How Heating Changes a Metal Wire’s Resistance | HSC Physics

At \(20^\circ\text{C}\), a metal wire has a voltage of \(2.00\text{ V}\) across it and carries a current of \(0.500\text{ A}\). You heat the same wire to \(80^\circ\text{C}\), keeping the voltage across it at \(2.00\text{ V}\). The current falls to \(0.403\text{ A}\).

Why?

The wire hasn’t become shorter, thinner, or changed material. Yet something has changed inside it.

Calculate its resistance before and after heating:

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

At \(20^\circ\text{C}\):

\[
R_{20}=\frac{2.00}{0.500}
=4.00\,\Omega
\]

At \(80^\circ\text{C}\):

\[
R_{80}=\frac{2.00}{0.403}
\approx4.96\,\Omega
\]

Its resistance has increased by approximately 24%.

Heating an ordinary metal wire increases its electrical resistance. The reason is that the metal’s atoms vibrate more strongly as temperature increases, causing more frequent scattering of conduction electrons.

But there’s an important detail.

If you keep the voltage across the wire constant, its current decreases. If you keep the current constant instead, the voltage across the wire must increase.

That difference is central to understanding temperature and resistance in HSC Physics.

01Why does heating a metal increase its resistance?

To understand the change, we need to look inside the metal.

Metals contain mobile electrons that can move through a lattice of positively charged metal ions. These electrons are responsible for electrical conduction.

At a lower temperature, the metal ions vibrate around their usual positions. When the temperature increases, they vibrate more strongly.

This changes how easily conduction electrons move through the material.

What happens to the electrons?

Imagine a puck sliding through a field of small obstacles.

If the obstacles barely move, the puck can travel through with relatively few interruptions. If the obstacles start moving around, the puck’s path is disrupted more frequently.

The analogy gives us a useful picture of what happens in a heated metal.

• The metal ions vibrate more strongly.
• Conduction electrons experience more frequent scattering.
• Their average drift mobility decreases.
• The material becomes less effective at conducting current for the same electric field.

The electrons aren’t being used up, and the wire isn’t running out of charge. The main change is that electron movement through the lattice becomes more strongly impeded.

The analogy has limits: electrons are quantum particles, not tiny pucks bouncing off stationary objects. However, it captures the important idea that increased lattice vibrations produce more electron scattering.

Resistance and resistivity

For a uniform wire, resistance is related to its resistivity by

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

where:

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

When an ordinary metal is heated, its resistivity generally increases.

This is the main reason its resistance increases.

Strictly speaking, heating also causes a metal wire to expand slightly, changing its length and cross-sectional area. For the usual HSC model, these dimensional changes are neglected unless the question specifically asks you to consider them. The increase in resistivity is the dominant effect.

If you need to revise how length, thickness, and material affect resistance, see Why Wire Resistance Depends on Length, Thickness and Material.

02How is the effect measured experimentally?

Suppose you want to investigate how a metal wire’s resistance changes as its temperature increases.

You need to measure two things:

1. The potential difference across the wire.
2. The current flowing through the wire.

You can then calculate its resistance using

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

The challenge is making sure that heating the wire doesn’t interfere with the measurement.

The initial experimental design

Start with a metal wire connected to a regulated low-voltage power supply.

An ammeter is connected in series with the wire, while a voltmeter is connected in parallel across it.

The power supply maintains a constant voltage across the wire.

What should happen when the wire heats up?

If the voltage across the wire stays constant, increasing resistance causes the current to decrease.

This follows directly from Ohm’s law:

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

The expected observation is a falling ammeter reading.

However, the initial design has a weakness.

If the current itself heats the wire, you don’t have independent control over its temperature. You also need to consider how much heat is lost to the surroundings.

As resistance increases, the electrical power supplied at constant voltage decreases:

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

This means the wire’s heating rate changes as its resistance changes.

You might observe a falling current, but you won’t have a reliable set of measurements showing exactly how resistance depends on temperature.

Improving the experimental design

A better approach is to heat the wire independently of the measuring circuit.

Place the metal wire inside a water bath. Use a separate heater to warm the water, and position a thermometer close to the wire.

The measuring circuit remains separate:

• A low-voltage regulated supply provides the test voltage.
• An ammeter is connected in series with the metal wire.
• A voltmeter is connected in parallel across the wire.
• A thermometer measures the water temperature close to the wire.

