Calculating Efficiency from Input and Output Power or Energy

Good to see you again. In the previous lesson, you calculated electrical energy in $\mathrm{kWh}$ from power and time, then found an operating cost from a tariff. Efficiency uses the same underlying idea of energy transfer, but asks a different question: how much of what was supplied became the output the device was meant to provide?

For this lesson, you will calculate efficiency from given input and useful output energy or power, present it as a percentage, and use a quick reasonableness check to catch reversed fractions. This is a frequent short calculation in electrical exams, so clean formula setup matters.

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Useful output versus wasted output

Every operating electrical device takes in electrical energy or electrical power. It transfers that input into one or more outputs.

Whether an output is useful depends on the device’s purpose:

Device	Useful output	Common wasted output
Motor	Mechanical movement	Heat and sound
Lamp	Light	Heat
Kettle	Heating the water	Heating the kettle body and surroundings
Fan	Air movement	Heat and sound
Loudspeaker	Sound	Heat

Do not automatically label all heat as wasted. For a heater or kettle, heating is the intended result, so heat transferred to the room or water is useful. Heat lost into unwanted parts of the appliance or the surroundings is not useful.

The energy-flow diagram below represents a water-heating device. Of the supplied electrical work, $180,000\mathrm{J}$ increases the water’s internal energy, which is useful here. The remaining $20,000\mathrm{J}$ is dissipated to the surroundings as infrared radiation.

The full input has not disappeared. It has been transferred between useful and non-useful outputs:

$$E_{\mathrm{in}}=E_{\mathrm{useful}}+E_{\mathrm{wasted}}$$

This is why an ordinary device cannot have an efficiency greater than $100\%$. A result above $100\%$ means the output has been divided by the input in the wrong order, the units have not been converted correctly, or the values have been copied incorrectly.

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The efficiency formula

Efficiency is the proportion of the input that becomes useful output. It has no unit, because the input and output units cancel.

When the question gives energies:

$$\eta =\frac{E_{\mathrm{useful}}}{E_{\mathrm{in}}}$$

When the question gives powers:

$$\eta =\frac{P_{\mathrm{useful}}}{P_{\mathrm{in}}}$$

The symbol $\eta$, pronounced “eta,” represents efficiency as a decimal. Exams often ask for efficiency as a percentage, so use:

$$\eta (\%)=\frac{\text{useful output}}{\text{total input}}\times 100\%$$

The most important memory cue is:

Useful output goes on top; total input goes underneath.

You may use either energy values or power values, but do not mix them in one fraction. For example, this is not valid:

$$\frac{120\mathrm{J}}{200\mathrm{W}}$$

Energy and power are different quantities. Use energy divided by energy, or power divided by power.

Also make the units match before dividing:

$$\frac{750\mathrm{W}}{1.0\mathrm{kW}}$$

is not yet ready for calculation. First convert one value:

$$1.0\mathrm{kW}=1000\mathrm{W}$$

Then the ratio is valid:

$$\frac{750\mathrm{W}}{1000\mathrm{W}}=0.75$$

The unit cancellation shows why efficiency has no unit:

$$\frac{\mathrm{W}}{\mathrm{W}}=1$$

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Reading an energy-flow diagram

Use the water-heating diagram as an exam-style example.

Given:

$$E_{\mathrm{in}}=200,000\mathrm{J}$$ $$E_{\mathrm{useful}}=180,000\mathrm{J}$$

Required: efficiency as a percentage.

Write the formula first:

$$\eta (\%)=\frac{E_{\mathrm{useful}}}{E_{\mathrm{in}}}\times 100\%$$

Substitute the useful energy on top and input energy below:

$$\eta =\frac{180,000}{200,000}\times 100\%$$ $$\eta =90\%$$ $$\boxed{\eta =90\%}$$

A useful extra check is to find the wasted percentage:

$$\text{wasted percentage}=100\%-90\%$$ $$\text{wasted percentage}=10\%$$

That agrees with the diagram: $20,000\mathrm{J}$ is one tenth of $200,000\mathrm{J}$.

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Worked power example

Question: A motor takes an input power of $2.0\mathrm{kW}$. Its useful mechanical output power is $1500\mathrm{W}$. Calculate its efficiency.

