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

In summary, the conversation involves finding the AC current in a circuit with a 0.197 H inductor and an applied AC voltage of peak value 166 V and frequency 60.5 Hz. The formula for inductive reactance is mentioned and used to calculate peak AC current, but the answer is incorrect due to a phase shift between voltage and current in an inductor. The correct effective current can be found by considering the phase shift.
  • #1
kiwikahuna
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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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  • #2
You have obtained peak AC current, because you used peak voltage in Ohm's law. Probably you're asked to find effective current.
 
  • #3
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?
 

1. How do you calculate the AC current for a 0.197 H inductor with a peak voltage of 166 V?

To calculate the AC current in an inductor, you can use the formula I = Vpeak / XL, where I is the current, Vpeak is the peak voltage, and XL is the inductive reactance. In this case, XL can be calculated as XL = 2πfL, where f is the frequency and L is the inductance. Plugging in the given values, the equation becomes I = 166 V / (2πf * 0.197 H).

2. What is the inductive reactance of a 0.197 H inductor at a frequency of 60 Hz?

To calculate the inductive reactance, you can use the formula XL = 2πfL, where f is the frequency and L is the inductance. Plugging in the given values, the equation becomes XL = 2π * 60 Hz * 0.197 H = 23.45 Ω.

3. How does the current in a 0.197 H inductor change with increasing frequency?

The current in an inductor is directly proportional to the frequency of the alternating current passing through it. This means that as the frequency increases, the current in the inductor will also increase.

4. Can the current in a 0.197 H inductor ever reach zero?

No, the current in an inductor cannot reach zero as long as there is an alternating current passing through it. This is because the inductor will always have some amount of stored energy that is continuously being exchanged with the alternating current, resulting in a non-zero current.

5. How does the inductance of a 0.197 H inductor affect the current in a circuit?

The inductance of an inductor affects the current in a circuit by creating an opposition to any changes in current. This means that the higher the inductance, the greater the opposition to changes in current, resulting in a lower current in the circuit. Additionally, a higher inductance also means a higher inductive reactance, which can further limit the current in a circuit.

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