How to Compare Power Ratings of Devices on the Same Supply

Two devices are connected in parallel across the same 12.0 V supply. Both are designed to operate at 12.0 V. One is rated at 6.0 W, and the other at 3.0 W.

Which device draws more current?

The 6.0 W device draws twice as much current. It transfers electrical energy at twice the rate of the 3.0 W device.

That comparison works because both devices receive the same voltage. Change the connection to series, or use devices with different rated voltages, and the answer may change.

The reliable method is to identify what the circuit keeps constant before comparing power ratings.

01First identify what the circuit keeps constant

The phrase “connected to the same supply” doesn’t automatically tell you whether two devices have the same voltage or current.

You need to inspect the circuit arrangement.

Parallel connections: same voltage

Two devices are connected in parallel when both are connected across the same pair of supply terminals.

Each device receives the full supply voltage.

For example, if two devices are connected in parallel across a 12.0 V supply:

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

The current divides between the branches. Each device can draw a different current depending on its electrical properties.

This is how ordinary household appliances are connected. Each appliance receives the supply voltage independently of the others, assuming the supply maintains its terminal voltage.

Series connections: same current

Two devices are connected in series when there is only one path for charge to flow through both.

The same current passes through each device.

However, the supply voltage is divided between them:

\[ I_A=I_B \]

The voltage across each device depends on its resistance and operating characteristics.

The difference determines which power equation is most useful.

These equations assume resistive behaviour. The more general relationship is \(P=VI\).

02What does a power rating actually tell you?

Power measures how quickly energy is transferred.

A device with a power of 6 W transfers 6 joules of energy every second under the stated operating conditions.

Mathematically,

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

where:

• \(P\) is power, measured in watts (W).
• \(\Delta E\) is the energy transferred, measured in joules (J).
• \(\Delta t\) is the time taken, measured in seconds (s).

Electrical power depends on both voltage and current.

\[ \boxed{P=VI} \]

Here:

• \(P\) is electrical power in watts.
• \(V\) is the potential difference across the device in volts (V).
• \(I\) is the current through the device in amperes (A).

This relationship explains why you must consider the circuit before comparing power ratings.

Where the two resistance equations come from

For an ohmic device, Ohm’s law gives

\[ V=IR \]

where \(R\) is resistance in ohms (\(\Omega\)).

Substitute this relationship into the electrical power equation.

If current is held constant:

\[ P=VI \]

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

If voltage is held constant:

\[ P=VI \]

Substituting \(I=V/R\):

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

Both equations describe the same physical quantity. They simply express power in terms of different variables.

The important distinction is the condition under which you use them.

Fixed voltage: higher power means lower resistance

Suppose two ohmic devices are connected in parallel across the same 12 V supply.

Their voltages are identical.

From

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

the voltage is constant, so power is inversely proportional to resistance.

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

Therefore:

• Lower resistance means greater current.
• Greater current means greater power at the same voltage.
• Higher resistance means lower power at the same voltage.

For two devices at the same voltage, the higher-power device has the lower operating resistance, provided the devices behave ohmically.

Fixed current: higher resistance means higher power

Now consider two resistors connected in series.

The same current passes through both.

Using

\[ P=I^2R \]

current is constant, so power is directly proportional to resistance.

\[ P\propto R \]

Therefore:

• Higher resistance means a greater voltage drop.
• The same current passes through both devices.
• The higher-resistance device dissipates more power.

This is the opposite resistance relationship from the fixed-voltage situation.

It is a common HSC mistake to assume that higher resistance always means lower power. It doesn’t. You must identify whether voltage or current is fixed.

03What is happening inside the device?

There is a useful physical way to understand why these relationships work.

