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It is known that connecting a voltmeter directly to the terminals(without any other load) of a battery will show you the value of the battery's EMF.However,I don't really undrerstand how this happens.The electomotive force is equal with:E=U+u(E=EMF,u=internal voltage drop,U=external voltage drop).The voltage the battery supplies is U.My question:How does a voltmeter measure EMF of a battery and why doesn't it show the value of U when connected directly to the battery?

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  • \$\begingroup\$ The voltmeter can only ever measure the voltage dropped over its internal impedance, nothing else. The rest is interpretation \$\endgroup\$ – PlasmaHH Mar 2 '16 at 13:52
  • \$\begingroup\$ If you can ensure that u = 0, then E = U. More practically, if the current drawn from the battery is tiny, then u will be as well. \$\endgroup\$ – Simon B Mar 2 '16 at 15:30
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In your model, the battery is a voltage source with a resistance in series. When the battery is delivering no current, there is no current thru this internal resistance, and therefore the voltage it drops is 0. The open circuit voltage of a battery is therefore the same as its internal voltage. When you put a load on the battery, current flows, the internal resistance drops some voltage, and the external voltage is then less than the internal voltage.

Voltmeters are designed to have high impedance. They present such a tiny load to the battery that we can ignore it for practical purposes. The internal and external voltages of the battery are therefore the same for practical purposes when measured with a voltmeter. Put another way, measuring a battery only with a voltmeter is measuring it under open circuit conditions.

I am referring to the internal voltage as being that produced by the cell directly, and the external voltage as that minus whatever is dropped across the inevitable and unavoidable internal resistance of the battery.

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  • \$\begingroup\$ You said that "measuring a battery only with a voltmeter is measuring it under open circuit conditions".Wouldn't this mean that the voltmeter always reads 0 volts no matter what battery you test? \$\endgroup\$ – Daniel Tork Mar 2 '16 at 19:03
  • \$\begingroup\$ @Dan: No, it means the voltmeter draws 0 Amps from the battery. \$\endgroup\$ – Olin Lathrop Mar 2 '16 at 19:28
  • \$\begingroup\$ I know this question has aged a bit,but I am thinking of accepting your answer.Howerver,I don't understand what you referred to with "and the the external voltage is less than the internal voltage" and "The internal and external voltages of the battery are therefore the same for practical purposes".Isn't U always bigger than u? \$\endgroup\$ – Daniel Tork Sep 10 '16 at 8:57
  • \$\begingroup\$ @Dan: I was referring to the internal voltage as being that produced by the cell directly, and the external voltage as that minus whatever is dropped across the inevitable and unavoidable internal resistance of the battery. \$\endgroup\$ – Olin Lathrop Sep 10 '16 at 10:49
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To measure the actual EMF, you measure U when the current supplied from the battery is precisely 0.

To do this, traditionally, you don't use a voltmeter but an ammeter ... and a device for measuring potential - that is, a potentiometer.

This "potentiometer" is fed from some higher (unknown) voltage, and adjusted until the current drawn from the battery is zero, and yields the ratio of two resistances.

That alone isn't very useful, until you also measure the EMF of a standard cell (Daniell or Weston cell) using the same technique, which allows you to calibrate the unknown driving voltage, and finally the EMF of the cell you're testing.

In practice, of course, the current drawn by an ordinary voltmeter is small enough that u is close enough to 0 not to matter, and U = EMF.

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    \$\begingroup\$ Potentiometer. Weston cell. In a nice wood box. I hate it when test equipment I've used is displayed in a museum setting. \$\endgroup\$ – Spehro Pefhany Mar 2 '16 at 17:03
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    \$\begingroup\$ True, but I don't miss taking fifteen minutes to measure a voltage! \$\endgroup\$ – Brian Drummond Mar 2 '16 at 17:12

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