Wheatstone bridge

Discussion in 'Homework Help' started by hitmen, Sep 22, 2008.

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  1. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    I have a qn about the wheatstone bridge. If the two ends are parallel, then where does the voltage flow? Is it in a forever loop?

    Also can anyone provide me with a link on how to calculate the pd across each resistor. I know how to simplify the circuit blindly. However, I do not know how to calculate their respective pdf.

    Thanks for helping.:D
     
    Last edited: Sep 24, 2008
  2. mik3

    Senior Member

    Feb 4, 2008
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    The voltage does not flow, the one that flows is the current through the resistors due to the applied voltage.

    By two ends you mean the two branches of series resistors?

    The power dissipation for a resistor equals:

    P=V*I=(V^2)/R=(I^2)*R

    where
    P=power dissipation
    V=voltage across the resistor
    I=current through the resistor
    R=resistance of the resistor

    You can choose on of the three above equations!
     
  3. studiot

    AAC Fanatic!

    Nov 9, 2007
    5,005
    513
    The whole point of a wheatstone bridge is that there is no current through the detector.

    This occurs because the potential (voltage) is the same at both nodes connected to it.

    So the signal, ac or dc, is applied across two of the four corners (nodes) and the output (equals zero) is taken across the other two.

    This configuration is called a bridge circuit in general terms. In general terms each of four sides has an impedance or an active element.

    Specifically for a wheatstone bridge there are only resistors, and the signal is dc.

    Thus if the signal is applied across (r1+r3) in parallel with (r2+r4) and the detector is across the junction of (r1,r3) and the junction of (r2,r4):

    the voltage at the junction of r1 and r3 equals the voltage at the junction of r2 and r4

    This voltage = Vin x r1/(r1+r3) but it also equals Vin x r2/(r2+r4)

    This is the condition for balance.
     
  4. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    I guess i should upload a sample of the question I dont understand. The example has 2 parts, both of which I do not understand. Firstly, In which direction does the current from the 156V flow? Also, how do I know which resistor is in parallel or series with another.

    I will gladly appreciate anyone explaining the concepts.

    The circuit looks like a very complicated Wheatstone bridge.

    In part 2 of the question, my lecturer gives only 2 loop equation. Isnt a three one needed for current flowing through the source?:confused:

    Many thanks for answering.

    Also, is there a better way of uploading 1 PAGE from a PDF instead of cutting and pasting the image onto paint?
     
  5. silvrstring

    Active Member

    Mar 27, 2008
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    hitmen,

    You do need to use two equations because you have two variables----I1 and I2. Two loops, two equations. You don't need a 3rd eq for the source. It is accounted for in the loop equations.

    Your first equation is off. It should read (I1 - I2)*4 + (I1)*2 + (I1)*4 - 52 = 0
    Your second equation (loop) will be (I2 - I1)*4 + 52 + (I2)*4 + (I2)*4 = 0.
    Simplify them and solve for each loop current, then you can find your E(th) accross those two 4ohms resistors closest to the R(L).

    For part c), the maximum power transfer theorem proves that the greatest amt of power transferable to R(L) will happen when R(L) is equal to R(th). So I think you need to find the R(th) before you can go further. You will need to do a delta-wye conversion so they will be in a more workable form. Then you will easily be able to determine the R(th).

    If there is an easier way, let me know. But this will work.
     
  6. hitmen

    Thread Starter Active Member

    Sep 21, 2008
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    I dont understand part of of how the Rth is obtained. The answers were provided by the professor.

    You said that the third loop linking to the source is "taken care of". how is that so?

    Thanks for answering.
     
  7. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,283
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    Replace the 156 volt source with a short; connect a 1 volt source to the A-B terminals. Solve for the current drawn from the 1 volt source.

    I think if you choose carefully, 2 loops will be enough.
     
    Last edited: Oct 7, 2008
  8. kai88

    New Member

    Jan 26, 2010
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    Since,Output voltage = Vin x r1/(r1+r3) but it also equals Vin x r2/(r2+r4),

    So if i change the Input voltage(due to instability) ,will it affect output voltage(Indicator)?

    I juz wan to know whether wheatstone bridge is power-source dependent or power source independent(like ratiometer type)?
     
  9. kai88

    New Member

    Jan 26, 2010
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    So if i change the Input voltage(due to instability) ,will it affect output voltage(Indicator)?

    I juz wan to know whether wheatstone bridge is power-source dependent or power source independent(like ratiometer type)?
     
  10. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    You are hijacking an old thread, and have another thread open on this subject elsewhere.
     
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