Why a Filament Lamp Has a Curved Current-Voltage Graph

In an illustrative test, a small filament lamp draws \(0.32\text{ A}\) when the voltage across it is \(2.0\text{ V}\). Increase the voltage to \(4.0\text{ V}\), and the current rises to just \(0.43\text{ A}\).

Why doesn’t the current double?

If the lamp behaved like an ordinary resistor with constant resistance, doubling the voltage would double the current to \(0.64\text{ A}\). Instead, the relationship is curved because the tungsten filament heats up as electrical power increases. As its temperature rises, its resistance increases, making it harder for additional current to flow.

This is the reason a filament lamp is called a non-ohmic conductor.

01Why the current-voltage graph curves

Start by thinking about what happens when you increase the voltage across a lamp.

At a low voltage, the filament is relatively cool. Its resistance is comparatively low, so current can flow relatively easily.

As you increase the voltage, more electrical energy is transferred to the filament. Its temperature rises, and its resistance increases.

The important consequence is that current still increases as voltage increases, but it increases less rapidly than it would in a component with constant resistance.

Consider the following illustrative steady-state measurements for a small tungsten filament lamp.

These are illustrative values, not measurements from a particular commercial lamp. Each reading represents an operating condition after the filament has had time to stabilise.

Resistance is calculated using:

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

For example, at \(2.0\text{ V}\):

\[ R=\frac{2.0}{0.32}=6.25\,\Omega \]

At \(4.0\text{ V}\):

\[ R=\frac{4.0}{0.43}=9.30\,\Omega \]

The resistance has increased substantially.

Notice what happens when the voltage doubles from \(2.0\text{ V}\) to \(4.0\text{ V}\):

• Voltage doubles.
• Current increases from \(0.32\text{ A}\) to \(0.43\text{ A}\).
• Resistance increases from approximately \(6.3\,\Omega\) to \(9.3\,\Omega\).

The current does not double because the resistance is no longer constant.

What does the graph actually look like?

For the standard current-voltage graph, put voltage \(V\) on the horizontal axis and current \(I\) on the vertical axis.

The graph rises from the origin but gradually becomes flatter.

The flattening has a specific meaning: each additional volt produces a smaller increase in current than the previous volt did.

Importantly, the current never needs to decrease. It continues increasing, just at a progressively slower rate.

If you reverse the axes and plot voltage against current, the same relationship appears as a curve that becomes progressively steeper.

This is why you should always check which variable is on each axis before interpreting a graph.

02What is happening inside the filament?

The filament is usually made from tungsten, a metal that can operate at extremely high temperatures.

To understand why its resistance increases, we need to look inside the metal.

Tungsten contains positively charged metal ions arranged in a lattice. Mobile electrons move through this structure and carry electrical current.

At relatively low temperatures, the metal ions vibrate around their usual positions. As the filament heats up, these vibrations become more intense.

The increased lattice vibrations cause conduction electrons to experience more frequent scattering.

This makes it harder for the electrons to maintain their average drift motion through the metal.

The result is an increase in electrical resistance.

A useful way to picture this is to imagine walking through a corridor.

If the people around you are standing relatively still, maintaining a forward path is straightforward. If everyone starts moving unpredictably, making forward progress becomes harder.

The analogy captures the idea of increased scattering, but it has limits. Electrons are quantum particles, not tiny people walking through a metal. The physical explanation involves interactions between conduction electrons and the vibrating metal lattice.

For HSC Physics, the important relationship is:

\[ \begin{aligned} \text{Higher temperature} &\Rightarrow \text{greater lattice vibrations}\\ &\Rightarrow \text{more electron scattering}\\ &\Rightarrow \text{greater resistance} \end{aligned} \]

This is closely related to the behaviour of an ordinary metal wire. The difference is that a filament lamp is deliberately operated at a temperature where its self-heating has a major effect on its electrical behaviour.

For more detail on the microscopic explanation, see How Heating Changes a Metal Wire’s Resistance | HSC Physics.

03How do we experimentally obtain the curved graph?

Suppose you wanted to investigate whether a filament lamp obeys Ohm’s law.

You need to measure two quantities:

• The potential difference across the lamp.
• The current flowing through the lamp.

