Which Resistor Gets Hotter in a Parallel Circuit?

A 4 Ω resistor and a 12 Ω resistor are connected in parallel across a 12 V battery. Both resistors have the same voltage across them. So which one gets hotter?

It is tempting to pick the 12 Ω resistor because it has greater resistance. But that prediction is wrong for this circuit.

The 4 Ω resistor dissipates three times as much electrical power as the 12 Ω resistor. If both resistors have the same physical construction and cooling conditions, the 4 Ω resistor will generally heat up faster and reach a higher temperature.

The key is understanding what happens to electrical power when the voltage across each resistor is fixed.

01Which resistor gets hotter in a parallel circuit?

For two resistors connected in parallel across an ideal voltage supply:

• Both resistors have the same voltage across them.
• The resistor with lower resistance draws more current.
• The resistor with lower resistance dissipates more electrical power.

The relationship is:

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

where:

• \(P\) is electrical power, measured in watts (W).
• \(V\) is the voltage across the resistor, measured in volts (V).
• \(R\) is the resistance, measured in ohms (Ω).

Notice that resistance is in the denominator. If voltage stays constant, increasing resistance decreases power.

Why does the lower-resistance resistor dissipate more power?

Imagine two water pipes connected between the same two points in a plumbing system. Both experience the same pressure difference, but the wider pipe allows more water to flow.

A similar relationship exists in a parallel circuit.

Both resistors experience the same potential difference. However, the lower-resistance branch allows a greater current to flow.

Electrical power depends on both voltage and current:

\[ P=VI \]

Using Ohm’s law, \(V=IR\), we can rewrite this as:

\[ P=I^2R \]

Or, substituting \(I=V/R\):

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

These are three forms of the same power equation. The correct one to use depends on what the question tells you is held constant.

The water-pipe analogy helps explain why the currents differ, but it has limits. Electric charge does not flow through a circuit in exactly the same way as water through pipes. The equations describe the electrical relationship precisely.

02A numerical comparison: power in parallel resistors

Consider two resistors connected across an ideal 12 V supply.

For Resistor A:

\[ P_A=\frac{12^2}{4} \]

\[ P_A=36\text{ W} \]

For Resistor B:

\[ P_B=\frac{12^2}{12} \]

\[ P_B=12\text{ W} \]

Therefore:

\[ \frac{P_A}{P_B}=\frac{36}{12}=3 \]

Resistor A converts electrical energy into other forms of energy at three times the rate of Resistor B.

If both resistors have the same thermal properties and cooling conditions, Resistor A will generally become hotter.

Does three times the power mean three times the temperature?

No. This is an important distinction.

Power tells us how quickly electrical energy is transferred, not the exact temperature a resistor will reach.

Temperature also depends on:

• The mass and material of the resistor.
• Its specific heat capacity.
• How effectively it loses heat to its surroundings.
• Its surface area and physical construction.
• How long the circuit operates.

For example, a small resistor dissipating 3 W may become much hotter than a large resistor dissipating 6 W if the smaller resistor has a much lower ability to absorb and release heat.

At thermal equilibrium, the rate of heat loss balances the rate of electrical energy input. Different resistors can therefore reach different temperatures even when their electrical power is known.

03What if the current is fixed instead?

This is where many students make mistakes.

You might remember that electrical power can also be calculated using:

\[ P=I^2R \]

This equation appears to suggest that a higher resistance produces more power.

That is correct when the current through each resistor is held constant.

The important question is: what quantity is fixed?

The two results are not contradictory. They describe different conditions.

Fixed current: the series circuit comparison

Imagine a 2 Ω resistor and a 6 Ω resistor connected in series to a current source that maintains a constant current of 2 A.

The same current flows through both resistors.

For the 2 Ω resistor:

\[ P_1=I^2R_1 \]

\[ P_1=2^2\times2=8\text{ W} \]

For the 6 Ω resistor:

\[ P_2=I^2R_2 \]

\[ P_2=2^2\times6=24\text{ W} \]

The 6 Ω resistor dissipates three times as much power.

This is the opposite of the fixed-voltage parallel example.

However, be careful with one subtle point: a parallel circuit connected to a fixed-total-current source does not have the same current through every resistor. The current divides between the branches.

For two parallel resistors, the lower-resistance branch still dissipates more power when the total current is fixed, assuming ideal components. The source voltage adjusts to maintain the total current.

So the rule is not simply that parallel means lower resistance gets hotter and series means higher resistance gets hotter. You must identify the actual electrical constraint.

For a more complete treatment of branch currents and equivalent resistance, see Parallel Circuits: Voltage, Current, Resistance and Power.

