# Proof of Wheatstone bridge equation

• whatdoido
In summary: Now you have an equation for the change in the voltage across A and B, in terms of the changes in the resistor values. What does that have to do with the Wheatstone bridge?Looks good. Now you have an equation for the change in the voltage across A and B, in terms of the changes in the resistor values. What does that have to do with the Wheatstone bridge?In summary, the equation ##\Delta U_{AB}=\frac{R_1R_4}{(R_1+R_4)^2}(\frac{\Delta R_1}{R_1}-\frac{\Delta R_2}{R_2}+\frac{\Delta R_3}{R_
whatdoido

## Homework Statement

Prove the following equation:

## \Delta U=\frac {R_1R_4}{(R_1+R_4)^2}(\frac {\Delta R_1}{R_1}-\frac {\Delta R_2}{R_2}+\frac{\Delta R_3}{R_3}-\frac{\Delta R_4}{R_4})E##

This is used in Wheatstone bridge

[/B]
U=RI

## The Attempt at a Solution

This has been a real head-scratcher

Two voltage dividers can be found for starters. Voltage's direction is assumed to be clockwise

##V_{in1}=I_2(R_2+R_3)##

##I_2=\frac{V_{in1}}{R_2+R_3}##

##V_{out1}=I_2R_3##

##V_{out1}=V_{in1}\frac{R_3}{R_2+R_3}##

Similarly:

##V_{out2}=V_{in1}\frac{R_4}{R_1+R_4}##

##V_G## is voltage between A and B

##V_{out1}-V_{out2}=V_G##

##V_{in1}\frac{R_3}{R_2+R_3}-V_{in1}\frac{R_4}{R_1+R_4}=V_G##

##V_{in1}(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})=V_G##

##V_{in1}=E##

##V_G=\Delta U## so then

##E(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})=\Delta U##

I have calculated voltages in different circuits and tried to think this problem in different ways, but the real problem is that how is ##\Delta R_i## inserted into equations. Assumption goes that it is added by ##R_i+\Delta R_i##. Maybe that is incorrect?

Help is very much appreciated!

edit: Misspelling corrected

Also particularizing that ##\Delta R_i## is a change in one resistance

#### Attachments

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whatdoido said:
that ##\Delta R_i## is a change in one resistance
Which suggests that the Δ in ΔU refers to the consequent change in U, not to the potential difference between A and B at a given set of R values.

haruspex said:
Which suggests that the Δ in ΔU refers to the consequent change in U, not to the potential difference between A and B at a given set of R values.
Yes that is true, ##\Delta U## is zero before the change of resistances.

whatdoido said:
Yes that is true, ##\Delta U## is zero before the change of resistances.
So this equation:
##E(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})=\Delta U##
##E(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})=U##
and you need a different expression for ##\Delta U##.

Right now I'm trying to figure out why this would not be possible:

##E(\frac{R_3+\Delta R_3}{R_2+\Delta R_2+R_3+\Delta R_3}-\frac{R_4+\Delta R_4}{R_1+\Delta R_1+R_4+\Delta R_4})=\Delta U##

I can simplify it a bit, but is this the right way to go

whatdoido said:
Right now I'm trying to figure out why this would not be possible:

##E(\frac{R_3+\Delta R_3}{R_2+\Delta R_2+R_3+\Delta R_3}-\frac{R_4+\Delta R_4}{R_1+\Delta R_1+R_4+\Delta R_4})=\Delta U##

I can simplify it a bit, but is this the right way to go
I do not see how you get that. It looks wrong.
You have an equation for U (second eqn in post #4). Write out the corresponding eqn for U+ΔU.

haruspex said:
I do not see how you get that. It looks wrong.
You have an equation for U (second eqn in post #4). Write out the corresponding eqn for U+ΔU.
Now that I thought about it, simply adding the change does not make so much sense.

But then I got an idea to take partial derivates since it is about change. Adding those partial derivates together should give the overall change in voltage.

