Art of Electronics Exercise 1.14 - Charging a capacitor - Page 1
Forum section: electronics / beginners

author: donturner
author_rank: Contributor
timestamp: February 14, 2017, 04:35:30 pm
content: Hello. I am currently learning electronics and am stuck on an exercise from The Art of Electronics book. Here's the exercise (copied verbatim):Exercise 1.14Take the energy challenge: imagine charging up a capacitor of capacitanceC, from 0V to some final voltageVf. If you do it right, the result won't depend on how you get there, so you don't need to assume constant current charging (though you're welcome to do so). At any instant the rate of flow of energy into the capacitor isVI(joules/s); so you need to integratedU = VIdtfrom start to finish. Take it from there.So my first problem is that I don't quite understand what the exercise is. Is it asking for the total charge on the capacitor for Vfwhen time is infinity? If so, this is given by:U = 1/2CVf2I'm really puzzled by the "integrate...from start to finish" part. I understand differentiation (converting from y=f(x) to dy/dx) and integration (from dy/dx to y=f(x)) but I don't see how that applies todU = VIdt. That could be because it's 20 years since I did A-level maths.If someone could just point me in the right direction, or tell me "you need to learn more about X" I'd really appreciate it.

author: Andy Watson
author_rank: Super Contributor
timestamp: February 14, 2017, 06:03:40 pm
content: Quote from: donturner on February 14, 2017, 04:35:30 pmI think it is asking for the total energy put into the capacitor.QuoteYou start with zero voltage and finish with the capacitor charged to \$V_f\$. As it says, the result does not depend on how you get there - so time is not a factor. I would write-out the integral for energy and then make the substitution \$i = C \frac{d V}{d t}\$

author: Doc Daneeka
author_rank: Contributor
timestamp: February 14, 2017, 06:39:08 pm
content: The point of the question is that to change the voltage across a capacitor you need to do work: put energy in or take it out. So imagine the voltage starts at v(0) and some time later is v(T). The instantaneous flow of energy is v(t)×i(t) (i.e the power - rate of work). So the total change in energy (total work done) over that time is the integral (over that time) of v(t)×i(t). You can then use the fact that i(t) =Cdv/dt for a capacitor and the fundamental theorem of calculus (integration by substitution) and you will see that the difference in energy only depends on v(0) and v(T). This is what it means by 'doesnt matter how you get there'

author: donturner
author_rank: Contributor
timestamp: February 14, 2017, 07:04:23 pm
content: Thank you Andy and Doc, really appreciate your help. So here goes....\$dU = Vidt\$substitute \$i = C\frac{d V}{d t}\$\$dU = VC\frac{d V}{d t}dt\$cancel the \$dt\$s on the right hand side\$dU = VCdV\$integrate both sides\$\int dU = \int VCdV\$Left hand side is easy. C is a constant which can be brought outside of the integration.\$U = C \int VdV\$Add the limits\$U = C \int_{0}^{f} VdV\$Solve the right hand side integration (k is the integration constant)\$U = (\frac{1}{2}CV_f^2 + k) - (\frac{1}{2}CV_0^2 + k)\$And since \$V_0 = 0V\$\$U = \frac{1}{2}CV_f^2\$Which matches the formula given for stored energy in a capacitor. It looks right, but did I miss something?

author: rstofer
author_rank: Super Contributor
timestamp: February 14, 2017, 08:23:19 pm
content: Looks ok to me!http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html