Polytechnic Diploma — Electrical Engineering

Welcome to Polytechnic Diploma — Electrical Engineering. We will study Single-Phase AC Circuits: RMS Values, Phasors, Impedance, Power Factor and Resonance, work through an example, and practise using the course's learning material. Semester-aligned learning support for a recognised diploma pathway in Electrical Engineering, including theory, calculations, drawing/design, laboratory concepts and examination revision. Sign in and use the study header to open your authorised course. Select this published lesson. Start Course Tutor with Auto audio on: it teaches continuously, answers a typed question, then resumes. Behaviour of the three elements. Pure resistance R: current is in phase with voltage. Power is consumed as heat. Pure inductance L: current lags voltage by 90 degrees. Opposition is the inductive reactance XL = 2 pi fL ohms, which rises with frequency. Example 2 — series R-L circuit. A coil of resistance 30 omega and inductive reactance 40 omega is connected to a 200 V, 50 Hz supply. Find Z, I, power factor, P, Q and S. Z = square root of (30 squared + 40 squared ) = square root of (900 + 1600) = square root of 2500 = 50 omega . I = 200 / 50 = 4 A. cos φ = R/Z = 30/50 = 0.6 lagging (inductive); sin φ = 40/50 = 0.8; φ = 53.13 degrees. P = VI cos φ = 200 times 4 times 0.6 = 480 W. Q = VI sin φ = 200 times 4 times 0.8 = 640 VAR. S = VI = 200 times 4 = 800 VA. Check by a second route: P = I squared R = 16 times 30 = 480 W; Q = I squared XL = 16 times 40 = 640 VAR; square root of (480 squared + 640 squared ) = square root of (230400 + 409600) = square root of 640000 = 800 VA. All three agree. Which mistake would change the reasoning in this example? Adding VR and VL arithmetically. In Example 2, VR = 120 V and VL = 160 V, but the supply is 200 V, not 280 V. Correction: add phasors, V = square root of (VR squared + VL squared ). Using peak values in power formulas. Ask Course Tutor: please explain this step from the example. P = VI cos φ = 200 times 4 times 0.6 = 480 W. Why does it follow? Add what you tried, so the explanation addresses your difficulty. Now try the source practice. A 230 V RMS supply is applied to a pure resistor. What is the peak voltage across it? Pause before the answer. The source answer is: Vm = 230 times 1.414 = 325.2 V, about 325 V. Compare your reasoning as well as the result. If your response differs, return to the worked example and identify the step that needs correction. Explain the key idea in your own words, revisit any mistaken step, then continue to the next lesson. Use Stop to end Course Tutor when you finish.