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)?

Could you also please use numbers in your explanation?

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


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