Calculating AC Current for a 0.197 H Inductor with 166 V Peak Voltage

AI Thread Summary
To calculate the AC current through a 0.197 H inductor with a peak voltage of 166 V and a frequency of 60.5 Hz, the inductive reactance (XL) is determined using the formula XL = ωL, where ω is the angular frequency. The current is initially calculated as I = V/XL, yielding a peak current of 2.22 A. However, this value represents the peak current, not the effective (RMS) current, which is typically required. The discussion highlights the importance of considering the phase shift between voltage and current in an inductor, indicating that the effective current must be calculated differently. Understanding these concepts is crucial for accurate AC circuit analysis.
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Homework Statement


A circuit contains only 0.197 H inductor. An AC voltage of peak value 166 V and frequency 60.5 Hz is applied to the circuit. When the voltage is 166 V, what is AC current? Answer in units of A.


Homework Equations





The Attempt at a Solution



we know that inductive reactance is XL= ωL
AC voltage peak value IS V = 166 Volts
then AC current is I = V/XL
ω = 2π(FREQUENCY) = 2π(60.5) RAD/SEC
L = INDUCTANCE = 0.197 H
THEN I = 166/ 2π(60.5)*0.197 = 2.22 A

The answer is wrong, could someone tell me what I'm doing wrong? Thank you!
 
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You have obtained peak AC current, because you used peak voltage in Ohm's law. Probably you're asked to find effective current.
 
Remember that V = L dI/dt, so there is a phase shift between voltage and current in an inductor. Do you know how many degrees the current waveform is off of the voltage excitation waveform, and in which direction?
 
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