You can then investigate whether current is directly proportional to voltage.

Setting up the experiment

A suitable experimental circuit contains:

• A regulated, variable DC power supply.
• A filament lamp.
• An ammeter connected in series with the lamp.
• A voltmeter connected in parallel across the lamp.
• A variable resistor in series, if needed, to control the current.

The positive terminal of the supply connects through the switch, ammeter, variable resistor, and lamp before returning to the negative terminal.

The voltmeter connects directly across the two lamp terminals.

Conventional current is taken as flowing from the positive supply terminal towards the negative terminal. The voltage is taken as positive at the lamp terminal where conventional current enters.

What should the original experiment show?

Imagine that the filament maintains a constant temperature throughout the experiment.

If the lamp behaves as an ohmic conductor under that condition, Ohm’s law predicts:

\[ V=IR \]

where:

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

If resistance remains constant, doubling the voltage should double the current.

The current-voltage graph should therefore be a straight line passing through the origin.

However, the actual filament does not maintain a constant temperature when the voltage increases.

As the electrical power increases, the filament heats up. Its resistance changes, and the measurements no longer follow the straight-line prediction.

This is the experimental evidence that the lamp is non-ohmic under normal self-heating conditions.

Why does the measurement method need care?

There is an important experimental complication.

The filament does not heat up instantly when the voltage changes. It has thermal inertia, meaning its temperature takes time to respond to changes in electrical power.

Imagine increasing the voltage from \(2.0\text{ V}\) to \(4.0\text{ V}\).

Immediately after the change, the filament may still be relatively cool. Its resistance is therefore lower than its eventual resistance at \(4.0\text{ V}\).

If you record the current immediately, you may obtain a higher value than the steady-state current.

A better experimental method is to:

1. Begin with a low voltage.
2. Increase the voltage in small steps.
3. Wait for the current and filament temperature to stabilise at each setting.
4. Record the voltage across the lamp and its current.
5. Repeat the measurements while decreasing the voltage.

If the increasing-voltage and decreasing-voltage measurements differ, thermal lag may be affecting the results.

Waiting for the filament to stabilise gives a more reproducible steady-state characteristic.

There is an important distinction here: when measuring the normal operating characteristic of a lamp, self-heating is part of the behaviour you want to observe. If you instead want to investigate the electrical behaviour at a fixed temperature, you need to control the filament temperature independently and minimise additional heating from the measuring current.

Why does the filament eventually stop heating?

The filament does not continue increasing in temperature indefinitely under normal operation.

Electrical power is transferred to the filament, increasing its internal energy. It also loses energy through thermal radiation and heat transfer to its surroundings.

Electrical power is calculated using:

\[ P=VI \]

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

As the filament heats, the rate at which it loses energy increases.

Eventually, it reaches thermal equilibrium:

\[ \text{Rate of electrical energy input} = \text{Rate of energy loss} \]

Its temperature then becomes approximately constant.

This is why the readings in a steady-state experiment can be reproducible. Each voltage setting produces a different equilibrium temperature and therefore a different resistance.

The relationship between voltage and current is consequently affected by both electrical behaviour and thermal energy transfer.

04Fixed voltage versus fixed current

This distinction is particularly important when answering HSC questions about heating and resistance.

Suppose you increase the temperature of a filament, causing its resistance to increase.

What happens to the current?

The answer depends on what the external circuit holds constant.

Case 1: Voltage is held constant

Suppose an ideal voltage supply maintains a constant potential difference across the lamp.

Ohm’s law gives:

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

If resistance increases while voltage remains constant, current decreases.

Electrical power is:

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

Therefore, power also decreases as resistance increases.

This is the situation when a lamp is connected directly across an ideal constant-voltage supply.

Case 2: Current is held constant

Now imagine a current-regulated supply that maintains the same current through the lamp.

Using:

\[ V=IR \]

an increase in resistance means the supply must increase the voltage to maintain the current.

Power is:

\[ P=I^2R \]

Therefore, electrical power increases as resistance increases.

Compare the two conditions

Notice that the resistance increases in both situations.

The difference is how the external circuit responds.

There is another subtle point worth remembering.

At fixed voltage, the electrical power supplied to the lamp can decrease while the filament temperature is still increasing.

