Why the Gradient Depends on Which Axis Contains Voltage

A student measures the current through a resistor as the voltage increases. They plot the results and calculate the gradient. Another student plots the same measurements, but swaps the axes. Both get straight lines, yet their gradients are different.

Which student has calculated the resistance correctly?

The answer is: both might have, depending on which variable is on which axis and what they do with the gradient.

This is because resistance is defined by the ratio of voltage to current, not simply by the steepness of a line. Put voltage on the vertical axis and the gradient represents resistance. Put current on the vertical axis and the gradient represents conductance, which is the reciprocal of resistance.

Let’s work through why this happens, how to recognise each case, and what changes when voltage or current is held constant.

01Why changing the axes changes the gradient

Imagine a resistor connected to a variable power supply. You gradually increase the supply voltage and measure the current flowing through the resistor.

For an ideal ohmic resistor, the current increases in direct proportion to the voltage, provided its physical conditions, particularly temperature, remain constant.

Suppose you collect the following illustrative data.

Notice that every 2 V increase produces a 0.5 A increase.

What happens if you plot voltage against current?

Both graphs show the same physical behaviour. The resistor has not changed. The measurements have not changed. Only the positions of the variables have changed.

However, the gradient is calculated differently.

Remember:

\[ \text{Gradient}=\frac{\text{vertical change}}{\text{horizontal change}} \]

This means:

\[ \text{Gradient}=\frac{\Delta y}{\Delta x} \]

The vertical axis determines the numerator. The horizontal axis determines the denominator.

That is the entire reason the gradient changes when you swap the axes.

02Voltage on the vertical axis: gradient is resistance

Start with the familiar relationship:

\[ V=IR \]

where:

• \(V\) is potential difference, measured in volts (V).
• \(I\) is current, measured in amperes (A).
• \(R\) is resistance, measured in ohms (\(\Omega\)).

Rearranging gives:

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

Now consider a graph with voltage on the vertical axis and current on the horizontal axis.

Its gradient is:

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

For an ohmic resistor with constant resistance:

\[ \boxed{\text{Gradient}=R} \]

The gradient is resistance because the graph measures the change in voltage divided by the change in current.

Using our illustrative data, select two points:

• Point A: \(I=0.5\text{ A}\), \(V=2.0\text{ V}\).
• Point B: \(I=2.5\text{ A}\), \(V=10.0\text{ V}\).

Therefore:

\[ \begin{aligned} R&=\frac{\Delta V}{\Delta I}\\ &=\frac{10.0-2.0}{2.5-0.5}\\ &=\frac{8.0}{2.0}\\ &=4.0\ \Omega \end{aligned} \]

The resistance is \(4.0\ \Omega\).

A steeper voltage-against-current graph means greater resistance.

Think about what this means physically.

For the same increase in current, a resistor with greater resistance requires a larger increase in voltage.

It takes more potential difference to drive the same additional current through it.

03Current on the vertical axis: gradient is reciprocal resistance

Now swap the axes.

Put voltage on the horizontal axis and current on the vertical axis.

The gradient becomes:

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

But Ohm’s law tells us:

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

For a constant resistance:

\[ \boxed{\text{Gradient}=\frac{1}{R}} \]

This is the reciprocal of resistance.

It is also called conductance, represented by \(G\).

\[ G=\frac{1}{R} \]

Conductance measures how readily a component allows electric current to flow for a given potential difference.

Its unit is the siemens (S), equivalent to \(\text{A V}^{-1}\).

Let’s use the same resistor data again.

\[ \begin{aligned} G&=\frac{\Delta I}{\Delta V}\\ &=\frac{2.5-0.5}{10.0-2.0}\\ &=\frac{2.0}{8.0}\\ &=0.25\text{ S} \end{aligned} \]

To recover resistance:

\[ R=\frac{1}{G} \]

\[ R=\frac{1}{0.25}=4.0\ \Omega \]

We get exactly the same resistance.

Nothing about the resistor has changed. We have simply calculated the reciprocal quantity.

Notice the important reversal:

• On a voltage-current graph, a steeper gradient means greater resistance.
• On a current-voltage graph, a steeper gradient means lower resistance.

This is a common HSC examination trap. You cannot judge resistance from steepness until you have checked the axes.

04What happens when voltage or current is held constant?

There is another reason the axis choice matters: it changes how you interpret the effect of changing resistance.

Consider two separate experiments.

Experiment A: Constant voltage

Connect different resistors, one at a time, to the same ideal voltage supply.

The potential difference across each resistor stays constant.

From Ohm’s law:

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

If voltage is fixed, increasing resistance decreases current.

For example, use a constant 12 V supply.

Doubling resistance halves the current.

