Question

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.
A. `t =0.2 s`
B. `t =0.32 s`
C. `t =0.5 s`
D. `t =1 s`

Answer 1

Correct Answer - A
Magnetic field energy stored `= 1/2 Li^(2)`
`32 = 1/2 L(4)^(2) or L=4H`
Power dissipated as heat is `P=i^(2)R`
`320 =4^(2)R or R=20 Omega`
Time constant of circuit =`tau=L/R 4/20 = 0.2 s`.