A-LevelPhysicsD.C. circuits9702/21 · May/June 2018 · Q610 Marks

Define the volt.

(a)Define the volt.[1]
(b)(i)A battery of electromotive force (e.m.f.) 4.5V and negligible internal resistance is connected to two filament lamps P and Q and a resistor R, as shown in Fig. 6.1. 4.5 V R P Q Fig. 6.1 The current in lamp P is 0.15A. The I–V characteristics of the filament lamps are shown in Fig. 6.2. 0.20 I/A 0.15 P Q 0.10 0.05 0 0 1.0 2.0 3.0 4.0 V/V Fig. 6.2 (i) Use Fig. 6.2 to determine the current in the battery. Explain your working. current =[2]
(b)(ii)(ii) Calculate the resistance of resistor R. resistance =[2]
(b)(iii)(iii) The filament wires of the two lamps are made from material with the same resistivity at their operating temperature in the circuit. The diameter of the wire of lamp P is twice the diameter of the wire of lamp Q. Determine the ratio length of filament wire of lamp P length of filament wire of lamp Q ratio =[3]
(b)(iv)(iv) The filament wire of lamp Q breaks and stops conducting. State and explain, qualitatively, the effect on the resistance of lamp P.[2]

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Mark scheme

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

[B1] joule/coulomb
[B1] lamps have same p.d./lamps have p.d. of 2.7 V
[A1] current = 0.15 + 0.090 = 0.24 A
[C1] R = (4.5 – 2.7)/0.24
[A1] R = 7.5Ω
[C1] Rp = 18 (Ω) and RQ = 30 (Ω) I/RT = 1/18 + 1/30 and so RT = 11.25 4.5 = 0.24 × (R + 11.25)
[C1] R = ρl/A
[C1] Rp/RQ = [(2.7/0.15)/(2.7/0.09)] (= 0.60)
[A1] ratio = 0.60 × 2² = 2.4
[B1] less p.d. across resistor/greater p.d. across P
[B1] greater current through P and so resistance (of P) increases

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