The physical interpretation is straightforward: the supply maintains the same energy transferred per coulomb, but the more resistive component allows less charge to flow each second.

Experiment B: Constant current

Now imagine an adjustable current source that maintains the same current through different resistors.

The current stays fixed while the resistance changes.

Rearranging Ohm’s law:

\[ V=IR \]

If current is fixed, increasing resistance increases voltage.

For example, use a constant current of 2.0 A.

Doubling resistance doubles the required voltage.

These two experiments reveal why you must identify what is held constant before describing the effect of resistance.

There is a subtle point here.

In a voltage-controlled experiment, the current changes as resistance changes. In a current-controlled experiment, the voltage changes instead.

You cannot assume both voltage and current remain fixed when resistance changes. Ohm’s law would then require resistance to remain fixed as well.

05Why the microscopic picture supports this relationship

So far, we have described resistance using measurements. But what is actually happening inside the resistor?

In a metal wire, electric current involves the movement of electrons through a lattice of positive metal ions.

An applied electric field causes the electrons to acquire a net drift motion. They also undergo frequent interactions with the vibrating lattice and imperfections in the material.

These interactions contribute to electrical resistance.

A useful way to picture this is a crowded corridor.

People can move through the corridor, but obstacles and interactions make the overall movement less efficient. Adding more obstacles makes it harder to maintain the same flow.

The analogy is not exact: electrons are quantum particles, not tiny people physically bumping into fixed objects. However, it gives a useful starting picture of why microscopic interactions affect resistance.

For a metal conductor, increasing temperature generally increases lattice vibrations. This makes electron scattering more frequent and usually increases resistance.

That is why the resistance of a filament lamp changes as it heats up.

Importantly, this microscopic explanation is not needed to derive the graph gradient. The gradient relationship comes directly from the measured voltage-current relationship.

The microscopic model helps explain why the resistance might change between measurements.

For more on this effect, see How Heating Changes a Metal Wire’s Resistance.

06What if the graph is not a straight line?

The simple gradient relationships are easiest to use when resistance remains constant.

However, not every electrical component behaves this way.

A filament lamp is a useful contrasting example.

As current increases, the filament heats up. Its resistance generally increases with temperature.

Consequently, its voltage-current relationship is no longer a straight line.

For a curved graph, there is an important distinction between two types of resistance.

1. Resistance at an operating point

The ratio of voltage to current at a particular operating point is:

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

This is sometimes called the static resistance.

2. Resistance associated with a small change

For a curved voltage-current graph, the gradient of a tangent represents:

\[ r=\frac{dV}{dI} \]

This is the differential resistance.

For a straight-line ohmic resistor, these values are equal.

For a non-ohmic component, they do not have to be equal.

This matters because you cannot automatically use the gradient of a curved graph as the resistance \(V/I\).

For example, imagine a filament lamp operating at:

\[ V=8.0\text{ V}, \qquad I=2.0\text{ A} \]

Its static resistance is:

\[ R=\frac{8.0}{2.0}=4.0\ \Omega \]

Suppose the tangent to its voltage-current curve at that operating point has a gradient of \(5.0\ \text{V A}^{-1}\).

Its differential resistance is \(5.0\ \Omega\), not \(4.0\ \Omega\).

Both calculations are correct. They describe different things.

The first describes the ratio of the operating voltage to current. The second describes how much additional voltage is needed for a small additional current change at that operating point.

This distinction is particularly useful when analysing non-ohmic conductors.

07A practical method for interpreting voltage-current graphs

When you encounter a graph in an HSC Physics question, follow this sequence.

Step 1: Identify the vertical axis.

Is it voltage or current?

Step 2: Identify the horizontal axis.

Is it current or voltage?

Step 3: Write the gradient using the actual axes.

Do not start with a memorised resistance formula.

Step 4: Connect the gradient to Ohm’s law.

• Voltage vertically, current horizontally: gradient represents resistance.
• Current vertically, voltage horizontally: gradient represents reciprocal resistance.

Step 5: Check whether the graph is straight or curved.

A straight line through the origin supports a constant-resistance model over the measured range.

A curved line indicates that the ratio \(V/I\) changes with the operating point.

Step 6: Check the experimental conditions.

Was voltage held constant? Was current held constant? Was temperature controlled?

These conditions help explain the behaviour, particularly when comparing different measurements.

If you want a broader explanation of why gradients and areas have physical meanings in Physics, read HSC Physics Graphs: How to Use Gradient and Area.

08Worked examples

09Questions and solutions

The following questions progress from identifying a graph’s gradient to reasoning about experimental conditions and changing resistance.

Question 1

An ohmic resistor produces a straight-line graph with voltage on the vertical axis and current on the horizontal axis.

The graph passes through the origin and the point \((2.0\text{ A}, 10\text{ V})\).

