Steady State Circuit Analysis with Phasors

[@robb1324]

Very very very clear and organized, best video on this subject on youtube. My only gripe is that your mathmatica screen is cut off... plus most of us will only have calculators for tests, so using a calculator is preferred!

[@kursatsaimtuna7163]

I liked how he says good luck at the end. He knows this is a hard job as well. But thank you very much Im so lucky to find this channel.

[@DailyClipsInternets]

You have amazing teaching skills that enabled me to pass my midterms :) i appreciate your work.

[@anujregmi4582]

you open a new dimension of circuit analysis for me man...U took me a different universe...thanks a lot....!!!!!!!!!!!!!!!

[@kevinacemusic]

It's awesome how you write the steps out, my exam allows for a 2 sided cheat sheet, so it's very nice to have it laid out. Thank you!

[@Operator87]

In case you didn't already figure it out, he used source transformation in order to go from a current source with an element in parallel to a voltage source with an element in series (using ohms law) so that he could use voltage division to find the voltage across the resistor.

[@stijnmattaar559]

An easier way to solve, even without using mathematica and the source transformation. Calculate the replacement impedance using (Zc*Zr)/(Zc+Zr) and you'll find that it is as easy as 8-4j. Then by using Ohm's law U=IR, you can multiply the source sqrt(2)/2-j*sqrt(2)/2 with this complex impedance 8-4j to find U = 2sqrt(2)-j*6sqrt(2) Now, the magnitude is found with the pythagorean theorem over the real and imaginary part: it is 4sqrt(5) which is approximately 8.94 [V] and the phase (or argument) is a bit more complicated using arctan(-6sqrt(2)/2sqrt(2)) which is arctan(-3) = -71.57 degrees. Final answer: 8.94 < -71.57 (using < as an 'angle' sign)

[@Lphills]

So much easier, thanks

[@clarkgriswold5842]

Thanks for this, we wouldn't be able to use mathematica for our exam

[@thetarasbulba1]

"Don't get stunned [by phasors]"... very funny. Thanks for the video, it helped a lot

[@AraujoMatt]

That comes from basic trigonometry for a 45 deg triangle. The magnitude of the phasor (1) is the hypotenuse. When converting to rectangular coordinates, we use the length of the sides.

[@sohailjanjua123]

Hi Metthew, I like your lecture.Thanks

[@LT1]

Intro was very helpful. Good stuff.

[@loganmasters3507]

Very helpful, well done and concise.

[@AM-bk2pe]

thanks this was super helpful - loved the steps being written out for how to solve it! Wish you used a calculator though since most of us won't use mathmatica tbh

[@2mjz84]

Thanks man, 2024 and still helped me :D

[@sridharchitta7321]

When you connect a circuit by making the final connection, feedback forces a rapid rearrangement of surface charges leading to the steady-state. This period of adjustment before establishing the steady-state is called the “initial transient”. What is meant by steady-state in a capacitive circuit subject to a sinusoidal voltage input? A sinewave depicts movements in the form of waves. It has a datum, rapid, slowing and steady growth in one direction for a quarter cycle with reference to a datum (a reference), and then slow and faster decay in one direction for the next quarter cycle, and all these elements again in the opposite direction (reversal) of the forward movement for the next half-cycle. It has peaks and valleys. In essence, the sinewave is a perfect embodiment of oscillatory movements like springs and quantities like voltage. It retains its waveshape when added to another sinewave of the same frequency and arbitrary phase and is the only periodic waveform which has this property. Imagine that you and your friend are playing a game of “swing”. When your friend sits on a stationary swing and you begin pushing it, it will take a few hard pushes initially to overcome inertia when the swing moves with a small displacement. You then synchronize your pushes by progressively moving slightly away from the stationary position of the swing, initially, pushing gently, and then pushing harder as you move away from the central stationary position of the swing. The point of pushing the swing is usually at the top of the swing cycle at one end. It takes a while of pushing before you are able to establish a rhythmic swing. The period before the rhythmic swing is established is the “transient”, and the rhythmic swing that is established after the few transient cycles elapse is the “steady-state”. This is analogous to establishing the steady-state in a capacitive circuit subject to a sinusoidal input. The capacitor being initially uncharged, will cause the current during the transient period to assume a value that will be quite different from that at the same voltage angle after steady-state is established. Electrostatics and circuits belong to one science, not two. These are discussed usually separately in textbooks and science and engineering courses. It is not possible to discuss the circuit processes which produce a sinusoidal current when a sinusoidal voltage is applied to a capacitor….the changing rates of change of the applied voltage ….the surface charge set up changing at every instant….the applied field changing in the wires ….and the current at each and every instant in time. Watch the two videos listed below to learn about current and the conduction process and surface charges (using a unified approach to electrostatics and circuits at a fundamental level). 1. kzbin.info/www/bejne/ioXXpWVul5aXj9E 2. kzbin.info/www/bejne/bnO0fpKurJeFnNE The last frame of video #1 lists textbook 4 which discusses the sinusoidal steady-state in capacitors and inductors with the help of sequential diagrams and animated power-point presentations of the varying field components in the circuit elements in more detail.

