Three capcitors of capacitances `4 muF`,
.
`4 muF`, and `8 muF`, are connected as show across a battery of emf `12 V`.
i. Find the equivalent capacitane.
ii. Find the potential difference and charge on each capacitor.
iii. Find the energy stored in each capacitor and the total energy stored in the system of capacitors.

Explanation

`C_(2)` and `C_(3)` are in parallel, so their equivalent capacitance is
`C'=C_(2)=4+8=12 muF`
This `C'` will be in series with `C_(1)`, so the net equilvent capacitance is
`C_(eq) =(C_(1)C')/(C_(1)+C')=(4xx12)/(4+12)=3 muF`
ii. `q_(1)=C_(eq)V=3xx12=36 mu C`
`q_(1)` will be divded into `q_(2)` and `q_(3)`, So
`q_(2)=(C_(2)q_(1))/(C_(2)+C_(3))=(4xx36)/(4+)=12 muC`
`q_(3)=(C_(3)q_(1))/(C_(2)+C_(3))=(8xx36)/(4+8)=24 muC`
`V_(1)=q_(1)/C_(1)=(36)/4=9 V`
`V_(2)=V_(3)=q_(2)/C_(2)=Q_(3)/C_(3)=(36)/4=3V`

Alternatively, potentials can also be calculated by dividing ` 12 V` and the inverse ratio of ` C_1` and `C'`
`V_1 = ( C'V) /(C_1 +C') = (12 xx 12) /(4 +12 ) =9 V`
`V_2 =V_3 = (C_1 V) /( C_1 +C') = (4 xx 12) /(2 + 12) =3 V`
(iii) ` U_i = 1/2 C_1 V_1^2 = 1/2 xx 4 xx 9^2 = 162 mu J`
`U_2 + 1/2 C_2 V_2^2 = 1/2 xx 4 xx 3^3 =18 mu J`
`U_3 = 1/2 C_3V_3 ^2 = 1/2 xx 8 xx 3^3 =36 mu J`
Total energy is `U_1 + U_2 +U_3 =216 muJ`
Alternatively
` U = 1/2 C_(eq)V^(2) = 1/2 xx 3 xx (12)^2 =216 muJ` .