Can you even measure reactive power? - Page 1
Forum section: electronics / beginners

author: cvriv
author_rank: Frequent Contributor
timestamp: July 30, 2016, 11:18:19 am
content: I'm having some trouble here. I'm doing old homework again to prep myself for upcoming classes. I know the book I'm using has many wrong answers in the back, but this one particular chapter has many wrong answers. I just wanted to make sure that I'm doing it right, even though I'm very sure I am, so I decided to simulate and probe the circuit via Multisim; I wanted to see the results with my own eyes. Well, I'm not seeing what I was hoping to see. I'm not doing something right here.I am to calculate reactive, resistive, and apparent power. I got papp = 35.8mVA, px = 34.3mVAR, and pr = 10.17mWI attached a photo of the Multisim circuit and resistive result. What am I doing wrong here?Can you even measure reactive power? Multisim? If I probe the inductors, I would basically be seeing the power dissipated by the winding resistance right? The components are ideal so that would explain why the reading are bouncing around from negative to positive pW.

author: Ian.M
author_rank: Super Contributor
timestamp: July 30, 2016, 11:29:34 am
content: I'm not familiar with that particular SPICE package, but it isn't unusual to either have to set up dependent sources and low-pass filtering into a unit load resistor, or to post-process the RAW file in an external application when you need to read out VA or VAR in a SPICE program.

author: joeqsmith
author_rank: Super Contributor
timestamp: July 30, 2016, 01:19:50 pm
content: Quote from: cvriv on July 30, 2016, 11:18:19 amFor starts, publish the actual problem from the book.

author: Arjan Emm
author_rank: Regular Contributor
timestamp: July 30, 2016, 02:10:11 pm
content: You are measuring current through and voltage acros an ideal resistor in the image you posted.In that case PF is always 1 regardless of the circuitry around it or the frequency of the signal. And the power is always real power in watts.

author: RES
author_rank: Regular Contributor
timestamp: July 30, 2016, 02:30:45 pm
content: Think you have to calculate back it from Watts (true power, resistive)Watts is what you measure with a voltmeter, VAR and VA are the other two vectors. (inductive, capacitive vectors)

author: rstofer
author_rank: Super Contributor
timestamp: July 30, 2016, 02:44:44 pm
content: Well, the calculation looks reasonable. Watts^2 + VARS^2 = VA^2Maybe this will help:http://home.earthlink.net/~jimlux/hv/varvom.htm

author: cvriv
author_rank: Frequent Contributor
timestamp: July 31, 2016, 04:30:03 am
content: Quote from: joeqsmith on July 30, 2016, 01:19:50 pmThat is the problem. The actual problem is the same circuit except it's drawn in a book. I am to calculate resistive, reactive, and apparent power. I attached my calculations. Can you just verify them? Thanks.The answers in the back of the book for this problem are PR = 1.56mW and PX = 1.66mVAR. The book didn't actually want me to calculate apparent power, but I did, and for this problem I converted apparent power to rectangular notation to get the PR and PX vectors. Also, What I think happened in the back of the book is that whoever calculated the answers for that circuit had to calculate the impedance, voltages, and current for a previous problem and got those wrong. Then two problems later was asked to calculate the power for that circuit, so instead of calculating the answers from scratch, they used all the wrong info from the previous problem to calculate more wrong answers.

author: Morgoroth
author_rank: Regular Contributor
timestamp: August 01, 2016, 04:10:25 am
content: Just to clarify conceptsEnviado desde mi SM-T700 mediante Tapatalk

author: R005T3r
author_rank: Frequent Contributor
timestamp: September 08, 2016, 01:18:04 pm
content: You don't have to eliminate all the foam, you can decrease it by re-phasing the loads(put a capacitor bank in parallel with the load by a given value), but you never have to eliminate it otherwise you will damage your loads.

author: CoffinDodger
author_rank: Regular Contributor
timestamp: September 08, 2016, 09:22:26 pm
content: You won't damage much by eliminating it but you'll definitely damage your source if you tip over to leading. And power generation equipment will tend to have kVAR meters or PF meters. Check the meter selection in Multisim for a VAR meter. If it does AC circuits I would think it would have one for education purposes.some other power formulas would be:P = IVcos(angle between current and voltage)P = IV(power factor)S = IVQ = IVsin(angle between current and voltage)pf = cos(angle between current and voltage)orpf = P/SAnd from my calculations off your circuit there your answers are correct.