This is possible because the filament may initially be receiving energy faster than it loses it. Its temperature continues rising until the input and output energy rates balance.

Similarly, increasing electrical power at fixed current does not automatically mean thermal runaway. The actual temperature depends on the balance between energy supplied and energy lost.

A useful contrast: an ordinary resistor

Consider an ordinary resistor operated at approximately constant temperature.

If you double the voltage across it, its current doubles. Its voltage-current graph is a straight line through the origin.

A filament lamp behaves differently because its operating temperature changes significantly as the electrical input changes.

The distinction is not that the lamp somehow ignores Ohm’s law. It is that its resistance does not remain constant.

At a particular operating point, you can still calculate its resistance using \(R=V/I\). What you cannot do is assume that the same resistance applies at every other operating point.

05Worked examples

06Questions and solutions

Question 1

An incandescent lamp operates at a voltage of \(5.0\text{ V}\) and carries a steady current of \(0.47\text{ A}\).

a) Calculate its resistance at this operating point.

b) Calculate its electrical power.

c) Explain why these measurements alone cannot establish that the lamp is an ohmic conductor.

Solution 1

The lamp has an operating-point resistance of approximately \(10.6\,\Omega\) and an electrical power of \(2.35\text{ W}\). These measurements do not establish whether it is ohmic because only one operating point has been measured.

Part a: Resistance

Use:

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

Substitute:

\[ \begin{aligned} R&=\frac{5.0}{0.47}\\ &=10.64\,\Omega \end{aligned} \]

Therefore:

\[ R\approx10.6\,\Omega \]

Part b: Electrical power

Use:

\[ P=VI \]

\[ \begin{aligned} P&=5.0\times0.47\\ &=2.35\text{ W} \end{aligned} \]

Part c: Ohmic behaviour

One measurement gives the resistance at one particular operating point.

To establish whether a component is ohmic, we need to investigate whether voltage is directly proportional to current while temperature remains constant.

A single value of \(V/I\) cannot establish that relationship.

For a filament lamp, resistance changes as its temperature changes, so its normal operating characteristic is non-ohmic.

Question 2

A filament lamp and a fixed \(12\,\Omega\) resistor are connected in parallel across an ideal \(6.0\text{ V}\) supply.

The lamp has a cold resistance of \(2.0\,\Omega\). After reaching thermal equilibrium, its current at \(6.0\text{ V}\) is \(0.50\text{ A}\).

Calculate:

a) The initial current through each branch immediately after switching on.

b) The total current supplied by the source immediately after switching on.

c) The total current supplied after the lamp reaches thermal equilibrium.

d) Explain why the current through the fixed resistor does or does not change.

Assume the ideal supply maintains exactly \(6.0\text{ V}\) across both branches throughout.

Solution 2

The initial total current is \(3.5\text{ A}\), falling to \(1.0\text{ A}\) after the lamp reaches thermal equilibrium. The current through the fixed resistor remains unchanged at \(0.50\text{ A}\).

Part a: Initial branch currents

Both branches have the same potential difference because they are connected in parallel.

Therefore:

\[ V_{\text{lamp}}=V_{\text{resistor}}=6.0\text{ V} \]

For the cold lamp:

\[ \begin{aligned} I_{\text{lamp}}&=\frac{V}{R}\\ &=\frac{6.0}{2.0}\\ &=3.0\text{ A} \end{aligned} \]

For the fixed resistor:

\[ \begin{aligned} I_{\text{resistor}}&=\frac{6.0}{12}\\ &=0.50\text{ A} \end{aligned} \]

Part b: Initial total current

Apply conservation of charge at the junction:

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

\[ \begin{aligned} I_{\text{total}}&=3.0+0.50\\ &=3.5\text{ A} \end{aligned} \]

Part c: Total current at thermal equilibrium

The lamp’s steady-state current is given as \(0.50\text{ A}\).

The resistor still carries \(0.50\text{ A}\).

Therefore:

\[ \begin{aligned} I_{\text{total}}&=0.50+0.50\\ &=1.0\text{ A} \end{aligned} \]

Part d: Why does the resistor current remain constant?

The fixed resistor remains connected across the same ideal \(6.0\text{ V}\) supply.

Its resistance does not change, and its voltage remains constant.

