# Intuition behind swapping the known and unknown resistance in a slide wire meter bridge

Today I was learning about wheatstones and balanced meter bridges. But while learning about the sources of error while working with the meter bridge (measuring unknown resistance), I couldn't understand why the known and the unknown resistance was swapped. I even googled up a lot and got answers like "It helps to get the error due to stray resistances out". But unfortunately, this didn't help me even a little.

Can you please tell me whether it is a good practice to do so and how does it reduce the error (the reason behind doing it)?

• You might be luckier asking this question in Electrical Engineering community: electronics.stackexchange.com – TheAverageHijano Feb 13 at 21:28
• won't this site help me? – user8718165 Feb 13 at 21:33
• Hopefully someone will give you an answer, but engineers have more experience with wheatstones bridges than physicists. – TheAverageHijano Feb 13 at 21:35
• won't my question be marked as duplicate if i ever do that? – user8718165 Feb 13 at 21:38
• I don't know, I'm relatively new to the site. If you contact a moderator he should be able to help you. – TheAverageHijano Feb 13 at 21:45

A metre bridge consists of a wire which is supposed to be exactly $$100.0\,\rm cm$$ long with the wire ends termination at $$0.0 \,\rm cm$$ and at $$100.0\,\rm cm$$.

However there may be end correction $$e$$ and $$e'$$ at the end of the wires.

The standard way to find these end corrections is to use two known resistors $$R_1$$ and $$R_2$$ and measure two balance lengths $$x_1$$ and $$y_1$$.

So the balance condition is $$\dfrac{R_1}{R_2} = \dfrac{x_1+e}{y_1+e'}$$

Now the two known resistors are interchanged and a new balance position found $$\dfrac{R_2}{R_1} = \dfrac{x_2+e}{y_2+e'}$$.

You have two equations so you can solve for the two unknowns $$e$$ and $$e'$$.

Your question has one of the resistors as an unknown and I will try and illustrate what might happen numerically.

Suppose $$R_1=4.000 \, \Omega$$ and $$R_2$$ is unknown and allow me to quote the values to an extra significant figure.

The following readings are obtained $$x_1=568.9\,\rm mm$$ and $$y_1 = 428.1\,\rm mm$$, and $$x_2=426.1\,\rm mm$$ and $$y_2 = 573.9\,\rm mm$$

The first set of readings gives $$R_2 = 3.031\,\Omega$$ and the second set gives $$R_2 = 2.974\,\Omega$$ giving an average of $$2.970 \,\Omega$$.

In fact I reverse engineered the calculation and started with $$R_1=4.000 \, \Omega$$ and $$R_2=3.000 \, \Omega$$ with end corrections $$e=+2 \, \rm mm$$ and $$e'=-3 \, \rm mm$$.

So if the end corrections had been known the two values of $$R_2$$ found by interchanging the resistors would have been the same and equal to $$2.999 \,\Omega$$ with there being rounding errors in the calculations.

Not knowing the end corrections and interchanging the resistors gave an average value of $$3.001 \,\Omega$$ which is closer to the actual value than either of the two individual values.