Phasor Diagram RLC Series AC Circuit


by SPYazdani
Tags: circuit, diagram, phasor, series
SPYazdani
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#1
Mar19-11, 08:18 PM
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1. The problem statement, all variables and given/known data

I'm struggling to figure out what vectors I use for my phasor diagram for an RLC series circuit.

2. Relevant equations

VC = 1/j[tex]\omega[/tex]C
VL = j[tex]\omega[/tex]L
VR = IZ

3. The attempt at a solution

I know how to add vectors, so that's not a problem, transfer my circuit from the time domain to frequency domain. I understand that VC leads the current and VL lags the current by [tex]pi/2[/tex] radians.

If I have say, r[tex]\angle[/tex][tex]\theta[/tex] as my input voltage, then I can draw that on my phasor diagram because it is given to me as a vector (easy). In contrast, VC, VL and VR are not vectors because after transferring them to the frequency domain, their unit is [tex]\Omega[/tex] which I think is a scalar quantity.

My question is, how do I get a vector out of VC, VL and VR in the frequency domain to plot onto my phasor diagram?
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vela
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Mar20-11, 02:06 AM
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Quote Quote by SPYazdani View Post
1. The problem statement, all variables and given/known data

I'm struggling to figure out what vectors I use for my phasor diagram for an RLC series circuit.

2. Relevant equations

VC = 1/j[tex]\omega[/tex]C
VL = j[tex]\omega[/tex]L
VR = IZ
The RHS of the first two equations are the impedances, ZC and ZL, not voltages. The Z in the last equation should be R.
3. The attempt at a solution

I know how to add vectors, so that's not a problem, transfer my circuit from the time domain to frequency domain. I understand that VC leads the current and VL lags the current by [tex]pi/2[/tex] radians.
You have that backwards. VC lags the current, and VL leads the current.
If I have say, r[tex]\angle[/tex][tex]\theta[/tex] as my input voltage, then I can draw that on my phasor diagram because it is given to me as a vector (easy). In contrast, VC, VL and VR are not vectors because after transferring them to the frequency domain, their unit is [tex]\Omega[/tex] which I think is a scalar quantity.
You're talking about impedances, not voltages. The impedances are complex quantities that relate the complex voltage to the complex current through Ohm's Law: V = IZ.
My question is, how do I get a vector out of VC, VL and VR in the frequency domain to plot onto my phasor diagram?
For the series RLC circuit, the current is common to all of the elements, so use that as your reference. At an instant in time, it will have some amplitude and phase. Using Ohm's Law, you can calculate the amplitude and phase of the voltage across each element at that instant in time. Those are the phasors that you draw on your diagram. Their vector sum then equals the source voltage.
SPYazdani
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Mar20-11, 08:00 AM
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Quote Quote by vela View Post
For the series RLC circuit, the current is common to all of the elements, so use that as your reference. At an instant in time, it will have some amplitude and phase. Using Ohm's Law, you can calculate the amplitude and phase of the voltage across each element at that instant in time. Those are the phasors that you draw on your diagram. Their vector sum then equals the source voltage.
Thanks heaps.

Can you please check if I'm on the right track? It looks good to me.

http://img585.imageshack.us/img585/117/55843578.jpg

vela
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Mar20-11, 11:44 AM
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Phasor Diagram RLC Series AC Circuit


You've made several minor errors, but you're getting there. One error is that you left the factor of π out when calculating ω. Also, the sign of ZC is incorrect. You can't just bring the j up from the denominator to the numerator. Finally, I'm not sure where you got X from, but the value you used for X is not consistent with your other calculations.

When you, for instance, calculate the voltage across the capacitor, don't write V=IR when you really mean V=IZC. It's sloppy and, from a mathematical perspective, plainly wrong. It's obvious what you mean, but it's the kind of thing that can annoy some instructors and graders.

The more serious conceptual error is that you're not keeping track of the phases correctly. In the frequency domain, the voltages, the current, and the various impedances are all complex values, related by Ohm's Law:

[tex]V = I Z[/tex]

This equation concisely expresses the relationship between both the amplitude and phase of all the quantities. The equation

[tex]|V| = |I| |Z|[/tex]

is similar, but it only accounts for the amplitudes. In your calculations, the impedance you calculated is actually |Z|, not Z. So when you wrote that the current I is V with a phase divided by |Z|, that's not really correct. It's evidenced in the fact that if you add up the individual voltages you calculated, you'll find they don't add up to 12 V with phase of 0 degrees as it should. That's because the current I you calculated is missing the phase information.
SPYazdani
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Mar20-11, 08:02 PM
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Quote Quote by vela View Post
The more serious conceptual error is that you're not keeping track of the phases correctly. In the frequency domain, the voltages, the current, and the various impedances are all complex values, related by Ohm's Law:

[tex]V = I Z[/tex]

This equation concisely expresses the relationship between both the amplitude and phase of all the quantities. The equation

[tex]|V| = |I| |Z|[/tex]

is similar, but it only accounts for the amplitudes. In your calculations, the impedance you calculated is actually |Z|, not Z. So when you wrote that the current I is V with a phase divided by |Z|, that's not really correct. It's evidenced in the fact that if you add up the individual voltages you calculated, you'll find they don't add up to 12 V with phase of 0 degrees as it should. That's because the current I you calculated is missing the phase information.
I'm really sorry, but I'm not understanding the difference between X and Z.
If I take the formula Z=[tex]\sqrt{R^2+X^2}[/tex], it appears strangely similar to the distance formula. Whereas XC and XL both have formulae -1/[tex]\omega[/tex]C and [tex]\omega[/tex]L respectively, yet they both have the unit of ohms.