Therefore, its current remains:

\[ I=\frac{6.0}{12}=0.50\text{ A} \]

The lamp’s changing resistance affects its own branch current, but it does not take voltage away from the other branch.

This is an important feature of parallel circuits. The branches share voltage, not a fixed amount of current.

For a more detailed explanation of this principle, see Parallel Circuits: Voltage, Current, Resistance and Power.

Question 3

A student investigates the current-voltage characteristic of a filament lamp.

In the first trial, the lamp is initially at thermal equilibrium at \(2.0\text{ V}\). The student rapidly increases the voltage to \(4.0\text{ V}\) and immediately measures a current of \(0.49\text{ A}\).

In a second trial, the lamp is initially at thermal equilibrium at \(6.0\text{ V}\). The student rapidly decreases the voltage to \(4.0\text{ V}\) and immediately measures a current of \(0.40\text{ A}\).

After holding the voltage at \(4.0\text{ V}\) for long enough, both trials produce a steady current of \(0.43\text{ A}\).

a) Calculate the resistance of the lamp in each of the two immediate measurements.

b) Which trial has the hotter filament at the moment of measurement?

c) Explain why the same voltage produces different currents in the two trials.

d) Explain why these measurements do not represent two different steady-state current-voltage characteristics.

Solution 3

The resistances are approximately \(8.16\,\Omega\) and \(10.0\,\Omega\), respectively. The filament in the second trial is hotter because it has the greater resistance and lower current at the same voltage.

Part a: Calculate the two resistances

For the first trial:

\[ \begin{aligned} R_{\text{up}}&=\frac{V}{I}\\ &=\frac{4.0}{0.49}\\ &=8.16\,\Omega \end{aligned} \]

For the second trial:

\[ \begin{aligned} R_{\text{down}}&=\frac{V}{I}\\ &=\frac{4.0}{0.40}\\ &=10.0\,\Omega \end{aligned} \]

Part b: Identify the hotter filament

The second trial has the greater resistance.

For a tungsten filament, resistance increases with temperature over its normal operating range.

Therefore, the filament in the second trial is hotter.

Part c: Explain the different currents

In the first trial, the filament was initially operating at \(2.0\text{ V}\).

When the voltage increases to \(4.0\text{ V}\), the filament does not immediately reach its new equilibrium temperature. It remains relatively cool for a short time.

Its resistance is therefore lower, allowing a larger current of \(0.49\text{ A}\).

In the second trial, the filament was initially operating at \(6.0\text{ V}\).

When the voltage decreases to \(4.0\text{ V}\), the filament remains hotter than its eventual equilibrium temperature.

Its resistance is higher, so the current is smaller at \(0.40\text{ A}\).

Part d: Explain the steady-state behaviour

Both trials eventually reach the same conditions:

• Voltage: \(4.0\text{ V}\).
• Current: \(0.43\text{ A}\).
• Resistance: approximately \(9.30\,\Omega\).

The initial difference is caused by thermal lag, not by two different permanent electrical characteristics.

The filament takes time to gain or lose thermal energy. Its temperature therefore depends temporarily on its previous operating conditions.

This is why an experiment must allow sufficient time for thermal equilibrium before recording steady-state measurements.

Question 4

The resistance of a filament lamp increases from \(8.0\,\Omega\) to \(10.0\,\Omega\) because the lamp is heated externally.

Two separate experiments are performed.

Experiment A: An ideal voltage supply maintains \(4.0\text{ V}\) across the lamp.

Experiment B: An ideal current-regulated supply maintains \(0.40\text{ A}\) through the lamp.

For each experiment:

a) Calculate the current and electrical power before and after the resistance increases.

b) Calculate the voltage and electrical power before and after the resistance increases.

c) Explain why electrical power changes in opposite directions in the two experiments.

d) Can the change in electrical power alone establish whether the filament temperature will continue increasing? Explain.

Solution 4

In Experiment A, electrical power decreases from \(2.0\text{ W}\) to \(1.6\text{ W}\). In Experiment B, it increases from \(1.28\text{ W}\) to \(1.60\text{ W}\).

The difference occurs because the controlled electrical quantity is different.

