How to Find Useful Output Power from Efficiency and Input Power

An electric pump draws \(1.20\,\text{kW}\) of electrical power from a supply. Its efficiency is \(75\%\). How much power is actually transferred usefully to the water?

Before calculating, consider what the \(75\%\) represents. Does it mean that \(75\%\) of the input power is lost, or that \(75\%\) becomes useful output?

It’s the second. The pump delivers \(75\%\) of its input power as useful output:

\[ P_{\text{useful}}=0.75\times1200 \]

\[ \boxed{P_{\text{useful}}=900\,\text{W}} \]

The remaining \(300\,\text{W}\) is transferred through non-useful pathways, such as heating, sound, and friction.

This is the central calculation:

\[ \boxed{P_{\text{useful}}=\eta P_{\text{input}}} \]

If you know the efficiency and input power, multiply them to find the useful output power. The important part is understanding what efficiency represents, using the correct percentage convention, and keeping power separate from energy.

01The relationship between efficiency and output power

Efficiency describes how much of a device’s input energy is converted into the useful form of energy that the device is designed to produce.

For example:

• An electric motor converts electrical energy into useful mechanical energy.
• A light bulb converts electrical energy into useful light.
• A pump converts electrical energy into useful mechanical energy that moves water.
• A heater converts electrical energy into useful thermal energy.

Not all the input energy necessarily becomes useful output. Some is transferred into unwanted forms, such as sound, vibration, and heating.

The efficiency equation is:

\[ \eta=\frac{P_{\text{useful output}}} {P_{\text{input}}} \]

To express efficiency as a percentage, multiply by \(100\%\):

\[ \boxed{ \eta=\frac{P_{\text{useful output}}} {P_{\text{input}}}\times100\% } \]

Rearranging gives the equation you need when the question supplies efficiency and input power:

\[ \boxed{ P_{\text{useful output}} = \frac{\eta}{100}\times P_{\text{input}} } \]

Here:

• \(P_{\text{useful output}}\) is the useful output power, measured in watts (W).
• \(P_{\text{input}}\) is the total power supplied to the device, measured in watts (W).
• \(\eta\) is the efficiency expressed as a percentage.

There is one important convention to remember: efficiency can be expressed either as a percentage or as a decimal.

If the question gives efficiency as \(75\%\), you must use \(0.75\) in the calculation, not \(75\).

What happens to the remaining power?

For a device operating steadily, with no net change in its stored energy, the input power is divided between useful output and non-useful transfers.

\[ P_{\text{input}} = P_{\text{useful output}} + P_{\text{non-useful}} \]

Therefore:

\[ P_{\text{non-useful}} = P_{\text{input}}-P_{\text{useful output}} \]

Consider a motor receiving \(800\,\text{W}\) with an efficiency of \(75\%\).

\[ P_{\text{useful}}=0.75\times800 =600\,\text{W} \]

The remaining power is:

\[ P_{\text{non-useful}}=800-600 =200\,\text{W} \]

Notice that the motor has not destroyed \(200\,\text{W}\) of energy. Energy is conserved. The motor simply transfers some of its input energy into forms that are not useful for its intended purpose.

For a closer explanation of how electrical energy becomes thermal energy, see Resistive Heating in HSC Physics: Power and Energy.

02Power is a rate, not an amount of energy

This distinction is essential when solving HSC Physics questions.

Power measures how quickly energy is transferred. Energy measures the total amount transferred.

The relationship is:

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

where:

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

One watt is equivalent to one joule per second:

\[ 1\,\text{W}=1\,\text{J/s} \]

Suppose a motor has a useful output power of \(600\,\text{W}\).

This means it transfers \(600\,\text{J}\) of useful energy every second, assuming its output power remains constant.

If it operates for \(10\,\text{s}\):

\[ E_{\text{useful}}=Pt \]

\[ E_{\text{useful}}=600\times10 \]

\[ E_{\text{useful}}=6000\,\text{J} \]

The useful output power is still \(600\,\text{W}\). The \(6000\,\text{J}\) is the total useful energy transferred during those ten seconds.

Why efficiency can use either power or energy

You might notice that efficiency is also commonly written as:

\[ \eta= \frac{E_{\text{useful output}}} {E_{\text{input}}} \]

This is equally valid.

