How to Measure Resistance with a Voltmeter and Ammeter

A resistor is labelled 12 Ω, but the label isn’t a measurement. How would you check its resistance yourself? Measure the potential difference across it with a voltmeter and the current flowing through it with an ammeter, then calculate \(R=V/I\).

The basic method is simple: connect the ammeter in series with the resistor and the voltmeter in parallel across it. However, getting a reliable result involves more than connecting two meters. You need to use the correct voltage reading, control the current, and understand how the meters themselves can affect the measurement.

01The ammeter-voltmeter method

To measure an unknown resistance:

1. Connect the ammeter in series with the test resistor.
2. Connect the voltmeter in parallel across the test resistor.
3. Switch on the low-voltage DC supply and record both readings.
4. Calculate the resistance using \(R=V/I\).
5. Repeat at several different current settings to check whether the resistance remains constant.

The circuit also contains a variable resistor, which allows you to adjust the current without changing the test resistor.

Why take both measurements?

The resistance of a component describes how much it opposes electric current.

Suppose you increase the potential difference across a resistor. What should happen to the current?

For an ordinary ohmic resistor at constant temperature, doubling the potential difference doubles the current. This is the relationship described by Ohm’s law.

Mathematically:

\[ V=IR \]

Rearranging gives the equation used to measure resistance:

\[ \boxed{R=\frac{V}{I}} \]

Where:

• \(R\) = resistance, measured in ohms (\(\Omega\)).
• \(V\) = potential difference across the resistor, measured in volts (V).
• \(I\) = current through the resistor, measured in amperes (A).

One ohm is equivalent to one volt per ampere.

\[ 1\ \Omega = \frac{1\ \mathrm{V}}{1\ \mathrm{A}} \]

The important detail is that you must use the potential difference across the test resistor, not automatically the voltage displayed on the power supply.

If another component is connected in series, some of the supply voltage will be across that component instead.

02Setting up the experiment correctly

Follow the circuit from the positive terminal of the power supply:

• Connect the positive terminal to a switch.
• Connect the switch to a variable resistor.
• Connect the variable resistor to the ammeter.
• Connect the ammeter to the unknown resistor.
• Return the resistor to the negative terminal of the supply.
• Connect the voltmeter across the two terminals of the unknown resistor.

The ammeter measures the current in the circuit, while the voltmeter measures the potential difference between the two ends of the test resistor.

For the standard ideal-meter model, the voltmeter draws negligible current, so the ammeter reading is treated as the current through the resistor.

Before switching on:

• Set the variable resistor to its maximum series resistance.
• Check that the ammeter and voltmeter are connected with the correct polarity.
• Select appropriate DC measurement modes and ranges.
• Make sure the switch is open while connecting the circuit.

Use a low-voltage supply, not mains electricity.

For the reasoning behind the series and parallel connections, see Why Ammeters Go in Series and Voltmeters in Parallel.

For practical guidance on selecting a multimeter range, see How to Choose a Multimeter Range Before Taking a Reading.

Taking reliable measurements

Once the circuit is connected, follow this sequence:

1. Close the switch and record the ammeter and voltmeter readings at the same time.
2. Open the switch if the resistor is likely to heat up.
3. Adjust the variable resistor to change the current.
4. Close the switch and record another pair of readings.
5. Repeat until you have several sets of measurements.

Why take several readings instead of just one?

A single pair of readings gives you an estimate of resistance. Several pairs let you check whether the resistance remains constant and reduce the influence of measurement uncertainty.

Keep the current reasonably low. A resistor heats up when electrical energy is transferred into thermal energy. For many metal resistors, increasing temperature also increases resistance.

If the resistor heats significantly between readings, you might calculate different resistance values even though you haven’t changed the component.

03Worked examples

04Practical extension: What happens when the meters aren’t ideal?

So far, we’ve treated the ammeter and voltmeter as perfect instruments.

In reality, neither meter is perfect.

An ideal ammeter has zero resistance, while an ideal voltmeter has infinite resistance. Real instruments have:

• A small internal resistance in the ammeter.
• A large, but finite, input resistance in the voltmeter.

These properties can affect the resistance you calculate.

This is a practical extension of the standard HSC measurement method. It becomes particularly important when questions provide the internal resistance of the meters.

Arrangement 1: Voltmeter connected directly across the resistor

Consider the standard arrangement:

• The voltmeter is connected across the unknown resistor.
• The ammeter is positioned before the parallel branches.

The voltmeter has a finite resistance, so a small amount of current flows through it.

Think of the voltmeter as an additional current pathway alongside the resistor. The ammeter measures the combined current entering both pathways.

