1) La corriente de un inductor de 90.0 mH cambia con el transcurso del tiempo de la forma I = 1.00 t 2 – 6.00 t (en unidades del SI). Determine la magnitud de la fem inducida en a) t = 1.00 s y b) t = 4.00 s. C) ¿En qué momento será la fem igual a cero?

a)   E_{L} = -360 V , b)      t = 3 s

Explanation:

The electromotive force in an inductor is

            = - L dI/ dt

in the exercise they give us the relation of i (t)

            i (t) = 1.00 t² -6.00t

we carry out the derivative and substitute

          E_{L} = - L (2.00 2t - 6.00 1)

a) the electromotive force at t = 1.00 s

          E_{L} = - 90.0 (4.00 1 - 6.00)

         E_{L} = -360 V

b) for t = 4 s

         E_{L}= - 90 (2 4 2 - 6 4)

         E_{L} = - 720 V

c) for the induced electromotive force to zero, the amount between paracentesis must be zero

           (2.00 t2 - 6.00t) = 0

            t (2.0 t-6.00) = 0

the solutions of this equation are

            t = 0

            2 t -6 = 0

            t = 3 s

to have a different solution the trivial (all zero) we must total t = 3 s

anchor = archer //

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.