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Find Voltage in RC circuit

  1. Mar 27, 2012 #1
    I have a circuit I need to find a voltage on. Is there a way to figure it out on the provided picture in here?


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  2. jcsd
  3. Mar 27, 2012 #2


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    Staff: Mentor

    Maths can usually give the answer.

    • The voltage between which two points?

    • Are you wanting to know the amplitude of a sinewave, omitting the transient component?

    • Is this a homework exercise?
  4. Mar 27, 2012 #3
    1) I want to find out the source voltage.

    2) I don't know, maybe...

    3) No, I am trying to figure this out for my job. I am an Mechanical Engineer that is trying to figure out EE stuff and I am a bit lost. ha
  5. Mar 27, 2012 #4


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    Staff: Mentor

    Typically, you apply a known source voltage. Apparently not in your case, so you'd better explain what this circuit applies to.

    You must know the value of a current or voltage somewhere in this circuit, in order to be able to determine the source voltage. Having marked it as 50Hz, I suspect you do know the source voltage.

    Anyway, explain how this came about.
  6. Mar 27, 2012 #5


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    Gold Member

    Without a source voltage....you will not be finding any output voltages.

    Once you do find the source voltage.....you simply apply V=IR

    The resistance (reactance) of your capacitors will be (1/JWC)......W in this case will be 2∏*50.

    You will need to use vector math because of the "J" in the equation. Combine resistances of each parallel load...then do a voltage divider to find the voltage across each load.

    Or.....you can find the combined total impedance for all the loads.......find the current using V=IR....then use a current divider thru each parallel load. First idea will be the easier of the two.

    For each prospective parallel branch.....the voltage will be the same across the resistor and capacitor. The voltages across the two prosective branches will be different however.
    Last edited: Mar 27, 2012
  7. Mar 27, 2012 #6

    How about knowing the max voltage that will be accross both parallel branches seperatly? Would that give me enough knowledge to then find the source voltage?

    This isnt exactly a circuit equation. I am actually trying to find breakdown voltages across multiple layers of materials and I came accross an article that had this as a way to solve it but it didnt go into much detail about solving it.
  8. Mar 27, 2012 #7
    Yes, the voltages across both parallel branches will be related to the voltage coming from the source. You can work backwards to solve for the source voltage. Note that the maximum voltage across any individual parallel branch does not necessarily correspond to the maximum voltage from the source at the same time. You have reactive components, so you will have phase differences.

    But after reading your other information, I think a breakdown voltage will essentially change the circuit you have drawn. Can you provide the reference you were using?
  9. Mar 27, 2012 #8
    Ok well the first branch has a voltage of 30 kV and the second branch has a voltage of 4 kV.
  10. Mar 27, 2012 #9
    Then your voltage source is 34 kV
  11. Mar 27, 2012 #10
    Its really just adding the 2 voltages together?
  12. Mar 27, 2012 #11
    Yes, but you could be omitting information if you are not so sure what is going on and where you got the 30kV and 4kV numbers.

    You just use kirchoff's voltage law (second item in the link):


    Its just like pressure drops in a hydraulic or thermodynamic system.
  13. Mar 27, 2012 #12
    What I meant by omitting information is that because your voltage source is AC, its voltage is 34kV only at the instant in time that the voltage across both // groups is 30kV and 4kV. This does not necessarily correspond to the maximum voltage from the AC source. The maximum voltage from the AC source is when the sum of the two // groups voltages are a maximum.
  14. Mar 27, 2012 #13
    I got the 30 and 4 from the dielectric strength of the materials. The dielectric strength of the dielectric layer is 30 kV/mm and it is going through 1 mm and the air gap is 4 mm and the dielectric strength of air is about 1 kV
  15. Mar 27, 2012 #14
    Dielectric strength just says what voltage is needed to breakdown the dielectric. It can't be used for circuit analysis unless you have more information.

    I think we'd be able to help you a lot more if you provide the reference you are using for this method.
  16. Mar 27, 2012 #15


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    Not buying the 30 and 4 numbers.

    Give us a voltage source and the problem is easily solved.

    Let's say the voltage source is 480 volts. Then you will get something like 240 volts accross the first load and 200 volts across the second load....for example.

    Let's now say the voltage is 13,200. You will then get something like 7200 volts across first branch and 6000 volts across second branch....for example.

    Huge difference.

    Caps are defined by C*(dv/dt)=i(t)

    In other words....the changed in voltage over time multiplied by the capacitance equals the current. The voltage source and frequency determine the behavior of the capacitors...not the other way around.
  17. Mar 27, 2012 #16
    hmm... well im trying to find what voltage it should take to break through multiple materials so I am unsure what the final voltage should be. from my initial testing i am getting 33-38 kV. I am just trying to find a way to calculate it to verify my results.
  18. Mar 27, 2012 #17
    And I can't seem to find my source of the information I got. I thought I had saved the article but I guess I didn't
  19. Mar 27, 2012 #18
    What is the purpose of the voltage divider and two different dielectrics? Are you trying to compare one to another?

    If you just want to test what voltage it takes to break through the materials, can't you just place the materials in series with a variable voltage source and increase it til they break down and write down that voltage?
  20. Mar 27, 2012 #19


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    Hmmmm.....I don't think you are looking at things correctly.

    Incidentally.....why do you want to find the voltage across the loads? Why does that interest you? If you tell us why you are interested in voltage across load....that might help us solve your question.
  21. Mar 27, 2012 #20
    Well I can do that and I have been doing that. I just want to see if I can verify my results with Math since I am being told that the numbers I am getting seem to be a bit high.

    We are trying to test our parts dielectric strength and we need to have an air gap so we can use less tooling so we dont have to have a different test fixture for every part.
  22. Mar 27, 2012 #21


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    Staff: Mentor

    Your breakdown tests appear to be spot on. If the dielectric were rated with a breakdown of 30kV then you'd expect this to be conservative, so finding it can reliably withstand at least 33kV would confirm this.

    Examining your RC model at 50Hz, we can overlook the presence of both resistances because they have little effect in comparison with capacitances of the values you give. Your air gap capacitance is so much smaller than the dielectric capacitance, that the air gap will carry 98% of the applied voltage until the air breaks down and conducts. This will occur at around 4kV.

    For applied voltages greater than 4kV the air gap is effectively conducting, leaving all of the applied voltage impressed across your dielectric. With your dielectric's rated breakdown strength of 30kV then for applied voltages exceeding 30kV you will find the dielectric is subject to breaking down.

    https://www.physicsforums.com/images/icons/icon2.gif [Broken] Essentially, the air gap is not interfering with your voltage tests.

    When the applied voltage is removed, the air gap capacitor repairs itself; alas, the plastic dielectric does not. :smile:
    Last edited by a moderator: May 5, 2017
  23. Mar 28, 2012 #22
    Ok, I think I get what you are saying.

    So long as the air gap is not large enough to have a higher breakdown voltage then the Dielectric, the air gap doesn't matter?
  24. Mar 28, 2012 #23


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    Do you have any idea what this means?

    This means that I have no idea what I'm talking about:)

    Interesting way to look at things....
  25. Mar 28, 2012 #24
    Won't the air gap have some resistance at this point when it is ionized, rather than acting strictly like a short?
    Last edited by a moderator: May 5, 2017
  26. Mar 28, 2012 #25
    Well I removed the air gap and my results didnt get any lower so I'm thinking that the air gap doesnt really do anything....
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