How to Find Resistance from a Voltage-Current Graph

A component gives \(1.0\,\mathrm{V}\) at \(0.20\,\mathrm{A}\), then \(2.0\,\mathrm{V}\) at \(0.40\,\mathrm{A}\). Both readings suggest a resistance of \(5.0\,\Omega\). But does that resistance stay constant as you increase the current?

A voltage-current graph helps you find out. To calculate resistance from a graph, put voltage on the vertical axis and current on the horizontal axis, then calculate the gradient:

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

If the axes are reversed, the gradient gives the reciprocal of resistance instead. Getting the axes right is the first step.

01Set up the measurement

Before calculating a gradient, you need reliable measurements of voltage and current.

Imagine testing an unknown resistor using a DC power supply. The circuit contains:

• An ammeter connected in series with the resistor.
• A voltmeter connected in parallel across the resistor.
• A variable resistor that allows you to adjust the current.
• A switch and a low-voltage power supply.

The current flows from the positive terminal of the supply, through the ammeter and variable resistor, through the test resistor, and back to the negative terminal. The voltmeter measures the potential difference directly across the test resistor.

Why are the meters connected differently?

• The ammeter measures the current flowing through the component, so it must be in series.
• The voltmeter measures the potential difference between two points, so it must be in parallel across the component.

If these connections are unfamiliar, the fundamentals are covered in Parallel Circuits: Voltage, Current, Resistance and Power.

What should the experiment show?

Suppose the test resistor is an ohmic conductor, and its temperature remains constant.

What would happen if you doubled the current?

You should expect the voltage to double as well. For an ohmic conductor, voltage is directly proportional to current:

\[
V\propto I
\]

This means the voltage-current graph should be a straight line passing through the origin.

That prediction gives us something to test. We take several pairs of voltage and current readings, plot them, and check whether the relationship is actually linear.

One reading can give you a value of \(V/I\). Several readings reveal whether that value stays constant.

02How to calculate resistance from the gradient

For a graph with voltage on the vertical axis and current on the horizontal axis:

\[
\text{Gradient}=\frac{\text{rise}}{\text{run}}
\]

Therefore:

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

Here:

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

The gradient tells you how many volts are needed for each additional ampere of current.

Choosing points for the gradient

There is a small detail here that can make a noticeable difference to your answer.

You should:

1. Draw a straight line of best fit through the plotted data.
2. Choose two points that lie on this line.
3. Make the points as far apart as practical.
4. Calculate the change in voltage divided by the change in current.

You do not have to use the original experimental data points. In fact, using two well-separated points on the best-fit line is often more accurate because it reduces the effect of small measurement errors.

For example, suppose two points on your best-fit line are:

• Point A: \(I=0.10\,\mathrm{A}\), \(V=0.50\,\mathrm{V}\).
• Point B: \(I=0.40\,\mathrm{A}\), \(V=2.00\,\mathrm{V}\).

Then:

\[
\begin{aligned}
R&=\frac{\Delta V}{\Delta I}\\
&=\frac{2.00-0.50}{0.40-0.10}\\
&=\frac{1.50}{0.30}\\
&=5.0\,\Omega
\end{aligned}
\]

The resistor has a resistance of \(5.0\,\Omega\).

Why do the units work?

The gradient has units of volts divided by amperes:

\[
\frac{\mathrm{V}}{\mathrm{A}}=\Omega
\]

This is exactly the unit of electrical resistance.

If your answer has units of amperes per volt, you have probably calculated the gradient of a graph with the axes reversed.

03Check which variable is on each axis

This is one of the easiest mistakes to make in an exam.

Not every voltage-current graph places voltage on the vertical axis. Some graphs put current on the vertical axis and voltage on the horizontal axis.

The title alone does not tell you which variable is on which axis. Always check the labels.

Conductance is the reciprocal of resistance. It measures how easily current flows for a given voltage.

Its unit is the siemens (S).

