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Refer to the figure below to answer the following questions about an electrical circuit - NSC Electrical Technology Electronics - Question 2 - 2017 - Paper 1

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Refer to the figure below to answer the following questions about an electrical circuit. 1. Calculate the line current $I_L$ if the total power $P$ is 30 kW. 2. De... show full transcript

Worked Solution & Example Answer:Refer to the figure below to answer the following questions about an electrical circuit - NSC Electrical Technology Electronics - Question 2 - 2017 - Paper 1

Step 1

1. Calculate the line current $I_L$ if the total power $P$ is 30 kW.

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Answer

To find the line current, we use the formula: IL=P3VI_L = \frac{P}{\sqrt{3} \cdot V} Using P=30,000P = 30,000 W and assuming the voltage VV is the line voltage, we can rearrange it to find the line current as follows: IL=30,0003380I_L = \frac{30,000}{\sqrt{3} \cdot 380} Calculating gives: IL=30,000657.9845.6AI_L = \frac{30,000}{657.98} \approx 45.6 A

Step 2

2. Determine the impedance $Z$ if the voltage $V_{p}$ is 380 V.

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Answer

We know that the impedance can be calculated using the formula: Z=VpIpZ = \frac{V_p}{I_p} Here, we found the phase current from the previous step, Ip=IL3I_p = \frac{I_L}{\sqrt{3}}. Thus: Z=38017.3221.93ΩZ = \frac{380}{17.32} \approx 21.93 \Omega

Step 3

3. Explain what happens to the current drawn by the load if the power factor is improved.

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Answer

If the power factor is improved, the load will draw less current. This is because improving the power factor means that the load becomes more efficient, thus reducing the reactive power in the circuit.

Step 4

4. Discuss any cost savings that could result from consumers using less current.

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Answer

Cost savings arise as consumers use less current, leading to lower electricity bills. Additionally, using less current reduces the load on the electrical system, which can decrease strain on infrastructure and minimize operational costs related to energy production and distribution.

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