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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. |
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Answer» Magnetic field energy ` = 32 = (1)/(2) LI^(2) = (1)/(2) L (4)^(2)` `:. L = 4 H` Power dissipated as heat, `P = I^(2) R` `320 = 4^(2) R R = 20 Omega` Time constant `tau = (L)/(R ) = (4)/(20) = 0.2 s` |
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