Problem with capacitor voltage example

[sshoptaugh1991]

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.

[Ratch]

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

[Fgrir]

where did the + 8(10^-3) come from?

I 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

[sshoptaugh1991]

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)?

[Ratch]

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)?

Still 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

[sshoptaugh1991]

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 <= 4ms

Step 1) Integrate the term -8(10^-6) to get a result of  -8(10^-6)t,  but 4(10^-3) remains the same
Step 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 remains
Step 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?

[sshoptaugh1991]

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!

[sshoptaugh1991]

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)?

[Ratch]

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)?

The vertical axis of the graph contains a misprint.  It should be labeled as microamps.

Ratch

[Dave]

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.

[MrAl]

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.

The 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-6

where 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.