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Figure 11 (a) Electricity supplied to homes has a frequency of ☐ A 0.02 Hz ☐ B 20 Hz ☐ C 50 Hz ☐ D 500 Hz (b) Explain why the National Grid uses high voltages with small currents to transfer electricity from power stations - Edexcel - GCSE Physics - Question 5 - 2023 - Paper 2

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Figure-11--(a)-Electricity-supplied-to-homes-has-a-frequency-of--☐-A-0.02-Hz-☐-B-20-Hz-☐-C-50-Hz-☐-D-500-Hz--(b)-Explain-why-the-National-Grid-uses-high-voltages-with-small-currents-to-transfer-electricity-from-power-stations-Edexcel-GCSE Physics-Question 5-2023-Paper 2.png

Figure 11 (a) Electricity supplied to homes has a frequency of ☐ A 0.02 Hz ☐ B 20 Hz ☐ C 50 Hz ☐ D 500 Hz (b) Explain why the National Grid uses high voltages wit... show full transcript

Worked Solution & Example Answer:Figure 11 (a) Electricity supplied to homes has a frequency of ☐ A 0.02 Hz ☐ B 20 Hz ☐ C 50 Hz ☐ D 500 Hz (b) Explain why the National Grid uses high voltages with small currents to transfer electricity from power stations - Edexcel - GCSE Physics - Question 5 - 2023 - Paper 2

Step 1

Electricity supplied to homes has a frequency of

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Answer

The frequency of electricity supplied to homes is typically 50 Hz. Therefore, the correct answer is C 50 Hz.

Step 2

Explain why the National Grid uses high voltages with small currents to transfer electricity from power stations.

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Answer

The National Grid uses high voltages for electricity transmission in order to minimize energy loss over long distances. Higher voltage enables the use of lower currents, which reduces the energy lost as heat due to the resistance in the wires. This method is efficient for transporting large amounts of electrical energy from power stations to consumers.

Step 3

Calculate the power in the primary coil.

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Answer

To calculate the power in the primary coil, we use the formula:

P=V×IP = V × I

where P is power, V is voltage, and I is current. To provide a specific numerical answer, the values for voltage and current must be known. Assuming some values: If V = 200 V and I = 0.1 A:

P=200extV×0.1extA=20extWP = 200 ext{ V} × 0.1 ext{ A} = 20 ext{ W}

Thus, power in the primary coil = 20 W.

Step 4

For the transformer in Figure 12, evaluate, in its simplest form, the ratio secondary voltage : primary voltage

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Answer

To evaluate the ratio of secondary voltage to primary voltage for the transformer, we must know the number of turns in the primary and secondary coils. Assuming the turns ratio is represented as:

rac{V_s}{V_p} = rac{N_s}{N_p}

where VsV_s is the secondary voltage, VpV_p is the primary voltage, NsN_s is the number of turns in the secondary coil, and NpN_p is the number of turns in the primary coil. Without specific values, we cannot compute the exact ratio.

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