Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Apply theorems (Thevenin's, Norton's, Superposition) to solve circuit problems.
• Analyze and design single-phase transformer circuits.
• Explain the principles of operation and control of DC machines.
• Differentiate between various types of protection devices.
• Solve practical problems related to power factor correction.

Lesson summary

This week's focus is on intensive revision and examination preparation, specifically targeting topics crucial for success in the Grade 12 Electrical Technology final examination. Solid exam preparation is vital for South African learners because it provides the opportunity to demonstrate the knowledge and skills acquired throughout the year, opening doors to further education, vocational training, and ultimately, contributing to the skilled workforce that our country needs to address its electrification challenges and grow its economy.

Lesson notes

This week, we'll delve into several crucial areas that often feature prominently in the final examination. 2.1 Circuit Theorems: Thevenin's, Norton's, and Superposition These theorems are powerful tools for simplifying complex circuits and finding specific circuit variables (current, voltage).

Thevenin's Theorem: Any linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source (Vth) in series with a resistor (Rth).

Vth (Thevenin Voltage): The open-circuit voltage at the terminals.

Rth (Thevenin Resistance): The resistance looking back into the terminals with all independent sources deactivated (voltage sources short-circuited, current sources open-circuited). Worked

Example: Consider a circuit with a voltage source of 12V in series with a 4-ohm resistor, connected to a parallel combination of a 6-ohm resistor and an 8-ohm resistor. Find the Thevenin equivalent circuit as seen by the 8-ohm resistor.

Find Vth: Remove the 8-ohm resistor. The voltage across the 6-ohm resistor is found using voltage division: Vth = (6/(4+6)) * 12V = 7.2V Find Rth: Deactivate the 12V source (short-circuit it). The 4-ohm and 6-ohm resistors are now in parallel: Rth = (4 * 6) / (4 + 6) = 2.4 ohms.

Thevenin Equivalent: A 7.2V voltage source in series with a 2.4-ohm resistor.

Norton's Theorem: Any linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source (In) in parallel with a resistor (Rn).

In (Norton Current): The short-circuit current at the terminals.

Rn (Norton Resistance): The resistance looking back into the terminals with all independent sources deactivated (same as Rth). Worked

Example: Using the same circuit as above, find the Norton equivalent circuit.

Find In: Remove the 8-ohm resistor and short-circuit the terminals. The total current flowing from the 12V source is 12V / (4 + (4||6)) = 12V / (4 + 2.4) = 1.875A. Using current division, In = 1.875A * (6/(6+4)) = 1.125

A. The internal resistance R = 6||4 = 2.4 Ohms.

Find Rn: Rn is the same as Rth, so Rn = 2.4 ohms.

Norton Equivalent: A 1.125A current source in parallel with a 2.4-ohm resistor.

Superposition Theorem: In a linear circuit with multiple independent sources, the voltage or current for any element is the algebraic sum of the contributions from each source acting independently, with all other independent sources deactivated. (Voltage sources shorted, current sources opened.) Worked

Example: Consider a circuit with a 6V voltage source in series with a 2-ohm resistor, and a 2A current source in parallel with the 2-ohm resistor. Find the current through the 2-ohm resistor.

Due to the voltage source: Short-circuit the current source. The current through the 2-ohm resistor is 6V / (2+2) = 1.5

A. Due to the current source: Open-circuit the voltage source. The current from the current source splits equally between the two 2-ohm resistors, so the current through the 2-ohm resistor is 2A / 2 = 1

A. Total Current: The total current through the 2-ohm resistor is 1.5A + 1A = 2.5A. 2.2 Single-Phase Transformers Transformers are essential for stepping up or down voltage levels in power systems.

Turns Ratio (a): a = Np/Ns (Np = number of primary turns, Ns = number of secondary turns).

Voltage Transformation: Vp/Vs = Np/Ns = a Current Transformation: Ip/Is = Ns/Np = 1/a Ideal Transformer Power: Vp Ip = Vs * Is Efficiency (η): η = (Pout / Pin) 100% = (Vs Is cos(θs) / (Vp Ip cos(θp)) * 100% Where cos(θ) is the power factor.

Voltage Regulation: Voltage Regulation = ((Vno-load - Vfull-load) / Vfull-load) 100% Worked

Example: A single-phase transformer has 500 turns on the primary and 100 turns on the secondary. The primary voltage is 220V, and the secondary load is 4 ohms. Assuming an ideal transformer, calculate the secondary voltage, primary current, and secondary current.

Secondary Voltage: Vs = Vp (Ns/Np) = 220V (100/500) = 44V Secondary Current: Is = Vs / Rload = 44V / 4 ohms = 11A Primary Current: Ip = Is (Ns/Np) = 11A (100/500) = 2.2A 2.3 DC Machines DC machines convert electrical energy into mechanical energy (DC motors) or vice-versa (DC generators).

Back EMF (Eb): Eb = k Φ * ω (k is a constant, Φ is the magnetic flux, ω is the angular speed). The back EMF opposes the applied voltage.

Torque (T): T = k Φ * Ia (Ia is the armature current).

Speed Control: Speed of a DC motor can be controlled by varying the armature voltage, field flux, or armature resistance.

Armature Voltage Control: Increasing the armature voltage increases the speed.

Field Flux Control: Decreasing the field flux increases the speed (field weakening).

Armature Resistance Control: Increasing the armature resistance decreases the speed. Worked

Example: A 220V DC shunt motor has an armature resistance of 0.5 ohms and a field resistance of 110 ohms. At no load, the motor runs at 1000 rpm and takes a current of 5A. Calculate the speed when the motor is loaded and takes a current of 50A.