For example, suppose a graph has current on the vertical axis and voltage on the horizontal axis, with a gradient of \(0.050\,\mathrm{A/V}\).

The resistance is:

\[
\begin{aligned}
R&=\frac{1}{0.050}\\
&=20\,\Omega
\end{aligned}
\]

Do not report \(0.050\,\Omega\). The gradient has units of conductance, not resistance.

04What if the graph is curved?

So far, we have assumed that the component is ohmic.

But not every electrical component behaves this way.

A filament lamp is a useful example. As current increases, the filament heats up. Its resistance changes, so the voltage-current relationship is no longer a straight line.

For a curved graph, there are two different quantities worth understanding.

1. Resistance at an operating point

The resistance at a particular operating point is calculated using:

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

This is sometimes called the static resistance or DC resistance.

You use the voltage and current at the same point on the graph. Geometrically, this is the gradient of a straight line drawn from the origin to that point.

2. Gradient at an operating point

The gradient of the tangent to a curved graph tells you how quickly voltage changes for a small change in current near that point.

It is written as:

\[
r=\frac{\Delta V}{\Delta I}
\]

where \(r\) represents the local or differential resistance.

For an ohmic resistor, the static resistance and tangent gradient are equal because the graph is a straight line through the origin.

For a non-ohmic component, they generally differ.

The important distinction: if a question asks for the resistance at a particular operating point, use \(V/I\). If it asks for the gradient at that point, draw a tangent and calculate its gradient.

Do not automatically use the tangent gradient as the component’s resistance in \(V=IR\).

05Why does a component’s resistance change?

Consider what happens inside a metal wire when current flows.

The wire contains positive metal ions arranged in a lattice, with mobile electrons moving through it. As the wire gets hotter, the ions vibrate more strongly around their fixed positions.

These stronger vibrations cause more frequent scattering of conduction electrons. This makes it harder for the electrons to move through the material, increasing its resistance.

This is a useful model for understanding why a metal wire’s resistance generally increases with temperature.

However, be careful with the explanation. Current does not automatically increase resistance just because more electrons are flowing. The important factor here is the heating caused by electrical energy transfer.

Fixed voltage versus fixed current

The conditions of an experiment matter.

Suppose the resistance of a metal wire increases as it heats up. What happens next depends on whether the voltage or current is being controlled.

At fixed voltage, the supply maintains the same potential difference, so the current falls as resistance increases.

At fixed current, the supply must provide a greater voltage to maintain the same current through the hotter wire.

This also affects electrical power. At fixed voltage, \(P=V^2/R\), whereas at fixed current, \(P=I^2R\). The heating behaviour therefore depends on the electrical conditions as well as heat lost to the surroundings.

Improving the experiment

There is a potential problem with the original measurement method.

If you gradually increase the current through a metal resistor, the resistor may heat up during the experiment. Later readings are then taken at a different temperature from earlier readings.

The graph might curve upwards, even though you initially expected a straight line.

How could you improve the experiment?

• Use relatively low currents to reduce heating.
• Take readings quickly rather than leaving the component connected at high current.
• Allow the component to cool between measurements.
• Repeat the measurements while decreasing the current.
• Monitor or control the temperature if accurate constant-temperature measurements are required.

If the increasing-current and decreasing-current measurements produce different voltages at the same current, the component’s state has probably changed during the experiment. Heating is one possible explanation.

The graph itself does not directly measure temperature. You need additional measurements or experimental controls to establish that heating caused the difference.

For a more detailed explanation of the temperature effect in metal conductors, see How Heating Changes a Metal Wire’s Resistance | HSC Physics.

06Worked examples

07Questions and solutions

The following questions build from straightforward gradient calculations to more challenging situations involving non-ohmic components, experimental limitations, and circuit conditions.

Question 1

An ohmic resistor is maintained at a constant temperature.

