Question

A constant voltage of 3.00 V has been observed over a certain time interval across a 3.00 H inductor. The current through the inductor, measured as 3.00 A at the beginning of the time interval, was observed to increase at a constant rate to a value of 8.00 A at the end of the time interval. How long was this time interval?

Answer 1

Answer:

Explanation:

If L be inductance and i be current , flux associated with it

Φ = L i

emf induced = dΦ / dt = L di / dt

Putting the given values

emf induced = 3 V , L = 3 H

di = 8 - 3 = 5 A

dt = ?

3 = 3 x 5 / dt

dt or time duration = 3 x 5 / 3

= 5 sec .