Experiment A: Fixed voltage

The supply maintains:

\[ V=4.0\text{ V} \]

Before heating, the resistance is \(8.0\,\Omega\).

Calculate the current:

\[ \begin{aligned} I_1&=\frac{4.0}{8.0}\\ &=0.50\text{ A} \end{aligned} \]

Calculate the power:

\[ P_1=VI \]

\[ P_1=4.0\times0.50=2.0\text{ W} \]

After heating, resistance increases to \(10.0\,\Omega\).

The new current is:

\[ \begin{aligned} I_2&=\frac{4.0}{10.0}\\ &=0.40\text{ A} \end{aligned} \]

The new power is:

\[ P_2=4.0\times0.40=1.6\text{ W} \]

Therefore, electrical power decreases.

Experiment B: Fixed current

The supply maintains:

\[ I=0.40\text{ A} \]

Before heating, resistance is \(8.0\,\Omega\).

Calculate the voltage:

\[ \begin{aligned} V_1&=IR\\ &=0.40\times8.0\\ &=3.2\text{ V} \end{aligned} \]

Calculate the power:

\[ \begin{aligned} P_1&=I^2R\\ &=0.40^2\times8.0\\ &=1.28\text{ W} \end{aligned} \]

After heating, resistance increases to \(10.0\,\Omega\).

The new voltage is:

\[ \begin{aligned} V_2&=0.40\times10.0\\ &=4.0\text{ V} \end{aligned} \]

The new power is:

\[ \begin{aligned} P_2&=0.40^2\times10.0\\ &=1.60\text{ W} \end{aligned} \]

Therefore, electrical power increases.

Part c: Explain the opposite power changes

The two experiments follow different power relationships.

At fixed voltage:

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

Increasing resistance decreases power.

At fixed current:

\[ P=I^2R \]

Increasing resistance increases power.

Part d: Consider the total energy balance

No. Electrical power is only the rate at which electrical energy is supplied.

The filament also receives energy from the external heater and loses energy through radiation and other heat-transfer processes.

To determine whether its temperature continues increasing, we need to consider the total energy supplied and the total energy lost.

A change in electrical power alone is not enough.

Question 5

An incandescent filament lamp is connected in series with a fixed \(12\,\Omega\) resistor across an ideal \(12.0\text{ V}\) supply.

Immediately after switching on, the cold lamp has a resistance of \(2.0\,\Omega\).

After reaching thermal equilibrium, the lamp operates at a measured steady-state point of \(6.0\text{ V}\) and \(0.50\text{ A}\).

Assume the fixed resistor remains at \(12\,\Omega\), the supply voltage remains constant, and the lamp’s resistance increases as its temperature rises.

a) Calculate the current immediately after switching on.

b) Calculate the initial voltage across the lamp and its initial electrical power.

c) Calculate the steady-state current, voltage across the fixed resistor, and electrical power of the lamp.

d) Explain how the lamp’s current can decrease while its electrical power increases.

e) The lamp continues heating, so imagine its resistance could increase beyond \(12\,\Omega\). Would its electrical power continue increasing? Justify your answer using the circuit relationships.

Solution 5

The initial current is approximately \(0.86\text{ A}\), and the initial lamp power is approximately \(1.47\text{ W}\). At thermal equilibrium, the current is \(0.50\text{ A}\), and the lamp dissipates \(3.0\text{ W}\).

The important result is that the lamp’s current decreases while its electrical power increases. This happens because the voltage across the lamp increases as its resistance rises.

Part a: Calculate the initial current

Immediately after switching on, the lamp is cold.

The total series resistance is:

\[ R_{\text{total}}=R_{\text{resistor}}+R_{\text{lamp}} \]

\[ R_{\text{total}}=12+2=14\,\Omega \]

The same current flows through both components.

Using Ohm’s law:

\[ \begin{aligned} I_{\text{initial}}&=\frac{V}{R_{\text{total}}}\\ &=\frac{12.0}{14}\\ &=0.857\text{ A} \end{aligned} \]

Therefore, the initial current is approximately \(0.86\text{ A}\).

Part b: Calculate the initial lamp voltage and power

The initial lamp resistance is \(2.0\,\Omega\).