Consequently:

\[ I_A=I_R+I_V \]

Where:

• \(I_A\) is the ammeter reading.
• \(I_R\) is the current through the resistor.
• \(I_V\) is the current through the voltmeter.

Because the voltmeter and resistor have the same potential difference, the calculated resistance becomes smaller than the actual resistance.

The apparent resistance is:

\[ R_{\mathrm{measured}}= \frac{RR_V}{R+R_V} \]

Here, \(R_V\) is the voltmeter’s internal resistance.

This arrangement therefore underestimates the true resistance.

It is generally preferable when measuring low resistances, where the voltmeter current is negligible compared with the resistor current.

Arrangement 2: Voltmeter connected across the resistor and ammeter

There is another way to arrange the meters.

• Connect the ammeter in series with the unknown resistor.
• Connect the voltmeter across the combination of the ammeter and resistor.

Now the ammeter measures the current through the resistor alone, but the voltmeter measures the potential difference across both components.

The measured voltage therefore includes the voltage drop across the ammeter’s internal resistance.

The apparent resistance becomes:

\[ R_{\mathrm{measured}}=R+r_A \]

Where \(r_A\) is the internal resistance of the ammeter.

This arrangement overestimates the actual resistance.

It is generally preferable when measuring high resistances, because the ammeter’s internal resistance is small compared with the resistance being measured.

Which arrangement should you use?

The decision depends on the relative sizes of the unknown resistance and the meter resistances.

For the first arrangement, the relative error is approximately \(R/R_V\) when \(R\) is much smaller than \(R_V\).

For the second arrangement, the relative error is approximately \(r_A/R\) when \(r_A\) is much smaller than \(R\).

The best arrangement is the one with the smaller expected relative error.

Remember that these corrections are only necessary when you are considering real meter resistances. If a question assumes ideal meters or provides no internal resistance values, use the standard ideal-meter model unless instructed otherwise.

05Temperature and resistance measurements

There is another reason why repeated readings might not give exactly the same resistance.

Consider a metal resistor carrying current. As current flows, electrical energy is transferred into thermal energy, heating the resistor.

For many metal conductors, increasing temperature increases resistance.

This creates an important limitation when measuring resistance: the act of taking a measurement can change the property you are trying to measure.

For an ordinary ohmic resistor, Ohm’s law assumes constant temperature.

If the temperature changes, the resistance may change too. The current might therefore increase by less than expected when you increase the potential difference.

To reduce this problem:

• Use a relatively low current.
• Take readings quickly.
• Avoid leaving the switch closed unnecessarily.
• Allow the resistor to cool between measurement sets.
• Keep the experimental conditions as consistent as possible.

This is especially important when measuring components such as filament lamps.

A filament lamp is not an ohmic conductor because its resistance changes significantly as its filament heats up.

For these components, you should not expect a single constant resistance across all operating conditions.

06Common mistakes to avoid

A good experimental result is not just a number. You should be able to explain how the circuit produced that number and why the measurement is reliable.

07Questions and solutions

Question 1

An unknown resistor is connected to a circuit containing an ammeter and voltmeter. The voltmeter reads 3.0 V, and the ammeter reads 0.25 A.

Assuming ideal meters and constant temperature, calculate the resistance of the resistor.

Solution 1

The resistance is \(12\ \Omega\).

Use Ohm’s law:

\[ R=\frac{V}{I} \]

Substitute the measured values:

\[ R=\frac{3.0}{0.25} \]

\[ \boxed{R=12\ \Omega} \]

The resistor has a resistance of 12 Ω because a potential difference of 3.0 V produces a current of 0.25 A.

The key is to use the voltage across the resistor, rather than automatically using the power supply’s voltage.

Question 2

A student measures the potential difference across an unknown resistor and the current through it.

The following data are collected:

(a) Plot a graph of potential difference against current.

(b) Estimate the resistance using the gradient of the best-fit line.

(c) Explain whether the data provide evidence that the resistor is ohmic.

Solution 2

The resistance is approximately \(5.0\ \Omega\), and the measurements are consistent with ohmic behaviour over the tested range.

(a) Plot current on the horizontal axis and potential difference on the vertical axis. Mark all four data points and draw a straight line of best fit.

(b) The gradient of the graph gives the resistance:

\[ R=\frac{\Delta V}{\Delta I} \]

Using the first and last measurements as an estimate:

\[ R=\frac{3.60-0.90}{0.72-0.18} \]

\[ R=\frac{2.70}{0.54} \]

\[ \boxed{R=5.0\ \Omega} \]

(c) The resistor is consistent with ohmic behaviour because the measurements follow an approximately linear relationship between voltage and current.

