Question

The resistanceless inductor is connected across the ac source whose voltage amplitude is 29.0Vand angular frequency is 1000rad/s . Find the current amplitude if the self-inductance of the inductor is

A) 1.00E^-2 H? (A)

B) 1.00 H? (mA)

C) 100 H? (mA)

Answer 1

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The resistanceless inductor is connected across the ac source whose voltage amplitude is 25.0 and angular frequency is 1000 . Find the current amplitude if the self-inductance of the inductor is
a. 1.00*10^-2 H
b. 1.00 H
c. 1.00 H
d. 100 H

Given

maximum   voltage   V = 25 V

angular frequency ? =   1000

  maximum current at L = 1*10^-2 H

     a)   I = V / X_L

              =   25 / 1000 * 0.01

               = 2.5 A

    b)   I = V / 1000*1

               = 0.025 A

    c) I = V / 1000*10

             = 25 / 10000

             = 0.0025A

    d) I = 25 / 1000*100

              = 0.00025 A