Find tme constant for discharging capacitor

[UselessLadder]

Homework Statement

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.

Homework 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.

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.

 

[Simon Bridge]

Homework Statement

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.

Homework Equations

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

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

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

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

The relation you have is for t=1s.

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

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.

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.

 

[Chestermiller]

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

 

[UselessLadder]

^ That's it, I just had a brain freeze trying to derive the time constant from that exponent expression. Simon's method would produce a more precise value, but my manual specifically asked for an exponential trendline so I just did that and it gave me a close enough estimate for the purposes of this lab. Thanks both of you!

 

[HalfLostAndSearching]

As a scientist, it is important to understand the concepts and equations used in your experiments. The time constant (τ) for a discharging capacitor can be calculated using the equation V = V0e^(-t/τ), where V is the voltage at time t, V0 is the initial voltage, and τ is the time constant. In your experiment, you have obtained the equation y = 19.89e^(-0.02x) from your data. To calculate the time constant, you can equate this equation to the general equation for discharging capacitors and solve for τ. This would give you τ = -1/0.02 = -50 seconds. However, it is important to note that the negative sign indicates that the voltage is decreasing over time. You can also use the natural logarithm (ln) to solve for τ, as ln(e^(-t/τ)) = -t/τ, which would give you the same result of -50 seconds. It is important to carefully interpret your data and equations to obtain accurate and meaningful results.