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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 calculate the resistance of the lamp, we can use Ohm's law, which states:

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

Here, the voltage (V) provided by the power supply is 4.5 volts, and the current (I) flowing through the lamp is 0.30 A.

Substituting the values:

R=4.5 V0.30 A=15 ΩR = \frac{4.5 \, \text{V}}{0.30 \, \text{A}} = 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 calculate the power supplied to the lamp, we can use the power 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 \, \text{V} \times 0.30 \, \text{A} = 1.35 \, \text{W}

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

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