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An inductor coil stores 32 J of magnetic field energy and dissiopates energy as heat at the rate of 320 W when a current of 4 A is passed through it. Find the time constant of the circuit when this coil is joined across on ideal battery.A. `t =0.2 s`B. `t =0.32 s`C. `t =0.5 s`D. `t =1 s` |
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Answer» Correct Answer - A Magnetic field energy stored `= 1/2 Li^(2)` `32 = 1/2 L(4)^(2) or L=4H` Power dissipated as heat is `P=i^(2)R` `320 =4^(2)R or R=20 Omega` Time constant of circuit =`tau=L/R 4/20 = 0.2 s`. |
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