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Current in Wheatstone Bridge

  1. Apr 17, 2015 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    The ammeter in the Wheatstone bridge of measures zero current when the resistance Rvar of the variable resistor is set to 193Ω .
    What is the current IL through the left side of the bridge?
    2. Relevant equations
    V1 + V2 + V3 + V4 + ..... + VN = E
    Rx = Rvar*R3/R1
    m0lzz0W.png
    3. The attempt at a solution
    To start, I calculated the equivalent resistance of the diamond part in the center, which is [itex] (\frac{1}{70+139} + \frac{1}{210+579} )^{-1} [/itex], where I calculated the R6 to be 579 Ω.

    I used this and V1 + V2 + V3 + V4 + ..... + VN = E to find the voltage across the whole diamond part (using Kirchoff's Current Rule). The current into the diamond should be equivalent to the one out of it, 0.04 A. Using this knowledge, and the voltage I calculated across the diamond part, and divided it by the equivalent resistance on the left side, which should give me current. I found the current on the right side the same way, and sure enough they added to 0.04 A. I got an answer of 0.02382 A for the left side, which is wrong.

    Another thing I tried is considering only the loop through the left side, it's voltages still will sum to zero, so I found the voltage across the left side in a similar manner and divided by the resistance and got 0.0324 A, a different number. But I have one attempt left and don't want to try it. Any EEs here to help?
     
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  3. Apr 17, 2015 #2

    ehild

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    Is the variable resistance 193 ohm or 139 ohm?

    How did you get 0.04 A for the current through the diamond?
     
  4. Apr 17, 2015 #3

    B3NR4Y

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    It is 193 ohms, to get 0.04 A for the current through the diamond I calculated the equivalent resistance of the diamond, then added it to the two other resistors that form a loop with the diamond (50 and 150), then I divided the Emf of the battery by this resistance and got my answer.
     
  5. Apr 17, 2015 #4

    ehild

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    The diamond is not connected in series with the other two resistors to the left.
     
  6. Apr 18, 2015 #5

    B3NR4Y

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    Then it's connected in parallel?
     
  7. Apr 18, 2015 #6

    SammyS

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    No. Not parallel either.

    The diamond is in parallel with the 350 Ω resistor.
     
  8. Apr 18, 2015 #7

    B3NR4Y

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    I understand that, but the only two things we have learned about are parallel and series...
     
  9. Apr 18, 2015 #8

    ehild

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    You can solve the problem with series and parallel resistors, but you need to know what series connection and parallel connection means. What do you know about the current flowing through series resistors? What do you know about the voltage across parallel resistors?
    You calculated the resultant resistance of the diamond. You can replace it with the resultant.
     
  10. Apr 18, 2015 #9

    B3NR4Y

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    Current through series resistors is the same, voltage through parallel resistors are the same.
     
  11. Apr 18, 2015 #10

    ehild

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    Splendid. Is the current through the diamond equal to I1? Is the voltage across the 350 ohm resistor equal to the voltage between b and d?
     
  12. Apr 18, 2015 #11

    B3NR4Y

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    I-1 is not the current through the diamond, because there's a junction there. And the voltage across the 350-ohm resistor is equal to the voltage across b and d.... I think. Circuits confuse me.
     
  13. Apr 18, 2015 #12

    ehild

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    So how are the diamond (Rd) and the 350 ohm resistor connected?

    wheatstone.jpg
     
  14. Apr 18, 2015 #13

    B3NR4Y

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    The diamond and the 350 ohm resistor are connected in parallel, and therefore should have the same voltage across them? And from there I can find the current in the left side of the diamond because I know the voltage across it?
     
  15. Apr 18, 2015 #14

    ehild

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    How do you know the voltage between b and d?
     
  16. Apr 18, 2015 #15

    B3NR4Y

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    Because voltages in parallel resistors are the same, and I can find the voltage through the 350-ohm resistor fairly simply, which is in parallel with the diamond.
     
  17. Apr 18, 2015 #16

    ehild

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    Yes, how do you find the voltage across the 350 ohm resistor? Show it, please.
     
  18. Apr 18, 2015 #17

    B3NR4Y

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    Firstly, the 350 ohm resistor is in series with the two resistors in the left, if you ignore the diamond, so sum the resistances and get 550 ohms. Now divide the emf of the battery by this resistance to get the current through this loop, 0.027-A. Now, by kirchoff's law, emf = V-1 + V-2 + V-3, or E-I(R-1 + R-3) = R-2 (where R-2 is the 350 ohm and R-1 is the 50, R-3 being the 150). This is equivalent to the voltage across the 350 ohm resistor. 9.5454 volts
     
  19. Apr 18, 2015 #18

    ehild

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    No, it is not series with the two resistors on the left. Check the currents. Is I1 the same as I2?

    Resistors are connected in series, if nothing else is connected to their common point. They must look as pearls of a necklace.

    Determine the parallel resultant of the 350 ohm with the diamond. That resultant is in series with the 150 ohm and 50 resistors.
     
  20. Apr 18, 2015 #19

    B3NR4Y

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    If they're not in series, and they're not in parallel, the only thing I can think of is 0 V across the 350 ohm resistor, and I don't think that makes sense.
     
  21. Apr 18, 2015 #20

    ehild

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    Resistors in a circuit are not all parallel or all series. Why do you think that there is 0 V across the 350 Ω resistor? Does current flow through it? If the current is not zero, is the voltage across it zero?
     
    Last edited by a moderator: Apr 18, 2015
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