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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. `0.1sec`B. `0.2sec`C. `0.3sec`D. `0.4sec` |
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Answer» Correct Answer - B `(1)/(2)Li^(2)=32 implies (1)/(2)Lxx(4)^(2)=32` `L=4H` `i^(2)R=320 implies (4)^(2)R=320` `R=20QOmega` `tau=(L)/(R)=(4)/(20)=0.2sec` |
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