2.1 Given the circuit diagram in Figure 1, answer the following questions: 2.1.1 The actual power consumed by the load 2.1.2 The power lost in the form of heat 2.1.3 The power lost to overcome the reactive component of the load 2.2 Explain the significance of the voltage waveforms shown in the graph - NSC Electrical Technology Power Systems - Question 2 - 2017 - Paper 1

The power lost in the form of heat is referred to as resistive power loss (I²R losses). It can be calculated using:

$P_{ ext{loss}} = I^2 R$ where:

• $I$ is the current flowing through the resistor.
• $R$ is the resistance of the load.

The power lost to overcome the reactive component is known as reactive power (Q). It does not perform any useful work but is essential in voltage regulation. It can be calculated using:

$Q = VI imes ext{Sin}( heta)$ where $V$ is the voltage and $heta$ is the phase angle.

The voltage waveforms for R, Y, and B represent the three phases in a three-phase system. Each phase is 120 degrees apart in time, ensuring that the power delivered is constant. This arrangement minimizes the overall power variation and increases system reliability. The balanced loads across phases reduce the neutral current and enhance efficiency.

Given the line current $I_L = 30 ext{ A}$, the phase current ($I_f$) can be calculated as:

I_f = rac{I_L}{\sqrt{3}} = \frac{30}{\sqrt{3}} \approx 17.32 ext{ A}

The impedance ($Z$) of the load can be calculated using:

$Z = \frac{V_P}{I_f}$ Substituting the given values: $Z = \frac{380}{17.32} \approx 21.93 \Omega$

If the power factor is improved, the load will draw less current for the same amount of active power consumption. This reduction in the current minimizes the resistive losses in the system, leading to more efficient power delivery.

Cost savings can be realized through improved power factor as it reduces the total current drawn from the supply. This lowers the electricity bill since utilities often charge higher rates for low power factor loads. Additionally, improved power factor can decrease energy losses in the electrical distribution system.

The total power ($P_t$) can be calculated as the sum of individual powers:

$P_t = P_1 + P_2 = 100 + 250 = 350 ext{ W}$