The following two questions refer to the circuit shown. Assume that the capacitors are initially uncharged.

After the switch has been closed for a long time, how much charge is on the positive plate of the `3muF` capacitor?

Options

A. `24muC`
B. `32muC`
C. `56muC`
D. `72muC`

Answer

C

Explanation

Initially capacitors act as conducting wire. So initially all the three resistor will be in parallel.
After long time, resistors will be in series,
`R_(eq) = 2+3+4 =9Omega`
`I = (24)/(9) = (8)/(3)A`
Potential difference `3muF =` potential difference across `(3Omega+4Omega)`
`rArr V_(3muF) = (3 +4)I = (56)/(3)v`
`q = CV_(3muF) = (3xx56)/(3) = 56muC`