Voltage describes the energy transferred per unit charge:

\[ V=\frac{\Delta E}{Q} \]

Current describes the amount of charge passing a point each second:

\[ I=\frac{\Delta Q}{\Delta t} \]

Multiplying them gives:

\[ P=VI \]

In other words:

\[ \text{Power} = \frac{\text{energy per charge}}{} \times \text{charge per second} \]

At fixed voltage, each coulomb transfers the same amount of energy through the device. A device drawing more current has more charge passing through it each second, so its energy transfer rate is greater.

At fixed current, the same amount of charge passes through each device per second. However, a higher-resistance device has a larger voltage drop. Each coulomb therefore transfers more energy in that device.

In a metal conductor, the electric field transfers energy to mobile charge carriers, which transfer energy to the material through interactions within the conductor. This can increase its thermal energy.

Don’t imagine electrons being consumed as they travel through a device. Charge is conserved. What changes is the energy transferred to the device.

04How to test which device uses more power

Suppose you want to check whether two devices with different power ratings actually draw different amounts of power.

You can investigate this using a low-voltage, current-limited DC power supply.

The aim is to measure the voltage across each device and the current through each branch.

Setting up the investigation

Use:

• A regulated 12.0 V DC power supply.
• Two devices rated for 12.0 V.
• An ammeter in series with each device.
• A voltmeter connected across the devices.
• Connecting wires and suitable switches.

Connect the two devices in parallel.

Current leaves the positive terminal of the supply, divides between the two branches, passes through each device, and then recombines before returning to the negative terminal.

The voltmeter measures the potential difference across the devices. The ammeters measure the individual branch currents.

What should the investigation show?

Suppose device A is rated at 6.0 W and device B at 3.0 W, with both rated for 12.0 V.

If the supply maintains 12.0 V across both devices, their expected currents are:

For A:

\[ I_A=\frac{6.0}{12.0}=0.50\text{ A} \]

For B:

\[ I_B=\frac{3.0}{12.0}=0.25\text{ A} \]

The higher-rated device should draw twice the current.

Why the first measurement attempt is insufficient

Imagine placing just one ammeter in the main supply wire before the current divides.

The ammeter measures the total current:

\[ I_{\text{total}}=I_A+I_B \]

In this example:

\[ I_{\text{total}}=0.50+0.25 \]

\[ I_{\text{total}}=0.75\text{ A} \]

But that single reading doesn’t tell you how much current passes through either device individually.

You cannot simply assume that the current divides equally.

The measurement needs to be revised by placing an ammeter in each branch, or by measuring the branches separately with one ammeter.

Comparing brightness alone is also insufficient. Different devices can convert electrical energy into light with different efficiencies.

The revised experiment gives you the actual voltage and current for each device. You can then calculate its power using \(P=VI\).

05The contrasting case: What changes when devices are in series?

The previous comparison relied on both devices receiving the same voltage.

Now connect two resistors in series.

The current is identical through both, so the relevant relationship becomes:

\[ P=I^2R \]

A higher resistance now means greater power dissipation.

06When comparing power ratings becomes misleading

There are two important limitations to the simple comparison.

Different rated voltages

Consider two devices:

• Device A: rated 12 V, 24 W.
• Device B: rated 6 V, 9 W.

You cannot simply connect both across a 9 V supply and assume their power consumption will be 24 W and 9 W.

Those ratings describe different operating voltages.

For an ohmic device, you can estimate its operating resistance from its rated values:

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

Then use the actual supply voltage to calculate its operating power, provided its resistance remains constant.

For a real lamp, heating element, or electronic device, resistance may change with temperature or operating conditions. In that case, the simple constant-resistance model may not accurately predict its actual power.

The most reliable measurement is still:

\[ P=V_{\text{actual}}I_{\text{actual}} \]

The supply voltage might not remain constant

An ideal voltage supply maintains the same terminal voltage regardless of how much current the devices draw.

A real battery or source has internal resistance.

When the total current increases, some voltage is lost inside the source. Its terminal voltage can therefore decrease.

For a source with electromotive force \(\mathcal{E}\), internal resistance \(r\), and total current \(I\):

\[ V_{\text{terminal}}=\mathcal{E}-Ir \]

This matters when devices are added or removed.

