Homework Solution Understanding Phasor Diagrams for Circuit Analysis

In summary, understanding phasor diagrams is crucial for circuit analysis in homework solutions. These diagrams help visualize the behavior and relationships of voltage, current, and impedance in AC circuits. By representing these quantities as vectors, phasor diagrams allow for easy calculation of circuit parameters such as impedance and power. Additionally, they aid in analyzing and solving complex circuits with multiple sources and components. Overall, a thorough understanding of phasor diagrams is essential for successful circuit analysis and completion of homework assignments.
  • #1
rock.freak667
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Phasor Diagrams??

Homework Statement



Say I am given this circuit diagram. I have the recorded values on the voltmeter and ammeter. I can verify the power factor rating using these two values.
But I also need to draw a phasor diagram for this circuit as well as two others which just have other devices added on.

The problem is that I have no idea what is a phasor diagram, my notes don't have anything about it. From what I read online, all usually have something to do with the equation [itex]Asin(\omega t + \phi)[/itex]. I just don't know how to get that equation given what I know.
 
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  • #2


There is an example of a phasor diagram here:
http://en.wikipedia.org/wiki/Phasor_diagram#Phasor_diagrams
(scroll up half a screenfull).

I haven't done this for many years so can't help much, but I expect you will have two arrows on your diagram, one for the current and one for the potential, both beginning at the origin. If you know the phase angles for them, you are all set.
 
  • #3


Delphi51 said:
There is an example of a phasor diagram here:
http://en.wikipedia.org/wiki/Phasor_diagram#Phasor_diagrams
(scroll up half a screenfull).

I haven't done this for many years so can't help much, but I expect you will have two arrows on your diagram, one for the current and one for the potential, both beginning at the origin. If you know the phase angles for them, you are all set.

But I don't know the phase angles,nor the frequency...:confused:, I was not given such information and I don't think I calculate such things from the diagram.
 
  • #5


Delphi51 said:
The phase angle between current and potential is directly related to the power factor.
You'll find a formula relating them at http://en.wikipedia.org/wiki/Power_factor

Thanks..I now realized that one (First time seeing these electrical engineering things)
 
  • #6


None of it meant a thing to me until I worked nights at a control center - by myself - and saw very large currents on the lightly loaded rural lines due to a large phase difference.
Good luck in EE!
 
  • #7


I'm having a bit of trouble understand what is going on in the notes for the phasor diagrams. The lecture didn't happen yet and I might need to submit this before the lecturer teaches it.
But according to what I understand so far:

pf= P/Q = cos([itex]\phi[/itex])

Where P= the real power and Q=Apparent power.
From the results I got, P=0.98*111=108.78W and Q=22*5=110W (from the pf meter).

so the pf=108.78/110=0.98890909090909090909090909090909
so that the phase angle is arccos(0.9889)=8.541 degrees.

I suspect in the diagram the axes are just Imaginary and Real axes

and from this http://people.sinclair.edu/nickreeder/EET155/mod06.htm", it says this
Phasor Form of Ohm's Law (Floyd, p. 614)

Once you know a component's resistance or reactance, you can use Ohm's law to find the component's current if you knows its voltage, or to find the component's voltage if you knows its current.
To use Ohm's law in AC circuits, we'll treat voltages, currents, resistances, and reactances as phasors.
For resistors, V = I × R.
For capacitors, V = I × XC.
For inductors, V = I × XL.
Recall that impedance is the general term applied to resistance, capacitive reactance, inductive reactance, or any combination of these. The symbol for impedance is Z, so we can write the most general form of Ohm's law as:
V = I × Z

Of course, you can also rearrange this equation (and the ones given above) to solve for current if you know voltage and impedance, or to solve for impedance if you know current and voltage:
I = V ÷ Z

Z = V ÷ I

Remember, in each case, all quantities are phasors, not ordinary numbers.

Unfortunately, I read that...but didn't understand what was happening in it.
 
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  • #8


Phasors are complex numbers.
V = I × Z
where all three quantities are complex numbers, Z being R + jX
where X is the reactance.

I think you have the correct phase difference angle. The question remaining is, at what angle will you draw V and I knowing only the angle difference between them?
It seems to me it would be equally correct to take either the current or potential to have an angle of zero. There is a nagging doubt in my mind as to which way the angle goes from there - clockwise or not? Perhaps you ought to use V = IZ on your circuit and see how it works out in complex number land. Take I to be purely real and see how V comes out.
 