Calculate:

a) The gradient of the graph.

b) The resistance of the resistor.

c) The current that would flow if the potential difference were increased to 15 V, assuming temperature remains constant.

Solution 1

The gradient is \(5.0\ \Omega\), the resistance is \(5.0\ \Omega\), and the current at 15 V is \(3.0\text{ A}\).

Part a: Calculate the gradient.

Voltage is on the vertical axis, so:

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

Using the origin and the given point:

\[ \text{Gradient}=\frac{10}{2.0} =5.0\ \text{V A}^{-1} \]

Part b: Identify the resistance.

For a voltage-against-current graph:

\[ R=\text{Gradient} \]

Therefore:

\[ R=5.0\ \Omega \]

Part c: Calculate the new current.

Use Ohm’s law:

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

\[ I=\frac{15}{5.0} =3.0\text{ A} \]

The current increases because the potential difference increases while resistance remains constant.

Question 2

An experimental graph has current on the vertical axis and voltage on the horizontal axis.

The best-fit line passes through the points:

• \((4.0\text{ V}, 0.80\text{ A})\)
• \((10.0\text{ V}, 2.0\text{ A})\)

a) Calculate the gradient.

b) Determine the resistance.

c) A student claims that the resistor has a resistance of \(0.20\ \Omega\) because this is the value of the gradient. Explain the error.

Solution 2

The gradient is \(0.20\text{ S}\), and the resistance is \(5.0\ \Omega\). The student’s error is treating the reciprocal resistance as resistance itself.

Part a: Calculate the gradient.

Current is on the vertical axis, so:

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

\[ \begin{aligned} G&=\frac{2.0-0.80}{10.0-4.0}\\ &=\frac{1.2}{6.0}\\ &=0.20\text{ S} \end{aligned} \]

Part b: Determine resistance.

The gradient represents reciprocal resistance:

\[ G=\frac{1}{R} \]

Rearrange:

\[ R=\frac{1}{G} \]

\[ R=\frac{1}{0.20}=5.0\ \Omega \]

Part c: Explain the error.

The student has correctly calculated the gradient but incorrectly interpreted its physical meaning.

Because current is on the vertical axis, the gradient is measured in amperes per volt. This is conductance, not resistance.

Resistance is the reciprocal of the gradient.

Question 3

Two ohmic resistors, A and B, are connected separately to the same ideal 12 V supply.

Resistor A has a resistance of \(3.0\ \Omega\), while resistor B has a resistance of \(6.0\ \Omega\).

a) Calculate the current through each resistor.

b) If the experiment is repeated using an ideal constant-current source set to 2.0 A, calculate the potential difference required across each resistor.

c) Explain why the effect of doubling resistance appears different in the two experiments.

Solution 3

With the constant-voltage supply, the currents are 4.0 A and 2.0 A. With the constant-current source, the required voltages are 6.0 V and 12 V.

Part a: Constant voltage.

Use:

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

For resistor A:

\[ I_A=\frac{12}{3.0}=4.0\text{ A} \]

For resistor B:

\[ I_B=\frac{12}{6.0}=2.0\text{ A} \]

Doubling resistance halves the current because voltage is fixed.

Part b: Constant current.

Use:

\[ V=IR \]

For resistor A:

\[ V_A=(2.0)(3.0)=6.0\text{ V} \]

For resistor B:

\[ V_B=(2.0)(6.0)=12\text{ V} \]

Doubling resistance doubles the voltage because current is fixed.

Part c: Explain the difference.

The two experiments hold different variables constant.

In the first experiment, voltage cannot change, so current must decrease when resistance increases.

In the second experiment, current cannot change, so voltage must increase when resistance increases.

The mistake would be to say that increasing resistance always decreases current without checking whether voltage is actually fixed.

Question 4

A student investigates a filament lamp and obtains the following illustrative measurements.

a) Calculate the resistance at each operating point.

b) Explain what happens to the resistance as the voltage increases.

c) The student plots voltage on the vertical axis and current on the horizontal axis. Explain why the gradient of the resulting curved graph cannot be treated as one constant resistance.

d) Suggest a physical explanation for the observed behaviour.

Solution 4

The calculated operating-point resistances are \(4.0\ \Omega\), \(5.0\ \Omega\), \(6.0\ \Omega\), and approximately \(7.0\ \Omega\). The resistance increases as the filament heats up.

Part a: Calculate the resistance.

Use:

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

For each measurement:

For the final measurement:

\[ R=\frac{8.0}{1.15} \approx 6.96\ \Omega \]

Part b: Describe the trend.

Resistance increases as voltage increases.

The current does not increase in direct proportion to voltage. For example, increasing voltage from 2.0 V to 4.0 V doubles the voltage, but current increases from 0.50 A to only 0.80 A.