The reason is that power is energy transferred per unit time. If input and output energies are measured over the same time interval, their ratio is equal to the ratio of their average powers.

Therefore:

\[ \boxed{ \eta= \frac{P_{\text{useful output}}}{P_{\text{input}}} = \frac{E_{\text{useful output}}}{E_{\text{input}}} } \]

This equivalence assumes that both measurements refer to the same interval.

If a device’s power changes during operation, use the average input and output powers over the same interval, or calculate the total energies directly.

03How an experiment measures input and useful output power

Imagine testing an electric motor that lifts a mass.

You can measure the electrical power supplied to the motor, but that does not automatically tell you how much mechanical power reaches the load.

You need two separate measurements.

Measuring the input power

Connect an ammeter in series with the motor to measure current, and a voltmeter across the motor terminals to measure potential difference.

For a DC motor operating under steady conditions, the electrical input power is:

\[ P_{\text{input}}=VI \]

where:

• \(V\) is the potential difference across the motor in volts (V).
• \(I\) is the current through the motor in amperes (A).

For example, if the motor operates at \(24\,\text{V}\) and draws \(4.0\,\text{A}\):

\[ P_{\text{input}}=24\times4.0 \]

\[ P_{\text{input}}=96\,\text{W} \]

The motor receives \(96\,\text{J}\) of electrical energy every second.

Measuring the useful output power

Now consider the mechanical work done by the motor.

The motor winds up the cord, raising a mass against gravity. The useful energy transferred to the mass is its increase in gravitational potential energy:

\[ \Delta E_p=mgh \]

where:

• \(m\) is the mass in kilograms (kg).
• \(g\) is gravitational acceleration in metres per second squared (\(\text{m/s}^2\)).
• \(h\) is the vertical height gained in metres (m).

Since power is energy transferred per unit time:

\[ P_{\text{useful}}=\frac{mgh}{t} \]

If the mass rises at constant speed \(v\), this can also be written as:

\[ P_{\text{useful}}=mgv \]

This gives the rate at which the motor transfers gravitational potential energy to the load.

For a more detailed treatment of the mechanical calculation, see How to Calculate Lifting Power at Constant Speed | HSC Physics.

What should the experiment show?

Suppose an experiment establishes that a motor has an efficiency of \(70\%\).

If its electrical input power is \(700\,\text{W}\), its useful mechanical output should be:

\[ P_{\text{useful}}=0.70\times700 \]

\[ P_{\text{useful}}=490\,\text{W} \]

The remaining \(210\,\text{W}\) is transferred through non-useful pathways.

The important observation is that the useful mechanical output is smaller than the electrical input, even though the motor is still transferring energy to the load.

An illustrative experimental measurement

Consider these illustrative measurements from a loaded motor test. They are example values, not measurements from a reported experiment.

First, calculate the input power:

\[ P_{\text{input}}=VI \]

\[ P_{\text{input}}=24.0\times4.0 =96\,\text{W} \]

Next, calculate the useful mechanical output:

\[ P_{\text{useful}}=mgv \]

\[ P_{\text{useful}}=5.0\times9.8\times1.0 \]

\[ P_{\text{useful}}=49\,\text{W} \]

The efficiency is therefore:

\[ \eta=\frac{49}{96}\times100\% \]

\[ \boxed{\eta\approx51\%} \]

The experiment shows how the input and useful output powers can be measured independently, then compared to determine efficiency.

Why the measurement conditions matter

There is a potential flaw in a simple version of this experiment.

Suppose you measure the motor’s voltage and current while it is running without a load, then attach the mass and measure its lifting speed.

You cannot automatically use the original electrical readings. The motor’s current can change when the load is attached, so the unloaded input power may not represent its input power during lifting.

A better design measures voltage and current while the motor is actually lifting the mass. The mechanical output is also measured over that same interval.

If the mass moves at constant speed, \(P_{\text{useful}}=mgv\) is straightforward to apply. If its speed changes, use the change in gravitational potential energy over the measured interval to find the average useful output power.

The general lesson is simple: efficiency compares input and useful output under matching operating conditions. A percentage measured at one load should not automatically be applied to a completely different operating condition.

04Worked calculations

The method itself is short. Most mistakes happen when students interpret the percentage incorrectly or use the wrong power in the calculation.

