The charge on $4 \mu \mathrm{F}$ capacitor, in the given circuit is

Question

Aicharya · Accepted Answer

The circuit can be redrawn as, $C_{\mathrm{cq}}=\frac{4 \times 6}{4+6}=2.4 \mu \mathrm{F}$ As we know, potential drop across parallel branch $A B$ is $V_{A B}=10 \mathrm{~V}$ So, charge $Q=C_{\mathrm{eq}} V=2.4 \times 10 \times 10^{-6} \mathrm{C}$ $Q=24 \mu \mathrm{C}$ Since, in a series capacitor branch, there is equal storage of charge in each capacitor. So, $4 \mu \mathrm{F}$ capacitor store a charge of $24 \mu \mathrm{C}$. Hence, the correct option is (a).

The charge on \(4 \mu \mathrm{F}\) capacitor, in the given circuit is

Solution