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

Problem: PS and PP are rates of energy dissipation in two resistors, when they are connected in series and in parallel respectively with some source of constant voltage. Let lower resistance is known, r, what is resistance of another resistor?

Problem

P_S and P_P are rates of energy dissipation in two resistors, when they are connected in series and in parallel respectively with some source of constant voltage. Let lower resistance is known, r, what is resistance of another 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=12 W, P_P=54 W, r=1Ω.

Solution:

V²/R=P

R_S=R+r

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

R_P=Rr/(R+r)

V²/R_P=P_P

V²/R_S=P_S

R_S/R_P=P_P/P_S

(R+r) / (Rr/(R+r)) = (R+r)²/Rr=P_P/P_S=a

(R+r)²/Rr=a

(R+r)²=aRr

(R/r+1)²=aR/r

R/r = x

(x+1)²=ax

x²+2x+1=ax

x²+2x-ax=-1

x²+2(1-a/2)x=-1

x²+2(1-a/2)x+(1-a/2)²=-1+(1-a/2)²

(  x+(1-a/2)  )² = -1+(1-a/2)²

x+(1-a/2) = ± (  -1+(1-a/2)²  )^½

x=-(1-a/2) ± (  -1+(1-a/2)²  )^½

x=(a/2 - 1) ± ( (a/2-1)²-1  )^½

R/r =    (P_P/P_S)/2 – 1) ±  ((P_P/P_S)/2-1)²-1  )^½

R = r·(   (P_P/P_S)/2 - 1) ± ( ((P_P/P_S)/2-1)²-1  )^½ )

(a)        R = r·(   (P_P/P_S)/2 – 1) + ( ((P_P/P_S)/2-1)²-1  )^½ )

(P_P/P_S)/2 - 1) = (54 W/12 W)/2-1

(P_P/P_S)/2 - 1) = (  9 / 2) /2-1 = 9/4 – 4/4 = 5/4

R = 1Ω ·(5/4 + (25/16-1) ^½) = 1Ω ·(5/4 + (25/16-16/16) ^½) = 1Ω ·(5/4 + (9/16) ^½) = 1Ω ·(5/4 + ¾) = 1Ω ·(8/4) =2Ω

(b)  R = 2Ω