Three capacitors `2muF, 3muF` and `5muF` can withstand voltages to `3V,2V` and `1V` respectively. Their series combination can withstand a maximum voltage equal to

Three capacitors 2 mu F, 3 muF and 5 muF can withstand voltages to 3 V, 2V and 1V respectively. Their series combination can withstand a maximum voltage equal to

Three capacitors of 2muF, 3muF and 5muF are independently charged with batteries of emf’s 5V, 20V and 10V respectively. After disconnecting from the voltage sources. These capacitors are connected as shown in figure with their positive polarity plates are connected to A and negative polarity is earthed. Now a battery of 20V and an uncharged capacitor of 4muF capacitance are connected to the junction A as shown with a switch S. When switch is closed, find : (a) the potential of the junction A. (b) final charges on all four capacitors.

A capacitor of capacitance C_(1)=1 muF can withstand maximum voltage V_(1)=6 kV and another capacitor of capacitance C_(2)=3 muF can withstand maximum voltage V_(2) 4 kV . When the two capacitors are connected in series, the combined system can withstand a maximum voltage of

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

Two capacitors 3muF and 4muF have same maximum voltage rating V_(0) . When they are connected in series, they can withstand a maximum potential difference of 700 V . What is V_(0) ?

Two capacitors 1muF and 3muF are charged individually to voltages of 100V and 200V respectively. These are then connected in a parallel combination. Find the loss in the energy stored in the capacitors.

Two capacitors 3muF and 5muF are charged to 18V and 25V respectively. These are then connected in series. Calculate the net voltage and net charge. Find the loss of energy after connecting them in series.