The improved method is:

1. Set the water bath to a chosen temperature, such as \(20^\circ\text{C}\).
2. Allow the wire and surrounding water to reach thermal equilibrium.
3. Apply a small test voltage across the wire.
4. Record the voltage and current.
5. Calculate the resistance using \(R=V/I\).
6. Repeat at higher temperatures, keeping the same wire and electrical connections.

Use a sufficiently small test current to minimise additional heating from the measurement itself. You can also open the switch between readings.

Stirring the water gently helps maintain a more uniform temperature.

The key experimental improvement is that the water bath controls the wire’s temperature independently of the electrical measurements.

What would the results look like?

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

Suppose the test voltage is maintained at \(0.200\text{ V}\) across a metal wire.

Notice the pattern.

As temperature increases:

• The voltage remains constant.
• The current decreases.
• The calculated resistance increases.

For example, at \(60^\circ\text{C}\):

\[
R=\frac{0.200}{0.0431}
\]

\[
R\approx4.64\,\Omega
\]

The wire’s resistance has increased from \(4.00\,\Omega\) at \(20^\circ\text{C}\) to \(4.64\,\Omega\) at \(60^\circ\text{C}\).

The approximately straight-line relationship is useful. It allows us to estimate how much a metal wire’s resistance changes for a given temperature increase.

03The temperature coefficient of resistance

For an ordinary metal over a modest temperature range, resistance can be approximated using

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

where:

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

The temperature coefficient tells us how strongly a material’s resistance changes with temperature.

For ordinary metals, \(\alpha\) is positive.

This means:

• Heating the wire increases its resistance.
• Cooling the wire decreases its resistance.

In the equation, the temperature change is

\[
\Delta T=T-T_0
\]

A positive \(\Delta T\) represents heating, while a negative \(\Delta T\) represents cooling.

Remember that a temperature difference of \(1^\circ\text{C}\) is equal to a temperature difference of \(1\text{ K}\).

Using the illustrative experimental data

We can estimate the temperature coefficient using the measurements at \(20^\circ\text{C}\) and \(60^\circ\text{C}\).

The reference resistance is \(4.00\,\Omega\), and the resistance at \(60^\circ\text{C}\) is \(4.64\,\Omega\).

Rearranging the temperature equation:

\[
\alpha=
\frac{R(T)-R(T_0)}
{R(T_0)(T-T_0)}
\]

Substituting:

\[
\alpha=
\frac{4.64-4.00}
{4.00(60-20)}
\]

\[
\alpha=\frac{0.64}{160}
\]

\[
\boxed{\alpha=0.0040\,^\circ\text{C}^{-1}}
\]

This tells us that the wire’s resistance increases by approximately \(0.40\%\) of its reference resistance for each degree Celsius of temperature increase, within the range where the linear approximation applies.

The coefficient is not universal. Different metals have different temperature coefficients, and the value can vary with temperature.

04Fixed voltage versus fixed current: the important distinction

This is where students often make mistakes.

A common statement is:

Heating a wire increases its resistance, so its current decreases.

That statement is incomplete.

The current only necessarily decreases if the voltage across the wire is held constant.

Consider two different situations.

Case 1: Voltage is held constant

Suppose the voltage across a metal wire is maintained at \(6.0\text{ V}\).

As temperature increases, resistance increases.

Using

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

the current decreases.

The electrical power also decreases because

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

As resistance increases, the same voltage supplies less electrical power.

Case 2: Current is held constant

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

As temperature increases, resistance increases.

Using

\[
V=IR
\]

the voltage across the wire must increase.

Electrical power is

\[
P=I^2R
\]

Since current remains constant and resistance increases, electrical power increases too.

Compare the two cases:

The resistance of the wire increases in both situations. The difference is how the external circuit responds.

There is another interesting consequence.

At constant voltage, the reduction in electrical power tends to oppose further heating. At constant current, the increase in electrical power can reinforce heating.

However, this doesn’t automatically mean the wire will overheat under constant current. Its actual temperature depends on the balance between electrical energy supplied and heat lost to the surroundings.

A useful contrasting case: thermistors

Don’t apply the positive temperature coefficient rule to every electrical component.

An NTC thermistor is made from semiconductor material rather than an ordinary metal wire. Its resistance generally decreases as temperature increases because the number of available charge carriers increases significantly.

