According to the statements' of the question, pnor"to the inclusion ofR2in
,
(           the circui~ the current through the two resistances then in the circuit and
connected in series was 50~ and maintained by an e.m.fof 12V. Therefore~ it follows from Ohm's low that

J .
. ,
	50		10-3			..L	, .
		x		=
						2Rl	:n
	RI	=	.w:n		=	200
			50			'j'. '

,Rz is connected in parallel as shown, the current in the, cirqJit increases to
62.SinA, which means that the total resistance in the circuit iS,now by Ohm's law
                           RT=        '2              = 320
                                        62.S x 10.3'
It f'OlIows therefore, the parallel combmation ofR} and R2 will have a re~istance
   of(32 - 20).0 or 12n. Therefore,                                                                ,

,R2 = 240 0 = 300 8
, The, current through R2 is readily computed from the current divider rule as

	IR2	=	2Q..	x	62.SA	=	25mA
; ;(			50
(
	IR2 =		25mA

(b)', As thoroughly calculated in part (a), R2 = 200
This question was basic electrical theory associated with resistors in circuits when connected in series/parallel. The question seemed quite unpopular with the candidates, with the perfonnance very poor.
, The poor performance was not unconnected with inadequate preparation instead of full coverage of the syllabus.