Applied Electricity Paper 2 , May/June 2010  
Questions: 1 2 3 4 5 6 7 8 Main
General Comments

Question 7

(a) State two differences between primary cells and secondary cells.
(b) A battery has a 9 V output on an open circuit which drops to 8.5 V
with a load current of 1 A. Calculate the:
  (i)   internal resistance of the battery;
  (ii) power dissipated by the internal resistance.


The expected answers were:

Primary cell

Secondary cell



Cannot be recharged


Can be recharged



once the chemicals


and used many



are exhausted.


times each time the





energy is used.



Converts chemical

(i i i)

Chemical action is



energy into electrical









(b)      Vopen circuit = 9 V
           Vload            = 8.5 V
           Iload           = 1 A
           (i)        Vdropon load = (9 - 8.5) V = 0.5 V
Rinternal = V drop/Il = 0.5/1 = 0.50
I = E/(R + r) where r = internal resistance
r = E/I-R = 9.0 - 8.5 = 0.50
           (ii)       Power dissipated = 12Rj

= 1²x 0.5 = 0.5W
The question was on Electrical Energy Supply and Direct Current Circuit Theory. The question was less unpopular. The performance was average.

Many of the candidates gave at least one difference that was correct. However, those who did not understand the part (b) of the question could not do this satisfactorily due to lack of practice and exposure as reported.
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