How is the Wheatstone Bridge equation affected when VG is not zero?

In summary, the student is asking for help with a homework equation but is not sure how to solve it if the voltage in the middle, VG, is not zero. If VG is not zero, then the equation cannot be solved.
  • #1
sciencegirl1
30
0
I need help! WHEATSTONE!

Homework Statement


http://en.wikipedia.org/wiki/Wheatstone_bridge

My pic. is almost same as this one, but R2 on the pic. is Rx at mine, Rx at the pic is R4 at mine. Pleas draw it up and help me:


Homework Equations



VE(voltage at left) is 9 V.
VG(voltage in the middle) is 3V
R1=R3=R4=10 KOhm.

What is Rx?

The Attempt at a Solution



I do not know how to solve it if VG is not zero.

 
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  • #2


Not a very well asked question! If you are asking people to help, make the small effort yourself of leaving the problem as in the diagram and, if you need to, converting the names of the things yourself after you have the answer! :smile: The names don't change the problem. Also it is not clear to me what you mean by E and G, left and right.

However a point you might have missed is that you are supposed to understand that there is no current through the voltmeter, i.e. through the path DB even if there is a potential difference across DB. This is because voltmeters have high resistance and take very small current so this can be ignored in solving the problem. (That is how most practical voltmeters work - they are really not voltmeters at all but high-resistance ammeters.)

The other point is, not to look at it in unnecessarily complicated fashion, the voltage drop across two resistors in series (when the same current is flowing through both i.e. none in the sideways branch in the diagram - as said above) is proportional to the resistances i.e. VAB/VBC = RAB/RBC (i.e. R3/Rx) in the wiki diagram.) That follows from conservation of current (KCL) and the definition of resistance.

Quite soon you will also learn to calculate everything even if there is current DB if you are given enough information.
 
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  • #4


But I did use the formula I gave you here above and got Rx=1000
Is that what you get?
 
  • #5


If you find it simpler. :biggrin:

Seems to me they make rather heavy weather of it, probably because it is a computer thing. Their starting VB= Vin*[Rx/(R3+Rx)- R2/(R1+R2)] is the same as mine essentially.

I am now confused between the various diags but you should know you are not going to get 1000V or anything greater than 9V out of that circuit!
 
  • #6


I´m trying to find Rx (the resistance) not the volts.
If you do understand this, could you please do me a favour and come with the correct formulas and show me how you finish calculating the example. It would help alot.
thanks
 
  • #7


Unfortunately, the question as you've stated it is ambiguous. "The voltage in the middle" isn't enough. Voltage is relative, so you need to specify voltage as the left node with respect to the right node, or the other way around.

Find out which way it's to be, and see if you can rearrange the equation and solve for Rx.
 
  • #8


The voltage at the left is Vs (where the + is)

http://www.daycounter.com/Calculato...WheatStone- --> Bridge-Calculator.phtml

this link might help you understand what i mean. The pictures shows the circuit
 
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  • #9


As phrak confirms you make it too hard for us to be sure what the question is.

Taking your explained changes of the wiki pic, I will assume also you mean VAC = 9V , and VDB = 3V. (i.e. that D is positive wrt B).

I am free to let VC = 0. Then VA = 9 V.

From my type of ratio formula in first post, since R3 = R4 it is particularly simple, the voltage drop A-B is same as B-C. So VB is half way - is 4.5V. VD is (given data) 3V higher than VB so is 7.5V. So the voltage drops VAD and VDC are 1.5 and (subtract this from 9) 7.5 V respectively. Simple ratio 5:1. The resistances as I said are in that same ratio so Rx should be 50 K ohm I think. If the problem is not, or with other probs, other data, that you should be able to do it along same lines.
 
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  • #10


I can´t see what´s so difficault for you to understand because I explain this perfectly.
I´m simply looking for a way to calculate Rx.
This is not the way.
 
  • #11


sciencegirl1 said:
I can´t see what´s so difficault for you to understand because I explain this perfectly.
I´m simply looking for a way to calculate Rx.

:confused: I did it in last post! And the rules are roughly we are not supposed to do that!

sciencegirl1 said:
This is not the way.
It's wrong?
It might be, well seem, a different way through than you did in class, but there are different ways. In this case not much different and all equivalent.

(In #5 I confused V with ohms late at night. However your 10 K ohm would make all four resistances equal. Then there would be 0 voltage across DB, contrary to what you told us, so 10 K ohm must be wrong.).
 
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  • #12


I´m sorry 50 was the correct answer. You were right.

You seem really smart. Can you beat this:


If VE is 9V and R4 is changeable and we let it make VD=0. What would you let R1 , R3 and R4 be to maximize the sensitiveness of the measurement dUD/dRx ?
 
  • #13


sciencegirl1 said:
I´m sorry 50 was the correct answer. You were right.

You seem really smart. Can you beat this:


If VE is 9V and R4 is changeable and we let it make VD=0. What would you let R1 , R3 and R4 be to maximize the sensitiveness of the measurement dUD/dRx ?

I have never had occasion to think about this and had a quick look, but I cannot be here for a few days, not even for flattery, perhaps someone else will intervene. In my quick look I was a bit surprised but am maybe rationalising the results. I have no bit advantage over you. They give you the mathematical definition of sensitivity. You have an equation such as that in #5. You have probably done elementary differentiation so can apply that formula. It will make it more interesting to you if you make a conjecture of what you think qualitatively will be the factors, to have the various resistances larger or smaller. Then try to do the math and see what comes out. It doesn't matter whether you get anything wrong as you are among friends.:smile:

On the side what do you think if I say, they say 9V but in measuring the Rx by this method it doesn't matter what the overall voltage is and you never need to know what it is, except that increasing the voltage will increase the sensitivity as defined. And that the accuracy of the result for Rx will never be better than the accuracy of your knowledge of the other resistances R1 R2.. etc.?

If you do come back here I recommend you to use and refer to just the Wiki Wheatstone bridge and its labels to make it easy for people to follow.
 
  • #14


sciencegirl1 said:
You seem really smart. Can you beat this:

You are missing the point. We are not here to 'beat' whatever you throw at us. We can help you solve it by yourself.
 

1. How does a Wheatstone Bridge work?

A Wheatstone Bridge is a circuit used for measuring electrical resistance. It consists of four resistors connected in a diamond-shaped formation. By varying the resistances of the known and unknown resistors, the bridge can be balanced and the unknown resistance can be calculated.

2. What is the purpose of a Wheatstone Bridge?

The purpose of a Wheatstone Bridge is to accurately measure unknown resistances. It is commonly used in electrical and electronic circuits for troubleshooting and calibration purposes.

3. How accurate is a Wheatstone Bridge?

A Wheatstone Bridge is a highly accurate instrument, with the ability to measure resistances with precision up to four decimal places. However, the accuracy also depends on the quality of the components used in the bridge and the skill of the operator.

4. What are the advantages of using a Wheatstone Bridge?

One of the main advantages of using a Wheatstone Bridge is its high accuracy in measuring resistances. It is also a simple and straightforward circuit, making it easy to use and understand. Additionally, it can be used to measure a wide range of resistances, from milliohms to megaohms.

5. Are there any limitations to using a Wheatstone Bridge?

While a Wheatstone Bridge is a reliable and accurate instrument, it does have some limitations. It requires a power source and cannot be used for measuring resistances in live circuits. It also cannot measure resistances that are too low or too high for the range of the known resistors in the bridge.

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