Two capacitors of capacitance `C_(1)=2muF and C_(2) = 8 muF` are connected in series and the resulting combination is connected acorss 300 V. Calculate the charge, potential difference and energy stored in the capacity separately.

Explanation

If C is th equivalent capacitance
then `(1)/(C ) = (1)/(C_(1))+ (1)/(C _(2)) = (1)/(2) +(1)/(8) = ( 5)/( 8)`
`:. C = ( 8)/( 5) = 1.6 mu F`
Charge `q= CV = 1.6 xx 10^(-6) xx 300 = 4.8 xx 10^(-4) C`
Potential across `C_(1) = V_(1) = (q)/(C_(1)) = 240 ` volt
Potentail across `C_(2) = V_(2) = ( q)/(C_(2)) = 60` volt
Energy stored in `C_(1) = (1)?(2) C_(1)V^(2) _(1) = 5.76 xx 10^(-2)` joule
Energy stored in `C_(2) = (1)/(2) C_(2)V^(2) _(2)=1.44 xx 10^(-2)` joule