The Power Formula defines the rate at which work is done or energy is transferred, expressed as P = W/t, where P is power in Watts, W is work in Joules, and t is time in seconds. This fundamental concept is covered in NCERT Class 9 Chapter 11 (Work and Energy) and Class 10 Chapter 12 (Electricity), and it appears regularly in CBSE board exams, JEE Main, and NEET. This article covers every form of the Power Formula, a complete formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Power Formula Expressions at a Glance
Quick reference for the most important Power Formula variants used in CBSE and competitive exams.
- Basic Power Formula: \( P = \dfrac{W}{t} \)
- Power in terms of force and velocity: \( P = F \cdot v \)
- Electric Power: \( P = V \times I \)
- Electric Power (resistance form): \( P = I^2 R \)
- Electric Power (voltage form): \( P = \dfrac{V^2}{R} \)
- Average Power: \( P_{avg} = \dfrac{\Delta W}{\Delta t} \)
- Instantaneous Power: \( P = \vec{F} \cdot \vec{v} \)
What is the Power Formula?
The Power Formula quantifies how quickly energy is used or work is performed. In physics, power is not just about how much work is done — it is about how fast that work happens. A machine that lifts a heavy load slowly has the same work output as one that lifts it quickly, but the faster machine has greater power.
Formally, power is defined as the work done per unit time. The SI unit of power is the Watt (W), named after Scottish engineer James Watt. One Watt equals one Joule of work done per second.
In NCERT Class 9 (Chapter 11 — Work and Energy), students first encounter the mechanical Power Formula. In Class 10 (Chapter 12 — Electricity), the electric Power Formula is introduced. Both forms are essential for CBSE board exams. In Class 11 (Chapter 6 — Work, Energy and Power), the concept is extended to instantaneous power and the dot product form involving force and velocity. Understanding every variant of the Power Formula is critical for scoring full marks in board exams and competitive tests like JEE and NEET.
Power Formula — Expression and Variables
The most fundamental expression of the Power Formula is:
\[ P = \frac{W}{t} \]
For mechanical systems involving a constant force, the Power Formula can also be written as:
\[ P = F \cdot v \]
For electric circuits, the Power Formula takes three equivalent forms:
\[ P = V \times I \quad = \quad I^2 R \quad = \quad \frac{V^2}{R} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | Power | Watt (W) = J/s |
| W | Work Done | Joule (J) |
| t | Time | Second (s) |
| F | Force | Newton (N) |
| v | Velocity | metre per second (m/s) |
| V | Voltage (Potential Difference) | Volt (V) |
| I | Electric Current | Ampere (A) |
| R | Resistance | Ohm (Ω) |
Derivation of the Power Formula
The mechanical Power Formula is derived directly from the definition of work. Work done is given by \( W = F \times d \), where d is displacement. Dividing both sides by time t gives:
\[ P = \frac{W}{t} = \frac{F \times d}{t} \]
Since displacement divided by time equals velocity \( (v = d/t) \), the expression simplifies to:
\[ P = F \times v \]
For electric power, using Ohm’s Law \( V = IR \), we substitute into \( P = VI \) to get \( P = I^2R \) or \( P = V^2/R \). These three electric power expressions are all equivalent and interchangeable depending on which quantities are known.
