When a current of 4a flows in a coil when it is connected to a 12v dc source.when the same coil is connected to 12v,8h a current of 2.4 a flows in the coil. calculate self inductance
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2.78
Energy stored in the inductor is given by E = 1/2(LI^2)
energy stored is given as 8 Henry. so self inductance
L= 8*2/(2.4)*2 = 16/5.76 = 2.78
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