0
$\begingroup$

Experimentally, I have seen how hooking up a battery to a simple circuit just with a high-resistance voltmeter raises the voltage reading (allegedly to a level equal to the EMF of the battery).

However, I find the explanation for why the reading rises, much less to an EMF, very unconvincing. We were told that the internal resistance and the necessary potential drop is ignored, because there is no current in the said circuit, hence why the voltmeter measures an EMF. How can this make sense? There clearly must be some current, albeit very little, flowing, for the high-resistance voltmeter to even have a reading, and that little current will still experience resistive forces from the internal resistance of the cell - so the EMF should not be attainable. Or is there a mechanism by which, when there is very little current, resistors are ignored, hence no work has to be done to traverse them?

Clearly I am wrong, as experimentally I saw the voltage rise in that super simple cell. My point of view suggests that there shouldn't be a difference between the reading in said circuit and the potential difference in a circuit consisting of 3 resistors (the voltmeter in this case measures the drop between these 3 resistors, note there is no other significant source of resistance other than the internal resistance). Really, in my theoretical understanding they should produce an equal reading, but they don't.

So, to be honest, not only do I believe that the reading we saw shouldn't have been the EMF, but not even any different from the reading in a normal circuit, as described (the latter belief clearly conflicts with reality).

I an eager to know what I am thinking wrongly about. Please ask if I can help clarify anything!

Thank you very much :)!

$\endgroup$
3
$\begingroup$

You're quite correct that there will be some current flowing, so there must be a voltage drop due to the internal resistance of the battery. The EMF measured by any voltmeter will always be less than the true EMF.

If the internal resistance of the battery is $R_b$ and the resistance of your voltmeter is $R_m$ then the voltage you measure will be:

$$ V = \frac{R_m}{R_m + R_b} E $$

The internal resistance of, for example a manganese alkaline battery is less than an ohm, while the resistance of a voltmeter is going to be 10s if not 100s of KOhms, so in practice the measured voltage is going to be less than the true EMF by less than one part in $10^4$. Since you probably can't read the meter that accurately the internal resistance can be ignored.

$\endgroup$
  • $\begingroup$ Thank you for the clarification! Just wondering, is there a way to explain why the voltage reading still significantly rose when the parallel circuit (with the 3 resistors one row and the voltmeter on another) was changed into the solo voltmeter circuit? $\endgroup$ – Just_a_fool Aug 28 '14 at 7:26
  • $\begingroup$ Actually, that's not correct. Both field effect transistors and vacuum tubes can be used to measure voltages without ANY current flowing. The only energy that is used in these measurements is that needed to overcome the input capacitance of these devices, which is finite, but can be made very small. A noise analysis shows, that the input capacitance is the limiting factor for the precision of the measurement that can be achieved for a given averaging time, even if there is no net charge flowing trough the device, at all, once its charged. $\endgroup$ – CuriousOne Aug 28 '14 at 7:36
  • $\begingroup$ @Just_a_fool: you'll have to draw a diagram of the three resistor circuit as I'm not sure what you're measuring in it. Incidentally CuriousOne is correct that there exist voltmeters that have effectively infinite resistance. $\endgroup$ – John Rennie Aug 28 '14 at 7:40
  • $\begingroup$ It may be easier to see, that voltage measurements do not require any current flow, by using a 19th century electrometer, which does also not have any current flow, except for the initial displacement current that charges the device. The internal resistance of cheap voltmeters is an artifact of their construction. Professional electrometer amplifiers and calibrated electrometers do not suffer from having an internal resistance that is caused by their internal design. $\endgroup$ – CuriousOne Aug 28 '14 at 7:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.