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.

[@anujregmi4582]

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

[@01HANKING]

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

[@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

[@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.

[@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

[@2mjz84]

Thanks man, 2024 and still helped me :D

[@loganmasters3507]

Very helpful, well done and concise.

[@YaJohny_]

You are the best!

[@LT1]

Intro was very helpful. Good stuff.

[@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.

[@AraujoMatt]

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

[@teezo3afana]

YOU my man are amazing .

[@jordanbendler8205]

Very helpful brotato. Thanks.

[@ComputerAtSea]

Nice job, chief!

[@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

[@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

[@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

[@crashonthehumble]

Outstanding lecture

[@chrissantoso8280]

Super helpful. Thanks a ton!

[@loviisamaenpaa6452]

Thank you!!

[@oscarlaruta2008]

which book are u using? good video

[@emintorabi7115]

thank you very much sir, it really helped a lot

[@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.

[@senoritasheh]

thanks alot for this lecture

[@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.

[@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!

[@AraujoMatt]

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

[@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.

[@nommernomz]

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

[@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 ??

[@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?

[@elviscardoz7280]

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

[@sijokuriakose3895]

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

[@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?

[@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.

[@amirmahdi_s6343]

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

[@ejiroojirevwe2258]

thanks it really helped

[@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*.

[@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....

[@jeffsam5495]

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

[@Qrockpot]

How did you get -20j?

[@j0mezzy]

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

[@DescartesRenegade]

werent you looking for i(t)?

[@asdf35644]

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

[@ishamael104]

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

[@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

[@leezhihoong]

why sqrt2/2

[@jessmac1893]

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

[@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