Physics · 4 Electricity and magnetism

4.2.4 Resistance

Before you start

Current $I = Q / t$ and p.d. $V = W / Q$ (4.2.2, 4.2.3).
How to connect an ammeter in series and a voltmeter in parallel.
Direct and inverse proportion, and reading a gradient from a graph.

Syllabus objectives covered (5)

Recall and use the equation for resistance R = V / I
Describe an experiment to determine resistance using a voltmeter and an ammeter and do the appropriate calculations
State, qualitatively, the relationship of the resistance of a metallic wire to its length and to its cross-sectional area
Sketch and explain the current–voltage graphs for a resistor of constant resistance, a filament lamp and a diode
Recall and use the following relationship for a metallic electrical conductor: (a) resistance is directly proportional to length (b) resistance is inversely proportional to cross-sectional area

Part 1 of 4

What resistance is

Resistance measures how strongly a component opposes the current through it. High resistance, small current for a given p.d.
It is defined by the equation, with $R$ in ohms, $V$ in volts and $I$ in amps:
$R = \frac{V}{I}$
The unit is the ohm, symbol $Ω$ . One ohm is one volt per amp.
The rearrangement $V = I R$ is the one you will use most often, and $I = V / R$ is worth having ready too. Practise all three.
In a metal, resistance arises because the drifting free electrons collide with the vibrating metal ions, losing energy to them. That energy appears as heating — which is why a resistor gets warm.
None of these equations is given in Paper 4. Write them out from memory as part of your revision, not just when you need them.

Part 2 of 4

Measuring a resistance

The method follows straight from $R = V / I$ : measure the p.d. across the component and the current through it, then divide.
Build a series circuit containing the power supply, the ammeter and the component under test. Connect the voltmeter in parallel across the component.
Ammeter in series, voltmeter across. Getting these the wrong way round is the most common circuit-diagram error in the whole unit.
The ammeter counts the current through R; the voltmeter measures the p.d. across it.
Include a variable resistor, or use a variable supply, so you can change the current and take several pairs of readings rather than just one.
Record $V$ and $I$ for each setting and calculate $R = V / I$ each time, then take a mean. Repeating reduces the effect of random error.
Better still, plot $V$ against $I$ . For a resistor of constant resistance the points lie on a straight line through the origin, and the gradient is the resistance — which uses all your data rather than one point.
One practical caution: keep the current low and switch off between readings. A current warms the component, and for most conductors warming changes the resistance you are trying to measure.

Part 3 of 4

What the resistance of a wire depends on

For a metallic wire, two dimensions matter: how long it is and how thick it is.
Resistance is directly proportional to length. Double the length and you double the resistance.
That makes sense from the electron picture — twice the length means twice as far to travel and twice as many collisions to make.
Resistance is inversely proportional to cross-sectional area. Double the area and you halve the resistance.
A thicker wire is like a wider road: more room for the electrons to move through, so less opposition.
Written together, $R \propto l$ and $R \propto 1 / A$ . So a wire twice as long and twice as thick has exactly the same resistance as the original — the two effects cancel.
One trap worth watching: if a question gives the diameter rather than the area, doubling the diameter quadruples the area, because area depends on the square of the radius. So the resistance falls to a quarter, not a half.

Part 4 of 4

Current–voltage graphs

Three components must be recognised from their $I$ – $V$ graphs, and you should be able to sketch and explain all three.
A resistor of constant resistance. A straight line through the origin, extending into the negative quadrant as a mirror image.
Straight through the origin means $I$ is directly proportional to $V$ , so $V / I$ is the same everywhere — the resistance is constant. The line is symmetrical because a resistor behaves identically whichever way the current flows.
A filament lamp. A curve through the origin that becomes less steep as the current increases, again symmetrical about the origin.
Straight, flattening, and one-way — the three shapes to recognise instantly.
The explanation: as the current increases, the filament gets hotter. The metal ions vibrate more, the electrons collide with them more often, and the resistance rises. A rising resistance means the current grows less and less for each extra volt.
Be careful with the wording. The graph flattening means the resistance is increasing, because resistance is $V / I$ , not the gradient of this graph.
A diode. Almost no current for negative p.d., and effectively none for small positive p.d.; then the current rises very steeply once the p.d. exceeds about 0.7 V.
The explanation: a diode has a very high resistance in the reverse direction and a very low resistance in the forward direction. It lets current flow one way only.
A diode's graph is the only one of the three that is not symmetrical, and that asymmetry is the entire point of the component.

