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A student investigates the resistance of a lamp - Edexcel - GCSE Physics - Question 5 - 2016 - Paper 1

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A student investigates the resistance of a lamp. She obtains these readings for the potential difference (voltage) across the lamp and the current in the lamp. | vo... show full transcript

Worked Solution & Example Answer:A student investigates the resistance of a lamp - Edexcel - GCSE Physics - Question 5 - 2016 - Paper 1

Step 1

Calculate the resistance, R, of the lamp.

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Answer

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

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

Substituting the given values:

R=6.0V0.40A=15ΩR = \frac{6.0 \, \text{V}}{0.40 \, \text{A}} = 15 \, \Omega

Thus, the resistance is 15 Ω.

Step 2

Calculate the power supplied to the lamp.

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Answer

The power supplied to the lamp can be calculated using the formula:

P=V×IP = V \times I

Substituting the known values:

P=6.0V×0.40A=2.4WP = 6.0 \, \text{V} \times 0.40 \, \text{A} = 2.4 \, \text{W}

Therefore, the power supplied is 2.4 W.

Step 3

Calculate the amount of energy transferred by the lamp in 40 s.

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Answer

The energy transferred by the lamp can be calculated using the formula:

E=P×tE = P \times t

where:

  • EE is the energy,
  • PP is the power, and
  • tt is the time in seconds.

Substituting the known values:

E=2.4W×40s=96JE = 2.4 \, \text{W} \times 40 \, \text{s} = 96 \, \text{J}

Thus, the energy transferred by the lamp in 40 seconds is 96 J.

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