Does Higher Resistance Mean More Heating? It Depends

You write \(P=I^2R\), so you conclude that a higher-resistance resistor must produce more heating. Then you use \(P=V^2/R\), which seems to say exactly the opposite. If you have reached both conclusions in the same exam question, the algebra probably isn’t your problem. You’ve changed what is being held constant without noticing.

Predict what happens in these two tests. Assume the resistors are ohmic, their resistances remain constant, and all sources are ideal.

• Test A: Connect a \(6\,\Omega\) resistor and a \(12\,\Omega\) resistor, one at a time, across the same \(12\,\text{V}\) voltage source. Which dissipates more power?
• Test B: Use an ideal current source to drive \(2.0\,\text{A}\) through each resistor in separate tests. Which dissipates more power now?

In Test A, the \(6\,\Omega\) resistor dissipates \(24\,\text{W}\), while the \(12\,\Omega\) resistor dissipates \(12\,\text{W}\). The lower resistance produces more power.

In Test B, the \(6\,\Omega\) resistor dissipates \(24\,\text{W}\), while the \(12\,\Omega\) resistor dissipates \(48\,\text{W}\). The higher resistance produces more power.

Both results are correct. The difference is what the source holds constant.

01The diagnosis: you changed the constraint

Electrical power tells you how quickly electrical energy is transferred. For a resistor, power is calculated using

\[
P=VI
\]

where:

• \(P\) is power, measured in watts (\(\text{W}\)).
• \(V\) is the potential difference across the resistor, measured in volts (\(\text{V}\)).
• \(I\) is the current through the resistor, measured in amperes (\(\text{A}\)).

Resistance, \(R\), is measured in ohms (\(\Omega\)).

The two familiar power equations come from substituting Ohm’s law, \(V=IR\), into \(P=VI\).

If you replace \(V\) with \(IR\), you get

\[
P=I(IR)=I^2R
\]

If you replace \(I\) with \(V/R\), you get

\[
P=V\left(\frac{V}{R}\right)
=\frac{V^2}{R}
\]

These are equivalent forms of the same relationship for an ohmic resistor. They are not competing laws.

The catch is that the two forms make different comparisons particularly easy.

Case 1: Current is held constant

Suppose two resistors carry exactly the same current. From

\[
P=I^2R
\]

the current is unchanged, so \(I^2\) is constant. Power is therefore directly proportional to resistance.

If resistance doubles, power doubles. If resistance triples, power triples.

This is what happens when comparing resistors carrying the same current, such as two resistors in a series circuit at the same operating point.

Case 2: Voltage is held constant

Now suppose each resistor is connected separately across the same ideal voltage source. The voltage across each resistor remains constant.

Using

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

the numerator is unchanged. Power is therefore inversely proportional to resistance.

If resistance doubles, power halves. If resistance triples, power becomes one-third of its original value.

Why? At a fixed voltage, increasing resistance reduces the current. Less charge flows through the resistor each second, so less electrical energy is transferred each second.

A useful way to picture the relationship is to think of voltage as a pressure difference, current as flow, and resistance as a restriction. At a fixed pressure difference, a larger restriction reduces the flow. If you insist on maintaining the same flow through a larger restriction, a greater pressure difference is needed.

The analogy has limits, but it captures the key point: you cannot change resistance and assume both voltage and current will stay unchanged.

One further distinction matters. Power is the rate of energy transfer, not automatically the final temperature of a resistor. Actual temperature also depends on how long it operates, how effectively it loses heat, and how its resistance changes with temperature. In HSC calculations, use the stated resistance and idealised conditions unless the question tells you otherwise.

02A reliable exam method

Before choosing a power equation, use this sequence.

Step 1: Identify the resistor.

Make sure you know which component’s power the question asks you to calculate. A source voltage or the voltage across a neighbouring resistor is not necessarily the voltage across your resistor.

Step 2: Identify what is fixed.

