# Learning Objectives

### Learning Objectives

By the end of this section, you will be able to do the following:

• 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 consider a few specific types of null measurements because they are common and interesting and further illuminate principles of electric circuits.

# 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 Figure 4.37. 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 $V,V, size 12{V} {}$ which is related to the emf of the battery by $V=emf−Ir,V=emf−Ir, size 12{V="emf" - ital "Ir"} {}$ where $II size 12{I} {}$ is the current that flows and $rr size 12{r} {}$ is the internal resistance of the battery.

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

Figure 4.37 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 Figure 4.38). A voltage source is connected to a resistor $R,R,$ say, a long wire, and passes a constant current through it. There is a steady drop in potential (an $IRIR size 12{ ital "IR"} {}$ drop) along the wire so that a variable potential can be obtained by making contact at varying locations along the wire.

Figure 4.38(b) shows an unknown $emfxemfx size 12{"emf" rSub { size 8{x} } } {}$ (represented by script $ExEx size 12{"emf" rSub { size 8{x} } } {}$ in the figure) connected in series with a galvanometer. Note that $emfxemfx size 12{"emf" rSub { size 8{x} } } {}$ 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, $emfx=IRx,emfx=IRx, size 12{"emf" rSub { size 8{x} } = ital "IR" rSub { size 8{x} } } {}$ where $RxRx size 12{R rSub { size 8{x} } } {}$ 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, so $emfxemfx size 12{"emf" rSub { size 8{x} } } {}$ is directly sensed.

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

4.71 $emfxemfs=IRxIRs=RxRs.emfxemfs=IRxIRs=RxRs. size 12{ { {"emf" rSub { size 8{x} } } over {"emf" rSub { size 8{s} } } } = { { ital "IR" rSub { size 8{x} } } over { ital "IR" rSub { size 8{s} } } } = { {R rSub { size 8{x} } } over {R rSub { size 8{s} } } } } {}$

Solving for $emfxemfx size 12{"emf" rSub { size 8{x} } } {}$ gives

4.72 $emfx=emfsRxRs.emfx=emfsRxRs. size 12{"emf" rSub { size 8{x} } ="emf" rSub { size 8{s} } { {R rSub { size 8{x} } } over {R rSub { size 8{s} } } } } {}$
Figure 4.38 The potentiometer, a null measurement device. (a) A voltage source connected to a long wire resistor passes a constant current $II size 12{I} {}$ through it. (b) An unknown emf (labeled script $ExEx$ in the figure) is connected as shown, and the point of contact along $RR size 12{R} {}$ is adjusted until the galvanometer reads zero. The segment of wire has a resistance $RxRx size 12{R rSub { size 8{x} } } {}$ and script $Ex=IRx,Ex=IRx, size 12{E rSub { size 8{x} } = ital "IR" rSub { size 8{x} } } {}$ where $II size 12{I} {}$ 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 $R,R, size 12{R} {}$ the ratio of resistances $Rx/RsRx/Rs size 12{R rSub { size 8{x} } /R rSub { size 8{s} } } {}$ 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 $emfxemfx size 12{"emf" rSub { size 8{x} } } {}$ 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 $Rx/RsRx/Rs size 12{R rSub { size 8{x} } /R rSub { size 8{s} } } {}$ and in the standard $emfs.emfs. size 12{"emf" rSub { size 8{s} } } {}$ Furthermore, it is not possible to tell when the galvanometer reads exactly zero, which introduces error into both $RxRx size 12{R rSub { size 8{x} } } {}$ and $RsRs size 12{R rSub { size 8{s} } } {}$ and may also affect the current $I.I. size 12{I} {}$

# Resistance Measurements and the Wheatstone Bridge

### Resistance Measurements and the Wheatstone Bridge

There are a variety of so-called ohmmeters that purport to measure resistance. What the most common ohmmeters actually do is to apply a voltage to a resistance, measure the current, and calculate the resistance using Ohm’s law. Their readout is this calculated resistance. Two configurations for ohmmeters using standard voltmeters and ammeters are shown in Figure 4.39. Such configurations are limited in accuracy because the meters alter both the voltage applied to the resistor and the current that flows through it.

