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  • Explain why a null measurement device is more accurate than a standard voltmeter or ammeter.
  • Demonstrate how a Wheatstone bridge can be used to accurately calculate the resistance in a circuit.

Standard measurements of voltage and current alter the circuit being measured, introducing uncertainties in the measurements. Voltmeters draw some extra current, whereas ammeters reduce current flow. Null measurements balance voltages so that there is no current flowing through the measuring device and, therefore, no alteration of the circuit being measured.

Null measurements are generally more accurate but are also more complex than the use of standard voltmeters and ammeters, and they still have limits to their precision. In this module, we shall consider a few specific types of null measurements, because they are common and interesting, and they further illuminate principles of electric circuits.

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 [link] . (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 size 12{V} {} , which is related to the emf of the battery by V = emf Ir size 12{V="emf" - ital "Ir"} {} , where I size 12{I} {} is the current that flows and r size 12{r} {} is the internal resistance of the battery.

The emf could be accurately calculated if r size 12{r} {} were very accurately known, but it is usually not. If the current I size 12{I} {} could be made zero, then V = emf size 12{V="emf"} {} , and so emf could be directly measured. However, standard voltmeters need a current to operate; thus, another technique is needed.

The diagram shows equivalence between two circuits. The first circuit has a cell of e m f script E and an internal resistance r connected across a voltmeter. The equivalent circuit on the right shows the same cell of e m f script E and an internal resistance r connected across a series combination of a galvanometer with an internal resistance r sub G and high resistance R. The currents in the two circuits are shown to be equal.
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 [link] .) A voltage source is connected to a resistor R, say, a long wire, and passes a constant current through it. There is a steady drop in potential (an IR size 12{ ital "IR"} {} drop) along the wire, so that a variable potential can be obtained by making contact at varying locations along the wire.

[link] (b) shows an unknown emf x size 12{"emf" rSub { size 8{x} } } {} (represented by script E x size 12{"emf" rSub { size 8{x} } } {} in the figure) connected in series with a galvanometer. Note that emf x size 12{"emf" rSub { size 8{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, emf x = IR x size 12{"emf" rSub { size 8{x} } = ital "IR" rSub { size 8{x} } } {} , where R x size 12{R rSub { size 8{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 emf x size 12{"emf" rSub { size 8{x} } } {} is directly sensed.

Now, a very precisely known standard emf s size 12{"emf" rSub { size 8{s} } } {} is substituted for emf x size 12{"emf" rSub { size 8{x} } } {} , and the contact point is adjusted until the galvanometer again reads zero, so that emf s = IR s size 12{"emf" rSub { size 8{s} } = ital "IR" rSub { size 8{s} } } {} . In both cases, no current passes through the galvanometer, and so the current I size 12{I} {} through the long wire is the same. Upon taking the ratio emf x emf s size 12{ { {"emf" rSub { size 8{x} } } over {"emf" rSub { size 8{s} } } } } {} , I size 12{I} {} cancels, giving

Questions & Answers

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º20.0º with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate an
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Practice Key Terms 5

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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