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Resistance of a component

Key idea: Measure voltage across a component and current through it together, then divide voltage by current to find resistance. Repeat at several settings and switch off between readings to limit heating.

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  • SEC G2 Science Physics component 2027
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No meter reads resistance directly in this investigation. Instead you measure the potential difference across a component and the current through it, at several settings, and find R = V/I from a graph of the readings. The method uses the meter connections from D.C. circuits, and its main problem is that the current warms the component and changes the very resistance you are measuring. Use only the low-voltage supply provided, have the circuit checked before switching on, and switch off before changing any connection.

Set up the circuit

R = V/I

V must be the potential difference across the component and I the current through it, measured at the same time.

Measure voltage across the component and current in the loopA low-voltage source, open switch, rheostat, ammeter and tested resistor form a series circuit. The voltmeter is connected only across the tested resistor, excluding the ammeter and rheostat. The switch is shown open for wiring. The voltmeter current is assumed negligible, so the ammeter reading closely approximates the component current.The voltmeter spans only the component under test+−low-voltagesourceswitch openrheostatAammeterin seriesVvoltmetertested resistorRead V and I at the same setting.Keep temperature approximately constant.
Measure voltage across the component and current in the loopA low-voltage source, open switch, rheostat, ammeter and tested resistor form a series circuit. The voltmeter is connected only across the tested resistor, excluding the ammeter and rheostat. The switch is shown open for wiring. The voltmeter current is assumed negligible, so the ammeter reading closely approximates the component current.The voltmeter spans only the component under test+−low-voltagesourceswitch openrheostatAammeterin seriesVvoltmetertested resistorRead V and I at the same setting.Keep temperature approximately constant.
Wire with the switch open. The ammeter goes in the loop; the voltmeter spans only the tested component. Use a suitable low-voltage supply, small currents and safe meter ranges. This model neglects voltmeter current.

Assume the voltmeter draws negligible current, so the ammeter reads the component current.

  1. Connect the cells, a switch, a rheostat, the ammeter and the component in one series loop.
  2. Connect the voltmeter across the component only, in parallel with it.
  3. Check that both analogue meters read zero before switching on. Use a range that safely covers the expected value; choose a more sensitive suitable range when needed, following the meter instructions.
  4. Switch on, set the rheostat, and read V and I together. Switch off.
  5. Take several paired readings within the component and meter ratings. Begin with a small current, switch off between readings and allow cooling if readings drift. A wide range is useful only while the component temperature stays approximately constant.

Check your understanding 1 Which meter must carry the component current?

Which meter is connected in series with the component?
Show answer

Connect the ammeter in series so it carries the component’s current. Connect the voltmeter in parallel across the component only. Wire with the supply switched off; an ammeter directly across a supply would make an unsafe low-resistance path.

These are constructed example readings for a resistor, read with an ammeter of full-scale deflection 1 A (to 0.01 A) and a voltmeter of full-scale deflection 3 V (to 0.05 V).

V / VI / AR = V/I / Ω
0.600.125.0
1.000.205.0
1.400.294.8
1.800.365.0
2.200.445.0
2.600.525.0

Potential difference against current for a resistor

Plotted points of potential difference V in V against current I in A for currents from 0.12 A to 0.52 A, with a dashed best-fit straight line through the origin. The data table lists every plotted value.

Scroll across the figure to read all labels.

Plotted points of potential difference V in V against current I in A for currents from 0.12 A to 0.52 A, with a dashed best-fit straight line through the origin. The data table lists every plotted value.Plotted points of potential difference V in V against current I in A for currents from 0.12 A to 0.52 A, with a dashed best-fit straight line through the origin. The data table lists every plotted value.
Six constructed example pairs of current and potential difference for a resistor at approximately constant temperature. The dashed line is an illustrative straight fit.
Open full-size graph
What to noticeConstructed practice readings, not reported measurements. Values are supplied to 0.01 A and 0.05 V; recording steps do not establish the full uncertainty. A straight V-against-I fit through the origin has gradient R.
View figure data
Figure data: Potential difference against current for a resistor
SeriesCurrent I (A)Potential difference V (V)
Example readings0.120.6
Example readings0.201.0
Example readings0.291.4
Example readings0.361.8
Example readings0.442.2
Example readings0.522.6
Best-fit line0.000.0
Best-fit line0.562.8

Worked example 1

Resistance from the graph

Problem

Use the graph of V against I to find the resistance of the resistor.

Show full solution

  1. Check the shape

    Method

    The points lie close to a straight line through the origin.

    Reason

    A straight pattern through the origin is consistent with approximately constant V/I over this range, provided the temperature stays constant.

    Working

    Answer

    At 1.40 V and 0.29 A, R = 4.83 Ω, about 3% below 5.0 Ω. It is close to the pattern, so check repeated readings before deciding its cause.

  2. Find the gradient

    Method

    Read (0.10 A, 0.50 V) and (0.50 A, 2.50 V) from the line.

    Reason

    For a straight V-against-I line through the origin, V = RI, so the gradient Δ V/Δ I equals the constant resistance.