06What happens inside a resistor when it heats up?

So far, we’ve treated resistance as a fixed number. That is useful for solving HSC circuit questions, but a real resistor changes as its temperature changes.

To understand why, consider what happens inside a metal conductor.

A metal contains mobile electrons that drift through a lattice of positive ions. When a potential difference is applied, an electric field causes the electrons to drift.

As they move, electrons interact with the lattice and transfer energy to it. These interactions contribute to the heating of the resistor.

As the metal heats up, its lattice vibrations generally become stronger. This makes it harder for conduction electrons to move through the material without scattering.

For an ordinary metallic resistor, the resistance generally increases with temperature.

This is called a positive temperature coefficient of resistance.

For a limited temperature range, the relationship can be approximated by:

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

where:

• \(R\) is the final resistance in ohms.
• \(R_0\) is the initial resistance in ohms.
• \(\alpha\) is the temperature coefficient of resistance in \(\text{°C}^{-1}\).
• \(\Delta T\) is the temperature change in °C.

This relationship is an approximation, not a universal law for every resistor or every temperature.

Does this change which resistor gets hotter?

It can affect the final temperature, but it does not reverse the initial power comparison for two ordinary resistors with the same fixed voltage and constant resistance assumptions.

For a metallic resistor connected to a constant-voltage supply, an increase in temperature causes its resistance to increase.

Since:

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

its electrical power then decreases.

This provides a form of negative feedback: as the resistor heats up, its increasing resistance reduces the power supplied to it.

However, this does not mean the resistor immediately cools down. Its temperature depends on the balance between electrical heating and heat loss to its surroundings.

The actual behaviour of a circuit also depends on the supply and other components. For more detail, see How Heating Changes a Metal Wire’s Resistance.

07A practical experiment: testing which resistor heats up more

You can investigate the prediction using two resistors with different resistance values.

Connect them in parallel across a low-voltage, regulated DC supply. Measure the voltage across each resistor and the current through each branch.

For example, suppose the measured values are:

The initial power calculations predict that Resistor A receives electrical energy at twice the rate of Resistor B.

To investigate temperature, attach suitable temperature sensors to the two resistors. Keep their surroundings, mounting, airflow, and initial temperatures as similar as possible.

Record the temperature of each resistor at regular time intervals.

What should the experiment show?

Initially, Resistor A should dissipate twice as much electrical power.

If both resistors have comparable thermal properties, Resistor A should generally heat up faster.

However, the experiment may not produce a temperature ratio of 2:1.

Why?

Because electrical power and temperature are different physical quantities. The resistors may have different thermal masses, surface areas, and heat-loss rates.

This is an important experimental distinction:

• Calculated power is determined by the electrical measurements.
• Measured temperature depends on both electrical heating and thermal behaviour.

A difference between the measured temperatures and the ideal prediction does not automatically mean Ohm’s law is wrong. It may indicate that the assumptions about the components or their surroundings were too simple.

For more on this distinction, read Why Circuit Practical Results Differ from Ideal Calculations.

08Questions and solutions

Question 1

A 3 Ω resistor and a 9 Ω resistor are connected in parallel across a 12 V ideal battery.

Calculate the power dissipated by each resistor and identify which dissipates more power.

Solution 1

The 3 Ω resistor dissipates 48 W, while the 9 Ω resistor dissipates 16 W. Therefore, the 3 Ω resistor dissipates three times as much power.

Because the resistors are connected in parallel, each has the full 12 V supply voltage.

Using:

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

For the 3 Ω resistor:

\[ P_1=\frac{12^2}{3} =48\text{ W} \]

For the 9 Ω resistor:

\[ P_2=\frac{12^2}{9} =16\text{ W} \]

The power ratio is:

\[ \frac{P_1}{P_2}=\frac{48}{16}=3 \]

The lower-resistance resistor dissipates more power because both resistors have the same voltage across them.

Question 2

Two resistors, 4 Ω and 12 Ω, are connected in parallel across an ideal 12 V supply.

The supply is replaced with an ideal constant-total-current source supplying 6 A.

Calculate:

a) The equivalent resistance of the parallel combination.

b) The voltage across each resistor.

c) The current through each resistor.

d) The power dissipated by each resistor.

Explain whether the lower-resistance resistor still dissipates more power.

Solution 2

The 4 Ω resistor dissipates 48 W, while the 12 Ω resistor dissipates 16 W. The lower-resistance resistor still dissipates three times as much power, even though the total current rather than the voltage is fixed.

Step 1: Calculate the equivalent resistance.