Parallel devices still share the same terminal voltage at any given moment, but that shared voltage can change when the total load changes.

The final question explores this less obvious situation.

For ordinary AC appliances, there is another limitation: with non-resistive loads, voltage multiplied by current is not necessarily the real power in watts. The power factor must also be considered. Basic HSC resistive-circuit comparisons generally assume resistive loads.

07A reliable decision process

When comparing devices connected to the same supply, use this sequence.

Keep the central distinction clear:

• Same voltage: higher current means higher power.
• Same current: higher resistance means higher power.
• Different operating voltages: calculate actual power rather than comparing wattage labels alone.

08Questions and solutions

Question 1

Two resistive devices are connected in parallel across an ideal 12.0 V supply.

Their operating ratings are:

• Device A: 12.0 V, 18 W.
• Device B: 12.0 V, 9.0 W.

Calculate:

a. The current through each device.

b. The total current supplied by the source.

c. Which device transfers electrical energy at the greater rate, and by what factor?

Solution 1

Device A draws 1.5 A, device B draws 0.75 A, and the total supply current is 2.25 A. Device A transfers energy twice as quickly as device B.

The devices are connected in parallel, so both receive 12.0 V.

Use:

\[ P=VI \]

Rearrange:

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

For device A:

\[ I_A=\frac{18}{12.0} \]

\[ I_A=1.50\text{ A} \]

For device B:

\[ I_B=\frac{9.0}{12.0} \]

\[ I_B=0.750\text{ A} \]

The total current is the sum of the branch currents:

\[ I_{\text{total}}=I_A+I_B \]

\[ I_{\text{total}}=1.50+0.750 \]

\[ I_{\text{total}}=2.25\text{ A} \]

Finally, compare their powers:

\[ \frac{P_A}{P_B}=\frac{18}{9.0}=2 \]

Device A transfers energy at twice the rate of device B.

The key reason is that both devices operate at the same voltage. Their power ratings can therefore be compared directly.

Question 2

Two ohmic resistors of resistance 3.0 Ω and 9.0 Ω are connected in series across an ideal 24.0 V supply.

Calculate:

a. The current through the circuit.

b. The voltage across each resistor.

c. The power dissipated by each resistor.

A student argues that the 3.0 Ω resistor must dissipate more power because \(P=V^2/R\), and lower resistance produces higher power.

Explain why this argument is incorrect.

Solution 2

The 3.0 Ω resistor dissipates 12 W, while the 9.0 Ω resistor dissipates 36 W. The higher-resistance resistor dissipates three times as much power.

The student has applied the fixed-voltage relationship without checking whether each resistor actually receives the full supply voltage.

Part a: Find the current.

The resistors are in series, so their resistances add:

\[ R_{\text{total}}=3.0+9.0 \]

\[ R_{\text{total}}=12.0\ \Omega \]

Using Ohm’s law:

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

\[ I=\frac{24.0}{12.0} \]

\[ I=2.0\text{ A} \]

Part b: Find the voltage across each resistor.

The current is the same through both.

For the 3.0 Ω resistor:

\[ V_1=IR_1 \]

\[ V_1=2.0\times3.0=6.0\text{ V} \]

For the 9.0 Ω resistor:

\[ V_2=IR_2 \]

\[ V_2=2.0\times9.0=18.0\text{ V} \]

The voltage drops add to 24.0 V, as required.

Part c: Find the power.

Since current is fixed, use:

\[ P=I^2R \]

For the 3.0 Ω resistor:

\[ P_1=(2.0)^2(3.0) \]

\[ P_1=12\text{ W} \]

For the 9.0 Ω resistor:

\[ P_2=(2.0)^2(9.0) \]

\[ P_2=36\text{ W} \]

The 9.0 Ω resistor dissipates three times the power.

The student’s mistake was treating the 24.0 V supply voltage as though it appeared across each resistor individually.

