A 240V battery is connected In series with a resistor and an

A 24.0-V battery is connected In series with a resistor and an inductor, with R =5.80 Ohm and L = 2.40 H respectively. (a) Find the energy stored in the inductor when the current reaches Its maximum value. ? (b) Find the energy stored in the inductor at an instant that is a time interval of one time constant after the switch is closed.

Solution

1) energy stored= 1/2 L I^2

I=V/R=24/5.8

energy = 1/2 *2.40*24*24/5.8/5.8 =20.546 J

2) after one time constant current would be 63.2 % of its max value.

so energy stored woud be = 1/2*L * (63.2/100)^2 * I^2 = 8.206 J