. (a) either (If in parallel) when one bulb fails, other
bulbs stay on
or (If in parallel) can identify which bulb has failed; (1) 1
or (If in parallel) can identify which bulb has failed; (1) 1
(b) (i) P = VI (1)
0.5 = 240 I
I = 2.1 × 10–3 A 1 s.f. in answer (–1) once only (1) 2
0.5 = 240 I
I = 2.1 × 10–3 A 1 s.f. in answer (–1) once only (1) 2
(ii) R
= V/I (1)
= 240/(2.1 × 10–3)
= 1.14 × 105 Ω or 1.15 × 105 Ω ans
accept (1.1 to 1.2) × 105 Ω. (1) 2
= 240/(2.1 × 10–3)
= 1.14 × 105 Ω or 1.15 × 105 Ω ans
accept (1.1 to 1.2) × 105 Ω. (1) 2
(iii) A
= ρ l / R (1)
= 1.1 × 10–6 × 6.0 × 10–3 / (1.14 × 105) (= 5.79 × 10–14 m2)
A = πr2 (1)
5.79 × 10–14 = πr2 so r = 1.4 × 10–7 m (1) 3
= 1.1 × 10–6 × 6.0 × 10–3 / (1.14 × 105) (= 5.79 × 10–14 m2)
A = πr2 (1)
5.79 × 10–14 = πr2 so r = 1.4 × 10–7 m (1) 3
(iv) filament
too thin / fragile to be manufactured / used without damage;
allow ecf from (iii). (1) 1
allow ecf from (iii). (1) 1
4
[20]
2. (a) current µ p.d / voltage (for a metallic conductor) M1
as
long as temperature is constant / physical conditions remain constant A1
(b) (i) (R
=) (= 0.0349) B1
(ii) R = (Allow
any subject ) C1
C1
resistivity =
5.6 ´ 10–5 A1
unit:
ohm metre / W
m (Allow V m A–1) A1
(5.6 ´ 10–n without unit or incorrect unit and n ¹ 5 or 3 – can
score 2/4)
(5.6
´ 10–3 W m – can score 3/4)
(5.6
´ 10–3 W cm – can score 4/4)
[7]
