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An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made from the power source and a 47.2 H inductor, determine the inductive reactance and the rms current through the inductor.

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Answer:

The inductance is 17784.96 ohm and rms current is 4.77 mA.

Explanation:

Voltage, V = 120 V

frequency, f = 60 Hz

Inductance, L = 47.2 H

The rms  voltage is

[tex]V_{rms}=\frac{V_o}{\sqrt 2}\\\\V_{rms}=\frac{120}{\sqrt 2}\\\\V_{rms} = 84.87 V[/tex]

The reactance is given by

[tex]X_L = 2\pi f L\\\\X_L = 2\times 3.14\times 60\times 47.2 \\\\X_L = 17784.96 ohm[/tex]

The rms current is

[tex]I_{rms} =\frac{V_{rms}}{X_L}\\\\I_{rms}=\frac{84.87}{17784.96}\\\\I_{rms} = 4.77\times 10^{-3} A = 4.77 mA[/tex]

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