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Effective Resistance in Circuit

  1. Jan 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the effective resistances between P and Q in circuit-1 and A and B in circuits 2 and 3 in each of the following circuit diagrams.
    In circuits 2 and 3 the resistance R has the value 1Ω.

    2. The attempt at a solution
    1. Why does the AB branch get removed in circuit-1 when we simplify the circuit further? Is it because A and B are points at equal potential (equipotential)? But then why are A and B equipotential points?

    2. Why do the CE and DF branches get removed in circuit-2 when we simplify the circuit further? It s because points C,E and D,F are equipotential pair of points. But why are they equipotential points?

    3. Circuit-3 simplifies to circuit-2 with CD resistance being 10/9Ω and EF resistance being 2/3Ω. All the rest of the resistances are 1Ω. But how does this happen? How does it get simplified to this?
     

    Attached Files:

  2. jcsd
  3. Jan 21, 2015 #2
    For the first circuit:

    You are right. Points A and B are points of equal potential. The first circuit has three branches arranged in parallel. All the branches have two resistances of the same value. The potential drop across all braches is the same in a parallel circuit. Considering the potential difference between points P and Q is 'V', what do you think will be the potential drop from P to A or P to B?
     
    Last edited: Jan 21, 2015
  4. Jan 21, 2015 #3
    I think it would be the same. But how does that explain why P and Q are equipotential?
     
  5. Jan 21, 2015 #4
    If P and Q were equipotential, no current would flow through the circuit. P and Q aren't equipotential. A and B are.
     
  6. Jan 21, 2015 #5
    Oh yes. I meant A and B. Sorry for the typo. So why is the potential difference from P to A and P to B the same? That is what the question essentially reduces down to.
     
  7. Jan 21, 2015 #6
    Because :
    1) The potential drop across all three branches is the same
    2) All the branches have 2 resistances of the same value

    So if the potential drop across two resistances is 'V', across one resistance it will be 'V/2'.

    Read up on the basics of division of current and voltage.
     
  8. Jan 21, 2015 #7
    The second circuit can be solved by Delta-Star conversion.
     
  9. Jan 21, 2015 #8
    Suppose the potential at P is VP, potential at A is VA and potential at B is VB. So you are saying that VA - VP = VB - VP which means VA = VB. But what is the logical explanation behind VA - VP = VB - VP?

    What is the Delta-Star Convention?
     
  10. Jan 21, 2015 #9
    And the potential drop across all the 3 branches will be the same only if all the 3 branches are in parallel, but are they?

    Two resistors are said to be in parallel if the ends of one resistor are connected across the ends of the other. So each of the two resistors can be a summation of a large number of resistors in parallel too, in this case two.

    But the two resistors are not in series because for two resistors to be in series they need to be connected end-to-end and the same current should flow through each of them. How can we be sure that the same current is flowing through each of the two resistors in each branch?
     
    Last edited: Jan 21, 2015
  11. Jan 21, 2015 #10
    Can someone please explain this more clearly?
     
  12. Jan 21, 2015 #11

    BvU

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    Would it be easier to understand if you imagine taking away the two 2R resistors drawn vertically between A and B ? Then you conditions are fulfilled and all three intermediate points are at V/2, so re-inserting the two 2R resistors has no effect.
     
  13. Jan 21, 2015 #12
    Yeah! Never thought of it that way. But the points would be at the same potential only until we haven't inserted the two 2R resistors. Once we insert them it again becomes unclear whether the 3 branches are in parallel or not and uncertain whether we can divide the potential difference like V/2 and V/2 as we used to do earlier. So inserting the the 2 resistors complicates things and makes it confusing. And that is why I am not able to fully understand how to go about solving this circuit.
     
  14. Jan 21, 2015 #13

    BvU

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    Why ? No potential difference ##\Rightarrow## no current, so no influence at all. You could even short-circuit between A and B in this symmetric case.
     
  15. Jan 21, 2015 #14
    Potential difference is 0 because of the V/2 logical explanation. V/2 logic exists because each of the 2 resistances in each of the branches are in series and because all the branches are in parallel. But the "no potential difference" condition remains only until we haven't added the 2R resistances. After we add the 2R resistors we can't be sure that the 2 resistors in each branch are in series. And so we can't divide the potential as V/2 to each of them. And finally we can't be sure that the potential difference between points A and B is 0. This is what is confusing me.
     
  16. Jan 21, 2015 #15

    BvU

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    Don't know what to say. You could look at the Wheatstone bridge : the top half and the bottom half of circuit 1 are similar to that; you know the various R so you can determine ##I_G## in the derivation there is zero.

    Or you could do a lot of work and use Kirchoff's law for the whole thing to come to the same conclusion.

    Using symmetry really is the easiest: assume these 2R in the middle carry a current. Flipping upside down would flip the current sign, right ?

    But the flipped picture is identical ##\rightarrow## same current. So current = - current ##\rightarrow## 2 current = 0 ##\rightarrow## current = 0.
     
  17. Jan 21, 2015 #16
    Wheatstone bridge is a good explanation. Thanks for that. So circuit-1 is done now. What about circuit-2 and circuit-3?
     
  18. Jan 21, 2015 #17

    BvU

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    Well, you know all you need to know to solve 2 and 3.
     
  19. Jan 21, 2015 #18
    Uhmmm..no..I don't think so. I would like some help.
     
  20. Jan 21, 2015 #19

    BvU

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    Well, what can I do (short of your work) ?
     
  21. Jan 21, 2015 #20
    Atleast get me on the track to solving 2 and 3 the correct way.

    I am not understanding why do they get simplified the way they are and its not at all related to the explanation of the first (wheatstone bridge logic). So even in the first, the logic of equal potential and the reasoning behind it didn't make sense to me (as I have stated in Post #14) and so I sought for another explanation and finally came up with wheatstone bridge. Similarly, even in 2, I am not able to get the logic of equal potential because it doesn't make sense to me(again, as stated in Post #14). And so I am looking for some other explanation like in circuit-1 but this time I am unable to think of it.

    As for 3, I have no idea how is this going to work because I don't see the possibility of wheatstone bridge as well as equipotential points in this.
     
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