Question

A 8.4-mH inductor carries a current I = Imaxsin ωt, with Imax = 4.00 A and f = ω/2π = 60.0 Hz. What is the self-induced emf as a function of time? (Express your answer in terms of t where e m f is in volts and t is in seconds.)

Answer 1

E= -3.166 cosωt V

Step-by-step explanation:

Given that

I = Imax sinωt

L= 8.4 m H

Imax= 4 A

f = ω/2π = 60.0 Hz

ω = 120π rad/s

We know that self induce E given as

`E=-L(dI)/(dt)`

`(dI)/(dt)= Imax \ \omega\ cos\omega t`

`E=-L* Imax \ \omega\ cos\omega t`

`E=-8.4* 120* \pi \ cos\omega t`

E= -3166.72 cosωt m V

E= -3.166 cosωt V

This is the induce emf.