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Figure 3 shows a lamp connected to a d.c - Edexcel - GCSE Physics - Question 2 - 2022 - Paper 1

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Figure 3 shows a lamp connected to a d.c. power supply. The power supply provides a potential difference (voltage) of 4.5V. The current in the lamp is 0.30 A. (i) ... show full transcript

Worked Solution & Example Answer:Figure 3 shows a lamp connected to a d.c - Edexcel - GCSE Physics - Question 2 - 2022 - Paper 1

Step 1

Calculate the resistance of the lamp

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Answer

To find the resistance of the lamp, we can use the formula for resistance:

R=VIR = \frac{V}{I}

Where:

  • VV (voltage) = 4.5 V
  • II (current) = 0.30 A

Substituting the values into the formula gives:

R=4.50.30=15  ΩR = \frac{4.5}{0.30} = 15 \; \Omega

Thus, the resistance of the lamp is 15 Ω.

Step 2

Calculate the power supplied to the lamp

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Answer

The power supplied to the lamp can be calculated using the formula:

Power=V×I\text{Power} = V \times I

Substituting the values:

Power=4.5 V×0.30 A=1.35 W\text{Power} = 4.5 \, V \times 0.30 \, A = 1.35 \, W

Thus, the power supplied to the lamp is approximately 1.4 W.

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