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Calculate current/equivalence resistance of this circuit!

  1. Aug 13, 2012 #1
    Hi guys! I'm new here. I've been lurking around for a while though and I have my first question that I can't seem to answer without finally making an account so I look to you all for help. Love the site!

    The following question is from my second semester physics class at UC davis.

    1. The problem statement, all variables and given/known data

    IMAG0041.jpg


    2. The attempt at a solution

    The reason I'm having trouble with this is because usually with circuits, we can usually tell right away if the problem is in series or parallel. This is confusing to me because I have no idea how to tell if it's in series or parallel. Does it matter which way I view it from? I tried to draw this out schematically but it still just doesn't make sense. I assumed that the middle three resistors were in series while the outer three resistors were in parallel but the answer I got doesn't match up to the answer on the answer key.

    If anyone could please help me with this problem and not simply tell me what the answer is, but HOW you got the answer, it would be greatly appreciated! THANK YOU!
     
  2. jcsd
  3. Aug 13, 2012 #2

    CWatters

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    Two separate hints:

    Look carefully at the bottom resistor. What's special about it?

    Are all the resistors the same value? If so look at the voltage on each end of the vertical resistor. What does that tell you about the current flowing in it?
     
  4. Aug 13, 2012 #3
    Hey there and thanks for your reply. I really apologize but I'm really not very good at circuits so I don't really know how to answer your two questions completely, so I will try my best.

    The bottom resistor: As far as I know, is what makes this resistor special is that it's connected to nothing but the battery? I'm really not sure :/

    And I don't know if it's specified that all of the resistors have the same value, the only thing that's given to me in the problem is that the voltage is 10V and current flow is going towards the left from the point the battery is at.

    I'm sorry, I'm just a bit confused and clueless right now as to how to even begin to tackle this problem.
     
  5. Aug 13, 2012 #4

    CWatters

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    Ok the bottom resistor is indeed connected in parallel with the battery. That means the current through it is fixed and changing the value of that resistor has no effect on the current through the other resistors. So that resistor can be ignored for the moment. Concentrate on the others...

    If the other resistors all have the same value then (by symetry) the voltage at each end of the vertical resistor is the same. If the voltage drop across a resistor is allways zero then no current flows through it. You could remove it or replace it with a wire and it would have no effect on the rest of the circuit. So remove it from the diagram and see what you are left with.
     
  6. Aug 13, 2012 #5
    Instead of drawing it as a triangle, think of the top 5 resistors as connected in two horizontal lines, with the vertical resistor bridging the middle. The bottom resistor directly connects with both terminals of the battery. So if you can collapse the top 5 resistors and find its equivalent resistance, you'll be able to find the equivalent resistance of the entire circuit.

    Looking at it that way, what do you notice about how the upper left two resistors and the upper right two resistors are connected? Can they be collapsed?
     
  7. Aug 13, 2012 #6
    CWatters this is what i'm left with. I'm confused as to the significance of leaving out the middle resistor.

    IMG_1015.jpg

    Arghzoo:

    IMG_1406.jpg

    what i notice about the other 4 resistors is that they are in parallel circuits. What do you mean by collapsed?


    A critical piece of information I left out was that the resistance of each resistor is 10 Ohms. I'm not sure what to do from here really.
     
  8. Aug 13, 2012 #7
    CWatters brought up the most important part: if the voltage at the ends are equal, no current can flow, and there might as well be empty space there. So, then the two top resistors and the two bottom resistors... Can be effectively "collapsed" into one top and one bottom.
     
  9. Aug 13, 2012 #8
    By "collapsed" I'm talking about how the resistances of two or more resistors connected in a particular way can be made into one resistor with a resistance that represents the resistance of all of them
     
  10. Aug 13, 2012 #9
    So is what you're saying is that I can collapse resistors 1/2 and 4/5? and the resistance of them altogether would still be 10 ohms? what about resistor 3 and the initial resistor that was connected to the battery? i'm having a hard time understanding what my overall picture would look like. also, in advance, I appreciate the help.
     
  11. Aug 13, 2012 #10
    If you've gotten to the point where you know the rules for resistors, they should tell you how to find the equivalent resistance of two resistors in series and in parallel. 1 and 4 are series, and so are 2 and 5. They're directly connected to each other. In this process, you can reduce the entire problem into one equivalent resistance.
     
  12. Aug 13, 2012 #11
    Look at top 2 branches of the circuit.
    If all resistors of equal value, the symmetry makes it equal current flow to each branch. That makes the potential difference of first 2 resistors equal.
    Thus no current flows in the vertical resistor.
    It is equivalent removing it from the circuit since it has no effect on total current, which imply no different in total resistors, with or without it.
     
    Last edited: Aug 13, 2012
  13. Aug 13, 2012 #12

    AGNuke

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    Click this link : WHEATSTONE BRIDGE

    The thing is, the resistors R1,R2,R3 and R4 are satisfying the balanced condition of wheatstone bridge (Their arrangement). Therefore, the potential difference across R5 becomes zero and no current flows through it. Therefore, it can easily be removed without affecting the overall circuit. Since there is no current, it is of no use in circuit. That's what other meant to "collapse" it.

    Wheatstone Bridge makes it easy for some problems to be solved. Better grasp it, as identifying it can be a pain sometimes.
     
  14. Aug 14, 2012 #13
    you guys are all great. I finally figured it out. very much appreciated. I look forward to coming back to this website again in the future. thanks again!
     
  15. Aug 14, 2012 #14

    CWatters

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    Just to clarify...

    What I (and a few others) have said only applies if the resistors have the same value. If they are all different values then the circuit is not symetrical and you have to use another approach.

    IF they are the same value then your last diagram is correct. The vertical resistor has no function and can be removed or shorted out. Then the circuit reduces to

    (two resistors in series) in parallel with (two resistors in series) in parallel with (one resistor).
     
  16. Aug 14, 2012 #15

    AGNuke

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    Not necessary. Its in the ratio. If those 4 resistors satisfied the balanced wheatstone condition, the 5th resistor can still be kicked out. No need of symmetry.
     
  17. Aug 14, 2012 #16

    CWatters

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    I agree. Would be nice to hear from the OP on the issue. Are/were values specified?
     
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