So here's a picture of a potentiometer from my book. And my book says that the potentiometer is used for measuring internal resistance (of a cell), emf (of a cell) etc. But I cannot understand the concept behind this as how to find the emf of a cell? Also how to find the internal resistance of the cell? Could someone please explain this concept?

  • \$\begingroup\$ Is the book written in English? I'm wondering if it's a mistranslation because I've never heard of potentiometer being used to refer to anything other than a variable resistor (which the above isn't). Edit - see that it can have a different meaning en.wikipedia.org/wiki/Potentiometer_%28measuring_instrument%29 \$\endgroup\$
    – PeterJ
    Commented Jan 19, 2014 at 6:23
  • \$\begingroup\$ Yes, it is a kind of variable resistor because the wire between A and B is not just a wire, actually it is a resistor whose resistance can be adjusted by adjusting the length(because length is directly proportional to the resistance) AJ and JB using the jockey which acts as the VIPER of potentiometer. It means that you can also use a variable resistor(the three legged one) which we also say as a potentiometer. Now can anyone explain the concept of potentiometer?? \$\endgroup\$ Commented Jan 19, 2014 at 6:34
  • \$\begingroup\$ Maybe your book means its a resistance which can be introduced in parallel, so that by measuring voltage across it we can measure the voltage of the cell. \$\endgroup\$
    – Sherby
    Commented Jan 19, 2014 at 6:43
  • \$\begingroup\$ @TheDreamCoder17 Is there an explanation of what K1 and K2 are? Also is there an explanation for R2? Are the show components meant to be ideal or not? If they are, I can't really see what R2 would be doing. R1 could be useful, since it would prevent the potentiometer from burning out in extreme settings. \$\endgroup\$
    – AndrejaKo
    Commented Jan 19, 2014 at 9:15
  • \$\begingroup\$ Anyway, we could measure the EMF of the bottom cell in in following way: First, connect a voltage which we expect to be higher than the cell voltage in place of the top cell. If the bottom cell is disconnected, the resistance which the top cell sees is equal to R1+Rpot. Note that Rpot is constant. Then we connect the bottom cell and slide J until G shows zero current. At that point, I think we can use the voltage divider rule to calculate the voltage of the bottom cell: Vcell=(Raj)/(R1+Rbj+Raj)*Vtopcell \$\endgroup\$
    – AndrejaKo
    Commented Jan 19, 2014 at 9:20

2 Answers 2


Yes, indeed while "potentiometer" nowadays means "variable resistor" - a fresh glance at the word itself will show that it really means "potential measuring device".

And that was its original purpose.

I have seen and used a real Potentiometer exactly once, at a six hundred year old university, as an exercise. (The lab equipment was much newer, and quite splendid!)

EDIT : the form of potentiometer described here is also known as a "metre bridge".

It is a length of resistance wire, stretched out along a board (mahogany, of course) over a metre ruler, with a very solid brass (what else!) bar alongside it, and a sliding right-angle knife-edge contact between the bar and the resistance wire. It is connected as your illustration shows, to the top cell, an unknown but steady voltage source (usually a 2V lead acid cell) with an unknown series resistance. The precise values of V,R don't matter.

The metre rule can be engraved on the brass bar, and with a Vernier scale on the sliding contact, the instrument can give four digit accuracy, better than many a digital multimeter today.

I believe K1 and K2 are brass tapered pegs which fit into tapered gaps in the brass bar, i.e. switches.

Now carefully connect up your unknown cell via G, which is preferably one of Lord Kelvin's fine mirror galvanometers, moving a focussed spot of light to amplify the smallest motion, and adjust the knife edge contact for zero current. At this point you have a length (measured from the ruler) representing the potential of the unknown cell.

Because you don't know the driving voltage, you also need to make the same measurement with a "standard cell" usually a Weston cell to get the length representing 1.01864 volts; the ratio between them gives you your actual cell voltage.

CAUTION : if you are careless in this step and draw significant current, you will wreck your valuable standard cell.

Now you can return to the unknown cell. Re-establish its open-circuit voltage (length) - any change shows that the driving cell needs to be recharged.

Then insert peg K2, connecting known resistance R2, and establish the new voltage (length) when driving current through R2.

The rest is mere arithmetic...


Brian's answer is quite thorough.

I'll just contribute a photo because I don't think many people are aware of what these things looked like.

Here is an instrument called a "portable potentiometer". It is a bit disturbing to find an instrument that I have used to make serious measurements and calibration of thermocouple instrumentation as a museum display.

enter image description here


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