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

Given the power output of each lamp is 0.75W, we can use the formula relating power, voltage, and resistance:

$P = \frac{V^2}{R}$

Rearranging this formula to find the resistance gives us:

$R = \frac{V^2}{P}$

Substituting in our values (V = 1.5V and P = 0.75W):

$R = \frac{(1.5)^2}{0.75} = \frac{2.25}{0.75} = 3.0 \Omega$

Thus, the resistance of each lamp is 3.0Ω.

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

1. Components: Use a power supply, an ammeter, a voltmeter, a filament lamp, and a variable resistor (rheostat).
2. Circuit Diagram:
   • Connect the power supply to a series circuit with an ammeter and the filament lamp.
   • Connect the voltmeter in parallel with the lamp to measure the potential difference across it.
   • Use the variable resistor to adjust the voltage.
3. Method:
   • Vary the potential difference by adjusting the rheostat.
   • Record the current from the ammeter and the voltage from the voltmeter.
   • Calculate the resistance using Ohm's Law, $R = \frac{V}{I}$, for each set of measurements obtained from the voltmeter and ammeter.
4. Analysis: Plot a graph of voltage (V) against current (I), and analyze the slope to determine how the resistance varies with voltage.