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Equivalent resistance between two nodes homework

  1. Feb 7, 2014 #1
    1. The problem statement, all variables and given/known data

    Determine the equivalent resistance Req as seen by the voltage source VS (the resistance between a and b)

    http://i.imgur.com/A1ueQnd.jpg


    2. Relevant equations

    Series-Parallel combinations

    3. The attempt at a solution

    http://i.imgur.com/K1ItpF0.jpg

    The only resistor that have both a and b on its terminals is R2 (12 ohms)

    So, is Req=R2=12 ohms?
     
  2. jcsd
  3. Feb 7, 2014 #2

    gneill

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    Staff: Mentor

    Nope. There are other paths that current can take starting at a and ending at b. Start simplifying the easy parts first, working from the left side of the circuit.
     
  4. Feb 7, 2014 #3

    CWatters

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    Science Advisor
    Homework Helper

    No. The Req is what you get if you resolve all the resistors (including R1) down to one new resistor.

    For example if you were to measure the current going from the voltage source into node "a" and called that "Is" then..

    Req = Vs/Is

    edit: Sorry my post crossed with the one from gneil.
     
  5. Feb 7, 2014 #4
    Thank you gneil, CWatter for your quick responses

    and it's better to hear (or read actually) as many opinions as possible

    So basically what I am gonna do is to find the equivalent resistance of all the resistors.

    So the idea is to follow the current's path from a until it gets to b

    but if I was asked to the find the equivalent resistance between f to e, the answer would be R2+R3+R4, isn't?
     
  6. Feb 7, 2014 #5

    gneill

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    Staff: Mentor

    Your best strategy is to systematically combine resistors where series or parallel opportunities arise, reducing the circuit complexity as you go. So yes, you could reduce R2, R3, and R4 to a single resistance value, replacing the three with a single resistor that is their sum.

    Your original diagram did not include labels e and f, but I presume you meant something like this:

    attachment.php?attachmentid=66384&stc=1&d=1391798952.gif

    So imagining that the circuit was severed at those points, the equivalent resistance seen "looking into" those terminals to the left would be Req = R2 + R3 + R4.
     

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  7. Feb 7, 2014 #6
    He's referring to e and f in his solution attempt:

    http://i.imgur.com/K1ItpF0.jpg

    The OP is really struggling with this; he's had help in another forum:

    http://forum.allaboutcircuits.com/showthread.php?t=94166

    Jason, the equivalent resistance seen between ANY two nodes in this circuit will include the effect of ALL the resistors in the entire circuit.

    The Req between a and b is not just the value of R1. The Req between e and f is not just R2+R3+R4; it involves ALL the resistors in the circuit.

    Imagine that you replace R2, R3 and R4 with a single resistor (call that single resistor Rx); what would be the value of that resistor? Now Rx is in parallel with R5; you could replace the parallel combination of Rx and R5 with another equivalent resistor (call it Ry).

    Ry is now in series with R6 and that series combination could be replaced with an equivalent resistor (call it Rz). Now Rz is in parallel with R8; replace that with another equivalent resistor (call it Ra). Ra is now in series with R7, which can be replaced with an equivalent resistor (call it Rb). Rb is in parallel with R1, and that is your answer.
     
  8. Feb 7, 2014 #7
    You got me :P

    It's true, I'm really really struggling with this, but I guess I'm getting the point.

    As long as the current flows through a resistor, it should be considered when calculating Req.

    Thank you, everyone :)
     
  9. Feb 7, 2014 #8
    Before you go away, try the calculation I described and post your result. We really want to help you and if you get the right answer, we'll feel like we were a help. If you don't, we can help you with your stumbling block.
     
  10. Feb 8, 2014 #9
    OK, where R1=12, R2=15, R3=25, R4=25, R5=30, R6=7, R7=4, R8=15

    The total resistance would be equal to 6.42 ohms?
     
  11. Feb 8, 2014 #10
    If I solve with exact rational arithmetic, I get Req = 132924/20773 = 6.39888

    I suspect you did everything correctly, but probably didn't carry enough digits in your calculations.

    Congratulations!
     
  12. Feb 8, 2014 #11
    wheeew (A long long one)

    Thank you for your concerning :)
     
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