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Experiment using a Wheatstone Bridge

  1. Dec 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Determine the unknown resistance using a Wheatstone Bridge.

    2. Relevant equations
    Unknown Resistance = known resistance (100-Length) / Length

    3. The attempt at a solution
    I've connected the set up as shown in the pictures but my only concern is, I don't have a resistance box so I'm using those blue boxes which seem like a resistance box to me but with just one resistance instead of many. I connected the wire of unknown resistance as shown, I stripped off the insulation of the copper wire and then wrapped it around. The null point I'm getting it not in the middle, it's closer to the higher potential and I get the resistance of the wire to be approximately 2.2 ohms when I use the 10 ohm blue resistance box. Is that correct?

    Secondly, the circuit diagram provided to us also has a rheostat, what use is a rheostat in this experiment?

    Thanks
     

    Attached Files:

  2. jcsd
  3. Dec 10, 2016 #2

    Nidum

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    It is not very clear what you are doing .

    Please post the circuit diagram that you have been given and show what components you have used in each part of the circuit .
     
    Last edited: Dec 10, 2016
  4. Dec 11, 2016 #3
    The second picture shows the complete set up which most definitely resembles a wheatstone bridge. Two ends are connected to the power supply. Left gap to a known resistance, right gap to an unknown resistance and the connection in between the left and right gap is made to the galvanometer which is connected to the jockey. Switch on the power supply, touch the meter string on one end with a jockey, it should show a positive deflection of the galvanometer (the end at the higher potential) and the other end should show negative deflection. This will ensure your wheatstone bridge is working. Then find the null point where the deflection is zero and measure that length. Finally substitute it in the equation provided above to find the unknown resistance.
     
  5. Dec 11, 2016 #4

    cnh1995

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    Is it this experiment?
    images (4).png
     
  6. Dec 11, 2016 #5
    Yes, but the problem as I was mentioning earlier is that we don't have a resistance box in the lab. So we used those little blue boxes as shown in the picture which are we're guessing fixed resistance boxes in place of the known resistance and a copper wire for the unknown resistance.
     
  7. Dec 11, 2016 #6

    cnh1995

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    Why are you using two blue boxes in the setup in the second image? You only need one, in the left gap. Have you connected the unknown resistance wire in the right gap? Is that the right gap in the last image?
     
  8. Dec 11, 2016 #7
    Yes I was only using it to test whether I get the other resistance or not. I later replaced it with a copper wire, yes I connected it in the right gap. So the blue box is fine?

    Secondly, they suggest we should use a rheostat in the set up, what use will a rheostat be? Thanks
     
  9. Dec 11, 2016 #8

    cnh1995

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    Yes. The null point should be on the right side of the meter scale, at about 18cm from the right end if you are getting 2.2 ohms as the unknown resistance.
    Looks ok but verify it with a digital multimeter. I think the length and resistivity of the wire are too small to have a resistance of 2.2 ohm. I may be wrong, I'm not sure.
     
  10. Dec 11, 2016 #9

    cnh1995

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    I see nothing special in it.
    Using a rheostat, you can take several readings for the null deflection point by varying the rheostat's position and find the unknown resistance using each reading. The final value for the unknown resistance will be the average of all the calculated values. This method will be more accurate.
     
  11. Dec 11, 2016 #10
    Should I use higher known resistances or 10 ohms is fine? I have another box of 5 ohms and 20 ohms so I can use those too or I have a colour band resistor I can use by attaching it with crocodile clips. Thanks
     
  12. Dec 11, 2016 #11

    cnh1995

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    Well, I think you should use even smaller known resistance (5 ohm would work) for more accuracy in this case since the unknown resistance is very small.
     
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