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phasor diagrams and reactance

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IamSmooth

Wed Oct 11 2006, 03:55AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

I hope someone can clear up a point for me about currents and phase shifting:

I order to keep things simple, let's assume that inductive reactance L equals the resistance R.

The current at point 2 is phase shifted 45 degrees and the current at point 1 has a phase shift of zero? If we put a capacitor in parallel with R, would one measure the combined phase shift of the current of L, R and C at point 2?

Simon

Wed Oct 11 2006, 04:29AM

Registered Member #32 Joined: Sat Feb 04 2006, 08:58AM
Location: Australia
Posts: 549

I'm assuming your voltages are being measured relative to the bottom rail (as a ground).

Since this is a series circuit the current is constant throughout. The voltage at 1 is the voltage across the source. There is a 45 degree difference between the current and the voltage from the source so that's the phase angle at 1. At 2 the voltage is simply being measured across the resistor and the voltage across a resistor is always in phase with the current.

More complicated circuits are treated like any other complicated circuit - you try to break it down to simpler equivalent circuits. This can be done using geometry and phasor diagrams or you can use complex number notation. Once you get the hang of complex numbers this is the easier way.

As a general rule, the phase angle between the voltage from A to B and the current flowing from A to B is the same as the phase angle in the impedance from A to B. This is the angle in the following (probably bad) diagram:

/|
  / | |Xi - Xc|
 /  |
/|  |
----
  R

IamSmooth

Wed Oct 11 2006, 05:28AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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If there is a 45 phase difference between voltage and current at 1, how would one see this with an oscilloscope?

Here's what I tried to do:
I put a probe at 1 (potential from 1 to gnd) and a probe at 2 (2 to gnd).
At close to the resonant frequency it seemed to me that the voltage wave at 2 was 45 degrees behind the voltage at 1.

Steve Conner

Wed Oct 11 2006, 09:33AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Measuring the voltage across the resistor amounts to the same as measuring the current in the circuit. Measuring the voltage at point 1 gives you the vector sum of the voltage across the resistor and the voltage across the inductor. (or the supply voltage, it's the same thing) So you're never seeing the voltage across the inductor on its own.

The voltage at "2" is phase shifted 45 degrees compared to "1", provided that you choose the circuit constants and drive frequency such that the impedance of L (XL) is equal to the resistance of R. If XL is large compared to R, the voltage at 2 will be phase shifted almost 90 degrees (and very small.) If XL is small compared to R... I'll let iamsmooth figure that one out himself

Note: when I say "the voltage at point 2" etc. I assumed everything to be with respect to the rail at the bottom.

IamSmooth

Wed Oct 11 2006, 11:45AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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The original question assumes that XL and XR are equal so the phase shift will be 45 degrees.

If I remove R from the circuit so it is just an inductor and voltage source, where will voltage and current be in phase and where will the current lag 90deg? Will 1 be lagging 90 and 2 be 0 degrees?

Simon and Steve, you both mention in the ciruit the voltage and current are out of phase at 1. How would one show this will an oscilloscope? What is the reference waveform? What if it was an inductor and capacitor?

Steve Conner

Wed Oct 11 2006, 11:48AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Because you replaced R with a short circuit, the voltage at 2 will be zero.
The current in the circuit will be the same everywhere (series circuit) and it will lag the voltage at 1 by 90 degrees.

IamSmooth

Wed Oct 11 2006, 11:50AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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That was a quick reply Steve.

What can be used as a reference waveform to show this, other than theory? I take it that with just an inductor the current and voltage are out of phase at points 1 and 2?

WaveRider

Wed Oct 11 2006, 12:53PM

Registered Member #29 Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500

Assuming inductive reactance is equal to the resistance:

V across resistor: Vr = V * (1-j) / 2 (time domain: Vr = V * cos(omega * t - 45 deg) / sqrt(2))

V across inductor Vl = V * (1+j) / 2 (time domain: Vl = V * cos(omega * t + 45 deg) / sqrt(2))

Phase shifts are with respect to the phase of the voltage source (V, phase = 0)
Since the voltage across the resistor is propostional to the current (and has the same phase), we can see that the current phase lags the voltage across the inductor by 90 degrees.

Steve Conner

Wed Oct 11 2006, 01:10PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

If you wanted to see two practical waveforms to compare phases, in the absence of a resistor, you would need to make a current sensing probe for your oscilloscope. (Either use a current transformer, or scope the voltage across a resistor whose value is small compared to XL)

IamSmooth

Wed Oct 11 2006, 02:05PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

Again, for all answering this thread, this is not about the mathematics or complex algebra. This is about the ability and means to see the current phase shift on an oscilloscope for various LRC circuits.

What if the function generator is floating and the ground is placed at point 2? Assume R is replaced by C. One should be able to see 180 shift if probe A is put at point 1 and probe B is put on the NEG end of the function generator?

EDIT:
I connected my setup to a scope. When connected at points 2 and 1 with the probes I could show myself 0deg shift at spot 2; with probe 1 I could see the net shift of 45deg. When I used the add/invert feature of my scope I could get the voltage across the inductor as being 90ahead of the resistor voltage. I plan to connect a LC circuit to a synchronous motor and use a stroboscope to see the effect of varying the current phase with the location of a marker on a flywheel. I will vary the phase with a variable inductor.

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