2.9 Required practical: resistance of wires and combinations
9 exam questions worth 19 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
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State how the resistance of a wire depends on its length, and describe the graph that shows this.
Mark scheme
resistance is directly proportional to length (1); the graph of resistance against length is a straight line through the origin (1)
Model answer
The resistance is directly proportional to the length. A graph of resistance against length is a straight line through the origin.
Explain why the current should be kept small and the circuit switched off between readings in this experiment.
Mark scheme
a large current / leaving it on makes the wire heat up (1); resistance changes with temperature so the results would not be valid / reliable (1)
Model answer
A large current, or leaving the circuit on, heats the wire. The resistance changes as the temperature changes, so the results would not be valid.
A student measures a potential difference of 1.5 V and a current of 0.30 A for a 50 cm length of wire. Calculate the resistance of the wire.
Mark scheme
R = V / I = 1.5 / 0.30 (1); 5.0 Ω (1)
Model answer
R = V / I = 1.5 / 0.30 = 5.0 Ω.
The wire in the previous question has a resistance of 5.0 Ω at 50 cm. Predict its resistance at 100 cm and give a reason.
Mark scheme
10 Ω (1); resistance is directly proportional to length (so doubling the length doubles the resistance) (1)
Model answer
10 Ω, because resistance is directly proportional to length, so doubling the length doubles the resistance.
A student connects two 10 Ω resistors in series and then in parallel. Describe how the total resistance of the combinations compares with 10 Ω.
Mark scheme
in series the total resistance is greater / 20 Ω (1); in parallel the total resistance is less than 10 Ω (1)
Model answer
In series the total resistance is bigger than 10 Ω (20 Ω). In parallel the total resistance is smaller than 10 Ω.
Which variable is the dependent variable when investigating the resistance of a wire of different lengths?
Choose one answer.
Correct Not quite: the answer is The resistance of the wire
Mark scheme
B: the resistance of the wire
Model answer
The resistance of the wire (B), because it is the quantity that is measured.
The resistance of 100 cm of a wire is 5.0 Ω. Calculate the resistance of 35 cm of the same wire.
Mark scheme
resistance is directly proportional to length: 5.0 × 35 ÷ 100 (1); 1.75 Ω (1)
Model answer
Resistance = 5.0 × 35 / 100 = 1.75 Ω (1.8 Ω to 2 s.f.).
Explain, in terms of electrons, why a thicker wire has a lower resistance than a thinner wire of the same length and material.
Mark scheme
a thicker wire has more space for electrons to move through (1); so more electrons carry charge at the same time / a larger current for the same p.d. (1); so the resistance (V ÷ I) is lower (1)
Model answer
A thicker wire gives the free electrons more room to flow, so more charge can flow every second for the same potential difference. The current is larger, so the resistance, V / I, is lower.
A student connects a 10 Ω resistor and a 15 Ω resistor in series, and then in parallel. State the total resistance in series, and state and explain whether the parallel total is more or less than 10 Ω.
Mark scheme
series total = 25 Ω (1); parallel total is less than 10 Ω (1); because the current has more than one path / more charge can flow for the same potential difference (1)
Model answer
In series, the total is 25 Ω. In parallel, the total is less than 10 Ω, because there are two paths for the current, so more current flows for the same p.d.
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