Its voltage-current graph has voltage on the vertical axis and current on the horizontal axis. The best-fit line passes through the following two points:

• Point A: \((0.10\,\mathrm{A},0.60\,\mathrm{V})\)
• Point B: \((0.50\,\mathrm{A},3.00\,\mathrm{V})\)

a) Calculate the resistance of the resistor.

b) Calculate the current through the resistor when connected to a \(9.0\,\mathrm{V}\) supply.

Solution 1

The resistance is \(6.0\,\Omega\), and the current at \(9.0\,\mathrm{V}\) is \(1.5\,\mathrm{A}\).

Part a: Calculate resistance.

Step 1: Identify the graph gradient.

Since voltage is on the vertical axis and current is on the horizontal axis:

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

Step 2: Substitute the two points.

\[
\begin{aligned}
R&=\frac{3.00-0.60}{0.50-0.10}\\
&=\frac{2.40}{0.40}\\
&=6.0\,\Omega
\end{aligned}
\]

Part b: Calculate current.

Step 1: Rearrange Ohm’s law.

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

Step 2: Substitute the values.

\[
\begin{aligned}
I&=\frac{9.0}{6.0}\\
&=1.5\,\mathrm{A}
\end{aligned}
\]

The current is \(1.5\,\mathrm{A}\).

This calculation assumes the resistor remains at the same temperature and its resistance stays constant when the voltage increases.

Question 2

A resistor is tested at constant temperature. Its graph has current on the vertical axis and voltage on the horizontal axis.

Two points on the best-fit line are:

• Point A: \((2.0\,\mathrm{V},0.10\,\mathrm{A})\)
• Point B: \((8.0\,\mathrm{V},0.40\,\mathrm{A})\)

A student calculates the gradient and reports that the resistance is \(0.050\,\Omega\).

a) Explain the student’s mistake.

b) Calculate the actual resistance.

c) Calculate the current when the resistor is connected to a fixed \(12.0\,\mathrm{V}\) supply.

d) Calculate the voltage required to maintain a fixed current of \(0.30\,\mathrm{A}\).

Solution 2

The correct resistance is \(20\,\Omega\), not \(0.050\,\Omega\).

Part a: Identify the mistake.

The student has calculated the gradient correctly but interpreted it incorrectly.

Because current is on the vertical axis and voltage is on the horizontal axis, the gradient represents conductance rather than resistance.

Part b: Calculate resistance.

Step 1: Calculate the gradient.

\[
\begin{aligned}
G&=\frac{\Delta I}{\Delta V}\\
&=\frac{0.40-0.10}{8.0-2.0}\\
&=\frac{0.30}{6.0}\\
&=0.050\,\mathrm{A/V}
\end{aligned}
\]

Here, \(G\) represents conductance.

Step 2: Take the reciprocal.

\[
\begin{aligned}
R&=\frac{1}{G}\\
&=\frac{1}{0.050}\\
&=20\,\Omega
\end{aligned}
\]

Part c: Calculate current at fixed voltage.

The supply voltage is held constant at \(12.0\,\mathrm{V}\).

Using Ohm’s law:

\[
\begin{aligned}
I&=\frac{V}{R}\\
&=\frac{12.0}{20}\\
&=0.60\,\mathrm{A}
\end{aligned}
\]

The current is \(0.60\,\mathrm{A}\).

Part d: Calculate voltage at fixed current.

This time, the current is held constant at \(0.30\,\mathrm{A}\).

Use:

\[
V=IR
\]

Substituting:

\[
\begin{aligned}
V&=0.30\times20\\
&=6.0\,\mathrm{V}
\end{aligned}
\]

The required voltage is \(6.0\,\mathrm{V}\).

Notice that changing which quantity is held constant changes what you calculate. The resistance remains \(20\,\Omega\) because the resistor is assumed to stay at constant temperature.

Question 3

A filament lamp has a curved voltage-current characteristic.

At one operating point, the voltage across the lamp is \(4.0\,\mathrm{V}\), and the current is \(0.50\,\mathrm{A}\).

