How to Calculate Heat Produced by a Resistor in a Given Time

A \(6.0\,\Omega\) resistor is connected to a \(12.0\,\text{V}\) power supply for \(30.0\,\text{s}\). How much heat does it produce?

One student calculates \(12.0\times30.0=360\,\text{J}\). Another calculates the current first, then uses the resistor’s resistance. Only the second approach gives the correct answer.

The missing quantity is current. The resistor carries \(I=V/R=12.0/6.0=2.0\,\text{A}\), so its power is \(24\,\text{W}\). Over 30 seconds, it transfers \(720\,\text{J}\) into thermal energy.

The key idea is that the heat energy produced by a resistor equals the electrical energy it converts into thermal energy. To calculate it, you need to know the power dissipated and how long the resistor operates.

01The formula for calculating heat produced by a resistor

Electrical power measures how quickly electrical energy is transferred.

Power is calculated using:

\[
P=\frac{E}{t}
\]

Rearranging gives the general energy equation:

\[
\boxed{E=Pt}
\]

where:

• \(E\) is the electrical energy converted into thermal energy, measured in joules (J).
• \(P\) is the electrical power dissipated, measured in watts (W).
• \(t\) is the operating time, measured in seconds (s).

Remember that \(1\,\text{W}=1\,\text{J\,s}^{-1}\). A resistor dissipating \(10\,\text{W}\) converts 10 joules of electrical energy into thermal energy every second.

For a resistor, electrical power is also given by:

\[
P=VI
\]

where \(V\) is the potential difference across the resistor in volts (V), and \(I\) is the current through it in amperes (A).

Substituting into the energy equation gives:

\[
\boxed{E=VIt}
\]

This is a useful starting point. However, you can simplify it further using Ohm’s law.

For an ohmic resistor:

\[
V=IR
\]

Substituting this relationship gives two alternative forms:

\[
\boxed{E=I^2Rt}
\]

and

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

Here, \(R\) is the resistance in ohms (\(\Omega\)).

All three expressions calculate the same electrical energy. The best one depends on which quantities are provided and which circuit conditions remain constant.

Which formula should you use?

For these simplified formulas, assume the resistor is ohmic and its resistance remains constant during the interval.

If resistance changes as the resistor heats up, you must account for the changing power. We’ll examine that shortly.

02Why fixed voltage and fixed current give different results

This is one of the most important distinctions when calculating electrical heating.

Imagine connecting the same metal resistor to two different power supplies:

• A regulated voltage supply keeps the potential difference constant.
• A regulated current supply keeps the current constant.

Both supplies deliver electrical energy, but the way power changes as the resistor heats up is different.

Case 1: Constant voltage

Suppose a resistor is connected to a supply that maintains a constant voltage across it.

Its power is:

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

If the resistance increases, the power decreases because voltage is fixed.

For an ordinary metal resistor, resistance generally increases as its temperature rises. Therefore, under constant voltage, the resistor draws less current as it heats up.

Case 2: Constant current

Now imagine that the supply maintains a constant current instead.

The power equation becomes:

\[
P=I^2R
\]

This time, increasing resistance increases power because current is fixed.

As the metal resistor heats up and its resistance increases, the voltage across it must increase to maintain the same current.

The difference can be summarised:

This is why you must identify the circuit conditions before choosing a formula.

A common mistake is to assume that a resistor with greater resistance always produces more heat. That is not true. It depends on whether voltage or current is being held constant.

For a deeper explanation of how temperature changes the resistance of a metal, read How Heating Changes a Metal Wire’s Resistance.

03What actually happens inside a resistor?

When current flows through a resistor, electrical energy is transferred into the material.

At the microscopic level, an electric field transfers energy to conduction electrons moving through the metal. Interactions between these electrons and the metal lattice transfer energy to lattice vibrations.

The increased internal energy raises the resistor’s temperature.

There is an important distinction here.

