A-LevelPhysicsD.C. circuits9702/23 · May/June 2018 · Q68 Marks
A wire X has a constant resistance per unit length of 3.0 Ω m⁻¹ and a diameter of 0.48 mm.
- (a)Calculate the resistivity of the metal of wire X. resistivity = ............................................................................. Ω m[3]
- (b)The wire X is connected into the circuit shown in Fig. 6.1. [Figure 6.1] The battery has an electromotive force (e.m.f.) of 5.0 V and an internal resistance of 2.0 Ω. The wire X and a resistor R of resistance 4.5 Ω are connected in parallel. The current in the battery is 1.6 A.
- (b)(i)Calculate the potential difference across resistor R. potential difference = ......................................................................... V[1]
- (b)(ii)Determine, for wire X,
- (b)(ii)1.its resistance, resistance = ......................................................................... Ω[3]
- (b)(ii)2.its length. length = ................................................................................... m[1]
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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.
- [C1] R = ρL/A
- [C1] 3.0 = ρ/[π × (0.48 × 10⁻³/2)²]
- [A1] ρ = 5.4 × 10⁻⁷ Ω m
- [A1] p.d. = 5.0 – (2.0 × 1.6) = 1.8 V
- [C1] current in resistor = 1.8/4.5 (= 0.40A)
- [C1] current in wire = 1.6 – 0.40 (= 1.2A)
- [A1] Rₓ = 1.8/1.2 = 1.5 Ω
- [C1] R_T = 1.8/1.6 or (5.0/1.6) – 2.0 (= 1.125Ω)
- [C1] (1/1.125) = (1/4.5) + (1/Rₓ)
- [A1] Rₓ = 1.5Ω
- [A1] length = 1.5/3.0 or 1.5 × 1.8 × 10⁻⁷ / (5.4 × 10⁻⁷) = 0.50 m
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