The tangent to the graph at this point passes through:

• \((0.40\,\mathrm{A},3.0\,\mathrm{V})\)
• \((0.60\,\mathrm{A},5.4\,\mathrm{V})\)

a) Calculate the resistance of the lamp at the operating point.

b) Calculate the gradient of the tangent.

c) A student argues that the two answers must be equal because both are measured in ohms. Explain why this is incorrect.

d) If the lamp is operating at the stated point, what current flows when its voltage is \(4.0\,\mathrm{V}\)?

Solution 3

The static resistance is \(8.0\,\Omega\), while the tangent gradient is \(12\,\Omega\). The current at the stated operating point is \(0.50\,\mathrm{A}\).

Part a: Calculate operating-point resistance.

Step 1: Use the voltage and current at the operating point.

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

Step 2: Substitute.

\[
\begin{aligned}
R&=\frac{4.0}{0.50}\\
&=8.0\,\Omega
\end{aligned}
\]

Part b: Calculate tangent gradient.

Step 1: Use the two points on the tangent.

\[
r=\frac{\Delta V}{\Delta I}
\]

Step 2: Substitute.

\[
\begin{aligned}
r&=\frac{5.4-3.0}{0.60-0.40}\\
&=\frac{2.4}{0.20}\\
&=12\,\Omega
\end{aligned}
\]

Part c: Explain why the values differ.

The student is confusing two different measurements.

The static resistance is calculated using the ratio of voltage to current at the operating point. The tangent gradient measures the rate of change of voltage for a small change in current.

For an ohmic resistor, these values are equal because its graph is a straight line through the origin.

For a filament lamp, the resistance changes as the filament heats up. The graph curves, so the tangent gradient and the ratio \(V/I\) are generally different.

Having the same units does not mean two quantities represent the same physical relationship.

Part d: Determine current.

The stated operating point already gives a voltage of \(4.0\,\mathrm{V}\) and a current of \(0.50\,\mathrm{A}\).

Therefore:

\[
\boxed{I=0.50\,\mathrm{A}}
\]

Using the tangent gradient as though it were a constant resistance would give an incorrect prediction of the operating current.

Question 4

A student investigates the resistance of a metal wire by gradually increasing the current, then immediately decreasing it.

The following measurements are obtained:

a) Calculate the apparent resistance for each measurement.

b) Which measurement is likely to have been taken when the wire was hotter? Explain your reasoning.

c) Explain what happens to the current if the voltage is held constant while the wire heats up. What happens to the voltage if the current is held constant instead?

d) Suggest an improvement to the experimental method that would help determine the wire’s resistance at a constant temperature.

Solution 4

The increasing-current sweep gives \(4.75\,\Omega\), while the decreasing-current sweep gives \(5.50\,\Omega\). The decreasing-current measurement was probably taken when the wire was hotter.

Part a: Calculate both resistances.

Use:

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

For the increasing-current measurement:

\[
\begin{aligned}
R_1&=\frac{1.90}{0.40}\\
&=4.75\,\Omega
\end{aligned}
\]

For the decreasing-current measurement:

\[
\begin{aligned}
R_2&=\frac{2.20}{0.40}\\
&=5.50\,\Omega
\end{aligned}
\]

Part b: Identify the hotter measurement.

The decreasing-current measurement has the greater resistance.

Both measurements have the same current, but the second requires a greater voltage.

For a metal wire, an increase in temperature generally increases resistance because stronger lattice vibrations cause more frequent scattering of conduction electrons.

The wire has already experienced higher currents during the increasing-current sweep. It may therefore still be hot when the current is reduced.

This is called thermal lag: the temperature does not immediately return to its original value when the electrical conditions change.

The measurements suggest heating, although a temperature measurement would provide stronger evidence.

Part c: Compare fixed voltage and fixed current.

At fixed voltage:

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

As resistance increases, current decreases.

At fixed current:

\[
V=IR
\]

As resistance increases, the required voltage increases.