Complete Physics Power Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Basic Power Formula | \( P = W/t \) | W = work, t = time | Watt (W) | Class 9, Ch 11 |
| Power from Force and Velocity | \( P = F \cdot v \) | F = force, v = velocity | Watt (W) | Class 11, Ch 6 |
| Electric Power (VI form) | \( P = VI \) | V = voltage, I = current | Watt (W) | Class 10, Ch 12 |
| Electric Power (I²R form) | \( P = I^2 R \) | I = current, R = resistance | Watt (W) | Class 10, Ch 12 |
| Electric Power (V²/R form) | \( P = V^2/R \) | V = voltage, R = resistance | Watt (W) | Class 10, Ch 12 |
| Instantaneous Power | \( P = \vec{F} \cdot \vec{v} \) | F = force vector, v = velocity vector | Watt (W) | Class 11, Ch 6 |
| Average Power | \( P_{avg} = \Delta W / \Delta t \) | ΔW = change in work, Δt = time interval | Watt (W) | Class 11, Ch 6 |
| Power in terms of kinetic energy | \( P = \Delta KE / t \) | ΔKE = change in kinetic energy, t = time | Watt (W) | Class 11, Ch 6 |
| Horsepower (practical unit) | \( 1 \text{ hp} = 746 \text{ W} \) | Conversion factor | Watt (W) | Class 11, Ch 6 |
| Energy from Power | \( E = P \times t \) | P = power, t = time | Joule (J) or kWh | Class 10, Ch 12 |
Power Formula — Solved Examples
Example 1 (Class 9-10 Level — Basic Mechanical Power)
Problem: A worker lifts a box of mass 20 kg to a height of 5 m in 10 seconds. Calculate the power developed by the worker. (Take g = 10 m/s²)
Given: Mass m = 20 kg, Height h = 5 m, Time t = 10 s, g = 10 m/s²
Step 1: Calculate the work done against gravity using \( W = mgh \).
\( W = 20 \times 10 \times 5 = 1000 \) J
Step 2: Apply the Power Formula \( P = W/t \).
\( P = \dfrac{1000}{10} = 100 \) W
Answer
The power developed by the worker is 100 Watts.
Example 2 (Class 10-11 Level — Electric Power)
Problem: A household electric iron operates at a voltage of 220 V and draws a current of 5 A. Calculate (a) the power consumed, (b) the resistance of the heating element, and (c) the energy consumed in 2 hours.
Given: V = 220 V, I = 5 A, t = 2 hours = 7200 s
Step 1: Calculate power using \( P = VI \).
\( P = 220 \times 5 = 1100 \) W
Step 2: Find resistance using \( P = V^2/R \), so \( R = V^2/P \).
\( R = \dfrac{220^2}{1100} = \dfrac{48400}{1100} = 44 \) Ω
Step 3: Calculate energy consumed using \( E = P \times t \).
\( E = 1100 \times 7200 = 7{,}920{,}000 \) J = 7.92 MJ
In kilowatt-hours: \( E = 1.1 \text{ kW} \times 2 \text{ h} = 2.2 \) kWh
Answer
(a) Power = 1100 W, (b) Resistance = 44 Ω, (c) Energy consumed = 2.2 kWh
Example 3 (JEE/NEET Level — Power with Variable Force)
Problem: A car of mass 1200 kg accelerates from rest to 20 m/s in 10 seconds on a level road. The frictional force opposing the motion is 400 N. Calculate the average power delivered by the engine during this acceleration.
Given: m = 1200 kg, u = 0 m/s, v = 20 m/s, t = 10 s, Friction force f = 400 N
Step 1: Calculate acceleration using \( a = (v – u)/t \).
\( a = (20 – 0)/10 = 2 \) m/s²
Step 2: Find the net force needed for acceleration using Newton's second law.
\( F_{net} = ma = 1200 \times 2 = 2400 \) N
Step 3: The engine force must overcome both inertia and friction.
\( F_{engine} = F_{net} + f = 2400 + 400 = 2800 \) N
Step 4: Calculate the distance covered using \( s = ut + \frac{1}{2}at^2 \).
\( s = 0 + \frac{1}{2} \times 2 \times 100 = 100 \) m
Step 5: Calculate work done by the engine.
\( W = F_{engine} \times s = 2800 \times 100 = 280{,}000 \) J
Step 6: Apply the Power Formula \( P = W/t \).
\( P = \dfrac{280000}{10} = 28000 \) W = 28 kW
Alternative check using \( P = F \cdot v_{avg} \): Average velocity \( v_{avg} = (0 + 20)/2 = 10 \) m/s. So \( P = 2800 \times 10 = 28000 \) W. Both methods agree.
Answer
The average power delivered by the engine is 28,000 W (28 kW).
CBSE Exam Tips 2025-26
- Know all three electric power forms: CBSE Class 10 boards frequently ask you to use \( P = VI \), \( P = I^2R \), and \( P = V^2/R \). We recommend memorising which form to use based on the quantities given in the problem.
