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The diagram below is a simplified representation of an AC generator - NSC Physical Sciences - Question 9 - 2021 - Paper 1

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The diagram below is a simplified representation of an AC generator. The coil is rotated in a clockwise direction in the magnetic field. 9.1 Write down the name of... show full transcript

Worked Solution & Example Answer:The diagram below is a simplified representation of an AC generator - NSC Physical Sciences - Question 9 - 2021 - Paper 1

Step 1

9.1 Write down the name of component X.

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Answer

Component X is known as the slip rings.

Step 2

9.2 Write down the function of component Y.

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Answer

Component Y allows the slip rings to rotate while maintaining contact with the external circuit, thus facilitating the transfer of current to the external circuit.

Step 3

9.3 Use the relevant principle to explain why an emf is induced in the coil when the coil is rotated in the magnetic field.

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Answer

According to the principle of electromagnetic induction, an emf is induced in the coil as a result of the change in the magnetic field. When the coil rotates, there is a change in the magnetic flux linked with the coil, causing a current to flow. This principle is governed by Faraday's law of electromagnetic induction, which states that the induced emf is directly proportional to the rate of change of magnetic flux.

Step 4

9.4 The coil rotates CLOCKWISE from the position shown in the diagram. In which direction will current be induced in segment PQ of the coil?

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Answer

When the coil rotates clockwise, the induced current in segment PQ will flow from 'P to Q'.

Step 5

9.5 Calculate the time t indicated in the above graph.

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Answer

From the graph, the time t indicated is 0.02 s, which corresponds to the period of the wave, calculated as follows: T = 1/f = 1/50 Hz = 0.02 s.

Step 6

9.6 Calculate the energy dissipated when the appliance is in operation for ONE minute.

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Answer

The energy dissipated can be calculated using the formula:

W=PimestW = P imes t

Where power P can be calculated using Ohm's law: P=Irms2imesRP = I_{rms}^2 imes R.

Given that the RMS voltage Vrms=Vmax2=3112219.71 VV_{rms} = \frac{V_{max}}{\sqrt{2}} = \frac{311}{\sqrt{2}} \approx 219.71 \text{ V}, we can find the current IrmsI_{rms}:

Irms=VrmsR=219.71100=2.1971 AI_{rms} = \frac{V_{rms}}{R} = \frac{219.71}{100} = 2.1971 \text{ A}.

Thus, P=Irms2×R=(2.1971)2×100483.605 WP = I_{rms}^2 \times R = (2.1971)^2 \times 100 \approx 483.605 \text{ W}.

For ONE minute (60 seconds), the energy is: W=483.605 W×60 s28983.3 JW = 483.605 \text{ W} \times 60 \text{ s} \approx 28983.3 \text{ J}.

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