AC waveform and reactance

  • Thread starter fedaykin
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  • #1
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Main Question or Discussion Point

Say we have an inductor connected to an AC voltage source. Current will then lag voltage by 90 degrees.

Does the current maintain it's previous value (as if the inductor was not there) but 90 degrees out of phase, or is it still dependent only on voltage?

Rephrasing (in case that was poorly worded):

Current is a function of voltage and resistance. Does reactance change this relationship?

If the relationship doesn't change, how can current and voltage be out of phase with each other?
 

Answers and Replies

  • #2
berkeman
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Say we have an inductor connected to an AC voltage source. Current will then lag voltage by 90 degrees.

Does the current maintain it's previous value (as if the inductor was not there) but 90 degrees out of phase, or is it still dependent only on voltage?

Rephrasing (in case that was poorly worded):

Current is a function of voltage and resistance. Does reactance change this relationship?

If the relationship doesn't change, how can current and voltage be out of phase with each other?
The fundamental equation relating current and voltage in an inductor is:

[tex]v(t) = L \frac{di(t)}{dt}[/tex]

All of the behavior of an inductor in a circuit is defined by that equation. The voltage developed across an ideal inductor (with no internal series resistance) depends only on the change in the current with respect to time. So, you can define the complex impedance of the inductor to be what you would get if you wanted to be able to think of the impedance as a "resistance" in the equation V = IR. What would that complex impedance be?
 
  • #3
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Thank you very much.

Hehe, I got this straight out of my books. It's been a while since I've done much reading on electricity.

Z = R + 2j(pi)fL
 
  • #4
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I've used some graphs to help me understand what's happening.

Since voltage is the derivative of current with respect to time, the local maxima for voltage are the minima for current; when current is changing most is when voltage is at its peak and is changing the least.

'Competition' from change in current reversing affects the voltage from the source ... ? This may be entirely incorrect.
Or is there a better way of putting it?
 
  • #5
berkeman
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56,664
6,560
I've used some graphs to help me understand what's happening.

Since voltage is the derivative of current with respect to time, the local maxima for voltage are the minima for current; when current is changing most is when voltage is at its peak and is changing the least.

'Competition' from change in current reversing affects the voltage from the source ... ? This may be entirely incorrect.
Or is there a better way of putting it?
You've got it pretty much correct. The only other thing to add into your mental picture is to not ignore the source impedance of the AC voltage source. The voltage source is not infinitely stiff -- remember to model the voltage source as an ideal AC voltage source, in series with the output impedance Zout. Zout is generally around 50 Ohms real for most lab signal generators, but can be different values (including complex) for other AC voltage sources.
 

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