In the circuit shown, the potential difference across the $4.5 \mu \mathrm{F}$ capacitor is
In the circuit shown, the potential difference across the $4.5 \mu \mathrm{F}$ capacitor is
$\frac{8}{3}$ volt
8 volt
6 volt
4 volt
Solution
$3 \mu F$ and $6 \mu F$ capacitors are in parallel.
Their equivalent capacitance is given by
$C_{1}=3+6=9 \mu F$
$4.5 \mu F$ and $9 \mu F$ capacitors are in series.
For series combination, the charge on the capacitors is the same.
$\begin{aligned}
& \therefore C_{1} V_{1}=C_{2} V_{2} \\
& \therefore 9 V_{1}=4.5 V_{2} \\
& \therefore V_{2}=2 V_{1}
\end{aligned}$
$\begin{aligned}
& \text { Also } V_{1}+V_{2}=12 V \\
& V_{1}+2 V_{2}=12 \\
& 3 V_{1}=12 \\
& V_{1}=4 V \\
& \therefore V_{2}=8 V
\end{aligned}$
