The Efficiency Formula gives the ratio of useful output energy (or work) to the total input energy, expressed as a percentage: η = (Useful Output Energy / Total Input Energy) × 100%. This concept is covered in NCERT Physics for Class 9, Class 10, and Class 11, and it forms an essential part of the Work, Energy and Power chapter. For JEE Main and NEET aspirants, efficiency problems appear regularly in mechanics and thermodynamics. This article covers the formula, derivation, a complete formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Efficiency Formulas at a Glance
Quick reference for the most important efficiency-related formulas.
- Basic efficiency: \( \eta = \dfrac{W_{useful}}{E_{input}} \times 100\% \)
- Power-based efficiency: \( \eta = \dfrac{P_{output}}{P_{input}} \times 100\% \)
- Carnot efficiency: \( \eta_{Carnot} = \left(1 – \dfrac{T_L}{T_H}\right) \times 100\% \)
- Mechanical efficiency: \( \eta = \dfrac{M.A.}{V.R.} \times 100\% \)
- Energy lost: \( E_{lost} = E_{input} – W_{useful} \)
- Efficiency from losses: \( \eta = \left(1 – \dfrac{E_{lost}}{E_{input}}\right) \times 100\% \)
What is the Efficiency Formula?
The Efficiency Formula measures how effectively a machine, engine, or process converts input energy into useful output energy. No real machine is 100% efficient, because some energy is always lost as heat, sound, or friction. Efficiency is therefore always less than or equal to 100%.
In NCERT Physics Class 9 (Chapter 11 — Work and Energy) and Class 11 (Chapter 6 — Work, Energy and Power), efficiency is introduced as a dimensionless ratio. The symbol used is the Greek letter eta (η). A higher value of η means the device wastes less energy.
Efficiency applies to a wide range of systems: electric motors, heat engines, simple machines (pulleys, levers, inclined planes), power plants, and even biological processes like muscle contraction. Understanding efficiency helps engineers design better machines and helps students solve a broad class of CBSE and competitive exam problems.
The NCERT definition states: Efficiency is the ratio of useful work done by a machine to the total energy supplied to it. This is fully consistent with the law of conservation of energy, which says energy is neither created nor destroyed — only converted from one form to another.
Efficiency Formula — Expression and Variables
The general form of the efficiency formula is:
\[ \eta = \frac{W_{useful}}{E_{input}} \times 100\% \]
When expressed using power (rate of energy transfer), the formula becomes:
\[ \eta = \frac{P_{output}}{P_{input}} \times 100\% \]
For a heat engine (Class 11 Thermodynamics), the Carnot efficiency formula is:
\[ \eta_{Carnot} = \left(1 – \frac{T_L}{T_H}\right) \times 100\% \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \eta \) | Efficiency | Dimensionless (expressed as %) |
| \( W_{useful} \) | Useful work output | Joule (J) |
| \( E_{input} \) | Total input energy | Joule (J) |
| \( P_{output} \) | Useful output power | Watt (W) |
| \( P_{input} \) | Total input power | Watt (W) |
| \( T_L \) | Temperature of cold reservoir | Kelvin (K) |
| \( T_H \) | Temperature of hot reservoir | Kelvin (K) |
| \( E_{lost} \) | Energy lost (heat, friction, etc.) | Joule (J) |
Derivation of the Efficiency Formula
Start with the law of conservation of energy. All input energy either becomes useful output or is wasted:
\[ E_{input} = W_{useful} + E_{lost} \]
Efficiency is the fraction of input energy that becomes useful output. Multiply by 100 to express it as a percentage:
\[ \eta = \frac{W_{useful}}{E_{input}} \times 100\% \]
Substituting \( W_{useful} = E_{input} – E_{lost} \) gives the loss-based form:
\[ \eta = \left(1 – \frac{E_{lost}}{E_{input}}\right) \times 100\% \]
This derivation shows clearly that efficiency increases as energy losses decrease. It also confirms that η can never exceed 100%, because \( W_{useful} \leq E_{input} \) always.