For two resistors in parallel:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1}+\frac{1}{R_2} \]

Substituting:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{4}+\frac{1}{12} \]

\[ \frac{1}{R_{\text{eq}}} = \frac{4}{12} = \frac{1}{3} \]

Therefore:

\[ R_{\text{eq}}=3\ \Omega \]

Step 2: Calculate the supply voltage.

Using Ohm’s law for the entire parallel combination:

\[ V=I_{\text{total}}R_{\text{eq}} \]

\[ V=6\times3=18\text{ V} \]

Both resistors have 18 V across them.

Step 3: Calculate the branch currents.

For the 4 Ω resistor:

\[ I_1=\frac{18}{4}=4.5\text{ A} \]

For the 12 Ω resistor:

\[ I_2=\frac{18}{12}=1.5\text{ A} \]

Check using Kirchhoff’s current law:

\[ I_{\text{total}}=4.5+1.5 =6.0\text{ A} \]

The branch currents add to the total current supplied.

Step 4: Calculate the power dissipated.

For the 4 Ω resistor:

\[ P_1=\frac{18^2}{4}=81\text{ W} \]

For the 12 Ω resistor:

\[ P_2=\frac{18^2}{12}=27\text{ W} \]

The ratio is:

\[ \frac{P_1}{P_2}=3 \]

The 4 Ω resistor still dissipates three times as much power.

The tempting mistake is to assume that fixing the total current means the current through each branch is fixed. It does not. The lower-resistance branch takes a greater share of the current.

Question 3

A 5 Ω resistor and a 15 Ω resistor are connected in parallel across an ideal 10 V supply.

Both resistors initially have a resistance independent of temperature.

After several minutes, the temperature of the 5 Ω resistor rises significantly. Assume it is an ordinary metallic resistor with a positive temperature coefficient.

Explain:

a) Why the 5 Ω resistor initially dissipates more power.

b) How its increasing temperature affects its resistance.

c) What happens to its electrical power if the supply voltage remains constant.

d) Why you cannot determine the final temperature from the initial resistance values alone.

Solution 3

The 5 Ω resistor initially dissipates three times as much power. As it heats up, its resistance increases, reducing its electrical power at constant voltage. Its final temperature cannot be determined from the initial resistance values alone.

Step 1: Calculate the initial power ratio.

Both resistors have the same voltage across them.

\[ P_1=\frac{10^2}{5}=20\text{ W} \]

\[ P_2=\frac{10^2}{15} =6.67\text{ W} \]

Therefore:

\[ \frac{P_1}{P_2}=3 \]

The 5 Ω resistor initially dissipates three times as much power.

Step 2: Explain the temperature effect on resistance.

For an ordinary metallic resistor, increasing temperature generally increases lattice vibrations.

These stronger vibrations increase electron scattering, making electrical conduction less effective.

The resistance therefore increases.

Step 3: Determine the effect on power.

The supply voltage remains constant, so:

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

As resistance increases, power decreases.

The resistor therefore receives less electrical power than it did initially.

Step 4: Explain why the final temperature is unknown.

The final temperature depends on the balance between electrical power input and heat loss to the surroundings.

We would need additional information, such as the resistor’s thermal properties, temperature coefficient, and cooling conditions.

The initial resistance and supply voltage are not enough to calculate the final temperature.

Question 4

Two resistors, A and B, are connected in parallel across a constant-voltage supply.

Initially:

• Resistor A has a resistance of 8 Ω.
• Resistor B has a resistance of 16 Ω.

After operating for some time, the resistance of A increases to 12 Ω due to heating. Assume B remains at 16 Ω.

The supply voltage remains constant at 12 V.

Calculate:

a) The initial power dissipated by each resistor.

b) The new power dissipated by each resistor.

c) The percentage decrease in A’s power.

d) Whether A still dissipates more power than B.

Solution 4

After heating, Resistor A dissipates 12 W and Resistor B dissipates 9 W. A still dissipates more power, but its power advantage has decreased.

Step 1: Calculate the initial powers.

For A:

\[ P_A=\frac{12^2}{8} =18\text{ W} \]

For B:

\[ P_B=\frac{12^2}{16} =9\text{ W} \]

Initially, A dissipates twice as much power as B.

Step 2: Calculate the new powers.

The voltage is still 12 V.

For A:

\[ P_A=\frac{12^2}{12} =12\text{ W} \]

For B:

\[ P_B=\frac{12^2}{16} =9\text{ W} \]

Step 3: Calculate the percentage decrease in A’s power.

