Question

The resultant resistance of two resistance wires in series combination is 108 ohm and in parallel combination is 27 ohm. find the value of individual resistance.

Answer 1

Let the resistance be R1 and r2
in series.....r1+ r2= 108 ohm
in parallel....r1r2/r1+r2 = 27 ohm
r1r2 = 27 * 108 = 2916
R1-r2 = √ ( r1+r2)² - 4 r1r2
          = √ 11664 - 11664 = 0
r1 = r 2
r1+r2 = 108
2r = 108
r = 54 ohm

Answer 2

Here you have to solve this question by solving quadratic equation...!

Given that when connected in parallel, the resultant resistance R(p) = r1 + r2 = 27 ohms
and when connected in series, the resultant resistance R(e) = r1 + r2 = 108 ohms ..(1)

1/R(p) = 1/r1 + 1/r2 = (r1+r2)/r1.r2
R(p) = r1.r2/(r1 + r2) = r1.r2/108              (r1 + r2 = 108 ohms, from eqn (1))
Therefore, R(p) = 27 = r1.r2/108 => r1.r2 = 2916  ........(2)

Now we have got two eqn (1) and (2), Now we can find the value of r1 and r2
we have:-

r1.r2 = 2916                                      ........(3)
r1 + r2 = 108 ohms => r1 = 108 - r2
Substituting the value of r1 in eqn 3, we get:-
(108 - r2)r2 = 2916
108r2 - r2^2 = 2916      ..............(4)
let r1 = x and r2 = y    (for sake of conviene)
108y - y^2 = 2916            ........(rewrite of eqn (4) )
-y^2 +108y - 2916 = 0
we can also write above eqn as:-
y^2 -108y +2916 = 0

Afterv solving above eqn you willl get its two roots.. That roots will be your answer..