Complete Physics Efficiency and Energy Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Basic Efficiency | \( \eta = \dfrac{W_{useful}}{E_{input}} \times 100\% \) | W = useful work, E = input energy | % (dimensionless) | Class 9, Ch 11; Class 11, Ch 6 |
| Power-based Efficiency | \( \eta = \dfrac{P_{out}}{P_{in}} \times 100\% \) | P = power (W) | % (dimensionless) | Class 11, Ch 6 |
| Carnot Efficiency | \( \eta = \left(1 – \dfrac{T_L}{T_H}\right) \times 100\% \) | T in Kelvin; L = cold sink, H = hot source | % (dimensionless) | Class 11, Ch 12 |
| Mechanical Efficiency (Simple Machine) | \( \eta = \dfrac{M.A.}{V.R.} \times 100\% \) | M.A. = Mechanical Advantage, V.R. = Velocity Ratio | % (dimensionless) | Class 9, Ch 11 |
| Work Done | \( W = F \cdot d \cdot \cos\theta \) | F = force (N), d = displacement (m), θ = angle | Joule (J) | Class 9, Ch 11; Class 11, Ch 6 |
| Kinetic Energy | \( KE = \dfrac{1}{2}mv^2 \) | m = mass (kg), v = velocity (m/s) | Joule (J) | Class 9, Ch 11; Class 11, Ch 6 |
| Potential Energy | \( PE = mgh \) | m = mass (kg), g = 9.8 m/s², h = height (m) | Joule (J) | Class 9, Ch 11; Class 11, Ch 6 |
| Power | \( P = \dfrac{W}{t} \) | W = work (J), t = time (s) | Watt (W) | Class 9, Ch 11; Class 11, Ch 6 |
| Heat Engine Efficiency | \( \eta = \dfrac{Q_H – Q_L}{Q_H} \times 100\% \) | Q_H = heat absorbed, Q_L = heat rejected | % (dimensionless) | Class 11, Ch 12 |
| Energy Conservation | \( E_{input} = W_{useful} + E_{lost} \) | All quantities in Joules | Joule (J) | Class 9, Ch 11; Class 11, Ch 6 |
Efficiency Formula — Solved Examples
Example 1 (Class 9-10 Level) — Basic Efficiency Calculation
Problem: A machine receives 500 J of energy and performs 350 J of useful work. Calculate the efficiency of the machine and the energy wasted.
Given:
- Total input energy, \( E_{input} = 500 \) J
- Useful work output, \( W_{useful} = 350 \) J
Step 1: Write the efficiency formula: \( \eta = \dfrac{W_{useful}}{E_{input}} \times 100\% \)
Step 2: Substitute values: \( \eta = \dfrac{350}{500} \times 100\% = 0.70 \times 100\% = 70\% \)
Step 3: Calculate energy wasted: \( E_{lost} = E_{input} – W_{useful} = 500 – 350 = 150 \) J
Answer
Efficiency of the machine = 70%. Energy wasted = 150 J.
Example 2 (Class 11-12 Level) — Power-based Efficiency
Problem: An electric motor draws 2.5 kW of electrical power from the mains. It lifts a load of 200 kg to a height of 10 m in 12 seconds. Calculate the efficiency of the motor. (Take g = 10 m/s²)
Given:
- Input power, \( P_{input} = 2.5 \) kW \( = 2500 \) W
- Mass, \( m = 200 \) kg
- Height, \( h = 10 \) m
- Time, \( t = 12 \) s
- \( g = 10 \) m/s²
Step 1: Calculate the useful work done against gravity: \( W_{useful} = mgh = 200 \times 10 \times 10 = 20000 \) J
Step 2: Calculate the useful output power: \( P_{output} = \dfrac{W_{useful}}{t} = \dfrac{20000}{12} \approx 1666.7 \) W
Step 3: Apply the power-based efficiency formula: \( \eta = \dfrac{P_{output}}{P_{input}} \times 100\% = \dfrac{1666.7}{2500} \times 100\% \approx 66.7\% \)
Answer
Efficiency of the electric motor ≈ 66.7%.
Example 3 (JEE/NEET Level) — Carnot Engine Efficiency
Problem: A Carnot engine operates between a hot reservoir at 627°C and a cold reservoir at 27°C. (a) Find the Carnot efficiency. (b) If the engine absorbs 8000 J of heat per cycle from the hot reservoir, how much useful work does it perform per cycle, and how much heat is rejected to the cold reservoir?
Given:
- Hot reservoir temperature, \( T_H = 627 + 273 = 900 \) K
- Cold reservoir temperature, \( T_L = 27 + 273 = 300 \) K
- Heat absorbed per cycle, \( Q_H = 8000 \) J
Step 1: Apply the Carnot efficiency formula: \( \eta_{Carnot} = \left(1 – \dfrac{T_L}{T_H}\right) \times 100\% \)
Step 2: Substitute temperatures: \( \eta = \left(1 – \dfrac{300}{900}\right) \times 100\% = \left(1 – \dfrac{1}{3}\right) \times 100\% = \dfrac{2}{3} \times 100\% \approx 66.7\% \)
Step 3: Calculate work done per cycle. Use \( \eta = \dfrac{W}{Q_H} \): \( W = \eta \times Q_H = \dfrac{2}{3} \times 8000 = \dfrac{16000}{3} \approx 5333 \) J
Step 4: Calculate heat rejected: \( Q_L = Q_H – W = 8000 – 5333 = 2667 \) J
Answer
(a) Carnot efficiency ≈ 66.7%. (b) Work done per cycle ≈ 5333 J. Heat rejected to cold reservoir ≈ 2667 J.
