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Ohm's Law Previous - Date Grade Physics 205L Ohms Law Name...

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Date: 08/03/07 Physics 205L Ohm’s Law Name: Section: 8 Partner’s Name: Instructor: Diala A. Haidar Ahmad Part I. Computer-aided measurements A- Ohmic devices- passive resistor: Determination of the resistance using the I-V curve: R = 10 R = 33 I (A) V (V) I (A) V (V) 0.031 0.327 0.024 0.801 0.080 0.806 0.050 1.675 0.135 1.362 0.099 3.301 0.243 2.456 0.147 4.893 Use linear regression to determine the experimental value for each resistance along with its root-mean-square error. Compare the experimental value to the actual value of resistance The I-V curve is linear hence it is in the form of y = Ax + B where A is the slope and B is the y-intercept. In this case B = 0 hence y = Ax. Since Ohm’s law states that V = RI, we can say by comparison that A=R For the 10Ω resistor: - = ∑ ∑ i i i i y x y x N A ) ( Where: N=4, x i =I, y i = V, 2 2 ) ( - = i i x x N ∑ (x i y i )= 0.031×0.327 + 0.080×0.806 + 0.135×1.362 + 0.243 ×2.456 = 0.8553 ∑ x i = 0.031 + 0.080 + 0.135 + 0.243 = 0.4890 ∑ y i = 0.327 + 0.806 + 1.362 + 2.456 = 4.951 ∑ x i 2 = 0.031 2 + 0.080 2 + 0.135 2 + 0.243 2 = 0.08463 (∑ x i ) 2 = 0.489 2 = 0.2391 Δ = 4×0.08463 − 0.2391 = 0.09942 A = [(4×0.8553) − (0.489×4.951)] / 0.09942 = 10.06 1 Grade:
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A = 10.06 Hence the experimental value for the 10Ω resistor is 10.06Ω but it has some error associated to it which will be determined from σ A - = 2 2 2 i A e N N σ Where: e i = y-y i such that y is the value of V obtained by replacing the value of I into the equation V = RI, where R= 10.06Ω and y i is the experimental value of V corresponding to I obtained on the computer. For I = 0.031 and V = 0.327, e i = 10.06×0.031 − 0.327 = −0.01514 and e i 2 = 2.292×10 -4 Similarly, for I = 0.080 and V = 0.806; e i = −0.0012 and e i 2 = 1.440×10 -6 For I = 0.135 and V = 1.362; e i = −0.0039 and e i 2 = 1.521×10 -5 For I = 0.243 and V = 2.456; ei =−0.01142 and e i 2 = 1.304×10 -4 ∑ e i 2 = 2.292×10 -4 +1.44×10 -6 + 1.521×10 -5 + 1.304×10 -4 = 3.762×10 -4 Δ = 0.09942 σ A 2 = [4/ 4-2] × 3.762×10 -4 / 0.09942 = 7.568×10 -3 σ A = √ σ A 2 = √ 7.568×10 -3 = 0.087 ≈ 0.09 R = [R exptal ± σ A ] = [10.06 ± 0.09] hence R falls in the interval [9.97; 10.15] The actual value of the resistance is 10; it lies in the interval which shows that our measurements were accurate and the experiment was accurately set and organized. % Error = [ | R given − R experimental | ÷ R given ] ×100 = [ |10 − 10.06 | ÷ 10] × 100 = 0.6% For the 33Ω resistor: - = ∑ ∑ i i i i y x y x N A ) ( Where: N=4, x i =I, y i = V, 2 2 ) ( - = i i x x N ∑ (x i y i )= 0.024 ×0.801 + 0.050 ×1.675 + 0.099 ×3.301 + 0.147 ×4.893 = 1.149 ∑ x i = 0.024 + 0.050 + 0.099 + 0.147 = 0.3200 ∑ y i = 0.801 + 1.675 + 3.301 + 4.893 = 10.67 ∑ x i 2 = 0.024 2 + 0.050 2 + 0.099 2 + 0.147 2 = 0.03449 2
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(∑ x i ) 2 = 0.320 2 = 0.1024 Δ = 4×0.03449 − 0.1024 = 0.03556 A = [(4×1.149) − (0.320×10.67)] / 0.03556 = 33.23 A = 33.23 Hence the experimental value of the 33 Ω is 33.23Ω but it has some error associated to it which will be determined from σ A - = 2 2 2 i A e N N σ Where: e i = y-y i such that y is the value of V obtained by replacing the value of I into the equation V = RI, where R= 33.23Ω and y i is the experimental value of V corresponding to I obtained on the computer. For I = 0.024 and V = 0.802, e i = 33.23×0.024 − 0.801 = −0.00348 and e i 2 = 1.211×10 -5 Similarly, for I = 0.050 and V = 1.675; e i = −0.0135 and e i 2 = 1.822×10 -4 For I = 0.099 and V = 3.301; e i = −0.01123 and e i 2 = 1.261×10 -4 For I = 0.147 and V = 4.893; e i =−0.00819 and e i 2 = 6.708×10 -5 ∑ e i 2 = 1.211×10 -5 + 1.822×10 -4 + 1.261×10 -4 + 6.708×10 -5 = 3.874×10 -4 Δ = 0.03556 σ A 2 = [4/ 4-2] × 3.874×10 -4 / 0.03556= 0.02179 σ A = √ σ A 2 = √ 0.02179= 0.148 ≈ 0.15 R= [R exptal ± σ A ] = (33.23 ± 2×0.15) hence R falls in the interval [32.93; 33.53] The actual value of R is 33 and it falls into the interval which shows that our measurements are accurate and the experiment was carefully set.
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