A-LevelPhysicsD.C. circuits9702/22 · Oct/Nov 2018 · Q69 Marks
(a) Define the volt. (b) A battery of electromotive force (e.m.f.) 7.0V and negligible internal resistance is connected in series with three components, as shown in Fig. 6.1. [Figure 6.1] Resistor X has a resistance of 5.2Ω. The resistance of the filament wire of lamp Y is 6.0Ω. The potential difference across resistor Z is 1.4V.
- (a)Define the volt.[1]
- (b)(i)Calculate the current in the circuit. current = A[2]
- (b)(ii)Determine the resistance of resistor Z. resistance = Ω[1]
- (b)(iii)Calculate the percentage efficiency with which the battery supplies power to the lamp. efficiency = %[3]
- (b)(iv)The filament wire of the lamp is made of metal of resistivity 3.7 × 10⁻⁷Ωm at its operating temperature in the circuit. Determine, for the filament wire, the value of α where α = cross-sectional area / length α = m[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.
- [B1] joule/coulomb
- [C1] 7.0 = (I × 5.2) + (I × 6.0) + 1.4
- [A1] I = 0.50 A
- [A1] R = 1.4/0.50
- [A1] = 2.8Ω
- [C1] P = EI or P = VI or P = I²R or P = V²/R
- [C1] efficiency = [(0.50²×6.0)/(7.0×0.50)] (×100)
- or
- [C1] efficiency = [(0.50×3.0)/(7.0×0.50)] (×100)
- or
- [C1] efficiency = [(3.0²/6.0)/(7.0×0.50)] (×100)
- [A1] efficiency = 43%
- [C1] R = ρL/A
- [C1] α = ρ/R
- [C1] = 3.7 x 10⁻⁷/6.0
- [A1] = 6.2 x 10⁻⁸m
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