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Parallel Resistors

  1. Oct 11, 2015 #1
    1. The problem statement, all variables and given/known data
    52681b4468.jpg
    Combine the resistors into one single resistor

    2. Relevant equations
    Resistors in parallel = 1/1/R1 + 1/R2 +1/R3....
    Resistors in series = add

    3. The attempt at a solution
    I have two ways of getting the resistance.
    One is by saying that the 80ohm is parallel 20ohm and then adding the 16ohm which gives me 32ohm(which is the correct given answer) (1/1/80 + 1/20)+16 = 32ohm (correct)


    Another way I thought of is to say that the 16ohm and 80ohm are in parallel and then adding the 20ohm which gives me 33.3ohms which isn't correct. (1/1/80 +1/16)+20 = 33.3ohm (not correct)

    What I don't understand is how is the second method not giving me the same answer?
     
  2. jcsd
  3. Oct 11, 2015 #2

    phinds

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    If you treat the voltage source as a short circuit, which apparently you are doing, and you have a short, as drawn, across the output, then the three resistors are in parallel. Your second method treats one of them as being in series, which is not correct.
     
  4. Oct 11, 2015 #3
    I'm not treating the voltage as a short circuit.
     
  5. Oct 11, 2015 #4

    phinds

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    In that case, how do you ever get the 20 ohm resistor to be in parallel with anything?
     
  6. Oct 11, 2015 #5
    Good point. The way ive been told is that the resistors connected across the same two nodes are in parallel. Thats how I am getting the parallel sum of 16 and 80 ohm resistors and then adding that with the 20ohm resistor which is giving me 33.3 which is incorrect.
     
  7. Oct 11, 2015 #6

    phinds

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    correct
    But the 16 and 80 ohm resistors don't share both nodes.

    OOPS ... sorry. I got myself confused. Yes, that's how you got that far, but how did you then (in the first part) get the 80 and 20 to be in parallel, since THEY don't share both nodes
     
  8. Oct 11, 2015 #7

    CWatters

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    The expression "Combine the resistors into one single resistor" does not compute without further information. Why are you trying to combine the resistors?

    For example if you were trying to calculate the load seen by the voltage source (with the output short circuited as drawn) THEN method 2 would be correct eg (16//80)+20=33.3. On the other hand if you were trying to calculate the output impedance then method 1 would be correct.

    They give different answers because they are the resistance at different places in the circuit.
     
  9. Oct 11, 2015 #8
    Yeah. I looked through my notes again and realized that was wrong. after I posted it. I think the answer of 32ohms given may be wrong.

    This is MY final solution(idk if correct):
    b70a04234d.jpg

    The 16ohm and 80ohm resistor both share the two red nodes thus they are in parallel. So I apply the resistors in parallel formula.

    (1/1/16 +1/80) + 20 = 33.3ohms
     
  10. Oct 11, 2015 #9
    The question is actually from here http://people.clarkson.edu/~jsvoboda/eta/dcWorkout/thevenin.pdf
    However I don't need to find the voltage as I know how to do that. My issue is in the understanding of the parallel parts. So I adapated that example 1 from that link to show you guys what I am having trouble with exactly with no extra stuff. They get the thevenin resistance to be 32ohms and Im getting it as 33.3. The method they use of changing voltage sources into current etc. is something that we haven't done so I don't understand their solution fully.
     
  11. Oct 11, 2015 #10

    CWatters

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    They are correct the answer is 32R. All will be clear once you have covered Thévenin and Norton Equivalent Circuits.

    Note that their final circuit has a 16V voltage source rather than the original 20V so they haven't just combined the resistors. There is more to it.
     
  12. Oct 11, 2015 #11
    We've done thevenin but not norton. im just trying to find more examples since I havent really been given many examples for circuits
     
  13. Oct 11, 2015 #12
    If I short
    Can you tell how you would do this? getting the equivalent resistance here.
    73c46ad054.jpg
     
  14. Oct 11, 2015 #13

    phinds

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    What can you tell about the shared nodes of each of the resistors?
     
  15. Oct 11, 2015 #14

    gneill

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    So, it seems that what you're really looking for is the Thevenin Equivalent for your circuit (without the short circuit on the output, which I presume was added as part of the implied circuit analysis)?

    Fig1.png

    Otherwise, what would be the point since the short circuit would impose a 0 Ω result no matter what, right?

    The usual approach is to suppress the sources and determine the resistance presented to the output terminals by the remainder of the network. That would mean the the 20 V source is replaced with a short, thus making parallel the 20 and 80 Ohm resistors as you first surmised.
     
  16. Oct 11, 2015 #15

    donpacino

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    That black line on the far right of the circuit. why is it a different thickness and style?
    should it be there
     
  17. Oct 12, 2015 #16

    CWatters

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    See the reply by phinds. Mark up the nodes with a red dot.
     
  18. Oct 12, 2015 #17

    CWatters

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    Can I suggest you practice simplifying some other series/parallel resistor circuits and return to the problem in that PDF another time. That particular problem is somewhat different. It's not really about simplifying a resistor network on it's own and using it as the basis for that has ended up confusing you.
     
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