A student investigated how the current in a filament lamp varies with the potential difference across the filament lamp - AQA - GCSE Physics - Question 6 - 2023 - Paper 1

To find the resistance (R) at a potential difference (V) of +3.0 V:

From Figure 6, look at the graph to find the corresponding current (I) for V = 3.0 V. Let's assume I = 0.16 A.

Using Ohm's Law: $R = \frac{V}{I}$ Substituting the values: $R = \frac{3.0\, \text{V}}{0.16\, \text{A}}$ Therefore, $R = 18.75\, \Omega$

To calculate energy (E) transferred:

1. First, calculate the power (P) using the formula: $P = V \times I$ For the given values: $P = 6.0\, \text{V} \times 0.21\, \text{A} = 1.26\, \text{W}$

2. Next, use the power to find the energy transferred over time (t) using: $E = P \times t$ Where t = 30 minutes = 1800 seconds: $E = 1.26\, \text{W} \times 1800\, \text{s} = 2268\, \text{J}$

The student's prediction is incorrect because, for power to quadruple with a doubled potential difference, the current would also need to double. However, in a filament lamp, the resistance changes with temperature which means the relationship between current and potential difference is not linear. Thus, doubling the potential difference does not result in a proportional increase in current, leading to a different power output than predicted.