Why Doubling a Wire’s Diameter Cuts Its Resistance to One-Quarter

A copper wire has its diameter doubled, while its length, material, and temperature stay the same. You might expect its resistance to halve because the wire is twice as thick. But the resistance actually falls to one-quarter of its original value.

The reason is easy to miss: doubling the diameter gives the wire four times the cross-sectional area, not twice the area. That changes how much current the wire can carry for a given potential difference.

Let’s work through the geometry, the physics, and an experiment that would let you test the prediction.

01Why doubling the diameter gives four times the area

Picture two circular wires viewed straight on. The second wire has twice the diameter of the first.

A common prediction is that twice the diameter means twice the space available for charge to flow. The problem is that a wire has a circular cross-section, so its conducting area depends on the square of its radius.

For a circle,

\[ A=\pi r^2 \]

where:

• \(A\) is the cross-sectional area in square metres (\(\mathrm{m^2}\)),
• \(r\) is the radius in metres (\(\mathrm{m}\)),
• \(\pi\) is approximately 3.14.

The radius is half the diameter:

\[ r=\frac{d}{2} \]

Substituting this into the area equation gives:

\[ A=\pi\left(\frac{d}{2}\right)^2 =\frac{\pi d^2}{4} \]

This is the important relationship: cross-sectional area is proportional to diameter squared.

If the diameter doubles, the new area is:

\[ A_{\text{new}}=\frac{\pi(2d)^2}{4} \]

\[ A_{\text{new}}=\frac{4\pi d^2}{4} =4A_{\text{old}} \]

The larger wire has four times the cross-sectional area.

That is the first half of the explanation. Now we need to connect area to electrical resistance.

02How cross-sectional area affects resistance

Imagine a stream of people walking through a corridor. If the corridor becomes wider, more people can pass through side by side without needing to squeeze into a narrow space.

A wire behaves somewhat similarly. A larger cross-sectional area provides more conducting material through which charge can move.

For a uniform wire, resistance is described by:

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

where:

• \(R\) is resistance, measured in ohms (\(\Omega\)),
• \(\rho\) (rho) is the resistivity of the material, measured in ohm metres (\(\Omega\,\mathrm{m}\)),
• \(L\) is the wire’s length, measured in metres (\(\mathrm{m}\)),
• \(A\) is its cross-sectional area, measured in square metres (\(\mathrm{m^2}\)).

Notice that area is in the denominator. Increasing the area decreases the resistance.

If the length, material, and temperature remain constant, then:

\[ R\propto\frac{1}{A} \]

Since doubling the diameter quadruples the area,

\[ R_{\text{new}}=\frac{R_{\text{old}}}{4} \]

So the resistance becomes 25% of its original value. This is a 75% decrease, not a 50% decrease.

A useful way to remember the relationship is:

The square relationship works in both directions. Halving the diameter does not halve the resistance. It makes the resistance four times larger.

03What would an experiment actually show?

Suppose you want to test whether doubling a wire’s diameter quarters its resistance. You need to compare two wires without accidentally changing another factor.

Consider an experiment using two copper wires:

• Both wires are 2.0 m long.
• Both are at the same temperature.
• Both have the same material and purity.
• Wire A has a diameter of 0.50 mm.
• Wire B has a diameter of 1.00 mm.

Connect a DC power supply, an ammeter, and the test wire in series. Connect a voltmeter in parallel across the test wire. The ammeter measures the current through the wire, while the voltmeter measures the potential difference across it.

The resistance can then be calculated using:

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

where \(V\) is the potential difference in volts (V), and \(I\) is the current in amperes (A).

What should the experiment show?

If the resistance of Wire A is four times the resistance of Wire B, the prediction is supported.

The following is an illustrative data set, not a record of an actual experiment.

For Wire A:

\[ R_A=\frac{1.00}{0.20}=5.0\ \Omega \]

For Wire B:

\[ R_B=\frac{1.00}{0.80}=1.25\ \Omega \]

The ratio is:

\[ \frac{R_A}{R_B}=\frac{5.0}{1.25}=4 \]

The thicker wire has one-quarter the resistance, exactly as predicted by the ideal model.

The experiment also illustrates something important: at the same potential difference, the thicker wire carries four times the current.

But be careful. This result assumes the wires remain at the same temperature and behave ohmically. If one wire heats significantly, its resistance may change during the measurement.

