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.