CBSE Exam Tips 2025-26
- Always convert temperatures to Kelvin before using the Carnot efficiency formula. Adding 273 to the Celsius value is the most common first step. Missing this conversion is a frequent source of lost marks.
- Write the formula before substituting values. CBSE marking schemes award one mark for writing the correct formula. Never skip this step, even if the calculation seems straightforward.
- State the unit clearly. Efficiency itself is dimensionless, but always write the % symbol in your final answer. Leaving it out may cost you half a mark in board exams.
- Distinguish between efficiency and percentage loss. If a question asks for energy wasted, first find \( E_{lost} = E_{input} – W_{useful} \), then calculate the loss percentage separately. We recommend writing both values clearly.
- For simple machines (Class 9), remember that mechanical efficiency uses Mechanical Advantage (M.A.) and Velocity Ratio (V.R.). Do not confuse these with energy-based efficiency in Class 11.
- Check your answer for physical sense. Efficiency must be between 0% and 100%. If you get a value above 100%, recheck your input and output values — they may have been swapped.
Common Mistakes to Avoid
- Swapping input and output: Students sometimes write \( \eta = E_{input} / W_{useful} \) instead of the correct form. Remember: useful output always goes in the numerator. Input energy is always the denominator.
- Using Celsius instead of Kelvin in Carnot formula: The Carnot efficiency formula requires absolute temperatures in Kelvin. Using °C directly gives a completely wrong answer. Always add 273 (or 273.15 for precision) to convert.
- Forgetting to multiply by 100: The ratio \( W_{useful}/E_{input} \) gives a decimal fraction (e.g., 0.75). You must multiply by 100 to express it as a percentage (75%). Leaving out this step is a very common error in Class 9 and Class 10 exams.
- Assuming efficiency can exceed 100%: A perpetual motion machine — one with efficiency greater than 100% — violates the first law of thermodynamics. No real machine achieves 100% efficiency either, due to friction and heat losses. If your calculated η exceeds 100%, your data is wrong.
- Confusing power efficiency with energy efficiency: Both give the same numerical result only when measured over the same time interval. In problems where power is given for one device and energy for another, convert everything to the same quantity (either both to energy or both to power) before applying the formula.
JEE/NEET Application of the Efficiency Formula
In our experience, JEE aspirants encounter the efficiency formula most frequently in three contexts: Carnot engines (JEE Main and Advanced), simple machines (JEE Main), and energy conversion problems in mechanics (both JEE and NEET).
Pattern 1 — Carnot Engine and Second Law (JEE Main, JEE Advanced)
JEE problems often combine Carnot efficiency with the concept of coefficient of performance (COP) of a refrigerator. The key relationship is:
\[ COP = \frac{T_L}{T_H – T_L} \]
A common JEE question type gives you the efficiency of a Carnot engine and asks for the COP of a Carnot refrigerator operating between the same temperatures. Our experts suggest memorising the relationship \( COP = (1 – \eta) / \eta \) (where η is expressed as a decimal) for quick calculations under exam conditions.
Pattern 2 — Multi-step Energy Conversion (NEET, JEE Main)
NEET problems frequently describe a chain of energy conversions — for example, a power plant where coal burns to produce steam, which drives a turbine, which drives a generator. Each stage has its own efficiency. The overall efficiency of a series of processes is the product of individual efficiencies:
\[ \eta_{overall} = \eta_1 \times \eta_2 \times \eta_3 \times \ldots \]
This formula is not explicitly in NCERT but is derived directly from the basic efficiency formula. It appears in both NEET Biology (muscle efficiency) and NEET/JEE Physics (engine chains).
Pattern 3 — Efficiency and Friction in Simple Machines (JEE Main)
For a simple machine like a pulley or inclined plane, the velocity ratio (V.R.) is fixed by geometry. However, friction reduces the mechanical advantage (M.A.). The efficiency formula \( \eta = M.A. / V.R. \) therefore quantifies the effect of friction. JEE Main questions may ask you to find the effort required to lift a load given the efficiency and V.R. — a direct rearrangement of the mechanical efficiency formula.
FAQs on Efficiency Formula
For more related formulas, explore our detailed articles on the Capacitance Formula, the Refractive Index Formula, and the Uniform Circular Motion Formula. You can also browse the complete Physics Formulas hub for a full list of NCERT-aligned formula articles for Classes 9 to 12. For the official NCERT syllabus reference, visit ncert.nic.in.