Formula to Calculate Efficiency

This calculator works out the efficiency of any process or machine when you enter the useful energy output, the total energy input, and (optionally) the time taken. You get back the efficiency as a percentage, plus the power output and power input in watts if you supplied a time. The core formula is Efficiency (%) = (Useful Output / Total Input) × 100.

Formula

The efficiency formula is written as:

η = (E_out / E_in) × 100%

Where:

• η (eta) is the efficiency expressed as a percentage.
• E_out is the useful energy output, measured in joules (J) or kilowatt-hours (kWh).
• E_in is the total energy input, measured in the same unit as E_out.

If you are working with power instead of energy, the same ratio applies: Efficiency = (Power Output / Power Input) × 100%. Power is energy per unit time, so the time cancels out.

Worked Example

Using the default values from the calculator: an electric motor draws 1000 J of electrical energy, and delivers 800 J of mechanical work. What is its efficiency?

1. Write down the formula: η = (E_out / E_in) × 100%
2. Insert the values: η = (800 J / 1000 J) × 100%
3. Divide: 800 / 1000 = 0.8
4. Multiply by 100: 0.8 × 100 = 80%

The motor is 80% efficient. The remaining 200 J is lost, usually as heat and sound. If the process took 2 seconds, the power output would be 800 J / 2 s = 400 W, and the power input 1000 J / 2 s = 500 W.

When to Use This Formula

The efficiency formula applies anywhere energy is converted from one form to another. Common uses include:

• Electric motors and generators
• Light bulbs (luminous efficiency is much lower than total efficiency)
• Car engines (thermal efficiency)
• Power supplies and chargers
• Human metabolism (work output vs. food energy)
• Solar panels (electrical output vs. sunlight energy)

In every case, you need two numbers: the useful output and the total input. Both must be in the same unit. If you have power in watts, use watts for both; if you have energy in joules, use joules for both. Mixing units is the most common error.

How to Read the Result

The result is a percentage between 0% and 100% (or higher for heat pumps, which we’ll discuss later). A higher percentage means less energy is wasted. For example:

• LED light bulbs: about 20-30% electrical-to-light efficiency
• Incandescent bulbs: about 5%
• Modern gas car engines: about 25-30%
• Large electric generators: about 90-98%
• Lithium-ion batteries: about 90-95% round-trip

No real process reaches 100% because of friction, heat loss, electrical resistance, and other inevitable losses. The second law of thermodynamics states that some energy is always dissipated as low-grade heat.

Common Mistakes and Pitfalls

Several errors trip up beginners and even experienced engineers:

• Mixing energy and power: Efficiency must compare energy to energy or power to power. Never divide joules by watts without converting.
• Forgetting to multiply by 100: The ratio 0.8 is not the percentage; 80% is.
• Using the wrong output: For a light bulb, the useful output is light, not total heat. For a car, useful output is mechanical work at the wheels, not the chemical energy in fuel.
• Ignoring units: If one value is in kilojoules and the other in joules, convert first. The calculator assumes consistent units.
• Rounding too early: Round only at the final step to avoid accumulating errors.

Limits of the Formula

The simple ratio works for steady-state processes where input and output are measured over the same period. It does not capture:

• Time-varying processes: If a machine warms up or cools down, instantaneous efficiency changes.
• Quality of energy: A joule of electricity is more useful than a joule of low-temperature heat, but the formula treats them equally.
• Exergy: Thermodynamic efficiency (second-law efficiency) compares actual output to the theoretical maximum, which is often more meaningful.
• Heat pumps and refrigerators: They can have a coefficient of performance (COP) above 100% because they move heat rather than convert it. For example, a heat pump with a COP of 3 delivers 3 kWh of heat for every 1 kWh of electricity. The efficiency formula above would give 300%, which is not wrong but is usually called COP instead.

Frequently Asked Questions

What is the difference between efficiency and effectiveness?

Efficiency is a ratio of useful output to total input, always a number between 0% and 100% (except for heat pumps). Effectiveness is a broader concept: it measures how well a goal is achieved, regardless of resources used. A machine can be effective (it does the job) but inefficient (it wastes energy).

Can efficiency be greater than 100%?

For ordinary energy conversion, no. That would violate conservation of energy. The exception is heat pumps and refrigerators, which move heat from a cold place to a hot place using external work. Their coefficient of performance (COP) can be 3 or 4, meaning they deliver more heat than the electrical energy they consume. But they are not creating energy; they are transporting it.

How do I calculate efficiency if I only have power values?

Use the same formula with power instead of energy: Efficiency = (Power Output / Power Input) × 100%. Since power is energy per second, the time cancels out. For example, if a device draws 500 W and outputs 400 W of useful power, efficiency = (400/500) × 100 = 80%.

What is the efficiency of the human body?

For mechanical work (like cycling or lifting), the human body is about 20-25% efficient. That means only about a quarter of the food energy you consume is converted to mechanical work; the rest is released as heat. This is why you get warm during exercise.

Why do manufacturers sometimes quote efficiency as a decimal instead of a percentage?

Both are common. A decimal (0.8) is the same as a percentage (80%). In engineering calculations, decimals are often easier to use in formulas because you don’t have to divide by 100 again. Just be consistent: if you multiply by 100, you get a percentage; if you don’t, you get a fraction.

Does efficiency change with load?

Yes, for most machines. Motors, engines, and power supplies usually have a peak efficiency at some partial load, often around 70-80% of full load. At very low load, fixed losses (like bearing friction) dominate; at very high load, resistive losses increase. The nameplate efficiency is often given at the rated load.

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in e_out | Useful energy output | J | 800 | 0
in e_in | Total energy input | J | 1000 | 0
in t_time | Time (optional) | s | - | 0
out eff | Efficiency | % | (e_out / e_in) * 100 | 1
out p_out | Power output | W | e_out / t_time | 1
out p_in | Power input | W | e_in / t_time | 1