Ohm's Law Previous - Date Physics 205L Ohm’s Law Name Section 8 Partner’s Name Instructor Diala A Haidar Ahmad Part I Computer-aided

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Unformatted text preview: 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 Ω R = 33 Ω 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: 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 • (∑ x i ) 2 = 0.320 2 = 0.1024 • Δ = 4×0.03449 − 0.1024 = 0.03556Δ = 4×0....
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This note was uploaded on 01/06/2012 for the course PHYS 205L taught by Professor Antar,ghassan during the Spring '11 term at American University of Beirut.

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Ohm's Law Previous - Date Physics 205L Ohm’s Law Name Section 8 Partner’s Name Instructor Diala A Haidar Ahmad Part I Computer-aided

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