The equation \(P=V^2/R\) is still valid, but the correct voltage must be used for each component. The larger resistor receives 18.0 V, while the smaller resistor receives only 6.0 V.

Question 3

Two resistive devices are connected in parallel across a regulated 9.0 V supply.

Their ratings are:

• Device A: 12.0 V, 24 W.
• Device B: 6.0 V, 9.0 W.

Assume both devices behave as ohmic resistors over the operating range, with constant resistance. Ignore temperature changes and assume the devices remain intact.

Calculate:

a. The operating resistance of each device from its rated values.

b. The actual power dissipated by each device when connected to the 9.0 V supply.

c. Whether either device exceeds its rated power.

d. Explain why comparing the 24 W and 9.0 W labels alone would give the wrong prediction about their actual power consumption.

Solution 3

Device A dissipates 13.5 W, while device B dissipates 20.25 W under the stated assumptions. Device B exceeds its rated power even though its nominal wattage is lower.

The important difference is that the devices have different rated voltages.

Part a: Calculate their operating resistances.

For an ohmic device:

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

For device A:

\[ R_A=\frac{12.0^2}{24} \]

\[ R_A=6.0\ \Omega \]

For device B:

\[ R_B=\frac{6.0^2}{9.0} \]

\[ R_B=4.0\ \Omega \]

Part b: Calculate actual power at 9.0 V.

Both devices are connected in parallel, so each receives 9.0 V.

For device A:

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

\[ P_A=\frac{9.0^2}{6.0} \]

\[ P_A=13.5\text{ W} \]

For device B:

\[ P_B=\frac{9.0^2}{4.0} \]

\[ P_B=20.25\text{ W} \]

Part c: Check the rated powers.

Device A:

\[ 13.5\text{ W}<24\text{ W} \]

It operates below its rated power.

Device B:

\[ 20.25\text{ W}>9.0\text{ W} \]

It dissipates:

\[ \frac{20.25}{9.0}=2.25 \]

or 2.25 times its rated power.

Part d: Explain the difference.

The tempting prediction is that device A must use more power because its label says 24 W, compared with device B’s 9.0 W.

That prediction assumes both devices are operating at their rated voltages.

They aren’t.

Device A is receiving less than its rated voltage, while device B is receiving more than its rated voltage.

Under the constant-resistance assumption, the actual voltage across each device determines its power.

In a real device, exceeding the rated operating voltage can cause overheating or damage. These calculations describe the specified idealised model, not a recommendation to operate a real device outside its rating.

Question 4

Two ideal ohmic components have resistances:

\[ R_A=12\ \Omega \]

\[ R_B=36\ \Omega \]

They are connected to the same ideal 12.0 V supply.

Consider two arrangements.

Arrangement 1: The components are connected in parallel.

Arrangement 2: The components are connected in series.

Calculate the power dissipated by each component in both arrangements.

Then explain why the component with the greater power changes when the circuit is rearranged.

Solution 4

In parallel, component A dissipates 12 W and component B dissipates 4 W. In series, component A dissipates 0.75 W and component B dissipates 2.25 W.

The ranking reverses because the fixed quantity changes from voltage to current.

Arrangement 1: Parallel

Both components receive the full 12.0 V.

For component A:

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

\[ P_A=\frac{12.0^2}{12} \]

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

For component B:

\[ P_B=\frac{12.0^2}{36} \]

\[ P_B=4.0\text{ W} \]

Component A dissipates three times as much power as component B.

The total power is:

\[ P_{\text{total}}=12+4 \]

\[ P_{\text{total}}=16\text{ W} \]

Arrangement 2: Series

The same current passes through both components.

First, calculate the total resistance:

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

\[ R_{\text{total}}=48\ \Omega \]

The circuit current is:

\[ I=\frac{12.0}{48} \]

\[ I=0.25\text{ A} \]

Now calculate the power of each component using \(P=I^2R\).

