# AC circuits with complex numbers

1. Sep 26, 2010

### Markel

Hello,

My prof showed a way of dealing with difficult circuits using complex numbers. But I have no idea what he was on about, and can't find that method in any book.

Does someone know what I'm talking about, and can someone point me in the way of some materials for this?

Thanks!

2. Sep 26, 2010

### Gerenuk

Hmm, it shouldn't be hard to find information because it's a standard topic. I haven't checked myself. Maybe
http://www.phy.duke.edu/~fortney/physics_52/AC_circuit_note.pdf [Broken]
? I'm sure someone on this forum will suggest a good book. I can't remember what I used.

To get you started I can outline the concept:
Complex number are used as a tool to solve the differential equations at work behind passive AC circuits. You examine the currents and voltages when a circuit is driven by a voltage at a single given (circular) frequency omega. Note that in this case all value at each point in the circuit will oscillate with the same frequency. So (co)sinusoidal input voltage can be written as the real part of the complex exponential:
$$V(t)=V_0\cos(\omega t)=\Re(V_0\exp j\omega t))$$
The term inside the brakets is the complex voltage (which basically has an auxiliary imaginary part) The point is that it's messy to transform from cosine to sine if for example the current is out of phase. But it's easy with complex numbers because all you have to do
is to multiply the complex voltage by a phase factor.
At a capacitor the differential equations would show that the current is out of phase with the voltage.
$$V(t)=V_0\cos(\omega t)$$
$$I(t)=I_0\sin(\omega t)$$
(or the other way round :) ) So Ohm's law R=V/I=const doesn't seem to hold. However, it can be regained if you apply it to the complex counterparts
$$V(t)=\Re(V_0\exp(j\omega t))=\Re(\tilde{V})$$
$$I(t)=\Re(I_0\exp(j\omega t)\cdot j)=\Re(\tilde{I})$$
And so Ohm's law still holds
$$Z=\frac{\tilde{V}}{\tilde{I}}=\frac{1}{j\omega C}=\text{const}$$
Z is the impedance and represents the complex equivalent for resistance. For a normal resistor it's the usual real number, but for capacitors or coils it's an imaginary number (this will automatically make the required differential equations hold)
http://en.wikipedia.org/wiki/Electrical_impedance#Capacitor
Now you can start calculations equivalent to Ohm's law and in the end just take the real parts of your solution to get real voltages and currents. Btw, you have to examine all frequencies of the driving voltage individually since this method works for a single driving frequency only.
I haven't explained everything, but maybe you can figure it out from here. Or just ask questions :)

Last edited by a moderator: May 4, 2017
3. Sep 26, 2010

### Markel

Wow, thanks for the help.

I'm studying for an exam, and haven't started with AC circuits yet, so right now I don't fully understand what you're talking about. I need to do some preperations first. I just wanted to try and get some materials organized so I'd be ready when I come to that section. Once I start that chapter I'll probably post in here again for some clarifications. I just kind of panicked when I saw that this topic wasn't in most standard books.

4. Sep 26, 2010

### Gerenuk

It's no more than Ohm's law with complex numbers. So don't worry.

5. Sep 26, 2010

### Markel

ah ok. I feel much better now.

I remember seeing this whiz by in the lecture and thinking. 'why can you just ignore the imaginary parts?' this didn't make sense to me at the time.

6. Sep 26, 2010

### schip666!

The imaginary component -- j or i depending on who you are -- allows one to represent two-dimensional vectors in your calculations by using complex numbers. The current and voltage in a circuit with reactive components (capacitors and inductors) do not change in tandem -- like they do with a resistors and DC circuits -- but we can track them in two dimensions. The magic of it all is that the complex vector calculations work to represent the actual circuit behavior AND they are easy (if you have that kind of brain) to visualize. Start with this wiki to get a little more:
http://en.wikipedia.org/wiki/Complex_number

7. Sep 26, 2010

### Born2bwire

More specifically along these lines, DC circuits deal with the amplitude of the signal. However, in AC signals we also have phase and the real and imaginary parts now encode both the amplitude and phase of the signal. Capacitive and inductive components cause the phase of the current and voltage to shift. That is, the points in time where the current peaks can change from that of the voltage. That is why it is important to retain the real and imaginary parts for the analysis of the circuit. Once you have found the complex currents and voltages, you can then convert them to the correct time dependent forms.

