How to Compare the Resistance of Wires Made from Different Materials

Two wires are made from different materials. They have the same length and diameter, and both are connected to a 1.20 V power supply. Wire A carries a current of 0.40 A, while wire B carries only 0.10 A.

Which material has the greater electrical resistance?

Wire B does. Its resistance is four times greater than wire A’s. But there is an important condition hidden in this comparison: both wires must have the same dimensions and temperature if we want to attribute the difference to their materials alone.

To compare the resistance of wires made from different materials, measure the voltage across each wire and the current through it, then calculate \(R=V/I\). For a fair comparison of the materials themselves, keep the wires’ lengths, cross-sectional areas, and temperatures the same. If their dimensions differ, calculate their resistivities instead.

Let’s work through how to make this comparison experimentally, interpret the results, and avoid the mistakes that can make a perfectly reasonable-looking conclusion incorrect.

01What makes a fair comparison?

Imagine you have a copper wire and a nichrome wire. You want to investigate whether the material affects electrical resistance.

There is one obvious problem: what if the copper wire is twice as thick as the nichrome wire?

A thicker wire generally has lower resistance. If the copper wire produces a larger current, you cannot immediately conclude that copper has a lower resistivity. Its greater thickness might explain the result.

To isolate the effect of material, control the other variables.

The ideal experiment uses two uniform wires with identical lengths and cross-sectional areas, held at the same temperature.

Only their materials should differ.

If you need to revise why length and thickness affect resistance, see Why Wire Resistance Depends on Length, Thickness and Material.

02Setting up the experiment

The aim is to measure the resistance of each wire using a circuit containing:

• A regulated DC power supply.
• An ammeter to measure current.
• A voltmeter to measure potential difference.
• A switch or selector to connect either wire.
• Two wire samples with controlled dimensions and temperature.

The ammeter is connected in series with the test wire. This ensures that the current passing through the wire also passes through the ammeter.

The voltmeter is connected in parallel across the test wire. It measures the potential difference between the two ends of the wire.

What should the experiment show?

Start by connecting wire A to the circuit.

Record the voltage across it and the current through it.

Next, replace wire A with wire B. Keep the power supply voltage unchanged and record the new readings.

For example, suppose the experiment produces these illustrative measurements:

The voltage is identical for both wires.

Wire B carries less current, suggesting that it has greater resistance.

We can calculate the resistance using Ohm’s law:

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

For wire A:

\[ R_A=\frac{1.20}{0.40} =3.0\,\Omega \]

For wire B:

\[ R_B=\frac{1.20}{0.10} =12\,\Omega \]

Therefore:

\[ \frac{R_B}{R_A} =\frac{12}{3} =4 \]

Wire B has four times the resistance of wire A.

Because the two wires also have identical lengths, cross-sectional areas, and temperatures, wire B has four times the resistivity.

This is the key experimental result: when the geometry and temperature are controlled, a difference in measured resistance can be attributed to the material.

03Fixed voltage versus fixed current

There are two useful ways to compare wire resistance. Both work, but the quantity you measure changes.

Understanding this distinction prevents a common HSC mistake.

Method 1: Keep the voltage constant

Connect each wire separately to the same regulated voltage supply.

The potential difference across each wire remains constant.

From Ohm’s law:

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

If \(V\) stays constant, increasing resistance causes the current to decrease.

For example:

The resistances are:

\[ R_A=\frac{2.0}{0.50}=4.0\,\Omega \]

\[ R_B=\frac{2.0}{0.25}=8.0\,\Omega \]

Wire B has twice the resistance.

At constant voltage, the wire carrying the smaller current has the greater resistance.

Method 2: Keep the current constant

Alternatively, use a regulated current source to pass the same current through each wire.

Measure the potential difference across each sample.

Ohm’s law can be rearranged:

\[ V=IR \]

If the current stays constant, increasing resistance causes the voltage across the wire to increase.

