Figure 1 shows a mobile phone with its battery removed - AQA - GCSE Physics Combined Science - Question 1 - 2021 - Paper 1

Using Ohm's Law, we can calculate the resistance (R) as follows:

$3.9 = 0.12 \times R$

To find R, rearranging the equation gives:

$R = \frac{3.9}{0.12}$

Calculating this yields:

$R \approx 32.5 \; \Omega$

The equation that links energy (E), power (P), and time (t) is:

$E = P \times t$

The energy transferred can be calculated using the equation derived above:

1. Convert time from minutes to seconds: $2500 \; \text{minutes} = 2500 \times 60 \; \text{seconds} = 150000 \; \text{seconds}$

2. Using the average power output: $E = 0.46 \; \text{W} \times 150000 \; \text{s} \approx 69000 \; J$

The circuit symbol shown in Figure 2 represents a thermistor.

As the temperature of the thermistor increases, the resistance decreases. This results in an increase in current, due to the constant potential difference across the component. According to Ohm's Law, this can be expressed as:

$I = \frac{V}{R}$ Where a lower R (resistance) will lead to a higher I (current).