The important point is that the electrical response depends on which quantity the supply controls.

Part d: Improve the experiment.

The student should reduce heating by using lower currents, taking measurements quickly, and allowing the wire to cool between readings.

A temperature sensor could also be used to monitor the wire throughout the experiment.

The student should repeat the measurements while increasing and decreasing the current. If the results become consistent when temperature is controlled, there is stronger evidence that the original differences were caused by heating.

Question 5

A filament lamp has the following illustrative steady-state voltage-current characteristic.

The characteristic is smooth between the measured points. The gradient of the tangent near \(0.50\,\mathrm{A}\) is approximately \(15\,\Omega\).

The lamp is connected in series with a \(10.0\,\Omega\) resistor to an ideal \(10.0\,\mathrm{V}\) supply.

Assume:

• The tabulated values represent the lamp’s steady-state characteristic, including temperature effects.
• The fixed resistor maintains a resistance of \(10.0\,\Omega\).
• The supply voltage remains constant.
• Connecting wires have negligible resistance.
• Linear interpolation between adjacent data points is sufficiently accurate.

a) Determine the steady current in the circuit and the voltage across the lamp.

b) A student treats the tangent gradient of \(15\,\Omega\) as the lamp’s constant resistance and predicts a circuit current of:

\[
I=\frac{10.0}{10.0+15.0}=0.40\,\mathrm{A}
\]

Explain precisely why this prediction is incorrect. Calculate the lamp’s actual resistance at the correct operating point.

c) The supply voltage is increased to \(12.0\,\mathrm{V}\). Using linear interpolation between the relevant data points, estimate the new circuit current and the lamp’s resistance at its new operating point.

Solution 5

The circuit operates at \(0.50\,\mathrm{A}\) with a \(10.0\,\mathrm{V}\) supply. When the supply increases to \(12.0\,\mathrm{V}\), the current rises to approximately \(0.58\,\mathrm{A}\), and the lamp’s static resistance increases to approximately \(10.8\,\Omega\).

Part a: Find the original operating point.

Step 1: Consider the voltage across the fixed resistor.

The resistor is connected in series with the lamp, so the same current flows through both components.

The voltage across the fixed resistor is:

\[
V_R=IR
\]

Step 2: Test the available lamp data.

At a current of \(0.50\,\mathrm{A}\), the lamp requires \(5.00\,\mathrm{V}\).

The fixed resistor requires:

\[
\begin{aligned}
V_R&=0.50\times10.0\\
&=5.00\,\mathrm{V}
\end{aligned}
\]

The total voltage required is:

\[
\begin{aligned}
V_{\text{total}}&=V_R+V_{\text{lamp}}\\
&=5.00+5.00\\
&=10.0\,\mathrm{V}
\end{aligned}
\]

This exactly matches the supply voltage.

Therefore, the steady current is:

\[
\boxed{I=0.50\,\mathrm{A}}
\]

The voltage across the lamp is:

\[
\boxed{V_{\text{lamp}}=5.00\,\mathrm{V}}
\]

Part b: Explain the incorrect prediction.

The student has used the tangent gradient as though it were the lamp’s constant resistance.

This is the central mistake.

The tangent gradient of \(15\,\Omega\) describes the local change in voltage for a small change in current. It does not represent the ratio of voltage to current at the operating point.

The actual resistance is:

\[
R_{\text{lamp}}=\frac{V_{\text{lamp}}}{I}
\]

Substituting:

\[
\begin{aligned}
R_{\text{lamp}}&=\frac{5.00}{0.50}\\
&=10.0\,\Omega
\end{aligned}
\]

The lamp’s actual static resistance at this operating point is \(10.0\,\Omega\).

We can also test the student’s predicted current of \(0.40\,\mathrm{A}\).

At this current, the table gives a lamp voltage of \(3.60\,\mathrm{V}\).