05Common mistakes to avoid

A few mistakes appear repeatedly in efficiency calculations.

A useful contrasting case: More efficient does not always mean more powerful

Consider two machines:

• Machine A receives \(1000\,\text{W}\) and operates at \(60\%\) efficiency.
• Machine B receives \(500\,\text{W}\) and operates at \(90\%\) efficiency.

Which produces more useful output power?

For Machine A:

\[ P_A=0.60\times1000=600\,\text{W} \]

For Machine B:

\[ P_B=0.90\times500=450\,\text{W} \]

Machine B is more efficient, but Machine A produces more useful output power.

This is because efficiency measures the proportion of input power converted usefully, not the absolute amount of useful power produced.

A smaller percentage of a large input can be greater than a larger percentage of a small input.

What if the question gives useful output power instead?

Sometimes an exam question works backwards.

Suppose a motor produces \(480\,\text{W}\) of useful output power and has an efficiency of \(80\%\).

You are now finding the input power, not the output power.

Rearrange the efficiency equation:

\[ P_{\text{input}}= \frac{P_{\text{useful output}}}{\eta} \]

\[ P_{\text{input}}=\frac{480}{0.80} \]

\[ \boxed{P_{\text{input}}=600\,\text{W}} \]

Notice the difference:

• Finding useful output from input: multiply by efficiency.
• Finding input from useful output: divide by efficiency.

Keep the direction of the calculation clear.

06Questions and solutions

Try each question before reading its solution. The questions progress from a direct efficiency calculation to problems involving multiple devices, experimental measurements, and changing operating conditions.

Question 1

An electric motor receives \(500\,\text{W}\) of input power and has an efficiency of \(72\%\).

Calculate:

a. the useful output power of the motor

b. the power transferred through non-useful pathways.

Solution 1

The motor produces \(360\,\text{W}\) of useful output power, with \(140\,\text{W}\) transferred through non-useful pathways.

First, convert the percentage into decimal form:

\[ \eta=\frac{72}{100}=0.72 \]

The useful output power is:

\[ P_{\text{useful}}=\eta P_{\text{input}} \]

\[ P_{\text{useful}}=0.72\times500 \]

\[ \boxed{P_{\text{useful}}=360\,\text{W}} \]

For part b, subtract the useful output from the input:

\[ P_{\text{non-useful}}=500-360 \]

\[ \boxed{P_{\text{non-useful}}=140\,\text{W}} \]

The common mistake is to calculate \(500\times0.28\) and report that as useful output. That gives the non-useful power instead. The \(72\%\) represents the useful fraction.

Question 2

An electric water pump receives \(2.40\,\text{kW}\) of electrical input power. Its efficiency is \(65\%\).

The pump operates continuously for \(90\,\text{s}\), transferring energy to water as it moves from a lower reservoir into an elevated storage tank.

Calculate:

a. the useful output power delivered to the water

b. the useful energy transferred to the water over the \(90\,\text{s}\) interval

c. the total energy supplied to the pump over the same interval.

Solution 2

The useful output power is \(1.56\,\text{kW}\), the useful energy transferred is \(140.4\,\text{kJ}\), and the total input energy is \(216\,\text{kJ}\).

Part a: Useful output power

Convert the efficiency into decimal form:

\[ \eta=0.65 \]

Apply the efficiency equation:

\[ P_{\text{useful}}=\eta P_{\text{input}} \]

\[ P_{\text{useful}}=0.65\times2.40 \]

\[ \boxed{P_{\text{useful}}=1.56\,\text{kW}} \]

Part b: Useful energy transferred

Convert the output power into watts:

\[ P_{\text{useful}}=1560\,\text{W} \]

Use:

\[ E_{\text{useful}}=P_{\text{useful}}t \]

Substitute:

\[ E_{\text{useful}}=1560\times90 \]

\[ \boxed{E_{\text{useful}}=140\,400\,\text{J}} \]

Therefore:

\[ \boxed{E_{\text{useful}}=140.4\,\text{kJ}} \]

Part c: Total input energy

The input power is \(2400\,\text{W}\).

\[ E_{\text{input}}=P_{\text{input}}t \]

\[ E_{\text{input}}=2400\times90 \]

\[ \boxed{E_{\text{input}}=216\,000\,\text{J}} \]

Therefore:

\[ \boxed{E_{\text{input}}=216\,\text{kJ}} \]

The pump receives \(216\,\text{kJ}\) of electrical energy, of which \(140.4\,\text{kJ}\) becomes useful energy transferred to the water.

