Figure 8 shows identical filament lamps connected together to a 12V power supply - Edexcel - GCSE Physics Combined Science - Question 6 - 2018 - Paper 1

We can use the formula for power, which is given by:

$P = I imes V$

Where:

• $P$ is the power (0.75 W)
• $I$ is the current, and
• $V$ is the potential difference across each lamp (1.5 V found in part (a)(i)).

Rearranging the formula to find current:

$I = \frac{P}{V} = \frac{0.75W}{1.5V} = 0.5A$

Now, using Ohm's law, which states:

$V = I imes R$

Rearranging to find resistance:

$R = \frac{V}{I} = \frac{1.5V}{0.5A} = 3 \Omega$

Thus, the resistance of each lamp is 3Ω.

To investigate how the resistance of a lamp changes with potential difference, the student should set up a circuit that includes the following components:

1. A power supply (battery)
2. An ammeter to measure current
3. A voltmeter to measure potential difference
4. The filament lamp under investigation
5. A means of varying the potential difference (such as a rheostat)

Method Steps:

1. Connect the circuit as shown in the circuit diagram:
   • The power supply should be connected to the circuit, ensuring that the ammeter is in series while the voltmeter is in parallel with the lamp.
2. Start with the lowest potential difference and turn on the power supply.
3. Record the potential difference across the lamp as indicated by the voltmeter.
4. Record the current passing through the lamp using the ammeter.
5. Increase the potential difference in increments (using the rheostat) and repeat steps 3 and 4, recording all data.
6. Calculate the resistance for each reading using Ohm's law: $R = \frac{V}{I}$
7. Analyze the results to see how resistance changes with varying potential difference.