The Potentiometer

The Potentiometer

Suppose you wish to measure the emf of a battery. Consider what happens if you connect the battery directly to a standard voltmeter as shown in this figure. (Once we note the problems with this measurement, we will examine a null measurement that improves accuracy.) As discussed before, the actual quantity measured is the terminal voltage \(V\), which is related to the emf of the battery by \(V=\text{emf}-\text{Ir}\), where \(I\) is the current that flows and \(r\) is the internal resistance of the battery.

The emf could be accurately calculated if \(r\) were very accurately known, but it is usually not. If the current \(I\) could be made zero, then \(V=\text{emf}\), and so emf could be directly measured. However, standard voltmeters need a current to operate; thus, another technique is needed.

An analog voltmeter attached to a battery draws a small but nonzero current and measures a terminal voltage that differs from the emf of the battery. (Note that the script capital E symbolizes electromotive force, or emf.) Since the internal resistance of the battery is not known precisely, it is not possible to calculate the emf precisely.

A potentiometer is a null measurement device for measuring potentials (voltages). (See this figure.) A voltage source is connected to a resistor \(\mathrm{R,}\) say, a long wire, and passes a constant current through it. There is a steady drop in potential (an \(\text{IR}\) drop) along the wire, so that a variable potential can be obtained by making contact at varying locations along the wire.

This figure (b) shows an unknown \({\text{emf}}_{x}\) (represented by script \({E}_{x}\) in the figure) connected in series with a galvanometer. Note that \({\text{emf}}_{x}\) opposes the other voltage source. The location of the contact point (see the arrow on the drawing) is adjusted until the galvanometer reads zero. When the galvanometer reads zero, \({\text{emf}}_{x}={\text{IR}}_{x}\), where \({R}_{x}\) is the resistance of the section of wire up to the contact point. Since no current flows through the galvanometer, none flows through the unknown emf, and so \({\text{emf}}_{x}\) is directly sensed.

Now, a very precisely known standard \({\text{emf}}_{s}\) is substituted for \({\text{emf}}_{x}\), and the contact point is adjusted until the galvanometer again reads zero, so that \({\text{emf}}_{s}={\text{IR}}_{s}\). In both cases, no current passes through the galvanometer, and so the current \(I\) through the long wire is the same. Upon taking the ratio \(\cfrac{{\text{emf}}_{x}}{{\text{emf}}_{s}}\), \(I\) cancels, giving

\(\cfrac{{\text{emf}}_{x}}{{\text{emf}}_{s}}=\cfrac{{\text{IR}}_{x}}{{\text{IR}}_{s}}=\cfrac{{R}_{x}}{{R}_{s}}.\)

Solving for \({\text{emf}}_{x}\) gives

\({\text{emf}}_{x}={\text{emf}}_{s}\cfrac{{R}_{x}}{{R}_{s}}.\)

The potentiometer, a null measurement device. (a) A voltage source connected to a long wire resistor passes a constant current \(I\) through it. (b) An unknown emf (labeled script \({E}_{\text{x}}\) in the figure) is connected as shown, and the point of contact along \(R\) is adjusted until the galvanometer reads zero. The segment of wire has a resistance \({R}_{\text{x}}\) and script \({E}_{\text{x}}={\text{IR}}_{\text{x}}\), where \(I\) is unaffected by the connection since no current flows through the galvanometer. The unknown emf is thus proportional to the resistance of the wire segment.

Because a long uniform wire is used for \(R\), the ratio of resistances \({R}_{\text{x}}/{R}_{\text{s}}\) is the same as the ratio of the lengths of wire that zero the galvanometer for each emf. The three quantities on the right-hand side of the equation are now known or measured, and \({\text{emf}}_{\text{x}}\) can be calculated. The uncertainty in this calculation can be considerably smaller than when using a voltmeter directly, but it is not zero. There is always some uncertainty in the ratio of resistances \({R}_{\text{x}}/{R}_{\text{s}}\) and in the standard \({\text{emf}}_{s}\). Furthermore, it is not possible to tell when the galvanometer reads exactly zero, which introduces error into both \({R}_{\text{x}}\) and \({R}_{\text{s}}\), and may also affect the current \(I\).