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

In the circuit with two identical filament lamps, the current I₁ entering the junction is equal to the sum of the currents through the lamps, I₂ and I₃. Therefore, we can state:

$I₁ = I₂ + I₃$

Additionally, since both lamps are identical and connected in parallel, the currents through them will be equal, meaning:

$I₂ = I₃$

Thus, we can conclude:

$I₁ = 2I₂$

To calculate the charge that flows through the cell, we start with the power relationship:

$P = IV$

Using Ohm's law, we can express voltage in terms of current and resistance:

$V = IR$

From the power equation, substituting for V gives:

$P = I(IR) = I^2 R$

We can rearrange to find current I:

I = rac{P}{R} = rac{3 ext{ W}}{12 ext{ Ω}} = 0.25 ext{ A}

To find the charge Q that flows in 1 minute (60 seconds), we use:

$Q = I imes t$

Substituting in:

$Q = 0.25 ext{ A} imes 60 ext{ s} = 15 ext{ C}$

The unit of charge is coulombs (C).

The readings on both meters will change depending on environmental factors such as light intensity and temperature:

• Light Intensity: If the light intensity increases, the resistance of the LDR (Light Dependent Resistor) decreases, causing more current to flow through the circuit, resulting in a higher reading on the ammeter.
• Temperature: As the temperature increases, the filament lamp's resistance also increases, causing a decrease in current flow, which can lead to a lower ammeter reading. Conversely, if the temperature decreases, the resistance lowers, allowing more current to flow.

Thus, both the LDR and filament lamps react to environmental changes that affect their resistance, subsequently altering the readings on the meters.