Introduction to Phasors

[@lordtez1]

Finally, someone who can explain WHY Phasors work. Excellent presentation; thank you.

[@bassk94]

it tools me hours to find a video as thorough as this one. Nicely done, Thanks for being on the camera, standing up strait and speaking clearly. I can't stand people behind the camera in a dark room who mumble no matter who smart they are

[@MrExspectator]

Great teaching style, very clear! Thank you, and I hope you upload more lessons!

[@NoseBleedrummer]

Great video, I liked how you gave a strong motivation for actually using phasors.

[@robertfaney4148]

Thank u for this great material , really learnt a lot being 4th year Electrical in USA - fantastic !

[@ssssssssssurvey]

your lecture is very clear, very good. thanks

[@jakeanderson6253]

Thank you! I finally understand this:)

[@skiptaker]

Amazing comprehensive Lecture...!! Thanks a lot!

[@great_India3344]

Extremely good video

[@Rgrazia1]

Excellent discussion

[@akinakinyilmaz2]

Thank you very much,that is very beneficial for understanding this topic

[@AI-ro2bm]

very helpful lecture, thank you very much sir.

[@pnichols6929]

At 14:30 you have a string of complex numbers that you are adding together. There appears to be a mistake from one line to the next...Shouldn't 12L-90 be 12e^-90i? Also, When adding the string of complex numbers on Wolfram/Alpha, why doesn't the website reach the same answer as you?

[@glassofwater3767]

Could you upload the PowerPoint you used during this video?

[@sammao8478]

Very good lecture, helped me a lot! Thank you! I thought at 29:21 there was an error, purely a calculation one. The total admittance should be: 0.5 - j19.25 = 19.26 ⦟ - 88.51°. Therefore the total impedance = 0.05192 ⦟ 88.51°.

[@sohailjanjua123]

Hi Douglas, I like your lecture

[@wenxin4234]

Wow! He explained it well!!

[@DANGJOS]

brilliant explanation!

[@DeltaSigma16]

Very good presentation : Thank you.

[@NauticalAdventure]

Well done. Thank you. However, I do think a little bit more of the lecture should go into adding two phasors and the math that goes into finding the real and imaginary components. Also drawing the triangle of the real and imaginary component to find the Vmax or the hypotenuse of the triangle. Like i said though, well done, it was very clear and easy to understand. I am just inputting advice on where I had to pause the video and dig a little deeper to understand.

[@douglasdeboer2259]

Good video. A correction: At 21:58 in two instances, the word "conductive" in blue text should be replaced by "capacitive" in blue text.

[@andrewzhang]

Very helpful video :) Just another small correction. At around 35:20 you say that the angles are orthogonal and that they differ by 45 degrees. Should be 90 degrees

[@munteanumichelle]

Super helpful, thank you!

[@syarifromadhon3026]

great video...I owe you so much...:)

[@valoianm]

a good video..thank you.

[@HaroWorld1]

phasors start at 9:05 if you want to skip the technicals and stuff

[@bassk94]

13:40 in the video You do not explain how to go from 12+j9 to 15phasor36.87 Nor does my book. That is the answer i have been searching for since 8am this morning. So frustrating that everyone skips that step. You can not use an inverse trig function because it is UNDEF

[@pfile]

pythagorean theorem: sqrt(12^2 + 9^2) = sqrt(144+81) = sqrt(225) = 15. arctan 9/12 = arctan 3/4 = 36.87 degrees

[@lloydlucin528]

I agree. Why in the world... anyways thanks pfile for the answer.

[@rrgiri]

fantastic one :)

[@eduardojreis]

Hi, first I want to thank you for this video. Really cool. Here is my question. At 10:15, it says that: -> nu*cos(omega*t + phi) = nu phase phi. But later, by the Euler's Formulas, it says: -> nu phase phi = nu*exp(j*phi) = nu(cos(phi) + j*sin(phi)). But here (14:30), you just go from the trigonometric notation that has only cosine, to the phasor one; then to the complex exponential one, that this one has also a sin in it. I got confused: Where did the j*sin(phi) came from, or where did it go in the first case?

