Grade: Date: March 06, 2012 Physics 205L Ohm’s Law Name: Aya Al Sayed Partner’s Name: Crystal Bablkiau Part I. Computer-aided measurements Section number: 4 Instructor: 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) 0.099 1.014 0.017 0.507 0.198 1.993 0.038 1.284 0.233 -2.500 0.058 1.993 0.270 2.872 0.076 2.703 Use linear regression to determine the experimental value for each resistance along with its root-mean-square error. Show your calculations. Compare the experimental value to the actual value of resistance. 1) For R=10 Ω , find the slope∧ y intersept of the function V =aI +b Xi Xi^2 Yi Y=Bx+A ei 0.099 0.198 0.233 0.270 =0.6029 9.801*10^-3 0.039201 0.054 0.0729 =0.1759 1.014 1.993 2.500 2.872 Y=10.938X-0.0929 0.989962 2.078 2.455 2.86036 -0.024 0.085 -0.045 -0.0684 ei^2 5.76*10^-4 7.225*10^-3 2.025*10^-3 4.68*10^-3 =0.01393 ∆=N x i 2−( xi )2 ∆=4∗0.1759−( 0.6029 )2 ¿ 0.696 α 2= 2 N ei ( N −2 ) ∆ =0.04 α =0.2 R=(10.9 ±0.2)Ω To check for accuracy we check if the real R is within the interval: [10.9−0.2,10 .9+0.2] that is [10.7,11.1] that is untrue for given R=10Ω so we ifˇ R ∈ [ 10.9−2 ( 0.20 ) , 10.9+2 ( 0.2 ) ]=[ 10.5,11.3 ] 1 2) For Xi R=33 Ω . The inaccuracy of the used instruments.364*10^-3 5.703 0.78.0576 α =0.26] that is not true for given R=33Ω ˇ the second interval [ R± 2 α ] so for 2 .37.189 r=0.284 1.02−0.2 So according to the study we get : 4 R=(37.02 2.We found that R=10 Ω belong ¿ neither of the intervals so our results are not accurate And this may be because of systematic errors such as Fluctuation of the ohmmeter due to wind or any other external condition that affects resistance.444*10^-3 3.24] that is [36.045 N e i2 α = ( N −2 ) ∆ 2 =0. Or may be due to random errors of unknown causes .9998 2.37 .2969*10^-3 ∆=N x i 2−( xi )2 −3 2 ∆=4∗(10.24) Ω To check for accuracy we check if the real R is within the interval: [37. Or observational errors in measuring.873∗10 )− ( 0.68 0.027 0.50 1.02X-0.9*10^-5 1.29*10^-4 =1.776*10^-3 =10.023 4.038 0.02± 0. consi der the function V =aI + b Xi^2 Yi Y=Bx+A ei ei^2 Y=37.189 ) ¿ 0.017 0.873*10^-3 0.02+ 0.127 0.507 1.004 0.993 2.89*10^-4 1.058 0.28 2.6*10^-5 7029*10^-4 5.007 0.24.076 =0. [37.356*10^-3 =65.48] that is true for given Req=43Ω 3 .721*10^-3 9*10^-4 0.139 r=0.023 0.33X+0.54.779 α=6 So according to the study we get : R=(42 ± 6)Ω Req (theoretical) = R1+ R2= 33+10=43Ω ( R1 and R2 in series) To check for accuracy we check if the real R is within the interval: [R ± α ] [42−6.48.049 0.405 0.08 0.37.10802 ∆=N x i 2−( xi )2 4 −¿ 65. Resistors in series I (A) V (V) 0.9999 10^-4 5.135 0. Xi Xi^2 Yi Y=Bx+A ei ei^2 Y=42.297 -0.066 =0. Compare these values to those that can be calculated from the previous measurements.02 0.066 2.48]=[36.088 0.28 2.139 )2 ∆=4∗¿ ¿ 7.723 2.98 0.01 0.236 0.015 0.642 Using linear regression determine the experimental value for each resistance along with its root-mean-square error.68 0.439 0.42+6 ] that is [36. thus Our results are inaccurate and this may be due to systematic errors or due random errors of unknown causes.723 0.29*10^-4 1.040 1.50 1.85∗10 ¿− ( 0.5] in which R=33 Ω doesn’t belong to the interval.85*10^-4 0.03 0.804 Resistors in parallel I (A) V (V) 0.6*10^-3 4.02-0.37.023 0.033 0.124 3.439 0.061 0.04 0.02 2.02+0.01 0.019∗10−3 α 2= N e i2 ( N −2 ) ∆ =30.804 0.0154 =0.980 1. 