitors.
Total working resistance in the circuit will be ${{R}_{2}}+r$. Given, ${{R}_{2}}=4\Omega $, $E=5V$, $r=1\Omega $
According to Ohm’s law,
$R=\dfrac{V}{I}$ - (1)
Here, $R$ is the resistance
$V$ is the potential difference
$I$ is the current in the circuit
Substituting given values in the above equation to get,
$\begin{align}
  & 4+1=\dfrac{5}{I} \\
 & \Rightarrow I=\dfrac{5}{5} \\
 & \therefore I=1A \\
\end{align}$
Therefore, the current in the circuit is $1A$.
The potential difference across ${{R}_{2}}$ from eq (1) will be-
$\begin{align}
  & 4=\dfrac{V}{1} \\
 & \Rightarrow V=4V \\
\end{align}$
Therefore, the potential difference across ${{R}_{2}}$ will be $4V$.
Arms HG and DC are connected in parallel to ${{R}_{2}}$, so their potential will be equal to ${{R}_{2}}$.
Equivalent capacitance in the arm DC is calculated by using the equivalent capacitance connected in series, therefore,
$\dfrac{1}{C}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}$
Given, $C=3\mu F$. Substituting given values in the above equation to get,
$\begin{align}
  & \dfrac{1}{C}=\dfrac{1}{3}+\dfrac{1}{3} \\
 & \Rightarrow \dfrac{1}{C}=\dfrac{2}{3} \\
 & \therefore C=\dfrac{3}{2}mF \\
\end{align}$
Therefore, the equivalent capacitance is $\dfrac{3}{2}\mu F$ and the potential across each capacitor is $\dfrac{4}{2}=2V$.
We know that,
$Q=CV$
Here, $Q$ is the charge on the capacitor
$C$ is the capacitance of the capacitor
$V$ is the value of the potential difference between plates of capacitor
Substituting given values in the above equation to get,
$\begin{align}
  & Q=\dfrac{3}{2}\mu F\times 2V \\
 & \Rightarrow Q=3\mu C \\
\end{align}$
The charge on each capacitor will be the same and the magnitude of charge is $3\mu C$.
Therefore, the magnitude of charge across each capacitor is $3\mu C$.

Hence, the correct option is (D).

Note:
In series, the charge across each capacitor is the same whereas in parallel, the potential across each capacitor is the same. When the capacitor gets fully charged, it does not allow current to flow through it and hence acts like an open circuit. Since, the capacitance is the same on all capacitors, the potential in series will be distributed uniformly. That is why, potential on each capacitor is half of the total potential.