waecE-LEARNING
Applied Electricity Paper 2 , May/June 2010  
Questions: 1 2 3 4 5 6 7 8 Main
General Comments
Weakness/Remedies
Strength


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.

_____________________________________________________________________________________________________
Observation

The expected answers were:
(a)

Primary cell

Secondary cell

--

(i)

Cannot be recharged

(i)

Can be recharged

 

 

once the chemicals

 

and used many

 

 

are exhausted.

 

times each time the

 

 

 

 

energy is used.

 

(ii)

Converts chemical

(i i i)

Chemical action is

 

 

energy into electrical

 

reversible.

 

 

energy.

 

 

 

(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
OR
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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