Figure 4.39 Two methods for measuring resistance with standard meters. (a) Assuming a known voltage for the source, an ammeter measures current, and resistance is calculated as $R=VI.R=VI. size 12{R= { {V} over {I} } } {}$ (b) Since the terminal voltage $VV size 12{V} {}$ varies with current, it is better to measure it. $VV size 12{V} {}$ is most accurately known when $II size 12{I} {}$ is small, but $II size 12{I} {}$ itself is most accurately known when it is large.

The Wheatstone bridge is a null measurement device for calculating resistance by balancing potential drops in a circuit (see Figure 4.40). The device is called a bridge because the galvanometer forms a bridge between two branches. A variety of bridge devices are used to make null measurements in circuits.

Resistors $R1R1 size 12{R rSub { size 8{1} } } {}$ and $R2R2 size 12{R rSub { size 8{2} } } {}$ are precisely known, while the arrow through $R3R3 size 12{R rSub { size 8{3} } } {}$ indicates that it is a variable resistance. The value of $R3R3 size 12{R rSub { size 8{3} } } {}$ can be precisely read. With the unknown resistance $RxRx size 12{R rSub { size 8{x} } } {}$ in the circuit, $R3R3 size 12{R rSub { size 8{3} } } {}$ is adjusted until the galvanometer reads zero. The potential difference between points b and d is then zero, meaning that b and d are at the same potential. With no current running through the galvanometer, it has no effect on the rest of the circuit. So the branches abc and adc are in parallel, and each branch has the full voltage of the source. That is, the $IRIR size 12{ ital "IR"} {}$ drops along abc and adc are the same. Since b and d are at the same potential, the $IRIR size 12{ ital "IR"} {}$ drop along ad must equal the $IRIR size 12{ ital "IR"} {}$ drop along ab. Thus,

4.73 $I1R1=I2R3.I1R1=I2R3. size 12{I rSub { size 8{1} } R rSub { size 8{1} } =I rSub { size 8{2} } R rSub { size 8{3} } } {}$

Again, since b and d are at the same potential, the $IRIR size 12{ ital "IR"} {}$ drop along dc must equal the $IRIR size 12{ ital "IR"} {}$ drop along bc. Thus,

4.74 $I1R2=I2Rx.I1R2=I2Rx. size 12{I rSub { size 8{1} } R rSub { size 8{2} } =I rSub { size 8{2} } R rSub { size 8{x} } } {}$

Taking the ratio of these last two expressions gives

4.75 $I1R1I1R2=I2R3I2Rx.I1R1I1R2=I2R3I2Rx. size 12{ { {I rSub { size 8{1} } R rSub { size 8{1} } } over {I rSub { size 8{1} } R rSub { size 8{2} } } } = { {I rSub { size 8{2} } R rSub { size 8{3} } } over {I rSub { size 8{2} } R rSub { size 8{x} } } } } {}$

Canceling the currents and solving for Rx yields

4.76 $Rx=R3R2R1.Rx=R3R2R1. size 12{R rSub { size 8{x} } =R rSub { size 8{3} } { {R rSub { size 8{2} } } over {R rSub { size 8{1} } } } } {}$
Figure 4.40 The Wheatstone bridge is used to calculate unknown resistances. The variable resistance $R3R3 size 12{R rSub { size 8{3} } } {}$ is adjusted until the galvanometer reads zero with the switch closed. This simplifies the circuit, allowing $RxRx size 12{R rSub { size 8{x} } } {}$ to be calculated based on the $IRIR size 12{ ital "IR"} {}$ drops as discussed in the text.

This equation is used to calculate the unknown resistance when current through the galvanometer is zero. This method can be very accurate (often to four significant digits), but it is limited by two factors. First, it is not possible to get the current through the galvanometer to be exactly zero. Second, there are always uncertainties in $R1,R1, size 12{R rSub { size 8{1} } } {}$ $R2,R2, size 12{R rSub { size 8{2} } } {}$ and $R3,R3, size 12{R rSub { size 8{3} } } {}$ that contribute to the uncertainty in $Rx.Rx. size 12{R rSub { size 8{x} } } {}$