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

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

Step 1

(i) Calculate the resistance of the lamp.

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Answer

To calculate the resistance of the lamp, we use the formula:

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

Where:

  • V = 4.5 V (voltage across the lamp)
  • I = 0.30 A (current through the lamp)

Substituting the values into the equation:

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

Thus, the resistance of the lamp is 15 Ω.

Step 2

(ii) Calculate the power supplied to the lamp.

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Answer

To find the power supplied to the lamp, we use the formula:

P=V×IP = V \times I

Again, substituting the known values:

  • V = 4.5 V
  • I = 0.30 A

Thus, the power can be calculated as follows:

P=4.5×0.30=1.35 WP = 4.5 \times 0.30 = 1.35 \ W

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

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