Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Recall key concepts of single-phase AC circuits.
• Solve numerical problems involving RLC circuits.
• Analyze and interpret circuit diagrams.
• Develop effective exam time management skills.
• Recognize typical exam question formats.

Lesson summary

This week focuses on a crucial aspect of your Grade 12 Electrical Technology studies: effective revision and examination preparation. Excelling in Electrical Technology opens doors to diverse career paths, from becoming a registered electrician contributing to South Africa's growing infrastructure to designing renewable energy solutions addressing our energy challenges. Mastering exam techniques is essential to demonstrate your knowledge and skills effectively, ensuring you achieve the best possible results and unlock future opportunities.

Lesson notes

This week, we focus on single-phase AC circuits, a fundamental building block in electrical systems. 2.1 Single-Phase AC Circuits Review: Alternating Current (AC): AC is an electrical current that periodically reverses direction and changes magnitude continuously with time. This is in contrast to Direct Current (DC), which flows in one direction only. The alternating current is described by the sinusoidal function.

Sinusoidal Waveform: The most common form of AC is the sinusoidal waveform, characterized by its amplitude (peak voltage or current), frequency (f), and period (T). Remember that T = 1/f. In South Africa, the standard AC frequency is 50 Hz.

Voltage (V) and Current (I): In AC circuits, voltage and current are also sinusoidal functions. We often use RMS (Root Mean Square) values for voltage and current, as these represent the effective values that deliver the same power as a DC voltage or current of the same magnitude. Vrms = Vpeak / √2 and Irms = Ipeak / √2 Phasor Representation: Phasors are a graphical and mathematical way to represent sinusoidal quantities. Voltage and current phasors are vectors with magnitude and angle (phase). 2.2 RLC Circuits and Impedance: Resistor (R): A resistor opposes the flow of current. In AC circuits, the voltage and current are in phase (phase angle = 0 degrees).

Inductor (L): An inductor opposes changes in current due to the phenomenon of inductance. In AC circuits, the voltage leads the current by 90 degrees. Inductive reactance, XL = 2πfL, is the opposition to current flow offered by an inductor.

Capacitor (C): A capacitor opposes changes in voltage due to the phenomenon of capacitance. In AC circuits, the voltage lags the current by 90 degrees. Capacitive reactance, XC = 1 / (2πfC)*, is the opposition to current flow offered by a capacitor.

Impedance (Z): Impedance is the total opposition to current flow in an AC circuit, combining resistance and reactance. It is a complex quantity, represented as Z = R + j(XL - XC), where j is the imaginary unit (√-1). The magnitude of impedance is |Z| = √(R² + (XL - XC)²) ohms. The phase angle (θ) between voltage and current is θ = arctan((XL - XC) / R)*.

Series RLC Circuits: In a series RLC circuit, the current is the same through all components. The total impedance is calculated as above.

Parallel RLC Circuits: In a parallel RLC circuit, the voltage is the same across all components. We typically calculate the admittances (Y = 1/Z) of each branch and add them to find the total admittance. Y = √(G² + B²)* where G is conductance (1/R) and B is susceptance (1/XL - 1/XC). 2.3 Power in AC Circuits: Instantaneous Power (p(t)): The power at any instant in time is p(t) = v(t) i(t).

Average or Real Power (P): The average power is the power actually consumed by the circuit and dissipated as heat or used to do work. P = Vrms Irms cos(θ) = I²rms R (Watts).

Reactive Power (Q): Reactive power is the power that is exchanged between the source and the reactive components (inductors and capacitors). It is not consumed but circulates in the circuit. Q = Vrms Irms sin(θ)* (VARs - Volt-Ampere Reactive).

Apparent Power (S): Apparent power is the total power supplied by the source, including both real and reactive power. S = Vrms Irms (VA - Volt-Amperes). S = √(P² + Q²).

Power Factor (PF): The power factor is the cosine of the phase angle between voltage and current. PF = cos(θ) = P/S. A power factor of 1 indicates that all the power is real power (resistive load), while a power factor of 0 indicates that all the power is reactive power (purely inductive or capacitive load). A low power factor is undesirable as it means a larger current is required to deliver the same real power, leading to increased losses and higher electricity bills. Eskom charges penalties for low power factors in industrial installations.

Example 1: A series RLC circuit has R = 10 ohms, L = 0.1 H, and C = 100 μF, connected to a 220V, 50Hz AC source. Calculate the impedance, current, and power factor. XL = 2πfL = 2π 50 0.1 = 31.42 ohms XC = 1 / (2πfC) = 1 / (2π 50 100 10^-6) = 31.83 ohms Z = R + j(XL - XC) = 10 + j(31.42 - 31.83) = 10 - j0.41 ohms |Z| = √(10² + (-0.41)²) = 10.01 ohms Irms = Vrms / |Z| = 220 / 10.01 = 21.98 A θ = arctan((XL - XC) / R) = arctan((-0.41) / 10) = -2.34 degrees PF = cos(θ) = cos(-2.34) = 0.999 (leading)* Example 2: A parallel circuit consists of a resistor of 100 ohms and an inductor of 0.2H connected in parallel across a 240V 50 Hz supply. Calculate the total current drawn from the supply and the power factor. XL = 2πfL = 2π 50 0.2 = 62.83 ohms IR = V/R = 240/100 = 2.4 A IL = V/XL = 240/62.83 = 3.82 A IT = √(IR² + IL²) = √(2.4² + 3.82²) = 4.51 A θ = arctan(IL/IR) = arctan(3.82/2.4) = 57.85 degrees PF = cos(θ) = cos(57.85) = 0.53 (lagging)* 2.4 Examination Techniques: Read Carefully: Understand the question fully before attempting to answer. Highlight key information.