EC4.docx
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Section VI-1 Measurement about resistor by DC power. Resistor A B C D l1 (cm) 49.9 49.8 49.9 49.9 l2 (cm) 50.1 50.1 50.1 50.1 R3 (Ω) 148 213 382 980 Rm (DMM) 148 211 378 961 π π΄,π₯ = π2 50.1ππ × π 3 = 148Ω = 148.6Ω π1 49.9ππ π π΅,π₯ = π2 50.1ππ × π 3 = 213Ω = 214.3Ω π1 49.8ππ π πΆ,π₯ = π2 50.1ππ × π 3 = 382Ω = 383.53Ω π1 49.9ππ π π·,π₯ = π2 50.1ππ × π 3 = 980Ω = 983.93Ω π1 49.9ππ 1 π2 ππ΄π₯ = π 3 × ( + 2 ) × π1 = 0.59 π1 π1 1 π2 ππ΅π₯ = π 3 × ( + 2 ) × π1 = 0.86 π1 π1 1 π2 ππΆπ₯ = π 3 × ( + 2 ) × π1 = 1.53 π1 π1 1 π2 ππ·π₯ = π 3 × ( + 2 ) × π1 = 3.94 π1 π1 So, π π΄ = 148.6 ± 0.6 Ω π π΅ = 214.3 ± 0.9 Ω π πΆ = 383.5 ± 2 Ω π π· = 983.9 ± 4 Ω For resistors A, B, C, and D, Rm is not in the range determined by Rx, but the values are relatively close. This discrepancy could be caused by inaccurate measurements or by using the same resistance and just moving farther away from the 50 cm mark. Section VI-2 Measurement about resistor by AC power. Resistor A B C D l1 (cm) 49.1 49.4 50.4 50.2 l2 (cm) 50.9 50.6 49.6 49.8 R3 (Ω) 133 203 400 1000 Rm (DMM) 148 211 378 961 π π΄,π₯ = π2 50.9ππ × π 3 = 133Ω = 137.9 Ω π1 49.1ππ π π΅,π₯ = π2 50.6ππ × π 3 = 203Ω = 207.9 Ω π1 49.4ππ π πΆ,π₯ = π2 49.6ππ × π 3 = 400Ω = 393.7 Ω π1 50.4ππ π π·,π₯ = π2 49.8ππ × π 3 = 1000Ω = 992.0 Ω π1 50.2ππ 1 π2 ππ΄π₯ = π 3 × ( + 2 ) × π1 = 0.55 π1 π1 1 π2 ππ΅π₯ = π 3 × ( + 2 ) × π1 = 0.83 π1 π1 1 π2 ππΆπ₯ = π 3 × ( + 2 ) × π1 = 1.57 π1 π1 1 π2 ππ·π₯ = π 3 × ( + 2 ) × π1 = 3.97 π1 π1 So, π π΄ = 138 ± 0.6 Ω π π΅ = 207.9 ± 0.8 Ω π πΆ = 393.7 ± 1.6 Ω π π· = 992 ± 4 Ω For every resistor, no Rm lies in the range determined by Rx, but all of the values are relatively close. This again could be caused by inaccurate measurements or using the same R3 value for each resistor and just changing the distance. The results from VI-1 are very similar if not the same as the results using AC Section VI-3 Capacitor A B C3 (f) 2.2×10-9 3.7×10-9 Cm (f) 2.298×10-9 3.871×10-9 πΆ1 = πΆ2 = 0.022 ± 0.0022 ππ ππΆ3 = 0.05 ππ΄,π₯ = √( πΆπ΄,π₯ = πΆ2 πΆ = 2.2 × 10−9 π πΆ1 3 πΆπ΅,π₯ = πΆ2 πΆ = 3.7 × 10−9 π πΆ1 3 ππΆπ΄ 2 ππΆπ΄ 2 ππΆπ΄ 2 ) × ππΆ1 2 + ( ) × ππΆ2 2 + ( ) × ππΆ3 2 = 0.30 × 10−9 ππΆ1 ππΆ2 ππΆ3 ππΆπ΅ 2 ππΆπ΅ 2 ππΆπ΅ 2 ππ΅,π₯ = √( ) × ππΆ1 2 + ( ) × ππΆ2 2 + ( ) × ππΆ3 2 = 0.52 × 10−9 ππΆ1 ππΆ2 ππΆ3 So, πΆπ΄,π₯ = (2.2 ± 0.3) × 10−9 π πΆπ΅,π₯ = (3.7 ± 0.5) × 10−9 π Cm is equal to the determined Cx for both capacitors. This is because C1 is equal to C2.