The key distinction is that \(1.56\,\text{kW}\) describes the rate of useful energy transfer, while \(140.4\,\text{kJ}\) describes the total useful energy transferred over the \(90\,\text{s}\) interval.

Question 3

Two industrial compressors are being considered for the same task.

a. Calculate the useful output power of each compressor.

b. Determine which compressor produces more useful output power.

c. Calculate the minimum input power Compressor B would need to produce the same useful output power as Compressor A, assuming its efficiency remains \(86\%\).

Solution 3

Compressor A produces \(936\,\text{W}\), while Compressor B produces \(860\,\text{W}\). Compressor A therefore produces \(76\,\text{W}\) more useful power. Compressor B would need an input power of approximately \(1.09\,\text{kW}\) to match Compressor A.

Part a: Calculate both useful output powers

For Compressor A:

\[ P_A=0.78\times1200 \]

\[ \boxed{P_A=936\,\text{W}} \]

For Compressor B:

\[ P_B=0.86\times1000 \]

\[ \boxed{P_B=860\,\text{W}} \]

Part b: Compare the results

The difference is:

\[ \Delta P=936-860 \]

\[ \boxed{\Delta P=76\,\text{W}} \]

Compressor A produces more useful power, despite being less efficient.

This is because its greater input power more than compensates for its lower efficiency.

Part c: Find the required input power for Compressor B

We know that:

\[ P_{\text{useful}}=\eta P_{\text{input}} \]

Rearrange:

\[ P_{\text{input}}= \frac{P_{\text{useful}}}{\eta} \]

Substitute Compressor A’s useful output and Compressor B’s efficiency:

\[ P_{\text{input,B}}=\frac{936}{0.86} \]

\[ \boxed{P_{\text{input,B}}\approx1088\,\text{W}} \]

Or:

\[ \boxed{P_{\text{input,B}}\approx1.09\,\text{kW}} \]

Compressor B needs approximately \(1.09\,\text{kW}\) of input power to match Compressor A.

The tempting assumption is that the more efficient machine must always deliver more useful power. However, efficiency is a ratio, not a measure of the total power produced. Both the efficiency and the available input power determine the result.

Question 4

A student investigates the efficiency of an electric motor that raises a \(4.0\,\text{kg}\) mass vertically through \(1.5\,\text{m}\) in \(2.0\,\text{s}\).

The student initially measures the motor’s voltage and current without a load attached. These readings are \(24\,\text{V}\) and \(2.0\,\text{A}\).

During the actual lifting test, the motor terminal voltage and current are measured simultaneously and remain steady at \(20\,\text{V}\) and \(2.5\,\text{A}\).

Assume the mass begins and ends at rest, gravitational acceleration is \(9.8\,\text{m/s}^2\), and the only useful output being measured is the increase in gravitational potential energy.

a. Calculate the useful average output power.

b. Calculate the correct average input power during the lifting test.

c. Determine the efficiency of the motor.

d. Explain why using the original unloaded voltage and current readings gives an incorrect efficiency.

Solution 4

The useful average output power is \(29.4\,\text{W}\), the loaded input power is \(50\,\text{W}\), and the correct efficiency is approximately \(59\%\).

Part a: Calculate the useful output power

The useful energy transferred is the increase in gravitational potential energy:

\[ \Delta E_p=mgh \]

Substitute:

\[ \Delta E_p=4.0\times9.8\times1.5 \]

\[ \Delta E_p=58.8\,\text{J} \]

The average useful output power is:

\[ P_{\text{useful}}=\frac{\Delta E_p}{t} \]

\[ P_{\text{useful}}=\frac{58.8}{2.0} \]

\[ \boxed{P_{\text{useful}}=29.4\,\text{W}} \]

Part b: Calculate the loaded input power

The voltage and current must be taken from the loaded operating condition.

\[ P_{\text{input}}=VI \]

\[ P_{\text{input}}=20\times2.5 \]

\[ \boxed{P_{\text{input}}=50\,\text{W}} \]

Part c: Calculate efficiency

Use the ratio of useful output power to input power:

\[ \eta=\frac{P_{\text{useful}}}{P_{\text{input}}} \times100\% \]

\[ \eta=\frac{29.4}{50}\times100\% \]

\[ \boxed{\eta=58.8\%\approx59\%} \]

Part d: Explain the experimental error

The original readings were taken without a load attached.

