Why Wire Resistance Depends on Length, Thickness and Material

Two copper wires are at the same temperature. Wire B is twice as long as wire A, but its diameter is also twice as large. Which wire has the greater resistance?

Predict before reading on. The longer wire seems as though it should resist current more. But doubling the diameter makes the wire’s cross-sectional area four times as large. These two effects combine to give wire B half the resistance of wire A. The shorter, thinner wire actually has the greater resistance.

This is the key to comparing wires: length, cross-sectional area, material, and temperature all matter, but they do not affect resistance in the same way.

01Start with what resistance measures

Resistance describes how much a conductor opposes electric current for a given voltage.

For an ohmic conductor at a fixed temperature, resistance is calculated using

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

where:

• \(R\) is resistance, measured in ohms (\(\Omega\)).
• \(V\) is the voltage across the conductor, measured in volts (V).
• \(I\) is the current through the conductor, measured in amperes (A).

For example, if a voltage of \(2.0\text{ V}\) produces a current of \(1.0\text{ A}\), the resistance is

\[
R=\frac{2.0}{1.0}=2.0\,\Omega
\]

If the same voltage produces only \(0.50\text{ A}\), the resistance is \(4.0\,\Omega\).

To compare two wires fairly, you can measure the voltage across each wire and the current through it, then calculate \(V/I\). However, you must control the conditions of the comparison. If you want to investigate length, for example, keep the material, cross-sectional area, and temperature the same.

A useful mental model

Imagine charge carriers moving through a crowded hallway. A longer hallway gives them more opportunities to scatter, while a wider hallway allows more people to move side by side.

This gives us a useful starting point:

• A longer wire generally has greater resistance.
• A wire with a larger cross-sectional area generally has lower resistance.

The analogy has limits. Electrons are not people, and a larger cross-sectional area does not literally create separate lanes. The microscopic behaviour of charge carriers is more complicated, but the analogy helps explain the trends.

We can now make those trends precise.

01Length: a longer wire has greater resistance

Imagine two copper wires with the same cross-sectional area, both at the same temperature. One wire is \(1.0\text{ m}\) long, and the other is \(2.0\text{ m}\) long.

Which has greater resistance?

The \(2.0\text{ m}\) wire has twice the resistance.

For a uniform wire, if the material, cross-sectional area, and temperature remain constant, resistance is directly proportional to length:

\[
R\propto L
\]

Here, \(L\) is the wire’s length.

This means:

• Double the length, and the resistance doubles.
• Triple the length, and the resistance triples.
• Halve the length, and the resistance halves.

Cutting a uniform wire into two equal pieces halves the resistance of each piece, provided the material, cross-sectional area, and temperature remain unchanged.

There is an important condition here. A longer wire does not automatically have greater resistance than every shorter wire. The comparison above only works when the other relevant properties are held constant.

A short, thin wire made from one material can easily have greater resistance than a long, thick wire made from another.

02Thickness: cross-sectional area matters

Now compare two wires of the same material and length, at the same temperature. One is thicker than the other.

The thicker wire has lower resistance because it has a larger cross-sectional area.

Cross-sectional area, represented by \(A\), means the area of the wire’s cut face, measured perpendicular to its length. It is not the surface area of the entire wire.

For a circular wire,

\[
A=\pi r^2
\]

where \(r\) is the radius.

Since the diameter \(d\) is twice the radius, we can also write

\[
A=\frac{\pi d^2}{4}
\]

Notice the square. This is where many students make mistakes.

Doubling the diameter quadruples the cross-sectional area, not doubles it.

For wires of the same material, length, and temperature, resistance is inversely proportional to cross-sectional area:

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

So, if the area doubles, resistance halves. If the area quadruples, resistance falls to one-quarter of its original value.

Why diameter changes resistance so quickly

Suppose the diameter increases by 50%.

The new diameter is \(1.5d\). Because area depends on diameter squared, the new area is

\[
A_{\text{new}}=(1.5)^2A_{\text{old}}
\]

\[
A_{\text{new}}=2.25A_{\text{old}}
\]

The resistance therefore becomes

\[
R_{\text{new}}=\frac{R_{\text{old}}}{2.25}
\]

The wire’s diameter has increased by 50%, but its resistance has decreased by about 56%.

