O-LevelPhysicsElectricity5054/21 · Oct/Nov 2018 · Q79 Marks

A student sets up the circuit shown in Fig. 7.1. The electromotive force (e.m.f.) of the battery is 6.0 V. [Figure 7.1]

(a)(i)The resistance of the variable resistor X is set to 1.8Ω. Determine the total resistance of the circuit, resistance = [3]
(a)(ii)the current measured by the ammeter. current = [2]
(b)(i)State what is meant by the potential difference (p.d.) across a component in a circuit.[1]
(b)(ii)Add to Fig. 7.1 the symbol for a component that is measuring the p.d. across the 2.8Ω resistor.[1]
(b)(iii)The resistance of X is increased. Explain why the p.d. across the 2.8Ω resistor decreases.[2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.

[C1] 1/R_p = 1/R1 + 1/R2 or 1/R_p = 1/3.6 + 1/1.8 or (R_p =) R1 R2 / (R1 + R2) or 1.8 x 3.6/(1.8 + 3.6)
[C1] 1.2 (Ω)
[A1] 4.0 Ω
[C1] (I =) V/ R or 6.0/4.0
[A1] 1.5 (A)
[B1] work done / energy (released) per unit charge (passed through component)
[B1] voltmeter symbol and across 2.8 Ω resistor
[B1] total resistance increases or resistance of circuit / parallel combination increases
[B1] current (in 2.8 Ω resistor) decreases

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