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11.1Potential difference, current and resistance

Potential difference

In Chapter 18 we will learn that there are electric forces between electric charges. The electric force is repulsive for same sign charges and attractive for opposite sign charges. So if we want to move a positive charge q closer to another positive charge, work must be done. If the work done in moving a charge q from A to B is W, the ratio W q is defined to be the potential difference the work done per unit charge in moving a point charge from one point to another between points A and B (Figure 11.1).

Figure 11.1: The potential difference between points A and B is the work done to move the charge from A to B divided by q. If the charge q is released at A it will accelerate to the right, gaining kinetic energy.

Conversely, if the charge at A is released, the electric force will push it away and the charge will gain kinetic energy. At point C the kinetic energy gained will be equal to qV where now V is the potential difference between points A and C.

KEY POINT

The potential difference V between two points is the work done per unit charge to move a point charge from one point to the other:

V=W q

The unit of potential difference is the volt, V, and 1 V = 1 J C−1.

Worked Example 11.1

The work done in moving a charge of 2.0 µC between two points in an electric field is 1.50 × 10−4 J. Determine the potential difference between the two points.

Answer

From the definition, the potential difference is:

V=W q V=1.50×10−4 2.0×10−6 V=75 V

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Worked Example 11.2

An electron is accelerated from rest through a potential difference of 75 V. What is the speed acquired by the electron?

Answer

From V=W q we find W = q V.

This work goes into the increase in kinetic energy,so we use the magnitude of the electron charge:

W=1.6×10−19×75=1.2×10−17 J

Hence: 1 2 m v 2=1.2×10−17 J

and so v=2×1.2×10−17 9.1×10−31=5.1×10 6 m s−1.

Electric current

Materials with many free electrons per unit volume are called conductors materials with many free electrons per unit volume, through which thermal energy and electric current can pass easily (such as metals).

In insulators materials with few free electrons per unit volume, through which thermal energy and electric current cannot readily pass (such as plastics) the number of free electrons per unit volume is very small.

In metals we have of order 1028 free electrons per metre cubed. These ‘free’ electrons move randomly, much like gas molecules in a container. They do so at high speeds, of order 105 m s–1. During this random motion, as many electrons move in one direction as in another (Figure 11.2) and so no charge is transferred in any one direction.

Figure 11.2: a The free electrons are not attached to any one particular atom. b The random electron motion does not carry net charge in any direction.

To transfer charge, the electrons have to move in the same direction. This motion of electrons in the same direction is known as a direct current (dc)rate of flow of charge through the cross sectional area of a conductor.

Links

The same materials that are good conductors of thermal energy are also good conductors of electricity. This suggests that the mechanisms of thermal energy and charge transfer are similar; they both involve electrons.

KEY POINT

We define electric current I in a wire as the rate of flow of charge through the cross-sectional area of a conductor:

I=Δ q Δ t

The unit of electric current is the ampere (A), which is a flow of one coulomb of charge per second (1 A = 1 C s−1).

CHECK YOURSELF 1

A current of 1 A is established in a conductor.

How many electrons move through the cross-sectional area of the conductor in 1 s?

To make the electrons move in the same direction requires a source of potential difference. One way to do this is to connect the conductor to a cell or a battery as will be explained later on.

Worked Example 11.3

Light is incident on a metallic plate. As a result, electrons are emitted from the surface at a rate of 2.6 × 1015 per second. What is the current leaving the surface?

Answer

The charge carried away from the surface in 1 s is 2.6 × 1015 × 1.6 × 10−19 = 4.2 × 10−4 C. Hence the current is 4.2 × 10−4 A.

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Electric resistance

When a given potential difference is established at the ends of a conductor, an electric current will be established in the conductor. The size of the current depends on a property of the conductor called its electric resistance the ratio of voltage across conductor to current through it.

KEY POINT

The electric resistance R of a conductor is defined as the potential difference V across its ends divided by the current I through it:

R=V I

The unit of electric resistance is the volt per ampere which is called the ohm, Ω.

