Which Resistor Gets Hotter in a Series Circuit? | HSC Physics

Connect a 10 Ω resistor and a 30 Ω resistor in series. Both carry exactly the same current. So which one gets hotter?

For two resistors with comparable construction and cooling, the 30 Ω resistor dissipates more power and heats up faster. In fact, it dissipates three times as much power.

But there’s an important catch: more electrical power does not automatically mean a greater temperature rise. The amount of material in the resistor, its heat capacity, and how quickly it loses heat also matter.

The key is to separate two questions:

• Which resistor converts electrical energy into thermal energy faster?
• Which resistor actually reaches the higher temperature?

These questions are related, but they aren’t identical.

01Why does the higher-resistance resistor heat up more?

First, make a prediction.

If the same current passes through both resistors, would you expect them to produce the same amount of heat?

It’s a reasonable guess. After all, current measures how much electric charge flows through a circuit each second. If the current is identical, surely the heating should be identical too?

Not quite.

Current tells us how much charge flows, but it doesn’t tell us how much electrical energy each coulomb transfers.

That depends on the potential difference across each resistor.

Start with the series circuit

In a series circuit, there is only one path for charge to follow. The current is therefore the same through every component.

Ohm’s law tells us:

\[ V=IR \]

where:

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

For the same current, a resistor with greater resistance has a larger potential difference across it.

Now consider electrical power.

Power is the rate at which electrical energy is transferred. For a resistor:

\[ P=VI \]

Substituting Ohm’s law, \(V=IR\):

\[ \boxed{P=I^2R} \]

This is the crucial relationship.

When comparing resistors carrying the same current, power is directly proportional to resistance.

\[ P\propto R \qquad \text{(constant current)} \]

So, in a series circuit:

• The higher-resistance resistor dissipates more power.
• The lower-resistance resistor dissipates less power.
• Both carry the same current.

A numerical comparison

Suppose a 10 Ω resistor and a 30 Ω resistor are connected in series. The current is 0.20 A.

For the 10 Ω resistor:

\[ P_1=I^2R_1 \]

\[ P_1=(0.20)^2(10) \]

\[ P_1=0.40\text{ W} \]

For the 30 Ω resistor:

\[ P_2=I^2R_2 \]

\[ P_2=(0.20)^2(30) \]

\[ P_2=1.20\text{ W} \]

The 30 Ω resistor dissipates three times as much power.

The higher-resistance resistor transfers electrical energy into thermal energy three times as quickly.

Notice something else: the voltage drops are different, even though the current is identical.

This is why equal current does not mean equal heating.

02What’s happening inside the resistor?

The electrical explanation becomes easier to understand when we consider what happens to charge carriers inside a metal.

An electric field acts on the electrons in the conducting material, causing them to drift through the metal. As they move, interactions and scattering transfer energy from the charge carriers to the material’s lattice.

This energy increases the internal energy of the resistor, usually raising its temperature.

Think of the potential difference as the amount of electrical energy transferred per unit of charge.

A potential difference of 1 V corresponds to 1 joule of energy transferred per coulomb of charge.

In our earlier example:

• The 10 Ω resistor has a 2 V potential difference. Each coulomb transfers 2 J of energy.
• The 30 Ω resistor has a 6 V potential difference. Each coulomb transfers 6 J of energy.

Since the same amount of charge passes through both resistors each second, the 30 Ω resistor transfers more energy per second.

That’s the physical reason for its greater power dissipation.

One important qualification: resistance isn’t simply a measure of how many times electrons collide with atoms. It depends on the material, geometry, temperature, and scattering processes within the conductor.

For metal resistors, heating can also increase resistance because stronger lattice vibrations affect electron scattering. This introduces another effect into a real circuit.

For a closer look, see How Heating Changes a Metal Wire’s Resistance.

03Does more power always mean a higher temperature?

No. This is where students often make an incorrect jump.

Power tells us how quickly electrical energy is transferred. Temperature tells us the thermal state of an object.

To work out how much a resistor’s temperature increases, we also need to consider its heat capacity.

