JEE MAIN - Physics (2021 - 26th August Morning Shift - No. 7)
An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds :
0.4
0.8
0.125
0.2
Explanation
$$U = {1 \over 2}L{i^2} = 64 \Rightarrow L = 2$$
$${i^2}R = 640$$
$$R = {{640} \over {{{(8)}^2}}} = 10$$
$$\tau = {L \over R} = {1 \over 5} = 0.2$$
$${i^2}R = 640$$
$$R = {{640} \over {{{(8)}^2}}} = 10$$
$$\tau = {L \over R} = {1 \over 5} = 0.2$$
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