Question

when two resistors are joined in series, the equivalent resistance is 90ohms, when the same resistors are joined in parallel, the equivalent resistance is 20ohms calculate the resistance of the two resistors.

Answer 1

let the two resistors be R1 and R2.
so,
R1 + R2 = 90. .....(i)
and,
(R1×R2)/R1 + R2 = 20
R1 × R2 = 20 (R1 + R2)
putting value of equation 1

   =20*90 = 1800

Convert this into a quadratic equation

Let R1 be α & R2 be β

Therefore, α β = 1800, α+β= 20

General form of quadratic eq. = x2 – x(α +β)+ α β

Substituting values we get the following eq.,

x2 – 20x + 1800

= x2 – 60x- 30x + 1800

= x(x-60) – 30(x-60)

=(x-30)(x-60)

(x-30), (x-60) are factors. Therefore zeroes are 30 & 60

Thus, α = R1= 30 ohms, β =R2 = 60 ohms

Keep on asking such questions

Answer 2

hello friend ,

• let the two given resistors be R1 and R2
when the two resistors are joined in series then the equivalent resistance ie 90 ohms
Rs = R1 + R2
R1 + R2 = 90 ..........(1)

when the resistors are joined in parallel , then the equivalent resistance is 20 ohms...
Rp = 1/ R1 + 1/ R2
20 = R1 + R2/ R1.R2

By eq.(1) , we get

20 = 90 / R1.R2
R1× R2 = 90/20 = 4.5 ........(2)

By eq (1) , we get

Now ,R1 ×( 90 - R1) = 4.5
90 R1 - R1 ² = 4.5
R1² - 90R1 + 4.5 = 0
this question seems to be wrong ....
I think the value is wrong ie 90 and 20
because I got negative values after solving the equation...
ie - 49.8 and - 80.4
hope you understand...