Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Define electrical power, energy, and efficiency and state their units.
• Calculate power using P = VI, P = I²R, and P = V²/R.
• Calculate energy consumed using E = Pt.
• Calculate efficiency using the formula.
• Apply these principles to common South African scenarios.

Lesson summary

Electrical power, energy, and efficiency are fundamental concepts in electrical technology. Understanding these principles is crucial, not just for excelling in this subject, but also for making informed decisions about energy usage in our daily lives and contributing to a more sustainable future for South Africa. From understanding how much electricity your home appliances consume to designing efficient electrical systems for businesses and communities, these concepts are essential. With South Africa's ongoing challenges with electricity supply and load shedding, a deep understanding of these concepts is more critical than ever.

Lesson notes

2.1 Electrical Power (P)

Definition: Electrical power is the rate at which electrical energy is transferred by an electric circuit. It is the amount of energy consumed or generated per unit of time.

Unit: Watt (W). 1 Watt = 1 Joule/second.

Formulas: P = VI (Power = Voltage x Current) P = I²R (Power = (Current)² x Resistance) P = V²/R (Power = (Voltage)² / Resistance)

Example 1: A kettle connected to a 230V supply draws a current of 5

A. Calculate the power consumed by the kettle.

Given: V = 230V, I = 5A Formula: P = VI Calculation: P = 230V x 5A = 1150W = 1.15 kW Answer: The power consumed by the kettle is 1150 Watts (1.15 Kilowatts).

Example 2: A resistor of 10 ohms has a current of 2A flowing through it. Calculate the power dissipated by the resistor.

Given: R = 10 ohms, I = 2A Formula: P = I²R Calculation: P = (2A)² x 10 ohms = 4 x 10 = 40W Answer: The power dissipated by the resistor is 40 Watts.

Example 3: A light bulb with a resistance of 484 ohms is connected to a 220V supply. Calculate the power consumed by the light bulb.

Given: R = 484 ohms, V = 220V Formula: P = V²/R Calculation: P = (220V)² / 484 ohms = 48400 / 484 = 100W Answer: The power consumed by the light bulb is 100 Watts. 2.2 Electrical Energy (E)

Definition: Electrical energy is the capacity to do work using electricity. It is the total amount of power consumed over a period of time.

Unit: Joule (J) or kilowatt-hour (kWh). 1 kWh = 3.6 x 10^6 J Formula: E = Pt (Energy = Power x Time) When using kWh, Power should be in kW and Time in hours. When using Joules, Power should be in Watts and Time in seconds.

Example 4: A 100W light bulb is switched on for 5 hours. Calculate the energy consumed in kWh.

Given: P = 100W = 0.1kW, t = 5 hours Formula: E = Pt Calculation: E = 0.1kW x 5 hours = 0.5 kWh Answer: The energy consumed by the light bulb is 0.5 kWh.

Example 5: A geyser with a power rating of 2kW is switched on for 3 hours daily. Calculate the energy consumed in one month (30 days) in kWh. Also, determine the cost if electricity costs R2.50 per kWh.

Given: P = 2kW, t = 3 hours/day, days = 30, cost = R2.50/kWh Total time per month: 3 hours/day 30 days = 90 hours Formula: E = Pt Calculation: E = 2kW x 90 hours = 180 kWh Total cost: 180 kWh R2.50/kWh = R450 Answer: The energy consumed by the geyser in one month is 180 kWh, and the cost is R

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5

0. This illustrates how geysers are often a major source of electricity consumption in South African households. 2.3 Efficiency (η)

Definition: Efficiency is a measure of how well an electrical device converts electrical energy into useful work. It is the ratio of power output to power input.

Unit: Percentage (%)

Formula: Efficiency (η) = (Power Output / Power Input) x 100% Example 6: An electric motor consumes 500W of power and delivers 400W of mechanical power. Calculate the efficiency of the motor.

Given: Power Input = 500W, Power Output = 400W Formula: Efficiency (η) = (Power Output / Power Input) x 100% Calculation: Efficiency (η) = (400W / 500W) x 100% = 0.8 x 100% = 80% Answer: The efficiency of the motor is 80%.

Example 7: A transformer has a power input of 10kW and a power output of 9.5k

W. Calculate its efficiency.

Given: Power Input = 10kW, Power Output = 9.5kW Formula: Efficiency (η) = (Power Output / Power Input) x 100% Calculation: Efficiency (η) = (9.5kW / 10kW) x 100% = 0.95 x 100% = 95% Answer: The efficiency of the transformer is 95%. Guided Practice (With Solutions)

Question 1: A cellphone charger uses 5W of power when charging a phone. If the charger is plugged in for 2 hours, how much energy is used in Joules?

Solution: Given: P = 5W, t = 2 hours = 2 x 60 x 60 seconds = 7200 seconds Formula: E = Pt Calculation: E = 5W x 7200s = 36000 J Answer: 36000 Joules.

Commentary: Here we converted hours to seconds as the question asked for energy in Joules. Remember to keep your units consistent.

Question 2: A television with a power rating of 150W is used for 4 hours a day. Calculate the daily energy consumption in kWh and the monthly cost (30 days) if the electricity price is R2.80 per kWh.

Solution: Given: P = 150W = 0.15kW, t = 4 hours/day, days = 30, cost = R2.80/kWh Daily Energy Consumption: E = Pt = 0.15kW x 4 hours = 0.6 kWh/day Monthly Energy Consumption: 0.6 kWh/day x 30 days = 18 kWh Monthly Cost: 18 kWh x R2.80/kWh = R50.40 Answer: Daily energy consumption is 0.6 kWh, and the monthly cost is R50.

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0. Commentary: We converted Watts to kilowatts before calculating energy consumption in kWh. This step is crucial for getting the correct answer.

Question 3: An electric stove element draws 8A from a 230V supply. Calculate the power consumed by the element.

Solution: Given: I = 8A, V = 230V Formula: P = VI Calculation: P = 230V x 8A = 1840W Answer: The power consumed by the element is 1840 Watts.

Commentary: This is a straightforward application of the P=VI formula. Ensure you understand the relationship between voltage, current, and power.