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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 4.5 V. 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

Calculate the resistance of the lamp.

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Answer

To determine the resistance of the lamp, we can use Ohm's Law, which is given by the equation:

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

Where:

  • R is the resistance in ohms (Ω)
  • V is the voltage in volts (V)
  • I is the current in amperes (A)

Substituting the values from the question:

  • V = 4.5 V
  • I = 0.30 A

Therefore, we have:

R=4.5 V0.30 A=15 ΩR = \frac{4.5 \text{ V}}{0.30 \text{ A}} = 15 \text{ Ω}

Thus, the resistance of the lamp is 15 Ω.

Step 2

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 for electrical power, which can be expressed as:

P=V×IP = V \times I

Where:

  • P is the power in watts (W)
  • V is the voltage in volts (V)
  • I is the current in amperes (A)

Substituting in the given values:

  • V = 4.5 V
  • I = 0.30 A

Thus, the power is calculated as follows:

P=4.5 V×0.30 A=1.35 WP = 4.5 \text{ V} \times 0.30 \text{ A} = 1.35 \text{ W}

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

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