Three capacitors of capacitances `2 muF`, `3mu F` and `6muF` are connected in series with a 12 V battery. All the connecting wire are disconnected, the three positive plates are connected together and the three negative plates are connected together. Find the charges on the three capacitors after the reconnection.

Answer

` (72)/(11) mu C ; (108)/(11) mu C ; (216)/(11) muC`

Explanation

`(1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3)) = (1)/(2) + (1)/(3) + (1)/(6) = (6)/(6)`
Charge on each capacitor `= C_(s) xx V`
`= 1xx12 = 12 mu C`
Thus charges of `+- 12 mu C` appear on the plates of each of the three capacitors. When connected in parallel, thier potentail becomes equal as the charges redistribute themselves,
Common potentail,
`V = ("total charge")/("total capacity") = (12xx3)/(2+3-6) = (36)/(11)`
`:. Q_(1) = C_(1) V = 2xx (36)/(11) = (72)/(11) mu C`
`Q_(2) = C_(2) V = 3 xx (36)/(11) = (108)/(11) mu C`
`Q_(3) = C_(3) V = 6 xx (36)/(11) = (216)/(11) muC`