Answer:
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
