Two capacitors of unknown capacitances `C_1` and `C_2` are connected first in series and then in parallel across a battery of 100 V . If the energy stored in the two combinations is 0.045 J and 0.25 J respectively , determine the value of `C_1` and `C_2` . Also calculate the charge on each capacitor in parallel combination .

Explanation

When capacitors , `C_1` and `C_2` are joined in series then the equivalent capacitance `C_(s) = (C_(1) C_2)/(C_(1) + C_(2))` and in parallel arrangement the equivalent capacitance `C_(p) = ( C_(1) + C_(2))` . As voltage V = 100 V , energy stored by series combination `U_(s) = 0.045` J and energy stored by parallel combination `U_(p) = 0.25 J` , hence we have
`U_(s) = (1)/(2) C_(s) V^(2) = (1)/(2) xx ((C_(1) C_(2))/(C_(1) + C_(2))) xx (100)^(2) = 0.045 " " ..... (i)`
and `U_(p) = (1)/(2) C_(p) V^(2) = (1)/(2) xx (C_(1) + C_(2)) xx (100)^(2) = 0.25 " " ..... (ii)`
On solving equations (i) and (ii) , we get :
`C_(1) = 38.2 mu F` and `C_(2) = 11.8 mu F`
In parallel combination voltage across each capacitor is same at V = 100 V
`therefore` Charge on 1st capacitor `Q_(1) = C_(1) V = 38.2 xx 100 mu C = 3.82 mC`
and charge on 2nd capacitor `Q_(2) = C_(2) V = 11.8 xx 100 mu C = 1.18 m C `