1. List the quantities and make units match

$$P_{\mathrm{in}}=2.0\mathrm{kW}$$ $$P_{\mathrm{useful}}=1500\mathrm{W}$$

Convert the input power to watts:

$$P_{\mathrm{in}}=2.0\times 1000$$ $$P_{\mathrm{in}}=2000\mathrm{W}$$

2. Select the power form of the formula

$$\eta (\%)=\frac{P_{\mathrm{useful}}}{P_{\mathrm{in}}}\times 100\%$$

3. Substitute

$$\eta =\frac{1500}{2000}\times 100\%$$ $$\eta =75\%$$ $$\boxed{\eta =75\%}$$

The motor wastes the remaining $25\%$ of the input power, mainly as heat and sound. In power terms, that is:

$$P_{\mathrm{wasted}}=P_{\mathrm{in}}-P_{\mathrm{useful}}$$ $$P_{\mathrm{wasted}}=2000-1500$$ $$P_{\mathrm{wasted}}=500\mathrm{W}$$

This is a useful physical check: useful output plus wasted output equals input power.

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Worked energy example

Question: During one operating period, a motor receives $12.0\mathrm{kJ}$ of electrical energy and delivers $9.6\mathrm{kJ}$ of useful mechanical energy. Calculate its efficiency.

Both quantities are energies and both are already in $\mathrm{kJ}$, so no conversion is needed.

$$\eta (\%)=\frac{E_{\mathrm{useful}}}{E_{\mathrm{in}}}\times 100\%$$ $$\eta =\frac{9.6}{12.0}\times 100\%$$ $$\eta =80\%$$ $$\boxed{\eta =80\%}$$

The non-useful energy transferred during that period is:

$$E_{\mathrm{wasted}}=E_{\mathrm{in}}-E_{\mathrm{useful}}$$ $$E_{\mathrm{wasted}}=12.0-9.6$$ $$E_{\mathrm{wasted}}=2.4\mathrm{kJ}$$

A quick estimate supports the answer: $9.6\mathrm{kJ}$ is a little less than $12\mathrm{kJ}$, so efficiency should be a little less than $100\%$. The exact result of $80\%$ is reasonable.

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A reliable exam setup

For a direct efficiency calculation, follow this sequence every time:

1. Identify the useful output.
   Ask what the device is designed to do: produce movement, light, sound, heat in a target material, and so on.

2. Identify the total input.
   In an electrical question, this is commonly supplied electrical energy or input electrical power.

3. Use matching quantities.
   Use either energy values or power values, not one of each.

4. Convert units if needed.
   Both values must be in the same unit, such as $\mathrm{W}$ and $\mathrm{W}$, or $\mathrm{kJ}$ and $\mathrm{kJ}$.

5. Write the formula before substituting.

$$\eta (\%)=\frac{\text{useful output}}{\text{total input}}\times 100\%$$

6. Put useful output on top.
   This is the most common exam error.

7. Check the result.
   For an ordinary device, the result should be greater than $0\%$ and no more than $100\%$.

For calculator entry, use brackets to make the intended order clear:

$$(1500÷2000)\times 100$$

Do not round during the division unless the question tells you to. Round only the final percentage if necessary.

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Common mistakes and how to catch them

Reversing the fraction

Incorrect:

$$\frac{P_{\mathrm{in}}}{P_{\mathrm{useful}}}\times 100\%$$

For the motor example, this would give:

$$\frac{2000}{1500}\times 100\%=133.3\%$$

That cannot be correct because the motor cannot supply more useful power than it receives as input.

Forgetting to multiply by $100\%$

$$\frac{1500}{2000}=0.75$$

This is a correct decimal efficiency, but if the question asks for a percentage, write:

$$0.75\times 100\%=75\%$$

Using total output rather than useful output

A device may produce several outputs. Only the output that meets the intended purpose belongs in the numerator. For a lamp, light is useful; unwanted heating of the room is not. For a room heater, heating the room is useful.

Dividing different units without converting

Before calculating, scan the units. A power ratio must be, for example, $\mathrm{W}/\mathrm{W}$ or $\mathrm{kW}/\mathrm{kW}$. An energy ratio must be $\mathrm{J}/\mathrm{J}$, $\mathrm{kJ}/\mathrm{kJ}$, or another matching pair.

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Key takeaways

Efficiency tells you the percentage of the total input that appears as the intended, useful output.

Use:

$$\eta (\%)=\frac{E_{\mathrm{useful}}}{E_{\mathrm{in}}}\times 100\%$$

or:

$$\eta (\%)=\frac{P_{\mathrm{useful}}}{P_{\mathrm{in}}}\times 100\%$$

Keep these checks in mind:

• useful output belongs on top;
• input belongs underneath;
• use energy with energy or power with power;
• convert units so they match;
• an ordinary device’s efficiency cannot exceed $100\%$.

You have now completed the core DC quantities, power, energy, cost, and efficiency section. Next, the course moves into series circuits, beginning with total resistance, circuit current, and individual voltage drops.