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Q.

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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a

1.125

b

0.8

c

0.2

d

0.4

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 =  =12LI2;64=12×L×8×8
L=2H
Also P = I2R

8×8×R=640;R=10Ω
Time constant is given by ,  τ=LR=210=0.2s
 

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