Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Solve single-phase transformer problems.
• Analyse and troubleshoot single-phase AC motors.
• Design and test basic electronic circuits.
• Apply SANS safety standards.
• Solve problems in digital electronics.

Lesson summary

This week focuses on intensive revision and examination preparation for Grade 12 Electrical Technology. Success in Electrical Technology is crucial, not just for academic achievement but also for future career opportunities in South Africa's growing energy sector, manufacturing, and technological industries. Understanding the principles of electrical technology allows you to contribute to solutions for South Africa's energy challenges and participate in the Fourth Industrial Revolution. This week's focused revision allows you to consolidate your knowledge and improve your examination performance.

Lesson notes

Single-Phase Transformers: A transformer is a static device that transfers electrical energy from one circuit to another through electromagnetic induction. In South Africa, transformers are vital for stepping up or stepping down voltage levels in the national grid for efficient power transmission and distribution.

Key Concepts: Voltage Ratio (Transformation Ratio): The ratio of the number of turns on the primary winding (N p ) to the number of turns on the secondary winding (N s ). V p / V s = N p / N s . A step-down transformer has N p > N s and V p > V s . A step-up transformer has N p s and V p s .

Current Ratio: The inverse of the voltage ratio. I p / I s = N s / N p Power: Ideally, power in equals power out. V p I p = V s * I s (ignoring losses). Efficiency (η): The ratio of output power to input power. η = (P out / P in ) 100%. Losses include core losses (hysteresis and eddy current losses) and copper losses (I 2 R losses).

Transformer Polarity: Important for paralleling transformers. Additive and subtractive polarity.

Worked example

A single-phase transformer has 500 turns on its primary winding and 100 turns on its secondary winding. If the primary voltage is 2200 V, and the primary current is 5 A, calculate:

a) The secondary voltage.

b) The secondary current (assuming ideal transformer).

c) The power output.

d) If the actual power output is 10 kW, what is the efficiency?

Solution:

a) V s = V p (N s / N p ) = 2200 V (100 / 500) = 440 V