For component A:

\[ P_A=(0.25)^2(12) \]

\[ P_A=0.75\text{ W} \]

For component B:

\[ P_B=(0.25)^2(36) \]

\[ P_B=2.25\text{ W} \]

Component B now dissipates three times as much power as component A.

The total power is:

\[ P_{\text{total}}=0.75+2.25 \]

\[ P_{\text{total}}=3.0\text{ W} \]

The supply power also confirms this:

\[ P=VI=12.0(0.25)=3.0\text{ W} \]

Why does the ranking reverse?

In parallel, voltage is fixed. Lower resistance produces higher power.

In series, current is fixed. Higher resistance produces higher power.

The components themselves haven’t changed. Their resistances are identical in both arrangements.

What changes is how the circuit distributes voltage and current.

This is why you should never compare resistance values without first identifying the circuit condition.

Question 5

A battery is modelled as an electromotive force of 12.0 V in series with an unknown internal resistance \(r\).

It supplies two fixed-ohmic devices connected in parallel.

Their rated operating values are:

• Device A: 12.0 V, 24 W.
• Device B: 12.0 V, 12 W.

When both devices are connected, a voltmeter across the battery terminals reads 8.0 V.

Assume the devices maintain constant resistance, the battery’s internal resistance is constant, and all calculations describe steady-state DC operation.

Answer the following:

a. Calculate the operating resistance of each device from its rated values.

b. Calculate the current through each device when both are connected.

c. Determine the battery’s internal resistance.

d. Calculate the actual power dissipated by each device. Check whether either exceeds its rated power.

e. Device A is then disconnected without changing the battery. Calculate the new terminal voltage and the new power dissipated by device B.

f. Explain why device B’s power increases even though the total power delivered to the loads decreases.

Solution 5

The internal resistance is 2.0 Ω. With both devices connected, A dissipates approximately 10.67 W and B dissipates 5.33 W. After A is disconnected, B’s power increases to approximately 8.82 W, even though total load power decreases.

This result exposes an important limitation of treating a battery as a perfectly constant-voltage supply.

Part a: Calculate the operating resistance of each device.

Both devices are rated at 12.0 V.

For an ohmic device:

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

For A:

\[ R_A=\frac{12.0^2}{24} \]

\[ R_A=6.0\ \Omega \]

For B:

\[ R_B=\frac{12.0^2}{12} \]

\[ R_B=12.0\ \Omega \]

Part b: Calculate the branch currents with both devices connected.

The voltmeter reads 8.0 V across the parallel combination.

Both devices therefore receive 8.0 V, not the full 12.0 V electromotive force.

For A:

\[ I_A=\frac{V}{R_A} \]

\[ I_A=\frac{8.0}{6.0} \]

\[ I_A=1.33\text{ A} \]

For B:

\[ I_B=\frac{V}{R_B} \]

\[ I_B=\frac{8.0}{12.0} \]

\[ I_B=0.667\text{ A} \]

The total battery current is the sum of the branch currents:

\[ I_{\text{total}}=1.33+0.667 \]

\[ I_{\text{total}}=2.0\text{ A} \]

Part c: Determine the internal resistance.

The battery has an electromotive force of 12.0 V but a terminal voltage of only 8.0 V.

The voltage lost inside the battery is:

\[ V_{\text{internal}}=12.0-8.0 \]

\[ V_{\text{internal}}=4.0\text{ V} \]

Using the internal resistance relationship:

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

Rearrange:

\[ r=\frac{V_{\text{internal}}}{I} \]

Substitute:

\[ r=\frac{4.0}{2.0} \]

\[ \boxed{r=2.0\ \Omega} \]

The battery’s internal resistance is 2.0 Ω.

Part d: Calculate the actual power of each device.

Both devices receive 8.0 V.