The electrons are not being used up as they travel through the resistor. Charge continues to flow around the circuit. What changes is the electrical energy transferred to the material.

Think of electrical energy as a quantity being transferred, rather than something the resistor creates from nothing.

The electrical supply does work on the charges, and the resistor converts that energy into thermal energy.

This explains why a resistor becomes warmer while current flows through it.

04An experimental way to understand resistor heating

How could you test whether the electrical energy calculated using \(E=VIt\) actually corresponds to the thermal energy gained by a resistor?

A calorimetry experiment provides a useful comparison.

The apparatus

Consider a resistor submerged in a known mass of water inside an insulated calorimeter.

The electrical circuit consists of:

• A regulated DC power supply.
• A switch connected in series with the resistor.
• An ammeter connected in series to measure current.
• A voltmeter connected in parallel across the resistor.

A thermometer is placed in the water without touching the resistor or the container’s sides.

A stopwatch measures how long the current flows.

What should happen?

Before switching on the supply, record the initial water temperature.

Then close the switch and allow current to flow for a measured time.

The resistor transfers electrical energy into thermal energy, which gradually warms the water and the surrounding apparatus.

If the resistor’s voltage and current remain constant, the electrical energy supplied is:

\[
E_{\text{electrical}}=VIt
\]

You can compare this with the thermal energy gained by the water:

\[
Q_{\text{water}}=mc\Delta T
\]

where:

• \(m\) is the mass of water in grams (g).
• \(c\) is the specific heat capacity of water, approximately \(4.18\,\text{J\,g}^{-1}\,^\circ\text{C}^{-1}\).
• \(\Delta T\) is the temperature increase in degrees Celsius.

If all electrical energy were transferred to the water, the two energy values would be equal.

In practice, they usually aren’t.

Why the first experimental design is incomplete

The water is not the only thing absorbing energy.

Some energy increases the temperature of the calorimeter, thermometer, and resistor itself. Some also escapes into the surrounding air.

There is another potential problem: as the resistor heats up, its resistance may change. If you use the initial voltage and current readings for the entire experiment, your calculated energy may be inaccurate.

Improving the experiment

A more reliable design would:

1. Insulate the calorimeter to reduce heat loss.
2. Include the heat capacity of the container, thermometer, and resistor in the thermal energy calculation.
3. Stir the water gently to maintain a more uniform temperature.
4. Record voltage and current throughout the experiment rather than relying on one initial measurement.
5. Calculate the electrical energy over short time intervals and add the results.

For the improved energy calculation:

\[
E_{\text{electrical}}\approx
\sum P_i\Delta t_i
\]

Here, \(P_i\) is the average power during an interval, and \(\Delta t_i\) is the duration of that interval.

The revised experiment accounts for changing power and provides a more meaningful comparison with the thermal energy gained by the apparatus.

The important conclusion is that the electrical energy calculation and the calorimetry measurement describe two sides of the same energy transfer. Any difference must be accounted for by energy stored elsewhere or transferred to the surroundings.

05When power changes during the experiment

So far, we’ve mostly considered a resistor operating at constant power.

But what happens when its resistance, current, or voltage changes during the experiment?

You cannot automatically multiply the initial power by the total operating time.

Instead, divide the experiment into intervals during which the power is approximately constant.

For example, suppose a resistor dissipates:

• \(20\,\text{W}\) for the first 30 seconds.
• \(15\,\text{W}\) for the next 30 seconds.

The total energy is:

\[
\begin{aligned}
E_1&=20\times30=600\,\text{J}\\
E_2&=15\times30=450\,\text{J}\\
E_{\text{total}}&=600+450\\
&=1050\,\text{J}
\end{aligned}
\]

You can also picture this as the area under a power-time graph.

This gives you a useful general rule:

When power is constant, multiply power by time. When power changes, calculate the energy for each interval and add the results.

For a continuously changing power, the total energy is the area under the power-time curve. At HSC level, questions can often be handled using rectangular or other familiar graph areas, or by adding energy over specified intervals.

