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

Using Ohm's law, which states that resistance (R) is equal to the potential difference (V) divided by the current (I):

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

Substituting in the values from the graph at a potential difference of +3.0 V, for which the corresponding current is 0.16 A:

$R = \frac{3.0 V}{0.16 A} \approx 18.75 \Omega$

To find the energy transferred (E), we can use the formula:

$E = P \times t$

First, calculate power (P) using the formula:

$P = V \times I$

Substituting values for V = 6.0 V and I = 0.21 A:

$P = 6.0 V \times 0.21 A = 1.26 W$

Next, convert 30 minutes into seconds:

$t = 30 \times 60 = 1800 s$

Now substitute for E:

$E = 1.26 W \times 1800 s = 2268 J$

The student's prediction is incorrect because doubling the potential difference does not necessarily double the power output of the filament lamp. The resistance of the filament lamp increases with temperature, and since the filament is not an ohmic conductor, the relationship between voltage and current is nonlinear. As a result, the output power does not scale linearly with an increase in potential difference.