This is why temperature sensors can behave differently from metal conductors.

For the metal-wire questions in this article, assume a typical metal with a positive temperature coefficient unless the question specifies otherwise.

05Worked examples

06Questions and solutions

The following questions build from straightforward temperature calculations towards experimental interpretation and more complex circuit reasoning.

Assume ordinary metal behaviour and use the linear temperature coefficient equation where appropriate. Unless stated otherwise, neglect thermal expansion and assume the wire remains within the range where the linear approximation is valid.

Question 1

A copper wire has a resistance of \(6.0\,\Omega\) at \(20^\circ\text{C}\).

It is heated to \(50^\circ\text{C}\). Its temperature coefficient is \(4.0\times10^{-3}\,^\circ\text{C}^{-1}\).

The voltage across the wire is maintained at \(12.0\text{ V}\).

Calculate:

a) The resistance at \(50^\circ\text{C}\).

b) The current before heating.

c) The current after heating.

Solution 1

The resistance increases to \(6.72\,\Omega\), and the current decreases from \(2.00\text{ A}\) to approximately \(1.79\text{ A}\).

Part a: Calculate the new resistance.

The temperature change is

\[
\Delta T=50-20=30^\circ\text{C}
\]

Using the temperature coefficient equation:

\[
R(T)=R_0[1+\alpha\Delta T]
\]

Substitute:

\[
R(T)=6.0[1+(0.0040)(30)]
\]

\[
R(T)=6.0(1.12)
\]

\[
\boxed{R(T)=6.72\,\Omega}
\]

Part b: Calculate the initial current.

At \(20^\circ\text{C}\):

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

\[
I_{20}=\frac{12.0}{6.0}
\]

\[
\boxed{I_{20}=2.00\text{ A}}
\]

Part c: Calculate the final current.

At \(50^\circ\text{C}\):

\[
I_{50}=\frac{12.0}{6.72}
\]

\[
\boxed{I_{50}=1.79\text{ A}}
\]

The current decreases because the voltage is held constant while the resistance increases.

The important detail is the condition of the circuit. We could not automatically conclude that current decreases if the question instead specified a constant-current supply.

Question 2

A metal wire is connected to a current-regulated power supply.

The current is maintained at \(0.60\text{ A}\) while the wire is heated.

The following measurements are recorded:

Calculate:

a) The resistance at each temperature.

b) The temperature coefficient of resistance.

c) The electrical power supplied to the wire at each temperature.

Explain why the voltage changes even though the current remains constant.

Solution 2

The resistance increases from \(4.00\,\Omega\) to \(4.80\,\Omega\), and the temperature coefficient is approximately \(3.33\times10^{-3}\,^\circ\text{C}^{-1}\).

The electrical power also increases, from \(1.44\text{ W}\) to \(1.73\text{ W}\).

Part a: Calculate the resistance at each temperature.

At \(20^\circ\text{C}\):

\[
R_{20}=\frac{2.40}{0.60}
\]

\[
\boxed{R_{20}=4.00\,\Omega}
\]

At \(80^\circ\text{C}\):

\[
R_{80}=\frac{2.88}{0.60}
\]

\[
\boxed{R_{80}=4.80\,\Omega}
\]

Part b: Calculate the temperature coefficient.

Rearrange:

\[
\alpha=
\frac{R_{80}-R_{20}}
{R_{20}(80-20)}
\]

Substitute:

\[
\alpha=
\frac{4.80-4.00}
{4.00(60)}
\]

\[
\alpha=\frac{0.80}{240}
\]

\[
\boxed{\alpha=3.33\times10^{-3}\,^\circ\text{C}^{-1}}
\]

Part c: Calculate the electrical power.

Use

\[
P=VI
\]

At \(20^\circ\text{C}\):

\[
P_{20}=2.40(0.60)
\]

\[
\boxed{P_{20}=1.44\text{ W}}
\]

At \(80^\circ\text{C}\):

\[
P_{80}=2.88(0.60)
\]

\[
\boxed{P_{80}=1.73\text{ W}}
\]

The voltage increases because the wire’s resistance increases while the supply maintains the same current.

The power also increases because the current remains constant while resistance increases.

This is an important contrast with a fixed-voltage circuit.

Question 3

A student investigates how temperature affects the resistance of a metal wire.