The original power was 18 W, and the new power is 12 W.

\[ \%\text{ decrease} = \frac{\text{original}-\text{new}} {\text{original}}\times100 \]

\[ = \frac{18-12}{18}\times100 \]

\[ =33.3\% \]

Step 4: Compare the new powers.

\[ \frac{P_A}{P_B}=\frac{12}{9}=1.33 \]

A still dissipates more power than B.

The important detail is that the power comparison changes as the resistance of A changes. It would be incorrect to keep using its original 8 Ω resistance after the resistor has heated up.

Question 5

Two identical metallic resistors, A and B, are connected in parallel across an ideal constant-voltage supply.

Initially, A has half the resistance of B.

Both resistors begin at the same temperature. Their thermal masses and cooling conditions are identical.

A student argues:

“Because A has half the resistance, it dissipates twice the power. Therefore, it must always be hotter than B. Even if its resistance increases as it heats, the same relationship will continue.”

Is the student’s conclusion necessarily correct?

Explain what happens during the heating process. Your answer must distinguish between:

• Initial electrical power.
• The changing resistance of each resistor.
• The rate of temperature increase.
• The possibility of thermal equilibrium.

Assume both resistors have the same positive temperature coefficient and that their resistance increases approximately linearly with temperature over the relevant range.

Solution 5

The student is correct about the initial power comparison but incorrect in claiming that A must always remain hotter and dissipate twice the power.

The crucial detail is that both resistors change their resistance as their temperatures change. Their electrical power is therefore not permanently fixed at the initial 2:1 ratio.

Step 1: Establish the initial power relationship.

Let the initial resistance of B be \(R\).

Then the initial resistance of A is:

\[ R_A=\frac{R}{2} \]

Both resistors have the same voltage, \(V\).

For A:

\[ P_A=\frac{V^2}{R/2} =\frac{2V^2}{R} \]

For B:

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

Therefore:

\[ P_A=2P_B \]

Initially, A dissipates twice as much power as B.

Because their thermal masses and initial temperatures are identical, A initially heats up faster.

Step 2: Consider what happens as A heats up.

A has the greater initial power, so its temperature begins increasing more rapidly.

Its metallic lattice vibrates more strongly, increasing electron scattering.

Its resistance therefore increases.

Since the supply voltage remains constant:

\[ P_A=\frac{V^2}{R_A} \]

An increase in \(R_A\) causes its power to decrease.

The original 2:1 power ratio no longer necessarily applies.

Step 3: Consider B’s temperature.

B also dissipates electrical power, so it heats up as well.

As its temperature increases, its resistance also increases.

Its electrical power therefore decreases.

The two resistors are both experiencing negative feedback: increased temperature causes increased resistance, which reduces electrical heating.

Step 4: Determine what happens to their temperature difference.

The temperature difference depends on the balance between electrical heating and heat loss.

A initially heats faster, but its increasing resistance reduces its power input.

B also heats, and its resistance increases in response.

The temperature difference may become smaller as the system approaches equilibrium.

Under the stated assumptions, the resistors have identical thermal properties, cooling conditions, and temperature-dependent resistance behaviour apart from their initial resistance scale.

Consequently, A retains twice the resistance-dependent power relationship in the ideal linear model only when their temperatures are equal. Once their temperatures differ, the ratio changes.

Step 5: Test the possibility of equal temperatures.

Suppose both resistors reach the same temperature.

Because they have the same positive temperature coefficient, their resistance ratio remains:

\[ \frac{R_A}{R_B}=\frac{1}{2} \]

Their power ratio is therefore still:

\[ \frac{P_A}{P_B} = \frac{R_B}{R_A} =2 \]

At the same temperature, A still receives twice the electrical power.

Since the resistors have identical thermal properties and cooling conditions, they would also lose heat at the same rate at the same temperature.

They cannot both be at the same thermal equilibrium temperature while receiving different electrical powers.

Therefore, the system cannot reach an equal-temperature equilibrium under these assumptions.

A must remain hotter at thermal equilibrium.

However, the student’s original explanation is still incomplete. The final temperature difference and the final power ratio require a thermal model. The initial 2:1 power ratio alone cannot determine either value.

The deeper lesson: electrical power determines the rate of energy input, but temperature depends on the entire heating and cooling process. A correct circuit calculation is only the starting point when the question asks what happens to a real resistor over time.

09The key decision to remember

When comparing resistors, identify the electrical condition before choosing a power equation.

For HSC questions, remember the distinction between electrical power and temperature. A resistor dissipating more power generally heats faster under comparable conditions, but its actual temperature also depends on its physical properties and heat loss.

The next useful step is learning how to calculate the energy transferred as heat over a given time. See How to Calculate Heat Produced by a Resistor in a Given Time.