If r is the total resistance of three resistors, connected in parallel, with resistances r1, r2, r3, then 1 r = 1 r1 + 1 r2 + 1 r3 . if the resistances are measured in ohms as r1 = 100 ω, r2 = 25 ω, and r3 = 10 ω, with a possible error of 0.5% in each case, estimate the maximum error in the calculated value of r. (round your answer to three decimal places.) ω

maximum error is 0.03333

Explanation:

given data

R1 = 100 Ω,

R2 = 25 Ω,

R3 = 10 Ω

1/ R = 1/ R1 + 1/ R2 + 1 /R3

possible error = 0.5%

to find out

maximum error

solution

we know

1/ R = 1/ R1 + 1/ R2 + 1 /R3

put all value R1, R2 and R3

1/ R = 1/ 100 + 1/ 25 + 1 /10

R = 20/3

now take derivative  

dR/dR(i) = R²/R(i)² for i = 1, 2, 3

we have given error 0.005

so dR(i) = 0.005×R(i)  for the i = 1,2,3

so the equation will be

dR = dR/dR(1) ×dR(1) +dR/dR(2) ×dR(2) + dR/dR(3) ×dR(3)

dR = R²/R²(1) ×dR(1) + R²/R²(2) ×dR(2) +  R²/R²(3) ×dR(3)

put the value dR(1) and dR(2) and dR(3) and R

dR =  (20/3)²/R²(1) ×0.005×R(1) +  (20/3)²/R²(2) ×0.005×R(2) +   (20/3)²/R²(3) ×0.005×R(3)

dR =  (20/3)²/R(1) ×0.005 +  (20/3)²/R(2) ×0.005 +   (20/3)²/R(3) ×0.005

dR =  (20/3)²/100 ×0.005 +  (20/3)²/20 ×0.005 +   (20/3)²/10 ×0.005

dR =  (20/3)² ( 0.005/100 + 0.005/25 + 0.005/10)

dR = 0.033333

maximum error is 0.03333

i believe the answer is a

see earlier answer