Alternating Current

JEE Main level, single correct, medium

104 of 190

Question 104 of 190

An inductor coil stores 64J64J of magnetic field energy and dissipates energy at the rate of 640W640W when a current of 8A8A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit (in ss).

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Answer: (B) 0.20.2


Energy stored in an inductor,
U=12LI2U=\dfrac{1}{2}L{I}^{2}
⇒64=12L(8)2⇒L=2H\Rightarrow 64=\dfrac{1}{2}L(8{)}^{2}\Rightarrow L=2H
and power dissipated in a resistor,
P=I2RP={I}^{2}R
⇒640=(8)2R⇒R=10Ω\Rightarrow 640=(8{)}^{2}R\Rightarrow R=10\Omega
Time constant of the
LRDCLRDC circuit is,
τ=LR=210=0.2s\tau =\dfrac{L}{R}=\dfrac{2}{10}=0.2s

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