06Worked examples

07A reliable method for HSC exam questions

When you encounter a question asking you to calculate heat produced by a resistor, use this sequence.

1. Identify the known quantities: voltage, current, resistance, and time.
2. Identify whether voltage or current is constant.
3. Check whether resistance is assumed constant or changes during the interval.
4. Choose the appropriate energy equation.
5. Convert time into seconds.
6. Substitute values with their units.
7. Check whether the result is physically reasonable.

A few common mistakes are worth avoiding:

08Questions and solutions

The following questions build from direct energy calculations to more challenging applications involving changing resistance, experimental measurements, and circuit behaviour.

Question 1

A \(15.0\,\Omega\) resistor is connected across a constant \(9.0\,\text{V}\) supply for \(2.00\) minutes.

Calculate the electrical energy converted into thermal energy.

Solution 1

The resistor converts 648 J of electrical energy into thermal energy.

Since voltage and resistance are given, use:

\[
E=\frac{V^2t}{R}
\]

First, convert the time:

\[
t=2.00\times60=120\,\text{s}
\]

Substitute:

\[
\begin{aligned}
E&=\frac{(9.0)^2(120)}{15.0}\\
&=\frac{81\times120}{15.0}\\
&=648\,\text{J}
\end{aligned}
\]

Therefore:

\[
\boxed{E=648\,\text{J}}
\]

The resistor dissipates \(648\,\text{J}\) over the two-minute interval.

The important detail is that the supply maintains constant voltage, making \(E=V^2t/R\) the most direct method.

Question 2

A \(5.0\,\Omega\) resistor operates under two different supply conditions.

• During the first 40 seconds, it is connected to a constant \(10.0\,\text{V}\) supply.
• During the next 50 seconds, it is connected to a constant-current supply delivering \(1.00\,\text{A}\).

Assume the resistance remains constant.

Calculate:

(a) The energy dissipated during the first interval.

(b) The energy dissipated during the second interval.

(c) The total energy dissipated over the 90 seconds.

Explain why multiplying the initial power by the full 90 seconds would give an incorrect answer.

Solution 2

The total energy dissipated is 1050 J. The two intervals must be calculated separately because the supply conditions change.

(a) First interval

Voltage and resistance are constant.

Use:

\[
E=\frac{V^2t}{R}
\]

Substitute:

\[
\begin{aligned}
E_1&=\frac{(10.0)^2(40)}{5.0}\\
&=\frac{4000}{5.0}\\
&=800\,\text{J}
\end{aligned}
\]

Therefore:

\[
\boxed{E_1=800\,\text{J}}
\]

(b) Second interval

The current is now constant.

Use:

\[
E=I^2Rt
\]

Substitute:

\[
\begin{aligned}
E_2&=(1.00)^2(5.0)(50)\\
&=250\,\text{J}
\end{aligned}
\]

Therefore:

\[
\boxed{E_2=250\,\text{J}}
\]

(c) Total energy

Add the energy from both intervals:

\[
\begin{aligned}
E_{\text{total}}&=E_1+E_2\\
&=800+250\\
&=1050\,\text{J}
\end{aligned}
\]

\[
\boxed{E_{\text{total}}=1050\,\text{J}}
\]

The initial power is:

\[
P_1=\frac{10.0^2}{5.0}=20\,\text{W}
\]

However, the second interval has a different current and therefore a different power:

\[
P_2=(1.00)^2(5.0)=5.0\,\text{W}
\]

Using the initial power for all 90 seconds would give:

\[
E=20(90)=1800\,\text{J}
\]

This is incorrect because the resistor does not dissipate \(20\,\text{W}\) throughout the experiment.

The supply conditions must be considered separately.

Question 3

A \(4.0\,\Omega\) heating resistor is submerged in \(100.0\,\text{g}\) of water inside a calorimeter.

A constant \(8.0\,\text{V}\) supply operates the resistor for \(100\,\text{s}\).

