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A coil having resistance `15 Omega` and inductance `10H` is connected across a `90` Volt `dc` supply. Determine the value of current after `2sec`, What is the energy stored in the magnetic field at that instant. |
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Answer» Give that , `R = 15 Omega, L = 10H, E = 90 "Volt"` Peak value of current `I_(0) = (E)/(R ) = (90)/(15)A = 6A` also, `tau_(L) = (L)/(R ) = (10)/(15) = 0.67 sec` Now, `I = I_(0)(I-e^((-Rt)/(L)))`, After `2 sec`, `I = 6[1-e^(-2//0.67)] = 6[1-0.05] = 5.7A` Energy stroed in the magnetic field `U = (1)/(2)LI^(2) = (1)/(2) xx 10 xx (5.7)^(2)J = 162.45 J`. |
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