# 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.

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by deepesh08

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by deepesh08

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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

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

The Brainliest Answer!

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..