Physics Paper 2, Nov/Dec. 2011
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
Weakness/Remedies
Strength

Question 14

(b)    Explain why High –Tension transmission of power is done with low current and high

voltage.
(c)

The diagram above illustrates a uniform potentiometer wire AB of length 100cm and resistance 2W.
A cell of e.m.f. 1.5 V and a variable resistor R are connected in series
with AB.
If R = 4Ω and the key K1 is closed, calculate the

(i)   current in the potentiometer circuit;
(ii)   length  AJ of the wire for which the galvanometer shows no deflection, when key
K2 is  also closed.

(d)    A heating coil is rated at 70 W.  Calculate the time it will take this coil to heat 1.4 kg of
water at 30o C to 100o C.
[Specific heat capacity of water = 4200Jkg-1 K-1]

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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.

•   supply a large current
•   durability (last longer)
•   can be left for a longer period in a discharged state

•  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