This is why you should never compare wire thickness by simply dividing or multiplying the diameters. Convert the diameter change into an area change first.

03Material: some substances resist current more than others

So far, we have compared wires made from the same material. What happens if two wires have identical dimensions but are made from different materials?

Their resistances can still be very different.

Copper and nichrome provide a useful example. Copper is commonly used for electrical wiring because it has relatively low resistivity. Nichrome, an alloy commonly used in heating elements, has much higher resistivity.

Resistivity, represented by the Greek letter \(\rho\) (rho), describes how strongly a material resists current under specified conditions. Unlike resistance, it does not depend on the particular length and cross-sectional area of a uniform sample.

For a uniform wire at a fixed temperature, the relationship between resistance, material, and geometry is

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

where:

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

The equation brings together everything we have discussed.

• Increasing \(L\) increases \(R\).
• Increasing \(A\) decreases \(R\).
• Increasing \(\rho\) increases \(R\).

The material affects the resistance through its resistivity. Changing the dimensions of a copper wire does not turn it into a material with a different resistivity.

Comparing resistivity

The following are approximate resistivity values at \(20^\circ\text{C}\).

Actual values vary with purity, alloy composition, and temperature.

If copper and nichrome wires have the same length, cross-sectional area, and temperature, their resistance ratio is equal to their resistivity ratio:

\[
\frac{R_{\text{nichrome}}}{R_{\text{copper}}}
=
\frac{\rho_{\text{nichrome}}}{\rho_{\text{copper}}}
\]

Using the values in the table,

\[
\frac{R_{\text{nichrome}}}{R_{\text{copper}}}
\approx 65
\]

The nichrome wire has about 65 times the resistance of the copper wire.

That difference comes from the material, not the geometry. If the wires had different lengths or cross-sectional areas, we would need to account for those differences as well.

04Temperature: the condition students often overlook

Suppose you measure a copper wire’s resistance while it is cool, then measure it again after a large current has heated it. The wire has the same material and almost the same dimensions. Should the resistance remain unchanged?

For an ordinary metal wire, the resistance generally increases as its temperature rises.

As the metal heats up, its atoms vibrate more strongly around their usual positions. These vibrations make it harder for charge carriers to move through the material, increasing its resistivity.

For a metal over a modest temperature range, the change can be approximated by

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

where:

• \(R(T)\) is the resistance at temperature \(T\).
• \(R(T_0)\) is the resistance at a reference temperature \(T_0\).
• \(\alpha\) is the temperature coefficient of resistance, measured in \({}^\circ\text{C}^{-1}\).
• \(T-T_0\) is the temperature change in degrees Celsius.

For common metals, \(\alpha\) is positive, so an increase in temperature generally produces an increase in resistance.

Temperature matters when measuring resistance because the measurement itself can change the wire’s condition. A large test current may heat the wire, meaning that a later value of \(V/I\) can be greater than the value measured when the wire was cool.

To compare wires fairly:

• Keep their temperatures the same.
• Use the same material when investigating geometry.
• Keep length and cross-sectional area the same when investigating material.
• Avoid unnecessary heating during resistance measurements, and allow wires to cool before repeating a comparison.

A difference in measured resistance does not automatically mean that a difference in length, thickness, or material caused it. Temperature may be the hidden variable.

06A quick rule for predicting resistance

Use this table when comparing wires before reaching for a calculator.

The general equation \(R=\rho L/A\) is the most reliable way to combine these effects. Just remember that resistivity itself depends on temperature.

Once you can predict which wire has the greater resistance, the next step is to use \(V=IR\) to predict the current through a circuit and the voltage drop across each component.

07Questions and solutions

For every question, rank the wires from lowest to highest resistance before doing any calculations. Assume the wires are uniform and the stated temperatures remain constant unless the question says otherwise.

Question 1

Two copper wires, P and Q, have the same length and temperature. Wire P has diameter \(d\), while wire Q has diameter \(2d\).

