Hey guys, I need you physics minded peoples skill and prowess, 40 hours w/o slee
ID: 1961323 • Letter: H
Question
Hey guys, I need you physics minded peoples skill and prowess, 40 hours w/o sleep and this question is last step before I can crash! Here's the question along with what I believe to be answered correctly so far in bold, just need to get those blanks filled.
The multi-part question:
You create a plot of voltage (in V) vs. time (in s) for an RC circuit as the capacitor is discharging, where V = V0(et/RC). You curve fit the data using the natural exponent function Y = A(eCx) + B and LoggerPro ( Computer program we use ) gives the following values for A, B, and C:
A = ( 43840 ± 345.1 )
B = ( 0.1821 ± 0.007469 )
C = ( 1.897 ± 0.2109 )
(a) What is the time constant for the circuit and its uncertainty?
( ______ ± _____ ) s
(b) If you used a 1 M (megaohm) resistor with a tolerance of 10%, what is the equivalent capacitance of your circuit (including the uncertainty)?
( ______ ± _____ ) F
(c) If the manufacturer has labeled the capacitor as 0.5 µF ± 10%, is this consistent with your result?
- *Yes* (I think it's yes)
- No
- It cannot be determined
Explanation / Answer
For (a) time constant is 1/C and the incertainty is ((1/C)*.2109)/1.897 for (b) its the time constant times 1.0E-6 and the uncertainty is .00000007 For (c) its yes