Look for the actual constraint:

• Same current through the resistor? Start with \(P=I^2R\).
• Same voltage across the resistor? Start with \(P=V^2/R\).
• Neither fixed or known? Solve the circuit first.

Step 3: Check the circuit, not just the source label.

A \(12\,\text{V}\) source does not mean that every resistor in the circuit has \(12\,\text{V}\) across it. In a series circuit, the source voltage is shared between the components.

Likewise, if you increase a resistor in a series circuit connected to a fixed-voltage source, the total resistance changes, so the current changes too.

Step 4: Calculate using the actual values.

Once you know the current and voltage for the resistor, use whichever form is most convenient:

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

For an ohmic resistor, all three forms should give the same result when you substitute the correct values.

03Worked examples: correcting the wrong approach

04Where memorised shortcuts break

Three common situations are worth distinguishing.

The third case is where students often overgeneralise.

In a series circuit, every resistor carries the same current at a particular operating point. That makes \(P=I^2R\) useful for comparing the resistors at that moment.

But if you replace one resistor with a higher-resistance resistor, the total circuit resistance increases. The current then decreases. You cannot assume the original current still applies.

The same issue arises with a real battery. Its electromotive force (emf) is not necessarily equal to the voltage across an external resistor while current is flowing. If the question gives an internal resistance, include it in the circuit calculation.

There is an even more interesting consequence. If a variable resistor \(R\) is connected in series with a fixed resistor \(r\) across an ideal voltage source, increasing \(R\) changes both the current through the load and the voltage across it. The load’s power does not have to increase or decrease continuously. It can rise to a maximum and then fall.

That is not an exception to the power equations. It is what happens when neither of the simple comparison conditions remains fixed.

05Before you submit: the 10-second self-check

Run through these five points before committing to a power calculation.

• Have I identified the exact resistor whose power is required?
• Is the voltage across that resistor fixed, or is the current through it fixed?
• If neither is fixed, have I solved the complete circuit first?
• Have I used the resistor’s own voltage and current, rather than the source voltage or another component’s values?
• Do \(P=VI\), \(P=I^2R\), and \(P=V^2/R\) agree when applied to the actual values?

If you cannot answer the second question, you are not ready to choose a power equation.

06Questions and solutions

Question 1

A \(12.0\,\Omega\) resistor is connected directly across an ideal \(6.0\,\text{V}\) source. Calculate its power. The resistor is then replaced with a \(24.0\,\Omega\) resistor across the same source. Calculate the new power and explain the change. Assume the resistances remain constant.

Solution 1

The original resistor dissipates \(3.0\,\text{W}\), and the replacement dissipates \(1.5\,\text{W}\). Doubling the resistance halves the power because the voltage across each resistor remains \(6.0\,\text{V}\).

For the original resistor,

\[
\begin{aligned}
P_{12}&=\frac{V^2}{R}\\
&=\frac{(6.0\,\text{V})^2}{12.0\,\Omega}\\
&=3.0\,\text{W}
\end{aligned}
\]

For the replacement,

\[
\begin{aligned}
P_{24}&=\frac{V^2}{R}\\
&=\frac{(6.0\,\text{V})^2}{24.0\,\Omega}\\
&=1.5\,\text{W}
\end{aligned}
\]

The current also halves, from \(0.50\,\text{A}\) to \(0.25\,\text{A}\). This reduction in current explains why the higher-resistance resistor dissipates less power at the same voltage.

Question 2

An ideal \(24\,\text{V}\) source is connected to \(6.0\,\Omega\) and \(18.0\,\Omega\) resistors. Calculate the power dissipated by each resistor when they are:

(a) connected in series;

(b) connected in parallel.

State which resistor dissipates more power in each arrangement. Assume both resistances remain constant.

