Input/Output Power Equation, Derivation & Rearrangements

Core idea

Overview

This equation quantifies energy conversion efficiency by comparing the useful power output to the total power input supplied to a system. It accounts for inevitable energy losses due to friction, heat, and sound, providing a percentage that represents mechanical or electrical effectiveness.

When to use: Use this formula when analyzing the performance of engines, motors, transformers, or any system where energy is transformed. It requires that both input and output power values are expressed in the same units, such as Watts or Horsepower.

Why it matters: Efficiency is fundamental to engineering because it dictates operational costs and environmental footprint. Improving efficiency reduces waste, allows for smaller cooling systems, and maximizes the utility derived from finite energy resources.

Symbols

Variables

\eta = Efficiency, P_{out} = Power Output, P_{in} = Power Input

η

Efficiency

P_{o u t}

Power Output

W

P_{in}

Power Input

W

Walkthrough

Derivation

Formula: Mechanical Power

Power is the rate of doing work or transferring energy. In mechanics it links work, force, distance, and time.

The force is constant over the time interval.
The motion is in the same direction as the force (so work is F $\cdot$ d).

1

Start with the Definition of Power:

Power (P) equals work done (W) divided by time (t). The unit is the watt (W).

P = \frac{W}{t}

2

Substitute Work Done:

Replace work with force times distance to link power to force and motion.

W = F \cdot d ⟹ P = \frac{F \cdot d}{t}

3

Link to Velocity:

Since v = d/t, mechanical power can also be found by multiplying force by velocity.

P = F \cdot v

Result

P = F \cdot v

Source: AQA GCSE Engineering — Energy and Power

Free formulas

Rearrangements

Solve for $η$

Make Eff the subject

η = \frac{P _{o u t}}{P _{in}} \times 100

Eff is already the subject of the formula.

Difficulty: 1/5

Solve for $P_{o u t}$

Make Pout the subject

P_{o u t} = \frac{η \times P _{in}}{100}

To make $P_{o u t}$ the subject of the efficiency formula, multiply both sides by $P_{in}$ and then divide by 100.

Difficulty: 2/5

Solve for $P_{in}$

Make Pin the subject

P_{in} = \frac{P _{o u t} \times 100}{η}

Rearrange the efficiency formula to solve for power input, ` $P_{in}$ `.

Difficulty: 2/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

The graph follows an inverse relationship where efficiency decreases as power input increases. Because power input is in the denominator, the curve approaches the horizontal axis as an asymptote but never touches it, remaining valid only for positive values. For an engineering student, this means that as power input grows very large for a fixed power output, the efficiency drops toward zero, while very small power inputs result in disproportionately high efficiency values. The most important feature is that the cur

Graph type: inverse

Why it behaves this way

Intuition

Visualize a system as a process where a certain amount of power flows in, and a smaller, useful portion flows out, with the remainder being dissipated internally, often as heat.

η

The proportion of total input power that is converted into useful output power.

A measure of how effectively a system uses its supplied energy to perform its intended function; a higher percentage means less waste.

P_{o u t}

The rate at which useful energy is delivered by the system.

The actual power that accomplishes the desired task or work.

P_{in}

The total rate at which energy is supplied to the system.

All the power put into the system, including what will be lost as heat, sound, or friction.

Signs and relationships

P_{out}/P_{in}: This ratio represents the fraction of input power that is successfully converted into useful output power. Due to the conservation of energy and inevitable losses, the useful output power ( $P_{o}$ ut)

Free study cues

Insight

Canonical usage

Efficiency is calculated as a dimensionless ratio or percentage by dividing output power by input power, requiring both quantities to share identical units.

Common confusion

Attempting to calculate efficiency using different unit scales for input and output (e.g., dividing Watts by kilowatts) or obtaining a value over 100%, which would violate the Law of Conservation of Energy.

Dimension note

Efficiency is the ratio of two identical physical quantities (Power/Power), resulting in the cancellation of all units.

Unit systems

$P_{o} u t$ W · The useful power transferred by the device.

$P_{i} n$ W · The total power supplied to the device; must be in the same units as P_out.

$η$ % · Efficiency is a dimensionless ratio, typically multiplied by 100 to be expressed as a percentage.

Ballpark figures

Quantity:

One free problem

Practice Problem

An industrial electric motor consumes 800 Watts of electrical power and produces 680 Watts of mechanical power. What is the efficiency of the motor?

Power Input800 W

Power Output680 W

Solve for: $E f f$

Hint: Divide the useful output power by the total input power and multiply the result by 100.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Motor efficiency rating.

Study smarter

Tips

Output power must always be less than or equal to input power due to the Law of Conservation of Energy.
Always convert your result to a percentage by multiplying the ratio by 100.
Check that units for Pin and Pout are consistent before beginning the calculation.

Avoid these traps

Common Mistakes

Inverting output/input.
Forgetting ×100.
Getting efficiency > 100%.

Keep going

Related Formulas

Common questions

Frequently Asked Questions

Power is the rate of doing work or transferring energy. In mechanics it links work, force, distance, and time.

Use this formula when analyzing the performance of engines, motors, transformers, or any system where energy is transformed. It requires that both input and output power values are expressed in the same units, such as Watts or Horsepower.

Efficiency is fundamental to engineering because it dictates operational costs and environmental footprint. Improving efficiency reduces waste, allows for smaller cooling systems, and maximizes the utility derived from finite energy resources.

Inverting output/input. Forgetting ×100. Getting efficiency > 100%.

Motor efficiency rating.

Output power must always be less than or equal to input power due to the Law of Conservation of Energy. Always convert your result to a percentage by multiplying the ratio by 100. Check that units for Pin and Pout are consistent before beginning the calculation.

References

Sources

Halliday, David, Resnick, Robert, Walker, Jearl. Fundamentals of Physics.
Bird, R. Byron, Stewart, Warren E., Lightfoot, Edwin N. Transport Phenomena.
Incropera, Frank P., DeWitt, David P., Bergman, Theodore L., Lavine, Adrienne S. Fundamentals of Heat and Mass Transfer.
Wikipedia: Energy conversion efficiency
IUPAC Gold Book
AQA GCSE Physics
Britannica
Halliday, Resnick, and Walker, Fundamentals of Physics

Input/Output Power

Overview

Variables

Derivation

Start with the Definition of Power:

Substitute Work Done:

Link to Velocity:

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Mechanical Efficiency

Frequently Asked Questions

Sources