8(a) A student investigates resistors connected in parallel using a number of resistors - Edexcel - GCSE Physics - Question 8 - 2020 - Paper 1

The potential difference across resistor R is the voltage reading from the voltmeter. The current in the circuit can be read from an ammeter connected in series with the resistor.

The resistance of a single resistor can be calculated from the student’s investigation using the values shown in Figure 20. From the data, we find that the resistance of a single resistor is 1 Ω.

The overall resistance decreases as the number of resistors in parallel increases. This indicates a non-linear relationship, where the reduction in overall resistance occurs at a decreasing rate. From Figure 20, for example, when moving from 1 to 2 resistors, the resistance drops significantly, but as more resistors are added, the change in overall resistance becomes less pronounced.

Using Ohm's law, the potential difference (V) across the 15Ω resistor can be calculated as:

$V = I × R$

where:

• $I = 0.20 A$
• $R = 15 Ω$

Thus, $V = 0.20 × 15 = 3 V$

The total resistance (R_total) in parallel can be calculated using:

$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{15} + \frac{1}{20}$

Calculating:

$\frac{1}{R_{total}} = \frac{4}{60} + \frac{3}{60} = \frac{7}{60}$

Thus, $R_{total} = \frac{60}{7} \approx 8.57 Ω$

Now, we use this to calculate power:

$P = I^2 × R_{total} = (0.20)^2 × \frac{60}{7}$

Calculating:

$P = 0.04 × 8.57 \approx 0.343 W$