Question

An induction coil stores $32\, J$ of magnetic energy and dissipates energy as heat at the rate of $320\, W$. A current of $4\, A$ is passed through it. Find the time constant (in second) of the circuit when the coil is joined across a battery.

Answer 1

$\frac{1}{2} L I^{2}=32$
$\because I=4 A$
$\therefore \frac{1}{2} \times L \times(4)^{2}=32$
$\Rightarrow L=4\, H$
Now, $P=I^{2} R$
$\therefore 320=(4)^{2} \times R$
$\Rightarrow R=\frac{320}{16}=20\, \Omega$
$\therefore $ Time constant $=\frac{L}{R}=\frac{4}{20}=\frac{1}{5}$
$=0.2\, s$