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

Two resistors connected in series have an equivalent resistance of 783.4 Ω. When they are connected in parallel, their equivalent resistance is 171.3 Ω. Find the resistance of each resistor.

1) small resistance (ohm)

2) large resistance (0hm)

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

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 9:27am
I 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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