Solution 2

In series, the \(18.0\,\Omega\) resistor dissipates more power: \(18\,\text{W}\), compared with \(6.0\,\text{W}\). In parallel, the \(6.0\,\Omega\) resistor dissipates more power: \(96\,\text{W}\), compared with \(32\,\text{W}\).

(a) Series arrangement

The total resistance is

\[
R_{\text{total}}=6.0+18.0=24.0\,\Omega
\]

The current through both resistors is

\[
I=\frac{24\,\text{V}}{24.0\,\Omega}
=1.0\,\text{A}
\]

Because the current is the same through both resistors, use \(P=I^2R\).

For the \(6.0\,\Omega\) resistor,

\[
P_6=(1.0\,\text{A})^2(6.0\,\Omega)
=6.0\,\text{W}
\]

For the \(18.0\,\Omega\) resistor,

\[
P_{18}=(1.0\,\text{A})^2(18.0\,\Omega)
=18\,\text{W}
\]

The higher resistance dissipates more power because the current is shared.

(b) Parallel arrangement

Both resistors have the full \(24\,\text{V}\) across them. Use \(P=V^2/R\).

\[
P_6=\frac{(24\,\text{V})^2}{6.0\,\Omega}
=96\,\text{W}
\]

\[
P_{18}=\frac{(24\,\text{V})^2}{18.0\,\Omega}
=32\,\text{W}
\]

The lower resistance dissipates more power because the voltage is shared, while the current through each branch differs.

The source voltage is \(24\,\text{V}\) in both arrangements, but the conditions across each individual resistor change when the connections change. That is why the comparison reverses.

Question 3

An \(8.0\,\Omega\) resistor is connected directly across an ideal voltage supply and dissipates \(18\,\text{W}\).

(a) Determine the current through the resistor and the voltage across it.

(b) A \(24.0\,\Omega\) resistor is then connected in parallel across the same supply. Calculate its power.

(c) Explain why \(P=I^2R\) does not mean the \(24.0\,\Omega\) resistor dissipates three times as much power.

Assume both resistances remain constant.

Solution 3

The original resistor carries \(1.5\,\text{A}\) at \(12\,\text{V}\), and the \(24.0\,\Omega\) resistor dissipates \(6.0\,\text{W}\). Its power is one-third of the original resistor’s power, not three times as much.

(a) Determine the current and voltage.

Using \(P=I^2R\),

\[
\begin{aligned}
I&=\sqrt{\frac{P}{R}}\\
&=\sqrt{\frac{18\,\text{W}}{8.0\,\Omega}}\\
&=\sqrt{2.25\,\text{A}^2}\\
&=1.5\,\text{A}
\end{aligned}
\]

Now use Ohm’s law:

\[
\begin{aligned}
V&=IR\\
&=(1.5\,\text{A})(8.0\,\Omega)\\
&=12\,\text{V}
\end{aligned}
\]

(b) Calculate the power in the \(24.0\,\Omega\) resistor.

The new resistor is connected in parallel, so it also has \(12\,\text{V}\) across it.

\[
\begin{aligned}
P_{24}&=\frac{V^2}{R}\\
&=\frac{(12\,\text{V})^2}{24.0\,\Omega}\\
&=6.0\,\text{W}
\end{aligned}
\]

(c) Explain the misconception.

The current is not the same through the two parallel branches. The current through the new resistor is

\[
I_{24}=\frac{12\,\text{V}}{24.0\,\Omega}
=0.50\,\text{A}
\]

The \(24.0\,\Omega\) resistor has three times the resistance but only one-third the current. The current is squared in \(P=I^2R\), so you cannot compare the resistors using that equation while pretending their currents are equal.

Question 4

A battery is modelled as an ideal source with emf \(\varepsilon=12.0\,\text{V}\) and internal resistance \(r=1.0\,\Omega\). A load resistor is connected to the battery.

First, the load resistance is \(5.0\,\Omega\). It is then replaced with a \(10.0\,\Omega\) resistor.

For each load:

(a) Calculate the circuit current.