Exam tip

You must be able to turn the equation R=V I into V=I R or I=V R instantly.

Worked Example 11.4

The potential difference across a conductor is 4.0 V. The resistance is 16 Ω.

What is the current through the conductor?

Answer

From R=V I we find I=V R=4.0 16=0.25 A.

The electric resistance of conducting wires is very small, so it is a good approximation to ignore this resistance. Conducting wires are represented by thin line segments in diagrams. Conductors whose resistance cannot be neglected are denoted by boxes and are called resistors (Figure 11.3).

Figure 11.3: a The potential difference V across the ends of the conductor creates an electric field within the conductor that forces a current I through the conductor. b How we represent a resistor and connecting wires in a circuit diagram.

In 1826, the German scientist Georg Ohm (1789–1854) discovered that when the temperature of most metallic conductors is kept constant, the current through the conductor is proportional to the potential difference across it:

I∝V

This statement is known as Ohm’s law at constant temperature the current through a metallic conductor is proportional to the voltage across the conductor.

Materials that obey Ohm’s law have a constant resistance at constant temperature. For these ohmic materials, a graph of I against V gives a straight line through the origin (Figure 11.4).

A filament light bulb will obey Ohm’s law as long as the current through it is small. As the current is increased, the temperature of the filament increases and so does the resistance (Figure 11.5).

Figure 11.4: The current–potential difference graph for a material that obeys Ohm’s law.

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Figure 11.5: The current–potential difference graph for a filament lamp; the resistance increases as the voltage increases.

In the graph for the ohmic material, no matter which point on the graph we choose (say the one with voltage 1.2 V and current 1.6 mA), the resistance is always the same:

R=1.2 1.6×10−3=750 Ω

However, looking at the graph in Figure 11.5 (the lamp filament), we see that at a voltage of 0.2 V the current is 0.8 mA and so the resistance is:

R=0.2 0.8×10−5=250 Ω

At a voltage of 0.3 V the current is 1.0 mA and the resistance is:

R=0.3 1.0×10−3=300 Ω

We see that as the current in the filament increases the resistance increases, so Ohm’s law is not obeyed. This is a nonohmic device.

Two other common components with variable resistance include light dependent resistors (LDRs)resistor where resistance decreases as light intensity increases and thermistors temperature dependent resistor where resistance decreases as temperature increases. The resistance of an LDR decreases as the intensity of light falling on it increases. The resistance of an NTC type thermistor decreases as the ambient temperature increases. Later on we will meet another component: the potentiometer.

Experiments show that three factors affect the resistance of a conductor kept at constant temperature. They are:

●the nature of the material

●the length of the conductor

●the cross-sectional area of the conductor.

It is found that the electric resistance R of a conductor (at fixed temperature) is proportional to its length L and inversely proportional to the cross-sectional area A:

R=ρ L A

The constant ρ is called resistivity the resistance of a conductor of unit length and unit cross-sectional area and depends on the material of the conductor and the temperature. The unit of resistivity is Ω m.

CHECK YOURSELF 2

The length and radius of a cylindrical conductor are both doubled. What happens to the resistance of the conductor?

A microscopic view of resistance

Electrons moving through a metal suffer collisions with the atoms of the metal. The collisions are inelastic which means the electrons lose energy which the atoms of the wire gain. Thus the atoms in the wire vibrate about their equilibrium positions with increased kinetic energy. We know from Chapter 7 that an increase in the random kinetic energy of molecules implies an increase in the temperature of the wire. This explains the heating effect of a current in a conductor.

The increased vibration of atoms means that we have an increased probability of collision between an electron and an atom. This means that the average speed of the electrons is reduced so the rate of flow of charge, i.e. the current, is reduced. This implies that resistance increases (R=V I). In other words, in most metallic conductors, an increase in temperature leads to an increase in resistance.

