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April 1, 2015

Posted by **gita** on Sunday, September 8, 2013 at 1:54am.

1) small resistance (ohm)

2) large resistance (0hm)

- Physics -
**Graham**, Sunday, September 8, 2013 at 4:07amGiven:

Series Resistance: S = 783.4 Ω

Parallel Resistance: P = 171.3 Ω

Let the resistances be r and q.

S = r + q

P = rq/(r + q)

.: rq = SP

and S = r + SP/r

.: r^2 - Sr + SP = 0

By symmetry if r is one root of the quadratic then q is the other. Let r be the lesser.

r = (S-√(S^2-4SP))/2

q = (S+√(S^2-4SP))/2

- Physics -
**Devron**, Sunday, September 8, 2013 at 9:27amI solved a problem exactly like this the other day, but I made a math mistake; I should have checked my math.

Equation 1.)

R1+R2=783.4

Equation 2.)

R1*R2/(R1+R2)=Req=783.4

Substitute equation 1 into 2:

R1=783.4-R2

and

R1*R2/(R1+R2)=Req=171.3

(783.4-R2)*R2/783.4-R2+R2=171.3

R2^2-783.4R2=-1.342 x 10^5

R2^2 -783.4R2+1.342 x 10^5=0

use the quadratic equation and solve for R

R=[-b + or - sqrt*(b^2-4ac)]/2a

a=1, b=-783.4, and c=1.342 x 10^5

R=[-(-783.4) + or - sqrt*((-783.4)^2-4(1)(1.342 x 10^5)]/2(1)

R=[783.4 + or - (6.1372 x 10^5-5.368 x 10^5)]/2

R=[783.4 + or - sqrt*(7.6916 x 10^4)]/2

R=(783.4-277.3/2) or (783.4+277.3/2)

R=253.05Ω or R=530.35Ω

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