Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Explain AC generation using Faraday's & Lenz's Law.
• Calculate instantaneous, peak, RMS, and average values of an AC waveform.
• Calculate frequency and period.
• Define phase difference.
• Solve basic AC circuit problems with resistance.

Lesson summary

This week, we delve into the fascinating world of Alternating Current (AC) electricity. AC is the backbone of our modern electrical grid, powering everything from lights and appliances in our homes to industries across South Africa. Understanding how AC is generated and the basic principles that govern its behaviour is crucial for any aspiring electrical technician or engineer. In South Africa, where reliable electricity supply is vital for economic growth and social well-being, knowledge of AC systems is indispensable. Load shedding and power grid maintenance are realities that require skilled professionals who understand AC principles.

Lesson notes

2. 1.

AC Generation: Faraday's Law and Lenz's Law AC generation is based on Faraday's Law of Electromagnetic Induction, which states that a changing magnetic field induces a voltage in a conductor. More specifically, the magnitude of the induced voltage is proportional to the rate of change of the magnetic flux linking the conductor. Mathematically, Faraday's Law is expressed as: `E = -N (dΦ/dt)` Where: `E` is the induced voltage (in Volts) `N` is the number of turns in the coil `dΦ/dt` is the rate of change of magnetic flux (in Webers per second) Lenz's Law complements Faraday's Law by specifying the direction of the induced voltage. It states that the induced voltage will create a current that produces a magnetic field opposing the change in the original magnetic flux. This "opposition" is represented by the negative sign in Faraday's Law.

How an AC Generator Works: A basic AC generator consists of a coil of wire (the armature) rotating within a magnetic field. As the coil rotates, the magnetic flux linking it changes continuously. This changing flux induces a voltage in the coil, according to Faraday's Law. Because the coil rotates through a complete circle, the induced voltage varies sinusoidally with time, producing an AC waveform.

The key components include: Stator: The stationary part of the generator, usually containing the magnetic field (either permanent magnets or electromagnets).

Rotor: The rotating part, consisting of coils of wire. This is the armature where the voltage is induced.

Slip Rings and Brushes: Conductors that allow the generated AC voltage to be transferred from the rotating rotor to the external circuit. 2.

2. AC Waveform Characteristics A sinusoidal AC waveform is characterized by several important parameters: Instantaneous Value (v(t)): The value of the voltage or current at a specific point in time. Represented mathematically as `v(t) = V_peak sin(ωt)`, where `V_peak` is the peak voltage, `ω` is the angular frequency (in radians per second), and `t` is time (in seconds). Peak Value (V_peak): The maximum value of the voltage or current. This is the amplitude of the sinusoidal waveform. Peak-to-Peak Value (V_p-p): The difference between the maximum positive peak and the maximum negative peak. `V_p-p = 2 V_peak` Period (T): The time taken for one complete cycle of the waveform (in seconds).

Frequency (f): The number of cycles per second (in Hertz). `f = 1/T` RMS (Root Mean Square) Value (V_rms): The effective value of the AC voltage or current. It's the equivalent DC voltage that would produce the same heating effect in a resistive load. `V_rms = V_peak / √2` Average Value (V_avg): The average value of the AC voltage or current over one complete cycle. For a pure sinusoidal waveform, the average value is zero.

However, the average value over half a cycle is not zero: `V_avg (half cycle) = (2 V_peak) / π` 2.

3. Frequency and Period The frequency (f) and period (T) are inversely related: `f = 1/T`. The frequency of the AC supply in South Africa is 50 Hz. This means that the AC waveform completes 50 cycles every second. The period is therefore T = 1/50 = 0.02 seconds or 20 milliseconds. The frequency of the generated AC voltage is directly related to the speed of rotation of the generator. The number of poles in the generator's magnetic field also affects the frequency.

The relationship is: `f = (P * N) / 120` Where: `f` is the frequency (in Hz) `P` is the number of poles `N` is the speed of rotation (in revolutions per minute - RPM) 2.

4. Phase Difference Phase difference refers to the difference in the timing of two AC waveforms. If two waveforms reach their peak values at the same time, they are said to be in phase (phase difference = 0 degrees). If they reach their peak values at different times, they are out of phase. Phase difference is typically measured in degrees or radians. For example, if waveform A leads waveform B by 90 degrees, it means that waveform A reaches its peak value 90 degrees of the cycle before waveform B. 2.

5. AC Circuits with Pure Resistance In a purely resistive AC circuit, the voltage and current are in phase.

Ohm's Law applies to AC circuits as well: `V = I * R` Where: `V` is the voltage (in Volts) `I` is the current (in Amperes) `R` is the resistance (in Ohms) When dealing with AC, we typically use RMS values for voltage and current in Ohm's Law calculations: `V_rms = I_rms * R` The instantaneous power in a resistive AC circuit is: `p(t) = v(t) * i(t)` The average power (P) dissipated in a resistive AC circuit is: `P = V_rms I_rms = I_rms^2 R = V_rms^2 / R`

Worked example

Example 1:

A sinusoidal AC voltage has a peak value of 325V. Calculate the RMS voltage and the average voltage over half a cycle.

Solution:

RMS Voltage: `V_rms = V_peak / √2 = 325V / √2 ≈ 230V`

Average Voltage (half cycle): `V_avg = (2 V_peak) / π = (2 * 325V) / π ≈ 207V`

Example 2:

A 100 Ohm resistor is connected to a 230V RMS AC supply. Calculate the RMS current and the average power dissipated by the resistor.

Solution:

RMS Current: `I_rms = V_rms / R = 230V / 100 Ohms = 2.3A`

Average Power: `P = V_rms I_rms = 230V 2.3A = 529W` Alternatively, `P = V_rms^2 / R = (230V)^2 / 100 Ohms = 529W` or `P = I_rms^2 R = (2.3A)^2 * 100 Ohms = 529W`

Example 3:

An AC generator has 4 poles and rotates at 1500 RPM. Calculate the frequency of the generated voltage.

Solution:

Frequency: `f = (P N) / 120 = (4 * 1500) / 120 = 50 Hz`

Guided Practice (With Solutions)

Question 1:

A sinusoidal voltage has an RMS value of 110V. What is its peak value?

Solution:

We know `V_rms = V_peak / √2`. Rearranging, we get `V_peak = V_rms √2`

`V_peak = 110V √2 ≈ 155.6V`