Electrodynamics: AC-circuits and Kirchhoff

[Niles]

Homework Statement

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 "+"? Or does the direction in this case even matter?

Thanks in advance.

Regards,
Niles.

 

[Hootenanny]

As for the case of a DC circuit, you may chose your reference direction arbitrarily. Just take care to stick to this direction throughout.

 

[Niles]

I have set CW to be positive, and I have solved it two ways:

1) Saying that the source has polarity as to send currect CCW

2) Same as #1, but CW.

When finding the current through the capacitor, the difference I get is in the phase angle. In #1 I get -1.25 and in #2 I get 1.25.

How is this to be interpreted? If anything, the difference should be 180 degrees?

 

[Hootenanny]

I have set CW to be positive, and I have solved it two ways:

1) Saying that the source has polarity as to send currect CCW

2) Same as #1, but CW.

When finding the current through the capacitor, the difference I get is in the phase angle. In #1 I get -1.25 and in #2 I get 1.25.

How is this to be interpreted? If anything, the difference should be 180 degrees?

Not necessarily. The reference direction that you chose determines the direction of the positive phase angle. So in the first case you chose your reference direction as counter-clockwise and obtained a negative phase angle, i.e. the phase angle is in the opposite direction to your reference direction. In the second case you chose your reference direction to be clockwise and therefore obtained a positive phase angle, i.e. the phase angle is in the same direction as your reference direction.

Does that make sense?

 

[Niles]

It does, but aren't I allowed to add 2pi, so both phase angles point in the same direction?

 

[Hootenanny]

It does, but aren't I allowed to add 2pi, so both phase angles point in the same direction?

Yes. So as I said in my original post, the choice of reference direction is arbitrary - it doesn't matter as your answers will be the same.

 

[Niles]

But the phase-angles only point in the same direction, they don't have the same value; so how can the answers be the same?

 

[Hootenanny]

But the phase-angles only point in the same direction, they don't have the same value; so how can the answers be the same?

Their value is relative the the initial direction that you chose. In the first case the phase angle is in the opposite direction to your reference direction (hence the negative value). However, in the second case the phase angle is in the same direction as your reference direction (and hence positive).

Note that both 'values' have the same magnitude and point in the same direction, therefore they are equal.

 

[Niles]

Ahh, I see. So I don't even have to add the 2pi to see it.

But when drawing a phasor diagram, projecting the phasor on the horizontal axis will give you the current. If I project the two phasors in this situation, they will give me the same current (i.e. that why I believed they had to differ by 180 degrees).

What is the explanation for this?

 

[Niles]

I don't mean to be annoying, but can you guys find the error in my reasoning?