In the given circuit, a charge of $+80 \mu C$ is given to the upper plate of the $4 \mu F$ capacitor. Then…
In the given circuit, a charge of $+80 \mu C$ is given to the upper plate of the $4 \mu F$ capacitor. Then in the steady state, the charge on the upper plate of the $3 \mu F$ capacitor is
$+32 \mu \mathrm{C}$
$+40 \mu \mathrm{C}$
$+48 \mu \mathrm{C}$
$+80 \mu \mathrm{C}$
Solution
The total charge on plate $A$ will be $80 \mu \mathrm{C}$.
$2 \mu F$ and $3 \mu F$ capacitors are in parallel. Therefore, $C_{e q}=2+3=5 \mathrm{HF}$
Charge on capacitor of $3 \mu \mathrm{F}$ capacitance $q=\frac{3}{5} \times 80=48 \mu C$
