An inductor of 2.00 H is connected to a battery. There is nevertheless resistanc

Question

An inductor of 2.00H is connected to a battery. There is nevertheless resistance in the circuit, in both the inductor and the battery. Assume that its value is 8.10? and it behaves just like a simple resistor in series with a pure inductor and an ideal battery of emf 6.40V .

Part A: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.

Answer: .249J

Part B: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?

Explanation / Answer

time constant=L/R=0.247 sec

in 1 time constant,

current=(6.4/8.1)*(1-exp(-1))=0.5 A

so energy=0.5*L*current^2=0.249 J

part B:

energy supplied=integral of 6.4*current*dt

with t=0 to t=0.247

answer=0.4594 J

part C:

0.4594-0.249=0.2104 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects