Why Stretching a Wire Increases Its Resistance

Stretch a metal wire to twice its original length, and its resistance can become four times as large. The surprising part is that the wire is made from exactly the same material. You haven’t changed copper into something else, or added another resistor. You’ve changed the wire’s geometry in two ways at once: it gets longer, and it gets thinner.

That second change is the one students often overlook. A longer wire has greater resistance, but a thinner wire also has greater resistance. When you stretch a wire, both effects work together.

The key result is that, if the wire’s volume and temperature remain constant and its resistivity is unchanged, doubling its length quadruples its resistance.

01Why stretching increases resistance

Imagine pulling a piece of copper wire from both ends. The wire becomes longer, but the same amount of copper must now occupy that longer shape.

Where does the extra length come from? The wire becomes thinner. Its cross-sectional area decreases as its length increases.

This gives us two changes:

• Length increases: charge carriers have a longer section of material to travel through.
• Cross-sectional area decreases: there is less conducting material available across each slice of the wire.

Both changes increase resistance.

The relationship between these quantities is

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

where:

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

Resistivity describes a property of the material. Length and cross-sectional area describe the shape of the conductor.

This distinction is important. Stretching a wire does not automatically change the material’s resistivity by a large amount. In the standard HSC model, we treat resistivity as constant and explain the increase in resistance through the changes in length and area.

For a deeper foundation on the separate effects of length, thickness, and material, see Why Wire Resistance Depends on Length, Thickness and Material.

02Why the increase can be much larger than expected

Let’s work out what happens when a wire is stretched without changing its volume.

The volume of a uniform wire is its length multiplied by its cross-sectional area:

\[ V_{\text{wire}}=LA \]

Assume the wire stretches uniformly, its volume stays constant, and its temperature and resistivity do not change.

If the length doubles, the cross-sectional area must halve to keep the volume constant.

\[ L'=2L \]

\[ A'=\frac{A}{2} \]

The new resistance is therefore

\[ R'=\rho\frac{2L}{A/2} \]

\[ R'=4\rho\frac{L}{A} \]

Since the original resistance is \(R=\rho L/A\),

\[ \boxed{R'=4R} \]

The resistance becomes four times its original value.

Notice what happened. Doubling the length alone would double the resistance. Halving the area then doubles it again. The two effects multiply.

For a general stretch, let the new length be \(k\) times the original length:

\[ L'=kL \]

Constant volume requires

\[ L'A'=LA \]

so

\[ A'=\frac{A}{k} \]

Substituting into the resistance equation gives

\[ R'=\rho\frac{kL}{A/k} \]

Therefore,

\[ \boxed{R'=k^2R} \]

This square relationship explains why even a moderate stretch can produce a noticeable increase in resistance.

For example, a 10% increase in length means \(k=1.10\):

\[ \frac{R'}{R}=(1.10)^2=1.21 \]

The resistance increases by 21%, not 10%.

That is the central idea: under the constant-volume model, resistance increases with the square of the length multiplier.

A useful contrast: adding wire is not the same as stretching wire

Consider two different ways to make a wire twice as long.

In the first situation, you’ve added more copper. The wire becomes longer, but its thickness stays the same.

In the second, you’ve redistributed the original copper along a greater length. The wire becomes both longer and thinner.

This is why you cannot simply say that doubling the length doubles the resistance whenever a question describes a stretched wire.

03What is happening inside the metal?

There is another detail worth getting right.

Metals contain mobile electrons that move through a lattice of positively charged metal ions. As electrons move through the material, they are scattered by interactions with the lattice, impurities, and defects.

These microscopic interactions contribute to the material’s resistivity, \(\rho\).

When a metal wire is stretched, the spacing between its atoms changes. The deformation can also affect defects in the metal. These changes can slightly alter how electrons move through the material, and therefore change its resistivity.

However, be careful with the explanation.

It is tempting to say that resistance increases simply because electrons collide more often as they travel through a longer wire. That is an oversimplification. The length and cross-sectional area already account for the main geometrical effect in the resistance equation.

For the standard HSC model, assume that the resistivity remains constant unless the question provides information suggesting otherwise.

There are also two important limitations:

• Constant volume is an approximation. During small elastic stretches, the sideways contraction of a real metal depends on its Poisson ratio, so its volume does not remain exactly constant.
• If the wire stretches unevenly and develops a narrow section, its cross-sectional area is no longer uniform. That narrow section can contribute disproportionately to the total resistance.

We’ll return to the second point in a challenging question later.

04How could we test the relationship experimentally?

We now have a prediction: if a wire is stretched uniformly while its volume, temperature, and resistivity remain constant, its resistance should increase according to the square of its length multiplier.

But how would you test that prediction?

The challenge is to change the wire’s length without accidentally changing other factors that affect resistance.

Setting up the investigation

Start with a uniform metal wire held between two clamps.

One clamp remains fixed. The other is attached to a screw mechanism that gradually pulls the wire, increasing its length.

Connect the electrical measuring equipment as follows:

• A low-voltage regulated power supply connects to the wire.
• An ammeter sits in series with the wire, measuring the current through it.
• A voltmeter connects in parallel across the wire, measuring the potential difference across it.
• A micrometer measures the wire’s diameter at several positions.
• A ruler or length scale measures the distance between the clamps.

The power supply is set to maintain a constant, low voltage across the wire.

At each stretch setting, record the voltage and current. Calculate the resistance using

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

You should also measure the wire’s length and diameter while it is held under tension.

What should happen?

Suppose the original wire has a length of \(0.500\text{ m}\) and a diameter of \(0.500\text{ mm}\).

As you stretch it, the length increases and the diameter decreases.

If volume remains constant, the expected diameter is

\[ d'=\frac{d}{\sqrt{k}} \]

where \(d\) is the original diameter and \(k\) is the length multiplier.

For example, if the length increases by 10%,

\[ k=1.10 \]

The predicted diameter becomes

\[ d'=\frac{0.500}{\sqrt{1.10}} \]

\[ d'\approx0.477\text{ mm} \]

Notice that the diameter decreases by approximately 4.7%, while the cross-sectional area decreases by approximately 9.1%.

That area reduction is what combines with the increased length to produce a 21% increase in resistance.

An illustrative set of measurements

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

Assume the power supply maintains a constant potential difference of \(0.500\text{ V}\) across the wire.

The resistance values are calculated using \(R=V/I\). Small differences from theoretical predictions are expected because the displayed measurements are rounded.

The pattern is clear.

As the wire stretches:

1. Its length increases.
2. Its diameter and cross-sectional area decrease.
3. Its resistance increases.
4. The current decreases because the potential difference is being held constant.

For the 10% stretch, the original resistance was \(2.00\ \Omega\), while the measured resistance became \(2.42\ \Omega\).

That is a 21% increase, matching the constant-volume prediction.

What could make the experiment misleading?

The first version of the experiment has a potential weakness: the electrical current can heat the wire.

Heating an ordinary metal wire increases its resistivity. If the wire warms during the measurements, some of the measured resistance increase could be caused by temperature rather than stretching.

This creates a problem. We want to measure the geometrical effect, but the measurement process itself can change the wire’s resistance.

Improve the method by:

• Using a low test voltage to minimise electrical heating.
• Switching off the supply while adjusting the stretching mechanism.
• Taking readings promptly and monitoring the wire’s temperature.
• Allowing the wire to return to the same temperature before comparing measurements.
• Measuring the diameter at several points to check that the wire has stretched uniformly.

The important experimental principle is to isolate the variable you’re investigating. If temperature changes significantly, you can no longer attribute the entire resistance change to geometry.

05Does the current increase or decrease?

So far, we’ve established that stretching increases resistance.

But what happens to the current?

The answer depends on how the wire is connected.

This is where students sometimes make a second mistake: assuming that an increase in resistance must always produce the same change in current or power.

It doesn’t.

Case 1: Constant voltage

Suppose the stretched wire is connected directly across a constant-voltage supply.

The potential difference remains unchanged.

Using Ohm’s law,

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

If resistance increases, current decreases.

The electrical power dissipated by the wire is

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

So power also decreases.

Case 2: Constant current

Now imagine a current-regulated power supply that maintains the same current through the wire as it stretches.

The current stays constant, so the potential difference must increase:

\[ V=IR \]

The power dissipated becomes

\[ P=I^2R \]

As resistance increases, power increases.

The difference is summarised below.

The equations for electrical power are particularly useful here.

Don’t decide what happens to power until you’ve identified which electrical quantity is being held constant.

For a more detailed explanation of when resistance can be treated as constant, see Ohm’s Law in HSC Physics: When Resistance Is Constant.

06Worked examples

07Questions and solutions

The following questions progress from straightforward calculations to situations where you need to question an assumption about the wire.

Question 1

A uniform copper wire has an initial resistance of \(1.80\ \Omega\) and a length of \(0.800\text{ m}\).

It is stretched uniformly until its length becomes \(0.880\text{ m}\).

Assuming constant volume, temperature, and resistivity, calculate its new resistance.

Solution 1

The new resistance is \(2.18\ \Omega\).

The key is to recognise that the wire’s length increases by 10%, but its cross-sectional area also decreases.

Step 1: Calculate the length multiplier.

\[ k=\frac{0.880}{0.800}=1.10 \]

Step 2: Apply the constant-volume relationship.

\[ R'=k^2R \]

\[ R'=(1.10)^2(1.80) \]

\[ R'=1.21\times1.80 \]

\[ \boxed{R'=2.18\ \Omega} \]

The resistance increases by 21%.

A common mistake is to multiply the original resistance by 1.10 and obtain \(1.98\ \Omega\). That accounts for the increased length but ignores the reduced cross-sectional area.

Question 2

A metal wire initially has a resistance of \(2.00\ \Omega\). It is connected across a constant \(0.600\text{ V}\) supply.

Its length is then increased by 10% through uniform stretching.

After stretching, the measured current is \(0.238\text{ A}\).

Assume the diameter measurements confirm the predicted constant-volume reduction in cross-sectional area.

a) Calculate the measured resistance after stretching.

b) Calculate the resistance predicted by the constant-volume model.

c) Determine the percentage by which the measured resistance exceeds the geometrical prediction.

d) If the wire’s temperature was not controlled, can you conclude that stretching itself changed the resistivity? Explain.

Solution 2

The measured resistance is approximately \(2.52\ \Omega\), which is greater than the geometrical prediction of \(2.42\ \Omega\). The difference is approximately 4.2%.

Step 1: Calculate the measured resistance.

Using Ohm’s law,

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

\[ R'=\frac{0.600}{0.238} \]

\[ \boxed{R'\approx2.52\ \Omega} \]

Step 2: Calculate the geometrical prediction.

The wire’s length multiplier is

\[ k=1.10 \]

Therefore,

\[ R_{\text{geometry}}'=k^2R_0 \]

\[ R_{\text{geometry}}'=(1.10)^2(2.00) \]

\[ \boxed{R_{\text{geometry}}'=2.42\ \Omega} \]

Step 3: Compare the measured and predicted values.

The percentage excess is

\[ \%\text{ excess} = \frac{R'_{\text{measured}}-R'_{\text{geometry}}} {R'_{\text{geometry}}}\times100 \]

\[ =\frac{2.52-2.42}{2.42}\times100 \]

\[ \boxed{\%\text{ excess}\approx4.2\%} \]

The measured resistance is approximately 4.2% greater than the geometrical prediction.

Step 4: Evaluate the possible explanation.

We cannot confidently attribute the additional increase to a change in resistivity caused by mechanical strain.

If the temperature had remained constant, the extra resistance would suggest that the effective resistivity had increased beyond the constant-resistivity model.

However, if the wire warmed during the measurement, its temperature-dependent resistivity could also have increased.

Measurement uncertainty and contact resistance are other possible explanations.

The tempting conclusion is that the constant-volume model must be wrong. But the data only show that the complete measurement differs from the simplified prediction. They do not identify the cause without further controls.

Question 3

Two identical metal wires are connected in parallel across a \(9.0\text{ V}\) supply.

Each wire initially has a resistance of \(3.00\ \Omega\).

One wire is stretched uniformly until its length increases by 50%. The other wire remains unchanged.

Assume constant volume, temperature, and resistivity.

Calculate:

a) The resistance of the stretched wire.

b) The current through each wire.

c) The fraction of the total current flowing through the stretched wire, expressed as a percentage.

Solution 3

The stretched wire has a resistance of \(6.75\ \Omega\). It carries approximately 30.8% of the total current.

The key is that parallel branches have the same potential difference across them, but their currents depend on their individual resistances.

Step 1: Calculate the resistance of the stretched wire.

A 50% increase in length gives

\[ k=1.50 \]

Therefore,

\[ R'=k^2R \]

\[ R'=(1.50)^2(3.00) \]

\[ \boxed{R'=6.75\ \Omega} \]

Step 2: Calculate the current through the unchanged wire.

The potential difference across each parallel branch is \(9.0\text{ V}\).

\[ I_1=\frac{V}{R_1} \]

\[ I_1=\frac{9.0}{3.00} \]

\[ \boxed{I_1=3.00\text{ A}} \]

Step 3: Calculate the current through the stretched wire.

\[ I_2=\frac{V}{R_2} \]

\[ I_2=\frac{9.0}{6.75} \]

\[ \boxed{I_2=1.33\text{ A}} \]

Step 4: Calculate the total current.

