Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Solve series and parallel RLC AC circuit problems.
• Analyze and troubleshoot single-phase transformer circuits.
• Describe the operating principles of electrical machines.
• Apply electrical safety regulations for installations.
• Solve basic automation problems using PLCs.

Lesson summary

This week focuses on consolidating your understanding of crucial Electrical Technology concepts required for your Grade 12 final examination. Effective revision and exam preparation are not just about memorizing formulas; it's about truly understanding the principles and being able to apply them to solve practical problems. Electrical technology plays a vital role in South Africa, powering our homes, industries, and infrastructure. Understanding these principles is essential not only for your studies but also for potential career opportunities in the field, contributing to South Africa's development and addressing challenges such as energy efficiency and sustainable power generation.

Lesson notes

2.1 AC Circuits (RLC Series and Parallel): Impedance (Z): Impedance is the total opposition to current flow in an AC circuit. It is the vector sum of resistance (R), inductive reactance (XL), and capacitive reactance (XC).

Inductive Reactance (XL): XL = 2πfL, where f is the frequency in Hertz and L is the inductance in Henries. Inductive reactance increases with frequency.

Capacitive Reactance (XC): XC = 1/(2πfC), where f is the frequency in Hertz and C is the capacitance in Farads. Capacitive reactance decreases with frequency.

Series RLC Circuit: Z = √(R² + (XL - XC)²) The current is the same through all components. The voltage drops across each component are calculated using Ohm's Law (V = IZ).

Resonance: Occurs when XL = XC. At resonance, the impedance is at its minimum (Z = R), and the current is at its maximum.

Parallel RLC Circuit: Admittance (Y) = 1/Z. It's often easier to calculate admittance by summing the admittances of each branch. Y = √(G² + (BL - BC)²) where G = 1/R (conductance), BL = 1/XL (inductive susceptance), and BC = 1/XC (capacitive susceptance). The voltage is the same across all components. The current through each component is calculated using Ohm's Law (I = V/Z).

Resonance: Occurs when BL = BC. At resonance, the impedance is at its maximum, and the current drawn from the source is at its minimum.

Power Factor (PF): PF = cos(θ), where θ is the phase angle between voltage and current. PF = R/Z (for a series circuit) PF = G/Y (for a parallel circuit) A power factor of 1 indicates a purely resistive load (voltage and current are in phase). A PF less than 1 indicates a reactive load (either inductive or capacitive). A low power factor increases current demand for a given power output, resulting in higher electricity bills and potential equipment overload. In South Africa, industries are often penalized for low power factors.

True Power (P): P = VIcos(θ) (measured in Watts)

Apparent Power (S): S = VI (measured in Volt-Amperes)

Reactive Power (Q): Q = VIsin(θ) (measured in VARs)

Example 1: Series RLC Circuit A series RLC circuit has R = 10 ohms, L = 20 mH, and C = 50 μ

F. The supply voltage is 230V at 50 Hz.

Calculate: a) XL b) XC c) Z d) Current (I) e) Power Factor (PF) f)

True Power (P)

Solution: a) XL = 2πfL = 2π 50 0.02 = 6.28 ohms b) XC = 1/(2πfC) = 1/(2π 50 50 * 10⁻⁶) = 63.66 ohms c) Z = √(R² + (XL - XC)²) = √(10² + (6.28 - 63.66)²) = 58.12 ohms d) I = V/Z = 230/58.12 = 3.96 A e) PF = R/Z = 10/58.12 = 0.172 (lagging – because XC > XL) f) P = VIcos(θ) = 230 3.96 0.172 = 156.5 W Example 2: Parallel RLC Circuit A parallel RLC circuit has R = 50 ohms, L = 100 mH, and C = 20 μF connected to a 240V, 50Hz supply.

Calculate: a) XL b) XC c) Branch Currents (IR, IL, IC) d) Total Admittance (Y) e) Total Current (I) f)

Power Factor (PF)

Solution: a) XL = 2πfL = 2π 50 0.1 = 31.42 ohms b) XC = 1/(2πfC) = 1/(2π 50 20 * 10⁻⁶) = 159.15 ohms c) IR = V/R = 240/50 = 4.8 A IL = V/XL = 240/31.42 = 7.64 A IC = V/XC = 240/159.15 = 1.51 A d) G = 1/R = 1/50 = 0.02 S BL = 1/XL = 1/31.42 = 0.0318 S BC = 1/XC = 1/159.15 = 0.0063 S Y = √(G² + (BL - BC)²) = √(0.02² + (0.0318 - 0.0063)²) = 0.0327 S e) I = VY = 240 * 0.0327 = 7.85 A f) PF = G/Y = 0.02/0.0327 = 0.612 (lagging - because IL > IC) 2.2 Single-Phase Transformers: Turns Ratio (a): a = Np/Ns = Vp/Vs = Is/Ip, where N is the number of turns, V is the voltage, and I is the current. Subscripts p and s refer to the primary and secondary windings, respectively.

Voltage Transformation: A step-up transformer increases voltage (Ns > Np), while a step-down transformer decreases voltage (Ns XL, indicating a capacitive-dominant circuit.

Question 2: A single-phase transformer has a primary voltage of 240 V and a secondary voltage of 12 V. If the primary has 400 turns, how many turns does the secondary have? What is the turns ratio? If the secondary current is 10 A, what is the primary current (assuming an ideal transformer)?

Solution: Calculate Ns: Vs/Vp = Ns/Np => Ns = (Vs/Vp) Np = (12/240) 400 = 20 turns Calculate turns ratio (a): a = Np/Ns = 400/20 = 20 Calculate Ip: Ip = Is (Ns/Np) = 10 (20/400) = 0.5 A

Commentary: This question focuses on transformer voltage and current relationships. The turns ratio is crucial for understanding the voltage transformation. Remember that in an ideal transformer, the power remains constant (VpIp = VsIs).

Question 3: Explain the purpose of earthing and bonding in an electrical installation according to SANS 10142-

1. Solution: Earthing provides a low-impedance path for fault currents to flow back to the source, causing protective devices (e.g., circuit breakers) to trip and disconnect the faulty circuit. This prevents dangerous voltages from appearing on exposed metal parts, reducing the risk of electric shock. Bonding connects different metal parts together to ensure they are at the same potential (equipotential).