QUESTION 3: AC THEORY \( \begin{array}{l}\text { The armature of an alternator rotates at } 3000 \text { rev/min between the poles of a } \\ \text { magnet. The maximum voltage of the generated wave is } 35,36 \mathrm{~V} \text { and the } \\ \text { waveform is sinusoidal. } \\ \text { Calculate the following: } \\ \begin{array}{ll}3.1 .1 \quad \text { The frequency } \\ 3.1 .2 & \text { The instantaneous } \mathrm{EMF} \text { after } 0,004 \text { seconds } \\ 3.1 .3 \quad \text { The time taken from the instant zero to reach } 20 \mathrm{~V} \text { for the first time } & \text { (5) } \\ 3.2 & \text { Define the term phasor diagram. }\end{array}\end{array} \begin{array}{l}\text { (2) }\end{array} \)

### **QUESTION 3: AC THEORY** **Given:** - **Rotational Speed (N):** 3000 revolutions per minute (rev/min) - **Maximum Voltage (Eₘₐₓ):** 35.36 V - **Waveform:** Sinusoidal Let's address each part of the question step-by-step. --- ### **3.1.1. Calculate the Frequency (f)** **Formula:** \[ f = \frac{N \times P}{120} \] Where: - \( N \) = Rotational speed in rev/min - \( P \) = Number of poles **Assumption:** - **Number of Poles (P):** 2 (since the alternator is rotating between the poles of a magnet) **Calculation:** \[ f = \frac{3000 \times 2}{120} = \frac{6000}{120} = 50 \, \text{Hz} \] **Answer:** \[ \boxed{50 \, \text{Hz}} \] --- ### **3.1.2. Calculate the Instantaneous EMF After 0.004 Seconds** **Formula for Instantaneous EMF:** \[ E(t) = Eₘₐₓ \times \sin(2\pi f t) \] **Given:** - \( Eₘₐₓ = 35.36 \, \text{V} \) - \( f = 50 \, \text{Hz} \) - \( t = 0.004 \, \text{seconds} \) **Calculation:** \[ E(t) = 35.36 \times \sin(2\pi \times 50 \times 0.004) \] \[ E(t) = 35.36 \times \sin(0.4\pi) \] \[ \sin(0.4\pi) \approx \sin(72^\circ) \approx 0.9511 \] \[ E(t) \approx 35.36 \times 0.9511 \approx 33.6 \, \text{V} \] **Answer:** \[ \boxed{33.6 \, \text{V}} \] --- ### **3.1.3. Calculate the Time Taken to Reach 20 V for the First Time** **Given:** - \( E(t) = 20 \, \text{V} \) - \( Eₘₐₓ = 35.36 \, \text{V} \) - \( f = 50 \, \text{Hz} \) **Steps:** 1. Set up the equation: \[ 35.36 \times \sin(2\pi \times 50 \times t) = 20 \] 2. Solve for the sine term: \[ \sin(2\pi \times 50 \times t) = \frac{20}{35.36} \approx 0.566 \] 3. Find the angle: \[ 2\pi \times 50 \times t = \arcsin(0.566) \approx 0.601 \, \text{radians} \] 4. Solve for \( t \): \[ t = \frac{0.601}{2\pi \times 50} \approx \frac{0.601}{314.16} \approx 0.001915 \, \text{seconds} \] \[ t \approx 1.915 \, \text{milliseconds} \] **Answer:** \[ \boxed{1.92 \, \text{milliseconds}} \] --- ### **3.2. Define the Term Phasor Diagram** **Definition:** A **phasor diagram** is a graphical representation used in alternating current (AC) circuit analysis to depict the magnitude and phase relationships between sinusoidal voltages or currents. In this diagram: - **Phasors** are represented as rotating vectors in a two-dimensional plane. - The **length** of each phasor corresponds to the amplitude (magnitude) of the waveform. - The **angle** each phasor makes with the reference axis represents the phase angle relative to a chosen reference phasor. **Purpose:** Phasor diagrams simplify the analysis of AC circuits by allowing engineers to visualize and calculate the resultant voltages and currents based on their magnitudes and phase angles. **Answer:** --- **Phasor Diagram**: *A graphical representation that uses rotating vectors (phasors) to illustrate the magnitudes and phase angles of sinusoidal voltages or currents in an AC circuit, showing their relative amplitudes and phase relationships.* ---