In the circuit shown, switch's is closed at t=0. then the ratio `(U_(1))/(U_(2))` of potential energy stored in the inductors `L_(1)` and `L_(2)` in steady state is

In the circuit shown, the switch 'S' is closed at t = 0 . Then the current through the battery steady state reached is

In the circuit shown, the switch 'S' is closed at t = 0 . Then the current through the battery steady state reached is

In the circuit as shown in figure the switch is closed at t = 0 . At the instant of closing the switch

In the circuit shown, switch S is closed at time t = 0 . Find the current through the inductor as a function of time t .

In the circuit shown, switch S is closed at time t = 0 . Find the current through the inductor as a function of time t .

Switch is closed at t = 0 then the current in the circuit at t = L/2R is

Switch is closed at t = 0 then the current in the circuit at t = L/2R is

In the circuit shown, what is the energy stored in the coil at steady state is closed to?

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 :