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3 years ago

11

A 13.4-mH inductor carries a current i =

{max}" align="absmiddle" class="latex-formula"> sin ωt, with $I_{max}$ = 4.80 A and f = ω/2π = 60.0Hz. 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. Do not include units in your expression.)

1 answer:

Digiron [165]3 years ago

6 0

The voltage across an inductor ' L ' is

V = L · dI/dt .

I(t) = I(max) sin(ωt)

dI/dt = I(max) ω cos(ωt)

V = L · ω · I(max) cos(ωt)

L = 1.34 x 10⁻² H

ω = 2π · 60 = 377 /sec

I(max) = 4.80 A

V = L · ω · I(max) cos(ωt)

V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)

<em>V = 24.25 cos(377 t)</em>

V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

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