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

Phasor V_RN represents a phase voltage because it is the voltage between R (Line 1) and the neutral (N), while line voltage is measured between two phase conductors.

Active power, or real power, refers to the capacity of a circuit for performing work in a particular time. It is measured in watts (W) and is the power consumed by a load to perform useful work.

The diagram should show three phases (L1, L2, L3) connected to a neutral point (N), and clearly indicate each line and the neutral line.

In the distribution stage, power is distributed to substations at high voltages (e.g., 22 kV) for efficiency. These voltages are then stepped down to lower levels (e.g., 380 V/220 V) for final distribution to consumers such as homes and businesses, reducing losses and improving safety.

The phase current can be calculated using the formula:

$I_{PH} = \frac{I_L}{\sqrt{3}}$

Substituting the values:

$I_{PH} = \frac{15}{\sqrt{3}} \\ I_{PH} \approx 8.66 A$

The impedance can be calculated using the formula:

$Z_{PH} = \frac{V_{PH}}{I_{PH}}$

Where

$V_{PH} = \frac{V_L}{\sqrt{3}} = \frac{400}{\sqrt{3}} \approx 231.7 V$

Thus,

$Z_{PH} = \frac{231.7}{8.66} \approx 26.8 \Omega$

The phase angle can be calculated using the formula:

$\cos(\theta) = pf$

Thus,

$\theta = \cos^{-1}(0.85) \approx 31.79^{\circ}$

Active power can be calculated using:

$P = \sqrt{3} V_L I_L \cos(\theta)$

Substituting the values:

$P = \sqrt{3} \times 400 \times 15 \times 0.85 \approx 8834.6 W \approx 8.83 kW$

The diagram should show a capacitor bank connected in parallel to the load on the three-phase supply with proper labeling of phases and components.

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

Three wattmeters are required for an unbalanced load. The terminals of the load must be available to connect the wattmeters if no neutral line exists.

For a balanced load:

$P_T = P_1 + P_2 + P_3 = 3 \times P_1$

Thus,

$P_T = 3 \times 450 W = 1350 W \approx 1.35 kW$