Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Explain AC generation using electromagnetic induction.
• Define and calculate key AC terms like frequency, period, RMS, and peak values.
• Calculate instantaneous voltage of a sine wave.
• Explain phase relationships in a resistive circuit.
• Convert between RMS and peak values for voltage and current.

Lesson summary

Alternating Current (AC) is the backbone of South Africa's electrical power system, from Eskom's power stations to the wall sockets in our homes and schools. Understanding how AC is generated and the fundamental principles behind it is crucial for any aspiring electrical technician or engineer in our country. Powering our industries, homes, and infrastructure relies heavily on the reliable generation, distribution, and utilisation of AC electricity. This week, we delve into the principles of AC generation and single-phase AC theory, providing the foundation for more advanced topics later in the year.

Lesson notes

2.1 AC Generation: Electromagnetic Induction AC voltage is generated through electromagnetic induction, as described by Faraday's Law. This law states that a changing magnetic field through a conductor induces a voltage in the conductor. In a simple AC generator (alternator), a coil of wire is rotated within a magnetic field. As the coil rotates, the magnetic flux linking the coil changes continuously. This changing magnetic flux induces an electromotive force (EMF), which is the AC voltage.

Imagine a simple alternator: Rotor: A coil of wire (armature) rotates within a magnetic field.

Stator: A stationary magnetic field is created by magnets or field windings.

As the rotor turns: When the coil is perpendicular to the magnetic field, the rate of change of magnetic flux is maximum, and the induced voltage is at its maximum value (peak voltage). When the coil is parallel to the magnetic field, the rate of change of magnetic flux is minimum (zero), and the induced voltage is zero. This continuous rotation creates a sinusoidal AC waveform. The frequency of the AC voltage is determined by the speed of rotation of the rotor (in revolutions per second or Hertz, Hz). Why sinusoidal? The voltage generated follows a sinusoidal pattern because the rate of change of magnetic flux is proportional to the sine of the angle between the coil and the magnetic field. 2.2 AC Waveform Parameters Frequency (f): The number of complete cycles of the AC waveform that occur in one second. Measured in Hertz (Hz). South Africa uses a standard frequency of 50 Hz. This means the AC voltage in our wall sockets completes 50 cycles every second.

Period (T): The time taken to complete one full cycle of the AC waveform.

It's the inverse of the frequency: T = 1/f.

Example: For South Africa's 50 Hz, T = 1/50 = 0.02 seconds (or 20 milliseconds).

Amplitude (V m or I m ): The maximum value of voltage or current reached during a cycle. Also known as the peak voltage (V p ) or peak current (I p ).

Instantaneous Value (v or i): The value of voltage or current at a specific instant in time. The instantaneous voltage (v) can be calculated using the formula: `v = Vm * sin(ωt)` Where: `Vm` is the peak voltage. `ω` (omega) is the angular frequency (ω = 2πf). `t` is the time in seconds.

RMS Value (V rms or I rms ): The Root Mean Square (RMS) value is the effective value of the AC voltage or current. It's the equivalent DC voltage or current that would deliver the same amount of power to a resistive load. The RMS value is related to the peak value by: `Vrms = Vm / √2 ≈ 0.707 Vm` `Irms = Im / √2 ≈ 0.707 Im` The voltage rating on our wall sockets (230V) is the RMS value.

Peak-to-Peak Value (V pp or I pp ): The difference between the maximum positive and maximum negative values of the AC waveform. `Vpp = 2 Vm` `Ipp = 2 Im` 2.3 Phase Angle and Phase Difference Phase Angle (θ): The angular position of a point on a waveform relative to a reference point (usually zero degrees). In the equation `v = Vm sin(ωt + θ)`, θ represents the phase angle.

Phase Difference: The difference in phase angle between two AC waveforms. If two waveforms reach their peak values at different times, they have a phase difference. In a purely resistive circuit, voltage and current are in phase. This means they reach their peak values and cross zero at the same time. The phase difference between voltage and current is zero degrees (θ = 0°). This simplifies calculations significantly. Ohms Law (V=IR) can be applied directly using RMS or peak values. 2.4 Worked Examples Example 1: A simple alternator generates a peak voltage of 340V at a frequency of 50 Hz.

Calculate: a) The RMS voltage. b) The period of the waveform. c) The instantaneous voltage at t = 0.005 seconds.

Solution: a)

RMS Voltage: `Vrms = Vm / √2 = 340V / √2 ≈ 240.4 V` b)

Period: `T = 1/f = 1 / 50 Hz = 0.02 s = 20 ms` c)

Instantaneous Voltage: `v = Vm * sin(ωt)` `ω = 2πf = 2 π 50 Hz ≈ 314.16 rad/s` `v = 340V sin(314.16 rad/s 0.005 s) ≈ 340V * sin(1.57 rad)` `v ≈ 340V * 1 = 340 V` Example 2: A 100Ω resistor is connected to a 230V (RMS), 50 Hz AC source.

Calculate: a) The RMS current flowing through the resistor. b) The peak current flowing through the resistor.

Solution: a)

RMS Current: Using Ohm's Law: `Irms = Vrms / R = 230V / 100Ω = 2.3 A` b)

Peak Current: `Im = Irms √2 = 2.3A √2 ≈ 3.25 A` Example 3: An AC voltage is described by the equation `v = 170sin(314t)` volts.

Determine: a) The peak voltage. b) The RMS voltage. c) The frequency.

Solution: a)

Peak voltage: Comparing to the general form `v = Vm * sin(ωt)`, we see that `Vm = 170V`. b)

RMS voltage: `Vrms = Vm / √2 = 170V / √2 ≈ 120.2 V` c)

Frequency: `ω = 314 rad/s`. Since `ω = 2πf`, then `f = ω / (2π) = 314 rad/s / (2π) ≈ 50 Hz`. Guided Practice (With Solutions)

Question 1: An AC generator produces a sinusoidal voltage with a frequency of 60 Hz and a peak voltage of 150

V. Calculate the period and the RMS voltage.

Solution: Period: `T = 1/f = 1/60 Hz ≈ 0.0167 s = 16.7 ms`.