For example:

The resistances are:

\[ R_A=\frac{0.60}{0.20} =3.0\,\Omega \]

\[ R_B=\frac{1.40}{0.20} =7.0\,\Omega \]

Wire B again has the greater resistance.

At constant current, the wire requiring the greater voltage has the greater resistance.

Comparing the two methods

You do not have to use the same method every time. The important thing is to know which electrical quantity is being controlled.

A common misconception

Suppose wire A carries 0.50 A, while wire B carries 0.25 A.

It is tempting to conclude that wire B has twice the resistance.

But what if the voltage across A is 1.0 V and the voltage across B is 0.50 V?

Then:

\[ R_A=\frac{1.0}{0.50}=2.0\,\Omega \]

\[ R_B=\frac{0.50}{0.25}=2.0\,\Omega \]

The wires have exactly the same resistance.

Wire B carries half the current because it also has half the voltage across it.

Never compare current readings alone unless the voltage is the same. Likewise, never compare voltage readings alone unless the current is the same.

If neither quantity is constant, calculate \(V/I\) separately for each wire.

04Why does the material change resistance?

So far, we have established how to measure resistance. But why do different materials produce different results?

The answer involves a property called electrical resistivity.

Resistivity describes how strongly a material opposes the flow of electric current. It is represented by the Greek letter \(\rho\) (rho) and measured in ohm metres (\(\Omega\,\text{m}\)).

For a uniform wire:

\[ R=\frac{\rho L}{A} \]

where:

• \(R\) is resistance, measured in ohms (\(\Omega\)).
• \(\rho\) is resistivity, measured in ohm metres (\(\Omega\,\text{m}\)).
• \(L\) is wire length, measured in metres (m).
• \(A\) is cross-sectional area, measured in square metres (\(\text{m}^2\)).

Rearranging:

\[ \rho=\frac{RA}{L} \]

This equation is particularly useful when comparing wires with different dimensions.

If two wires have identical lengths and cross-sectional areas, their resistance ratio is equal to their resistivity ratio:

\[ \frac{R_1}{R_2} = \frac{\rho_1}{\rho_2} \]

But if their dimensions differ, you must calculate resistivity separately.

What happens inside the material?

In a metal, electrical conduction is primarily carried by mobile electrons.

When an electric field is applied, these electrons acquire a net drift motion through the metal.

However, their movement is influenced by the material’s atomic structure.

The electrons can be scattered by lattice vibrations, impurities, and structural imperfections. Different materials have different electronic structures and different scattering behaviour.

For example, copper has relatively low resistivity, making it useful for electrical wiring. Nichrome, an alloy of several metals, has much higher resistivity, making it useful in heating elements.

The microscopic picture helps explain why the material matters, but resistivity is not determined by simply counting how many times an electron hits an atom.

The electronic structure of the material also matters. The scattering model is a useful explanation of the behaviour, not a complete microscopic theory.

The important HSC connection is:

• The material determines its resistivity under specified conditions.
• A higher resistivity produces a higher resistance for the same wire dimensions.
• Changing the length or thickness changes resistance, but does not change the material’s resistivity.

05Worked examples

06Controlling temperature during the comparison

There is another variable that can quietly interfere with your experiment: temperature.

Suppose you test two wires at room temperature, then leave them connected to a power supply for several minutes.

They begin to heat up.

For ordinary metal wires, increasing temperature generally increases resistivity. The metal ions vibrate more strongly, changing how conduction electrons move through the material.

As a result, the resistance you measure after heating may be different from the initial resistance.

This is particularly important when using a fixed-voltage supply.

The electrical power dissipated by a resistor is:

\[ P=\frac{V^2}{R} \]

At constant voltage, a lower-resistance wire initially dissipates more power.

With a fixed-current supply, the relationship is:

\[ P=I^2R \]

At constant current, a higher-resistance wire dissipates more power.

This means that the electrical conditions of your experiment can affect how quickly the samples heat up.

However, electrical power alone does not determine the final temperature. Heat capacity, thermal contact, surface area, and heat loss to the surroundings also matter.