04Why the experimental conditions matter

There are two different ways to compare the wires, and they produce different observations.

Case 1: Both wires have the same potential difference.

Using Ohm’s law,

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

If resistance falls to one-quarter while voltage stays constant, current becomes four times larger.

Case 2: Both wires carry the same current.

Rearranging Ohm’s law gives:

\[ V=IR \]

If current stays constant while resistance falls to one-quarter, the potential difference across the thicker wire also falls to one-quarter.

The resistance has not changed its relationship with diameter. The external condition has changed what you observe.

This distinction is useful in HSC questions. Always identify whether voltage or current is being held constant before predicting what happens to the other quantity.

05What changes inside the metal?

So far, we’ve used equations to predict what happens. But there’s another useful question: does a thicker wire make individual electrons move faster?

Not necessarily.

In a metal, mobile electrons are already distributed throughout the conductor. They drift through the metal when an electric field is established. The current depends on both the number of charge carriers moving through a cross-section and their average drift speed.

A simple microscopic model expresses this as:

\[ I=nqAv_d \]

where:

• \(I\) is conventional current in amperes (A),
• \(n\) is the number of mobile charge carriers per cubic metre (\(\mathrm{m^{-3}}\)),
• \(q\) is the magnitude of the charge on each carrier in coulombs (C),
• \(A\) is cross-sectional area in square metres (\(\mathrm{m^2}\)),
• \(v_d\) is the magnitude of the average drift velocity in metres per second (\(\mathrm{m\,s^{-1}}\)).

For electrons, their drift direction is opposite to conventional current. This equation uses magnitudes.

Let’s compare the two supply conditions.

When voltage is fixed

For two wires of equal length and material, the electric field along each wire is the same if the potential difference is the same.

At the same temperature, the average drift speed is therefore the same in the simple conduction model. However, the thicker wire has four times the cross-sectional area.

Four times as much charge passes through a cross-section each second, so the current is four times larger.

When current is fixed

The same amount of charge must pass through each wire per second.

The thicker wire has four times the area, so the average drift speed of its electrons is one-quarter as large, assuming the same mobile charge-carrier density.

The important point is that a thicker wire does not automatically make individual electrons travel faster. The relationship between current and drift speed depends on the conditions imposed on the circuit.

This microscopic model is useful for understanding the result, but it is a simplified description of conduction. It does not mean electrons travel in neat, straight paths through the metal.

06A useful distinction: resistance, current, and power

A common mistake is to assume that if resistance decreases, electrical power must also decrease.

That is not always true. The result depends on what the circuit holds constant.

Electrical power can be calculated using:

\[ P=VI \]

Combining this with Ohm’s law gives:

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

or

\[ P=I^2R \]

where \(P\) is power in watts (W).

Consider what happens when the wire’s resistance falls to one-quarter.

This is a useful HSC distinction.

At fixed voltage, the thicker wire draws four times the current and dissipates four times the electrical power.

At fixed current, the thicker wire requires only one-quarter of the potential difference and dissipates one-quarter of the power.

Neither result tells you directly how much hotter the wire will become. Temperature also depends on heat capacity, heat transfer, surface area, and the time allowed for heating.

There is another limitation. The equation \(R=\rho L/A\) assumes the resistivity remains constant. For an ordinary metal, heating generally increases resistivity. A wire carrying a large current may therefore develop a higher resistance than the constant-temperature calculation predicts.

07Worked examples

08A reliable approach for HSC questions

When a question involves changing a wire’s diameter, use this sequence.

1. Identify what stays constant: material, length, and temperature.
2. Convert the diameter change into an area change by squaring the diameter ratio.
3. Use the inverse relationship between resistance and area.
4. Check whether voltage or current is fixed before predicting the other quantity.
5. If the wire is part of a larger circuit, calculate the total circuit resistance rather than assuming the current changes by the same factor as the wire’s conductance.

For a comparison between two wires, the combined relationship is particularly useful:

\[ \frac{R_2}{R_1} = \frac{\rho_2}{\rho_1} \times \frac{L_2}{L_1} \times \left(\frac{d_1}{d_2}\right)^2 \]

This applies to uniform circular wires, with resistivity evaluated at the relevant temperatures.

For a broader explanation of how length, thickness, and material affect resistance, see Why Wire Resistance Depends on Length, Thickness and Material.

