A) The reactance of the inductor is given by:
X = 2πfL
X = reactance, f = source frequency, L = inductance
Given values:
f = 75.0Hz, L = 28.5×10⁻³H
Plug in and solve for X:
X = 2π(75.0)(28.5×10⁻³)
X = 13.4Ω
B) V = IZ
V = rms source voltage, I = rms current, Z = impedance
Given values:
V = 86.0V
With just the inductor in the circuit, Z = X = 13.4Ω
Plug in and solve for I:
86.0 = I(13.4)
I = 6.42A
C) The rms current is given by:
[tex]I_{rms} = I_{max}/\sqrt{2}[/tex]
We have [tex]I_{rms}[/tex] = 6.42A so plug that in and solve for [tex]I_{max}[/tex]:
[tex]6.42 = I_{max}/\sqrt{2}[/tex]
[tex]I_{max}[/tex] = 9.08A
