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2.1 Name the instrument that is used to measure electrical energy - NSC Electrical Technology Power Systems - Question 2 - 2016 - Paper 1

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2.1 Name the instrument that is used to measure electrical energy. 2.2 State TWO advantages of three-phase systems over single-phase systems. 2.3 Draw a neatly lab... show full transcript

Worked Solution & Example Answer:2.1 Name the instrument that is used to measure electrical energy - NSC Electrical Technology Power Systems - Question 2 - 2016 - Paper 1

Step 1

Name the instrument that is used to measure electrical energy.

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Answer

The instrument used to measure electrical energy is a Kilo-watt hour meter.

Step 2

State TWO advantages of three-phase systems over single-phase systems.

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Answer

  1. Three-phase systems can be operated in star or delta configurations, which allows for flexibility in applications and systems.

  2. Three-phase systems can supply both three-phase and single-phase installations, making them versatile for various electrical needs.

Step 3

Draw a neatly labelled sketch representing the voltage waveforms of a three-phase AC generation system.

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Answer

A diagram representing the voltage waveforms of a three-phase AC generation system would show three sine waves, each 120 degrees out of phase with each other. The peaks should be labeled +Vmax and -Vmax, and the voltage values should be shown clearly to differentiate each phase (e.g., Ph₁, Ph₂, Ph₃, etc.).

Step 4

Calculate the: Line voltage

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Answer

To calculate the line voltage (Vₗ), we use the formula:

S = rac{\sqrt{3} Vₗ Iₗ}{S}

Solving for Vₗ, we have:

Vl=S3Il=200003×25=461.88VVₗ = \frac{S}{\sqrt{3} Iₗ} = \frac{20000}{\sqrt{3} \times 25} = 461.88 \, V

Step 5

Calculate the: Phase voltage

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Answer

To find the phase voltage (Vₚ), we can use the relationship between line and phase voltage in a star connection:

Vp=Vl3=461.883=267.67VVₚ = \frac{Vₗ}{\sqrt{3}} = \frac{461.88}{\sqrt{3}} = 267.67 \, V

Step 6

Calculate the: Total input power

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Answer

The total input power (Pₜ) is calculated by summing the wattmeter readings:

Pt=P1+P2=8kW+4kW=12kWPₜ = P₁ + P₂ = 8 \, kW + 4 \, kW = 12 \, kW

Step 7

Calculate the: Line current.

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Answer

To find the line current (Iₗ), we use the formula:

P=3VlIlcosφP = \sqrt{3} Vₗ Iₗ \cos φ

Rearranging to solve for Iₗ gives us:

Il=P3Vlcosφ=120003×380×0.8=22.79AIₗ = \frac{P}{\sqrt{3} Vₗ \cos φ} = \frac{12000}{\sqrt{3} \times 380 \times 0.8} = 22.79 \, A

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