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 formulas like P = VI.
• Calculate energy consumption using E = Pt.
• Calculate the efficiency of electrical devices.
• Apply these concepts to simple circuits.

Lesson summary

Electrical power, energy, and efficiency are fundamental concepts that underpin the operation of virtually every electrical device we use daily. Understanding these principles is not only crucial for a career in electrical technology, but also empowers us to make informed decisions about energy consumption and its impact on our environment, particularly relevant in South Africa where energy security and sustainability are significant national concerns. From understanding the electricity bill in your household to designing efficient electrical systems in buildings and factories, these concepts are invaluable.

Lesson notes

2.1 Electrical Power (P): Definition: Electrical power is the rate at which electrical energy is transferred or consumed in a circuit. It essentially tells us how much energy is being used or generated per unit of time.

Unit: Watt (W). One Watt is defined as one Joule per second (1 W = 1 J/s).

Formulas: P = VI: Power = Voltage x Current. This formula is the most fundamental and applicable in most cases. P = I²R: Power = (Current)² x Resistance. This formula is useful when current and resistance are known. P = V²/R: Power = (Voltage)² / Resistance. This formula is useful when voltage and resistance are known. 2.2 Electrical Energy (E): Definition: Electrical energy is the total amount of work done or energy consumed by an electrical circuit or device over a period of time.

Unit: Joule (J) or Kilowatt-hour (kWh). While the Joule is the SI unit, the Kilowatt-hour is more commonly used for billing and large-scale energy consumption due to the Joule being a relatively small unit. 1 kWh = 3.6 x 10⁶ J (3.6 million Joules).

Formula: E = Pt: Energy = Power x Time. The time must be in seconds if energy is required in Joules, and in hours if energy is required in kilowatt-hours (kWh). 2.3 Efficiency (η): Definition: Efficiency is a measure of how effectively an electrical device or system converts input power into useful output power. It represents the ratio of useful output power to the total input power. In reality, no system is 100% efficient, as some energy is always lost due to heat, friction, or other factors.

Unit: Percentage (%). Efficiency is a dimensionless ratio, usually expressed as a percentage.

Formula: η = (Pout / Pin) x 100%: Efficiency = (Power Output / Power Input) x 100%. 2.4 Series and Parallel Circuits and Power Calculations: When dealing with series and parallel circuits, remember to first calculate the equivalent resistance and total current or voltage before calculating the power dissipated by individual components or the entire circuit.

Series Circuit: The current is the same through all components. Use P = I²R to calculate power dissipated by each resistor. The total power consumed by the circuit is the sum of the power consumed by each resistor.

Parallel Circuit: The voltage is the same across all components. Use P = V²/R to calculate power dissipated by each resistor. The total power consumed by the circuit is the sum of the power consumed by each resistor.

Worked example

Example 1: A kettle operating at 230V draws a current of 10A. Calculate the power consumed by the kettle.

Solution:

Given: V = 230V, I = 10A

Formula: P = VI

Calculation: P = 230V 10A = 2300W = 2.3kW

Answer: The kettle consumes 2300W (2.3kW) of power.

Example 2: A 100W light bulb is left on for 5 hours. Calculate the energy consumed in Joules and kWh.

Solution:

Given: P = 100W = 0.1kW, t = 5 hours

Formula: E = Pt

Energy in kWh: E = 0.1kW 5h = 0.5 kWh

Energy in Joules: First, convert time to seconds: 5 hours 60 minutes/hour * 60 seconds/minute = 18000 seconds

E = 100W 18000s = 1,800,000 J = 1.8 x 10⁶ J

Answer: The light bulb consumes 0.5 kWh or 1.8 x 10⁶ J of energy.

Example 3: An electric motor draws 5A from a 220V supply and delivers a mechanical power output of 880W. Calculate the efficiency of the motor.

Solution:

Given: Pin (electrical) = VI, Pout (mechanical) = 880W, V = 220V, I = 5A

Calculate Pin: Pin = 220V 5A = 1100W

Formula: η = (Pout / Pin) 100%

Calculation: η = (880W / 1100W) 100% = 80%

Answer: The efficiency of the motor is 80%. This means that 20% of the electrical energy is being lost as heat or other forms of energy during the conversion.

Example 4: Consider a series circuit with a 12V supply and two resistors, R1 = 10 ohms and R2 = 20 ohms. Calculate the power dissipated by each resistor and the total power.

Solution:

Given: V = 12V, R1 = 10 ohms, R2 = 20 ohms

Calculate the total resistance: R_total = R1 + R2 = 10 + 20 = 30 ohms

Calculate the current: I = V / R_total = 12V / 30 ohms = 0.4A

Power dissipated by R1: P1 = I²R1 = (0.4A)² 10 ohms = 1.6W

Power dissipated by R2: P2 = I²R2 = (0.4A)² 20 ohms = 3.2W

Total power: P_total = P1 + P2 = 1.6W + 3.2W = 4.8W

Alternatively, P_total = VI = 12V*0.4A = 4.8W

Answer: R1 dissipates 1.6W, R2 dissipates 3.2W, and the total power dissipated is 4.8W.

Guided Practice (With Solutions)