An inductor with L = 40.0 mH has a current I = at 2 + bt + c flowing through it.

Question

An inductor with

L = 40.0 mH

has a current

I = at2 + bt + c

flowing through it. Here

a = 3.00 A/s2,

b = 5.00 A/s,

and

c = 0.270 A.

(a) Find an expression for the induced emf in terms of the given variables. (Use the following as necessary: t. Do not enter any units. Assume SI units.)

An inductor with L 40.0 mH has a current 1 at bt c flowing through it. Here a 3.00 A/s2, b 5.00 A/s, and CH 0.27o A. induced emf in terms of the given variables. Use the following as necessary: t (a) Find an expression for the Do not enter any units Assume SI units g 0.04 (6t 5) x (b) What is the magnitude of the induced emf at the end of 8.00 s? 6.92 (c) Determine the time (after t 0) at which the magnitude of the emf will be three times its value at t 0. 1.6 X s

Explanation / Answer

a)

I = a* t^2 + b * t + c

dI/dt = 2at + b

induced emf = L * dI/dt

induced emf = L * (2*a * t +b)

b)

at t = 8 s

induced emf = 0.040 * (2 * 3 * 8 + 5)

induced emf = 2.12 V

c)

let the time is t

L * (2 * a * 0 + b) * 3 = L * (2 * a * t + b)

5 * 3 = 2 * t * 3 + 5

solving for t

t = 1.67 s

the time is 1.67 s

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