Improving the experimental design

A simple initial experiment uses a power supply to measure the resistance of each wire. Its weakness is that the measuring current can heat the samples.

A better design separates temperature control from electrical measurement.

Place the two wire samples in a temperature-controlled water bath. Use a separate heater to bring the bath to the required temperature.

Position a thermometer close to the wires so that their temperature can be monitored.

Keep the electrical measuring circuit separate from the heating circuit.

A more reliable procedure is:

1. Prepare two wires with equal lengths and cross-sectional areas.
2. Allow both wires to reach the same temperature.
3. Connect the first wire to the measuring circuit.
4. Use a small test current to minimise self-heating.
5. Record the voltage and current.
6. Repeat the measurement with the second wire under the same conditions.
7. Repeat the measurements and calculate average resistance values.

If the resistance is very small, the resistance of the connecting leads and electrical contacts can become significant compared with the wire itself. More advanced four-terminal measurement methods can reduce this problem.

For a closer look at how heating changes electrical resistance, see How Heating Changes a Metal Wire’s Resistance.

07Questions and solutions

Question 1

Two uniform wires, A and B, have identical lengths and cross-sectional areas. Both are maintained at \(20^\circ\text{C}\).

Each wire is connected separately across a 1.8 V power supply.

The measured currents are:

• Wire A: 0.60 A
• Wire B: 0.20 A

Calculate the resistance of each wire and determine which material has the greater resistivity.

Solution 1

Wire B has three times the resistance and three times the resistivity of wire A.

Because both wires have the same voltage, we can use Ohm’s law to calculate their resistances.

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

For wire A:

\[ R_A=\frac{1.8}{0.60} =3.0\,\Omega \]

For wire B:

\[ R_B=\frac{1.8}{0.20} =9.0\,\Omega \]

The resistance ratio is:

\[ \frac{R_B}{R_A} = \frac{9.0}{3.0} =3 \]

Since both wires have identical lengths, cross-sectional areas, and temperatures, their resistivity ratio is also three.

Wire B’s material therefore has three times the resistivity of wire A’s material.

The important condition is that the geometry is identical. Without that information, we could compare resistance but not directly attribute the difference to the materials.

Question 2

Two uniform wires, X and Y, have identical lengths and cross-sectional areas and are maintained at the same temperature.

A regulated current source passes 0.12 A through each wire separately.

The voltage across wire X is 0.30 V, while the voltage across wire Y is 0.84 V.

(a) Calculate the resistance of each wire.

(b) Determine the ratio of their resistivities.

(c) The wires are then connected in parallel across a 2.4 V supply. Calculate the current through each branch.

Solution 2

Wire Y has 2.8 times the resistance and resistivity of wire X. When connected in parallel, wire X carries the greater current.

(a) Calculate the resistances.

Use:

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

For wire X:

\[ R_X=\frac{0.30}{0.12} =2.5\,\Omega \]

For wire Y:

\[ R_Y=\frac{0.84}{0.12} =7.0\,\Omega \]

(b) Compare the resistivities.

Since the wires have identical dimensions and temperatures:

\[ \frac{\rho_Y}{\rho_X} = \frac{R_Y}{R_X} \]

\[ \frac{\rho_Y}{\rho_X} = \frac{7.0}{2.5} \]

\[ \frac{\rho_Y}{\rho_X}=2.8 \]

Wire Y has 2.8 times the resistivity of wire X.

(c) Calculate the parallel branch currents.

In a parallel circuit, the potential difference across each branch is the same.

Therefore:

\[ V_X=V_Y=2.4\,\text{V} \]

For wire X:

\[ I_X=\frac{V}{R_X} \]

\[ I_X=\frac{2.4}{2.5} =0.96\,\text{A} \]

For wire Y:

\[ I_Y=\frac{V}{R_Y} \]

\[ I_Y=\frac{2.4}{7.0} =0.343\,\text{A} \]

Wire X carries approximately 0.96 A, while wire Y carries approximately 0.34 A.

This makes sense because parallel branches have equal voltages, but their currents depend on their individual resistances.

