A-LevelPhysicsD.C. circuits9702/23 · May/June 2017 · Q611 Marks

Two identical filament lamps are connected in series and then in parallel to a battery of electromotive force (e.m.f.) 12V and negligible internal resistance, as shown in [Figure 6.1a] and [Figure 6.1b]. The I–V characteristic of each lamp is shown in [Figure 6.2].

(a)(i)a metallic conductor at constant temperature,[1]
(a)(ii)a semiconductor diode.[2]
(b)(i)1.the circuit of Fig. 6.1a,[1]
(b)(i)2.the circuit of Fig. 6.1b.[2]
(b)(ii)Calculate the total resistance in 1. the circuit of Fig. 6.1a, 2. the circuit of Fig. 6.1b.[3]
(b)(iii)Calculate the ratio power dissipated in a lamp in the circuit of Fig. 6.1a power dissipated in a lamp in the circuit of Fig. 6.1b[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

6(a)(i) straight line through the origin
6(a)(ii) zero current for one direction (-ve V) up to zero or a few tenths of volt (+ve V)
6(a)(ii) straight line positive gradient/increasing gradient (+ve V)
6(b)(i) 1. current = 2.8 A
6(b)(i) 2. 4(.0) A for each lamp
6(b)(i) current in circuit = 8(.0) A
6(b)(ii) use of R = V/I with correct values of V from graph for each arrangement
6(b)(ii) 1. series resistance (= 2.1 + 2.1) = 4.2 or 4.3 Ω
6(b)(ii) or (12/2.8) = 4.3 Ω
6(b)(ii) 2. parallel resistance 1.5Ω (each lamp 3.0 Ω)
6(b)(ii) or (12/8.0) = 1.5 Ω
6(b)(iii) power = IV or V²/ R or I²R
6(b)(iii) ratio = (2.8 × 6.0) / (4.0 × 12) = 0.35

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