The fixed resistor requires:

\[
V_R=0.40\times10.0=4.00\,\mathrm{V}
\]

The total voltage required would be:

\[
V_{\text{total}}=3.60+4.00=7.60\,\mathrm{V}
\]

But the supply provides \(10.0\,\mathrm{V}\).

Therefore, \(0.40\,\mathrm{A}\) cannot be the correct operating current. The proposed current does not satisfy the voltage constraint of the circuit.

The correct approach is to find the operating point that satisfies both the lamp’s voltage-current characteristic and the series-circuit relationship:

\[
V_{\text{supply}}=IR_{\text{fixed}}+V_{\text{lamp}}
\]

Part c: Find the new operating point.

The supply voltage is now \(12.0\,\mathrm{V}\).

The lamp’s resistance cannot simply be assumed to remain at \(10.0\,\Omega\), because its operating current and temperature have changed.

Step 1: Calculate the total voltage required at two nearby currents.

At \(0.50\,\mathrm{A}\):

\[
\begin{aligned}
V_{\text{total}}&=5.00+(0.50\times10.0)\\
&=10.0\,\mathrm{V}
\end{aligned}
\]

At \(0.60\,\mathrm{A}\):

\[
\begin{aligned}
V_{\text{total}}&=6.60+(0.60\times10.0)\\
&=12.6\,\mathrm{V}
\end{aligned}
\]

The required supply voltage of \(12.0\,\mathrm{V}\) lies between these two values.

Therefore, the operating current lies between \(0.50\,\mathrm{A}\) and \(0.60\,\mathrm{A}\).

Step 2: Use linear interpolation.

The total voltage increases from \(10.0\,\mathrm{V}\) to \(12.6\,\mathrm{V}\) over a current increase of \(0.10\,\mathrm{A}\).

The fraction of this interval needed to reach \(12.0\,\mathrm{V}\) is:

\[
\frac{12.0-10.0}{12.6-10.0}
=\frac{2.0}{2.6}
\]

Therefore:

\[
\begin{aligned}
I&=0.50+\frac{2.0}{2.6}(0.10)\\
&=0.5769\,\mathrm{A}
\end{aligned}
\]

The new current is approximately:

\[
\boxed{I=0.58\,\mathrm{A}}
\]

Step 3: Calculate the new voltage across the lamp.

The supply voltage is shared between the fixed resistor and the lamp.

\[
V_{\text{lamp}}=V_{\text{supply}}-IR_{\text{fixed}}
\]

Substituting:

\[
\begin{aligned}
V_{\text{lamp}}&=12.0-(0.5769\times10.0)\\
&=6.23\,\mathrm{V}
\end{aligned}
\]

Step 4: Calculate the new static resistance.

\[
\begin{aligned}
R_{\text{lamp}}&=\frac{V_{\text{lamp}}}{I}\\
&=\frac{6.23}{0.5769}\\
&\approx10.8\,\Omega
\end{aligned}
\]

The new operating-point resistance is:

\[
\boxed{R_{\text{lamp}}\approx10.8\,\Omega}
\]

This question brings together several ideas.

The lamp has a changing resistance, the fixed resistor has a constant resistance, and the supply voltage must equal the sum of the voltage drops across both components.

A tangent gradient cannot replace the lamp’s operating-point resistance, and even its static resistance should not automatically be treated as constant when the operating conditions change.

08What to remember in an exam

When you are asked to find resistance from a voltage-current graph, work through these decisions:

• Check which variable is on each axis.
• If voltage is vertical and current is horizontal, resistance is the gradient.
• If current is vertical and voltage is horizontal, resistance is the reciprocal of the gradient.
• Use two well-separated points on the best-fit line.
• Include units of ohms.
• For a curved graph, distinguish \(V/I\) from the gradient of a tangent.
• Check whether temperature is constant or whether heating could change the resistance.

Once you can interpret a voltage-current graph, the next useful step is to use its operating points to calculate electrical power, \(P=VI\), and investigate how changing resistance affects the energy transferred as heat.