Two condensers of capacities ' $\mathrm{C}^{\prime}$ and ${ }^{\prime} 2 \mathrm{C}^{\prime}$ are connected…
Two condensers of capacities ' $\mathrm{C}^{\prime}$ and ${ }^{\prime} 2 \mathrm{C}^{\prime}$ are connected in parallel and then in series with $3^{\text {rd }}$ condenser of capacity '3C'. The combination is charged to 'V' volt. The charge on the condenser of capacity 'C' is
$\frac{C V}{3}$
$\frac{C V}{2}$
$2 C V$
$C V$
Solution
$\begin{aligned}
& \text { Ceq }=3 C / 2 \\
& q=\text { Ceq }^* v=3 C v / 2
\end{aligned}$
As pair of $C, 2 C$ and $3 C$ are in series, charges are equal on both ,i.e, $3 \mathrm{CV} / 4$
Now as q1=1/2q2
So $q 1+q 2=3 C V / 4$
$\Rightarrow 3 q 1 / 2=3 C V / 4$
Thus, q1=CV/2