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11.4Terminal potential difference and the potential divider

Ammeters and voltmeters

The current through a resistor is measured by an instrument called an ammeter an instrument that measures the electric current through it, which is connected in series to the resistor as shown in Figure 11.46.

Figure 11.46: An ammeter measures the current in the resistor connected in series to it.

The ammeter itself has a small electric resistance. An ideal ammeter an ammeter with zero resistance has zero resistance. The potential difference across a resistor is measured with a voltmeter an instrument that measures the potential difference across its ends connected in parallel to the resistor (Figure 11.47).

Figure 11.47: A voltmeter is connected in parallel to the device we want to measure the potential difference across.

An ideal voltmeter a voltmeter with infinite resistance (takes no current when connected to a resistor) has infinite resistance, which means that it takes no current when it is connected to a resistor. Real voltmeters have very high resistance. Unless otherwise stated, ammeters and voltmeters will be assumed to be ideal.

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Thus, to measure the potential difference across and current through a resistor, the arrangement shown in Figure 11.48 is used.

Figure 11.48: The correct arrangement for measuring the current through and potential difference across a resistor. The variable resistor allows the current in the resistor R to be varied so as to collect lots of data for current and voltage.

Worked Example 11.15

In the circuit in Figure 11.49, the emf of the cell is 9.00 V and the internal resistance is assumed negligible. A nonideal voltmeter whose resistance is 500 kΩ is connected in parallel to a resistor of 500 kΩ.

Figure 11.49: For worked example 11.15.

aDetermine the reading of the (ideal) ammeter.

bA student is shown the circuit and assumes, incorrectly, that the voltmeter is ideal. Estimate the resistance the student would calculate if he were to use the current found in a.

cThe resistance R is changed to one of 50 Ω. What resistance would a student measure if she assumes the voltmeter is ideal?

Answer

aSince the two 500 kΩ resistances are in parallel, the total resistance of the circuit is found from:

1 R=1 500+1 500=1 250⇒R=250 kΩ

Using I=V R, the current that leaves the cell is:

I=9.0 250×10 3=3.6×10−5 A

I=36 μ A

This is the reading of the ammeter in the circuit.

bThe reading of the voltmeter is 9.0 V. If the student assumes the voltmeter is ideal, he or she would conclude that the current in the resistor is 36 μA. He or she would then calculate that:

R=V I=9.0 V 36 μ A=250 kΩ and would get the wrong answer for the resistance. The percent error is | measured − true true |×100%=|250−500 500|×100%=50%

cThe total resistance of the circuit is now found from:

1 R=1 500×10 3+1 50⇒R=49.9995≈50 Ω

This is what we would have if the voltmeter were ideal. The student will measure the correct resistance at 50 Ω. The voltmeter behaves as an ideal voltmeter if its resistance is much larger than the other resistances in the circuit.

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Worked Example 11.16

A student constructs the circuit shown in Figure 11.50 and incorrectly places an ideal voltmeter in series to the resistor.

Figure 11.50: For worked example 11.16.

What is the reading of the ammeter, and what is the reading of the voltmeter?

Answer

The voltmeter has infinite resistance so the total resistance of the circuit is infinite. Hence the current is zero.Since there is no current, there is no voltage drop across the ammeter and no voltage drop across the resistor,and so the voltmeter will read 12 V.

Worked Example 11.17

The graph (Figure 11.51a) shows the I–V characteristics of an ohmic resistor and a filament lamp.The lamp and the resistor are connected in series to a cell of emf 2.7 V and negligible internal resistance (Figure 11.51b).

Figure 11.51: For worked example 11.17.

What is the total power dissipated in the circuit?

Answer

The lamp and the resistor take the same current so we must look at horizontal lines in the graph (which correspond to the same current in the lamp and the resistor), find the corresponding voltages and see if they add up to 2.7 V.This is a trial and error method. Trying a few cases we see that a current of 0.15 A corresponds to a voltage of 1.2 V across the lamp and 1.5 V across the resistor, so this is the right choice. The total power is then ε I = 2.7 × 0.15 = 0.40 W.

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Terminal potential difference

In all the examples we have considered so far, the cell or battery was assumed to be ideal. In reality, some energy is always dissipated inside the cell or the battery which implies that cells and batteries have an internal resistance, the yellow box in Figure 11.52. What would a voltmeter read if it is connected across the terminals of a battery?

Figure 11.52: The potential difference across the terminals of a battery is equal to the emf when there is no current in the circuit. It is less than the emf when there is current.

If the current through the battery is zero, then there is zero potential difference across the internal resistance and the voltmeter would measure the emf of the battery. If there is current I through the battery then there is a potential difference Ir across the internal resistance. The internal resistance drops the potential from ε to ε – Ir and so the voltmeter would read

V=ε−I r

This confirms that when the current is zero (I = 0), V = ε.

CHECK YOURSELF 7

A cell with emf 6.0 V and internal resistance 0.50 Ω is connected in series to an external resistor of resistance 2.5 Ω. What is the power dissipated in:

athe cell

bthe external resistor?

Worked Example 11.18

The potential difference across the terminals of a battery is 9.6 V when the current is 1.2 A and 9.0 V when the current is 1.5 A. Determine the emf of the battery and the internal resistance.

Answer

We need to use V = ε − Ir twice:

9.6=ε−1.2 r

9.0=ε−1.5 r

Subtracting, we get 0.6 = 0.3r and so r = 2.0 Ω. Hence ε = 12 V.

Worked Example 11.19

A cell is connected to a variable resistor (Figure 11.53a). As the resistance is varied the terminal voltage and the current in the circuit are measured. The graph (Figure 11.53b) shows how the potential difference across the terminals of a cell varies with the current leaving the cell.

