Why are these capacitor voltages opposite polarity?

In summary, the textbook authors defined the voltages in a different way and didn't include a (-) sign in eqn 4, which causes the equation to be incorrect.
  • #1
kostoglotov
234
6

Homework Statement



JETj0Cq.png


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

Homework Equations



[tex]V_L = L\frac{dI_1}{dt}[/tex]

[tex]V_L + V_1 + V_2 + V_{R1} = 0[/tex]

The Attempt at a Solution


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

[tex]\frac{dI_1}{dt} = \frac{-V_2 - V_1 - V_{R1}}{L}[/tex] and not

[tex]\frac{dI_1}{dt} = \frac{V_2 - V_1 - V_{R1}}{L}[/tex]

which is what the question is asking us to find...it would appear then that the capacitors have opposing polarities...why?
 
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  • #2
I think you are right, the terms on the right hand side in (4) should all have negative signs.
 
  • #3
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.
 
  • #4
donpacino said:
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.
 
  • #5
kostoglotov said:
No, this is not from a textbook, this is from an assignment.
same thing, whoever defined the assignment did a bad job

kostoglotov said:
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.
 
  • #6
donpacino said:
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.
 
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  • #7
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?
 
  • #8
NascentOxygen said:
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.
 
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  • #9
NascentOxygen said:
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.
 
  • #10
NascentOxygen said:
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.
 

1. Why do capacitors have opposite polarity voltages?

Capacitors have opposite polarity voltages because they consist of two conductive plates separated by a dielectric material. When a voltage is applied, one plate becomes positively charged and the other becomes negatively charged, resulting in opposite polarity voltages.

2. What determines the polarity of a capacitor?

The polarity of a capacitor is determined by the orientation of the plates and the direction of the voltage applied to it. The plate connected to the positive side of the voltage source will have a positive polarity, while the plate connected to the negative side will have a negative polarity.

3. Can the polarity of a capacitor be changed?

Yes, the polarity of a capacitor can be changed by simply reversing the direction of the voltage applied to it. This will cause the plates to switch polarities, resulting in a change in the overall polarity of the capacitor.

4. Why is opposite polarity important in capacitors?

Opposite polarity in capacitors is important because it allows them to store electrical energy. When connected in a circuit, the capacitor's opposite polarity voltages create an electric field that stores energy in the form of an electric charge. This stored energy can then be released when needed.

5. Are there any exceptions to the opposite polarity rule in capacitors?

Yes, there are some types of capacitors, such as non-polarized and electrolytic capacitors, that do not have opposite polarity voltages. These types of capacitors are designed to be used in circuits where the polarity of the voltage may change, and they are able to function in either direction.

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