8. Oct 3, 2010

### Markel

Thanks for the help everyone.

I just reached this topic in the 'fundamentals of physics' book, and unfortunately it doesn't use this complex number method at all. It uses phasor diagrams. My questions is now: what is the difference between using phasor diagrams and the complex number method? is it that the complex number method is easier? (if so, why isn't it in the fundamentals of physics book?)

Also, I still haven't found a book that treats AC circuits with complex numbers. Does anyone know of one that gives a concise overview and maybe has a few worked out problems?

Thanks!

9. Oct 3, 2010

### Gerenuk

Complex numbers and phasors are more or less the same. Just plot the argument of a complex number as the angle in a diagram and you got a phasor.

I guess phasors are just a visual tool designed to confuse people :) It seems more visual and appealing at first, but then you never end asking questions.

Maybe you can just try out a calcuation with a simple recipe:
- pick any circuit with resistors, inductances and capacitors and a sinusoidal voltage driving it a circular frequency omega=w
- assign a real voltage V to the source
- assign complex resistances Z=R to resistors, Z=jwL to inductances and Z=1/(jwC) to capacitors (where w means omega)
- now do the usual circuit calculation for currents and voltage drops that you know from simple resistor networks (use Z=V/I, Kirchhoff's current and junction laws)
- after you've found the complex voltages and currents this way, multiply all these by exp(iwt)
- now take the real part of all of them and you got the circuit solved!

Finally you could try to figure out what the phasor business was about, by comparing it to you complex number solution.

Unfortunately I still haven't got access to a library to look for books or example problems :( Hope you can find some! :)

10. Oct 4, 2010

### Markel

Hmm,

Went to my professor to ask him about this today and he kind of outlined it for me. but I rarely understand something when someone is explaining (or from lectures) I need to have a book and go slowly. But my exam is in one day! yikes. And I still can't find any book or website that gives this a formal treatment.

He told me that phasors and complex numbers are more or less the same. But I would like to understand the complex number method fully. I think now there's not enough time though, and I should just stick with the phasor method for this exam and figure the other method out later. sigh

11. Oct 4, 2010

### Markel

Hmmm. Still going over this.

Can you use the phasor method for multiloop circuits as well?? Are they essentially ultimately identical in every way? If only I had some fully solved problems, then I could work out both methods and see if they give the same results.

12. Oct 4, 2010

### Markel

Ok, one more question.

If you have a large circuit, perhaps multiloops of different arangements of capacitors inductors and resistors, does the Emf in all of the capacitors all vary by Pi/2 ?

This can't be right. The chapter in my book is called "series RLC circuits"
But on my exam I'm sure there will be a multiloop RLC circuit. Help!

13. Oct 6, 2010

### Born2bwire

Phasor and complex numbers are equivalent. Phasor deals with the phase and amplitude of the signal and as I mentioned earlier complex numbers are simply another way of encoding the same information. Quite simply,
$$Ae^{i\phi} = A\left(\cos\phi + i\sin\phi \right)$$
where A is the amplitude and \phi is the phase in radians. I do not think it is worth working out how to do it in complex numbers at this point if you are still having trouble with phasors. The concepts are completely the same it's just different on how you work the math. Complex numbers is easier for more direct computation via calculator or computer.

As for multiloops, yes, either one is perfectly capable of describing any analog circuit. You just reduce the circuit down to its equivalent impedance to find the total current. Then, using the total current you can then use the Kirchoff voltage and current rules to find the voltages and currents of the branches. You treat it in the same manner as if it was all resistors, just replace the inductors and capacitors with their equivalent impedances.

14. Oct 6, 2010

### K^2

Phase of EMF on capacitor will always be shifted by Pi/2 relative to current flowing through that capacitor. In a multiloop circuit, however, you can have different phases of current through different capacitors, so the phase of EMF can all be different.

For example:

Code (Text):

C1       C2
+---||---+---||---+
|        |        |
|        z        z
V~       z R1     z R2
|        z        z
|        |        |
+--------+--------+

Here, phases on R1 and R2 will be different, as well as phases on C1 and C2, and depend on frequency of your power supply. However, EMF phases on C1 and R1 will be shifted by Pi/2 relative to each other. Same for C2 and R2. You can easily use complex impedances to figure out these phases and frequency dependence.