A scientist sets up an electrical circuit - OCR Gateway - GCSE Physics - Question 17 - 2022 - Paper 1

When the switch is open, the entire potential difference of the supply (6.0V) is across the entire circuit since no current flows through any other component. Therefore, the potential difference across the resistor is equal to the potential difference of the supply:

Potential difference = 6.0V

When the switch is closed, the total resistance of the circuit decreases because the current path is completed, allowing current to flow through both the lamp and the resistor. Hence, the current through the resistor increases due to the lesser total resistance:

$I = \frac{V}{R_{total}}$

With a lower total resistance, the current increases.

As the switch closes and current begins to flow, the potential difference across the resistor also changes. The potential difference across the resistor can be calculated using Ohm's Law:

$V_{R} = I \times R$

As the current increases when the switch is closed, the potential difference across the resistor increases correspondingly.

When the switch is closed, the potential difference across the lamp can be determined by considering the voltage drops in the circuit. Since the total voltage supplied is 6.0V and a portion of it will drop across the resistor, the remaining voltage will drop across the lamp. Assuming the current remains steady and the resistances are constant:

Potential difference = 6.0V - V_{R}

Thus, the potential difference across the lamp can be calculated once the voltage across the resistor is known.