I'm a pre-university student, during the lesson of internal resistance r, there's actually resistance in battery or a cell, and if we were to connect it up in a simple set up as below: enter image description here

The situation I would like to mention here is when the switch is open. From what I have learnt in my reference book, it said that there's no current flowing when the switch is opened and the value of the voltmeter shows is the emf of the cell. Besides, we know there's actually a little of current flowing through the voltmeter which has high resistance otherwise the pointer of voltmeter will not deflect. In other words, what I'm trying to ask are :

  1. If there's a little current flowing through the voltmeter during the switch is opened, shouldn't that the reading of voltmeter is the potential drops across the cell?
  2. I noticed that my reference books usually mention "The emf of a cell can be measured approximately by connecting a high resistance voltmeter across the cell as shown in the figure above. The voltmeter reading is approximately the emf, E of the cell.". So why is the value of voltmeter shows is approximately the actually value of emf, could it because the potential drops due to internal resistance of the cell is included?
  • $\begingroup$ The internal resistance of the voltmeter is specified, so you know how much current is flowing in addition to your load. For many electrochemical measurements that's good enough. For very high impedance sources like pH-probes etc. an electrometer amplifier is needed which has an extremely low bias current (in the range of sub-fA). If you lack that, you can always do a potentiometric measurement: en.wikipedia.org/wiki/Potentiometer_(measuring_instrument) $\endgroup$ – CuriousOne May 12 '16 at 9:36
  • $\begingroup$ @CuriousOne - Where is the internal resistance of the voltmeter specified? $\endgroup$ – DIYser May 12 '16 at 9:57
  • $\begingroup$ @DIYser: It's usually specified on the voltmeter, it's in the manual for sure. Sorry for not reacting lately. I will have more time over the next couple weeks. Give me a ring. $\endgroup$ – CuriousOne May 12 '16 at 10:01
  • $\begingroup$ @CuriousOne - ok, ty. $\endgroup$ – DIYser May 12 '16 at 10:08

Suppose the emf of a cell is 1.500 V and its internal resistance is 10 $\Omega$.
Connecting a voltmeter of resistance 10 M$\Omega$ across the cell will result in a voltage drop across the internal resistance of the cell of approximately $1.5 \times 10^{-6}$ volt which is hardly going to affect the reading on the voltmeter.
On the other hand if the resistance of the voltmeter is 1000 $\Omega$ then the voltage drop across the internal resistance of approximately 0.015 volts is possibly significant.

  • $\begingroup$ Yeah, so the fact that potential drops is due to the internal resistance of the cell when the switch is open although it's negligible, and during calculations, we assume that the 'approximate value of voltmeter' to be the actual value E for the equation E=I(R+r), am my explanations correct? $\endgroup$ – Acery May 12 '16 at 13:23

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