The small differences between individual ratios are reasonable for experimental measurements.

A common mistake is to expect every experimental reading to produce precisely the same resistance. Real measurements contain uncertainty, so a best-fit line provides a more reliable estimate than treating every small variation as evidence of non-ohmic behaviour.

Question 3

A student uses a 6.0 V DC power supply to measure the resistance of an unknown resistor.

A variable resistor is connected in series with the test resistor. The potential difference across the variable resistor is 2.0 V, and the ammeter reads 0.50 A.

Assume ideal meters and negligible internal resistance in the power supply.

(a) Calculate the potential difference across the unknown resistor.

(b) Calculate its resistance.

(c) Explain why a student who calculates resistance using \(6.0/0.50\) obtains an incorrect result.

Solution 3

The unknown resistor has a resistance of \(8.0\ \Omega\).

(a) The potential difference across the power supply is shared between the variable resistor and the unknown resistor.

Using conservation of energy in the circuit, or Kirchhoff’s voltage law:

\[ V_{\mathrm{supply}}=V_{\mathrm{variable}}+V_R \]

Rearrange:

\[ V_R=V_{\mathrm{supply}}-V_{\mathrm{variable}} \]

Substitute:

\[ V_R=6.0-2.0 \]

\[ \boxed{V_R=4.0\ \mathrm{V}} \]

(b) Use the measured voltage across the unknown resistor:

\[ R=\frac{V_R}{I} \]

\[ R=\frac{4.0}{0.50} \]

\[ \boxed{R=8.0\ \Omega} \]

(c) The calculation \(6.0/0.50=12\ \Omega\) is incorrect because it assumes the entire supply voltage is across the unknown resistor.

However, the variable resistor also experiences a potential difference.

Ohm’s law must be applied using the voltage and current associated with the same component. Using the total supply voltage would incorrectly attribute the variable resistor’s voltage drop to the unknown resistor.

Question 4

An unknown resistor has a true resistance of \(100\ \mathrm{k}\Omega\).

Two real meters are available:

• Voltmeter input resistance: \(1.0\ \mathrm{M}\Omega\).
• Ammeter internal resistance: \(2.0\ \Omega\).

Consider two measurement arrangements.

Arrangement A: The voltmeter is connected directly across the resistor, and the ammeter is positioned before the parallel branches.

Arrangement B: The ammeter is connected in series with the resistor, and the voltmeter is connected across the combination of the resistor and ammeter.

(a) Calculate the apparent resistance measured using arrangement A.

(b) Calculate the apparent resistance measured using arrangement B.

(c) Calculate the percentage error for each arrangement.

(d) Identify which arrangement gives the more accurate result and explain why.

Solution 4

Arrangement B gives the more accurate result, with an apparent resistance of approximately \(100.002\ \mathrm{k}\Omega\).

(a) Arrangement A

The voltmeter draws additional current, so the ammeter measures the combined current through both parallel branches.

The apparent resistance is the equivalent parallel resistance:

\[ R_{\mathrm{measured}} = \frac{RR_V}{R+R_V} \]

Convert the values into ohms:

\[ R=100\,000\ \Omega \]

\[ R_V=1\,000\,000\ \Omega \]

Substitute:

\[ R_{\mathrm{measured}} = \frac{100\,000\times1\,000\,000} {100\,000+1\,000\,000} \]

\[ \boxed{R_{\mathrm{measured}}=90\,909\ \Omega} \]

Or approximately:

\[ \boxed{90.9\ \mathrm{k}\Omega} \]

(b) Arrangement B

The voltmeter measures the potential difference across both the resistor and ammeter.

Therefore:

\[ R_{\mathrm{measured}}=R+r_A \]

Substitute:

\[ R_{\mathrm{measured}}=100\,000+2 \]

\[ \boxed{R_{\mathrm{measured}}=100\,002\ \Omega} \]

Or:

\[ \boxed{100.002\ \mathrm{k}\Omega} \]

(c) Percentage errors

Use:

\[ \%\text{ error}= \frac{|R_{\mathrm{measured}}-R_{\mathrm{true}}|} {R_{\mathrm{true}}}\times100 \]

For arrangement A:

\[ \%\text{ error} = \frac{|90\,909-100\,000|} {100\,000}\times100 \]

\[ \boxed{9.09\%\text{ low}} \]

For arrangement B:

\[ \%\text{ error} = \frac{|100\,002-100\,000|} {100\,000}\times100 \]

\[ \boxed{0.002\%\text{ high}} \]

(d) Comparison

Arrangement B is considerably more accurate.

The voltmeter’s input resistance is only ten times the unknown resistance, so its additional current creates a noticeable error in arrangement A.