The relationship between thermal energy transferred and temperature change is:

\[ Q=mc\Delta T \]

where:

• \(Q\) is the thermal energy transferred, in joules (J).
• \(m\) is the mass of the resistor, in kilograms (kg).
• \(c\) is its specific heat capacity, in J kg\(^{-1}\) K\(^{-1}\).
• \(\Delta T\) is its temperature change, in kelvin (K) or degrees Celsius (°C).

For constant power over a time interval:

\[ Q=Pt \]

Combining the two equations gives:

\[ \boxed{\Delta T=\frac{Pt}{mc}} \]

This equation reveals an important distinction.

A resistor’s temperature rise depends on both the energy it receives and how much energy is needed to increase its temperature.

A resistor with a larger thermal mass needs more energy for the same temperature rise.

Consider two different resistor designs

Imagine two resistors carrying the same current.

• Resistor A dissipates 1 W and has a heat capacity of 0.5 J K\(^{-1}\).
• Resistor B dissipates 3 W and has a heat capacity of 1.5 J K\(^{-1}\).

Assume both start at the same temperature and lose negligible heat to their surroundings.

Resistor B dissipates three times as much power, but it also has three times the heat capacity.

For the same time interval, both have the same temperature rise.

This is why you should never conclude that a resistor must become three times hotter just because it dissipates three times the power.

What about heat escaping into the surroundings?

The previous calculation assumes that no energy escapes.

Real resistors lose energy to their surroundings through conduction, convection, and thermal radiation.

As a resistor becomes hotter, the rate of heat loss generally increases.

Eventually, a resistor may reach thermal equilibrium, where:

\[ \text{electrical power input} = \text{rate of thermal energy loss} \]

At this point, its temperature stops increasing significantly.

Consequently, the final temperature depends on more than electrical resistance. The resistor’s size, material, surface area, mounting, and airflow all affect its thermal behaviour.

For a straightforward HSC calculation, assume constant resistance and negligible heat loss unless the question states otherwise.

04How could we test which resistor heats faster?

We can investigate the prediction with a controlled experiment.

The aim is to separate electrical power from thermal behaviour rather than simply touching two resistors and deciding which feels hotter.

Setting up the experiment

Use two resistors of similar physical construction but different resistance values.

The apparatus consists of:

• A low-voltage DC power supply with constant-current control.
• A switch and ammeter connected in series.
• A 10 Ω resistor followed by a 30 Ω resistor in the same circuit.
• A voltmeter connected across each resistor.
• Temperature probes attached to both resistor bodies and connected to a data logger.

The positive terminal of the power supply connects to the switch, then the ammeter, then the first resistor, then the second resistor, and finally back to the negative terminal.

The voltmeters connect across the individual resistors, not in series with them.

Both resistors should start at the same temperature and be mounted under comparable conditions.

What should the experiment show?

Before switching on the circuit, predict the readings.

Suppose the current is maintained at 0.10 A.

The expected voltage drops are:

\[ V_1=IR_1=0.10(10)=1.0\text{ V} \]

\[ V_2=IR_2=0.10(30)=3.0\text{ V} \]

The calculated powers are:

\[ P_1=0.10(1.0)=0.10\text{ W} \]

\[ P_2=0.10(3.0)=0.30\text{ W} \]

The following table gives an illustrative idealised data set.

The temperature rises assume equal heat capacities of 0.60 J K\(^{-1}\), identical initial temperatures, and negligible heat loss. These are calculated illustrative values, not reported experimental measurements.

The electrical measurements support the prediction: the 30 Ω resistor receives electrical energy at three times the rate of the 10 Ω resistor.

Under the matched thermal assumptions, it also experiences three times the temperature rise.

What could make the experiment misleading?

Suppose the 30 Ω resistor has a much larger ceramic body than the 10 Ω resistor.

Even though it dissipates more power, it might not experience a greater temperature rise because it has a larger thermal capacity.

Other problems include:

• Different airflow around the two resistors.
• Unequal contact between temperature probes and resistor surfaces.
• One resistor being mounted against a larger heat-conducting surface.
• Heat from one resistor affecting the temperature probe on the other.

