Figure 19 shows a lamp connected to a d.c - Edexcel - GCSE Physics - Question 8 - 2022 - Paper 1

To calculate the power (P) supplied to the lamp, we can use the formula:

$P = V \times I$

Substituting the known values:

$P = 4.5 \text{ V} \times 0.30 \text{ A}$

Calculating this yields:

$P = 1.35 \text{ W}$

In Figure 19, the lamp operates alone and receives the full potential difference from the power supply. In contrast, in Figure 20, two identical lamps are in parallel, meaning each lamp receives the same voltage but shares the total current. As a result, the brightness of each lamp in Figure 20 will be less than that of the lamp in Figure 19 because the total current provided by the power supply is divided among both lamps, leading to a lower current through each one. Thus, the lamp in Figure 19 will be brighter.

To measure the resistance of a section of wire, the student should use a circuit that includes:

• A low voltage power supply or battery,
• An ammeter in series to measure current,
• A voltmeter in parallel to measure voltage across the section of wire,
• And the resistance wire itself.

The circuit diagram should look like this:

   +----[Ammeter]----[Resistance Wire]----+
   |                                      |
[Power Supply]                          [Voltmeter]
   |                                      |
   +--------------------------------------+

This setup allows the student to measure the voltage and current to calculate resistance.

Direct current (d.c.) flows in one direction only, providing a constant voltage. In contrast, alternating current (a.c.) changes direction periodically, alternating the voltage over time. This is relevant in how each type of current powers devices: d.c. is typically used in batteries, while a.c. is commonly used in household power supplies.