Questions
An inductor coil stores 32 J of magnetic field energy and dissipates energy as heat at the rate of 320 W when a current of 4 amp is passed through it. Find the time constant of the circuit when it is formed across an ideal battery.
detailed solution
Correct option is A
Magnetic field energy stored=U=12Li232=12L(4)2 or L=4 Henry Power dissipated as heat is P=i2R320=42R or R=20ohm Time constant of circuit =τ=LR=420=0.2secTalk to our academic expert!
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Two different coils have self-inductance mH, . The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same rate. At a certain instant of time, the power given to the two coils is the same. At that time the current, the induced voltage and the energy stored in the first coil are and respectively. Corresponding values for the second coil at the same instant are and respectively. Then
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