The capacitor of capacitance C in the circuit shown in fully charged initially. Resistance is R.–

After the switch S is closed, the time taken to reduce the stored energy in the capacitor to half its initial value is

A capacitor of capacitance Chas initial charge Q_0 and connected to an inductor L as shown. At t=0 switch S is closed. The current through the inductor when energy in the capacitor is three times the energy of inductor is

In the circuit shown, the capacitor C_1 is initially charged with charge q_0 . The switch S is closed at time t = 0 . The charge on C_2 after time t is

In the circuit shown, the capacitor C_1 is initially charged with charge q_0 . The switch S is closed at time t = 0 . The charge on C_2 after time t is

The capacitor shown in the figure is initially unchanged, the battery is ideal. The switch S is closed at time t= 0, then the time after which the energy stored in the capacitor becomes one - fourth of the energy stored in it in steady - state is :

The capacitor shown in the figure is initially unchanged, the battery is ideal. The switch S is closed at time t= 0, then the time after which the energy stored in the capacitor becomes one - fourth of the energy stored in it in steady - state is :

In the circuit shown in figure, the battery is an ideal one, w/The emf V. The capacitor is initially uncharged. The switch s is closed at time t = 0 The charge Q on the capacitor at time t is

In the R-C circuit shown in the figure the total of 3.6 xx 10^(-3)J is dissipated in the 10 ohms resistor when the switch S is closed. The initial charge on the capacitor is

Initially, both capacitors are uncharged. A long time after the switch S is closed, the potential energy stored in the capacitor of capacitance 2muF is __________xx10^(-6)J .

A parallel pate capacitor of capacitance ‘C’ has charge on its plates initially as shown in the figure. Now at the switch ‘S’ is closed. Select the correct alternative (s) for this circuit diagram.