    Working

    Answer

    R = (2.50 - 0.50)/(0.50 - 0.10) = (2.00 V)/(0.40 A) = 5.0 Ω

  3. State the resistance

    Working

    Answer

    The resistance is 5.0 Ω over the measured range, with temperature approximately constant.

Check your understanding 2 Read another V-against-I gradient

Show the steps Optional working
Unit: V
Unit: A
Unit: Ω
Show answer

For this origin-passing constant-temperature line, R = ΔV/ΔI = (3.60 − 0.40)/(0.45 − 0.05) = 8.0 Ω. The axis order matters. An I-against-V gradient would be 1/R, not R.

Keep the component cool

A current warms the component it passes through, and the resistance of a typical metal wire generally rises as it warms up; the size of the change depends on the material. Leaving the current on can raise the temperature and change V/I. For a metal wire whose resistance rises with temperature, the ratio drifts upwards. Switch off between readings, allow cooling and keep currents small. Recheck an earlier setting to see whether the resistance is drifting.

Check your understanding 3 Readings that drift upwards

A metal wire initially carries 0.20 A at 1.0 V. After the current has been left on, it carries 0.16 A at the same p.d. After cooling, the current returns to 0.20 A. Calculate the resistance before and during the drift. Which observation supports heating rather than a permanent change to the wire?

Unit: Ω
Unit: Ω
Evidence for a reversible temperature effect
Show answer

Initially, R = 1.0 ÷ 0.20 = 5.0 Ω. During the drift, R = 1.0 ÷ 0.16 = 6.25 Ω, about 6.3 Ω. Returning to the original current after cooling also returns the resistance to 5.0 Ω. That reversibility supports a temperature effect rather than a permanent change. Switch off between readings and allow cooling to keep the temperature approximately constant.

Evaluate the method

Worked example 2

Evaluating the resistor experiment

Problem

The smallest current in the resistor table is 0.12 A, read on a 1 A meter to 0.01 A. Evaluate this reading and suggest an improvement.

Show full solution

  1. Source

    Method

    The reading uses only a small part of the scale.

    Reason

    An absolute current-reading error has a larger fractional effect on a smaller current. A display or recording step is only one contribution to uncertainty.

    Working

    Answer

    For illustration, 0.01 A is about 8% of 0.12 A, but about 2% of 0.52 A. They show why the same reading error matters more for a small reading.

  2. Effect

    Method

    For a given absolute current-reading error, the fractional effect on R = V/I is largest at the smallest currents.

    Reason

    A small error in a small current makes a large change in V/I.

    Working

    Answer

    A fit uses all the readings to judge the trend and can reduce the effect of random scatter. It cannot remove a common meter offset or a temperature change.

  3. Choose a suitable range

    Method

    Use a meter range, or a milliammeter, on which small currents give a large deflection.

    Reason

    A suitable lower range may give a finer absolute reading step, making the reading error a smaller fraction of the measured current.

    Working

    Answer

    Also look straight at the scale, so that there is no parallax, and check the needle reads zero before switching on.

  4. Give the evaluation

    Working

    Answer

    A small current reading has a large fractional sensitivity to a fixed reading error. Choose a safe lower meter range with finer divisions for small currents, read straight-on and check the zero.

SourceEffectImprovement
Current left on between readingsThe component warms up and its resistance changes, so later values of V/I driftSwitch off between readings and keep the current small.
Meter needle not at zero before switching onEvery reading is shifted by the same amountAdjust the zero, or subtract the zero reading.
Voltmeter connected across the ammeter and component togetherV includes the potential difference across the ammeterConnect the voltmeter across the component only.
Loose or dirty crocodile-clip contactsReadings flicker or change when a wire is movedClean and tighten the connections.

Check your understanding 4 Identify what the voltage measurement includes

A component carries 0.40 A with 1.80 V across it. The series ammeter has 0.20 V across it. Assume negligible voltmeter current. A voltmeter spanning both reads 2.00 V. Find the component resistance and choose where to reconnect the voltmeter.

Unit: Ω
Correct voltmeter connection
Show answer

The component resistance is 1.80/0.40 = 4.5 Ω. Using the combined 2.00 V gives 5.0 Ω, including the ammeter’s 0.20/0.40 = 0.50 Ω. Reconnect the voltmeter across the component only, with the supply switched off.

Check your understanding 5 Compare the corrected circuit

Draw the corrected circuit on paper with a supply, open switch, rheostat, ammeter, component and voltmeter. Compare its meter branches with the model.

Show answer

The supply, open switch, rheostat, ammeter and component form one series loop. The voltmeter spans the component’s own endpoints in parallel, excluding the ammeter and rheostat. Change connections with the supply off.

The same circuit is used to find how a wire’s resistance depends on its length in wire length and resistance.

Common mistakes

  • Connecting the ammeter across the component, or directly across the supply, instead of in series.
  • Connecting the voltmeter across the ammeter and the component together instead of the component only.
  • Taking V and I at different settings, or after the component has warmed up.
  • Leaving the current on between readings, so the resistance drifts upwards.
  • Taking the gradient of an I against V graph as R. On those axes the gradient is 1/R.

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