For parallel branches, the total current is the sum of the branch currents.

\[ I_{\text{total}}=I_1+I_2 \]

\[ I_{\text{total}}=3.00+1.33 \]

\[ I_{\text{total}}=4.33\text{ A} \]

Step 5: Calculate the fraction flowing through the stretched wire.

\[ \text{Fraction}= \frac{I_2}{I_{\text{total}}} \]

\[ =\frac{1.33}{4.33} \]

\[ \boxed{\text{Fraction}\approx0.308=30.8\%} \]

The stretched wire carries less current because its resistance is higher.

A tempting mistake is to assume that identical materials mean identical currents. But the wires no longer have identical dimensions. Their resistances are different, even though they are connected to the same voltage.

Question 4

A metal wire has an initial resistance of \(8.00\ \Omega\). It is connected in series with a separate \(4.00\ \Omega\) resistor across a constant \(12.0\text{ V}\) supply.

The wire is then stretched uniformly until its length increases by 50%.

Assume its temperature, volume, and resistivity remain constant. The \(4.00\ \Omega\) resistor is unaffected.

Calculate:

a) The initial current in the circuit.

b) The new current after stretching.

c) The potential difference across the wire before and after stretching.

d) The power dissipated by the wire before and after stretching.

Solution 4

After stretching, the wire has a resistance of \(18.0\ \Omega\). The circuit current decreases from \(1.00\text{ A}\) to approximately \(0.545\text{ A}\).

The important detail is that the supply voltage is fixed, but the voltage across the wire is not. The separate series resistor also takes part of the supply voltage.

Step 1: Calculate the new wire resistance.

The length multiplier is

\[ k=1.50 \]

Therefore,

\[ R'=k^2R \]

\[ R'=(1.50)^2(8.00) \]

\[ \boxed{R'=18.0\ \Omega} \]

Step 2: Calculate the initial circuit current.

Before stretching, the total series resistance is

\[ R_{\text{total}}=8.00+4.00 \]

\[ R_{\text{total}}=12.0\ \Omega \]

Using Ohm’s law,

\[ I_0=\frac{V}{R_{\text{total}}} \]

\[ I_0=\frac{12.0}{12.0} \]

\[ \boxed{I_0=1.00\text{ A}} \]

Step 3: Calculate the new circuit current.

After stretching, the total resistance becomes

\[ R_{\text{total}}'=18.0+4.00 \]

\[ R_{\text{total}}'=22.0\ \Omega \]

Therefore,

\[ I'=\frac{12.0}{22.0} \]

\[ \boxed{I'\approx0.545\text{ A}} \]

The current decreases because the total series resistance increases.

Step 4: Calculate the voltage across the wire.

Before stretching,

\[ V_{\text{wire},0}=I_0R_0 \]

\[ V_{\text{wire},0}=1.00\times8.00 \]

\[ \boxed{V_{\text{wire},0}=8.00\text{ V}} \]

After stretching,

\[ V_{\text{wire}}'=I'R' \]

\[ V_{\text{wire}}'=0.54545\times18.0 \]

\[ \boxed{V_{\text{wire}}'\approx9.82\text{ V}} \]

Notice that the voltage across the wire increases, even though the circuit current decreases.

The remaining voltage is across the unchanged \(4.00\ \Omega\) resistor.

Step 5: Calculate the power dissipated by the wire.

Use

\[ P=I^2R \]

Before stretching,

\[ P_0=(1.00)^2(8.00) \]

\[ \boxed{P_0=8.00\text{ W}} \]

After stretching,

\[ P'=(0.54545)^2(18.0) \]

\[ \boxed{P'\approx5.36\text{ W}} \]

The power decreases.

This is an important distinction: although the wire’s resistance increases, the current is not fixed. The increase in total circuit resistance reduces the current enough that the wire’s power also decreases.

A student who assumes that the entire \(12.0\text{ V}\) supply remains across the wire has ignored the series resistor.

Question 5

A uniform metal wire has an original length \(L\), cross-sectional area \(A\), and resistance \(R_0\).

Imagine dividing the wire into two equal sections, each containing half the original material.

The wire is then stretched non-uniformly:

• The first half becomes 1.6 times its original length.
• The second half remains at its original length.

The two sections remain joined end-to-end, forming one continuous wire. Assume each section remains uniform internally, and ignore any additional resistance at their junction.

