A 10 V battery is in series with a 50 Ohm resistor and a 10 Henry...
A 10 V battery is in series with a 50 Ohm resistor and a 10 Henry inductor and an open switch. What is the energy contained in the inductor 0.4 s after the switch is closed?
Homework Answers
I = Vo/R*(1-e-Rt/L )
I = 10/50*(1-e-50*0.4/10 )
I= 0.1729329 A
energy in Inductor = 1/2*L*I2 = 0.5*10*0.1729329^2 = 0.14952894 J answer
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