Problem. PS and PP are rates of energy dissipation in 3 resistors...

Problem

P_S and P_P are rates of energy dissipation in 3 resistors, when they are connected in series and in parallel respectively with some source of constant voltage. Let lowest resistance is known, r, 2^nd resistance is greater then the first on the factor of N. What is the resistance of the 3^rd resistor?
(a) Derive algebraic expression for the second resistance
and then
(b) use this expression to find numerical solution for the following data:
P_S=1 W, P_P=11 W, N=2, r=1Ω.

Solution:

V²/R_P=P_P

V²/R_S=P_S

R is 3^rd resistance.

R_S=r+Nr+R

1/R_P=1/r+1/(Nr) +1/R

R_P=1/( 1/r+1/(Nr) +1/R  )

P_P/P_S =R_S/R_P = R_S·(1/R_P)

P_P/P_S =R_S/R_P = (r+Nr+R)·( 1/r+1/(Nr) +1/R)

P_P/P_S=a

(r+Nr+R) (1/r+1/(Nr)+/R)=a

(1+N+R/r)(1+ 1/N + r/R) = a; Solve for x=R/r

R/r=x

P_P/P_S=a=11

11=(1+2+x)(1+1/2+1/x)

11=(3+x)(3/2+1/x)

11x=(3+x)(1.5x+1)

Solutions: x=3, x=2/3

R/r = 3

R=3·1Ω = 3Ω