The voltage across it is:

\[ \begin{aligned} V_{\text{lamp}}&=IR\\ &=0.857\times2.0\\ &=1.71\text{ V} \end{aligned} \]

Calculate its electrical power:

\[ P_{\text{lamp}}=I^2R \]

\[ \begin{aligned} P_{\text{initial}}&=0.857^2\times2.0\\ &=1.47\text{ W} \end{aligned} \]

The lamp initially receives approximately \(1.47\text{ W}\) of electrical power.

Part c: Calculate the steady-state values

At thermal equilibrium, the question gives the lamp’s operating point:

\[ V_{\text{lamp}}=6.0\text{ V} \]

\[ I=0.50\text{ A} \]

The current through the series resistor is also \(0.50\text{ A}\).

The voltage across the resistor is:

\[ \begin{aligned} V_{\text{resistor}}&=IR\\ &=0.50\times12\\ &=6.0\text{ V} \end{aligned} \]

Check conservation of energy around the circuit:

\[ V_{\text{supply}}=V_{\text{lamp}}+V_{\text{resistor}} \]

\[ 12.0=6.0+6.0 \]

The voltage drops are consistent with the supply voltage.

The lamp’s steady-state power is:

\[ \begin{aligned} P_{\text{final}}&=VI\\ &=6.0\times0.50\\ &=3.0\text{ W} \end{aligned} \]

Part d: Explain why current decreases while lamp power increases

This is the central difficulty in the question.

As the lamp heats, its resistance increases.

Because the lamp and fixed resistor are in series, the total circuit resistance increases.

The supply voltage remains constant, so the total current decreases.

However, the voltage across the lamp increases.

Initially, the lamp receives only \(1.71\text{ V}\). At thermal equilibrium, it receives \(6.0\text{ V}\).

The lamp’s electrical power depends on both its voltage and current:

\[ P_{\text{lamp}}=V_{\text{lamp}}I \]

Although the current decreases, the voltage across the lamp increases sufficiently to produce a larger power.

Comparing the two operating conditions:

The lamp’s electrical power has approximately doubled even though its current has decreased.

The tempting mistake is to assume that increasing resistance must always decrease electrical power.

That is only necessarily true when voltage across the component is held constant. Here, the lamp voltage is changing because it shares the supply with a series resistor.

Part e: Determine whether lamp power continues increasing

No. The lamp’s electrical power reaches a maximum when its resistance equals the resistance of the fixed series resistor.

To see why, let:

• \(V_s\) be the constant supply voltage.
• \(R_s\) be the fixed series resistance.
• \(R_L\) be the lamp resistance.

The circuit current is:

\[ I=\frac{V_s}{R_s+R_L} \]

The lamp’s power is:

\[ P_L=I^2R_L \]

Substituting the current equation:

\[ P_L=\frac{V_s^2R_L}{(R_s+R_L)^2} \]

For fixed \(V_s\) and \(R_s\), this expression reaches its maximum when:

\[ R_L=R_s \]

In this question:

\[ R_L=R_s=12\,\Omega \]

Therefore, the lamp reaches its maximum electrical power at the stated final operating point.

As the lamp resistance increases from \(2.0\,\Omega\) towards \(12\,\Omega\), its electrical power increases.

If its resistance increases beyond \(12\,\Omega\), its electrical power decreases.

The reason is that the current continues falling, and the increase in the lamp’s share of the supply voltage is no longer enough to compensate.

This is the important conceptual distinction: the effect of increasing resistance on electrical power depends on the rest of the circuit, not just the resistance itself.

07Where this leads next

A filament lamp’s current-voltage graph is curved because its electrical resistance changes with temperature.

The key relationship to remember is:

\[ \boxed{ \text{Higher filament temperature} \Rightarrow \text{Higher resistance} } \]

But when analysing a circuit, you must also identify what is being held constant.

• At fixed voltage, increasing resistance decreases current and electrical power.
• At fixed current, increasing resistance increases voltage and electrical power.
• In a series circuit, the same current flows through each component, but voltage is divided.
• In a parallel circuit, each branch has the same voltage, but branch currents can differ.

Once you understand why the lamp’s resistance changes, the next step is analysing circuits containing both ohmic resistors and non-ohmic components. You can no longer replace every component with one constant resistance and expect the circuit calculations to remain accurate.