Question

An inductor has a 50.0-Ω reactance when connected to a 60.0-Hz source. The inductor is removed and then connected to a 45.0-Hz source that produces a 85.0-V rms voltage. What is the maximum current in the inductor?

Answer 1

Given:

`X_(L) = 50.0 \ohm`

frequency, f = 60.0 Hz

frequency, f' = 45.0 Hz

`V_rms} = 85.0 V`

Solution:

To calculate max current in inductor,
`I_{L(max)`:

At f = 60.0 Hz

`X_(L) = 2\pi fL`

`50.0 = 2\pi* 60.0* L`

L = 0.1326 H

Now, reactance
`X_(L)` at f' = 45.0 Hz:

`X'_(L) = 2\pi f'L`

`X'_(L) = 2\pi* 45.0* 0.13263 = 37.5\ohm`

Now,
`I_{L(max)` is given by:

`I_{L(max) = \sqrt {(2V_(rms))/(X'_(L))}`

`I_{L(max) = \sqrt {(2* 85.0)/(37.5)} = 2.13 A`

Therefore, max current in the inductor,
`I_{L(max)` = 2.13 A