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A coil having resistance 15Omegaand inductance 10H is connected across a 90 Volt de supply. Determine the value of current after 2 sec. What is the energy stored in the magnetic field at that instant |
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Answer» Solution :Given that `R = 15 Omega ,L = 10 H, 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 (1 - e^((-Rt)/(L)) )` , after 2 sec `I = 6 [1-e^(-2//0.67) ] = 6[1-0.05] = 5.7A` Energy stored in the magnetic field `U= 1/2 LI^2 = 1/2 XX 10 xx (5.7)^2 J = 162.45J` |
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