  • #9


Delphi51 said:
Phasors are complex numbers.
V = I × Z
where all three quantities are complex numbers, Z being R + jX
where X is the reactance.
So if I should find V/I to get Z. But how would I find X if I do not have an inductor or capacitor connected to the circuit?

Delphi51 said:
It seems to me it would be equally correct to take either the current or potential to have an angle of zero. There is a nagging doubt in my mind as to which way the angle goes from there - clockwise or not?

How would it equally correct for either one to be entirely real?

Delphi51 said:
Perhaps you ought to use V = IZ on your circuit and see how it works out in complex number land. Take I to be purely real and see how V comes out.

I don't understand how I would go about doing that. I have a value for I and a value for V. Then won't Z work out to be entirely real as well?
 
  • #10


So if I should find V/I to get Z.
Oh, I misunderstood. I thought you had a circuit with known components. You must have a capacitor and/or inductor or you would have a power factor of zero. I was thinking z was known.

What measured values do you have? You mention that the apparent power is different from the real power so you must have more than just I and V measurements - or do you have them as complex numbers?
 
  • #11


Delphi51 said:
Oh, I misunderstood. I thought you had a circuit with known components. You must have a capacitor and/or inductor or you would have a power factor of zero. I was thinking z was known.

What measured values do you have? You mention that the apparent power is different from the real power so you must have more than just I and V measurements - or do you have them as complex numbers?

Well these are the circuits I need to draw the phasor diagrams for:
http://img237.imageshack.us/my.php?image=75445543ts3.jpg

http://img4.imageshack.us/my.php?image=88188298qm6.jpg
(For this one I have readings when the switch was opened and closed)

http://img9.imageshack.us/my.php?image=69108894ib1.jpg

For this circuits, all I have are the readings on the ammeter(s), voltmeter, resistor, the pf meter (which you have to multiply a reading by a factor to get the apparent power), the inductance and resistance of the inductor, and the values of the capacitors.


EDIT: I was supposed to also mention that I am not too clear on what was real/apparent power. I assumed it was that the real power is the product of the readingso on the ammeter and voltmeter while the apparent power was the reading on the pf meter (multiplied by the factor written on the meter).
 
  • #12


I can't seem to view those pictures - just get "error 404 page not found".

Having the values of the capacitors, etc. you ought to be able to find the impedence of the circuit as a phasor complex number and use V = IZ.

Alternatively, the power factor meter seems to be giving you the phase angle difference directly.

ammeter reading times voltmeter reading gives the apparent power in Volt-amps, not the power in watts.

Sorry I am not much help here! Hopefully someone familiar with this kind of question will respond.
 
  • #13


Do you happen to know how I would find the Real Power? I am a bit confused since I have 3 ammeters.
 
  • #14


rock.freak667 said:
EDIT: I was supposed to also mention that I am not too clear on what was real/apparent power. I assumed it was that the real power is the product of the readings on the ammeter and voltmeter while the apparent power was the reading on the pf meter (multiplied by the factor written on the meter).

Mixing up the definitions of real and apparent power might be your problem, hopefully this will clear it up a bit for you.

Real power is the product of the Vrms value, the Irms value, and the power factor.
Real Power (P) = (Vrms)(Irms)(pf)


Apparent power ( S , |S| ) is the magnitude of the Complex Power (S = P + jQ) or ( S∠cos-1(pf) )
|S| = S = (Vrms)(Irms) = [tex]\sqrt{(P^2 + Q^2)}[/tex]

**The values in BOLD type are phasors. **
 
  • #15


Then my real power P is really VI/2? Do voltmeters and ammeters measure the rms value or the max value of voltage and current?

so if pf=real power/apparent power

for S=P+jQ, I read that [itex]Q=I_{rms}V_{rms}sin\phi[/itex] (phi is the phase difference between V and I)

So the reading on pf meter gives what, if not the apparent power?
 
  • #16


rock.freak667 said:
Then my real power P is really VI/2?
If you are using peak values instead of rms values, then real power (P) would be:
P = [(VpkIpk)/2](pf)
What you have listed, VI/2, is the average power.