Notice the change in experimental conditions. In the first experiment, current was controlled, so the higher-resistance wire required a larger voltage. In the parallel circuit, voltage is controlled, so the higher-resistance wire carries a smaller current.

Question 3

Two uniform wires, P and Q, are made from different materials. Both are maintained at \(25^\circ\text{C}\) and connected separately across a 1.20 V supply.

Their properties and measured currents are:

(a) Calculate the resistance of each wire.

(b) Calculate the resistivity of each material.

(c) Determine the ratio of their resistivities.

Explain why comparing their resistances alone does not give the correct comparison of their materials.

Solution 3

Material Q has 2.25 times the resistivity of material P, even though its wire has only 1.5 times the measured resistance.

(a) Calculate resistance.

For wire P:

\[ R_P=\frac{1.20}{0.30} =4.0\,\Omega \]

For wire Q:

\[ R_Q=\frac{1.20}{0.20} =6.0\,\Omega \]

(b) Calculate resistivity.

First, calculate the cross-sectional areas.

For wire P:

\[ r_P=0.20\,\text{mm} =2.0\times10^{-4}\,\text{m} \]

\[ A_P=\pi(2.0\times10^{-4})^2 \]

\[ A_P=1.26\times10^{-7}\,\text{m}^2 \]

For wire Q:

\[ r_Q=0.30\,\text{mm} =3.0\times10^{-4}\,\text{m} \]

\[ A_Q=\pi(3.0\times10^{-4})^2 \]

\[ A_Q=2.83\times10^{-7}\,\text{m}^2 \]

Now use:

\[ \rho=\frac{RA}{L} \]

For wire P:

\[ \rho_P= \frac{(4.0)(1.26\times10^{-7})}{0.80} \]

\[ \rho_P=6.28\times10^{-7}\,\Omega\,\text{m} \]

For wire Q:

\[ \rho_Q= \frac{(6.0)(2.83\times10^{-7})}{1.20} \]

\[ \rho_Q=1.41\times10^{-6}\,\Omega\,\text{m} \]

(c) Calculate the resistivity ratio.

\[ \frac{\rho_Q}{\rho_P} = \frac{1.41\times10^{-6}} {6.28\times10^{-7}} \]

\[ \frac{\rho_Q}{\rho_P}=2.25 \]

Material Q has 2.25 times the resistivity of material P.

The measured resistance ratio is only:

\[ \frac{R_Q}{R_P} = \frac{6.0}{4.0} =1.5 \]

The difference arises because the wires have different geometries. Wire Q is longer, which increases its resistance, but it also has a larger cross-sectional area, which decreases its resistance.

These geometric effects alter the measured resistance independently of the material’s resistivity.

This is why resistivity is the more useful quantity when the aim is to compare different materials rather than individual wire samples.

Question 4

A student investigates the temperature dependence of copper and nichrome wires.

The wires have identical lengths and cross-sectional areas. They are initially both at \(20^\circ\text{C}\) and connected separately across a constant 1.00 V supply.

The initial currents are:

Both wires are then heated to \(70^\circ\text{C}\).

Their approximate temperature coefficients of resistance are:

• Copper: \(\alpha=0.0040\,^\circ\text{C}^{-1}\)
• Nichrome: \(\alpha=0.00040\,^\circ\text{C}^{-1}\)

Assume resistance varies linearly with temperature over this range, and ignore changes in wire dimensions.

(a) Calculate the initial resistance of each wire.

(b) Calculate the resistance of each wire at \(70^\circ\text{C}\).

(c) Calculate the new current through each wire.

(d) Explain why the copper wire experiences a greater percentage change in current, even though nichrome has the greater resistivity.

Solution 4

Nichrome has much greater resistivity at both temperatures. Copper experiences a larger percentage change in current because its temperature coefficient is greater.

(a) Calculate the initial resistances.

