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A student investigates how the potential difference across the secondary coil of a transformer changes with the number of turns on the secondary coil - OCR Gateway - GCSE Physics - Question 21 - 2023 - Paper 4

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Question 21

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A student investigates how the potential difference across the secondary coil of a transformer changes with the number of turns on the secondary coil. The diagram s... show full transcript

Worked Solution & Example Answer:A student investigates how the potential difference across the secondary coil of a transformer changes with the number of turns on the secondary coil - OCR Gateway - GCSE Physics - Question 21 - 2023 - Paper 4

Step 1

Describe a method that the student uses to obtain valid results.

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Answer

  1. Wrap coils around a soft iron rod: This creates a transformer setup where the magnetic field can easily induce voltage.
  2. Connect a voltmeter across the secondary coil: This allows for the measurement of the potential difference obtained from the secondary coil.
  3. Connect the power supply to the primary coil: This provides the necessary input current to generate a magnetic field.
  4. Change the number of turns in the secondary coil: By varying this parameter, the effect on potential difference can be studied.
  5. Keep the number of turns in the primary coil constant: This ensures that the experiment measures only the effect of secondary coil turns on potential difference.

Step 2

Calculate the p.d. across the secondary coil of the transformer.

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Answer

The potential difference across the secondary coil (V_s) can be calculated using the transformer equation:

VsVp=NsNp\frac{V_s}{V_p} = \frac{N_s}{N_p}

where:

  • V_s = potential difference across the secondary coil
  • V_p = potential difference across the primary coil = 230 V
  • N_s = number of turns in the secondary coil = 300
  • N_p = number of turns in the primary coil = 3540

Rearranging the equation gives:

Vs=Vp×NsNpV_s = V_p \times \frac{N_s}{N_p}

Substituting the known values:

Vs=230V×300354019.5VV_s = 230 V \times \frac{300}{3540} \approx 19.5 V

Step 3

Calculate the current in the primary coil of the transformer.

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Answer

The current in the primary coil (I_p) can be calculated using the relationship between power and current:

Using the transformer equation again: Pprimary=Vp×Ip=PsecondaryP_{primary} = V_p \times I_p = P_{secondary}

Where:

  • Current in secondary coil (I_s) = 4.62 A
  • Potential difference across secondary coil (V_s) = 19.5 V (from previous calculation)

So, Psecondary=Vs×Is=19.5V×4.62A90.15WP_{secondary} = V_s \times I_s = 19.5 V \times 4.62 A \approx 90.15 W

Now we set the power in the primary side equal to this value: 90.15W=230V×Ip90.15 W = 230 V \times I_p

Rearranging gives: Ip=90.15W230V0.392AI_p = \frac{90.15 W}{230 V} \approx 0.392 A

Step 4

Use data from the graph to explain why the student is incorrect.

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Answer

According to the graph, the relationship between the turns ratio and power loss is not linear. When the turns ratio doubles, the power loss does decrease, but it does not halve proportionally. This indicates that the student's claim does not accurately reflect the data shown in the graph, as power loss decreases at a rate slower than the doubling of the turns ratio.

Step 5

Explain why step-up transformers are used in the national grid.

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Answer

Step-up transformers are used in the national grid primarily to increase the voltage for efficient power transmission over long distances. Higher voltage reduces current and decreases resistive losses in the transmission lines, leading to more efficient energy transfer. This minimizes energy loss as heat due to resistance, allowing electricity to be transported over vast distances with minimal loss.

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