Question

A constant voltage of 12.00 V has been observed over a certain time interval across a 1.20 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 is 0.5 sec.

Step-by-step explanation:

Given that,

Voltage V= 12.00 V

Inductance L= 1.20 H

Current = 3.00 A

Increases rate = 8.00 A

We need to calculate change in current

`\Delta A = 8.00-3.00= 5.00\ A`

We need to calculate the time interval

Using formula of inductor

`V=L(\Delta A)/(\Delta t)`

`\Delta t =(L\Delta A)/(V)`

Where,
`\Delta A` = change in current

V = voltage

L = inductance

Put the value into the formula

`\Delta t=(1.20*5.00)/(12.00)`

`\Delta t=0.5\ sec`

Hence, The time is 0.5 sec.