##U=E(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})##

Seems like marking ##U## as ##U_{BA}## is needed since I took potential difference with ##V_{out1}-V_{out2}=V_G##

##U_{BA}=(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})E##

This should be legal: ##\Delta U_{BA}=dU_{BA}##

Thus ##dU_{BA}=\frac {\partial} {\partial R_1}U_{BA}\Delta R_1+\frac {\partial} {\partial R_2}U_{BA}\Delta R_2+\frac {\partial} {\partial R_3}U_{BA}\Delta R_3+\frac {\partial} {\partial R_4}U_{BA}\Delta R_4##

Solving partial derivates each:

##\frac {\partial} {\partial R_1}U_{BA}\Delta R_1=\frac {\partial} {\partial R_1}(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})E\Delta R_1=-\frac {R_4}{(R_1+R_4)^2}E\Delta R_1##

##\frac {\partial} {\partial R_2}U_{BA}\Delta R_2=\frac {\partial} {\partial R_2}(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})E\Delta R_2=\frac{R_3}{(R_2+R_3)^2}E\Delta R_2##

##\frac {\partial} {\partial R_3}U_{BA}\Delta R_3=\frac {\partial} {\partial R_3}(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})E\Delta R_3=-\frac{R_2}{(R_2+R_3)^2}E\Delta R_3##

##\frac {\partial} {\partial R_4}U_{BA}\Delta R_4=\frac {\partial} {\partial R_4}(\frac{R_3}{R_2+R_3}-\frac{R_4}{R_1+R_4})E\Delta R_4=\frac{R_1}{(R_1+R_4)^2}E\Delta R_4##

##dU_{BA}=-\frac {R_4}{(R_1+R_4)^2}E\Delta R_1+\frac{R_3}{(R_2+R_3)^2}E\Delta R_2-\frac{R_2}{(R_2+R_3)^2}E\Delta R_3+\frac{R_1}{(R_1+R_4)^2}E\Delta R_4##

##dU_{BA}=(-\frac{R_4}{R_1(1+\frac{R_4}{R_1})^2}\frac{\Delta R_1}{R_1}+\frac{R_3}{R_2(1+\frac{R_4}{R_1})^2}\frac{\Delta R_2}{R_2}-\frac{R_2}{R_3(1+\frac{R_1}{R_4})^2}\frac{\Delta R_3}{R_3}+\frac{R_1}{R_4(1+\frac{R_1}{R_4})^2}\frac{\Delta R_4}{R_4})E##

I just kept playing with the identity ##\frac{R_2}{R_3}=\frac{R_1}{R_4}## and I got:

##dU_{BA}=\frac{R_1R_4}{(R_1+R_4)^2}(-\frac{\Delta R_1}{R_1}+\frac{\Delta R_2}{R_2}-\frac{\Delta R_3}{R_3}+\frac{\Delta R_4}{R_4})E##

It has wrong signs because of ##V_{out1}-V_{out2}=V_G##

So I think the equation is about ##U_{AB}##

##\Delta U_{AB}=dU_{AB}=-dU_{BA}=-\frac{R_1R_4}{(R_1+R_4)^2}(-\frac{\Delta R_1}{R_1}+\frac{\Delta R_2}{R_2}-\frac{\Delta R_3}{R_3}+\frac{\Delta R_4}{R_4})E##

##=\frac{R_1R_4}{(R_1+R_4)^2}(\frac{\Delta R_1}{R_1}-\frac{\Delta R_2}{R_2}+\frac{\Delta R_3}{R_3}-\frac{\Delta R_4}{R_4})E##

Looks good.

## 1. What is the Wheatstone bridge equation?

The Wheatstone bridge equation is a mathematical formula used to determine the value of an unknown resistance in a circuit. It is based on the principle of balancing the arms of a bridge circuit to find the unknown resistance.

## 2. How is the Wheatstone bridge equation derived?

The Wheatstone bridge equation is derived from Ohm's law and Kirchhoff's laws. It involves using the ratio of the known resistances in the circuit to find the value of the unknown resistance.

## 3. What is the significance of the Wheatstone bridge equation?

The Wheatstone bridge equation is important because it allows for the accurate measurement of unknown resistances in a circuit. It is commonly used in various electronic devices and experiments in the field of electrical engineering and physics.

## 4. Can the Wheatstone bridge equation be used for non-linear circuits?

No, the Wheatstone bridge equation is only applicable to linear circuits where Ohm's law is valid. Non-linear elements, such as diodes and transistors, cannot be accurately measured using the Wheatstone bridge method.

## 5. Are there any limitations to using the Wheatstone bridge equation?

Yes, the Wheatstone bridge equation is only accurate if the resistances in the circuit are known and stable. Any changes in the resistance values can affect the accuracy of the measurement. Additionally, the Wheatstone bridge method is not suitable for measuring very small or very large resistances.

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