Rank the wires by resistance, then state the ratio \(R_Q/R_P\).

Solution 1

Wire Q has the lower resistance, with

\[
R_Q<R_P
\]

Doubling the diameter quadruples the cross-sectional area:

\[
\frac{A_Q}{A_P}=2^2=4
\]

Since the wires have the same material, length, and temperature, resistance is inversely proportional to area. Therefore,

\[
\frac{R_Q}{R_P}=\frac{A_P}{A_Q}
=\frac14
\]

Wire Q has one-quarter the resistance of wire P. The important point is that doubling the diameter does not merely halve the resistance; it reduces the resistance to one-quarter.

Question 2

Three copper wires are at the same temperature.

• Wire P has length \(L\) and diameter \(d\).
• Wire Q has length \(2L\) and diameter \(d\).
• Wire S has length \(3L\) and diameter \(2d\).

Rank the wires from lowest to highest resistance. Then express \(R_Q\) and \(R_S\) in terms of \(R_P\).

Solution 2

The ranking is

\[
R_S<R_P<R_Q
\]

Wire Q has twice the length of wire P but the same cross-sectional area. Therefore,

\[
R_Q=2R_P
\]

Wire S has three times the length of wire P. However, its diameter is doubled, so its cross-sectional area is quadrupled.

Combining the two changes gives

\[
R_S=\frac{3}{4}R_P
\]

Thus,

\[
\boxed{R_S=0.75R_P}
\]

Wire S is longer than both other wires, but its larger cross-sectional area more than compensates for that extra length. This is why ranking wires by length alone can give the wrong answer.

Question 3

Two copper wires, X and Y, have identical lengths and cross-sectional areas. Wire X is kept at \(20^\circ\text{C}\), while wire Y is warmed to \(80^\circ\text{C}\).

Rank the wires by resistance. A student argues that their resistances must be equal because their dimensions and material are identical. Explain why this argument is incomplete, and describe one way to make a fair experimental comparison.

Solution 3

Wire Y has the greater resistance:

\[
R_Y>R_X
\]

The student’s argument ignores temperature. For a typical metal such as copper, increasing temperature increases resistivity, so the warmer wire has greater resistance even though its geometry and material are unchanged.

A fair comparison of the wires’ geometrical effects would require both wires to be at the same temperature. You could use a small test current to minimise heating, measure the wires under the same temperature conditions, and allow them to cool before repeating the measurements.

The result illustrates an important experimental principle: keeping dimensions constant is not enough if another property affecting resistance changes.

Question 4

Wire P is made from aluminium and has a length of \(3.0\text{ m}\) and a cross-sectional area of \(0.60\text{ mm}^2\). Wire Q is made from copper and has a length of \(2.0\text{ m}\) and a cross-sectional area of \(0.20\text{ mm}^2\).

Both wires are at \(20^\circ\text{C}\). Use the following resistivities:

\[
\rho_{\text{Al}}=2.8\times10^{-8}\,\Omega\,\text{m}
\]

\[
\rho_{\text{Cu}}=1.7\times10^{-8}\,\Omega\,\text{m}
\]

First rank the wires by resistance without calculating. Then calculate both resistances and determine their ratio.

Solution 4

Wire P has the lower resistance:

\[
R_P<R_Q
\]

This prediction is not immediately obvious. Aluminium has greater resistivity than copper, and wire P is longer. Both changes tend to increase its resistance. However, wire P has three times the cross-sectional area of wire Q, which more than compensates for those disadvantages.

Convert the areas into square metres:

\[
A_P=6.0\times10^{-7}\text{ m}^2
\]

\[
A_Q=2.0\times10^{-7}\text{ m}^2
\]

For wire P,

\[
R_P=\frac{\rho_P L_P}{A_P}
\]

\[
R_P=\frac{(2.8\times10^{-8})(3.0)}
{6.0\times10^{-7}}\,\Omega
\]

\[
\boxed{R_P=0.14\,\Omega}
\]

For wire Q,

\[
R_Q=\frac{\rho_Q L_Q}{A_Q}
\]

\[
R_Q=\frac{(1.7\times10^{-8})(2.0)}
{2.0\times10^{-7}}\,\Omega
\]

\[
\boxed{R_Q=0.17\,\Omega}
\]

The ratio is

\[
\frac{R_P}{R_Q}
=\frac{0.14}{0.17}
\approx0.82
\]

Wire P has approximately 82% of wire Q’s resistance, or about 18% less resistance.