Worked examples

Example 1. A p.d. of $9.0$ V across a resistor produces a current of $0.30$ A. Calculate the resistance.

Show the working

Identify the equation from what is given: p.d. and current in, resistance out. So $R = V / I$ .
Substitute.
$R = \frac{9.0}{0.30} = 30 Ω$
Give the unit — the ohm, $Ω$ .
Sense-check by working backwards: $V = I R = 0.30 \times 30 = 9.0$ V. Correct.

Answer: $R = 30 Ω$ .

Example 2. A wire has resistance $12 Ω$ . Determine the resistance of a wire of the same material that is three times as long and has twice the cross-sectional area.

Show the working

Take the two effects one at a time, since they act independently.
Resistance is directly proportional to length, so tripling the length multiplies the resistance by 3: $12 \times 3 = 36 Ω$ .
Resistance is inversely proportional to cross-sectional area, so doubling the area halves the resistance.
Apply the second factor to the result of the first.
$R = 12 \times 3 \times \frac{1}{2} = 18 Ω$
Sense-check the direction: it is longer (more resistance) but thicker (less), and the lengthening wins by a factor of three against two, so the resistance should rise. It has.

Answer: $R = 18 Ω$ .

Example 3. Explain the shape of the current–voltage graph for a filament lamp.

Show the working

Describe the shape first: a curve through the origin that becomes less steep as the p.d. increases, symmetrical about the origin.
Give the physical cause: as the current increases, more energy is transferred in the filament and its temperature rises.
Explain what heat does at the particle level: the metal ions vibrate more, so the drifting electrons collide with them more frequently.
Link that to resistance: more frequent collisions mean greater opposition to the current, so the resistance increases.
Close the loop: because the resistance rises, each additional volt produces a smaller increase in current, which is exactly what the flattening curve shows.

Answer: The curve flattens because the filament heats up as the current increases. The metal ions vibrate more, the electrons collide with them more often, and the resistance rises — so each extra volt produces a smaller increase in current.

Example 4. Describe an experiment to determine the resistance of a fixed resistor, and explain how a graph can give a more reliable value than a single calculation.

Show the working

Set up a series circuit with a variable supply, an ammeter and the resistor, and connect a voltmeter in parallel across the resistor.
Adjust the supply to obtain a reading, then record the current $I$ and the p.d. $V$ together.
Repeat for at least six different settings, covering a good spread of currents, and switch off between readings so the resistor does not warm up.
Plot $V$ on the vertical axis against $I$ on the horizontal axis and draw the best straight line through the points and the origin.
The gradient of that line is the resistance. This is more reliable than one calculation because it uses every reading and averages out random errors, and an anomalous point is visible rather than hidden.

Answer: Measure $V$ and $I$ together for several supply settings, plot $V$ against $I$ , and take the gradient of the best straight line as the resistance. The graph uses all the data, averaging random error and exposing anomalous points, which a single $V / I$ calculation cannot do.

Where people go wrong

Mistake: Reading the gradient of an $I$ – $V$ graph as the resistance.

Instead: Resistance is $V / I$ . On a graph of $V$ against $I$ the gradient is the resistance; on a graph of $I$ against $V$ it is $1 / R$ .

Which axis is which decides the answer. Check the axes before taking any gradient.

Mistake: Saying the filament lamp's resistance decreases because the graph flattens.

Instead: The resistance increases. Flattening means less current per extra volt, which is greater opposition.

The flattening is caused by heating, and heating raises the resistance of a metal.

Mistake: Halving the resistance when the diameter of a wire is doubled.

Instead: Doubling the diameter quadruples the area, so the resistance falls to a quarter.

The proportionality is to cross-sectional area, which depends on the square of the radius.

Mistake: Drawing a diode's $I$ – $V$ graph symmetrically.

Instead: A diode conducts in one direction only, so its graph is not symmetrical.

