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

Answer:

The time is 0.5 sec.

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

$V=L\dfrac{\Delta A}{\Delta t}$

$\Delta t =\dfrac{L\Delta A}{V}$

Where, $\Delta A$ = change in current

V = voltage

L = inductance

Put the value into the formula

$\Delta t=\dfrac{1.20\times5.00}{12.00}$

$\Delta t=0.5\ sec$

Hence, The time is 0.5 sec.