Friday, October 9, 2015

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

Problem
PS and PP 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, 2nd resistance is greater then the first on the factor of N. What is the resistance of the 3rd resistor?
(a) Derive algebraic expression for the second resistance
and then
(b) use this expression to find numerical solution for the following data:
PS=1 W, PP=11 W, N=2, r=1Ω.

Solution:
/RP=PP
/RS=PS
R is 3rd resistance.
RS=r+Nr+R
1/RP=1/r+1/(Nr) +1/R
RP=1/( 1/r+1/(Nr) +1/R  )
PP/PS =RS/RP = RS·(1/RP)
PP/PS =RS/RP = (r+Nr+R)·( 1/r+1/(Nr) +1/R)
PP/PS=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
PP/PS=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

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