Respuesta :
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
The maximum error in calculating the value of equivalent resistance, R is 6.667 ohms.
The given parameters;
- first resistor, R₁ = 100 Ω
- second resistor, R₂ = 25 Ω
- third resistor, R₃ = 10 Ω
- error in each measurement; = 0.5% = 0.005
The equivalent resistance of the three parallel resistors is calculated as;
[tex]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \\\\\frac{1}{R} = \frac{R_2R_3\ + \ R_1R_3 \ + \ R_1R_2}{R_1R_2R_3} \\\\R = \frac{R_1R_2R_3}{(R_2R_3) + (R_1R_3)+(R_1R_2)} \\\\[/tex]
The maximum error in calculating the value of R is estimated as follows;
[tex]R = \frac{0.005(R_1R_2R_3)}{0.005(R_2R_3) + 0.005(R_1R_3)+0.005(R_1R_2)}[/tex]
[tex]R = \frac{0.005(100\times 25 \times 10)}{0.005(250) \ + \ 0.005(1000)\ +\ 0.005(2500)}\\\\R = \frac{125}{18.75} \\\\R = 6.667 \ ohms[/tex]
Thus, the maximum error in calculating the value of equivalent resistance, R is 6.667 ohms.
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