Refer to FIGURE 4.1 and answer the questions that follow - NSC Electrical Technology Power Systems - Question 4 - 2022 - Paper 1

Phasor Vrn represents a phase voltage because it is the voltage between R (Line 1) and neutral (N).

Active power, also known as real power, refers to the capacity of a circuit to do useful work over time, which is measured in watts (W). It represents the power consumed by resistive elements in an electrical circuit and is usually the power that is converted into work.

Stepping up the voltage in transmission lines reduces the current flowing through the lines, which minimizes power losses due to heat in the conductors. Higher voltage levels allow for more efficient long-distance electricity transmission, as lower current reduces the I²R losses.

A diagram for a four-wire three-phase star-connected system consists of three phases (R, Y, B) connected to a neutral wire. Each phase is connected to a load, and the neutral wire is connected to the star point where all phases meet.

In the distribution stage, power is initially distributed at high voltages (like 22 kV) to reduce losses. The voltage is then stepped down to lower levels (such as 380 V/220 V) for supply to homes and businesses, allowing for safe and efficient delivery of electricity.

The phase current can be calculated using the formula: $I_{PH} = \frac{I_L}{\sqrt{3}}$ Substituting the values gives: $I_{PH} = \frac{15 A}{\sqrt{3}} \approx 8.66 A$

The impedance of the load can be calculated using: $Z_{PH} = \frac{V_{L}}{I_{PH}}$ Given that V_L = 400 V: $Z_{PH} = \frac{400 V}{8.66 A} \approx 46.19 \Omega$

The phase angle can be derived using the power factor (pf): $\theta = \cos^{-1}(pf) = \cos^{-1}(0.85) \approx 31.79^ ext{o}$

Active power can be calculated using: $P = \sqrt{3} V_{L} I_{L} \cos(\theta)$ Substituting the values: $P = \sqrt{3} \cdot 400 V \cdot 15 A \cdot 0.85 \approx 8833.46 W$

The diagram will include a three-phase supply with capacitance connected in parallel with the load to correct power factor. Each phase will have a capacitor bank connected, labeled as C1, C2, and C3 for the respective phases.

1. It can measure both balanced and unbalanced loads.
2. The power consumption of each phase can be determined individually.

One disadvantage is that three wattmeters are needed for unbalanced loads, which may complicate the measurement process.

For a balanced load using the three-wattmeter method, the total power is given by: $P_{T} = P_{1} + P_{2} + P_{3}$ For three wattmeter readings, this can be expressed as: $P_T = 3 \cdot 450 W = 1350 W$