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Figure 6 shows a graph of current against potential difference for an electrical component - Edexcel - GCSE Physics Combined Science - Question 4 - 2018 - Paper 1

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Figure 6 shows a graph of current against potential difference for an electrical component. Which electrical component will show this variation of current with pote... show full transcript

Worked Solution & Example Answer:Figure 6 shows a graph of current against potential difference for an electrical component - Edexcel - GCSE Physics Combined Science - Question 4 - 2018 - Paper 1

Step 1

Which electrical component will show this variation of current with potential difference?

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Answer

The correct answer is D diode. A diode allows current to flow in one direction and has a non-linear relationship between current and potential difference, as illustrated in Figure 6.

Step 2

(i) Calculate the power of the lamp.

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Answer

To calculate the power of the lamp, we can use the equation:

P=IimesVP = I imes V

Substituting the given values:

  • Current, I=0.12extAI = 0.12 ext{ A}
  • Potential difference, V=0.24extVV = 0.24 ext{ V}

Now, performing the calculation:

P=0.12imes0.24=0.0288extWP = 0.12 imes 0.24 = 0.0288 ext{ W}

Thus, the power of the lamp is approximately 0.029 W.

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