The student records the following measurements:

The student concludes:

“As the temperature increases, the current decreases. Therefore, the resistance increases.”

a) Calculate the resistance at each temperature.

b) Calculate the percentage increase in resistance.

c) Explain why the decrease in current alone would not be sufficient evidence that resistance increased.

d) Suggest one improvement to the experimental design.

Solution 3

The student’s conclusion is supported by the calculated resistance values: resistance increases from \(4.0\,\Omega\) to \(4.5\,\Omega\), an increase of \(12.5\%\).

However, the decrease in current alone would not establish this. The voltage across the wire also changes.

Part a: Calculate the resistance at each temperature.

At \(25^\circ\text{C}\):

\[
R_{25}=\frac{0.60}{0.150}
\]

\[
\boxed{R_{25}=4.0\,\Omega}
\]

At \(75^\circ\text{C}\):

\[
R_{75}=\frac{0.54}{0.120}
\]

\[
\boxed{R_{75}=4.5\,\Omega}
\]

Part b: Calculate the percentage increase.

Use

\[
\%\text{ increase}=
\frac{\text{change in resistance}}
{\text{initial resistance}}\times100
\]

Substitute:

\[
\%\text{ increase}
=
\frac{4.5-4.0}{4.0}\times100
\]

\[
\boxed{12.5\%}
\]

Part c: Explain why current alone is insufficient.

The current decreases from \(0.150\text{ A}\) to \(0.120\text{ A}\), but the voltage also decreases from \(0.60\text{ V}\) to \(0.54\text{ V}\).

A decrease in current could result from a decrease in voltage rather than an increase in resistance.

We need to calculate \(V/I\) at each temperature to establish whether resistance has changed.

Part d: Suggest an experimental improvement.

Use a regulated power supply to maintain a constant voltage directly across the test wire.

Then record the current at each temperature.

Alternatively, continue measuring both voltage and current and calculate resistance at every temperature.

Using an independently heated water bath also helps control the wire’s temperature without relying on electrical self-heating.

The key experimental principle is that a change in one measured quantity does not automatically identify the cause. Other relevant variables must be measured or controlled.

Question 4

A metal wire has a resistance of \(5.0\,\Omega\) at \(20^\circ\text{C}\).

Its temperature coefficient is \(4.0\times10^{-3}\,^\circ\text{C}^{-1}\), and it is heated to \(80^\circ\text{C}\).

Consider two separate experiments.

Experiment A: The voltage across the wire is maintained at \(10.0\text{ V}\).

Experiment B: The current through the wire is maintained at \(1.00\text{ A}\).

a) Calculate the resistance at \(80^\circ\text{C}\).

b) For Experiment A, calculate the current and electrical power before and after heating.

c) For Experiment B, calculate the voltage and electrical power before and after heating.

d) Explain why the statement “a hotter wire always produces more electrical heating” is incomplete.

Solution 4

The resistance increases to \(6.2\,\Omega\).

In Experiment A, electrical power decreases from \(20\text{ W}\) to approximately \(16.1\text{ W}\).

In Experiment B, electrical power increases from \(5.0\text{ W}\) to \(6.2\text{ W}\).

Part a: Calculate the new resistance.

The temperature change is

\[
\Delta T=80-20=60^\circ\text{C}
\]

Using

\[
R(T)=R_0[1+\alpha\Delta T]
\]

\[
R(T)=5.0[1+(0.0040)(60)]
\]

\[
R(T)=5.0(1.24)
\]

\[
\boxed{R(T)=6.2\,\Omega}
\]

Part b: Experiment A, fixed voltage.

The voltage remains at \(10.0\text{ V}\).

At \(20^\circ\text{C}\):

\[
I_{20}=\frac{10.0}{5.0}
\]

\[
I_{20}=2.00\text{ A}
\]

The initial power is

\[
P_{20}=VI
\]

\[
P_{20}=10.0(2.00)
\]

\[
\boxed{P_{20}=20.0\text{ W}}
\]

At \(80^\circ\text{C}\):

\[
I_{80}=\frac{10.0}{6.2}
\]

\[
I_{80}\approx1.61\text{ A}
\]

The final power is

\[
P_{80}=\frac{V^2}{R}
\]

\[
P_{80}=\frac{10.0^2}{6.2}
\]

\[
\boxed{P_{80}\approx16.1\text{ W}}
\]

Therefore, increasing resistance reduces electrical power when voltage is held constant.

