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Resistance of a network between two points

  1. Sep 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the resistance of the network shown below: (a) between A and B, (b) between A and C.

    a1024cb4248b.jpg

    Answers: (a) 3 Ω, (b) 4 Ω

    2. The attempt at a solution
    (a) I would say that the resistance is 4 Ω between A and B.

    (b) For A and C the resistance is: (1/4 + 1/4)-1 = 2 Ω.

    In both cases the answers are wrong. What am I missing?
     
  2. jcsd
  3. Sep 27, 2016 #2

    kuruman

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    In (a) you treat the problem as if the three resistors along path ADCB were not there.
    In (b) you treat the 4-ohm resistors along AC as if they were in parallel. Are they?

    On edit: We cannot help you find out what you are missing if you don't show the relevant equations and how you used them.
     
  4. Sep 27, 2016 #3
    I looked in my studybook and in Google, but I couldn't find a similar problem, only something like this (from here). If we use 1/R = 1/R1 + 1/R2 + 1/Rn we'll get 1/R = 0.25 + 0.25 + 0.25 + 0.25, so R = 1 Ω. No idea what to do with it without V or I (current).
     
  5. Sep 27, 2016 #4

    kuruman

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    You need to recognize parallel and series combinations in each case, then use the equations to replace these combinations with their equivalent resistance. However, it seems you are confused about which is which. Can you answer the following 4 questions?

    1. What do parallel resistors share in common, voltage or current?
    2. What do series resistors share in common, voltage or current?
    3. 1/Req. = 1/R1 + 1/R2 + ... +1/Rn is for parallel or for series?
    4. Req. =R1 + R2 + ... +Rn is for parallel or for series?
     
  6. Sep 27, 2016 #5
    1. Parallel resistors have V/R = V/R1 + V/R2 + ..., the same V is across each resistor.
    2. Series resistors have V = V1 + V2 + ..., where V1 = IR1. The same current flows through each resistor.
    3. This is for parallel.
    4. This is for series.

    In our case I don't understand whether it's a parallel one or a series one. It looks like a series.
     
  7. Sep 27, 2016 #6

    kuruman

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    Good. Now let's look at (a). Imagine a battery connected between A and B. In view of your answers to the first two questions, can you recognize which combinations of resistors are in parallel and which are in series?
     
  8. Sep 27, 2016 #7
    This is how I see it:
    1aad6b53f8a9.jpg

    They are all in series because as in my book example all the parallel ones are separated with black dots (in every example in my book, not only in this one). Like in the example 10 and 15 Ohm are parallel because of the two black dots. However, 4 and 2 Ohm resistors aren't parallel but in series, even though they are under 90 degrees to each other.

    That's how I see it at this moment...
     
  9. Sep 27, 2016 #8

    kuruman

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    Your drawing is not correct. Imagine a battery connected at A and B. Start by drawing the square of resistors as given to you. Then draw one vertical line at point A going up and another vertical line at point B also going up. Draw black dots at A and B and a battery across the free ends of the vertical ends that you drew. Do you see what's going on now?
     
  10. Sep 27, 2016 #9
    Something like this?
     
  11. Sep 27, 2016 #10

    kuruman

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    Almost. You need to erase one battery, say the one on the right and connect the negative terminal of the battery on the left to point B. Then you will have a complete circuit. Now imagine current starting at the battery. It reaches point A and splits in two parts. Which resistors are in series and which are in parallel?
     
  12. Sep 27, 2016 #11
    Is this what you suggest?

    Calculations:
    (a) 1/R = 1/4 + 1/(4+4+4) so R = 3 Ohm.

    (b) 1/R = 1/(4+4) + 1/(4+4) so R = 4 Ohm

    Update: only one battery now here, I think this is what you meant.
     
  13. Sep 27, 2016 #12

    kuruman

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    You got it.
     
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