The charge deposited on `4muF` capacitor in the given circuit is:

Options

A. `6xx10^(-6)C`
B. `12xx10^(-6)C`
C. `24xx10^(-6)C`
D. `36xx10^(-6)C`

Answer

D

Explanation

The circuit reduces to
`Q=3muFxx12V=36muC`

Now `(Q_(1))/(Q_(2))=(C_(1))/(C_(2)) implies Q_(1)=2Q_(2)`
`Q=Q_(1)+Q_(2)`
Solve for `Q_(1) and Q_(2)`