This may seem like a redundant question but I was wondering why gradient calculations are preferred when calculating a value by experiment as opposed to just plugging in raw values me getting an Answer ? I know it's something to do with the fact that the gradient considers a greater number or points and this makes the value more accurate but can someone give me a proper textbook answer ?

For example, I am doing a capacitor lab where I find the time constant of a capacitor. I plotted a graph of V vs t and and an appropriate linear graph of ln V vs t. My teacher says its preferred to use the gradient of the linear graph of lnv vs t to calculate time constant rather than reading off from the exponential graph of V vs t. I wanted to know why is it more accurate to use the gradient calculations ?


closed as unclear what you're asking by Aaron Stevens, Mike, garyp, Bill N, ZeroTheHero Oct 23 '18 at 8:01

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Can you give a little context? $\endgroup$ – Gilbert Oct 23 '18 at 1:54
  • $\begingroup$ The question, as it is currently worded, is hard to answer without more details. Can you please edit to include a specific example of the process you are asking about? $\endgroup$ – Aaron Stevens Oct 23 '18 at 2:11
  • $\begingroup$ I have edited the question $\endgroup$ – user122343 Oct 23 '18 at 2:35
  • $\begingroup$ If you make several calculations using several values from the graph, you will arrive at several values for your result. Which o e is then the correct one from experiment? $\endgroup$ – Triatticus Oct 23 '18 at 2:50
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    $\begingroup$ If the physical quantity has an exponential dependence, using a log scale "shortens" the graph and displays the exponential behavior because the eye follows lines much better than general curves . $\endgroup$ – anna v Oct 23 '18 at 3:43

Experimental physics also incorporates uncertainty. You can never say the time constant = 3 ms, but it exists as a value +/- another value. Same goes every measurement you make. On a graph rather than a point you will have fuzzy regions of uncertainty.

For a linear relation you expect a linear fitted line.

For an exponential or logarithmic relationship, a linear line is possible if one axis is on logarithmic scale.

For a power relationship, a line is possible if both axes are logarithmic.

In all cases, slopes and intercepts allow you to determine two parameters of the equation. If you get a computer to do the fit and read off the value, it is doing all of this behind the scenes anyway.

Having a linear line is useful in that it allows you to visually assess the fit, e.g. for deviations from "linearity" that aren't readily apparent otherwise.

Now back to the errors. Lines of best fit are only the beginning. You can also define lines of worse fit. These help you determine a range of values possible for a given input. But wait. Your input also has an error. The resultant error can be determined by multiplying by the slope.


I am assuming by "time constant of the capacitor" you mean the time constant of an RC circuit $\tau=RC$. Let's assume we are looking at a discharging capacitor. Then the voltage across the capacitor as a function of time is given by $$V(t)=V_0e^{-t/\tau}$$ where $V_0$ is the initial potential across the capacitor.

Now, let's compare this to what you have discussed in your answer: $$\ln(V(t))=\ln\left(V_0e^{-t/\tau}\right)=-\frac{1}{\tau}t+\ln(V_0)$$

So, as you have said, the gradient, or slope, of $\ln(V(t))$ gives us the desired $\tau$ we want. Slopes are very easy to calculate from linear plots. Even if your data is not perfectly linear, linear models are probably the easiest models to fit data to. You could even get a good estimate of a line of best fit by just drawing one by hand and then determining the slope from there. Pretty simple.

But what about the original expression for $V(t)$? Well this is an exponential function. You would most likely need some sort of program to fit your data to the exponential function. And this is not as simple as a linear function. You could draw an exponential curve you think fits the data, but it would be harder than a line (and the curve you draw might not even be a true exponential function with base $e$). Even then, it's harder to pull the time constant from the exponential decay than it is to find the slope of a line.

Therefore, in this context the linear function and its slope are easier to work with. Lines are simple to visualize and work with. Although with today's technology, either method should be fine if you have the software to do it. In the labs I would TA for, we would fit directly to the exponential functions with very few issues.


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