Problem with capacitor voltage example - Page 1
Forum section: electronics / beginners

author: sshoptaugh1991
author_rank: Contributor
timestamp: September 08, 2017, 03:59:57 pm
content: Alright, I feel really stupid right now. I cannot figure out this example problem in my textbook. The problem is to find the waveform of the voltage in a capacitor by analyzing the current waveform. I understand how the i(t) equations are derived, as well as the first v(t) equation for the interval 0 <= t <= 2ms.The problem arises when I am at the step of the v(t) equation over the interval 2ms <= t <= 4ms. I understand how they got v(t) = -2t by integrating, but where did the + 8(10^-3) come from? Am I not seeing a step, or am I making a mistake in my math somewhere? The result I had thought would be correct is v(t) = -2t + 4(10^-3) for 2ms <= t <= 4ms.

author: Ratch
author_rank: Regular Contributor
timestamp: September 08, 2017, 05:38:47 pm
content: sshoptaugh1991,If you integrate the current between 2 to 4 mv, you will get the term shown below in the attachment. The integration shown in the book matches what I calculated using a math program, so you can assume it is correct. Try calculating again and see if you get the same answer.Ratch

author: Fgrir
author_rank: Regular Contributor
timestamp: September 08, 2017, 06:45:34 pm
content: Quote from: sshoptaugh1991 on September 08, 2017, 03:59:57 pmI think you have neglected the constant term that arises from the beginning of the integration interval for the last step.https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Corollary

author: sshoptaugh1991
author_rank: Contributor
timestamp: September 08, 2017, 07:08:53 pm
content: So just remove the -8(10^-6) as a constant, and thus the constant with the 1/C coefficient comes out to be -2 and just distributes to the integral of 1 and + 4(10^-3)?

author: Ratch
author_rank: Regular Contributor
timestamp: September 08, 2017, 08:17:01 pm
content: Quote from: sshoptaugh1991 on September 08, 2017, 07:08:53 pmStill don't get it? Look at my attachment above. One of the 1/250 mv = 4 mv comes from the capacitor being energized to 4 mV during the first 2 ms.. The other 1/250mv comes from the integration. Together they equal 8 mv, which is what you were asking about.Ratch

author: sshoptaugh1991
author_rank: Contributor
timestamp: September 09, 2017, 03:21:38 am
content: I still do not understand, I apologize. This must be something very elementary that I am not getting.So the expression below is what we start with on the interval of 2ms <= t <= 4msStep 1) Integrate the term -8(10^-6) to get a result of -8(10^-6)t, but 4(10^-3) remains the sameStep 2) Divide the result of the integral by the term 4(10^-6) to get a result of -2t and the 4(10^-3) still remainsStep 3) Since this is for the interval 2ms <= t <= 4ms, we still need to add the result of the interval of 0 <= t <= 2ms, 4(10^-3),resulting in v(t) = -2t + 8(10^-3)Am I on the right track here, or am I still off by a big step?

author: sshoptaugh1991
author_rank: Contributor
timestamp: September 09, 2017, 03:32:03 am
content: I got it! I can't believe it took me this long to see it, but I got it. I was forgetting to finish the integration of -8(10^-6)t by doing the definite integral calculation.After integration, the result is:-8(10^-6)t/4(10^-6) + 16(10^-9)/4(10^-6) + 4(10^-3)Simplifies to:-2t + 4(10^-3) + 4(10^-3) ===> v(t) = -2t + 8(10^-3)I feel really dumb now for not catching that crucial mistake.Thanks Ratch!

author: sshoptaugh1991
author_rank: Contributor
timestamp: September 09, 2017, 01:18:45 pm
content: One more question about this example problem:If the graph of the current is in amps (A), why are the values in the i(t) equations written as microamps (uA)?

author: Ratch
author_rank: Regular Contributor
timestamp: September 09, 2017, 01:52:36 pm
content: Quote from: sshoptaugh1991 on September 09, 2017, 01:18:45 pmThe vertical axis of the graph contains a misprint. It should be labeled as microamps.Ratch

author: Dave
author_rank: Super Contributor
timestamp: September 09, 2017, 05:49:23 pm
content: There is another more crucial mistake in the voltage graph. The left part of the voltage graph should be perfectly parabolic - zero slope at the beginning (corresponding to zero initial current) and progressively getting steeper and steeper.

author: MrAl
author_rank: Super Contributor
timestamp: September 09, 2017, 07:28:38 pm
content: Quote from: sshoptaugh1991 on September 08, 2017, 03:59:57 pmThe short answer is because the input current is a ramp, we have to solve for the slope so that we can integrate. That is where the 8e-3 comes from.For slope 'm' w have:x*m=16b-6where 16b-6 is the peak current at the top tip of the ramp, and we solve his for the slope 'm' at x=0.002 seconds.