For A:

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

\[ P_A=\frac{8.0^2}{6.0} \]

\[ P_A=10.67\text{ W} \]

For B:

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

\[ P_B=\frac{8.0^2}{12.0} \]

\[ P_B=5.33\text{ W} \]

Neither device exceeds its rated power.

The total power delivered to the external circuit is:

\[ P_{\text{load}}=10.67+5.33 \]

\[ P_{\text{load}}=16.0\text{ W} \]

Part e: Disconnect device A.

This is where the interesting change occurs.

Device B remains connected, and its resistance is still 12.0 Ω.

The battery’s electromotive force and internal resistance also remain unchanged.

The circuit now contains the 2.0 Ω internal resistance in series with the 12.0 Ω load.

The total resistance is:

\[ R_{\text{total}}=2.0+12.0 \]

\[ R_{\text{total}}=14.0\ \Omega \]

The new battery current is:

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

\[ I=\frac{12.0}{14.0} \]

\[ I=0.857\text{ A} \]

The new terminal voltage is the voltage across device B:

\[ V_B=IR_B \]

\[ V_B=0.857\times12.0 \]

\[ V_B=10.29\text{ V} \]

Now calculate the new power:

\[ P_B=\frac{V_B^2}{R_B} \]

\[ P_B=\frac{10.29^2}{12.0} \]

\[ \boxed{P_B\approx8.82\text{ W}} \]

Device B’s power has increased from 5.33 W to 8.82 W.

Part f: Explain the surprising result.

The tempting prediction is that device B should continue dissipating 5.33 W because its resistance has not changed.

That would be true if its voltage remained at 8.0 V.

But the battery has internal resistance.

When device A is disconnected, the total current drawn from the battery decreases.

This reduces the voltage drop across the internal resistance.

The terminal voltage therefore increases from 8.0 V to approximately 10.29 V.

Because device B remains ohmic and its resistance is constant:

\[ P_B=\frac{V_B^2}{R_B} \]

Its power increases as the terminal voltage rises.

The total load power, however, decreases.

Before A is disconnected:

\[ P_{\text{load}}=16.0\text{ W} \]

After A is disconnected:

\[ P_{\text{load}}=8.82\text{ W} \]

There is no contradiction. Device B receives more power, but the circuit no longer contains device A, so the total power delivered to the loads is smaller.

We can check energy conservation as well.

With both devices connected, the battery supplies:

\[ P_{\text{source}}=\mathcal{E}I \]

\[ P_{\text{source}}=12.0(2.0)=24.0\text{ W} \]

The internal resistance dissipates:

\[ P_{\text{internal}}=I^2r \]

\[ P_{\text{internal}}=(2.0)^2(2.0)=8.0\text{ W} \]

Therefore:

\[ 24.0=16.0+8.0 \]

The source power is divided between the external devices and the battery’s internal resistance.

After A is disconnected, the same accounting still works:

\[ P_{\text{source}}=12.0(0.857) \]

\[ P_{\text{source}}\approx10.29\text{ W} \]

The internal loss is:

\[ P_{\text{internal}}=(0.857)^2(2.0) \]

\[ P_{\text{internal}}\approx1.47\text{ W} \]

The remaining power is delivered to B:

\[ 10.29\approx8.82+1.47 \]

The key distinction is that parallel devices share a voltage, but that voltage is not necessarily fixed by the battery’s electromotive force.

If the source were an ideal 12.0 V supply with zero internal resistance, disconnecting A would not change the voltage across B. Its power would remain 12.0 W.

The increase in B’s power comes specifically from the battery’s internal resistance and the resulting change in terminal voltage.

09What to understand next

Comparing power ratings is really an exercise in identifying the electrical conditions experienced by each device.

Once you can distinguish fixed voltage from fixed current, you can handle parallel loads, series components, changing supply conditions, and device operating ratings without relying on memorised shortcuts.

The next useful connection is understanding how voltage and current change across a transformer while power is transferred between its primary and secondary circuits. See Ideal Transformer Current and Power for HSC Physics for that related comparison.