I've made some corrections, but now I have more errors :(

With my new values for VC and VL, if I add them, they add to negative 12, which can't be right because I have a negative magnitude.

http://img854.imageshack.us/img854/338/74199395.jpg
vela
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#6
Mar20-11, 10:32 PM
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The reactance X of an element is just its Z without the factor of j. The impedance of the circuit is

[tex]Z = Z_R + Z_L + Z_C = R + j\omega L - j\frac{1}{\omega C} = R + j\left(\omega L - \frac{1}{\omega C}\right)[/tex]

Note Z is a complex quantity. If you find the modulus of Z, you get

[tex]|Z| = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}[/tex]

The Z you calculated is actually |Z|. It doesn't contain phase information any more, which is why you got the puzzling results where everything had 0 phase.

The current I you calculated is actually |I|. It's the amplitude with which the current oscillates. Similarly, you found |VL| and -|VC|. (The minus sign for VC indicates that VC lags the current by 90 degrees.)

To find the phase, what you want to do is draw the phasor diagram with VL, VC, and VR. |VL| and |VC| are the lengths of two of the vectors. The vector sum of the three voltages then equals V of the source.
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#7
Mar20-11, 11:16 PM
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To elaborate a bit more...

The voltage V, the current I, and the impedance Z are all complex quantities, so we can write them in polar form

[tex]\begin{align*}
V & = |V| e^{j\phi_V} \\
I & = |I| e^{j\phi_I} \\
Z & = |Z| e^{j\phi_Z}
\end{align*}[/tex]

where |V| and |I| correspond to the amplitude of the oscillations. Now Ohm's Law says

[tex]V = IZ = (|I| e^{j\phi_I})(|Z| e^{j\phi_Z}) = |I||Z|e^{j(\phi_I + \phi_Z)}[/tex]

from which you can see that [itex]|V| = |I| |Z|[/itex] and [itex]\phi_V = \phi_I + \phi_Z[/itex].

So you have two common approaches to the problem. You can work with V, I, and Z as complex numbers, use complex algebra, and grind everything out. This method is straightforward, but it's prone to errors because you're dealing with j's all over the place.

The other approach is to work with |V|, |I|, and |Z|, which are real numbers, and figure out the phase relationship using the phasor diagram. It may seem a bit more complicated at first, but once you understand it, it's easier to remember and work with.
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#8
Mar22-11, 09:11 AM
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Quote Quote by vela View Post
To elaborate a bit more...

The voltage V, the current I, and the impedance Z are all complex quantities, so we can write them in polar form

[tex]\begin{align*}
V & = |V| e^{j\phi_V} \\
I & = |I| e^{j\phi_I} \\
Z & = |Z| e^{j\phi_Z}
\end{align*}[/tex]

where |V| and |I| correspond to the amplitude of the oscillations. Now Ohm's Law says

[tex]V = IZ = (|I| e^{j\phi_I})(|Z| e^{j\phi_Z}) = |I||Z|e^{j(\phi_I + \phi_Z)}[/tex]

from which you can see that [itex]|V| = |I| |Z|[/itex] and [itex]\phi_V = \phi_I + \phi_Z[/itex].

So you have two common approaches to the problem. You can work with V, I, and Z as complex numbers, use complex algebra, and grind everything out. This method is straightforward, but it's prone to errors because you're dealing with j's all over the place.

The other approach is to work with |V|, |I|, and |Z|, which are real numbers, and figure out the phase relationship using the phasor diagram. It may seem a bit more complicated at first, but once you understand it, it's easier to remember and work with.
I don't mind grinding out the complex algebra, its good practice.
I tried the second method you mentioned using my results from the link above. I think what I've done is correct. I'm not sure if I'm missing any more information

Although it isn't to scale or in a presentable format, I think it's all there.

http://img692.imageshack.us/img692/4722/21313865.jpg
vela
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#9
Mar22-11, 11:58 AM
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The image isn't loading here.
SPYazdani
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Mar22-11, 06:38 PM
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Quote Quote by vela View Post
The image isn't loading here.
Me either. That's weird.

Try the attachment
Attached Thumbnails
Untitled.jpg  
vela
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#11
Mar23-11, 12:24 AM
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Looks good!
SPYazdani
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Mar23-11, 03:29 AM
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Thank you very much for your help. My lab report is now complete and ready for submission :D

Also, a side question, in the lab report, he asked us to provide two phasor diagrams. One for 1000Hz and another for 3000Hz. They both appear to be roughly 1KHz above and below the resonance frequency which I calculated to be 1930.04 Hz. Apart from the relationship of the phase between the elements, I'm not sure what I'm supposed to learn from the two phasor diagrams.

Heres the second phasor diagram I did. As before, it's not to scale or in a presentable format.

http://img813.imageshack.us/img813/7845/82891587.jpg


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