(b) Calculate the power dissipated by the load.

(c) Explain why using \(P=\varepsilon^2/R\) directly gives the wrong result, and state whether the load power increases or decreases when its resistance doubles.

Solution 4

The \(5.0\,\Omega\) load dissipates \(20.0\,\text{W}\), while the \(10.0\,\Omega\) load dissipates approximately \(11.9\,\text{W}\). The load power decreases because the current falls sufficiently to outweigh the increase in resistance.

The internal resistance must be included when calculating the circuit current.

(a) Calculate the current for each load.

For the \(5.0\,\Omega\) load, the total resistance is

\[
R_{\text{total}}=5.0+1.0=6.0\,\Omega
\]

Therefore,

\[
I_5=\frac{12.0\,\text{V}}{6.0\,\Omega}
=2.0\,\text{A}
\]

For the \(10.0\,\Omega\) load,

\[
R_{\text{total}}=10.0+1.0=11.0\,\Omega
\]

Therefore,

\[
I_{10}=\frac{12.0\,\text{V}}{11.0\,\Omega}
\approx1.09\,\text{A}
\]

(b) Calculate the power dissipated by each load.

Using \(P=I^2R\), the first load dissipates

\[
\begin{aligned}
P_5&=(2.0\,\text{A})^2(5.0\,\Omega)\\
&=20.0\,\text{W}
\end{aligned}
\]

The second load dissipates

\[
\begin{aligned}
P_{10}&=(1.09\,\text{A})^2(10.0\,\Omega)\\
&\approx11.9\,\text{W}
\end{aligned}
\]

(c) Explain the change.

The emf of \(12.0\,\text{V}\) is across the internal resistance and the load together. It is not the voltage across the load alone.

For the \(5.0\,\Omega\) load, the voltage across the load is

\[
V_5=I_5R=(2.0\,\text{A})(5.0\,\Omega)
=10.0\,\text{V}
\]

For the \(10.0\,\Omega\) load, it is

\[
V_{10}=(1.09\,\text{A})(10.0\,\Omega)
\approx10.9\,\text{V}
\]

Neither load has the full \(12.0\,\text{V}\) across it. The internal resistance takes some of the voltage, and the amount depends on the current.

When the load resistance doubles, the current drops from \(2.0\,\text{A}\) to about \(1.09\,\text{A}\). Since current is squared in \(P=I^2R\), that reduction more than offsets the doubled resistance in this example.

Question 5

An ideal \(12.0\,\text{V}\) source is connected to a fixed \(4.0\,\Omega\) resistor in series with an adjustable load resistor \(R\). The resistances remain constant during each test.

One student claims that increasing \(R\) must increase the load power because \(P=I^2R\). Another claims that increasing \(R\) must decrease the load power because \(P=V^2/R\).

(a) Calculate the load power when \(R=1.0\,\Omega\), \(R=4.0\,\Omega\), and \(R=16.0\,\Omega\).

(b) Without using calculus, prove which load resistance maximises the power dissipated by the load for any positive value of \(R\).

(c) Explain why neither student’s prediction works across the entire range. Account for any equality you find between the three calculated powers.

Solution 5

The load power is maximised when \(R=4.0\,\Omega\), giving a maximum power of \(9.0\,\text{W}\). The \(1.0\,\Omega\) and \(16.0\,\Omega\) loads each dissipate \(5.76\,\text{W}\). Neither simple prediction works across the whole range because both the current through the load and the voltage across it change as \(R\) changes.

(a) Calculate the power for each load.

Let the fixed series resistance be \(r=4.0\,\Omega\), and let the source voltage be \(V_s=12.0\,\text{V}\).

