A student built a circuit using filament lamps - AQA - GCSE Physics Combined Science - Question 6 - 2018 - Paper 1

The relationship between the currents in a parallel circuit shows that:

• I₁ = I₂ + I₃
• Each lamp carries the same voltage, therefore the total current is the sum of the currents through each branch.

1. First, we need to calculate the current using the formula:

   $P = I imes V$

   Since the voltage across each lamp can be calculated using Ohm's law as:

   $V = I imes R$

   where R is the resistance of 12 Ω. By substituting and rearranging, we can find:

   $I = \frac{P}{V} = \frac{P}{I \times R}$

   Upon solving:

   $I = \frac{3}{3} = 0.25 \, A$

   Then, to find the charge (Q) that flows in 1 minute:

   $Q = I \times t$

   where t is in seconds (60 seconds for 1 minute).

   Thus:

   $Q = 0.25 \times 60 = 15 \, C$

   In conclusion:

   • Charge = 15 C
   • Unit = coulombs.

When environmental conditions change, such as light intensity affecting a Light Dependent Resistor (LDR), we observe the following:

• If the light intensity increases, the resistance of the LDR decreases, allowing a greater current to flow, which may lead to a higher reading on the ammeter.
• Conversely, if light intensity decreases, the resistance of the LDR increases, reducing current flow and yielding a lower reading on the ammeter.
• The voltmeter will reflect changes in voltage drop based on the current changes, as potential difference is shared between components in the circuit. Hence, increasing current will result in a higher voltage reading, while decreasing current will lower the voltage reading on the voltmeter.