# Finding and plotting the current graph from a voltage graph (piecewise function)

Problem description:

My attempt at making the piecewise function:

Then, in order to find i(t), I took the derivative of 80x/3 and -80x/3, with respect to x, and multiplied C. Obviously, I'm left with a unknown variable C, so I'm not sure how to proceed from here. Am I supposed to find C first and then take the derivative? (how do I do that? I've been searching the web for this possibility and can't seem to find anything)

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## 1 Answer

C stands for the capacitance of the capacitor. According to the circuit diagram, it is given by 5 F.

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I got 87.5 Amps (after integration over limits, and multiplying by 5 Farads)... which seems rather large. Is this correct? – chaver Jan 13 '13 at 5:32
You made a mistake. The piecewise function you gave does not match the plot. If you insert x=1.25, you do not get 20. Furthermore, your function takes on negative values in the third section. – Frederic Brünner Jan 13 '13 at 5:57
Okay, I made some re-calculations and got the function to be 80x/3 - 40/3 and -80x/3 + 160/3, respectively. I integrated according to the limits, got 15/2, then multiplied by 5 Farads, with the final result being 37.5 Amps from 0.5 to 2.0 (both were positive). Is this correct ? – chaver Jan 13 '13 at 16:09
You do not have to integrate anything. The current is given by the derivative of voltage times capacitance. Therefore you need to take the derivative (slope) of the plot you are given. – Frederic Brünner Jan 14 '13 at 14:04
In that case.. it would be (80/3)*5=133.3 Amps for 0.5<x<1.25 and -133.3 Amps for 1.25<x<2.0. – chaver Jan 16 '13 at 2:18