A better design uses matched resistor packages, identical mounting arrangements, and carefully positioned temperature probes.

Swapping the physical positions of the resistors and repeating the experiment can also help identify location-related effects.

It is particularly useful to compare the initial rates of temperature rise, before heat losses have had much time to affect the results.

The experiment therefore tests two separate predictions:

1. The higher-resistance resistor dissipates more electrical power at the same current.
2. Whether it becomes hotter depends on its thermal properties and heat losses.

05Series circuits versus parallel circuits: why the answer changes

Now consider a different arrangement.

Instead of connecting the resistors in series, connect them in parallel across the same power supply.

Would the higher-resistance resistor still dissipate more power?

No. The result reverses.

This is one of the most important comparisons to understand when answering HSC circuit questions.

Series circuits: same current

For resistors in series, the current is identical.

Starting with:

\[ P=I^2R \]

A larger resistance produces greater power dissipation.

\[ \boxed{P\propto R} \]

This is the fixed-current comparison.

Parallel circuits: same voltage

For resistors in parallel, each branch has the same potential difference.

Starting with:

\[ P=VI \]

Substitute \(I=V/R\):

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

Now power is inversely proportional to resistance when voltage is constant.

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

The lower-resistance resistor dissipates more power.

For example, connect 60 Ω and 120 Ω resistors in parallel across a 12 V supply.

For the 60 Ω resistor:

\[ P_1=\frac{12^2}{60}=2.4\text{ W} \]

For the 120 Ω resistor:

\[ P_2=\frac{12^2}{120}=1.2\text{ W} \]

The 60 Ω resistor dissipates twice as much power.

The difference comes from the quantity held constant:

A common mistake is to use \(P=I^2R\) to compare parallel resistors while assuming their currents are equal. They aren’t.

Always identify what the circuit makes equal before comparing power.

06Worked examples

07An important complication: What if the resistance changes?

So far, we’ve compared two resistors within the same circuit.

But there’s another situation that can catch you out in an exam.

Imagine a 10 Ω resistor connected in series with a 30 Ω resistor across a fixed 12 V supply.

The 30 Ω resistor dissipates more power.

Now replace the 30 Ω resistor with a 60 Ω resistor.

Would you expect its power to double?

It sounds reasonable if you’re thinking about \(P=I^2R\). However, there’s a problem with that prediction.

The supply voltage is fixed, so increasing the total circuit resistance decreases the current.

You cannot increase the resistance while assuming the original current remains unchanged.

For a fixed-voltage supply and two series resistors:

\[ I=\frac{V_s}{R_1+R_2} \]

The power in either resistor therefore depends on both resistances.

For a variable resistor \(R\) in series with a fixed resistor \(R_f\):

\[ P_R=I^2R \]

Substituting the circuit current:

\[ \boxed{ P_R=\frac{V_s^2R}{(R+R_f)^2} } \]

Increasing \(R\) changes the current as well as the resistance.

This means that doubling a resistor’s resistance does not necessarily double its power when the supply voltage is fixed.

You’ll see why this matters in Question 4.

08Questions and solutions

Question 1

A 12 V ideal battery is connected to two resistors, 10 Ω and 20 Ω, in series.

Calculate:

a) The current through the circuit.

b) The power dissipated by each resistor.

c) Which resistor has the greater electrical heating rate.

Assume both resistors have constant resistance.

Solution 1

The 20 Ω resistor dissipates twice the power of the 10 Ω resistor: 3.2 W compared with 1.6 W. It therefore has the greater electrical heating rate.

Step 1: Calculate the total resistance.

For a series circuit:

\[ R_{\text{total}}=10+20=30\text{ Ω} \]

Step 2: Calculate the current.

Using Ohm’s law:

\[ I=\frac{V}{R_{\text{total}}} \]

\[ I=\frac{12}{30} \]

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

The current through both resistors is 0.40 A.

Step 3: Calculate the power of each resistor.

