O-LevelPhysicsElectricity5054/22 · May/June 2022 · Q58 Marks

(a) A student connects a battery, thermistor X and resistor Y in parallel, as shown in Fig. 5.1. [Figure 5.1] The electromotive force (e.m.f.) of the battery is E. The current in X is I. The resistance of Y is greater than the resistance of X. Tick two boxes, one to show the correct statement about the potential difference across Y and the other to show the correct statement about the current in Y. The potential difference across Y: is less than E is equal to E is greater than E. The current in Y: is less than I is equal to I is greater than I. (b) The student connects thermistor X in series with the resistor Y and a battery of e.m.f. 6.0 V, as shown in Fig. 5.2. [Figure 5.2] In this circuit, at room temperature, the resistance of thermistor X is 600 Ω and the current in thermistor X is 0.0020 A.

(a)Tick two boxes, one to show the correct statement about the potential difference across Y and the other to show the correct statement about the current in Y. The potential difference across Y: is less than E is equal to E is greater than E. The current in Y: is less than I is equal to I is greater than I.[1]
(b)(i)Calculate the power produced in the thermistor.[2]
(b)(ii)Calculate the resistance of Y.[3]
(b)(iii)The thermistor is cooled. Explain why this causes the potential difference across Y to decrease.[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

[1] middle box of 1st three boxes and top box of 2nd three boxes
[1] (P =) VI or I²R algebraic or numerical seen in any form
[1] 0.0024 W
[1] V = IR algebraic or numerical in any form or potential divider formula seen
[1] 4.8 (V) seen or 6.0 – 1.2/0.002 or 3000 – 600 OR 6 = 0.002 × (600+Rᵞ) or 1.2 = 6 × 600 / (Rᵞ + 600)
[1] 2400 Ω
[1] resistance of thermistor / X increases (on cooling)
[1] current (in thermistor) decreases OR potential divider argument used to explain decrease in pd across Y OR pd across X increases

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