Part c: Explain the graph’s gradient.

The voltage-current graph is curved because the resistance is changing.

For a curved graph, the gradient of a tangent at a particular point represents differential resistance:

\[ r=\frac{dV}{dI} \]

It does not necessarily equal the operating-point resistance:

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

Therefore, one constant gradient cannot describe the entire graph.

Part d: Explain the physical behaviour.

As current increases, the filament heats up.

The higher temperature increases lattice vibrations in the metal. Electrons experience more frequent scattering interactions, increasing the resistance.

The filament therefore requires progressively greater voltage increases to produce further increases in current.

This explanation assumes the lamp has reached a sufficiently stable operating temperature for each recorded measurement.

Question 5

An ideal ohmic resistor is tested using two different experimental arrangements.

In the first arrangement, a variable voltage supply is used. The student increases the voltage from 4.0 V to 12 V while recording the current.

In the second arrangement, an ideal constant-current source is used. The student increases the current from 1.0 A to 3.0 A while recording the voltage.

The resistor remains at constant temperature, and both experiments use the same resistor.

A student makes the following argument:

“The first experiment should produce a voltage-current graph with a gradient three times larger than the second experiment because the voltage increases by a factor of three in the first experiment, whereas the current increases by a factor of three in the second. The resistor also has the same resistance, so both gradients must represent the same physical quantity.”

Is this argument correct?

Determine the resistance using both experiments, calculate the gradients for the two possible axis arrangements, and explain precisely why the student’s comparison is flawed.

Solution 5

The argument is incorrect. Both experiments measure the same resistance, but the gradients of the two different graph arrangements represent reciprocal quantities, not two measurements of the same quantity.

Step 1: Establish the resistor’s operating relationship.

The resistor is ohmic, and its temperature is constant.

Therefore:

\[ V=IR \]

Its resistance is constant throughout both experiments.

From the first experiment, the voltage increases from 4.0 V to 12 V.

The current increases from 1.0 A to 3.0 A.

Using either operating point:

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

\[ R=\frac{4.0}{1.0}=4.0\ \Omega \]

The second point confirms this:

\[ R=\frac{12}{3.0}=4.0\ \Omega \]

Step 2: Calculate the gradient for the voltage-against-current graph.

For the first arrangement, take current as the horizontal axis and voltage as the vertical axis.

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

\[ \begin{aligned} \text{Gradient} &=\frac{12-4.0}{3.0-1.0}\\ &=\frac{8.0}{2.0}\\ &=4.0\ \Omega \end{aligned} \]

The gradient represents resistance.

Step 3: Calculate the gradient for the current-against-voltage graph.

Now reverse the axes.

Voltage is horizontal, and current is vertical.

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

\[ \begin{aligned} \text{Gradient} &=\frac{3.0-1.0}{12-4.0}\\ &=\frac{2.0}{8.0}\\ &=0.25\text{ S} \end{aligned} \]

The gradient represents reciprocal resistance.

Step 4: Compare the results.

The two gradients are:

• Voltage against current: \(4.0\ \Omega\).
• Current against voltage: \(0.25\text{ S}\).

They are reciprocals:

\[ 0.25=\frac{1}{4.0} \]

Both experiments correctly describe the same resistor.

Step 5: Identify the student’s conceptual error.

The student has focused on the fact that voltage and current both increase by a factor of three.

However, the factor of three does not determine what the gradient represents.

The axes determine that.

A gradient is always the change in the vertical variable divided by the change in the horizontal variable.

Swapping the axes reverses the ratio:

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

unless the numerical value happens to be 1 in the chosen units.

There is another important point. The two experiments are not demonstrating two different resistance behaviours. They are simply two ways of controlling and measuring the same ohmic resistor.

In the first experiment, the voltage is adjusted and the current responds. In the second, the current is adjusted and the voltage responds.

The resistor’s physical properties remain unchanged.

The deeper lesson is that a graph gradient is not a physical quantity by itself. Its meaning comes from the variables and units assigned to the axes.

09Where to go next

Once you can interpret voltage-current gradients correctly, you can use the same reasoning to investigate power, energy transfer, and non-ohmic components.

For example, if voltage is fixed, increasing resistance decreases current and therefore changes the power transferred to a resistor. If current is fixed, increasing resistance increases both voltage and power.

These relationships follow from:

\[ P=VI \]

and, under the appropriate conditions:

\[ P=I^2R \]

The key habit is to identify what is being held constant before predicting what changes.

For further practice, explore How to Calculate Heat Produced by a Resistor in a Given Time and How to Find Resistance from a Voltage-Current Graph.

When you next see a voltage-current graph in an HSC question, check the axes before touching your calculator. That single habit prevents one of the most common mistakes in electrical graph interpretation.