A technician is testing a filament lamp from a car - Edexcel - GCSE Physics - Question 4 - 2017 - Paper 1

The appropriate symbol for an ammeter should be added in series with the lamp in the circuit.

Using Ohm's Law, the resistance can be calculated as:

$R = \frac{V}{I} = \frac{1.5 \text{ V}}{0.18 \text{ A}} \approx 8.33 \Omega$

Using the power formula,

$P = V \times I$

where:

• $V = 12 ext{ V}$
• $I = 800 ext{ mA} = 0.8 ext{ A}$,

thus:

$P = 12 \text{ V} \times 0.8 \text{ A} = 9.6 \text{ W}$

The current is not 400 mA because, when connected to a 6 V battery, the resistance in the circuit has changed, leading to a different voltage drop across the filament. Additionally, the temperature of the filament may have decreased, which can also affect the resistance and reduce the current further.