For the 10 Ω resistor:

\[ P_1=I^2R_1 \]

\[ P_1=(0.40)^2(10) \]

\[ P_1=1.6\text{ W} \]

For the 20 Ω resistor:

\[ P_2=I^2R_2 \]

\[ P_2=(0.40)^2(20) \]

\[ P_2=3.2\text{ W} \]

Step 4: Compare the results.

\[ \frac{P_2}{P_1}=\frac{3.2}{1.6}=2 \]

The 20 Ω resistor dissipates twice as much power.

The important principle is that the current is common to both resistors. Therefore, the higher-resistance resistor dissipates more power.

Question 2

Two resistors, A and B, are connected in series across a 9 V ideal power supply.

Their properties are:

The circuit operates for 40 seconds.

Assume:

• Both resistors start at the same temperature.
• Their resistances remain constant.
• No thermal energy escapes to the surroundings.
• Each resistor’s temperature is uniform.

Calculate:

a) The circuit current.

b) The power dissipated by each resistor.

c) The energy transferred to each resistor.

d) The temperature rise of each resistor.

Solution 2

Both resistors experience the same temperature rise of 56.25 K, despite resistor B dissipating three times as much power. Its thermal capacity is also three times greater.

Step 1: Calculate the total resistance.

\[ R_{\text{total}}=12+36 \]

\[ R_{\text{total}}=48\text{ Ω} \]

Step 2: Calculate the current.

\[ I=\frac{V}{R_{\text{total}}} \]

\[ I=\frac{9}{48} \]

\[ I=0.1875\text{ A} \]

Step 3: Calculate the power dissipated by each resistor.

For resistor A:

\[ P_A=I^2R_A \]

\[ P_A=(0.1875)^2(12) \]

\[ P_A=0.421875\text{ W} \]

For resistor B:

\[ P_B=I^2R_B \]

\[ P_B=(0.1875)^2(36) \]

\[ P_B=1.265625\text{ W} \]

Resistor B dissipates three times the power of resistor A.

Step 4: Calculate the energy transferred over 40 seconds.

Using:

\[ E=Pt \]

For resistor A:

\[ E_A=0.421875(40) \]

\[ E_A=16.875\text{ J} \]

For resistor B:

\[ E_B=1.265625(40) \]

\[ E_B=50.625\text{ J} \]

Step 5: Calculate the temperature rise.

Rearrange the thermal energy equation:

\[ \Delta T=\frac{E}{C} \]

Here, \(C\) represents the total thermal capacity of each resistor.

For resistor A:

\[ \Delta T_A=\frac{16.875}{0.30} \]

\[ \Delta T_A=56.25\text{ K} \]

For resistor B:

\[ \Delta T_B=\frac{50.625}{0.90} \]

\[ \Delta T_B=56.25\text{ K} \]

Conclusion: Both resistors experience the same temperature rise. The tempting prediction that resistor B must become three times hotter ignores its greater thermal capacity.

This is an important distinction between electrical power and actual temperature change.

Question 3

Two resistors, 30 Ω and 90 Ω, are connected in parallel across a 9 V ideal power supply.

The circuit operates for 60 seconds.

Calculate:

a) The current through each resistor.

b) The power dissipated by each resistor.

c) The electrical energy transferred to each resistor.

d) Which resistor has the greater heating rate.

Explain why the result differs from a series circuit.

Solution 3

The 30 Ω resistor dissipates three times as much power as the 90 Ω resistor. This happens because both parallel branches have the same voltage, making the lower-resistance branch dissipate more power.

Step 1: Identify the voltage across each resistor.

In a parallel circuit, the potential difference across each branch equals the supply voltage.

Therefore:

\[ V_1=V_2=9\text{ V} \]

Step 2: Calculate the branch currents.

For the 30 Ω resistor:

\[ I_1=\frac{V}{R_1} \]

\[ I_1=\frac{9}{30} \]

\[ I_1=0.30\text{ A} \]

For the 90 Ω resistor:

\[ I_2=\frac{9}{90} \]

\[ I_2=0.10\text{ A} \]

The lower-resistance branch carries three times the current.

Step 3: Calculate the power dissipated.