[@andrewzhang]

The phasor is a transform technique. We use transforms (like phasor, Laplace and Fourier) to switch between equivalent domains in order to perform operations or analysis more easily. In the case of phasors, as stated in the video, we are transforming time domain signals (in this case steady-state cosine and sine functions with same frequencies) into frequency domain signals which are represented by complex exponentials. Complex exponential operations and analysis in frequency domain are much easier compared to trig operations in time domain. So we need a way to convert from time domain to frequency domain and this is where Euler's formula comes in. Euler's formula is stated as: e^(j*phi) = cos(phi) + j*sin(phi) The thing with Euler's formula is it contains more than what we need. You can picture the complex exponential and sin/cos relationship as a vector rotating around the origin in the complex plane with some angular frequency omega. At any give time, the position of the vector can be determined by the angular frequency and the initial phase theta. Using basic trig, we can separate the vector into the real (x-axis) component and the imaginary (y-axis): x = cos(omega*t + theta) = cos(phi) y = sin(omega*t + theta) = sin(phi) Some pictures to hopefully help illustrate the point: upload.wikimedia.org/wikipedia/commons/7/71/Euler%27s_formula.svg upload.wikimedia.org/wikipedia/commons/8/89/Unfasor.gif This is essentially what Euler's formula does: "breaking down" a rotating vector into a real (x) component and an imaginary (y) component. Now although both the real and imaginary components are represented in Euler's formula, they can be considered separate entities when we are using phasors. For phasor representations, the convention is to represent all time domain sinusoidals in terms of the real component (in terms of cosine) of Euler's formula and simply ignore the imaginary component. So if we had sin(phi) in time domain and you want to represent it as a phasor, you would first convert to cos(phi-90) before expressing in phasor form. This may feel "uncomfortable" to just ignore the imaginary/sin component but it really has no physical meaning in the case of phasors. Hope that helps.

[@sheilacheruto3605]

thanks alot...helpfull tutorial

[@JoshBrownPhotography]

where does the pi over 2 come from? My unit circle has pi over 4 for 45 degrees

[@DouglasHarder]

Is this something I say--looking through the slides, I can't see what you're referring to.

[@DouglasHarder]

Found it. Added a comment.

[@JoshBrownPhotography]

Douglas Harder Sorry to make you search for it. I should've mentioned the time 5:45. So far I don't see the comment but it's probably KZbin delaying it. Thank you so much for replying. This is a GREAT video.

[@DouglasHarder]

Josh Brown Sorry, I added a comment in the description, with an acknowledgment. Thanks.

[@JoshBrownPhotography]

Douglas Harder Oh ok, well thank you for replying, and the update.

[@6san6sei6]

amaizing video thanks

[@bassionbean]

Wait.... Don't you also need to multiply the root mean square value (root 2) to the amplitude v if you're converting to a sinusoid? 15:47 is wrong I'm pretty sure... University of Nottingham's interactive gave me this piece of info.

[@douglasdeboer2259]

The video is correct, and bassionbean might also be correct, It depends. . . In the context of linear circuit analysis or electronics the phasor magnitudes usually have units of "Volts peak." In this system phasor 10L45(degrees) is equal to 10cos(wt + 45(degrees)), consistent with the video. To be strictly correct, the phasor should be shown as "10L45(degrees) V-peak" (Where I used "L" in place of the symbol for "at angle" and I used "(degrees)" in place of the symbol for "degrees.") In the context of power systems analysis (the electric power industry) phasor magnitudes usually are expressed in volts RMS. In this system phasor 10L45(degrees) is about equal to 14.14cos(wt + 45(degrees)), To be strictly correct, the phasor should be shown as "10L45(degrees) V-RMS" (Where I used "L" in place of the symbol for "at angle" and I used "(degrees)" in place of the symbol for "degrees.") It's kind of like the difference between U.S. statute miles and nautical miles. A full understanding of the units in play makes all the difference.

[@jarodolsen8955]

I must be dumb, clear as mud.

