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Current Electricity Concept

  1. Jul 29, 2006 #1

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    INTRODUCTION: I had been doing some current electricity froblems but now and then i'm having some problems with the distribution of current.I'm really confused. I have attached a figure with this thread, in which i've tried to combine all situations of current distribution in which i'm facing problems.
    THE EXACT PROBLEM: In the given figure, the current enters at A and leaves at B.What would be the current distribution in the wires in each of the following conditions:-
    (a)If all the wires, i.e. AB,AD,CD,BC have identical resistances.
    (b)If only AB has some resistance and the other three wires in the rectangle have 0 resistance.
    (c)If the resistance of AD is more than that of AB.
    (d)If the resistance of AD is less than that of AB.
    You may add if there are any more conditions which will differ from the above.
    WHAT MY BRAIN SUGGESTED: (a)I think current i will distribute as i/2 in each of the two wires AB and AD as they are symmetrical.
    (b) Here the whole current i will flow through AB.
    (c)&(d) I feel that the current will distribute accordingly, i.e. more current will flow through less resistance wire.But here i'm getting totally confused.
    CONFUSION: To explain why i'm getting confused, lets take an example. Let's say each of the wires in the rectangle, i.e.AB,AD,BC,CD has resistance 2 ohm each. Now according to (a) current i should distribute as i/2 in each of AB and AD. But again we note that the combined resistance of AD,CD and BC is 6 ohm. Now my question is:- "WHILE DISTRIBUTION OF CURRENT i AT THE JUNCTION A, SHOULD WE CONSIDER THE RESISTANCE OF AB AND AD ONLY OR SHOULD WE CONSIDER THE RESISTANCE OF AB AND THE COMBINED RESISTANCE OF (AD,CD,BC)?"
    CONCLUSION: I know that it's really difficult to understand someone's problem through text but i've tried my best to explain it.And believe me, I'm reall really confused.Please help.Thanks.
     

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  3. Jul 29, 2006 #2

    nrqed

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    It might take a long while to get the attachment appoved (especially given that in most of the americas it is night time right now!)

    Can you draw in Ascii or describe the way the wires are connected?
     
  4. Jul 29, 2006 #3

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    see there is a square with AB as the base such that ABCD forms a square in the anticlockwise direction. now current i enters at A and leaves at B.
     
    Last edited: Jul 29, 2006
  5. Jul 29, 2006 #4

    Hootenanny

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    I can't see your image as it has not yet been approved. However, you will do well to remember that the current in a parallel branch is equal to the resistance of the opposite branch divided by the total resistance multiplied by the total current. For example if we have a circuit which branches into to branches, one with a resistance (R1) of 2 ohms and one with a resistance (R2) of 3 ohms and the total current before the branch is 4A. Then the current in the first branch would be;

    [tex]I_{1} = \frac{R_{2}}{R_{1}+R_{2}}\cdot I_{T} = \frac{3}{2+3}\times4 = 2.4A[/tex]

    This page http://floti.bell.ac.uk/principles/Parallel_circuits/sld005.htm may help.

    Nice presentation btw!:smile:

    Edit: Pat got there before me.
     
  6. Jul 29, 2006 #5

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    ok. thanks but please try to see my image when it is approved. then u can answer according to the exact situation shown in the figure. anyway thanks a lot.
     
  7. Jul 29, 2006 #6

    Hootenanny

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    (a)If all the wires, i.e. AB,AD,CD,BC have identical resistances.
    In this case imagine you have placed a resistor on each length of wire. Now, can you sum of resistors at AD, DC & CD to represent a single resistance? Note that AD, DC & CD are in series and all are parallel to AB. In this case you have a simple parallel circuit.

    If only AB has some resistance and the other three wires in the rectangle have 0 resistance.
    You may wish to reconsider your answer here.
     
    Last edited: Jul 29, 2006
  8. Jul 29, 2006 #7

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    Current Is Teasing Me...

    CURRENT IS TEASING ME: I do get your answer but now let's consider another situation. lets have a cube with ABCD facing towards us. All edges have 2 ohm resistance each. Current enters at A and leaves at B as in the earlier case.
    WHAT MY BRAIN SUGGESTED: Here the junction at A has three paths for the current to move in. Since the three paths are symmetrical, current distributes as i/3 in each of the three edges connected to juction A.
    PROBLEM: Now my question is "WHY ARE WE CONSIDERING ONLY THE THREE EDGES CONNECTED TO JUNCTION A WHILE DISTRIBUTING CURRENT IN THIS CASE? WHY AREN'T WE CONSIDERING THE OTHER WIRES (i.e. AD,BC,CD,etc.) IN THE CASE OF A CUBE? WE HAD CONSIDERED THEM IN THE CASE OF THE SQUARE. BUT WHY AREN'T WE DOING SO IN THIS CASE? "
    CONCLUSION: I know i'm acting foolish but i'm really really confused. Thanks a lot for helping.
     
  9. Jul 29, 2006 #8

    Doc Al

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    What do you mean by saying that wires AB and AD are "symmetric"? That they have the same resistance? That's true, but what matters is the symmetry of the two paths between junctions A and B: path A-D-C-B versus path A-B. These paths are not symmetric: they have different resistances.
     
  10. Jul 29, 2006 #9

    nrqed

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    You *must* consider the entire path (AD, CD, BC). If a current can split at some point among two (or more) paths, you must compare the resistances along the *entire* paths.

    More answers below..


    No, for the reason given above. Actually, if you do the maths, if a current I enters pt A, one fourth of it will go through the branch AB and 3/4 will go through A-D-C-B (so that the current in the second branch is a third of the current in the first)
    :confused: Why?
    It's actually the opposite! The current has a possible path where there are no resistances at all, so it will go *all* along D,C,B!
    Yes, as long as "resistance of the wire" you are including the *entire* resistance of the path, so you must compare AB to AD+DC+CB. There will be more current in the one of the two branches that has the last resistance.

    Going back to your question, if the resistance of AD is more than AB, then it is obvious that the resistance of AD-DC-CB is more than the resistance of AB (since AD-DC-CB are in series). SO then we know for sure that there will be more resistance in the AB branch.

    But if we know that the resistance AD is smaller than the resistance AB, we can not tell in which branch there will be more current without knowing what the resistances of DC and CB are.

    Does that help?

    Patrick
     
  11. Jul 30, 2006 #10

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    Got It!

    THANKS: Ya now i got the current distribution as far as the square ABCD is concered only.
    MORE PROBLEM: What about the cube?(Please refer the quote above)
    LET'S PLAY: Now,what if we have to find the magnetic field at the centre of the square ABCD in the figure given below?
    WHAT MY BRAIN SUGGESTED: The current flowing through AB is (3/4)i while that through each of AD,CD and BC is (1/4)i. The magnetic field due to AB is vertically upward and that due to the other three is vertically downward and they add upto that produced by AB.So they cancel and the magnetic field at the centre is zero,right?
    I GOT IT: Now i got it.Thanks a lot.But if you do get time, please explain the current distribution in the cube as mentioned earlier.
     

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