A-LevelPhysicsD.C. circuits9702/23 · Oct/Nov 2017 · Q610 Marks

A filament lamp is rated as 30W, 120V. A potential difference of 120V is applied across the lamp.

(a)(i)For the filament wire of the lamp, calculate the current,[2]
(a)(ii)the number of electrons passing a point in 3.0 hours.[2]
(b)Show that the resistance of the filament wire is 480 Ω.[2]
(c)The filament wire has an uncoiled length of 580 mm and is made of metal. The metal has resistivity 6.1 × 10⁻⁷ Ω m at the operating temperature of the lamp. Calculate the diameter of the wire.[3]
(d)The potential difference across the lamp is now reduced. State and explain the effect, if any, on the resistance of the filament wire.[1]

Mark scheme

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

6(a)(i) P = VI C1 I = 30/120 A1 = 0.25 A
6(a)(ii) Q = 0.25 x 3.0 × 3600 (= 2700) C1 number = (0.25 × 3.0 × 3600) / 1.60 × 10⁻¹⁹ A1 = 1.7 x 10²²
6(b) R = V/I C1 or R = P/I² or R = V²/P = 120/0.25 or = 30/0.25² or = 120²/30 = 480 Ω A1
6(c) R = ρl / A C1 A = (6.1 × 10⁻⁷ × 580 × 10⁻³) / 480 (= 7.37 × 10⁻¹⁰) C1 d = [(4 × 7.37 × 10⁻¹⁰) / π]¹/² A1 = 3.1 x 10⁻⁵ m
6(d) temperature decreases and so resistance decreases B1

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.