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

Magnetic field energy ` = 32 = (1)/(2) LI^(2) = (1)/(2) L (4)^(2)`
`:. L = 4 H`
Power dissipated as heat, `P = I^(2) R`
`320 = 4^(2) R R = 20 Omega`
Time constant `tau = (L)/(R ) = (4)/(20) = 0.2 s`