Since voltage is common, use:

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

For the 30 Ω resistor:

\[ P_1=\frac{9^2}{30} \]

\[ P_1=2.7\text{ W} \]

For the 90 Ω resistor:

\[ P_2=\frac{9^2}{90} \]

\[ P_2=0.90\text{ W} \]

Step 4: Calculate the energy transferred over 60 seconds.

\[ E=Pt \]

For the 30 Ω resistor:

\[ E_1=2.7(60)=162\text{ J} \]

For the 90 Ω resistor:

\[ E_2=0.90(60)=54\text{ J} \]

Conclusion: The 30 Ω resistor has the greater electrical heating rate and receives three times as much energy.

The result differs from a series circuit because the common quantity has changed.

In series, current is common, so \(P=I^2R\) favours higher resistance.

In parallel, voltage is common, so \(P=V^2/R\) favours lower resistance.

Question 4

A 10 Ω resistor is connected in series with a replaceable resistor across a fixed 12 V ideal power supply.

Initially, the replaceable resistor has a resistance of 30 Ω.

It is then replaced with a 60 Ω resistor.

Calculate the power dissipated by each resistor:

a) Before the replacement.

b) After the replacement.

c) Determine whether the power dissipated by the replaceable resistor doubles, increases, or decreases.

Explain why the result might be different from your initial prediction.

Assume constant resistance and an ideal voltage supply.

Solution 4

The power dissipated by the replaceable resistor decreases from 2.70 W to approximately 1.76 W. Doubling its resistance does not double its power because the circuit current decreases.

Step 1: Calculate the initial current.

Initially, the resistors are 10 Ω and 30 Ω.

\[ R_{\text{total}}=10+30=40\text{ Ω} \]

\[ I=\frac{12}{40} \]

\[ I=0.30\text{ A} \]

Step 2: Calculate the initial power of each resistor.

For the 10 Ω resistor:

\[ P_{10}=(0.30)^2(10) \]

\[ P_{10}=0.90\text{ W} \]

For the 30 Ω resistor:

\[ P_{30}=(0.30)^2(30) \]

\[ P_{30}=2.70\text{ W} \]

Step 3: Calculate the new current after replacing the 30 Ω resistor.

The new total resistance is:

\[ R_{\text{total}}=10+60 \]

\[ R_{\text{total}}=70\text{ Ω} \]

Therefore:

\[ I=\frac{12}{70} \]

\[ I\approx0.1714\text{ A} \]

The current has fallen from 0.30 A to approximately 0.1714 A.

Step 4: Calculate the new power of each resistor.

For the 10 Ω resistor:

\[ P_{10}=(0.1714)^2(10) \]

\[ P_{10}\approx0.294\text{ W} \]

For the 60 Ω resistor:

\[ P_{60}=(0.1714)^2(60) \]

\[ P_{60}\approx1.76\text{ W} \]

Step 5: Compare the replaceable resistor’s power.

Before replacement:

\[ P_{30}=2.70\text{ W} \]

After replacement:

\[ P_{60}=1.76\text{ W} \]

The power has decreased by approximately 34.7%.

Conclusion: The 60 Ω resistor still dissipates more power than the fixed 10 Ω resistor in the new series circuit. However, its own power has decreased compared with the original 30 Ω resistor.

The tempting approach is to double the resistance in \(P=I^2R\) while keeping the original current of 0.30 A.

That is incorrect because the supply voltage is fixed. Increasing the total resistance reduces the current, and power depends on the square of that current.

This question highlights the difference between comparing resistors in one circuit and changing the resistance of the entire circuit.

Question 5

Two straight metal wire sections, X and Y, are made from the same uniform metal.

They have:

• The same cross-sectional area.
• The same initial temperature.
• The same specific heat capacity.
• Lengths of \(L\) and \(3L\), respectively.

The wires are connected in series to an ideal constant-current source. A current \(I\) flows through both wires for a time \(t\).

Assume:

• Both wires are heated uniformly.
• No thermal energy escapes to the surroundings.
• Resistivity and specific heat capacity remain constant.
• End effects are negligible.

A student argues:

“The longer wire has three times the resistance, so it dissipates three times the power. Therefore, it must experience three times the temperature rise.”

Evaluate this argument.