When the motor raises a mass, it must provide additional torque. This changes the electrical conditions, including the current drawn by the motor.

Using the unloaded readings would give:

\[ P_{\text{input,unloaded}}=24\times2.0 \]

\[ P_{\text{input,unloaded}}=48\,\text{W} \]

This is not the actual input power during the lifting test.

Using \(48\,\text{W}\) would produce an apparent efficiency of:

\[ \eta=\frac{29.4}{48}\times100\% \approx61.3\% \]

That result is incorrect because the input and output measurements do not describe the same operating condition.

The revised measurement uses the loaded voltage and current, giving the correct efficiency of approximately \(59\%\).

This is an important experimental principle: efficiency is meaningful only when input and output measurements refer to the same device conditions and time interval.

Question 5

An industrial machine operates through two stages of a production cycle.

During the first stage, it receives \(600\,\text{W}\) of input power for \(10\,\text{s}\), operating at an efficiency of \(80\%\).

During the second stage, it receives \(200\,\text{W}\) of input power for \(10\,\text{s}\), operating at an efficiency of \(40\%\).

A student calculates the average efficiency as:

\[ \eta_{\text{average}}=\frac{80+40}{2}=60\% \]

The student also calculates the average input power as:

\[ P_{\text{input,average}}=\frac{600+200}{2} =400\,\text{W} \]

They conclude that the average useful output power is:

\[ P_{\text{useful,average}}=0.60\times400 =240\,\text{W} \]

a. Determine whether the student’s calculation of average useful output power is correct.

b. Calculate the total input energy, total useful output energy, average useful output power, and overall efficiency for the complete 20-second cycle.

c. The operator then extends the second stage from \(10\,\text{s}\) to \(20\,\text{s}\), without changing either stage’s input power or efficiency. Calculate the new overall efficiency and explain why it changes.

Assume the stated efficiencies remain constant throughout their respective stages, and there is no net change in energy stored within the machine over the complete operating cycle.

Solution 5

The student’s \(240\,\text{W}\) result is incorrect. The actual average useful output power is \(280\,\text{W}\), and the overall efficiency is \(70\%\). When the second stage is extended to \(20\,\text{s}\), the overall efficiency falls to \(64\%\).

The important issue is that efficiencies cannot generally be averaged by simply taking the arithmetic mean when the input power changes between stages.

Part a: Evaluate the student’s calculation

The student has correctly calculated the arithmetic mean of the two efficiency percentages and the arithmetic mean of the two input powers.

However, multiplying those two averages does not necessarily give the average useful output power.

Why?

Because the machine receives different amounts of input energy during the two stages.

Efficiency describes the fraction of input energy converted usefully. Therefore, each stage’s efficiency must be applied to the input energy received during that stage.

Part b: Calculate the overall efficiency and average useful output power

Start by examining each stage separately.

Stage 1

The input power is \(600\,\text{W}\), the efficiency is \(0.80\), and the operating time is \(10\,\text{s}\).

The useful output power is:

\[ P_{\text{useful,1}}=0.80\times600 \]

\[ P_{\text{useful,1}}=480\,\text{W} \]

The input energy is:

\[ E_{\text{input,1}}=600\times10 \]

\[ E_{\text{input,1}}=6000\,\text{J} \]

The useful output energy is:

\[ E_{\text{useful,1}}=480\times10 \]

\[ E_{\text{useful,1}}=4800\,\text{J} \]

Stage 2

The input power is \(200\,\text{W}\), the efficiency is \(0.40\), and the operating time is \(10\,\text{s}\).

