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A student uses this circuit to investigate how the current in a filament lamp varies with the potential difference (voltage) across the lamp - Edexcel - GCSE Physics - Question 5 - 2016 - Paper 1

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A student uses this circuit to investigate how the current in a filament lamp varies with the potential difference (voltage) across the lamp. (a) Add a voltmeter to... show full transcript

Worked Solution & Example Answer:A student uses this circuit to investigate how the current in a filament lamp varies with the potential difference (voltage) across the lamp - Edexcel - GCSE Physics - Question 5 - 2016 - Paper 1

Step 1

Add a voltmeter to the circuit that can be used to measure the potential difference (voltage) across the lamp.

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Answer

To measure the potential difference across the lamp, the voltmeter should be connected in parallel with the lamp component in the circuit. This configuration allows for accurate voltage measurement without significantly affecting the circuit.

Step 2

Complete the sentence by putting a cross (X) in the box next to your answer.

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Answer

The correct answer is:

☑ A 12 joules per coulomb

This indicates the unit of electric potential difference, or voltage, is joules per coulomb.

Step 3

When the variable resistor is at the halfway position, the ammeter reads 0.37 A and the voltmeter reads 40.0 V. Show that the resistance of the filament in the lamp is about 11 Ω.

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Answer

To calculate the resistance of the filament, we can use Ohm's Law, which is defined as:

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

Substituting in the given values:

  • Voltage (V) = 40.0 V
  • Current (I) = 0.37 A

R=40.0 V0.37 AR = \frac{40.0 \text{ V}}{0.37 \text{ A}}

Calculating this gives:

R108.11 ΩR \approx 108.11 \text{ Ω}

Hence, rounding to the nearest whole number, the resistance of the filament in the lamp is about 11 Ω.

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