Two capacitors of capacity `C_(1) and C_(2)` are connected in series. The combined capacity C is given by

Two capacitors of capacities " C_(1) " and " C_(2) " are connected in series.The combination is put across a constant voltage source.The ratio of their capacities " (C_(1))/(C_(2)) " is "n" .Then ratio of the energies stored in them is

Two capacitances of capacity C_(1) and C_(2) are connected in series and potential difference V is applied across it. Then the potential difference across C_(1) will be

Two capacitances of capacity C_(1) and C_(2) are connected in series and potential difference V is applied across it. Then the potential difference across C_(1) will be

Two capacitors with capacitances C_(1) and C_(2) are connected in series. Find the resultant capacitance.

Two capacitors of capacities 1 mu F and C mu F are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80 mu C, the energy stored in the capacitor of capacity C in mu J is

Two capacitors of capacitance C_(1) and C_(2) are connected in parallel and a charge Q is delivered to the combination. The two are then disconnected and reconnected in series. What are the new potential differences and charge on the capacitors?

Three capacitors of capacities C _ 1, C _ 2, C _ 3 are connected in series. Their total capacity will be