[@crashonthehumble]

Outstanding lecture

[@AraujoMatt]

Arash - For these videos I'm using a Wacom Intuos tablet. I think the Wacom Bamboo tablets are fairly inexpensive and will work just as well for this.

[@tiagoatwi1768]

You are an attractive man Mr.Matthew

[@Kevn37]

You can also apply KCL to solve this after finding out the input current.

[@zhenisotarbay8162]

can you show how you get -14.14-14.14j just tired and having wrong numbers

[@przemysawwiczoek2542]

It's just multiplication (-20j) * (√ 2/2 - √ 2j/2) = -14.14j - 14.14

[@catherinegideon4689]

Zhenis Otarbay also he was trying to do source transformations and he changed that current source and parallel zc into voltage source in series with zc by multiply ,rmb V=IR

[@ComputerAtSea]

Nice job, chief!

[@jordanbendler8205]

Very helpful brotato. Thanks.

[@AraujoMatt]

Runback....the question for this problem is written beneath the circuit. We're looking for the voltage across the resistor. Vo(t) = ?

[@YaJohny_]

You are the best!

[@teezo3afana]

YOU my man are amazing .

[@AraujoMatt]

Sorry about the clipping. I do have a video or two where I use a TI 89 to solve a problem involving phasors

[@jessmac1893]

Kind of confused on the mathematica portion as it's cut off. Any chance you could post the full code as a comment or in the description please? Everything else was really well explained. Thanks!

[@oscarlaruta2008]

which book are u using? good video

[@mamanieto]

Could you please explain what happend when you multiplied the voltage (-14.14-14.14j)*(10/10-20j) I understand that 10-20j i the sum of the ohms but where does the 10 in the numerator come from ??

[@omarm.7068]

Is there anything wrong with using current division, using the equivalent impedance of both elements, to get the current at the resistor branch and then calculating the voltage using Ohm's Law? I seem to be getting a different answer.

[@quinnimon]

I'm still new to circuits, so I don't quite understand how you turned a parallel circuit into a series circuit so easily. Please help me here.

[@bradwatter45]

source transformation

[@bradwatter45]

Brad Watterworth using ohms law

[@ngvmanh7749]

source transformation easy to change current source to voltage source.

[@nommernomz]

so the v(t) that we found out is the voltage for the 10ohm resistor, is that right?

[@steechung]

Don't be too "stunned"... I see what you did there

[@AraujoMatt]

I really couldn't help myself

[@parkerflop]

Matthew Araujo Absolutely shocked. Electrified. Totally charged up for the test tomorrow. My Capacitance for learning is hopefully great. I am currently studying.

[@Neur0n911]

+steechung "May the force be with you." Harry Potter in Star Trek. I know my references. ;)

[@Avraham.Eisenberg]

I had stoned in mind....

[@christiankelleher4424]

Came to the comments for this

[@elviscardoz7280]

if the input source is given as a function of Sin, do we need to convert it into cosine function.???

