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

To determine the power supplied to the lamp, we can use the formula:

$\text{power} = V \times I$

Substituting the values:

$\text{power} = 4.5 \, \text{V} \times 0.30 \, \text{A} = 1.35 \, \text{W}$

This can be approximated to power = 1.4 W.

In Figure 19, there is one lamp connected to the power supply, while in Figure 20, there are two identical lamps in series connected to the same power supply.

Since there are two lamps in Figure 20, the voltage across each lamp will be less compared to the lamp in Figure 19. This results in each lamp in Figure 20 receiving less power, making them dimmer than the single lamp in Figure 19.

Thus, the brightness of the lamp in Figure 19 is greater than that of the lamps in Figure 20.

To measure the resistance of 50 cm of wire, the circuit should include:

• A low voltage power supply
• An ammeter to measure current
• A voltmeter to measure voltage across the wire
• The resistance wire itself
• To tap off the 50 cm of wire, it can be connected in series with the ammeter.

The circuit diagram should be drawn clearly showing these components.

Direct current (d.c.) flows in one direction only, meaning the electrons move in a continuous path.

Alternating current (a.c.), on the other hand, changes direction periodically, causing the flow of electrons to reverse at regular intervals.