Some students investigate electric circuits - Edexcel - GCSE Physics - Question 3 - 2012 - Paper 1

For the circuit, the ammeter should be connected in series with the lamp, while the voltmeter should be connected in parallel across the lamp. This configuration allows the ammeter to measure the current flowing through the lamp, and the voltmeter to measure the potential difference across it.

To calculate the potential difference across the lamp, we can use Ohm's Law:

$V = I imes R$

Where:

• $I = 0.5 \, A$ (current)
• $R = 8 \, \Omega$ (resistance)

Substituting the values:
$V = 0.5 \, A \times 8 \, \Omega = 4 \, V$
Therefore, the potential difference across the lamp is 4 V.

Only some of the electrical energy is transferred to light energy in the lamp due to inefficiencies related to heat production. When electric current flows through the filament of the lamp, most of the energy is converted into heat rather than light. The filament must reach a certain temperature to emit light, but not all energy is converted into the visible spectrum, resulting in energy loss primarily as thermal energy.

Power can be calculated using the formula:

$P = V imes I$

Where:

• $V = 5 \, V$ (voltage across the lamp)
• $I = 0.4 \, A$ (current through the lamp)

Substituting the values:
$P = 5 \, V \times 0.4 \, A = 2 \, W$
Thus, the power being supplied to the lamp is 2 W.