Part c: Experiment B, fixed current.

The current remains at \(1.00\text{ A}\).

At \(20^\circ\text{C}\):

\[
V_{20}=IR
\]

\[
V_{20}=1.00(5.0)
\]

\[
\boxed{V_{20}=5.0\text{ V}}
\]

The initial power is

\[
P_{20}=I^2R
\]

\[
P_{20}=(1.00)^2(5.0)
\]

\[
\boxed{P_{20}=5.0\text{ W}}
\]

At \(80^\circ\text{C}\):

\[
V_{80}=1.00(6.2)
\]

\[
\boxed{V_{80}=6.2\text{ V}}
\]

The final power is

\[
P_{80}=(1.00)^2(6.2)
\]

\[
\boxed{P_{80}=6.2\text{ W}}
\]

Part d: Explain the difference.

The statement is incomplete because electrical power depends on both resistance and the circuit condition.

At fixed voltage:

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

Increasing resistance decreases power.

At fixed current:

\[
P=I^2R
\]

Increasing resistance increases power.

The wire has exactly the same temperature-dependent resistance in both experiments. The different power responses arise from the different supply conditions.

Question 5

Challenge question

A metal wire is connected in parallel with a fixed resistor across a battery.

The battery is modelled as an ideal electromotive force of \(12.0\text{ V}\) in series with an internal resistance of \(4.0\,\Omega\).

Initially, at \(20^\circ\text{C}\):

• Metal wire A has a resistance of \(4.0\,\Omega\).
• Fixed resistor B has a resistance of \(4.0\,\Omega\).

Wire A is then heated to \(70^\circ\text{C}\), while resistor B remains at \(20^\circ\text{C}\).

The temperature coefficient of wire A is \(4.0\times10^{-3}\,^\circ\text{C}^{-1}\).

Assume the battery’s electromotive force and internal resistance remain constant.

a) Calculate the resistance of wire A after heating.

b) Before heating, calculate the total current supplied by the battery and the current through each parallel branch.

c) After heating, calculate:

• The equivalent resistance of the parallel combination.
• The total current supplied by the battery.
• The terminal voltage of the battery.
• The current through wire A.
• The current through resistor B.

d) A student argues:

“Heating wire A increases the total resistance of the circuit, so the current through both A and B must decrease.”

Is this correct? Explain why the current through resistor B behaves differently from what the student predicts.

e) How would the result for resistor B change if the battery were ideal, with zero internal resistance?

Solution 5

The student’s prediction is incorrect.

After heating, the total current supplied by the battery decreases, and the current through wire A decreases. However, the current through resistor B actually increases.

The reason is that the battery has internal resistance. As the total current decreases, the voltage lost inside the battery decreases, allowing its terminal voltage to rise.

This is the important interaction that changes the result.

Part a: Calculate the new resistance of wire A.

The temperature change is

\[
\Delta T=70-20=50^\circ\text{C}
\]

Using the temperature coefficient equation:

\[
R_A=R_0[1+\alpha\Delta T]
\]

Substitute:

\[
R_A=4.0[1+(0.0040)(50)]
\]

\[
R_A=4.0(1.20)
\]

\[
\boxed{R_A=4.8\,\Omega}
\]

Wire A’s resistance increases from \(4.0\,\Omega\) to \(4.8\,\Omega\).

Part b: Calculate the initial circuit currents.

Before heating, both parallel branches have resistance \(4.0\,\Omega\).

For two parallel resistors:

\[
\frac{1}{R_{\text{parallel}}}
=
\frac{1}{R_A}+\frac{1}{R_B}
\]

Substituting:

\[
\frac{1}{R_{\text{parallel}}}
=
\frac14+\frac14
\]

Therefore:

\[
R_{\text{parallel}}=2.0\,\Omega
\]

The battery’s internal resistance is in series with the parallel combination.

Therefore, total circuit resistance is

\[
R_{\text{total}}=4.0+2.0
\]

\[
R_{\text{total}}=6.0\,\Omega
\]

The total current supplied by the battery is

\[
I_{\text{total}}=\frac{\mathcal{E}}{R_{\text{total}}}
\]

where \(\mathcal{E}\) is the battery’s electromotive force.

\[
I_{\text{total}}=\frac{12.0}{6.0}
\]

\[
\boxed{I_{\text{total}}=2.00\text{ A}}
\]

The terminal voltage is

\[
V_{\text{terminal}}=
\mathcal{E}-I_{\text{total}}r
\]

where \(r=4.0\,\Omega\) is the internal resistance.

\[
V_{\text{terminal}}=12.0-(2.00)(4.0)
\]

\[
\boxed{V_{\text{terminal}}=4.0\text{ V}}
\]

Both parallel branches experience this same potential difference.

