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

Answer:

The limit on the series resistance is R ≤ 400μΩ

Explanation:

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

$i(t) = Iie^{\frac{R}{L} t}$ (li = initial current)

So, the initial energy stored in the inductor is

$Wi = \frac{1}{2} Li^{2}_{i}$

After 1 hour

$w(3600) = \frac{1}{2} Li_{i}e^{-\frac{R}{L} 3600 }$

Knowing it is equal to 75

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

This way we have,

R = $-10 \frac{ln 0.75}{2 * 3600} = 400$ μΩ

Than, the resistance is R ≤ 400μΩ