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A teacher investigates the resistance of a filament lamp - OCR Gateway - GCSE Physics - Question 17 - 2022 - Paper 1

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A teacher investigates the resistance of a filament lamp. Fig. 17.1 shows the circuit the teacher uses. **Fig. 17.1** (a) The teacher takes measurements of the cu... show full transcript

Worked Solution & Example Answer:A teacher investigates the resistance of a filament lamp - OCR Gateway - GCSE Physics - Question 17 - 2022 - Paper 1

Step 1

Ammeter position:

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Answer

The ammeter should be placed in position K to measure the current through the filament lamp directly.

Step 2

Voltmeter position:

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Answer

The voltmeter should be placed in position J to measure the potential difference across the lamp.

Step 3

Which two students have made a correct statement about the component?

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Answer

Students A and D made correct statements. Student A correctly identifies that the component can influence the current in the circuit, while Student D correctly states that it can change the total resistance in the circuit.

Step 4

Plot the two missing points on Fig. 17.2 and draw a line of best fit.

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Answer

To plot the missing points, identify the potential difference of 6.0 V corresponding to the current of 3.0 A. Draw a line of best fit that approximates the trend of the plotted data, ensuring it represents the linear relationship.

Step 5

Calculate the power dissipated by the filament lamp when the potential difference is 5.0 V.

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Answer

From Fig. 17.2, when the potential difference is 5.0 V, the corresponding current is 2.8 A. Using the formula:

extPower=extPotentialdifferenceimesextCurrent=5.0imes2.8=14.0extW ext{Power} = ext{Potential difference} imes ext{Current} = 5.0 imes 2.8 = 14.0 ext{ W}

Step 6

Calculate the energy transferred if the filament lamp is used for 2 minutes.

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Answer

Using the power calculated from part (b)(ii), which is 14.0 W, and the equation:

extEnergy=extPowerimesextTime ext{Energy} = ext{Power} imes ext{Time}

Convert 2 minutes to seconds (2 minutes = 120 seconds):

extEnergy=14.0imes120=1680extJ ext{Energy} = 14.0 imes 120 = 1680 ext{ J}

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