The water temperature increases from \(20.0^\circ\text{C}\) to \(22.5^\circ\text{C}\).

Use the specific heat capacity of water:

\[
c=4.18\,\text{J\,g}^{-1}\,^\circ\text{C}^{-1}
\]

Calculate:

(a) The electrical energy supplied to the resistor.

(b) The thermal energy gained by the water.

(c) The percentage of the electrical energy accounted for by the water’s temperature increase.

(d) Explain why the two energy values are different.

Solution 3

The electrical energy supplied is 1600 J, while the water gains 1045 J. Approximately 65.3% of the electrical energy is accounted for by the water’s temperature increase.

(a) Electrical energy

Use:

\[
E=\frac{V^2t}{R}
\]

Substitute:

\[
\begin{aligned}
E&=\frac{(8.0)^2(100)}{4.0}\\
&=\frac{64\times100}{4.0}\\
&=1600\,\text{J}
\end{aligned}
\]

Therefore:

\[
\boxed{E_{\text{electrical}}=1600\,\text{J}}
\]

(b) Thermal energy gained by the water

First, calculate the temperature change:

\[
\Delta T=22.5-20.0=2.5^\circ\text{C}
\]

Use the calorimetry equation:

\[
Q=mc\Delta T
\]

Substitute:

\[
\begin{aligned}
Q&=(100.0)(4.18)(2.5)\\
&=1045\,\text{J}
\end{aligned}
\]

Therefore:

\[
\boxed{Q_{\text{water}}=1045\,\text{J}}
\]

(c) Percentage accounted for by the water

Divide the energy gained by the water by the electrical energy supplied, then multiply by 100.

\[
\begin{aligned}
\text{Percentage}
&=\frac{1045}{1600}\times100\\
&=65.3125\%
\end{aligned}
\]

Therefore:

\[
\boxed{65.3\%}
\]

(d) Explain the difference

The remaining \(555\,\text{J}\) is not necessarily missing energy.

Some energy increases the temperature of the calorimeter, thermometer, and resistor itself. Some is also transferred to the surrounding environment.

The calculation of electrical energy accounts for the energy supplied to the resistor, whereas the water calculation measures only the energy gained by the water.

This is an important experimental distinction: the water’s temperature increase does not necessarily measure all the energy transferred by the resistor.

Question 4

A metal resistor is connected to a regulated \(12.0\,\text{V}\) supply.

As the resistor heats up, a data logger records the following current readings:

Assume the voltage remains constant and the current is approximately constant within each interval.

(a) Calculate the total electrical energy dissipated over the 60 seconds.

(b) Calculate the resistance of the resistor during each interval.

(c) A student uses the initial power to calculate the energy for the entire 60 seconds. Calculate their result and explain why it is incorrect.

Solution 4

The correct total electrical energy is 576 J. The resistance increases from \(12\,\Omega\) to \(20\,\Omega\), causing the power to decrease under the constant-voltage condition.

(a) Calculate the energy in each interval

Use:

\[
E=VIt
\]

For the first 20 seconds:

\[
\begin{aligned}
E_1&=12.0(1.00)(20)\\
&=240\,\text{J}
\end{aligned}
\]

For the second 20 seconds:

\[
\begin{aligned}
E_2&=12.0(0.80)(20)\\
&=192\,\text{J}
\end{aligned}
\]

For the final 20 seconds:

\[
\begin{aligned}
E_3&=12.0(0.60)(20)\\
&=144\,\text{J}
\end{aligned}
\]

Add the three intervals:

\[
\begin{aligned}
E_{\text{total}}&=240+192+144\\
&=576\,\text{J}
\end{aligned}
\]

\[
\boxed{E_{\text{total}}=576\,\text{J}}
\]

(b) Calculate resistance during each interval

Use Ohm’s law:

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

First interval:

\[
R_1=\frac{12.0}{1.00}=12.0\,\Omega
\]

Second interval:

\[
R_2=\frac{12.0}{0.80}=15.0\,\Omega
\]

Third interval:

\[
R_3=\frac{12.0}{0.60}=20.0\,\Omega
\]

Therefore:

The increasing resistance is consistent with the behaviour of an ordinary metal resistor as its temperature rises.

