2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V. 2.2 Draw a neat, labelled voltage phasor diagram that represents a three-phase delta-connected system. 2.3 A star-connected alternator generates 20 kVA at a power factor (p.f.) of 0,87 lagging. The line voltage is 380 V. Calculate the: 2.3.1 Current delivered by the alternator at full load 2.3.2 Power rating of the alternator 2.4 State the function of a kilowatt-hour meter. 2.5 State TWO methods used to improve the power factor of a resistive inductive load. 2.6 Refer to FIGURE 2.1 and answer the questions that follow. Given: P1 = 120 W P2 = 50 W 2.6.1 Calculate the power consumed by the load. 2.6.2 TWO advantages of using the two-wattmeter method when measuring the power of a balanced load can be determined.

NSC Electrical Technology Electronics Calculations: 2.1 Determine the value of the p

2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V - NSC Electrical Technology Electronics - Question 2 - 2016 - Paper 1

Question 2

2.1-Determine-the-value-of-the-phase-voltage-in-a-delta-connected-system-if-the-line-voltage-is-380-V-NSC Electrical Technology Electronics-Question 2-2016-Paper 1.png

2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V. 2.2 Draw a neat, labelled voltage phasor diagram that represe... show full transcript

Worked Solution & Example Answer:2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V - NSC Electrical Technology Electronics - Question 2 - 2016 - Paper 1

Step 1

2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V.

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Answer

In a delta-connected system, the phase voltage ( V_{ph}) is equal to the line voltage ( V_L). Therefore, given that the line voltage is 380 V, we have:

V_{ph} = V_L = 380 ext{ V}

Step 2

2.2 Draw a neat, labelled voltage phasor diagram that represents a three-phase delta-connected system.

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The voltage phasor diagram for a three-phase delta-connected system consists of three phasors, each separated by 120 degrees. The three voltages can be labeled as:

$V_{L1}$ (Phase R)
$V_{L2}$ (Phase Y)
$V_{L3}$ (Phase B)

Each phase voltage is represented as:

$V_{L1}$ at 0 degrees
$V_{L2}$ at 120 degrees
$V_{L3}$ at 240 degrees

This configuration illustrates the balanced nature of the system.

Step 3

2.3.1 Current delivered by the alternator at full load

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Answer

The current delivered by the alternator at full load can be calculated using the formula:

I_L = \frac{S}{\sqrt{3} V_L}

Substituting the known values:

$S = 20 kVA = 20 \times 10^3 VA$
$V_L = 380 V$

We find:

I_L = \frac{20 \times 10^3}{\sqrt{3} \times 380} \approx 30,39 A

Step 4

2.3.2 Power rating of the alternator

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Answer

To calculate the power rating of the alternator, the following formula is used:

P = \sqrt{3} V_L I_L \cos \theta

Where:

$\cos \theta$ is the power factor. Given a power factor of 0.87:

P = \sqrt{3} \times 380 \times 30,39 \times 0,87 \approx 17,4 kW

Step 5

2.4 State the function of a kilowatt-hour meter.

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Answer

The function of a kilowatt-hour (kWh) meter is to measure the amount of energy consumed over time. This device quantifies the electrical energy used by a load in kilowatt-hours, allowing for the billing of electricity by utility companies.

Step 6

2.5 State TWO methods used to improve the power factor of a resistive inductive load.

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Answer

Two common methods to improve the power factor of a resistive inductive load are:

Adding power factor correction capacitors in parallel with the load.
Using synchronous motors which can be operated under leading power factor conditions to compensate for lagging loads.

Step 7

2.6.1 Calculate the power consumed by the load.

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The power consumed by the load can be determined by summing the readings from the two wattmeters:

P_{total} = P_1 + P_2 = 120 W + 50 W = 170 W

Step 8

2.6.2 TWO advantages of using the two-wattmeter method when measuring the power of a balanced load.

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The advantages of the two-wattmeter method include:

It allows for the measurement of power in both balanced and unbalanced loads.
It can determine the power factor of a balanced load directly, providing valuable information about system efficiency.

2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V - NSC Electrical Technology Electronics - Question 2 - 2016 - Paper 1

Worked Solution & Example Answer:2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V - NSC Electrical Technology Electronics - Question 2 - 2016 - Paper 1

2.1 Determine the value of the phase voltage in a delta-connected system if the line voltage is 380 V.

2.2 Draw a neat, labelled voltage phasor diagram that represents a three-phase delta-connected system.

2.3.1 Current delivered by the alternator at full load

2.3.2 Power rating of the alternator

2.4 State the function of a kilowatt-hour meter.

2.5 State TWO methods used to improve the power factor of a resistive inductive load.

2.6.1 Calculate the power consumed by the load.

2.6.2 TWO advantages of using the two-wattmeter method when measuring the power of a balanced load.

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