In arrangement B, the ammeter’s internal resistance is tiny compared with the unknown resistance. Its additional voltage drop is therefore negligible.

The important lesson is that correct meter placement is not always enough to guarantee an accurate measurement. You must also consider the internal resistance of the instruments.

Question 5

A student measures the voltage-current relationship of a filament lamp using ideal meters.

The following readings are recorded after the lamp reaches a steady operating temperature at each voltage.

The student also performs a separate experiment.

Starting with a cold filament, they apply a potential difference of 4.0 V. The initial current is 0.80 A, but it gradually decreases to 0.44 A as the lamp warms up.

Assume the supply maintains a constant potential difference and the meter readings are accurate.

(a) Calculate the resistance of the lamp at 4.0 V when it has reached its steady operating temperature.

(b) Calculate the resistance of the cold filament immediately after the supply is connected.

(c) Use the steady-state data at 3.0 V and 5.0 V to estimate the gradient of the voltage-current graph around 4.0 V.

(d) A student claims that the gradient in part (c) must equal the resistance calculated in part (a), because both describe resistance. Explain why this conclusion is incorrect.

(e) Can the initial cold-filament reading be added to the same steady-state voltage-current graph? Justify your answer.

Solution 5

The steady-state operating resistance is approximately \(9.09\ \Omega\), the initial cold resistance is \(5.0\ \Omega\), and the estimated local graph gradient is approximately \(18.2\ \Omega\).

These are three different quantities, and understanding why they differ is the key to this question.

(a) Resistance at the steady operating temperature

Use the steady-state reading at 4.0 V:

\[ V=4.0\ \mathrm{V} \]

\[ I=0.44\ \mathrm{A} \]

The operating-point resistance is:

\[ R=\frac{V}{I} \]

\[ R=\frac{4.0}{0.44} \]

\[ \boxed{R=9.09\ \Omega} \]

This is the resistance of the lamp at its steady operating condition.

(b) Resistance of the cold filament

Immediately after switching on:

\[ V=4.0\ \mathrm{V} \]

\[ I=0.80\ \mathrm{A} \]

Therefore:

\[ R=\frac{V}{I} \]

\[ R=\frac{4.0}{0.80} \]

\[ \boxed{R=5.0\ \Omega} \]

The cold filament has a lower resistance than the hot filament.

As the filament heats up, its resistance increases. Since the supply maintains a constant potential difference, the current decreases.

There is no contradiction with Ohm’s law. The temperature is changing, so the resistance is not constant.

(c) Estimate the gradient around 4.0 V

The gradient of a voltage-current graph is:

\[ \text{Gradient}=\frac{\Delta V}{\Delta I} \]

Use the steady-state measurements at 3.0 V and 5.0 V:

\[ \text{Gradient} = \frac{5.0-3.0}{0.49-0.38} \]

\[ =\frac{2.0}{0.11} \]

\[ \boxed{\text{Gradient}\approx18.2\ \Omega} \]

This estimates the local gradient around the 4.0 V operating point.

(d) Why are the gradient and operating-point resistance different?

The student is confusing two different measurements.

The operating-point resistance is calculated using:

\[ R=\frac{V}{I} \]

This compares the voltage and current at a particular operating point. Graphically, it is the gradient of a line drawn from the origin to that point.

The gradient of the voltage-current graph is instead:

\[ \frac{\Delta V}{\Delta I} \]

It describes how much the voltage changes for a given change in current between nearby operating points.

For an ohmic resistor, the graph is a straight line through the origin, so these quantities are equal.

For a filament lamp, the graph is curved because the filament’s temperature and resistance change as the operating conditions change.

Consequently, the two quantities need not be equal.

The gradient of approximately 18.2 Ω is not the lamp’s operating-point resistance at 4.0 V. That value is approximately 9.09 Ω.

(e) Can the cold reading be added to the steady-state graph?

No. The cold reading does not belong on the same steady-state characteristic because it describes a different thermal condition.

The three tabulated readings were recorded after the filament reached a steady operating temperature at each voltage.

The initial 4.0 V reading was recorded while the filament was still cold and heating up.

Although both readings have the same potential difference, they correspond to different filament temperatures and therefore different resistances.

Adding the initial reading to the steady-state graph would mix two different experimental conditions.

This is a useful reminder: a voltage-current graph describes a component’s behaviour under specified conditions. If those conditions change, the relationship can change too.

Once you can measure resistance reliably, the next useful step is understanding how resistance affects an entire circuit. In particular, investigating voltage-current characteristics helps you distinguish ohmic resistors from non-ohmic components and predict how current changes when circuit conditions are altered.