(c) Assess the student’s calculation

The initial power is:

\[
P_1=VI
\]

\[
P_1=12.0(1.00)=12.0\,\text{W}
\]

If the student assumes this power remains constant for 60 seconds:

\[
\begin{aligned}
E&=Pt\\
&=12.0(60)\\
&=720\,\text{J}
\end{aligned}
\]

The student calculates \(720\,\text{J}\), which is \(144\,\text{J}\) greater than the correct result.

The mistake is assuming that current remains constant. The supply maintains constant voltage, not constant current.

As the resistor heats up, its resistance increases and the current decreases. Consequently, its power also decreases.

This question tests whether you can interpret changing circuit measurements rather than simply substitute the first available readings.

Question 5

Two identical metal resistors, A and B, each initially have a resistance of \(10.0\,\Omega\) at the same temperature.

Resistor A is then warmed externally until its resistance increases to \(12.0\,\Omega\). Resistor B remains at \(10.0\,\Omega\).

The resistors are connected to an ideal \(12.0\,\text{V}\) supply in two separate configurations:

• Configuration 1: A and B are connected in series.
• Configuration 2: A and B are connected in parallel.

Assume both resistors are ohmic at their respective temperatures, have positive temperature coefficients, and have negligible connecting-wire resistance.

(a) For the series configuration, calculate the circuit current and the power dissipated by each resistor.

(b) For the parallel configuration, calculate the current and power dissipated by each resistor, as well as the total circuit current.

(c) Before resistor A was heated, both resistors had equal resistance. Calculate the power dissipated by each resistor in the original equal-temperature series and parallel arrangements.

(d) Using your results, explain which circuit configuration creates electrical feedback that tends to reinforce the temperature difference and which tends to oppose it.

(e) Does the series configuration necessarily cause resistor A to undergo thermal runaway? Explain why the electrical power calculations alone cannot establish this.

Solution 5

The series configuration directs more electrical power into the warmer resistor relative to the cooler resistor, whereas the parallel configuration directs less power into it. However, neither result alone proves that the resistor will undergo thermal runaway.

This problem combines Ohm’s law, electrical power, circuit arrangements, and temperature-dependent resistance.

(a) Series configuration

The total resistance is the sum of the two resistances:

\[
\begin{aligned}
R_{\text{total}}&=R_A+R_B\\
&=12.0+10.0\\
&=22.0\,\Omega
\end{aligned}
\]

The same current flows through both resistors in series.

Using Ohm’s law:

\[
\begin{aligned}
I&=\frac{V}{R_{\text{total}}}\\
&=\frac{12.0}{22.0}\\
&=0.545\,\text{A}
\end{aligned}
\]

Now calculate the power dissipated by resistor A:

\[
\begin{aligned}
P_A&=I^2R_A\\
&=(0.545)^2(12.0)\\
&\approx3.57\,\text{W}
\end{aligned}
\]

For resistor B:

\[
\begin{aligned}
P_B&=I^2R_B\\
&=(0.545)^2(10.0)\\
&\approx2.98\,\text{W}
\end{aligned}
\]

Therefore:

\[
\boxed{P_A=3.57\,\text{W}}
\]

\[
\boxed{P_B=2.98\,\text{W}}
\]

The warmer resistor dissipates more power than the cooler resistor.

(b) Parallel configuration

In a parallel circuit, both resistors have the same potential difference across them.