- Always state the SI unit: Writing the answer without the unit (Watt) costs you half a mark. Always write W or kW as appropriate.
- Convert units before substituting: If time is given in minutes or hours, convert to seconds before using \( P = W/t \). For energy bills, convert to kilowatt-hours (kWh).
- Link power to energy: Many 3-mark questions combine \( P = W/t \) with \( W = mgh \) or \( W = \frac{1}{2}mv^2 \). Practise these combined problems regularly.
- Distinguish between power and energy: CBSE examiners often include a 1-mark question asking students to differentiate power (rate of energy use) from energy (total work done). Keep this distinction clear.
- Horsepower conversion: Remember that 1 hp = 746 W. This conversion appears in both Class 11 NCERT exercises and CBSE sample papers for 2025-26.
Common Mistakes to Avoid
- Confusing Power with Energy: Power is the rate of energy use, not the total energy. A 100 W bulb and a 200 W bulb both use energy, but the 200 W bulb uses it twice as fast. Never substitute energy values directly into power equations without accounting for time.
- Using the wrong electric power formula: Students often apply \( P = I^2R \) when only V and R are given. Always check which two quantities are provided and choose the matching formula. If V and R are given, use \( P = V^2/R \).
- Forgetting to convert time units: The Power Formula requires time in seconds. If the problem states 5 minutes, use t = 300 s, not t = 5.
- Ignoring friction in mechanical power problems: In JEE-level problems, the engine must overcome both the inertial force and friction. Forgetting the friction force leads to an underestimated power value.
- Mixing up kW and kWh: Kilowatt (kW) is a unit of power. Kilowatt-hour (kWh) is a unit of energy. Electricity bills use kWh, not kW. This distinction is tested in CBSE Class 10 and Class 11 exams.
JEE/NEET Application of the Power Formula
In our experience, JEE aspirants encounter the Power Formula in at least two to three questions per paper, spanning mechanics, electricity, and modern physics. NEET questions tend to focus on electric power and biological energy expenditure contexts. Here are the three most common application patterns:
Pattern 1: Power Delivered by a Variable Force (JEE Main)
JEE Main frequently tests the instantaneous power formula \( P = \vec{F} \cdot \vec{v} \). When a force acts at an angle θ to velocity, the formula becomes \( P = Fv\cos\theta \). Questions may give force as a function of time or displacement, requiring integration to find average power. Our experts suggest practising problems where \( F = F_0 + kt \) and velocity must be derived from Newton's second law before calculating power.
Pattern 2: Power Dissipation in Resistor Networks (JEE Main and NEET)
Both JEE and NEET test electric power in series and parallel circuits. In a series circuit, the same current flows through all resistors, so \( P = I^2R \) is the most convenient form. In a parallel circuit, the voltage across each branch is equal, so \( P = V^2/R \) is preferred. A common JEE question asks which resistor dissipates more power in a series combination — the answer is always the one with higher resistance, since \( P \propto R \) when current is constant.
Pattern 3: Efficiency and Power (JEE Advanced)
JEE Advanced introduces the concept of efficiency: \( \eta = P_{output}/P_{input} \times 100\% \). A pump lifting water to a height h in time t has output power \( P_{out} = mgh/t \). If the pump's efficiency is given, the input power can be calculated. These multi-step problems combine the Power Formula with fluid mechanics or thermodynamics, making them high-difficulty questions worth practising thoroughly.
For NEET, metabolic power problems appear occasionally. The human body's basal metabolic rate is approximately 80 W. Questions may ask how long a person must exercise at a given power output to burn a specific number of Calories. The same Power Formula \( P = E/t \) applies, with energy converted from Calories to Joules (1 Cal = 4200 J).
FAQs on the Power Formula
We hope this comprehensive guide to the Power Formula has clarified every aspect you need for CBSE boards and competitive exams. For related topics, explore our detailed articles on the Heat Load Formula, the Induced Voltage Formula, and the Critical Velocity Formula. You can also browse the full Physics Formulas hub for a complete list of NCERT-aligned formula guides. For official NCERT textbook references, visit ncert.nic.in.