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

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1 (a) 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.... show full transcript

Worked Solution & Example Answer:1 (a) 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 find the resistance of the lamp, we can use the formula: R=VIR = \frac{V}{I}

Where:

  • VV is the voltage, which is 4.5 V.
  • II is the current, which is 0.30 A.

Substituting the values: R=4.50.3=15  ΩR = \frac{4.5}{0.3} = 15 \; \Omega

Thus, the resistance of the lamp is 15 ohms.

Step 2

(ii) Calculate the power supplied to the lamp.

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Answer

To find the power supplied to the lamp, we can use the formula: P=V×IP = V \times I

We know:

  • Voltage (VV) = 4.5 V
  • Current (II) = 0.30 A

Substituting the values: P=4.5×0.3=1.35  WP = 4.5 \times 0.3 = 1.35 \; W

Therefore, the power supplied to the lamp is 1.35 watts.

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