Part (a) was well attempted by most responding candidates.
In part (b), most candidates wrote about the effects of transmission at high current and not why transmission was done at low current and high voltage as demanded in the question.
Part c(i) was fairly well attempted by most of the candidates that attempted this question however, in c(ii), most of the candidates did not realize that the potential difference (p.d) across the potentiometer wire AB was not 1.5 V.
Part (d) was satisfactorily attempted by most candidates.
The expected answers are:
(a)(i) Advantages of alkaline cells over lead –acid cells
- supply a large current
- durability (last longer)
- can be left for a longer period in a discharged state
(ii) Disadvantage of alkaline cells over lead-acid cells
- lower e.m.f
- Emf falls rapidly when being used
(b) Electrical Power = IV
For constant power, current is inversely proportional to voltage, hence high voltage implies low current. With low current in transmission cable, the effect of Joule heat generated is reduce and energy loss is minimized. Also the cable needed to carry low current is relatively thin hence cheaper to purchase
(c) Total resistance in the potentiometer circuit = 6Ω
(i) Current in the potentiometer circuit
= V/RT
= 1.5
6
= 0.25A
(ii) Let resistance of AJ be r
0.25r = 0.08
r = 0.08
0.25
= 0.32Ω
But resistance of 100cm of wire = 2 Ω
length of AJ = 100 x 0.32
2
= 16cm
(d) Pt = mc
70t = 1.4 x 4200 x (100 – 30)
t = 1.4 x 4200 x 70
70
= 58880s
or
= 5.9 x 103s
