Question

1) You calculated a equivalent capacitance of 0.42 μF ± 0.08 μF. If the manufacturer has labeled the capacitor as 0.5 μF ± 10%, is this consistent with your result? yes, no, or cannot be determined?

2) You calculated a value of 0.53 s ± 0.06 s for the time constant from your graph.  If you used a 1MΩ resistor with a tolerance of 10%, what is the equivalent capacitance of your circuit (including the uncertainty)?

Answer 1

Answers DE AC=104. of 0.5p2F =>(0.5MF) (100) =0.05 jef Cequ = 0.5 MF70.05 AF =70.55 MF Cool Cequ=0.5HF -0.05 MF-> 0,45MF the calculated value of escúvalent capacitance lies in the follong values Ceece = 0.49 MF +0.08 Mp => 0.50 MF. Ceque = 0.48MF-0.08MF = 0.34 MF . :: Both maximumiminimum equivalent capacitances as per the manufacturer's label are in between the maxima and minima of the calculated value of grevalent capacitance Yes, the manufacturer's reading is constant with the calculated values. 2) T-RC => => 0.53 a=0.53MF The fraction of change in the time constant of the circct is 46 - (9)+( 9 AC = (0.533166 p.)(6. for the AC = 0.08 X100F Ces = 0,53x106F 10.08xjos F = (0.53£0.08) MF .: The equavalent capacitance is (0.5350.08) PF