did i do this right?

Question

i performed a formula to see how much volts and amps a capacitor discharges in each millisecond and the question is if i did it right the capacitor had 150uf and the battery that charged it had 1.5v and 5.0a and i even made a chart of the results so did i do it right?
(note: the letters along the bottom are the milliseconds and the ones along the left side are the measurements)

both are based on these formulas
V=v(e^t/T)
the instantanious voltage is equal to the initial voltage times exponent to the power of time over tau

I=i(e^t/T)
the instantanious current is equal to the initial current times exponent to the power of time over tau

heres a link to the chart
http://img100.imageshack.us/img100/6939/capacitordischargews3.jpg

Answer 1

For charging and discharging, you have to use a negative exponent, as the voltage approaches an asymptotic value at an ever-decreasing rate (exponential decay). A positive exponent would describe change at an ever-increasing rate (exponential growth). So for discharge, you would write V=V0*(e^(-t/T)), and for charge V=Vs*(1-e^(-t/T)). (V0 is initial voltage, Vs is supply voltage.) This is because when t=0, e^(-t/T) = 1, and when t is large, e^(-t/T) approaches 0. Thus in the discharge equation, V starts at V0 and approaches 0, and in the charge equation, V starts at 0 and approaches Vs. I think you must have correctly used a negative exponent and the discharge equation to get that chart.
Given a capacitor C = 150 uF, and initial voltage V0 = 1.5 and current I = 5.0 (reading from the chart), you would have a resistor R = V0/I = 0.3 ohms. This means the R-C circuit should have a time constant T of 150^10^-6*0.3 = 0.045 ms. This is not consistent with the V-I plot in the link, which shows an apparent time constant of roughly 0.12 ms. (The time constant is the time it takes for voltage to decrease to 1/e or 0.37 * its current value.) Could you have made a measurement error?