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Wire length and resistance

Key idea: Vary the measured length of one wire and find each resistance from voltage divided by current. Keep the wire cool and its thickness unchanged to test resistance against length.

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  • SEC G2 Science Physics component 2027
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In a longer wire, moving electrons encounter more resistance along their path. For the same material, thickness and temperature, resistance increases with length. Moving a crocodile clip along a single wire lets you change its length while keeping the material and thickness the same, and the circuit from resistance of a component measures each resistance. Use only the low-voltage supply provided, switch off before moving a clip, and keep the current small: the wire and contacts can become hot.

Resistance and the length of a wire

The paired-meter method now tests a new variable: the length of wire between two contacts. Connect the voltmeter across exactly that section, and keep the material, cross-sectional area and temperature approximately constant.

Measure voltage and length between the same contactsA low-voltage source, open switch, rheostat and ammeter form a series circuit with the selected section of a resistance wire. A voltmeter connects between the two contacts that bound that section. The ruler runs from zero at the wire end; the fixed contact is at 0.10 m and the moving contact at 0.70 m, so selected length l is 0.60 m. Unused wire extends beyond both contacts. The switch is shown open for wiring. The voltmeter current is assumed negligible.Selected wire section: match the contacts and meter leads+−low-voltagesourceswitch openrheostatAammeterin seriesVvoltmeterfixed contactmoving contact0102030405060708090cml = 0.60 m between contacts
Measure voltage and length between the same contactsA low-voltage source, open switch, rheostat and ammeter form a series circuit with the selected section of a resistance wire. A voltmeter connects between the two contacts that bound that section. The ruler runs from zero at the wire end; the fixed contact is at 0.10 m and the moving contact at 0.70 m, so selected length l is 0.60 m. Unused wire extends beyond both contacts. The switch is shown open for wiring. The voltmeter current is assumed negligible.Selected wire section: match the contacts and meter leads+−low-voltagesourceswitch openrheostatAammeterin seriesVvoltmeterfixed contactmoving contact0102030405060708090cml = 0.60 m between contacts
Measure l between the contacts: here 0.70 − 0.10 = 0.60 m. The voltmeter must connect across that same section. Switch off before moving contacts; use a small current and check for heating.

For a uniform wire at constant temperature, resistance is expected to be directly proportional to length. Review resistance and wire dimensions.

  1. Tape a constantan wire along a metre rule. Connect one end of it into the circuit in place of the component.
  2. Make the second connection with a crocodile clip or jockey at a length l along the wire. Measure l between the two contact points.
  3. Connect the voltmeter across that length only.
  4. With the circuit switched off, set the contacts and voltmeter leads for the selected length. Switch on briefly, adjust to a small suitable current, read V and I together, then switch off. Keep the on-time brief to limit heating.
  5. Repeat across a range of lengths, using the same wire. Recheck readings at a smaller current if resistance drifts, and record repeats so you can judge the scatter.

Check your understanding 1 The wrong section

Current passes through 0.60 m of wire, but the voltmeter spans only 0.40 m. Which section’s resistance does V/I give? How should the voltmeter be connected for the 0.60 m section?

Section measured by V/I
Voltmeter connection for 0.60 m
Show answer

V/I gives the resistance of the 0.40 m section only. Switch off, then connect the voltmeter across the two contacts marking the 0.60 m length.

These constructed example readings use five lengths and the same nominal small current. The values are for practising analysis, not reported measurements.

l / mV / VI / A
0.2000.250.30
0.4000.550.30
0.6000.800.30
0.8001.050.30
1.0001.300.30

Resistance against selected wire length

Five wire lengths from 0.200 m to 1.000 m have calculated resistances from about 0.83 Ω to 4.33 Ω. The points lie close to a straight illustrative fit with gradient about 4.33 Ω/m and a small intercept of about 0.03 Ω. The table gives each plotted coordinate.

Scroll across the figure to read all labels.

Five wire lengths from 0.200 m to 1.000 m have calculated resistances from about 0.83 Ω to 4.33 Ω. The points lie close to a straight illustrative fit with gradient about 4.33 Ω/m and a small intercept of about 0.03 Ω. The table gives each plotted coordinate.Five wire lengths from 0.200 m to 1.000 m have calculated resistances from about 0.83 Ω to 4.33 Ω. The points lie close to a straight illustrative fit with gradient about 4.33 Ω/m and a small intercept of about 0.03 Ω. The table gives each plotted coordinate.
Constructed example wire readings. Resistance is calculated from paired V and I values for each selected length. The straight fit is not forced through the origin.
Open full-size graph
What to noticeThese are constructed analysis data, not measurements. A nearly straight pattern close to the origin supports approximate proportionality over the tested range, but does not prove an exact law or identify the cause of deviations. Calculated plotting values are rounded to three significant figures; this is a presentation choice, not an uncertainty estimate.
View figure data
Figure data: Resistance against selected wire length
SeriesSelected wire length l (m)Resistance R (Ω)
Example readings0.20.83300
Example readings0.41.83000
Example readings0.62.67000
Example readings0.83.50000
Example readings1.04.33000
Illustrative straight fit0.00.03333
Illustrative straight fit1.14.80000

The example fit is drawn freely rather than forced through zero. Repeat the measurements and check the meters before explaining a small offset.

