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

To calculate the resistance (R), we can rearrange the equation:

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

Substituting the values:

$R = \frac{3.9 \, \text{V}}{0.12 \, \text{A}} = 32.5 \, \Omega$

Thus, the resistance of the electronic circuit is approximately 32.5 Ω.

The equation that links energy, power, and time is:

$E = P \times t$

To calculate the energy transferred, first convert the time from minutes to seconds:

$2500 \, \text{minutes} = 2500 \times 60 \, \text{seconds} = 150000 \, \text{seconds}$

Now using the energy equation:

$E = P \times t = 0.46 \, \text{W} \times 150000 \, \text{s} = 69000 \, \text{J}$

Therefore, the energy transferred by the battery is 69000 J.

The circuit symbol shown in Figure 2 represents a thermistor.

As the temperature of the thermistor increases, its resistance decreases. Since the potential difference remains constant, according to Ohm's Law, the current will increase:

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

This means that as resistance decreases, current increases.