An inductance `L` and a resistance `R` are first connected to a battery. After some time the battery is disconnected but `L` and `R` remain connected in a closed circuit. Then the current reduces to `37%` of its initial value in

An inductance of 2 H and resistance of 10 Omega are connected to a battery of 5 V. the time constant of the circuits is:

A coil resistance R and inductance L is connected to a battery of emf E volt. The final current in the coil is

A coil having an inductance L and a resistance R is connected to a battery of emf E . Find the time taken for the magnetic energy stored in the circuit to change from one fourth of the steady-state value.

The current in an L - R circuit in a time t = 2L / R reduces to-

A Capacitor C and resister R are connected to a battery of 5V in series. Now battery is disconnected and a diode is connected as shown in figure (a) and (b) respectively. Then charge on the capacitor after time RC in (a) and (b) respectively is Q_A and Q_B . Their value are

A coil having an inductance L and a resistance R is connected to a battery of emf epsilon . Find the time elapsed before (a) the current reaches half its maximum value, (b) the power dissipated in heat reaches half its maximum value and © the magnetic field energy stored in the circuit reaches half its maximum value.

An inductor L and a resistance R are connected in series with a battery of emf E and a switch. Initially the switch is open. The switch is closed at an instant t = 0. Select the correct alternatives