Guided practice 1

Is resistance proportional to length?

About 5 min

Problem

For the 0.600 m and 0.400 m rows, calculate resistance divided by length. Use V/I for each resistance. The full solution compares all five rows.

Try this before viewing the solution

Unit: Ω
Unit: Ω
Show the steps Optional working
Unit: Ω
Unit: Ω/m
Show the steps Optional working
Unit: Ω
Unit: Ω/m

Hints

Hint 1

one row at a time

Use R = V/I with the readings in the same row.

Hint 2

what stays the same?

If R is directly proportional to l, what is the same for every row?

Show solution step by step

  1. Resistance of each length

    Method

    Divide each V by the current in the same row.

    Reason

    V and I were read together for that length.

    Working

    Answer

    R = 0.83 Ω, 1.83 Ω, 2.67 Ω, 3.50 Ω and 4.33 Ω.

  2. Test for proportionality

    Method

    Work out R/l for each row.

    Reason

    Direct proportion means R/l is the same for every length: a graph of R against l is a straight line through the origin.

    Working

    Answer

    R/l = 4.17, 4.58, 4.44, 4.38 and 4.33 Ω/m. The ratios are close to about 4.4 Ω/m.

  3. Assess the deviations

    Method

    State the conclusion and which readings are most affected by a given voltage-reading error.

    Reason

    The shortest lengths have the smallest potential differences. A given absolute voltage-reading error has a larger fractional effect there; repeated readings and the meter specifications are needed to assess uncertainty.

    Working

    Answer

    “The results are approximately consistent with resistance increasing in direct proportion to length over the tested range. The R-against-l pattern is nearly straight and close to the origin, with a gradient around 4.3–4.4 Ω/m. Repeat readings to check the scatter around the trend.”

  4. State the conclusion

    Working

    Answer

    The results are approximately consistent with R being proportional to l over the tested range. Repeats and meter specifications are needed to judge whether the departures are within measurement uncertainty.

Check your understanding 2 A line that misses the origin

A straight resistance–length graph gives 4.0 Ω at 0.50 m and 6.0 Ω at 1.00 m. Find the resistance ratio R(1.00 m)/R(0.50 m), the gradient and the intercept. Is resistance directly proportional to length?

Show the steps Optional working
Unit: Ω
Unit: m
Unit: Ω/m
Show the steps Optional working
Unit: Ω/m
Unit: Ω
Directly proportional?
Show answer

The resistance ratio is 6.0/4.0 = 1.5, while length doubles. Gradient = (6.0 − 4.0)/(1.00 − 0.50) = 4.0 Ω/m. Intercept = 4.0 − 4.0 × 0.50 = 2.0 Ω. A non-zero intercept means this straight line is not direct proportion.

Check your understanding 3 Use each trial’s own current

Contacts in trial A lie at 0.10 m and 0.40 m: V = 0.60 V, I = 0.20 A. In B they lie at 0.20 m and 0.80 m: V = 0.90 V, I = 0.15 A. Find each selected length and resistance.

Unit: m
Unit: m
Unit: Ω
Unit: Ω
Show answer

A: length = 0.40 − 0.10 = 0.30 m, resistance = 0.60/0.20 = 3.0 Ω. B: length = 0.80 − 0.20 = 0.60 m, resistance = 0.90/0.15 = 6.0 Ω.

Check your understanding 4 Explain the comparison

The same uniform wire was kept at the same temperature in both trials. Explain what the results suggest about resistance and length, despite the different currents. What further readings would strengthen the conclusion?

Show answer

Doubling length from 0.30 m to 0.60 m doubles resistance from 3.0 Ω to 6.0 Ω; R/length is 10 Ω/m in both trials. Different paired currents are suitable for comparison while material, thickness and temperature stay fixed. Test more lengths with repeats, then inspect a freely fitted resistance–length graph.

Common mistakes

  • Measuring the length from the end of the rule instead of between the two contacts.
  • Connecting the voltmeter across a different section from the one carrying the measured current.
  • Leaving the current on so long that the wire warms up and its resistance rises.
  • Forcing the best-fit line through the origin, or calling a line with a clear intercept direct proportion.

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