The useful output power is:

\[ P_{\text{useful,2}}=0.40\times200 \]

\[ P_{\text{useful,2}}=80\,\text{W} \]

The input energy is:

\[ E_{\text{input,2}}=200\times10 \]

\[ E_{\text{input,2}}=2000\,\text{J} \]

The useful output energy is:

\[ E_{\text{useful,2}}=80\times10 \]

\[ E_{\text{useful,2}}=800\,\text{J} \]

Combine both stages

The total input energy is:

\[ E_{\text{input,total}}=6000+2000 \]

\[ \boxed{E_{\text{input,total}}=8000\,\text{J}} \]

The total useful output energy is:

\[ E_{\text{useful,total}}=4800+800 \]

\[ \boxed{E_{\text{useful,total}}=5600\,\text{J}} \]

The overall efficiency is therefore:

\[ \eta_{\text{overall}}= \frac{E_{\text{useful,total}}} {E_{\text{input,total}}}\times100\% \]

\[ \eta_{\text{overall}}= \frac{5600}{8000}\times100\% \]

\[ \boxed{\eta_{\text{overall}}=70\%} \]

The complete cycle lasts:

\[ t_{\text{total}}=10+10=20\,\text{s} \]

Therefore, the average useful output power is:

\[ P_{\text{useful,average}} =\frac{E_{\text{useful,total}}}{t_{\text{total}}} \]

\[ P_{\text{useful,average}} =\frac{5600}{20} \]

\[ \boxed{P_{\text{useful,average}}=280\,\text{W}} \]

The student’s answer of \(240\,\text{W}\) is too low.

The exact problem is that the arithmetic mean efficiency of \(60\%\) does not represent the overall efficiency of the cycle.

The first stage receives \(6000\,\text{J}\) of input energy, while the second receives only \(2000\,\text{J}\). The first stage therefore contributes much more to the overall energy balance.

The correct overall efficiency is effectively weighted by the input energy received during each stage:

\[ \eta_{\text{overall}} = \frac{\sum E_{\text{useful}}} {\sum E_{\text{input}}} \]

This is why averaging the two percentages by time gives the wrong result.

Part c: Extend the second stage

The second stage now lasts \(20\,\text{s}\) instead of \(10\,\text{s}\).

Its input power remains \(200\,\text{W}\), and its efficiency remains \(40\%\).

The first stage is unchanged.

The first stage still receives:

\[ E_{\text{input,1}}=600\times10 =6000\,\text{J} \]

Its useful output energy remains:

\[ E_{\text{useful,1}}=480\times10 =4800\,\text{J} \]

For the extended second stage:

\[ E_{\text{input,2}}=200\times20 =4000\,\text{J} \]

Its useful output energy is:

\[ E_{\text{useful,2}}=80\times20 =1600\,\text{J} \]

The new total input energy is:

\[ E_{\text{input,total}}=6000+4000 \]

\[ E_{\text{input,total}}=10\,000\,\text{J} \]

The new total useful output energy is:

\[ E_{\text{useful,total}}=4800+1600 \]

\[ E_{\text{useful,total}}=6400\,\text{J} \]

Therefore, the new overall efficiency is:

\[ \eta_{\text{overall}}= \frac{6400}{10\,000}\times100\% \]

\[ \boxed{\eta_{\text{overall}}=64\%} \]

The efficiency has decreased from \(70\%\) to \(64\%\).

This makes physical sense. The machine spends more time operating in its less efficient second stage. That stage contributes a larger proportion of the total input energy, so it has a greater influence on the overall efficiency.

The deeper point is that overall efficiency depends on how much energy enters the system under each operating condition, not simply how long each condition lasts.

06What to remember for HSC Physics

When a question asks you to calculate useful output power from efficiency and input power, the essential relationship is:

\[ \boxed{ P_{\text{useful output}} = \frac{\eta}{100}\times P_{\text{input}} } \]

Use this decision process:

1. Identify the input power supplied to the device.
2. Identify the efficiency and convert its percentage into a decimal.
3. Multiply the input power by the decimal efficiency.
4. Express your answer in watts or kilowatts, matching the units of the input.
5. If required, calculate non-useful power by subtracting useful output power from input power.

Also remember three important distinctions:

• Efficiency is the useful fraction of the input, not the fraction lost.
• Power is the rate of energy transfer, while energy is the total amount transferred.
• A more efficient device does not necessarily produce more useful power. Its input power matters too.

Once you can calculate useful output power confidently, the next step is connecting that result to the physical work done by a machine. For a motor lifting a load, this means using \(P=Fv\) or the rate of increase of gravitational potential energy, \(P=mgh/t\), to check whether your calculated output power makes sense.