08 =0. B.11*10^-4 2 ∆=N x i −( xi ) 2 ∆=4∗(10.9979 6.4*10^-3 =10.1.8 .41*10^-4 0.021 0 1.007 0.21*10^-4 4.236 0.29∗10−3)− ( 0.7.182 r=0.384 0 0.049 4*10^-4 1.4)Ω Since R1 and R2 are in parallel.247 0.3 ] ∧we found that Req belongs¿ it so we fo so we can say that the measurement was accurate .135 0.Non-ohmic devices. 1) For the resistors∈¿ .649 0.9.9] that is untrue for given Req=7.5−0.4 So according to the study we get: R=(8.5+0. consider the function V =aI +b Xi Xi^2 Yi Y=Bx+A ei ei^2 Y=8.02 0.8.182 )2 ¿ 8.152 α =0.so themeasurement was accurate .03446 0.5 ± 0.011 -0. then 1 1 1 R 1 × R 2 33 × 10 = + .642 0.033 0.49*10^-4 =6.548X-0. Req ( theoratical ) = = =7.67 Ω Req R 1 R 2 R 1+ R 2 33+10 To check for accuracy we check if the real R is within the interval: [8.405 0.4.135 0.401*10^-3 0.29*10^-3 0.036*10^-3 α 2= N e i2 ( N −2 ) ∆ =0.67Ω ˇ r the secondinterval of [ R ±2 α ] =[ 7.4 ] that is [8.the Light Bulb: 4 . So .089*10^-3 2. 1 Hz.63 130.I-V curve of a light bulb From the I-V curve of the bulb for a ramp waveform at 0.1 Hz 50Hz 200 140 120 100 80 60 40 20 0 150 100 50 0 0 0 0.5 2.5 1.5 2 2.5 1 1. 1.5 1 1.5 I(mA) 30.4 19.5 Draw the I-V curve of the bulb for the saw tooth waveform at two different frequencies 0. As the temperature increases the resistivity of the filament increases ( increases since ( R= 1+α (T −T 0 ) ) and as the resistivity increases the resistance ρ=ρ0 ¿ ρl A ).5 2.5 14. calculate the resistance of the filament when the voltage is 0.1 R(Ω) 16.5 1.30 89 151.17 as the voltage increases. the resistance of the lamp increases.3 7.5V.5V.5 0. For f = 50Hz: V(V) 0.5V.8 16. 2. What pattern do you observe? How can you explain the change in resistance? 1) For f = 0.5 3 3 5 . This is because the lamp’s resistance is affected by the temperature of the filament that it increases with the rise in temperature.012 102.5 We observe that I(mA) R(Ω) 71.1 Hz V(V) 0.5 16.5 2 2. consider the function V =aI + b Xi Xi^2 Yi Y=Bx+A ei ei^2 Y=16.27 =0.1 Hz.23)Ω However.23)Ω But what is noticeable about this study at f=50Hz.What is the resistance of the light bulb filament at 50 Hz? Does it changes when frequency is decreased to 0. the period was 6 .52± 0. the value of R is quite variable with its values changing greatly to become 7Ω.02 0 0 4*10^-4 0 =4*10^-4 2 ∆=3∗(0.5 2.0127 0.48 2.0316 r=0. 15Ω.03 0.0316)− ( 0. at f = 0.0547 α=0. we notice that r=0.089 0.9998 that is about 1 which mean that the two variables V and I are linearly related which means that R is constant.5 0.52± 0.0079 0.9998 2 ∆=N x i −( xi ) 0 -0.0228 0. and 19Ω at different voltages that is different from that at 50 Hz. Explain each of these two I-V curves and deduce why do such curves depend on frequency for a light bulb? The main effect responsible for the change in resistance and shape of curve is the temperature and the heating of the light bulb.151 9*10^-4 0.5 =0. At a frequency of 50 Hz.0219 α 2= N e i2 ( N −2 ) ∆ =0.5 1.52X+0.27 )2 ¿ 0.5 1. So at 50 Hz R is constant at a value R= (16.23 So according to the study we get: R=(16.1 Hz? 1) For the f =50 Hz . 