Three capacitors each of capacitance of `2 muF` are Connected in parallel across 6V battery. Find the charge in each capacitor.

Explanation

`C_(1) = C_(2) =C_(3) =2 muF`
Let `C_s` be the net capacitance in series then
`1/C_(s) = 1/C_(1) + 1/C_(2) + 1/C_(3)`
`1/C_(s) =1/2 + 1/2 + 1/2 = 3/2`
`C_(s) = 2/3 = 0.67 muF`
`therefore V = 6V`
`q = C_(s)V = 6 xx 3/2 = 4 muC`
`therefore` In series combination, charge on each capacitor remains same
`therefore q_(1) = q_(2) = q_(3) = 4muC`