Paper 1 June 2023, Question 3 11 marks
In Figure 2 the cell has emf ε and internal resistance r.
The current in the circuit is I.
The potential difference (pd) across R1 is V1 and the pd across R2 is V2.
Explain how the law of conservation of energy applies in this circuit. You should consider the movement of one coulomb of charge around the circuit.
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Figure 3 shows a variable resistor made with a thin conducting layer on an insulating base.
The conducting layer has constant width and thickness and has connections at the ends A and B. C is a sliding contact that can move along the surface of the conducting layer between A and B.
Figure 4 shows a circuit that uses the variable resistor as a potential divider.
The variable resistor is connected to a battery of emf 3.00 V and internal resistance r The resistance of the conducting layer between A and B is 125 Ω.
The sliding contact C is moved to end B of the variable resistor. The switch is closed. The digital voltmeter reads 2.89 V.
Show that r is approximately 4.8 Ω.
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C is set at of the distance between A and B. The thickness of the conducting layer is uniform so the resistance between A and C is 25.0 Ω.
Determine the voltmeter reading at this setting.
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Figure 5 shows a variable resistor similar to the one shown in Figure 3 but with the following three manufacturing faults:
• at P the conducting layer changes in thickness so that AP is thinner than PB
• at Q there is a scratch into the surface of the conducting layer and across its full width
• from R to B the conducting connector is laid over the conducting layer.
The width of the conducting layer is constant.
A pd of 3.0 V is applied across A and B. C is moved from A to B.
Sketch, on the axes in Figure 6, a graph to show how the pd between A and C varies as C is moved from A to B.
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