Calculating real power of a circuit - Page 1
Forum section: electronics / beginners

author: nForce
author_rank: Frequent Contributor
timestamp: January 28, 2018, 06:18:03 pm
content: How do I calculate real power of this circuit (green):So do I write down the impedance, and use 1/2*(U^2/)Z?

author: pigrew
author_rank: Frequent Contributor
timestamp: January 28, 2018, 06:32:54 pm
content: Only the resistors will have real power, not the inductors or capacitors.You'll have to calculate the voltage across or current through the final resistor, and then use the appropriate formula.The first step I'd do is to substitute the source and two resistors with a Thévenin equivalent.

author: Benta
author_rank: Super Contributor
timestamp: January 28, 2018, 06:38:21 pm
content: There are several ways to attack this one.Personally, I'd calculate the total impedance [seen from ug(t)] using R, 1/sC and sL for the impedances (LaPlace notation). Substituting jw for s, you can calculate real and imaginary current (and thus power).I'm certain there are simpler ways, others will chime in, but this is usually foolproof for me.PS: how do you insert symbols? Using w instead of greek omega is sub-optimal.

author: nForce
author_rank: Frequent Contributor
timestamp: January 28, 2018, 07:21:45 pm
content: Sorry benta, but pigrew said that only R has real power, why bother then calculating impedance for L and C?

author: Benta
author_rank: Super Contributor
timestamp: January 28, 2018, 07:34:34 pm
content: Quote from: nForce on January 28, 2018, 07:21:45 pmBecause otherwise you won't know the currents through and voltages across your resistors. I'll give you my solution when you've tried it yourself.

author: dmills
author_rank: Super Contributor
timestamp: January 28, 2018, 08:34:28 pm
content: There is a bit of a trick hiding in here.You are given that omega = 1/sqrt(LC) which is nice, think about what you know about series resonance in LC networks, it makes this real easy to solve.Regards, Dan.

author: Benta
author_rank: Super Contributor
timestamp: January 28, 2018, 09:35:55 pm
content: Quote from: dmills on January 28, 2018, 08:34:28 pmAaah! Well spotted, Sir.I missed that little point, as the circuit analysis is not trivial at all. The impedance seen from the source works out as:[ s22CLR + s3CR2+ 2R ] / [ s2CL + s2CR + 1 ]Not nice at all.

author: Vtile
author_rank: Super Contributor
timestamp: January 28, 2018, 09:49:00 pm
content: If you remember the generator topic of yours..

author: Benta
author_rank: Super Contributor
timestamp: January 28, 2018, 09:55:38 pm
content: Quote from: Vtile on January 28, 2018, 09:49:00 pmCryptic.

author: Vtile
author_rank: Super Contributor
timestamp: January 28, 2018, 10:13:43 pm
content: Quote from: Benta on January 28, 2018, 09:55:38 pmHe/she made a topic of "reactive power in AC generators" @general section around at xmas. This topic relates to concepts discussed there.

author: Benta
author_rank: Super Contributor
timestamp: January 28, 2018, 10:22:06 pm
content: 'Nuff said. If you have a thing about referring to previous topics somewhere else that nobody understands, the Finnish winter is apparently too boring.

author: dmills
author_rank: Super Contributor
timestamp: January 28, 2018, 10:30:49 pm
content: Quote from: Benta on January 28, 2018, 09:35:55 pmI hate it when they do that in an exam, 40 minutes or so of fucking about with algebraic manipulation (and no marks to show for it), because you missed one little critical detail.I make the impedance seen by the generator 3R/2 at the specified frequency without any need for Laplace.Regards, Dan.

author: Vtile
author_rank: Super Contributor
timestamp: January 28, 2018, 10:33:07 pm
content: Quote from: Benta on January 28, 2018, 10:22:06 pmSometimes.Unfortunately I'm writing from phone so I can't easily link it here.

author: Benta
author_rank: Super Contributor
timestamp: January 28, 2018, 10:35:35 pm
content: QuoteYep - and I fell right into it- fail.

author: IanB
author_rank: Super Contributor
timestamp: January 28, 2018, 11:22:34 pm
content: Quote from: Benta on January 28, 2018, 06:38:21 pmYou can use MathJax. So for example:Code:\$u_g(t)\$, \$R\$, \$1/sC\$ and \$j\omega\$will produce:\$u_g(t)\$, \$R\$, \$1/sC\$ and \$j\omega\$

author: rstofer
author_rank: Super Contributor
timestamp: January 29, 2018, 01:10:17 am
content: Quote from: IanB on January 28, 2018, 11:22:34 pmThanks for that! I'll have to cut and paste that code into a file and save it somewhere.I found this:https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference

author: bson
author_rank: Unknown
timestamp: January 29, 2018, 06:58:04 am
content: There's also a handyMathJax demo. Just type there until you like the result, then insert the latex into a post.

author: tatus1969
author_rank: Super Contributor
timestamp: January 29, 2018, 09:48:04 am
content: you don`t need *any* complex number math to solve this one. Start with L, C, and w=1/sqrt(L*C). You can solve this one by looking at the picture only.

author: nForce
author_rank: Frequent Contributor
timestamp: January 29, 2018, 12:29:31 pm
content: Quote from: dmills on January 28, 2018, 08:34:28 pmYes, that green circuit is original series LC circuit with resonant frequency \$\omega= \frac{1}{\sqrt{LC}}\$But how do i use that for calculating real power?

author: dmills
author_rank: Super Contributor
timestamp: January 29, 2018, 01:07:08 pm
content: I read that as telling you to calculate the reactive power at frequency w=1/Sqrt(LC), which is trivial if you think about what impedance a series LC network has at resonance.Xc + Xl = 0 at resonance, so the green box degenerates into what?The rest is just trivial arithmetic.Regards, Dan.

author: nForce
author_rank: Frequent Contributor
timestamp: January 29, 2018, 01:37:04 pm
content: Quote from: dmills on January 29, 2018, 01:07:08 pmOh, I get it. In series resonance the impedance of L and C (both of them) is 0. So only the resistor remains. Because of that I can calculate with P = (U^2) / Z

author: nForce
author_rank: Frequent Contributor
timestamp: January 29, 2018, 05:43:19 pm
content: I have another question which is related to this.Why do we look at impedance in series RLC circuit and admitance in parallel RLC circuit?

author: dmills
author_rank: Super Contributor
timestamp: January 29, 2018, 05:47:15 pm
content: Impedances sum in series, admittances sum in parallel, it just makes the maths less annoying.Regards, Dan.