Physics
posted by gita on .
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)

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^24SP))/2
q = (S+√(S^24SP))/2 
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=783.4R2
and
R1*R2/(R1+R2)=Req=171.3
(783.4R2)*R2/783.4R2+R2=171.3
R2^2783.4R2=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^24ac)]/2a
a=1, b=783.4, and c=1.342 x 10^5
R=[(783.4) + or  sqrt*((783.4)^24(1)(1.342 x 10^5)]/2(1)
R=[783.4 + or  (6.1372 x 10^55.368 x 10^5)]/2
R=[783.4 + or  sqrt*(7.6916 x 10^4)]/2
R=(783.4277.3/2) or (783.4+277.3/2)
R=253.05Ω or R=530.35Ω