Figure 11.53: For worked example 11.19.

Determine the emf of the cell and its internal resistance.

Answer

We again need to use V = ε − Ir, from which we deduce that the emf is the vertical intercept and the internal resistance the negative of the gradient of the graph. Extending the straight line we find an intercept of 11 mV, which is the emf, and a (negative) gradient of 0.25 Ω, which is the internal resistance.

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Potential dividers

The circuit in Figure 11.54a shows a potential divider (which is also known as a potentiometer). XY is usually a wire of fixed length, and S (the slider) can be attached to any point of the wire (with a crocodile clip, for example). It can be used to investigate, for example, the current–voltage characteristic of some device denoted by resistance R. This complicated-looking circuit is simply equivalent to the circuit in Figure 11.54b. In this circuit, the resistance R 1 is the resistance of the wire from end X to the slider S, and R 2 is the resistance of the wire from S to end Y. The current that leaves the cell splits at point M. Part of the current goes from M to N, and the rest goes into the device with resistance R.The right end of the resistance R (the slider) can be connected to any point on the wire XY.

Figure 11.54: a This circuit uses a potential divider. The voltage and current in the device with resistance R can be varied by varying the point where the slider S is attached to the variable resistor. b The potential divider circuit is equivalent to this simpler-looking circuit.

By varying where the slider S connects to XY, different potential differences and currents are obtained for the device R. One advantage of the potential divider is that now the potential difference across the resistor can be varied from a minimum of zero volts, when the slider S is placed at X, to a maximum of ε, the emf of the battery (assuming zero internal resistance), by connecting the slider S to point Y.

Worked Example 11.20

Look at Figure 11.55. The cell has negligible internal resistance and emf 18 V.

What is the range of readings of the voltmeter as the slider is moved from X to Y?

Figure 11.55: For worked example 11.20.

Answer

The current leaving the cell is 2.0 A.

So the potential difference across the 6.0 Ω resistor is 12 V.

At X the voltmeter will read 0 V and at Y 12 V so the range is 0 to 12 V.

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NATURE OF SCIENCE

In 1825 in England, Peter Barlow proposed a law explaining how wires conducted electricity. His careful experiments using a constant voltage showed good agreement, and his theory was accepted. At about the same time in Germany, Georg Ohm proposed a different law backed up by experimental evidence using a range of voltages. The experimental approach to science was not popular in Germany, and Ohm’s findings were rejected. It was not until 1841 that the value of his work was recognised, first in England and later in Germany. In modern science, before research findings are published they are reviewed by other scientists working in the same area (peer review). This would have shown the errors in Barlow’s work and given Ohm recognition sooner.

TEST YOUR UNDERSTANDING

48A cell has emf 6.0 V. When connected to an external 5.0 Ω resistor the current leaving the cell is 1.0 A.

aWhat is the internal resistance of the cell?

bWhat is the terminal potential difference?

49The terminal potential difference across a cell is 5.8 V when the current leaving the cell is 0.50 A. When the current is 0.60 A the terminal potential difference is 5.7 V.

aWhat is the internal resistance of the cell?

bWhat is the emf of the cell?

50A cell of emf 6.0 V and internal resistance 2.0 Ω is connected to an external variable resistor R.

aBy making a graph (on your graphics calculator) of the power dissipated in R against R (or otherwise) determine the value of R for which the power dissipated is a maximum.

bEstimate this maximum power.

51A voltmeter is connected across the terminals of a cell. An ammeter is inserted as shown in Figure 11.56.

Figure 11.56: For question 51.

aWhat happens to the readings of the ammeter and of the voltmeter when the switch is closed?

bHow do the answers to part a change, if at all, if the cell had zero internal resistance?

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52Figure 11.57 shows a potential divider (potentiometer). The cell has negligible internal resistance and emf 12 V.

Figure 11.57: For question 52.

aWhat is the reading of the voltmeter when the slider is connected to i X and ii Y?

bSketch a graph to show how the reading of the voltmeter varies with the distance d from point X as the slider moves from X to Y.

53The circuit in Figure 11.58 has a variable resistor; its resistance can be varied from 0 to 6.0 Ω. The cell has emf 9.0 V and negligible internal resistance.

Figure 11.58: For question 53.

aWhat is the range of readings of the voltmeter?

bHow does the answer to part a change, if at all, if the cell had an internal resistance of 1.0 Ω?

54A source supplies electrical power to a device D at a distance of 2.0 km. The device operates properly at 120 V and current 25 A. The connecting wires have a resistance per unit length of 4.0 Ω km−1 (Figure 11.59).

Figure 11.59: For question 54.

aWhat is the minimum emf required at the source?

bWhat fraction of the power produced by the source gets wasted in the connecting wires?

55Look at Figure 11.60. In both circuits, the cell has emf 6.0 V and an internal resistance 2.0 Ω. The external resistance is 3.0 Ω.

Figure 11.60: For question 55.

What is the reading of the voltmeters in circuits A and B:

awhen the switch is open?

bwhen the switch is closed?

Links

The microscopic–macroscopic connection

If you are plumber, do you need to know the molecular structure of water to help you in your work? The flow of water in pipes is a macroscopic phenomenon whereas the detailed molecular structure of water is microscopic. We have a vast difference in scales of length in the two cases. In very many phenomena the presence of two different scales means that the detailed physics operating at one scale does not affect the physics at the other. This is also the case with current: it was possible to give detailed descriptions of the behaviour of current in circuits long before it was discovered that current is electrons moving in the same direction. (However, the most complicated problems in physics are those in which the physics at one length scale does affect the physics at the other scale.)