For more on when you can treat resistance as constant when applying \(V=IR\), see Ohm’s Law in HSC Physics: When Resistance Is Constant.

09Questions and solutions

Question 1

A uniform copper wire has a resistance of \(16\ \Omega\). It is replaced with another copper wire of the same length and temperature, but with twice the original diameter.

Calculate the resistance of the replacement wire.

Solution 1

The replacement wire has a resistance of \(4.0\ \Omega\).

The diameter doubles, so its cross-sectional area increases by a factor of four:

\[ \frac{A_{\text{new}}}{A_{\text{old}}} =2^2=4 \]

Resistance is inversely proportional to cross-sectional area:

\[ R_{\text{new}}=\frac{16}{4} \]

\[ \boxed{R_{\text{new}}=4.0\ \Omega} \]

The resistance falls to one-quarter of its original value, assuming the material, length, and temperature remain unchanged.

Question 2

Two uniform copper wires are at the same temperature.

Wire A has a length of \(2.0\ \mathrm{m}\), a diameter of \(0.40\ \mathrm{mm}\), and a resistance of \(6.0\ \Omega\).

Wire B has a length of \(3.0\ \mathrm{m}\) and a diameter of \(0.60\ \mathrm{mm}\).

Calculate the resistance of wire B. Then calculate the potential difference across each wire if a constant current of \(0.50\ \mathrm{A}\) passes through them separately.

Solution 2

Wire B has a resistance of \(4.0\ \Omega\), and its potential difference at \(0.50\ \mathrm{A}\) is \(2.0\ \mathrm{V}\).

First, compare the lengths:

\[ \frac{L_B}{L_A}=\frac{3.0}{2.0}=1.5 \]

Now compare the diameters:

\[ \frac{d_B}{d_A}=\frac{0.60}{0.40}=1.5 \]

The area ratio is:

\[ \frac{A_B}{A_A}=(1.5)^2=2.25 \]

Combine the effects of length and area:

\[ \frac{R_B}{R_A} = \frac{1.5}{2.25} =\frac{2}{3} \]

Therefore:

\[ R_B=6.0\times\frac{2}{3} \]

\[ R_B=4.0\ \Omega \]

Now calculate the potential differences using \(V=IR\).

For wire A:

\[ V_A=0.50\times6.0 =3.0\ \mathrm{V} \]

For wire B:

\[ V_B=0.50\times4.0 =2.0\ \mathrm{V} \]

Wire B is longer, which increases its resistance, but its larger cross-sectional area has a stronger effect. Its resistance is therefore lower overall.

Because the current is fixed, its lower resistance also means a smaller potential difference.

Question 3

Two copper wires are connected in separate parallel branches across an ideal \(4.0\ \mathrm{V}\) supply.

Both wires have the same length and temperature. Wire A has a resistance of \(12\ \Omega\), while wire B has twice the diameter of wire A.

Calculate:

a) The resistance of wire B.

b) The current through each wire.

c) The electrical power dissipated by each wire.

A student concludes that wire B must reach four times the temperature of wire A because it dissipates four times the power. Explain why this final conclusion is not justified.

Solution 3

Wire B has a resistance of \(3.0\ \Omega\), carries \(1.33\ \mathrm{A}\), and dissipates approximately \(5.33\ \mathrm{W}\). It has four times the electrical power dissipation of wire A, but this does not mean it reaches four times the temperature.

Part a: Resistance of wire B

Its diameter doubles, so its area quadruples.

Therefore:

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

\[ R_B=3.0\ \Omega \]

Part b: Current through each wire

Both wires are connected in parallel, so each has the full \(4.0\ \mathrm{V}\) supply potential difference.

For wire A:

\[ I_A=\frac{4.0}{12} \]

\[ I_A=0.333\ \mathrm{A} \]

For wire B:

\[ I_B=\frac{4.0}{3.0} \]

\[ I_B=1.33\ \mathrm{A} \]

Wire B carries four times the current of wire A.

Part c: Electrical power

Use \(P=VI\).

For wire A:

\[ P_A=4.0\times0.333 \]

\[ P_A\approx1.33\ \mathrm{W} \]

For wire B:

\[ P_B=4.0\times1.33 \]

\[ P_B\approx5.33\ \mathrm{W} \]

The power ratio is:

\[ \frac{P_B}{P_A}=4 \]

However, electrical power is the rate at which electrical energy is transferred. It is not a direct measurement of temperature.

