Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Define electrical power, energy, and efficiency.
• Calculate power using P = VI, P = I²R, and P = V²/R.
• Calculate energy consumption using E = Pt.
• Calculate the efficiency of an electrical system.
• Apply knowledge to solve real-world problems.

Lesson summary

This week, we're diving into the fundamental electrical principles of power, energy, and efficiency. Understanding these concepts is crucial, not just for passing exams, but for understanding how electrical devices work in your everyday life and making informed decisions about energy consumption. In a country like South Africa, where electricity supply can be unreliable and affordability is a concern for many households, knowing how to calculate power, energy usage, and appliance efficiency becomes incredibly important for managing your resources and minimizing costs. This understanding also fosters responsible citizenship as we grapple with our national energy challenges.

Lesson notes

2.1 Electrical Power (P): Electrical power is the rate at which electrical energy is transferred by an electric circuit. It is measured in watts (W). One watt is defined as one joule per second (1 W = 1 J/s). Power tells us how quickly an electrical device uses energy. A higher power rating indicates that the device consumes more energy in a given time. There are three key formulas for calculating power: P = VI (Power = Voltage x Current) - This formula is used when we know the voltage across and current through a component. P = I²R (Power = Current squared x Resistance) - This formula is useful when we know the current through and resistance of a component. P = V²/R (Power = Voltage squared / Resistance) - This formula is useful when we know the voltage across and resistance of a component.

Example 1: A cellphone charger operates at 220V and draws a current of 0.1

A. Calculate the power consumed by the charger.

Solution: P = VI P = 220V x 0.1A P = 22W Example 2: A heating element in a kettle has a resistance of 25 ohms and carries a current of 8

A. Calculate the power dissipated by the heating element.

Solution: P = I²R P = (8A)² x 25 ohms P = 64A² x 25 ohms P = 1600W or 1.6kW Example 3: A light bulb connected to a 220V supply has a resistance of 484 ohms. Calculate the power consumed by the light bulb.

Solution: P = V²/R P = (220V)² / 484 ohms P = 48400V² / 484 ohms P = 100W 2.2 Electrical Energy (E): Electrical energy is the capacity to do work. It is the total amount of power used over a period of time. The standard unit of energy is the joule (J).

However, for practical purposes, especially when dealing with household electricity consumption, we often use the kilowatt-hour (kWh). One kilowatt-hour is the energy consumed by a 1-kilowatt device operating for one hour. The formula for calculating electrical energy is: E = Pt (Energy = Power x Time)

Where: E is the energy in joules (J) or kilowatt-hours (kWh) P is the power in watts (W) or kilowatts (kW) t is the time in seconds (s) or hours (h)

Important Conversions: 1 kWh = 3.6 x 10⁶ J (3.6 million Joules) 1 kW = 1000 W Example 4: A 100W light bulb is left on for 5 hours. Calculate the electrical energy consumed in kWh.

Solution: First, convert watts to kilowatts: 100W = 100/1000 kW = 0.1 kW E = Pt E = 0.1 kW x 5 h E = 0.5 kWh Example 5: A geyser with a power rating of 3kW is switched on for 2 hours each day. Calculate the energy consumed by the geyser in a week, in kWh.

Solution: Energy consumed per day: E = Pt = 3kW x 2h = 6 kWh Energy consumed in a week: 6 kWh/day x 7 days = 42 kWh 2.3 Efficiency (η): Efficiency is a measure of how effectively an electrical device converts electrical energy into useful output. It is the ratio of output power to input power, usually expressed as a percentage. No device is 100% efficient because some energy is always lost as heat, sound, or other forms of energy.

The formula for efficiency is: Efficiency (η) = (Output Power / Input Power) x 100% Example 6: An electric motor has an input power of 500W and an output power of 400

W. Calculate the efficiency of the motor.

Solution: Efficiency = (Output Power / Input Power) x 100% Efficiency = (400W / 500W) x 100% Efficiency = 0.8 x 100% Efficiency = 80% Example 7: A transformer has an input power of 2k

W. If the efficiency of the transformer is 90%, calculate the output power.

Solution: Efficiency = (Output Power / Input Power) x 100% 90% = (Output Power / 2kW) x 100% 0.9 = Output Power / 2kW Output Power = 0.9 x 2kW Output Power = 1.8kW Guided Practice (With Solutions)

Question 1: A television set is rated at 150W and is used for 4 hours per day. Calculate the daily energy consumption in kWh. What is the cost of running the TV for a month (30 days) if electricity costs R2.50 per kWh?

Solution: Energy consumption per day: E = Pt = (150/1000)kW x 4h = 0.6 kWh Energy consumption per month: 0.6 kWh/day x 30 days = 18 kWh Cost per month: 18 kWh x R2.50/kWh = R45.00

Commentary: This question links energy consumption to cost, which is directly relevant to household budgets in South Africa.

Question 2: An electric iron has a resistance of 22 ohms and is connected to a 220V supply. Calculate (a) the current flowing through the iron, (b) the power consumed by the iron.

Solution: (a)

Using Ohm's Law: V = IR => I = V/R = 220V / 22 ohms = 10A (b) P = VI = 220V x 10A = 2200W or 2.2kW

Commentary: This question combines Ohm's Law with the power formula, reinforcing the relationship between voltage, current, resistance, and power.

Question 3: A motor delivers 5 horsepower of mechanical power when connected to a 220V supply. If the motor draws a current of 20A, calculate the efficiency of the motor.