# AC-circuits

Discussion in 'Homework Help' started by Niles, Dec 20, 2008.

1. ### Niles Thread Starter Active Member

Nov 23, 2008
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0
Hi all.

Please take a look at the attached circuit. I want to find the current through the capacitor, and of course I will use Kirchhoffs laws for AC-current.

In this case, ε(t) = ε_0*cos(ωt).

My question is: If this was an DC-circuit, I would define e.g. counterclockwise to be positive, and then go through the circuit. When I see an EMF which has the terminals the correct way (i.e. it wants to "send" the current through CCW), it gets a "+".

In this case with the AC-circuit, I don't know which way the EMF wants to send the current, because it changes. Should I always just give ε_0 "+"?

Regards,
Niles.

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2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
In this question it doesn't really matter but if you want you can define one side of the source as positive and the other as negative. It would matter if there were more than one voltage or current sources.

3. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
So I am allowed to treat the source as a DC-source? If there is more than 1 source, then I can just use the superposition-principle, and solve for each source individually, and add each solution, right?

Thanks in advance, and thanks for responding quickly.

Last edited: Dec 20, 2008
4. ### mik3 Senior Member

Feb 4, 2008
4,846
63
No, you don't treat the AC source as a DC source. You just assume a positive polarity to be able to deal with phase differences in case you have many AC sources. You can use superposition if you like.

5. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
If there are many AC sources, I would use superposition principle.

But in the case of one AC source: You wrote that it doesn't matter. How is that? I mean, when using Kirchhoff's law for one of the loops including ε(t), I need to define a positive direction and the direction that ε(t) wants to send the current it will make a difference then?

6. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
Ok, so according to the link you sent me, if I choose the "wrong" direction, my phase should come out with an extra 180 degrees. In my case I get a phase angle of -1.24905 choosing a "wrong" direction, and 1.24905 with the correct direction.

Adding 2∏ wont help either. Have I understood it correctly?