The final temperature depends on how effectively each wire transfers heat to its surroundings, as well as its mass, specific heat capacity, and initial temperature.

A thicker wire also contains more metal and has a different surface area. We cannot determine the final temperature ratio from electrical power alone.

The calculation tells us how rapidly electrical energy is being transferred, not how hot each wire must become.

Question 4

A student tests two copper wires of equal length and nominally equal temperature. Wire B has twice the diameter of wire A.

Both wires are connected separately to the same \(2.0\ \mathrm{V}\) supply.

The following illustrative measurements are recorded:

a) Calculate the resistance of each wire.

b) Calculate the ratio \(R_A/R_B\).

c) Explain why the measured ratio differs slightly from the ideal prediction.

d) Describe two improvements to the experiment that would make the comparison more reliable.

Solution 4

The measured resistance ratio is \(3.75\), rather than the ideal value of \(4.00\). The difference could result from temperature changes, measurement uncertainty, or variations in the actual wire dimensions.

Part a: Calculate both resistances

Use:

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

For wire A:

\[ R_A=\frac{2.0}{0.40} \]

\[ R_A=5.0\ \Omega \]

For wire B:

\[ R_B=\frac{2.0}{1.50} \]

\[ R_B=1.33\ \Omega \]

Part b: Calculate the resistance ratio

\[ \frac{R_A}{R_B} =\frac{5.0}{1.33} \]

\[ \frac{R_A}{R_B}\approx3.75 \]

The ideal prediction is a ratio of four because wire B has four times the cross-sectional area.

Part c: Explain the difference

The ideal model assumes that both wires have the same resistivity.

However, the thicker wire carries more current at the same potential difference. It also dissipates more electrical power.

If wire B becomes warmer, its resistivity may increase. This would increase its resistance above the ideal constant-temperature prediction and reduce its current.

That could help explain why its measured resistance is slightly greater than one-quarter of wire A’s resistance.

However, temperature is not the only possible explanation. The measured diameter may differ slightly from its nominal value, and contact resistance or instrument uncertainty may also affect the readings.

The result does not establish that the theoretical relationship is wrong.

Part d: Improve the experiment

Two useful improvements are:

• Control the temperature. Allow both wires to reach the same initial temperature, use a low potential difference, and minimise heating during measurements.
• Measure the actual diameters at several points using a suitable micrometer, rather than relying only on nominal values.

For a more accurate low-resistance measurement, separate current-carrying contacts from voltage-sensing contacts. This reduces the effect of resistance in the connecting leads and contacts.

The experiment should also be repeated, with the results averaged where appropriate.

The key is to isolate diameter as the independent variable while controlling the other factors that affect resistance.

Question 5

A copper wire has a resistance of \(3.0\ \Omega\) at \(20^\circ\mathrm{C}\). It is connected in series with a fixed \(3.0\ \Omega\) resistor and an ideal \(6.0\ \mathrm{V}\) battery.

The wire is then replaced with another copper wire of the same length and temperature, but with twice the original diameter.

Assume both wires are ohmic, the battery voltage remains constant, and all resistances remain at their stated values.

A student argues:

“The replacement wire has four times the cross-sectional area, so its resistance becomes one-quarter. Therefore, the current through the circuit becomes four times larger.”

a) Identify the error in the student’s reasoning.

b) Calculate the circuit current before and after the replacement.

c) Calculate the potential difference across the test wire before and after the replacement.

d) Calculate the electrical power dissipated by the test wire in both cases.

e) Using the microscopic model \(I=nqAv_d\), determine the ratio of the average electron drift speeds in the replacement wire and the original wire. Assume the mobile charge-carrier density is unchanged.

Explain why the result is consistent with your answers to the earlier parts.

Solution 5

The replacement wire carries \(1.6\ \mathrm{A}\), not four times the original current. Its resistance falls to \(0.75\ \Omega\), but the fixed series resistor prevents the total circuit resistance from falling to one-quarter.

The potential difference across the replacement wire falls to \(1.2\ \mathrm{V}\), its power dissipation falls to \(1.92\ \mathrm{W}\), and its average electron drift speed becomes 40% of the original value.

Part a: Identify the error

The student correctly predicts that the wire’s resistance falls to one-quarter.

