How to draw the graph

1. Put the right quantity on the right axis.The quantity you changed goes across the bottom — here that is the current. The quantity you measured because of it goes up the side — the voltage.
2. Label both axes with the quantity and its unit.current in R (A), not I and not amps. An unlabelled axis loses the mark every time.
3. Mark the origin 0.Both axes start at zero for a V–I graph, because no current means no voltage.
4. Choose a scale that fills the paper.Your points should spread over more than half the grid in both directions. Make each big square worth 1, 2, 5 or 10 — never 3, 7 or 15, because you will misread it.
5. Plot each point as a small neat cross.A fat blob is worth about three squares of uncertainty.
6. Draw one best-fit line, not dot-to-dot.A straight line if the points lie on one; a smooth curve if they clearly bend. Roughly as many points should sit above the line as below it.
7. Read your line, not your points, to get the resistance.Pick a point that lies exactly on the line — ideally where it crosses a heavy grid line so both values are easy to read — and use R = V ÷ I. Read a point off the line even if none of your measured crosses sits there; that is what the line is for.
8. Steeper line, bigger resistance.You will not be asked to calculate a gradient at National 5 in S3 — that comes in S4 — but you do need to say what a steeper line means: more volts are needed to push the same current through, so the resistance is larger.

What loses marks

• Axes swapped — voltage across the bottom. Read the question: it says plot voltage against current, and the first-named quantity goes up the side.
• A scale like 3 V per big square, so every reading is a guess.
• Points joined dot-to-dot in a zig-zag instead of one best-fit line.
• Using a measured cross that does not sit on the best-fit line. Read the line.
• Units left off the axes, or off the final answer.

The tasks

Task 1 — a resistor

A pupil measures the current in a resistor and the voltage across it, for six different settings of a variable resistor.

current in R (A)	0·02	0·04	0·06	0·08	0·10	0·12
voltage across R (V)	1·39	2·68	4·13	5·41	6·86	8·11

1. Plot a graph of voltage across R against current in R.
2. Draw the best-fit line.
3. Read a point that lies on your line and use R = V ÷ I to determine the resistance of R. Write down the point you used.
4. State one thing about the graph that tells you the resistor obeys Ohm's law.

Task 2 — a different resistor

The pupil repeats the experiment with a second resistor.

current in R (A)	0·05	0·10	0·15	0·20	0·25	0·30
voltage across R (V)	1·60	3·36	4·91	6·67	8·19	9·94

1. Plot this data on a new pair of axes and draw the best-fit line.
2. Read a point on your line and determine the resistance of this resistor.
3. Sketch, on one set of axes, how the two lines would compare. State whether the steeper line belongs to the larger or the smaller resistance, and say why.

Task 3 — a filament lamp

The pupil now replaces the resistor with a filament lamp and repeats the experiment again.

current in the lamp (A)	0·10	0·20	0·30	0·40	0·50	0·60
voltage across the lamp (V)	0·35	0·95	1·75	2·75	3·95	5·35

1. Plot the data and draw a smooth best-fit curve. Do not force a straight line through these points.
2. Determine the resistance of the lamp when the current in it is 0·20 A.
3. Determine the resistance of the lamp when the current in it is 0·60 A.
4. State what happens to the resistance of the lamp as the current increases, and explain why.

✅ Check your graphs — answers and marking notes

Task 1

Read a point that lies on your line — for example (0·10, 6·8) — and use R = V ÷ I: 6·8 ÷ 0·10 = 68 Ω. Anything from about 65 Ω to 71 Ω is a good answer, because you are reading a hand-drawn line.

The line is straight and passes through the origin, so the voltage is directly proportional to the current — that is Ohm's law, and it means the resistance is the same at every current.

Task 2

Using (0·20, 6·6) from the line: 6·6 ÷ 0·20 = 33 Ω (accept about 31 Ω to 35 Ω).

The steeper line is the larger resistance: a steeper line means more volts are needed to push the same current through, which is exactly what a bigger resistance does. So Task 1's resistor (68 Ω) gives the steeper line. You are not asked to calculate a gradient at National 5 in S3 — only to say what a steeper line means.

Task 3

At 0·20 A the voltage is 0·95 V, so R = V ÷ I = 0·95 ÷ 0·20 = 4·7 Ω.

At 0·60 A the voltage is 5·35 V, so R = 5·35 ÷ 0·60 = 8·9 Ω.

The resistance increases — by a factor of about 1·9. The larger current heats the filament, the metal ions vibrate more, the electrons collide with them more often, and the resistance rises. That is why the graph curves upwards instead of staying straight.

Watch out: for a curve there is no single answer to "what is the resistance" — it depends where you look. You must read one point off the curve and divide, and say which current it belongs to.

Sources & credits: the graphing method and the data on this page are original to this site, written to match the V–I experiment examined at National 5 — past papers and marking instructions © Qualifications Scotland (SQA). Graph paper and diagrams drawn for this site © R Stewart, 2026.