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A coil having resistance 15 and inductance 10 H is connected across a 90 Volt 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» Solution :Given 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)(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) xx10xx(5.7)^(2)J = 162.45` J. |
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