An inductor coil stores 32 J of magnetic field energy and dissiopates energy as heat at the rate of 320 W when a current of 4 A is passed through it. Find the time constant of the circuit when this coil is joined across on ideal battery.

Explanation

The magnetic field energy field energy stored in an inductor is ` u =(1)/(2) Li^2`
thus, `32J = (1)/(2) L(4 A)^2` or, L = 4 H
The power dissipated as heat is given by ` P = i^2 R` or, `320 W = (4A)^2 R, giving R = 20 Omega
The time constant of the circuit is `tau = (L)/(R) = (4H)/(20 Omega) = 0
2 s