Therefore:

\[
V_A=V_B=12.0\,\text{V}
\]

Calculate the current through resistor A:

\[
\begin{aligned}
I_A&=\frac{V}{R_A}\\
&=\frac{12.0}{12.0}\\
&=1.00\,\text{A}
\end{aligned}
\]

Calculate the current through resistor B:

\[
\begin{aligned}
I_B&=\frac{V}{R_B}\\
&=\frac{12.0}{10.0}\\
&=1.20\,\text{A}
\end{aligned}
\]

The total current is the sum of the branch currents:

\[
\begin{aligned}
I_{\text{total}}&=I_A+I_B\\
&=1.00+1.20\\
&=2.20\,\text{A}
\end{aligned}
\]

Now calculate the power in each resistor using \(P=V^2/R\).

For resistor A:

\[
P_A=\frac{12.0^2}{12.0}
\]

\[
\boxed{P_A=12.0\,\text{W}}
\]

For resistor B:

\[
P_B=\frac{12.0^2}{10.0}
\]

\[
\boxed{P_B=14.4\,\text{W}}
\]

The warmer resistor now dissipates less power than the cooler resistor.

(c) Compare with the original equal-temperature arrangement

Initially, both resistors have a resistance of \(10.0\,\Omega\).

For the series arrangement:

\[
R_{\text{total}}=10.0+10.0=20.0\,\Omega
\]

The current is:

\[
I=\frac{12.0}{20.0}=0.600\,\text{A}
\]

Each resistor dissipates:

\[
\begin{aligned}
P&=I^2R\\
&=(0.600)^2(10.0)\\
&=3.60\,\text{W}
\end{aligned}
\]

For the parallel arrangement, each resistor receives \(12.0\,\text{V}\).

Therefore:

\[
P=\frac{12.0^2}{10.0}
\]

\[
P=14.4\,\text{W}
\]

The comparison is:

Here is the subtle part.

In the series circuit, resistor A dissipates more power than B, but its power is actually slightly lower than its original \(3.60\,\text{W}\). The increase in total resistance has reduced the circuit current.

In the parallel circuit, A’s power falls from \(14.4\,\text{W}\) to \(12.0\,\text{W}\), while B’s power remains at \(14.4\,\text{W}\).

So even though the total electrical power supplied by the source decreases in both arrangements, the distribution of power between the resistors is different.

(d) Explain the electrical feedback

In the series configuration:

\[
P=I^2R
\]

Both resistors have the same current. Therefore, the resistor with greater resistance dissipates more power.

The warmer resistor A receives more electrical power than the cooler resistor B. This creates positive electrical feedback: the electrical heating tends to reinforce the existing temperature difference.

In the parallel configuration:

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

Both resistors have the same voltage. The resistor with greater resistance dissipates less power.

The warmer resistor A receives less electrical power than the cooler resistor B. This creates negative electrical feedback: the electrical heating tends to oppose the existing temperature difference.

The crucial distinction is how the circuit distributes electrical energy between its components.

(e) Does series connection guarantee thermal runaway?

No. The electrical feedback alone cannot establish thermal runaway.

The power calculations tell us how quickly electrical energy is supplied to each resistor. They do not tell us how quickly each resistor loses thermal energy to its surroundings.

A more complete thermal model would consider:

\[
P_{\text{net}}=
P_{\text{electrical}}-
P_{\text{thermal loss}}
\]

The temperature change also depends on each resistor’s heat capacity.

For example, the warmer resistor may dissipate more electrical power than the cooler one in the series arrangement, but it may also lose more energy to its surroundings because of its higher temperature.

Whether the temperature difference actually grows depends on the balance between electrical heating and thermal losses.

This is the important limitation: positive electrical feedback does not automatically mean unlimited temperature increase or thermal runaway.

The problem requires you to distinguish the electrical behaviour of a circuit from the complete thermal behaviour of the physical resistors.

09Where to go next

Once you can calculate the energy dissipated by a resistor, the next useful step is understanding how temperature changes its resistance.

That connection explains why the power of a heating element can change during operation, even when its supply voltage remains constant.

Read How Heating Changes a Metal Wire’s Resistance | HSC Physics to build that link between electrical energy, resistance, and temperature.