The asymmetry is the defining property of the component; a symmetrical graph describes a resistor.

Mistake: Taking a single pair of readings and calling it the resistance.

Instead: Take several pairs and plot a graph, or at least take a mean.

One reading carries the full random error and gives no way of spotting an anomalous result.

Check yourself

State the equation for resistance and the unit it is measured in.
Answer
$R = V / I$ , measured in ohms ( $Ω$ ).
A current of $2.5$ A flows through a $8.0 Ω$ resistor. What is the p.d. across it?
Answer
$V = I R = 2.5 \times 8.0 = 20$ V.
State how the resistance of a metallic wire depends on its length and its cross-sectional area.
Answer
Resistance is directly proportional to length and inversely proportional to cross-sectional area.
A wire of resistance $20 Ω$ is replaced by one of the same material, half the length and the same thickness. What is the new resistance?
Answer
$10 Ω$ , because resistance is directly proportional to length.
Sketch and describe the $I$ – $V$ graph for a resistor of constant resistance.
Answer
A straight line through the origin, symmetrical in both directions. It shows current is directly proportional to p.d., so the resistance is constant.
Why does a diode's $I$ – $V$ graph show almost no current for negative p.d.?
Answer
A diode has a very high resistance in the reverse direction, so it allows current to flow in one direction only.
In an experiment to find resistance, why should the current be kept low and the circuit switched off between readings?
Answer
A current warms the component, and a change in temperature changes its resistance, so the value being measured would not stay constant.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A student measures the p.d. across a component and the current through it, obtaining $6.0$ V and $0.40$ A.

(a) Calculate the resistance of the component. 2 marks

Mark scheme

C1 R = V / I used
A1 15 Ω; unit required

(b) Describe how the ammeter and voltmeter must be connected in this experiment. 2 marks No calculator

Mark scheme

B1 ammeter in series with the component
B1 voltmeter in parallel across the component

(c) Explain why plotting a graph of several readings gives a more reliable value of the resistance than a single calculation. 2 marks No calculator

Mark scheme

B1 the gradient uses all the readings, averaging out random errors
B1 an anomalous result can be identified and ignored, owtte

Question 2. Two wires, P and Q, are made from the same metal. Wire Q is twice as long as wire P and has half the cross-sectional area. Wire P has a resistance of $5.0 Ω$ .

(a) Calculate the resistance of wire Q. 3 marks

Mark scheme

C1 doubling the length doubles the resistance
C1 halving the area doubles the resistance
A1 20 Ω; unit required

(b) Explain, in terms of the movement of electrons, why a longer wire has a greater resistance. 2 marks No calculator

Mark scheme

B1 the electrons travel further through the metal
B1 so there are more collisions with the metal ions, opposing the current more, owtte

Question 3. A student obtains current–voltage graphs for a fixed resistor, a filament lamp and a diode.

(a) Sketch the graph obtained for the filament lamp, for both positive and negative values of p.d. 2 marks No calculator

Mark scheme

B1 curve through the origin becoming less steep as V increases
B1 symmetrical about the origin, with the same shape for negative V

(b) Explain the shape of this graph. 3 marks No calculator

Mark scheme

B1 as the current increases the filament becomes hotter
B1 the metal ions vibrate more, so electrons collide with them more frequently
B1 the resistance increases, so the current rises less for each extra volt

(c) Compare the diode's graph with the fixed resistor's graph. 2 marks No calculator

Mark scheme

B1 the resistor's graph is a straight line through the origin and symmetrical
B1 the diode conducts only in the forward direction, so its graph is not symmetrical; almost no current flows for reverse p.d.

Words to know

Resistance: The opposition of a component to the current through it, $R = V / I$ , measured in ohms.
Ohm ( $Ω$ ): The unit of resistance. One ohm is one volt per amp.
Filament lamp: A lamp whose resistance increases as its filament heats up, giving a flattening $I$ – $V$ curve.
Diode: A component with very low resistance in one direction and very high resistance in the other.
Cross-sectional area: The area of the face of a wire cut across its width; resistance is inversely proportional to it.

All Physics lessons Next: 4.2.5 Electrical energy and electrical power