Question:
A capacitor C is initially charged to 10V and then connected across a resistor R. The current through R is measured at 1 milisecond intervals is:
t : 0 1 2 3 4 ms +-0.5%
i : 4.5 2.8 1.5 1.0 0.6 mA +-2.5%
(Basically this is a table where i corresponds to the t value)
Using the method of least squares, find R and C.
I need help on what equation I should use. Is it i=ie-(t/RC)?
I0 is one of the pieces of information you want to calculate. Take the ln of both sides of the equation:
The advantage of this form is that the equation is now in the form y = mx + b where y = ln(I) and x = t. So to do the linear regression use the t values as your x data and ln(I) values as your y data. You will come out with a slope and an intercept. The slope will be equal to -1/(RC) and the intercept will be ln(I0).
-Dan
I cheated and did the regression on my calculator. My answers are close, but different from yours. If you are allowed to use a computer to do the problem I'd recommend it.
I get m = 0.505943 x 10^3 /s and c = 1.497552. Thus I'm getting 1/(RC) = 0.505943 x 10^3 /s and I0 = 4.47073 x 10^{-3} A.
Now we have to get a bit wily. We only have one number to get RC so we need another relationship. We get that through the definition of capacitance:
C = Q/V, where Q is the initial charge on the capacitor (unknown at this point) and V is the charging potential 10 V. So...
Now recall that V = I0*R when the capacitor starts to discharge:
and we have values of 1/(RC) and I0 from the regression. So you can find Q. From Q you can find C. From C you can find R.
Using my values I get your given answers. Your values give answers slightly off from that. I am at a loss as to what advice to give you to allow you to calculate better numbers.
-Dan
I must've done something wrong rounding up the numbers when calculating it manually (and no, we wouldn't be allowed to use computers during our exams )
Yes I was confused about what other equations I needed to use to compare with the RC-equation. ( I thought of using the equation of i = C (dv/dt) but I wasn't sure if the method of least squares would work with a differential equation.
But thank you for the step-to-step explanation! I got the answers