Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Recall and apply fundamental electrical principles (Ohm's Law, Kirchhoff's Laws).
• Analyze and evaluate the operation of electrical components and circuits.
• Design and implement basic control circuits.
• Understand and apply SANS codes and safety regulations.
• Confidently solve complex numerical problems.

Lesson summary

This week is dedicated to focused revision and exam preparation, a crucial stage for Grade 12 Electrical Technology learners. Our goal is to consolidate your understanding of key concepts covered throughout the year, improve problem-solving skills, and build confidence for the upcoming final examinations. Successfully navigating these examinations is pivotal, as it unlocks pathways to higher education, technical colleges, or direct entry into the electrical industry – a sector vital to South Africa's infrastructure development and economic growth.

Lesson notes

This week, we'll revisit core concepts. Let's start with circuit analysis. 2.1 DC Circuits: Ohm's Law: The foundation of circuit analysis. V = IR, where V is voltage (in volts), I is current (in amperes), and R is resistance (in ohms). It describes the linear relationship between voltage, current and resistance in a circuit.

Kirchhoff's Laws: These laws govern the conservation of energy and charge in electrical circuits.

Kirchhoff's Current Law (KCL):* The sum of currents entering a node (junction) is equal to the sum of currents leaving the node. Mathematically, ΣI in = ΣI out . This is about conservation of charge.

Kirchhoff's Voltage Law (KVL):* The algebraic sum of voltages around any closed loop in a circuit is zero. Mathematically, ΣV =

0. This is about conservation of energy. Always observe polarity when tracing a loop.

Power: The rate at which electrical energy is transferred. P = VI = I 2 R = V 2 /R, where P is power (in watts). Understanding power is critical for sizing components and ensuring circuit safety.

Example 1: A 12V battery is connected to a resistor of 10 ohms. Calculate the current flowing through the resistor and the power dissipated by the resistor.

Solution: Using Ohm's Law: I = V/R = 12V / 10Ω = 1.2A Using the power formula: P = VI = 12V 1.2A = 14.4W Example 2: (Applying KCL & KVL) Consider a simple series circuit with a 24V source, a 4Ω resistor, and an 8Ω resistor connected in series. Determine the current flowing through the circuit and the voltage drop across each resistor.

Solution: Calculate Total Resistance: R total = 4Ω + 8Ω = 12Ω Calculate Current: Using Ohm's Law: I = V/R = 24V / 12Ω = 2A Calculate Voltage Drops: Voltage across 4Ω resistor: V 4Ω = I R = 2A * 4Ω = 8V Voltage across 8Ω resistor: V 8Ω = I R = 2A * 8Ω = 16V Verification: V 4Ω + V 8Ω = 8V + 16V = 24V = Source Voltage (KVL is satisfied) 2.2 AC Circuits: RMS Values: AC voltages and currents are often expressed as Root Mean Square (RMS) values. V RMS = V peak / √2 and I RMS = I peak / √

2. RMS values represent the equivalent DC voltage or current that would produce the same heating effect. This is crucial because most meters read RMS values, and component ratings are based on RM

S. Impedance (Z): The total opposition to current flow in an AC circuit, consisting of resistance (R), inductive reactance (X L ), and capacitive reactance (X C ). Z = √(R 2 + (X L - X C ) 2 )

Inductive Reactance (X L ):* Opposition to current flow offered by an inductor. X L = 2πfL, where f is frequency (in Hertz) and L is inductance (in Henries).

Capacitive Reactance (X C ):* Opposition to current flow offered by a capacitor. X C = 1 / (2πfC), where C is capacitance (in Farads).

Power Factor (PF): The cosine of the angle between voltage and current in an AC circuit. PF = cos(φ) = R/Z. A power factor of 1 (or 100%) indicates that the voltage and current are in phase, and all the power is being used effectively. A lower power factor means more reactive power is being drawn, leading to inefficiency. ESKOM penalizes large industrial consumers for low power factors.

Example 3: A series RLC circuit consists of a resistor of 50Ω, an inductor of 0.2H, and a capacitor of 10μF connected to a 220V, 50Hz AC supply. Calculate the impedance, current, and power factor of the circuit.

Solution: Calculate Reactances: X L = 2πfL = 2 π 50Hz 0.2H = 62.83Ω X C = 1 / (2πfC) = 1 / (2 π 50Hz 10 * 10 -6 F) = 318.31Ω Calculate Impedance: Z = √(R 2 + (X L - X C ) 2 ) = √(50 2 + (62.83 - 318.31) 2 ) = √(2500 + 65371.6) = √67871.6 ≈ 260.52Ω Calculate Current: I = V/Z = 220V / 260.52Ω ≈ 0.84A Calculate Power Factor: PF = R/Z = 50Ω / 260.52Ω ≈ 0.19 2.3 Transformers: Turns Ratio (N): The ratio of the number of turns in the primary winding (N p ) to the number of turns in the secondary winding (N s ). N = N p / N s . This determines the voltage transformation ratio.

Voltage and Current Transformation: V p / V s = N p / N s = N I p / I s = N s / N p = 1/N (Ideal transformer)

Transformer Losses: Transformers are not perfectly efficient. Losses occur due to hysteresis, eddy currents, and copper losses (I 2 R losses in the windings).

Example 4: A transformer has 500 turns on its primary winding and 100 turns on its secondary winding. If the primary voltage is 220V, calculate the secondary voltage. Also, if the secondary current is 5A, calculate the primary current (assuming an ideal transformer).

Solution: Calculate Turns Ratio: N = N p / N s = 500 / 100 = 5 Calculate Secondary Voltage: V p / V s = N => V s = V p / N = 220V / 5 = 44V Calculate Primary Current: I p / I s = 1/N => I p = I s / N = 5A / 5 = 1A 2.4 Motors and Generators: Motor Principles: Motors convert electrical energy into mechanical energy using the interaction of magnetic fields. Understanding Fleming’s left-hand rule is essential.

Generator Principles: Generators convert mechanical energy into electrical energy using electromagnetic induction. Faraday’s Law and Lenz’s Law are fundamental.