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Find the resistance of a resistor using Ohm's Law?

  1. Jan 21, 2016 #1
    • Thread moved from the technical forums, so no HH Template is shown.
    Hi, I have been at this problem for days and I can't seem to see what I am doing wrong.

    Here is the circuit layout along with my work
    Screen Shot 2016-01-21 at 12.03.50 PM copy.png
    As you can see I am running into problem getting values of current that do not add up equal to zero. My professor has never discussed how adding a wire like this would effect the circuit. I am assuming the other 0.5A go through the wire? I don't know. Everything written in blue is what I have added (and the red arrow).
     
  2. jcsd
  3. Jan 21, 2016 #2

    berkeman

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    Ix - I3 is not necessarily zero, because of the wire.

    BTW, since you are given R2 and I2, you know the voltage on the shorting wire... :wink:
     
  4. Jan 21, 2016 #3

    cnh1995

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    I believe the simpler way to solve this circuit is just by inspection, without writing too many equations. Berkeman has given an important hint! Assume ground(0V) at the -ve terminal of the source.
     
  5. Jan 21, 2016 #4
    I thought I already found the voltage of V2 to be 4V? I am not quite sure how that specific element is responsible for the voltage of the wire?
     
  6. Jan 21, 2016 #5

    cnh1995

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    That is the voltage 'drop' across the 8Ω resistor. If you assumed ground at the -ve termianl of the source, then potential of the wire will be 4V. You'll have to fix the ground first.
     
  7. Jan 21, 2016 #6

    berkeman

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    Actually, the 4V drop across that resistor results in a voltage that is 4V below Vs... :smile:
     
  8. Jan 21, 2016 #7
    Hm, we have never discussed fixing a ground before. I am not understand how a wire could hold voltage or current if there isn't any element there. Does this mean all of my other calculations are wrong because I didn't account for the properties of the wire?
     
  9. Jan 21, 2016 #8

    cnh1995

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    Wire is assumed to be an ideal conductor. It's potential is same throughout it's length i.e potential difference across the wire is 0. So, whatever is the potential of the point to which the wire is connected, it is same on the wire everywhere. As the wire is connected to a 4V point(assuming ground at the -ve terminal), its potential is 4V.
     
  10. Jan 21, 2016 #9

    berkeman

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    Correct.

    It may help you to re-draw the circuit. Make the Vs source be vertical, with ground at the bottom and +10V at the top. Then re-draw the (positive side of Vs) resistors as 2 parallel resistors that connect to that shorting wire, and 2 more resistors in parallel below the shorting wire that connect to ground. You already know the voltage of that common shorting wire point...

    Edit -- sorry, I thought the Vs source had + to the right, but it is to the left. So when you re-draw the circuit, rotate the Vs source clockwise 90 degrees.
     
  11. Jan 21, 2016 #10

    cnh1995

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    Once you have fixed the ground at the -ve terminal, you have 3 points in the circuit with known potentials. That will be sufficient to find the currents in all the resistors. Plus, you know the voltage across Rx.
     
  12. Jan 21, 2016 #11
    Thank god for photoshop.
    Maybe?.png
    I am not sure what you meant by switching around vs? I hope I drew this correctly.
     
  13. Jan 21, 2016 #12

    cnh1995

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    Correct!
     
  14. Jan 21, 2016 #13
    So I can combine the resistance of R1 and R3 to equal Req=2Ω? Alright, I have to go to another class right now but I will sure to post my attempt of a solution by 5pm EST :)
     
  15. Jan 21, 2016 #14

    cnh1995

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    Yes.
     
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