Question

A constant voltage of 6.00 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

The time interval is 2.5 sec

Step-by-step explanation:

Given:

applied voltage
`V = 6` V

Inductance
`L = 3` H

Current change
`dI = 8 -3= 5` A

From the formula of magnitude of induced emf in terms of inductance,

`V = L(dI)/(dt)`

For finding the time interval,

`dt = (L dI)/(V)`

`dt = (3 * 5)/(6)`

`dt = 2.5` sec

Therefore, the time interval is 2.5 sec