Then consider a second arrangement in which X and Y are instead manufactured as equal-mass resistor components with equal thermal capacities, while retaining resistances in the ratio \(1:3\).

Would your conclusion change?

Solution 5

The student’s argument is incomplete. For the two wire sections, the temperature rises are equal, not in a ratio of 1:3. However, for equal-thermal-capacity resistor components, the higher-resistance component does experience three times the temperature rise under the stated ideal conditions.

The missing idea is that the longer wire contains more material to heat.

Step 1: Compare the electrical resistances.

The resistance of a uniform wire is:

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

where:

• \(\rho\) is the electrical resistivity of the metal.
• \(L\) is the wire length.
• \(A\) is its cross-sectional area.

Since both wires have the same material and cross-sectional area, resistance is directly proportional to length.

Therefore:

\[ R_Y=3R_X \]

Step 2: Compare the electrical power.

Both wires carry the same current because they are connected in series.

Using:

\[ P=I^2R \]

For wire Y:

\[ P_Y=I^2(3R_X) \]

\[ P_Y=3P_X \]

The longer wire dissipates three times the power.

So far, the student’s reasoning is correct.

Step 3: Compare the thermal capacities.

The wires are made from the same material and have identical cross-sectional areas.

The mass of a wire is proportional to its length.

Therefore:

\[ m_Y=3m_X \]

Since both wires also have the same specific heat capacity:

\[ C_Y=3C_X \]

The longer wire has three times the thermal capacity.

Step 4: Compare the energy transferred.

Over the same time interval:

\[ E=Pt \]

Therefore:

\[ E_X=P_Xt \]

For wire Y:

\[ E_Y=3P_Xt \]

Hence:

\[ E_Y=3E_X \]

Wire Y receives three times as much electrical energy.

Step 5: Calculate the temperature rises.

Using:

\[ \Delta T=\frac{E}{mc} \]

For wire X:

\[ \Delta T_X=\frac{E_X}{m_Xc} \]

For wire Y:

\[ \Delta T_Y=\frac{3E_X}{3m_Xc} \]

Simplifying:

\[ \boxed{\Delta T_Y=\Delta T_X} \]

The longer wire receives three times the energy, but it also has three times the thermal capacity.

Its temperature rise is therefore identical to that of the shorter wire.

Step 6: Consider the equal-mass resistor components.

Now suppose the resistors have equal thermal capacities instead of thermal capacities proportional to their lengths.

Their resistances remain in the ratio:

\[ R_Y=3R_X \]

The current and operating time are still identical.

Therefore:

\[ E_Y=3E_X \]

But now:

\[ C_Y=C_X \]

Using the thermal energy relationship:

\[ \Delta T=\frac{E}{C} \]

We obtain:

\[ \Delta T_Y=3\Delta T_X \]

The higher-resistance component now experiences three times the temperature rise.

Conclusion: The student’s original claim confuses power with temperature rise. Greater resistance produces greater power at the same current, but temperature rise also depends on thermal capacity.

For the same-material wire sections, both resistance and thermal capacity increase in proportion to length. Their temperature rises are therefore equal under the stated assumptions.

For equal-thermal-capacity resistor components, the higher-resistance component experiences the greater temperature rise.

This is why resistance alone cannot tell you exactly which physical resistor will reach the highest temperature.

09The key ideas to remember

When comparing heating in electrical circuits, follow this sequence:

• Series circuit: The current is the same through every resistor. Use \(P=I^2R\). The higher-resistance resistor dissipates more power.
• Parallel circuit: The voltage is the same across every branch. Use \(P=V^2/R\). The lower-resistance resistor dissipates more power.
• Fixed-voltage series circuit: If you change one resistance, recalculate the total current before comparing power.
• Temperature rise: Use \(\Delta T=Pt/(mc)\), assuming negligible heat loss. Consider the thermal capacity as well as electrical power.
• Real resistors: Cooling, construction, mounting, and temperature-dependent resistance can change the actual temperature reached.

If you want to extend this reasoning, the next useful step is understanding how electrical power, resistance, and operating time combine to determine the thermal energy produced by a resistor.