Question

A capacitor of capacitance $C_1=1\mu F$ charged up to a voltage $V=110V$ is connected in parallel to the terminals of a circuit consisting of two uncharged capacitors connected in series and possessing the capacitance $C_2=2\mu F$ and $C_3=3.0\mu F$. Then what is the charge flowing through the wire

• A. $30\mu C$
• B. $40\mu C$
• C. $50\mu C$
• D. $60\mu C$

Answer 1

The correct option is D $60\mu C$

Since we know the voltage and the capacitance then the charge on the plate will be,
$Q=CV$
$=1\mu F\times 110V$
$Q=110\mu C$
Now if we analyise the circuit then,
As charge is conserved,
$\begin{array}{}Q+q=110\mu C& \text{(1)}\end{array}$
where $q$ is charge across $C_2$ and $C_3$.

In betweeen plate $C_2$ and $C_3$
$-q+q^{\prime }=0$
$\begin{array}{}{q}^{\prime }=q& \text{(2)}\end{array}$
Between $C_1$ and $C_3$
$-Q+(-q^{\prime })=0$
$\begin{array}{}-Q=q& \text{(3)}\end{array}$ ($\because q=q^{\prime }$)

By Kirchhoff's Loop Law

$\frac{-Q}{1\times {10}^{-6}}+\frac{-q}{3\times {10}^{-6}}+\frac{q}{2\times {10}^{-6}}=0$
$-Q+\frac{q}{3}+\frac{Q}{2}=0$
$\begin{array}{}Q=\frac{5q}{6}& \text{(4)}\end{array}$
Put $Q$ in equation (1) we get,
$\frac{5q}{6}+q=110\mu C$
$q=60\mu C$
From equation (2) and (3),
$q=q^{\prime }=60\mu C$
Substituting the value of $q$ in equation (4),
$Q=\frac{5\times 60}{6}=50\mu C$
Hence charge flowing through the wire is $60\mu C$
For detailed solution watch the next video.