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Equivalent capacitance of circuit

  1. Jan 27, 2016 #1
    1. The problem statement, all variables and given/known data
    I am having trouble simplifying a circuit of capacitors that I was given. I am asked to determine the equivalent capacitance of the combination below (attached photo).

    2. Relevant equations
    Capacitors in series: 1/Ceq = 1/C1+1/C2+....

    Capacitors in parallel: Ceq = C1 + C2 +.....

    3. The attempt at a solution

    (1) Attempt #1 in my diagram would involve putting capacitor 1F and 2F in series, which would create equivalent capacitors of 2/3F each. Then I would find the equivalent capacitance in parallel: 2/3+2/3+3 = 4.33F

    (2) Attempt #2 in my diagram involves a different approach... where both 1F capacitors are in parallel, join again at 3F, and break into a parallel circuit again for the 2F capacitors.
    This would simplify into 1F, 3F, and 4F in series, giving Ctotal=1.583F

    Which simplified circuit would be correct? (If either of these...)
     

    Attached Files:

  2. jcsd
  3. Jan 27, 2016 #2

    Suraj M

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    Firstly, both your attempts are wrong.
    Are you familiar with what a Wheatstone bridge is?
     
  4. Jan 27, 2016 #3
    No I'm not, I don't think that content is covered in our introductory curriculum. Is that approach necessary to simplify this circuit?
     
  5. Jan 27, 2016 #4

    Suraj M

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    It would have made it simpler but it's still doable with a little logic.
    If you had labeled the junctions in your first image it would have helped.
    From the main terminal the current(charge) would split in two directions and then reach the next junction, these 2 (second junctions) on either side(upper and lower) are actually at the same potential.
    So now can you solve it? Do you think there would be any current(actually -charge) through the middle "bridge"?
     
  6. Jan 27, 2016 #5

    Suraj M

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    Last edited: Jan 28, 2016
  7. Jan 28, 2016 #6

    NascentOxygen

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    I think you should start over again and follow the suggestion by Suraj M of initially pretending C3 is not present. Without C3, determine the voltage at the point on the top line where C1 is joined to C2, and also determine the voltage at the point on the bottom line where C1 is joined to C2.

    It might be convenient to consider one of your black dots is 0 volts and the other black dot is V volts.

    See how you go with that step.
     
  8. Jan 28, 2016 #7
    I found the equivalent capacitance of C1 and C2 in series, to be 0.667F (in both upper and lower branches of the circuit).

    However, I have two issues that I can't quite wrap my head around.
    (i) I would not think it to be okay to combine C1 and C2 in series, since they are separated by that "vertical bridge". How do you know it's acceptable to do this?
    (ii) Based on how you framed the question, I am thinking that there wouldn't be any charge flowing through the middle bridge. If my assumption is correct, I am thinking that the logic behind it is based on "the path of least resistance". Since the equivalent capacitance of C1 and C2 is 0.667F is less than C3, the current would "preferably" flow through the C1-C2 path. However, can this "path of least resistance" be applied here, since current is flowing through capacitors instead of resistors?

    I'm sorry for all the trivial questions! I've always had such a hard time with the concepts behind electrical circuits.
     
  9. Jan 28, 2016 #8

    cnh1995

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    Did you read the link Suraj provided in an earlier post?
     
  10. Jan 28, 2016 #9
    I did read the link to the Wheatbridge material, but to be honest, it just went over my head. I'm currently even struggling with the basics of understanding circuits.

    Can you elaborate on the concept behind KVL (voltage divider)?
     
  11. Jan 28, 2016 #10

    cnh1995

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    Ok. Whenver you find such bridge in a circuit, always first check if the bridge is "balanced". Ratio of components on one side of the bridge=ratio of components on the other side of the bridge. This is the condition for bridge balance. Here, the bridge is balanced. Do you see it?
     
    Last edited: Jan 28, 2016
  12. Jan 28, 2016 #11

    cnh1995

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    Well, that comes after you remove C3. First, you need to know under which condition C3 is redundant.
     
    Last edited: Jan 28, 2016
  13. Jan 28, 2016 #12

    Suraj M

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    Wheat stone* bridge

    Have you officially been taught how to apply KVL and KCL?
    Just remember the condition for a Wheatstone bridge, as soon as you see it

    If you encounter any unbalanced circuit then you'll need KVL, so if KVL has been taught to you, you can solve any circuit ( if you have a reasonable amount of time).
     
  14. Jan 28, 2016 #13

    gneill

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    If you want to convince yourself that you can eliminate the "3" capacitor without an appeal to symmetry (which is a very powerful tool, by the way), then you can write loop equations and solve for the individual charges on the capacitors.

    With purely capacitive circuits, after a DC potential is applied you know that there is no ongoing current when transients have died down and the capacitors take on their steady-state charges (instantaneously with ideal components). What has happened though is that some amounts of charge have moved through the circuit (as transient currents) to place the charges on the capacitors.

    Since charge is conserved you can apply a "static" form of KCL at the junctions. After all, charge is just the integral of current, and integrating won't change the relationships between them at a junction.

    Suppose in your circuit that a potential is places across terminals AB so that some amount of charge Q is pushed into terminal A and the same amount of Q is removed from terminal B. The current Q will split into two paths at the first junction. Call the split charges q1 and q2. You can then use "charge KCL" at the junctions to write expressions for the charges that end up on the capacitors. Assume that some charge q3 ends up on the middle capacitor. Here's a pic:

    upload_2016-1-28_9-38-23.png

    Noting that the voltage across a capacitor is given by V = Q/C, you should be able to write KVL around three loops. Using those loop equations you should be able to show that q1 = q2, and q3 = 0.

    If q3 is zero then it can have no potential difference across it, and it passes no current. Effectively it doesn't exist as far as the rest of the circuit or its behavior are concerned. In fact you can choose to remove it or replace it with a wire! (Open circuit or short circuit).
     
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