Three capacitors of capacitances `3 muF`, `6muF`, and `4 muF` are connected as shown across a battery of emf `6 V`. .
i. Find the equivalent capacitance,
ii. Find the potential difference and charge on each capacitor.
iii. Find the energy stored in the system of capacitors and total energy stored in the system of capacitor.

Explanation

i. `C_(1)` and `C_(2)` are in series, so their equivalent capacitance is
`C=(C_(1)C_(2))/(C_(1)+C_(2))=(3xx6)/(3+6)=2 muF`
This `C'` will be in parallel `C_(3), So
`C_(eq)=C'+ C_(3)=2+4=6 muF`
ii. `6 V` will be in parallel across `C_(1)` and `C_(2)` in the inverse ratio of the capacitances. So potential defference across `C_(1)` is
`V_(1)=(C_(2)V)/(C_(1)+C_(2))=(6xx6)/(3+6)=4 V`.
and potential differnce across `C_(2)` is `V_(2)=(C_(1)V)/(C_(1)+C_(2))=(3xx6)/(3+6)=2 V`
Because `C_(3)` is connected directly aross the battery without any other capacitor in between , so potential difference across `C_(3)` is `V_(3) =V=6 V`. Also charge on `C_(1)` is
`q_(1)=C_(1)V_(1)=3xx4=12 muC`
and charge on `C_(2)` is
`1q_(2)=C_(2)V_(2)=6xx2-12 muC`
We see that `q_(1)q_(2)`, because is series, charge is the same.
chaege on `C_(3)` is
`q_(3)=C_(3)V_(3)=4xx6=24 muC`.
Alternative methode to find charge.

Total charge that will flow through the battery is
`Q = C_(eq)V = 6 xx 6 = 36 muC`
It will be divided inthe direct ratio of `C'` and `C_(3)`, so
`q_(1) = (C'Q)/(C'+C_(3)) = (2 xx 36)/(2+4) = 12 muC`
`q_(3) = (C'Q)/(C'+C_(3)) = (4 xx 36)/(2+4) = 24 muC`
Energy in `C_(1)` is
`U_(1) = (1)/(2)C_(1)V_(1)^(2) = (1)/(2)xx3xx4 = 24mu J`
Energy in `C_(2)` is
`U_(2) = (1)/(2)C_(2)V_(2)^(2) = (1)/(2)xx6xx2^(2) = 12mu J`
Energy is `C_(3)` is
`U_(3) = (1)/(2)C_(3)V_(3)^(2) = (1)/(2)xx4xx6 = 72mu J`
Total energy is
`U = U_(1) + U_(2) + U_(3) = 24 + 12 + 72 = 108 mu J`
Alternatively
`U = (1)/(2) xx 6 xx 6^(2) = 108 muJ`