An inductor of 250 H is connected to a battery There is neve

An inductor of 2.50 H is connected to a battery. There is nevertheless resistance in the circuit, in both the inductor and the battery. Assume that its value is 7.90 12 and it behaves just like a simple resistor in series with a pure inductor and an ideal battery of emf 6.00 V. Consider the instant one time constant after the inductor is connected to the battery. Calculate the energy that resides in the inductor\'s magnetic field at this instant. How much energy was supplied by the (ideal) battery from the time it was first connected to the inductor to the instant one time constant later? How much energy was dissipated in the circuit\'s resistance during the same time interval?

Solution

Maximum current

I_max=V/R =6/7.9

I_max=0.76 A

b)

Current flowing when battery was first connected to the inductor to the instant one time constant is

I=I_oe^-t/T

Given t=T

I=0.76e^-1 =0.28 A

Energy supplied by the battery is

U=(1/2)LI² =(1/2)*2.5*0.28²

U=0.098 J

c)

Energy dissipated in resistor is

U_R=I²Rt =I²R*(L/R)0.28²*7.9

U_R=I²L =0.28²*2.5 =0.196 Watts