Using:

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

For copper:

\[ R_{\text{Cu}}=\frac{1.00}{0.50} =2.0\,\Omega \]

For nichrome:

\[ R_{\text{Ni}}=\frac{1.00}{0.010} =100\,\Omega \]

Since the wires have identical dimensions and temperature, nichrome has 50 times the resistivity of copper.

(b) Calculate the resistances at \(70^\circ\text{C}\).

Use the approximate temperature relationship:

\[ R(T)=R(T_0)[1+\alpha(T-T_0)] \]

Here:

• \(R(T)\) is the resistance at the new temperature.
• \(R(T_0)\) is the initial resistance.
• \(\alpha\) is the temperature coefficient.
• \(T-T_0\) is the temperature increase.

The temperature increase is:

\[ \Delta T=70-20=50^\circ\text{C} \]

For copper:

\[ R_{\text{Cu},70} = 2.0[1+(0.0040)(50)] \]

\[ R_{\text{Cu},70}=2.4\,\Omega \]

For nichrome:

\[ R_{\text{Ni},70} = 100[1+(0.00040)(50)] \]

\[ R_{\text{Ni},70}=102\,\Omega \]

(c) Calculate the new currents.

The voltage remains constant at 1.00 V.

For copper:

\[ I_{\text{Cu}}=\frac{1.00}{2.4} \]

\[ I_{\text{Cu}}=0.417\,\text{A} \]

For nichrome:

\[ I_{\text{Ni}}=\frac{1.00}{102} \]

\[ I_{\text{Ni}}=0.00980\,\text{A} \]

(d) Explain the difference.

Copper’s current decreases from 0.50 A to approximately 0.417 A.

Its percentage decrease is:

\[ \frac{0.50-0.417}{0.50}\times100 \]

\[ =16.6\% \]

Nichrome’s current decreases from 0.010 A to approximately 0.00980 A.

Its percentage decrease is:

\[ \frac{0.010-0.00980}{0.010}\times100 \]

\[ =2.0\% \]

Copper experiences a much larger percentage change because its temperature coefficient is ten times greater.

However, nichrome still has a much greater absolute resistance at \(70^\circ\text{C}\).

This question demonstrates why you must distinguish between a material’s resistivity and how strongly that resistivity changes with temperature.

Question 5

Two different uniform metal wires, A and B, are being tested to determine which material has the greater resistivity at \(20^\circ\text{C}\).

Their properties and experimental readings are:

Assume the temperature relationship is linear:

\[ R(T)=R_{20}[1+\alpha(T-20)] \]

Assume the wires are uniform, their dimensions remain constant with temperature, and both behave as ohmic conductors.

(a) Calculate the resistance of each wire at its experimental temperature.

(b) Determine the resistance of each wire at \(20^\circ\text{C}\).

(c) Calculate the resistivity of each material at \(20^\circ\text{C}\), and determine their ratio.

(d) Both wires are now maintained at \(20^\circ\text{C}\).

First, each wire is connected separately across a 2.00 V supply. Determine which wire dissipates more power.

Next, the wires are connected in series to a regulated 0.20 A current source. Determine which wire now dissipates more power.

Explain why the wire dissipating more power changes between the two arrangements.

Solution 5

Material B has approximately 14.1 times the resistivity of material A at \(20^\circ\text{C}\). However, wire A dissipates more power under the fixed-voltage condition, while wire B dissipates more power under the fixed-current condition.

This problem requires us to separate three effects:

• The measured resistance depends on the material and the wire’s dimensions.
• The measured resistance also depends on temperature.
• The power dissipated depends on whether voltage or current is held constant.

(a) Calculate resistance at the experimental temperatures.

For wire A:

\[ R_{A,50}=\frac{2.00}{0.40} \]

\[ R_{A,50}=5.0\,\Omega \]

For wire B:

\[ R_{B,30}=\frac{2.00}{0.25} \]

\[ R_{B,30}=8.0\,\Omega \]

Wire B has the greater measured resistance.

However, this is not yet a fair comparison of resistivity because the wires are at different temperatures and have different dimensions.