87 5.005 cm Copper rod length: 61.15 9.17 To get the resistance.02 s f 50 .01 -0.06 3.98 6.6025 1.mV 2. consider the function V =aI +b Xi Xi^2 Yi Y=Bx+A ei ei^2 Y=0. II.1 1 ¿ = =0.03 1.87 038 1.10 7.32 2.06 4 2.02 5.32 2 1. the period was 1 1 = =10 s .57 4 2.94 0.6*10^-3 7 .1 Hz.98 5.1 which is large enough and can provide time for the heating of the filament of the bulb causing a change and variation in resistance and resistivity.08 6.05 3.005 cm Aluminum rod diameter: 0. However. at 0.05 4. we need to apply linear regression: 1) For the copper rod .99 3.05 cm Aluminum rod length: 62.92 0.05 Copper: Rod number ( 6 ) Aluminum: Rod number ( 6 ) Current (A) Voltage across .85 7.03 4 6.33 2. that’s why the bulb didn’t heat up and its resistivity and resistance in this small period didn’t change.01 5.98 63. which is quite small and provide not enough time for the heating of the filament of the bulb. Small resistance measurement (metal rods): Copper rod diameter: 0.8 0.03 4.01 -0.mV Current (A) Voltage across . 2.6 0.6804 5.1 8.1209 16 36.6308X+0.56 3.02 10^-4 1*10^-4 4*10^-4 0.07 0. f 0.57 3.37 10. 76 0.33 -0.9998 1.03 64.6194 α 2= N e i2 ( N −2 ) ∆ =3. 2 4 5.04 =30.15 0.1119 ∆=N x i 2−( xi )2 ∆=5∗( 220.04 ) 2 ¿ 197.36*10^-3 ∆=N x i 2−( xi )2 ∆=5∗(220.08 3.0225 10.6004 6.677*10^-5 α =¿ 6*10^-3 Then.9.06 2.208 2.5123X+0.0219 0. Copper Rod – Slope (resistance) is R=(0.15 4.48 4.0042)−( 30.31 0.15 0.6*10^-4 =4. consider the function V =aI +b Xi Xi^2 Yi Y=Bx+A ei ei^2 Y=0.847*10^-4 α =0.02 100.03 8 .631 ±0.23 3.9795 =220.3 0.88 1.37 6.76 )−( 30.06)mΩ 2) For the Aluminum rod .015 0.98 99.25 8.398 α 2= 2 N ei ( N −2 ) ∆ =9.04 =220.0225 0.0042 r=0.17 5.148 0. 16 35.0225 =30.04 )2 ¿ 201.0225 =0.4 5.99 4.04 r=0.10 1836 1.15 0. 2.618 m For copper rod : we have R=ρ ( ) Then we use the propagation of error technique to get the error: A=π r 2= π d2 =6. so ρ=R .So according to the study we get : R=(0.03)mΩ Then. =R .01]×10−7 Ω m =[2.72 ×10−8 Ω m so we check if this [2.83 2 ) ×10−10=10−9 Ωm ρ=(2.23 ±0.07∗10−5 m2 2 αA= A=(6.03 ) mΩ Calculate the resistivities of the two metals along with their respective rms error. =0.94 ± 0. R=0.24] 9 .94∗10−5 m 2 4 πd αd =0. Aluminum – Slope (resistance) : R=( 0.05 cm . compare with literature values and comment. =R . 1) l A π r2 .51 ±0.226 × 10−7 Ωm l l 4 0.07 ¿∗10−5 m2 AαR 2 RαA 2 −RAαl + + l l l2 ( ) ( )( 2 α ❑= α=9.006 mΩ 2 0.94 −2 × 10 m 2 A π r2 π d2 ρ=R .005 cm∧l=61.23 ± 0.631± 0.22.01) ×10−7 Ωm The actual resistivity of copper is: values lies in the interval of ρ=1.631× 10−3 Ω × π × =2.8± 0.94 ± 0. =R . A l l diameter=0.51± 0. 8x10-8 Ω m . =R .25]×10−7 and the actual values doesn’t belong to this interval too.92 ± 0.07∗10−5 m2 2 αA= A=(6. so ρ=R .6476× 10−5 m2 4 πd αd =0. Then our results are not accurate.3)×10−8 Ωm The actual resistivity of Aluminum is: ρ=2. R=0.so we check for the second interval [ ρ ±2 α ¿=[ 2.03 mΩ −5 ❑ ( 06.23± 0.−7 ×10 Ω m∧we found t h at it doesnt belong¿ t h is interval . 6 x10-8 Ω m ]) from the calculated value. 5.6 ± 0. =0.005 cm∧l=62.1x10-8 Ω m .4 ± 0. ρ=R .65 ± 0.so our results are not accurate 2) For Aluminum rod: we have R=ρ l A π r2 .7 x10-8 Ω m ]) or even 2 error deviation ([4.51± 0.2.21. A l l diameter=0.626 2 Then we use the propagation of error technique to get the error: A=π r 2= π d2 =6.82 ×10−8 Ω m but this value doesn’t occur within 1 error deviation ([5.3 ×10−8 Ωm ( ) ( )( 2 ) ρ=(5.05 cm .65 ×10 ) A πr =R .02 ] ×10−7=[2.51 ×10−3 × =5.417 × 10−8 Ωm l l 0.07 ¿∗10−5 m2 2 2 2 α ❑= AαR RαA −RAαl + + 2 l l l α=0. The inaccuracy for our results in both Cu and Al may be due to systematic errors such as Fluctuation of the ohmmeter due to wind or 10 . The inaccuracy of the used instruments. Or may be due to random errors of unknown causes 11 . Or observational errors in measuring.any other external condition that affects resistance.