However, the student incorrectly assumes that the potential difference across the wire remains constant.

The wire is connected in series with another resistor. The battery voltage is shared between the two components, so changing the wire’s resistance also changes the potential difference across it.

The relationship \(I=V/R\) only predicts a fourfold increase in current if the potential difference across the wire itself stays constant.

Part b: Calculate the circuit current

Initially, the wire resistance is \(3.0\ \Omega\).

The total series resistance is:

\[ R_{\text{total,old}}=3.0+3.0 \]

\[ R_{\text{total,old}}=6.0\ \Omega \]

Using Ohm’s law:

\[ I_{\text{old}}=\frac{6.0}{6.0} \]

\[ I_{\text{old}}=1.0\ \mathrm{A} \]

Now calculate the replacement wire’s resistance.

Its diameter doubles, so its cross-sectional area quadruples:

\[ R_{\text{new}}=\frac{3.0}{4} \]

\[ R_{\text{new}}=0.75\ \Omega \]

The new total resistance is:

\[ R_{\text{total,new}}=3.0+0.75 \]

\[ R_{\text{total,new}}=3.75\ \Omega \]

Therefore:

\[ I_{\text{new}}=\frac{6.0}{3.75} \]

\[ I_{\text{new}}=1.6\ \mathrm{A} \]

The current increases from \(1.0\ \mathrm{A}\) to \(1.6\ \mathrm{A}\), a factor of 1.6 rather than four.

Part c: Calculate the potential difference across each wire

Use \(V=IR\).

For the original wire:

\[ V_{\text{old}}=1.0\times3.0 \]

\[ V_{\text{old}}=3.0\ \mathrm{V} \]

For the replacement wire:

\[ V_{\text{new}}=1.6\times0.75 \]

\[ V_{\text{new}}=1.2\ \mathrm{V} \]

The potential difference across the replacement wire is only 40% of its original value.

The remaining potential difference is across the fixed resistor.

Initially, the fixed resistor has \(3.0\ \mathrm{V}\) across it. After replacement, it has:

\[ V=1.6\times3.0=4.8\ \mathrm{V} \]

The battery voltage is still \(6.0\ \mathrm{V}\):

\[ 4.8+1.2=6.0\ \mathrm{V} \]

This confirms that the voltage changes are consistent with the series circuit.

Part d: Calculate the electrical power

Use:

\[ P=I^2R \]

For the original wire:

\[ P_{\text{old}}=(1.0)^2\times3.0 \]

\[ P_{\text{old}}=3.0\ \mathrm{W} \]

For the replacement wire:

\[ P_{\text{new}}=(1.6)^2\times0.75 \]

\[ P_{\text{new}}=1.92\ \mathrm{W} \]

The replacement wire dissipates less electrical power, even though the total circuit current has increased.

This is because the wire’s resistance has fallen and its share of the battery’s potential difference has decreased.

Part e: Compare electron drift speeds

Start with the microscopic current relationship:

\[ I=nqAv_d \]

Rearrange to compare drift speeds:

\[ v_d=\frac{I}{nqA} \]

The material is unchanged, so \(n\) and \(q\) remain constant.

The replacement wire has four times the area:

\[ \frac{A_{\text{new}}}{A_{\text{old}}}=4 \]

Therefore:

\[ \frac{v_{\text{new}}}{v_{\text{old}}} = \frac{I_{\text{new}}}{I_{\text{old}}} \times \frac{A_{\text{old}}}{A_{\text{new}}} \]

Substitute:

\[ \frac{v_{\text{new}}}{v_{\text{old}}} = \frac{1.6}{1.0}\times\frac{1}{4} \]

\[ \frac{v_{\text{new}}}{v_{\text{old}}}=0.40 \]

The average drift speed is 40% of its original value.

This is the important conceptual result.

The replacement wire has four times the conducting area, but the circuit current increases by only 60%. The charge carriers therefore need a lower average drift speed to produce that current.

The student’s original prediction failed because it treated the wire as though it were connected directly across an unchanged voltage. The fixed resistor changes the conditions of the comparison.

The next useful step is to consider what happens when the wire’s temperature is no longer held constant. In a real metal wire, increased current can cause heating, which changes resistivity and makes the simple quarter-resistance prediction less exact. That is the connection between wire geometry, electrical power, and how heating changes a metal wire’s resistance.