The larger area is the decisive factor in this comparison. The material, length, and area must all be considered together; none should be judged in isolation.

Question 5

A uniform copper wire has length \(L\), cross-sectional area \(A\), and resistance \(R\) at a fixed temperature.

The wire is cut into two equal halves. One half is left unchanged. The other half is stretched uniformly until its length returns to \(L\). Assume that no material is lost, its volume remains constant, its resistivity is unchanged, and its temperature stays the same.

The two resulting pieces are then connected in parallel across the same two terminals. Ignore the resistance of the connecting leads and contacts.

Before calculating, rank the following from lowest to highest resistance:

• The original wire.
• The untouched half.
• The stretched half.
• The parallel combination of the two halves.

Then determine the resistance of each item in terms of \(R\).

Solution 5

The ranking, from lowest to highest resistance, is

\[
\boxed{
R_{\parallel}
<
R_{\text{half}}
<
R
<
R_{\text{stretched}}
}
\]

Their resistances are, respectively,

\[
R_{\parallel}=\frac25R,\qquad
R_{\text{half}}=\frac12R,\qquad
R,\qquad
R_{\text{stretched}}=2R
\]

The key is to track both length and cross-sectional area as the wire changes.

Step 1: Find the resistance of the untouched half.

The original wire has resistance

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

The untouched half has length \(L/2\) and cross-sectional area \(A\). Therefore,

\[
R_{\text{half}}=\frac{\rho(L/2)}{A}
\]

\[
\boxed{R_{\text{half}}=\frac12R}
\]

Halving the length halves the resistance because the cross-sectional area and resistivity remain unchanged.

Step 2: Find the resistance of the stretched half.

This is the critical step. The stretched half has the same final length as the original wire, but it does not have the same cross-sectional area.

Before stretching, its volume is

\[
V_{\text{wire}}=\frac{L}{2}A
\]

Since volume is conserved and its final length is \(L\),

\[
LA_{\text{new}}=\frac{LA}{2}
\]

Therefore,

\[
A_{\text{new}}=\frac{A}{2}
\]

The stretched half is now twice as long as it was before stretching, and its cross-sectional area has halved.

Its resistance is

\[
R_{\text{stretched}}
=\frac{\rho L}{A/2}
\]

\[
\boxed{R_{\text{stretched}}=2R}
\]

A tempting mistake is to say that the stretched half has the same resistance as the original wire because both have length \(L\). That ignores the halving of the cross-sectional area, which doubles the resistance relative to the original wire.

Step 3: Calculate the parallel combination.

The two branches have resistances \(R/2\) and \(2R\). For two resistors connected in parallel,

\[
\frac{1}{R_{\parallel}}
=
\frac{1}{R_{\text{half}}}
+
\frac{1}{R_{\text{stretched}}}
\]

Substituting,

\[
\frac{1}{R_{\parallel}}
=
\frac{2}{R}+\frac{1}{2R}
\]

\[
\frac{1}{R_{\parallel}}=\frac{5}{2R}
\]

Therefore,

\[
\boxed{R_{\parallel}=\frac25R}
\]

Step 4: Check the ranking.

The four resistances are

\[
0.4R,\quad 0.5R,\quad R,\quad 2R
\]

So the parallel combination has the lowest resistance, followed by the untouched half, the original wire, and finally the stretched half.

The result combines three ideas: halving a wire’s length halves its resistance, stretching a wire while conserving volume changes both its length and area, and connecting resistances in parallel produces an equivalent resistance lower than either branch.

That last point matters when you move from individual wires to complete circuits. The resistance of a circuit depends not just on the properties of each conductor, but also on how the conductors are connected.