(b) Correct both resistances to \(20^\circ\text{C}\).

Rearrange the temperature relationship:

\[ R_{20}= \frac{R(T)}{1+\alpha(T-20)} \]

For wire A:

\[ R_{A,20} = \frac{5.0}{1+(0.0040)(50-20)} \]

\[ R_{A,20}=\frac{5.0}{1.12} \]

\[ R_{A,20}=4.46\,\Omega \]

For wire B:

\[ R_{B,20} = \frac{8.0}{1+(0.0020)(30-20)} \]

\[ R_{B,20}=\frac{8.0}{1.02} \]

\[ R_{B,20}=7.84\,\Omega \]

Both resistances are now expressed at the same reference temperature.

(c) Calculate the resistivities.

First, calculate the cross-sectional areas.

For wire A:

\[ r_A=0.20\,\text{mm} =2.0\times10^{-4}\,\text{m} \]

\[ A_A=\pi(2.0\times10^{-4})^2 \]

\[ A_A=1.26\times10^{-7}\,\text{m}^2 \]

For wire B:

\[ r_B=0.40\,\text{mm} =4.0\times10^{-4}\,\text{m} \]

\[ A_B=\pi(4.0\times10^{-4})^2 \]

\[ A_B=5.03\times10^{-7}\,\text{m}^2 \]

Now use:

\[ \rho=\frac{RA}{L} \]

For wire A:

\[ \rho_A= \frac{(4.46)(1.26\times10^{-7})}{1.00} \]

\[ \rho_A=5.61\times10^{-7}\,\Omega\,\text{m} \]

For wire B:

\[ \rho_B= \frac{(7.84)(5.03\times10^{-7})}{0.50} \]

\[ \rho_B=7.88\times10^{-6}\,\Omega\,\text{m} \]

Compare the resistivities:

\[ \frac{\rho_B}{\rho_A} = \frac{7.88\times10^{-6}} {5.61\times10^{-7}} \]

\[ \frac{\rho_B}{\rho_A} \approx14.1 \]

Material B has approximately 14.1 times the resistivity of material A.

Notice that wire B is shorter and thicker than wire A. Both of these geometric differences would reduce its resistance if the materials were identical.

However, material B’s much greater resistivity more than compensates for those geometric advantages.

(d) Compare the power under different electrical conditions.

At \(20^\circ\text{C}\), the resistances are:

\[ R_A=4.46\,\Omega \]

\[ R_B=7.84\,\Omega \]

First, consider each wire connected separately across a constant 2.00 V supply.

Use:

\[ P=\frac{V^2}{R} \]

For wire A:

\[ P_A=\frac{2.00^2}{4.46} \]

\[ P_A=0.896\,\text{W} \]

For wire B:

\[ P_B=\frac{2.00^2}{7.84} \]

\[ P_B=0.510\,\text{W} \]

Wire A dissipates more power.

This is because both wires have the same voltage, but wire A has the smaller resistance and therefore draws more current.

Now connect the wires in series to a regulated 0.20 A current source.

The same current passes through both wires.

Use:

\[ P=I^2R \]

For wire A:

\[ P_A=(0.20)^2(4.46) \]

\[ P_A=0.179\,\text{W} \]

For wire B:

\[ P_B=(0.20)^2(7.84) \]

\[ P_B=0.314\,\text{W} \]

This time, wire B dissipates more power.

Its greater resistance produces a larger voltage drop when the current is held constant.

The result is not contradictory. Power depends on both voltage and current, and the relationship between them changes with the circuit conditions.

Also, these are electrical power comparisons at a controlled temperature. They do not automatically predict which wire will reach the higher final temperature, because that depends on heat capacity and heat loss as well.

The central lesson is that resistance, resistivity, and power describe different aspects of electrical behaviour. You need to identify which one an experiment is measuring before interpreting its results.

Once you can compare resistances reliably, the next useful step is to investigate how voltage and current are related across a wire using a voltage-current graph. The gradient provides another way to determine resistance and test whether a conductor behaves ohmically.