The diagram shows a circuit with an operational amplifier connected to a capacitor and a resistor - NSC Electrical Technology Power Systems - Question 2 - 2017 - Paper 1

The power loss in the form of heat can be evaluated using the resistive component of the load. It can be expressed by the formula:

$P_{loss} = I^2 imes R$

where:

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

This represents the power that is dissipated as heat due to the resistive elements within the circuit.

This part would typically explore the relationship of voltage waveforms across the three phases (R, Y, B) at specific time intervals (0°, 120°, 240°) on the axis of time/frequency. The graph illustrates the sinusoidal nature of the voltages in a three-phase system, emphasizing that they are phase-shifted by 120 degrees.

To find the phase current, we can utilize the formula:

$I_f = \frac{I_L}{\sqrt{3}}$

Given that the line current ($I_L$) is 30 A, we calculate:

$I_f = \frac{30}{\sqrt{3}} \approx 17.32 A$

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

$Z = \frac{V_p}{I_f}$

Substituting in the values:

• $V_p$ = 380 V,
• $I_f$ = 17.32 A,

we find:

$Z = \frac{380}{17.32} \approx 21.93 \Omega$

If the power factor is improved, it implies that the load will consume more real power and thereby draw less current. A higher power factor signifies that the load is utilizing power more efficiently.

Cost savings can be realized as consumers result in lower energy bills when the load draws less current due to improved power factor. This reduces energy loss across the distribution network as well.

To find the total power (Pt) consumed, we add the power components:

$P_t = P_1 + P_2$

Assuming $P_1$ is 100 W and $P_2$ is 250 W, we calculate:

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