Why are these capacitor voltages opposite polarity?

[kostoglotov]

Homework Statement

imgur link: http://i.imgur.com/JETj0Cq.png

Homework Equations

V_L = L\frac{dI_1}{dt}

V_L + V_1 + V_2 + V_{R1} = 0

The Attempt at a Solution

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Just using basic KVL, shouldn't V_L = -V_2 - V_1 - V_{R1} and so therefore we should get

\frac{dI_1}{dt} = \frac{-V_2 - V_1 - V_{R1}}{L} and not

\frac{dI_1}{dt} = \frac{V_2 - V_1 - V_{R1}}{L}

which is what the question is asking us to find...it would appear then that the capacitors have opposing polarities...why?

 

[NascentOxygen]

I think you are right, the terms on the right hand side in (4) should all have negative signs.

 

[donpacino]

It entirely depends on how you define the capacitor voltages. My guess, they (being the book authors) defined the voltages in a different way and didn't actually write it down through an oversight.

 

[kostoglotov]

It entirely depends on how you define the capacitor voltages. My guess, they (being the book authors) defined the voltages in a different way and didn't actually write it down through an oversight.

No, this is not from a textbook, this is from an assignment.

I1 and I3 are going in the same direction around the loop that we use in KVL though, no matter how you set it up.

I've only followed the current directions indicated. I haven't chosen my own directions independent of the problem.

 

[donpacino]

No, this is not from a textbook, this is from an assignment.

same thing, whoever defined the assignment did a bad job

I1 and I3 are going in the same direction around the loop that we use in KVL though, no matter how you set it up.

I've only followed the current directions indicated. I haven't chosen my own directions independent of the problem.

does not change what I said... saying they have V1 backwards without defining V1 is useless, same goes for V2 and VR1.

My guess, they defined V2 vertically with the positive end on the top and neglected to mention it.

 

[NascentOxygen]

It entirely depends on how you define the capacitor voltages.

It isn't up to us to arbitrarily define the voltages. Once you have shown eqn (2) is true, and eqn(3) is true, you have already been manipulated into defining the voltages the way the authors intend. The upshot is that eqn(4) needs that extra - sign.

 

[NascentOxygen]

kostoglotov, in future could you reduce the size of images. You can probably see how your over-size image in this thread causes the browser to shrink the page (and along with it, the text) so the full image fits onto the screen?

 

[donpacino]

It isn't up to us to arbitrarily define the voltages. Once you have shown eqn (2) is true, and eqn(3) is true, you have already been manipulated into defining the voltages the way the authors intend. The upshot is that eqn(4) needs that extra - sign.

good point... only [(2) and (3)] or [(4)] can be correct

then assuming op presumed 2 and 3 are correct, then yea 4 is missing a (-) sign.

 

[kostoglotov]

kostoglotov, in future could you reduce the size of images. You can probably see how your over-size image in this thread causes the browser to shrink the page (and along with it, the text) so the full image fits onto the screen?

No worries. I don't notice what you're seeing, perhaps because I'm using a large monitor?...I'll keep it in mind.

 

[kostoglotov]

It isn't up to us to arbitrarily define the voltages. Once you have shown eqn (2) is true, and eqn(3) is true, you have already been manipulated into defining the voltages the way the authors intend. The upshot is that eqn(4) needs that extra - sign.

Exactly, eqn 4 needs to have that missing - sign if we are to accept the previous equations. So, I need to take this back to the tutor or professor.