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

Answer:

E= -3.166 cosωt   V

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

$E=-L\dfrac{dI}{dt}$

$\dfrac{dI}{dt}= Imax \ \omega\ cos\omega t$

$E=-L\times Imax \ \omega\ cos\omega t$

$E=-8.4\times 120\times \pi \ cos\omega t$

E= -3166.72 cosωt  m V

E= -3.166 cosωt   V

This is the induce emf.