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An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8 A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit (in seconds) is

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0.4

0.2

1.125

0.8

answer is B.

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Detailed Solution

Energy stored in inductor , U= 64J
Current , I= 8A
Power , P = 640 W
Let the inductance is L and resistance is R.
So, U = $= \frac{1}{2} {LI}^{2}; 64 = \frac{1}{2} \times L \times 8 \times 8$
L=2H
Also P = I²R

$8 \times 8 \times R = 640; R = 10 Ω$
Time constant is given by , $τ = \frac{L}{R} = \frac{2}{10} = 0 .2 s$

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