In a decaying `L-R` circuit, the time after which energy stores in the inductor reduces to one fourth of its initial values is

In a L-R decay circuit, the initial current at t=0 is I. The total charge that has flown through the resistor till the energy in the inductor has reduced to one-fourth its initial value, is

Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t_1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t_2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t_1//t_2 will be

LetC be the capacitance of a capacitor discharging through a resistor R. Suppose t_1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t_2 is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t_1//t_2 will be

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 a chrging cirucit, after how many time constant will the energy stored in the capacitor reach one - half of its equilibrium value?

Energy Stored in an Indutor || Inductor in a Circuit

A simple LR circuit is connected to a battery at time t = 0 . The energy stored in the inductor reaches half its maximum value at time