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  • Brainly User
2014-06-02T14:09:45+05:30

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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
2 3 2
there are two individual resistances....u have to find each of them....in parallel they are r1+r2/r1r2 not r1r2/r1+r2
when a number of resistors are connected in parallel, the sum of the reciprocal of the resistance of the individual resistors is equal to the reciprocal of the effective resistance of the combination.
Hence reciprocal of resistance is only r1+r2/r1r2
The Brainliest Answer!
2014-06-02T14:13:19+05:30
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..

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