Two capacitor having capacitances `8muF` and `16 muF` have breaking voltage `20V` and `80V`. They are combined in series. The maximum charge they can store individually in the combination is

Two capacitors each having capacitance C and break down voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be

Two capacitor each having capacitance C and breakdown voltage V are joind in series .The capacitance and the breakdown voltage of the combination will be

Two capacitor each having capacitance C and breakdown voltage V are joind in series .The capacitance and the breakdown voltage of the combination will be

Two capacitor each having capacitance C and breakdown voltage V are joind in series .The capacitance and the breakdown voltage of the combination will be

Two capacitors of capacitance 10 muF and 20 muF are charged to potential 20 V and 10 V respectively. If they are connected to share charges, then loss of energy is

Two capacitors of capacitances 1muF and 3muF are charged to the same voltages 5V. They are connected in parallel with oppositely charged plates connected together. Then

Two capacitor each having capacitance C and breakdown voltage V are joined in parallel,the capacitance and the breakdown voltage of the combination will be

Three capacitors each of cpacitance C and breakdown voltage V are joined in series. The capacitance and break-down voltage of the combination will be

Three capacitors of capacitances 2muF,4muF and 8muF are joined in series and the combination is connected to a 12 V source. Calculate then energy stored in the combination when these are fully charged.

A capacitor of capacitance 1muF withstands a maximum voltage of 6kV, while another capacitor of capacitance 2muF , the maximum voltage 4kV. If they are connected in series, the combination can withstand a maximum of