A 15.0 mH inductor produces a peak voltage of 10.6 V. s connected across an AC...

Question

Question

A 15.0 mH inductor produces a peak voltage of 10.6 V. s connected across an AC generator that Part A What is the peak current

Answer 1

Answer #1

A)
Z = 2*3.14*(frequency)*(inductance)
= 2*3.14*100*(15.0*10^-3)
= 9.42 ohm

Imax = Vmax/Z
= 10.6/9.42
= 1.125 A
= 1.12 A

Answer: 1.12 A

B)
Z = 2*3.14*(frequency)*(inductance)
= 2*3.14*(100*10^3)*(15.0*10^-3)
= 9420 ohm

Imax = Vmax/Z
= 10.6/9420
= (1.125*10^-3) A
= (1.12*10^-3) A
= 1.12 mA

Answer 2

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