[@monoham1]

impossible to understand. doent show equations on how to get voltage. you appear to use VR/R+XL but ive never heard of that. is it meant to be XL-XC/R^2 or maybe ZL=sqrt(R^2+XL^2) ? and what about phase for current source?

[@masonhuffman3480]

I'm wondering why we use "j" instead of "i." My professor never told us why and I can't find a reason in the textbook.

[@AraujoMatt]

This is a convention for electrical engineering to avoid confusion between i - which is used for current, and i - the imaginary number.

[@masonhuffman3480]

Matthew Araujo appreciate your help.

[@ngvmanh7749]

Matthew Araujo a good answer, :D

[@dooterino]

+Mfaniseni Thusi I think the primary reason is that "i/I" appears as the symbol for current in EE, so they just use 'j' instead. Though it probably does come from vector notation and the vertical axis frequently being the 'j-hat' dimension.

[@chrissantoso8280]

Super helpful. Thanks a ton!

[@emintorabi7115]

thank you very much sir, it really helped a lot

[@codyswisher3062]

I am trying to solve an RLC circuit with phasors for an abitrary value of omega (by leaving omega as a symbolic term). I get an extremely ugly value for the output voltage. Anyone know any tricks?

[@sijokuriakose3895]

Is there a reason why img part was left out ? -jsin(2500t-71)

[@arashkhm2000]

What did you use to write so comfortably? I am trying to do similar thing but it is super difficult to actually write using a mouse....

[@asdf35644]

You switched from a parallel circuit to a seriers circuit...without warning... could you explain that?

[@robalviani7752]

One thing I'm confused about here. I thought i(t) takes the form of Acos(wt-x), x being theta, no? In this case, your phase shift would be 45* not -45*.

[@KiwiKiller88]

why is it -14.14J didnt you multiply a negative by a negative?

[@malditoguero]

+KiwiKiller88 j*j = -1 Therefore, you'll get -14.14j-14.14 which he then just rewrites as -14.14-14.14j.

[@j0mezzy]

So main reason why we use phasors is to make the math easier.? Right

[@DescartesRenegade]

werent you looking for i(t)?

[@senoritasheh]

thanks alot for this lecture

[@Qrockpot]

How did you get -20j?

[@loviisamaenpaa6452]

Thank you!!

[@ejiroojirevwe2258]

thanks it really helped

[@jeffsam5495]

hey i got -2.82+8.82j !!! whats goin on here!!!

[@ishamael104]

There is only one way to study for a final... Watch Matt talk about stunning

[@amirmahdi_s6343]

thanks 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏

[@leezhihoong]

why sqrt2/2

[@jessmac1893]

Because the hypotenuse of the right triangle is 1, thus each of the sides are sqrt2/2

[@michaelromo2437]

Not sure if you're still active, but there might be a mistake here, pm me if you care to look over my work

[@kmbstar]

I love when you say that you must convince yourself that 1/j is -j. Made me lol. I see "you must convince yourself of yadda yadda" in physics books and circuits books and it always makes me laugh. Everything so technical and precise and factual and then suddenly, you must convince yourself of this, lol! You had a good explanation of it though. I have convinced myself of this -j of course. I always think -j/wL.

[@jcfor3ver]

(-14.14-14.14j)(10/10-20j) = -8.484-2.828j That is what my calc gave me which means your calculation was incorrect?

[@AraujoMatt]

you might have a typo on your calculator. I rechecked on my ti89 and I get the same result I did in the video

[@jcfor3ver]

yea nvm you're correct

[@jeffsam5495]

hey i got -2.82+8.82j !!! whats goin on here!!!

[@alexaculp4537]

This video is great except for the fact that you do not show the math for finding those complex numbers which is a huge part of the problem. The lack of that math renders this whole video essentially useless.

[@salehs3952]

6:15 OMG it all makes sense now My professor is a lazy piece of crap

[@praveenkumar-hr2nb]

kkkkkkkkkk

[@belovedsandworm]

Mathematica elitism