Therefore:

\[
I_A=\frac{4.0}{4.0}=1.00\text{ A}
\]

\[
I_B=\frac{4.0}{4.0}=1.00\text{ A}
\]

Initially, both branches carry \(1.00\text{ A}\).

Part c: Calculate the circuit after heating.

First, calculate the new equivalent resistance of the parallel branches.

\[
\frac{1}{R_{\text{parallel}}}
=
\frac{1}{4.8}+\frac14
\]

\[
R_{\text{parallel}}
=
\frac{4.8\times4.0}{4.8+4.0}
\]

\[
\boxed{R_{\text{parallel}}=2.18\,\Omega}
\]

The total circuit resistance is

\[
R_{\text{total}}=4.0+2.18
\]

\[
R_{\text{total}}=6.18\,\Omega
\]

The total current becomes

\[
I_{\text{total}}=
\frac{12.0}{6.18}
\]

\[
\boxed{I_{\text{total}}\approx1.94\text{ A}}
\]

Notice that the total current has decreased from \(2.00\text{ A}\) to approximately \(1.94\text{ A}\).

Now calculate the battery’s terminal voltage.

\[
V_{\text{terminal}}=
\mathcal{E}-I_{\text{total}}r
\]

\[
V_{\text{terminal}}=
12.0-(1.94)(4.0)
\]

\[
\boxed{V_{\text{terminal}}\approx4.24\text{ V}}
\]

This is higher than the original terminal voltage of \(4.0\text{ V}\).

Now calculate the current through each branch.

For wire A:

\[
I_A=\frac{V_{\text{terminal}}}{R_A}
\]

\[
I_A=\frac{4.24}{4.8}
\]

\[
\boxed{I_A\approx0.882\text{ A}}
\]

For resistor B:

\[
I_B=\frac{V_{\text{terminal}}}{R_B}
\]

\[
I_B=\frac{4.24}{4.0}
\]

\[
\boxed{I_B\approx1.06\text{ A}}
\]

The results are:

Part d: Explain why the current through B increases.

The student’s mistake is assuming that a decrease in total circuit current means the current in every branch must decrease.

That isn’t necessarily true.

When wire A heats up, its resistance increases. This increases the equivalent resistance of the parallel load.

Consequently, the total current supplied by the battery decreases.

However, this also reduces the voltage lost across the battery’s internal resistance:

\[
V_{\text{internal}}=Ir
\]

Because the internal voltage drop becomes smaller, the battery’s terminal voltage increases.

Resistor B is connected directly across the battery terminals, so it experiences the increased terminal voltage.

Its resistance remains constant.

Therefore, from

\[
I_B=\frac{V_{\text{terminal}}}{R_B}
\]

its current increases.

The total battery current decreases, but the current in one of its parallel branches increases.

There is no contradiction: the total current is the sum of the two branch currents.

Part e: What if the battery were ideal?

If the battery had zero internal resistance, its terminal voltage would remain fixed at \(12.0\text{ V}\).

Before heating:

\[
I_B=\frac{12.0}{4.0}=3.0\text{ A}
\]

After heating:

\[
I_B=\frac{12.0}{4.0}=3.0\text{ A}
\]

The current through B would remain unchanged.

Only the current through wire A would decrease.

This final comparison is important. The response of a circuit depends not only on how a component’s resistance changes, but also on the characteristics of the power supply and the way the components are connected.

07Where this leads next

Heating a metal wire changes its resistance because the metal’s lattice vibrations increase electron scattering. That is the material-level explanation.

The next step is understanding how this changing resistance affects an entire circuit.

In particular, distinguish between:

• A component connected directly across a constant-voltage supply.
• A component supplied by a constant-current source.
• A component connected in series with other resistances.
• A component connected in parallel with other branches.
• A real battery whose terminal voltage changes because of internal resistance.

These distinctions turn a simple temperature question into a circuit-analysis problem. They are often what separate a correct HSC Physics calculation from an answer that applies Ohm’s law without considering the actual circuit.