Question

P4.36. Real inductors have series resistance associated with the wire used to wind the coil. Suppose that we want to store energy in a 10-H inductor. Determine the limit on the series resistance so the energy remaining after one hour is at least 75 percent of the initial energy.

Answer 1

The limit on the series resistance is R ≤ 400μΩ

Step-by-step explanation:

Considering the circuit has a series of inductance and resistance. The current current in the current in the circuit in time is

`i(t) = Iie^{(R)/(L) t}` (li = initial current)

So, the initial energy stored in the inductor is

`Wi = (1)/(2) Li^(2)_(i)`

After 1 hour

`w(3600) = (1)/(2) Li_(i)e^{-(R)/(L) 3600 }`

Knowing it is equal to 75

`w(3600) = 0.75Wi = 0.75 (1)/(2) Li^(2)_(i) = (1)/(2) Li_(i)e^{-(R)/(L) 3600 }\\`

This way we have,

R =
`-10 (ln 0.75)/(2 * 3600) = 400` μΩ

Than, the resistance is R ≤ 400μΩ