Auto-Electrics Any Five Questions: Q1a. Calculate P1, P2, P3 enter the values in the table. Q1b. Find all the unknown values of E, I, R and P for the parallel circuit above.

    $$
    P_1 = V_1 \times I_1 = 5.7 \text{ V} \times 0.57 \times 10^{-3} \text{ A} = 0.003249 \text{ W} = 3.249 \text{ mW}
    $$

• For R2:
  • Given: $V_2 = 0.74 \text{ V}$, $I_2 = 2.63 \text{ mA}$

    $$
    P_2 = V_2 \times I_2 = 0.74 \text{ V} \times 2.63 \times 10^{-3} \text{ A} = 0.0019462 \text{ W} = 1.946 \text{ mW}
    $$

• For R3:
  • Given: $V_3 = 3.21 \text{ V}$, $I_3 = 3.21 \text{ mA}$

    $$
    P_3 = V_3 \times I_3 = 3.21 \text{ V} \times 3.21 \times 10^{-3} \text{ A} = 0.0103041 \text{ W} = 10.304 \text{ mW}
    $$

• For Total:
  • Given: $V_T = 9 \text{ V}$, $I_T = 3.21 \text{ mA}$

    $$
    P_T = V_T \times I_T = 9 \text{ V} \times 3.21 \times 10^{-3} \text{ A} = 0.02889 \text{ W} = 28.89 \text{ mW}
    $$

Filled Table for Q6a:

Q6b: Find all the unknown values of E, I, R and P for the Parallel Circuit above.

Given a parallel circuit with:

• Total Voltage $E_T = 9 \text{ V}$
• Resistor $R_1 = 1 \text{ k}\Omega = 1000 \Omega$
• Resistor $R_2 = 10 \text{ k}\Omega = 10000 \Omega$

For a parallel circuit, the voltage across each component is the same as the total voltage.

• Step 1: Determine voltages.

  $$V_1 = V_2 = E_T = 9 \text{ V}$$

• Step 2: Calculate individual currents.

  $$I_1 = \frac{V_1}{R_1} = \frac{9 \text{ V}}{1000 \Omega} = 0.009 \text{ A} = 9 \text{ mA}$$ $$I_2 = \frac{V_2}{R_2} = \frac{9 \text{ V}}{10000 \Omega} = 0.0009 \text{ A} = 0.9 \text{ mA}$$

• Step 3: Calculate total current.

  $$I_T = I_1 + I_2 = 9 \text{ mA} + 0.9 \text{ mA} = 9.9 \text{ mA}$$

• Step 4: Calculate total resistance.

  $$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{1000 \Omega} + \frac{1}{10000 \Omega} = \frac{10}{10000 \Omega} + \frac{1}{10000 \Omega} = \frac{11}{10000 \Omega}$$ $$R_T = \frac{10000}{11} \Omega \approx 909.09 \Omega \approx 0.909 \text{ k}\Omega$$

• Step 5: Calculate individual powers.

  $$P_1 = V_1 \times I_1 = 9 \text{ V} \times 0.009 \text{ A} = 0.081 \text{ W} = 81 \text{ mW}$$ $$P_2 = V_2 \times I_2 = 9 \text{ V} \times 0.0009 \text{ A} = 0.0081 \text{ W} = 8.1 \text{ mW}$$

• Step 6: Calculate total power.

  $$P_T = P_1 + P_2 = 81 \text{ mW} + 8.1 \text{ mW} = 89.1 \text{ mW}$$

Filled Table for Q6b:

Q8a: Using Simple Sketches explain the two (2) types of Stator Winding of the alternator could be connected and state the operational difference between them.

The two main types of stator winding connections in an alternator are Star (Wye) connection and Delta connection.

• i) Star (Wye) Connection

  • Sketch Description: Imagine three separate windings. One end of each winding is connected together at a common point, called the neutral point. The other three ends of the windings are connected to the output terminals of the alternator.
  • Operational Difference:
    • Provides a higher output voltage for a given winding, as the line voltage (output voltage) is $\sqrt{3}$ times the phase voltage (voltage across a single winding).
    • Offers a neutral point, which can be grounded, providing stability and allowing for both phase and line voltage measurements.
    • Generally more efficient at lower engine speeds due to higher voltage output.

• ii) Delta Connection

  • Sketch Description: Imagine three separate windings connected end-to-end to form a closed triangular loop. The output terminals of the alternator are taken from the three connection points where the windings meet.
  • Operational Difference:
    • The line voltage (output voltage) is equal to the phase voltage (voltage across a single winding).
    • Does not provide a neutral point.
    • Can handle higher current loads for a given voltage compared to a Star connection, but typically produces a lower voltage output for the same winding design.

Q8b: State the category of Vehicle in which the two Stators windings connection in Q8ai above is applicable.

• Star (Wye) Connection: This connection is predominantly used in passenger cars and light commercial vehicles. Its ability to provide higher voltage output and better regulation at varying engine speeds makes it ideal for these applications.
• Delta Connection: While less common in modern passenger vehicles, the Delta connection can be found in some heavy-duty vehicles, commercial trucks, or specialized industrial equipment where high current output is a primary requirement, or in older alternator designs.