A second, identical wire is stretched uniformly until it reaches the same final total length as the non-uniformly stretched wire.

Both wires have the same material, total volume, and temperature.

Determine which wire has the greater resistance. Calculate both resistances in terms of \(R_0\), and explain why having the same total length and volume does not guarantee equal resistance.

Solution 5

The non-uniformly stretched wire has the greater resistance.

Its resistance is \(1.78R_0\), compared with \(1.69R_0\) for the uniformly stretched wire.

The tempting prediction is that both wires should have the same resistance because they have the same material, volume, and final length.

That prediction overlooks an important feature: resistance depends on the distribution of cross-sectional area along the wire, not just the total amount of material and its final length.

Step 1: Determine the final dimensions of the non-uniform wire.

The original wire has two equal sections, each with length

\[ \frac{L}{2} \]

The first section stretches by a factor of 1.6.

Its new length is

\[ L_1'=1.6\left(\frac{L}{2}\right) \]

\[ L_1'=0.8L \]

Because its volume remains constant,

\[ \frac{L}{2}A=L_1'A_1' \]

Substituting,

\[ \frac{L}{2}A=0.8LA_1' \]

Therefore,

\[ \boxed{A_1'=\frac{5A}{8}} \]

The second section is unchanged:

\[ L_2'=\frac{L}{2} \]

\[ A_2'=A \]

The total final length is

\[ L_{\text{total}}'=0.8L+0.5L \]

\[ \boxed{L_{\text{total}}'=1.3L} \]

Step 2: Calculate the resistance of the first section.

Using

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

the first section has resistance

\[ R_1'=\rho\frac{0.8L}{5A/8} \]

\[ R_1'=1.28\frac{\rho L}{A} \]

Since \(R_0=\rho L/A\),

\[ \boxed{R_1'=1.28R_0} \]

Step 3: Calculate the resistance of the second section.

The second section is unchanged from its original dimensions:

\[ R_2'=\rho\frac{L/2}{A} \]

\[ \boxed{R_2'=0.50R_0} \]

Step 4: Calculate the total resistance of the non-uniform wire.

The sections are connected end-to-end, so their resistances add.

\[ R_{\text{non-uniform}}=R_1'+R_2' \]

\[ R_{\text{non-uniform}}=1.28R_0+0.50R_0 \]

\[ \boxed{R_{\text{non-uniform}}=1.78R_0} \]

Step 5: Calculate the resistance of the uniformly stretched wire.

The second wire has the same final length, \(1.3L\), but it is stretched uniformly.

Its cross-sectional area must therefore be

\[ A'=\frac{A}{1.3} \]

Its resistance becomes

\[ R_{\text{uniform}}=\rho\frac{1.3L}{A/1.3} \]

\[ R_{\text{uniform}}=(1.3)^2\frac{\rho L}{A} \]

Therefore,

\[ \boxed{R_{\text{uniform}}=1.69R_0} \]

Step 6: Compare the results.

The difference is

\[ \Delta R=1.78R_0-1.69R_0 \]

\[ \boxed{\Delta R=0.09R_0} \]

The non-uniformly stretched wire has approximately 5.3% greater resistance than the uniformly stretched wire.

Why?

The narrow section has a smaller cross-sectional area, so it contributes more resistance per unit length. Since the sections are in series, that extra resistance cannot be bypassed.

The deeper mathematical reason is that the resistance contributions depend on the square of each section’s stretch factor.

For two equal sections with stretch factors \(k_1\) and \(k_2\),

\[ \frac{R_{\text{non-uniform}}}{R_0} =\frac{k_1^2+k_2^2}{2} \]

For a uniform stretch to the same final length,

\[ \frac{R_{\text{uniform}}}{R_0} =\left(\frac{k_1+k_2}{2}\right)^2 \]

The difference between these expressions is

\[ \frac{(k_1-k_2)^2}{4} \]

This is positive whenever the two sections stretch by different amounts.

So the result is not an accident of the chosen numbers. Under the assumptions in this question, non-uniform stretching produces greater resistance than uniform stretching to the same final length.

This is also why a narrow section, or a neck, can matter so much when analysing a real stretched conductor.

08Where to go next

When a question describes a stretched wire, start by finding its new resistance. Then identify the actual circuit condition before calculating current, voltage, or power.

The next useful skill is applying that new resistance in a complete circuit, especially when the wire is connected in series or parallel with other components. That’s where a correct geometrical prediction can still lead to an incorrect electrical answer if you assume the wrong quantity stays constant.