Physics Paper 2, Nov/Dec. 2010  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
General Comments

Question 14
Part II : Candidates are expected to answer any three form this part

(a) (i) What is a capacitor?
(ii) Explain the term short-circuit.
(b) Why is a 200 watt bulb brighter than a 60 watt bulb when connected to a 220 volt mains outlet?


Five identical capacitors each of capacitance 1 0 ~ are illustrated in the diagram above. Calculate the equivalent capacitance between X and Y.

(d) A 15 µF capacitor is connected in parallel across three capacitors of capacitances 30, 40 and 120 µF in series. The combination is connected across a battery of e.m.f. 220 V.
       (i)          Draw a circuit diagram for the arrangement
       (ii)         Calculate the charge in the circuit.



(a)(i) This question posed no problem to majority of the responding candidates.
(ii) Performance was poor.

(b) Candidates' were expected to recognize that more current passes through the 200W bulb hence higher brightness. Majority ofthe candidates performed satisfactorily.
 (c) This was a numerical problem and candidates' reactions were fairly satisfactory.
(d) This was another numerical problem. Quite a few candidates got the required circuit diagram right. However, the application ofthe formula linking Q, C and V posed no problem to many candidates. Performance was good.

The expected answers are:

(a)(i)A capacitor is a device for storing charges/electrical energy.
A capacitor is a device consisting of two parallel metal plates, carrying equal but opposite charges separated by a dielectric/an insulator.

(ii)           A short circuit
An electrical connection of relatively very low resistance made between two points (at different potentials) in a circuit
When two wires touch and the resistor in the circuit is by-passed/there is not enough resistance in the circuit.

(b) More current passes through the 200 watt bulb than the 60 watt bulb and the more current that passes through the bulb the brighter and more powerful the bulb.


Ca = 5 µF

1/Cb = 1/C3 + 1/C4 =1/10 + 1/10 + 1/5
Cb = 5 µF

Cc = Ca + Cb = 5+ 5 =10 µF

1/Cxy = 1/10 + 1/10 = 2/10
Cxy = 5 µF


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