Do voltmeters and ammeters measure the rms value or the max value of voltage and current?
Sorry, I'm not sure if voltmeters/ammeters give peak value or rms value. You may be able to get either value, depending on if there are options on the particular meters. I would guess that it will give rms values. I assume this only because I know you probably used those same meters in DC analysis, and rms values are referred to as 'DC equivalents'. So the DC reading settings would give you rms values.

So the reading on pf meter gives what, if not the apparent power?
I'm not exactly sure what you're asking, but here are a few relationships:
pf = (P / S) = (P / |S|) = cos(θv - θi) = cos(θz)
The power factor is the ratio between the real power and the apparent power. The cos-1(pf) value is the phase angle difference between the voltage and current. I would think that the pf in your circuit would be lagging, that is, the current is lagging the voltage. Your circuit is inductive.
 
  • #17


mplayer said:
Sorry, I'm not sure if voltmeters/ammeters give peak value or rms value. You may be able to get either value, depending on if there are options on the particular meters. I would guess that it will give rms values. I assume this only because I know you probably used those same meters in DC analysis, and rms values are referred to as 'DC equivalents'. So the DC reading settings would give you rms values.

I had to use an analogue voltmeter/ammeter, so I don't know if it measured rms or peak value

mplayer;2087864 I'm not exactly sure what you're asking.[/QUOTE said:
Well I assumed that the pf meter in the circuit measured the apparent power, since it gave me a reading of 22x5=110W. According to a formula on the lab handout, the real power was given by VI
 
  • #18


rock.freak667 said:
I had to use an analogue voltmeter/ammeter, so I don't know if it measured rms or peak value

Well I assumed that the pf meter in the circuit measured the apparent power, since it gave me a reading of 22x5=110W. According to a formula on the lab handout, the real power was given by VI

The real power is given by VrmsIrms only when there is a unity power factor. In other words when the circuit or circuit element is purely resistive, with 0 inductive or capacitive reactance.
P = VrmsIrms(pf)
Presistive = VrmsIrms ; (pf = 1)

Since the lab handout is giving you formulas with rms values, that's probably what you were measuring. You may want to email your lab TA to be sure though.
 
  • #19


Meters normally measure RMS - very unusual to find one with a peak to peak scale.
VI*cos(phase angle) is the real power.
 
  • #20


Delphi51 said:
Meters normally measure RMS - very unusual to find one with a peak to peak scale.
VI*cos(phase angle) is the real power.

So my values for the real and apparent powers , 108.78VA and 110W respectively are correct? The ratio of real:apparent would be 0.98

If the apparent power is 108.78VA and the real is 110W, the ratio of real:apparent is 1.01, which would mean that cos(phase angle)=1.01, which is wrong mathematically.
 
  • #21


Yes, that looks good to me.
 
  • #22


Delphi51 said:
Yes, that looks good to me.

So then for S=P+jQ, is to find Q by using S2=P2+Q2 (say I found Q=34)

and then just plotting S=108.78+34j on an argand diagram?
 
  • #23


I'll take a look at it thanks. :biggrin:
 

1. What is a phasor diagram and why is it used in circuit analysis?

A phasor diagram is a graphical representation of the amplitude and phase of a sinusoidal waveform. It is commonly used in circuit analysis to visualize the behavior of complex AC circuits and to determine the relationships between voltage and current in different components of the circuit.

2. How do I interpret a phasor diagram for circuit analysis?

To interpret a phasor diagram, you must understand the relationship between the length and direction of the phasor and the amplitude and phase of the corresponding sinusoidal waveform. The length of the phasor represents the amplitude of the waveform, while the direction represents the phase. By analyzing the relative positions of different phasors, you can determine the voltage and current relationships between different components in the circuit.

3. How do I use phasor diagrams to solve circuit problems?

Phasor diagrams can be used to solve circuit problems by converting the complex AC circuit into a simplified representation using phasors. By manipulating phasors using mathematical operations, you can determine the voltage and current relationships in the circuit and solve for unknown values.

4. Can phasor diagrams be used for non-sinusoidal waveforms?

No, phasor diagrams are only applicable for sinusoidal waveforms. For non-sinusoidal waveforms, other mathematical techniques such as Fourier analysis must be used to analyze the circuit.

5. Are there any limitations to using phasor diagrams for circuit analysis?

Yes, phasor diagrams only represent the steady-state behavior of a circuit and cannot account for transient or non-linear effects. Additionally, they are only applicable for linear circuits with passive components such as resistors, capacitors, and inductors.

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