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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 4.5 V. 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

(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, represented by the equation:

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

Where:

  • VV is the voltage (4.5 V)
  • II is the current (0.30 A)

Substituting the given 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

Power can be calculated using the formula:

P=V×IP = V \times I

Using the given voltage and current:

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

Alternatively, we could also use the resistance calculated in part (i):

P=V2RP = \frac{V^2}{R}

Substituting the values:

P=(4.5)215=20.2515=1.35  WP = \frac{(4.5)^2}{15} = \frac{20.25}{15} = 1.35 \; W

Hence, the power supplied to the lamp is 1.35 W.

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