[@groinBlaster31]

Thank you

[@Engr.MuhammadAbbasKhan]

Is impedance a phasor?

[@OswaldChisala]

Hi Muhammad! Short Answer: That's an interesting question :) . YES: The impedance of your circuit, sub-circuit, circuit element, or whatever, is a phasor. Long Answer: We can derive the impedance of a capacitor (or whatever-else really) to see what's going on & prove it to ourselves, so let's do that. Allow a purely capacitive load of capacitance C to be driven by an input voltage signal V = cos(wt). You can add a phase but it doesn't really change the outcome. Okay, so the current that flows through a capacitor is given by I(t) = C*dV/dt, suggesting that the IMPEDANCE we seek is calculated as V/I. All we need is I(t). I(t) = C*dV/dt = C*d(cos(wt))/dt = -wC*sin(wt). Thus, the quotient V/I = IMPEDANCE = [cos(wt)] / [-wC*sin(wt)] . One way to simplify the impedance function above is to take the trig function cos(wt)/sin(wt) and convert it to polar exponential form (you can learn how to do that at www.electronics-tutorials.ws/accircuits/complex-numbers.html). First step: convert sin(wt) to a cosine to that we can re-express it as cos(Arg)+j*sin(Arg). sin(wt) = cos(wt - pi/2). Second step, convert the trig function to polar exponential form using Euler's Identity. cos(wt)/cos(wt - pi/2) = exp(j*(wt + 0)) / exp(j*(wt - pi/2)) = [exp(jwt) * exp(j0)] / [exp(jwt) * exp(-j*pi/2)] = exp(j0) / exp(-j*pi/2). At this point, we're almost there. Notice that exp(j0) / exp(-j*pi/2) = [cos(0) + j*sin(0)] / [cos(-pi/2) + j*sin(-pi/2)] = [1] / [-1] = -1/j. Awesome! Notice how the frequency information dropped off? We have obtained a phasor representation of the trig function and have crossed over into the frequency domain i.e. s-domain. Okay, so [time domain] cos(wt)/sin(wt) -1/j [frequency domain]. Copying and pasting into the impedance expression from before then provides us: V/I = IMPEDANCE = [cos(wt)] / [-wC*sin(wt)] = (-1/(wC) * -1/j) = 1/(jwC). Letting s = jw then provides V/I = Xc = 1/(s*C), which is the impedance of the capacitor provided from quantities derived from a phasor representation by using basic manipulations of Euler's Identity. The j represents the phase shift, and in this case shows that the impedance is at -j to the input/driving signal i.e. it lags by 90 degrees. Yes, it was just fortunate that the wt terms canceled out for this calculation...we should have extracted the phasor information from each part of the V/I quotient, but I'm not a mathematician, and we got the right answer anyway. IF you want to be correct on a technical level, extract the phasor from the signal before the calculation even commences. Capacitors and inductors are duals, so whatever we did to show that the impedance of a capacitor is technically a phasor quantity will hold perfectly well for an inductor. Hopefully this helps! I'm typically more of a Quora person (www.quora.com/profile/Oswald-Kay-Chisala), but I guess today I commented on KZbin. Cool. Have a nice day, Muhammad Khan. Yours with good intent, Oswald Kay, The Wishful Thinker :)

[@Engr.MuhammadAbbasKhan]

Thanks for responding! Sir now i have one query some one answering me that "Voltages and currents, sinusoidal waves in time, can be represented by complex numbers and we call those numbers "Phasors" (short for "phase vector"). Impedance don't have phases, they are not sinusoidal functions of time, so their complex number representation is just that, a complex number, not a phasor. An actual phasor is multiplied by exp(jwt), that is, it is rotating in the complex plane with a speed of w rad/s " What you view ?

[@WangzGamez]

Phasors, so are they like the sparky from clash royale?

[@Semitable]

Thank you a lot.. :)

[@ntuthukosangweni5204]

thank u 3 much

[@edmilsonfilho]

Doulgas, I was reading Fortescue's Theorem and I have one doubt: what is a phasor rotating in opposite direction?

[@ndumisonkosi6818]

Confusing.......