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Figure 2 shows a circuit that a student built - AQA - GCSE Physics Combined Science - Question 4 - 2019 - Paper 1
Question 4
Figure 2 shows a circuit that a student built. The lamp has a resistance of 10 Ω Each resistor has a resistance of 10 Ω What is the total resistance of the circuit?... show full transcript
Worked Solution & Example Answer:Figure 2 shows a circuit that a student built - AQA - GCSE Physics Combined Science - Question 4 - 2019 - Paper 1
Step 1
The total resistance of the circuit?
Answer
The total resistance of the circuit is less than 20 Ω because when resistors are connected in parallel, the total resistance is given by the formula: In this case, if two resistors of 10 Ω each are in parallel, the total resistance is calculated as:
Step 2
Explain your answer to Question 04.1.
Answer
The reason the total resistance is less than 20 Ω is due to the nature of parallel circuits. In a parallel arrangement, the total resistance is always less than the smallest individual resistor. Here, since each resistor has a resistance of 10 Ω, the total resistance being calculated is significantly lower than 20 Ω.
Step 3
Step 4
Explain how the current in the thermistor changed.
Answer
As the temperature of the thermistor increases, the resistance of the thermistor decreases. According to Ohm's Law, as resistance decreases, the current increases. Thus, we can conclude that the current in the thermistor will increase as the temperature rises.
Step 5
Explain how the potential difference across the resistor will change when the switch is closed.
Answer
When the switch is closed, the potential difference across the resistor becomes 0 V because the circuit allows current to flow through an alternative path. This means there's no voltage drop across the resistor as it becomes short-circuited, leading to no current flow through it.
Step 6
Explain how the potential difference across the lamp will change when the switch is closed.
Answer
When the switch is closed, the potential difference across the lamp remains the same as the battery. Since the lamp is still part of the circuit, the full voltage from the battery is applied across it, while the potential difference across the resistor becomes negligible.