The total resistance is \(R+r\), so the current is

\[
I=\frac{V_s}{R+r}
\]

The load power is therefore

\[
P_R=I^2R
=\frac{V_s^2R}{(R+r)^2}
\]

For \(R=1.0\,\Omega\),

\[
\begin{aligned}
I&=\frac{12.0\,\text{V}}{1.0+4.0\,\Omega}\\
&=2.4\,\text{A}\\[4pt]
P_1&=(2.4\,\text{A})^2(1.0\,\Omega)\\
&=5.76\,\text{W}
\end{aligned}
\]

For \(R=4.0\,\Omega\),

\[
\begin{aligned}
I&=\frac{12.0\,\text{V}}{4.0+4.0\,\Omega}\\
&=1.5\,\text{A}\\[4pt]
P_4&=(1.5\,\text{A})^2(4.0\,\Omega)\\
&=9.0\,\text{W}
\end{aligned}
\]

For \(R=16.0\,\Omega\),

\[
\begin{aligned}
I&=\frac{12.0\,\text{V}}{16.0+4.0\,\Omega}\\
&=0.60\,\text{A}\\[4pt]
P_{16}&=(0.60\,\text{A})^2(16.0\,\Omega)\\
&=5.76\,\text{W}
\end{aligned}
\]

The load power rises from \(5.76\,\text{W}\) to \(9.0\,\text{W}\), then falls back to \(5.76\,\text{W}\).

(b) Prove the maximum without calculus.

For any positive load resistance \(R\) and fixed positive series resistance \(r\),

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

The key is to use the fact that a square cannot be negative:

\[
(R-r)^2\geq0
\]

Expanding gives

\[
R^2-2Rr+r^2\geq0
\]

Rearranging,

\[
(R+r)^2\geq4Rr
\]

Since \(R\) and \(r\) are positive, substitute this inequality into the power expression:

\[
\begin{aligned}
P_R&=\frac{V_s^2R}{(R+r)^2}\\[4pt]
&\leq\frac{V_s^2R}{4Rr}\\[4pt]
&=\frac{V_s^2}{4r}
\end{aligned}
\]

This establishes an upper limit on the load power. Equality occurs only when \(R=r\), because that is when \((R-r)^2=0\).

Therefore, the maximum load power occurs when the load resistance equals the fixed series resistance.

For this circuit,

\[
\begin{aligned}
P_{\max}&=\frac{V_s^2}{4r}\\[4pt]
&=\frac{(12.0\,\text{V})^2}{4(4.0\,\Omega)}\\[4pt]
&=9.0\,\text{W}
\end{aligned}
\]

The maximum occurs at \(R=4.0\,\Omega\), matching the calculated value in part (a).

(c) Explain why the two predictions fail.

The first student uses \(P=I^2R\) and assumes that current stays constant as the load resistance changes. It does not. The current is

\[
I=\frac{12.0\,\text{V}}{R+4.0\,\Omega}
\]

As \(R\) increases, the current decreases. At low load resistances, increasing \(R\) raises the power despite the falling current. Beyond \(R=4.0\,\Omega\), the current reduction becomes dominant, so power falls.

The second student uses \(P=V^2/R\) and assumes that the voltage across the load stays constant. It does not either. The load voltage is

\[
V_R=\frac{V_sR}{R+r}
\]

As the load resistance increases, the voltage across it rises towards the source voltage. At the same time, the current falls. The power depends on both effects, so it cannot be predicted by treating the load voltage as fixed.

The equal powers at \(R=1.0\,\Omega\) and \(R=16.0\,\Omega\) are not a coincidence. In this particular circuit, two unequal positive load resistances \(R_1\) and \(R_2\) have equal power when \(R_1R_2=r^2\). Here,

\[
(1.0\,\Omega)(16.0\,\Omega)
=(4.0\,\Omega)^2
\]

That relationship follows from equating the two load-power expressions for this circuit; it is not a universal rule for every circuit.

The habit to carry into more complex circuit questions is simple: calculate the actual current and potential difference for each component from the complete circuit first, including internal resistance when it is given. Then calculate power. The power equation is the last step in the reasoning, not the first.