Lesson Notes By Weeks and Term v3 - Senior Secondary 1
Electric Power
Download the Lessonotes Mobile Nigeria 2025 app for faster lesson access on Android and iPhone.
Electric Power
Download the Lessonotes Mobile Nigeria 2025 app for faster lesson access on Android and iPhone.
Subject: Basic Electronics
Class: Senior Secondary 1
Term: 3rd Term
Week: 6
Theme: Electrical Quantities
This page supports the lesson note with a companion video and a short classroom-ready summary.
For class groups and homework, share this lesson page so learners also get the summary, objectives, and full lesson context.
Students should be familiar with these three main formulae for electrical power: P = IV (when voltage and current are known) P = I2R (when current and resistance are known) P = V2/R (when voltage and resistance are known)
Unit of Electrical Power: The standard international (SI) unit for electrical power is the Watt (W). One Watt is defined as one Joule per second (1 W = 1 J/s). Larger units are often used for significant power consumption: Kilowatt (kW): 1 kW = 1000 W Megawatt (MW): 1 MW = 1,000,000 W (or 1000 kW)
Relevance: These units are commonly seen on appliance labels (e.g., a "1.5 kW electric iron") and in discussions about national power generation capacity (e.g., "Egbin Power Plant has a capacity of over 1320 MW").
Instrument for Measuring Electrical Power: The instrument used to measure electrical power in a circuit is a Wattmeter.
A wattmeter has two sets of coils: a current coil (connected in series with the load to measure current) and a voltage coil (connected in parallel across the load to measure voltage). It internally calculates the product of voltage and current, thereby displaying the power.
Distinction: It is important not to confuse a wattmeter with an ammeter (measures current), a voltmeter (measures voltage), or an energy meter (measures total energy consumed over time, typically in kilowatt-hours, kWh). To derive the primary formula for electrical power, P = IV, we begin with fundamental definitions: Voltage (V): Voltage, or potential difference, is the work done (W) per unit charge (Q) moved between two points in a circuit. V = W / Q Therefore, Work Done (W) = V × Q (Equation 1)
Current (I): Electric current is the rate of flow of electric charge (Q) past a point in a conductor per unit time (t). I = Q / t Therefore, Charge (Q) = I × t (Equation 2)
Power (P): As defined earlier, power is the rate at which work is done or energy is transferred. P = W / t (Equation 3) Now, substitute Equation 1 into Equation 3: P = (V × Q) / t Next, substitute Equation 2 into this modified Equation 3 (replacing Q with I × t): P = (V × (I × t)) / t P = V × I × t / t P = V × I or P = IV This fundamental formula states that the electrical power dissipated or supplied in a circuit is the product of the voltage across the circuit component and the current flowing through it.
Worked Example 1: An electric kettle commonly found in Nigerian homes is rated 240V and draws a current of 5A when switched on. Calculate the power consumed by the kettle.
Given: Voltage (V) = 240 V, Current (I) = 5 A Formula: P = IV Calculation: P = 240 V × 5 A = 1200 W Answer: The power consumed by the electric kettle is 1200 Watts or 1.2 Kilowatts.
Start with the fundamental power formula: P = IV Recall Ohm's Law: V = IR (where R is resistance in Ohms, Ω) Substitute the expression for V from Ohm's Law into the power formula: P = (IR) × I P = I2R This formula indicates that power dissipated is directly proportional to the square of the current and the resistance. This is particularly useful for understanding heat dissipation in resistive components, such as heating elements in electric irons or filament lamps.
Worked Example 2: A filament lamp has a resistance of 480 Ω and draws a current of 0.5
A. Calculate the power dissipated by the lamp.
Given: Resistance (R) = 480 Ω, Current (I) = 0.5 A Formula: P = I2R Calculation: P = (0.5 A)2 × 480 Ω = 0.25 A2 × 480 Ω = 120 W Answer: The power dissipated by the filament lamp is 120 Watts.
Start with the fundamental power formula: P = IV Recall Ohm's Law and rearrange it to express current (I) in terms of voltage and resistance: I = V/R Substitute the expression for I into the power formula: P = V × (V/R) P = V2/R This formula is useful when the voltage across a component and its resistance are known.
Worked Example 3: An electric heater, designed for a 240 V supply, has an internal resistance of 24 Ω. Calculate the power consumed by the heater.
Given: Voltage (V) = 240 V, Resistance (R) = 24 Ω Formula: P = V2/R Calculation: P = (240 V)2 / 24 Ω = 57600 V2 / 24 Ω = 2400 W Answer: The power consumed by the electric heater is 2400 Watts or 2.4 Kilowatts.
Household Electricity Consumption and Billing: Application: Explaining how monthly electricity bills (e.g., from EEDC, AEDC) are generated. The units on an electricity bill are in kilowatt-hours (kWh), which is a measure of energy, not power.
However, energy is directly related to power (Energy = Power × Time).
Integration: Students can be encouraged to check the power ratings of appliances in their homes (e.g., a 100W bulb, a 1500W iron, a 200W fan). Discuss how running multiple high-power appliances simultaneously can lead to higher electricity bills or even circuit overloads, a common problem in areas with unreliable power supply.
Example: Calculating the energy consumed by a 1.5 kW electric iron used for 2 hours (1.5 kW 2 h = 3 kWh). This translates directly to units on an electricity bill.
Appliance Selection and Energy Efficiency: Application: When purchasing appliances like refrigerators, air conditioners, or fans, consumers in Nigeria often consider their power ratings. Higher wattage generally means higher power consumption and potentially higher running costs.
Integration: Discuss the concept of energy-efficient appliances (e.g., inverter ACs, LED bulbs vs. incandescent bulbs). Students can compare the power ratings of different types of bulbs (e.g., 60W incandescent vs. 9W LED for similar light output) and deduce the economic and environmental benefits of choosing lower-power options, supporting sustainable living.
Generator Sizing and Circuit Protection: Application: Many Nigerian homes and businesses rely on generators during power outages. The capacity of a generator (e.g., 2.5 kVA, 5 kVA) is related to the total power it can supply. Similarly, fuse and circuit breaker ratings (e.g., 15A, 30A) are crucial for circuit protection.
Integration: Students can learn that the sum of the power ratings of all appliances connected to a generator or a circuit must not exceed the generator's capacity or the circuit's power limit (derived from its voltage and current rating, P_max = V I_max). This helps prevent generator overloading or circuit damage due to excessive current, which is a significant safety concern. ---