Chapter 26, Problem 85GP

A potentiometer is a device to precisely measure potential differences or emf, using a “null” technique. In the simple potentiometer circuit shown in Fig. 26–74. R′ represents the total resistance of the resistor from A to B (which could be a long uniform “slide” wire), whereas R represents the resistance of only the part from A to the movable contact at C. When the unknown emf to be measured, ${\mathcal{E}}_x$ placed into the circuit as shown, the movable contact C is moved until the galvanometer G gives a null reading (i.e., zero) when the switch S is closed. The resistance between A and C for this situation we call R_x. Next, a standard emf, ${\mathcal{E}}_{\text{s}}$, which is known precisely, is inserted into the circuit in place of ${\mathcal{E}}_x$ and again the contact C is moved until zero current flows through the galvanometer when the switch S is closed. The resistance between A and C now is called R_s. (a) Show that the unknown emf is given by

${\mathcal{E}}_x=\left(\frac{R_x}{R_s}\right){\mathcal{E}}_s$

where R_x, R_s, and ${\mathcal{E}}_{\text{s}}$ are all precisely known. The working battery is assumed to be fresh and to give a constant voltage. (b) A slide-wire potentiometer is balanced against a 1.0182-V standard cell when the slide wire is set at 33.6 cm out of a total length of 100.0 cm. For an unknown source, the setting is 45.8 cm. What is the emf of the unknown? (c) The galvanometer of a potentiometer has an internal resistance of 35 Ω and can detect a current as small as 0.012 mA. What is the minimum uncertainty possible in measuring an unknown voltage? (d) Explain the advantage of using this “null” method of measuring emf.

FIGURE 26–74 Potentiometer circuit. Problem 85.