Find tme constant for discharging capacitor

1. Mar 12, 2014

1. The problem statement, all variables and given/known data

Hi, I have a quick question about applying the RC time constant formula for a lab report. In the lab, we charged a capacitor to 20 V and then let them discharge, recording the voltage every 10 seconds, up to 240 sec. Now I was asked to graph the time vs. voltage in excel, fit an exponential trendline and obtain an equation.

2. Relevant equations

The equation I got from my line is: y = 19.89e-0.02x

My lab manual gives the time constant formula as V = V0e-1/t , with t being the time constant.

I have to use the experimentally obtained equation to calculate the time constant.

3. The attempt at a solution

I see the V0 is very close to 20, so that makes sense, but I don't understand how to combine the two equations to get the time constant. It seems that V should be the final voltage, but I have 24 values, all of which would give a different x. I'm confused how to proceed from here.

2. Mar 13, 2014

Simon Bridge

I'm guessing you mean: $$V=(19.89V)e^{-(0.02)t}$$

... they mean: $$V=V_0e^{-t/\tau}$$ where $\tau$ is the time constant.

The relation you have is for t=1s.

You can use the data point at t=1s - very sloppy:don't do this.

You can use the first few data points as an approximate straight line - the intersection of the best fit with the t axis will be close to the time constant.

Best approach is to linearize the graph ... plot y=ln(V) against x=t.
$$\ln|V|=\ln|V_0|-\frac{t}{\tau}$$ ... this is a straight line with intercept $V_0$ and slope $1/\tau$. Find the slope.

3. Mar 13, 2014

Staff: Mentor

Another way of doing this is to recognize that, in the excel fit to your data, that 0.02 is the reciprocal of the time constant (are there any additional significant figures?). So the time constant should be about 50 seconds. This should be roughly the same result you get using Simon's linear plot.

Chet

4. Mar 14, 2014