When two resistors of resistances R1 and R2 are connected in parallel, the net resistance is 32.
When connected in series, its value is 16 1. Calculate the values of R, and R.
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R1+R2 = 16
R2 .= 16-R1
1/R1+1/R2= 1/32
BY SUBSTITUTING R2 VALVE IN PARALLEL CONNECTION WE GET
1/R1+1/(161 -R1) = 1/32
16/R1 (16-R1) =1/32
16/16R1-R1^2=1/32
512=16R1-R1^2
R1^2-16R1+512 = 0
continue to solve this u will get it .........
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