Worked Example 11.5

The resistivity of copper is 1.68 × 10−8 Ω m. Calculate the length of a copper wire of diameter 4.00 mm that has a resistance of 5.00 Ω.

Answer

We use R=ρ L A to get L=R A ρ and so:

L=5.00×π×(2.00×10−3)2 1.68×10−8

L=3739 m

The length of copper wire is about 3.74 km.

Exam tip

Do not confuse diameter with radius.

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TEST YOUR UNDERSTANDING

1The work done in moving a charge of 5.0 nC across two points is 240 nJ. What is the potential difference between the points?

2An alpha particle has twice the charge of a proton and 4 times its mass. An alpha particle and a proton are accelerated from rest by a potential difference of 250 V. What are the ratios kinetic energy of alpha particle kinetic energy of proton and speed of alpha particle speed of proton ?

3The current in a wire is 2.0 A. How many electrons move through the cross-sectional area of a wire in a time of 5 s?

4The current in a wire is I. The charge of an electron is e. How many electrons go through the cross-sectional area of the wire in 1 s?

5The current in a device obeying Ohm’s law is 1.5 A when connected to a source of potential difference 6.0 V. What will the potential difference across the same device be when the current is 3.5 A?

6The current through a resistor obeying Ohm’s law is 0.80 A when the potential difference across the resistor is 240 V. What current would we have when the potential difference is decreased to 160 V?

7Electrons are moving in a metallic wire of cross-sectional area A. There are n free electrons per unit volume. The charge on one electron is e and the average speed of the electrons is v.

aShow that an expression for the electric current is I = nAve.

bEstimate the magnitude of the drift speed in a wire that carries a current of 1 A. The wire has radius 2 mm, and the number of electrons per unit volume (the number density) of free electrons is n = 1028 m−3.

cFigure 11.6 shows electric current that flows in a conductor of variable cross-sectional area. State and explain whether the electron drift speed at B is smaller than, equal to, or greater than that at A.

Figure 11.6: For question 7.

8 aWhy does the passage of current in a conductor result in heating of the conductor?

bWhy does the resistance of a metallic conductor increase when the temperature increases?

9The graph (Figure 11.7) shows the current–voltage characteristics of a device.

Figure 11.7: For question 9.

By finding how the resistance varies as the voltage increases, explain why the device does not obey Ohm’s law.

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10Figure 11.8 shows a light-dependent resistor connected to an ohmic resistor. The intensity of light increases. The current stays the same. What happens to the reading of the voltmeter?

Figure 11.8: For question 10.

11A filament lamp is most likely to burn out when it is turned on rather than when it is turned off. What could be the reason for this?

12The graphs (Figure 11.9) show the current as a function of voltage across the same piece of metal wire which is kept at two different temperatures.

Figure 11.9: For question 12.

aDiscuss whether the wire obeys Ohm’s law.

bSuggest which of the two lines on the graph corresponds to the higher temperature.

13A cylindrical copper wire has resistance R. The wire is melted and a new wire of double the length is made out of all the available material. What is the resistance of the new wire?

14A conductor has the shape of a rectangular block of size 10 cm × 20 cm × 40 cm as shown in Figure 11.10.

Figure 11.10: For question 14.

A potential difference is established across opposite faces of the block. In which case is the smallest current established in the block?

15The resistivity of a metal wire is 2.5 × 10−8 Ω m. The radius of the wire is 0.25 mm, and its length is 2.2 m. What is the resistance of the wire?

16What happens to the resistance of a cylindrical wire if the length and the radius are both tripled?

17The resistivity of a metal wire is 1.7 × 10−8 Ω m.The radius of the wire is 0.30 mm, and its resistance is 5.0 Ω. What is its length?

18The resistivity of a metal wire is 5.6 × 10−8 Ω m.The length of the wire is 3.5 km, and its resistance is 4.0 Ω. What is its radius?

19Two copper wires X and Y have the same resistance. X has double the length of Y. What is the ratio radius of X radius of Y?