AC Example-Nodal Analysis

Analyzes a circuit using nodal analysis find an output voltage using AC steady State techniques. More instructional engineering videos can be found at www.engineering....
This video is licensed under the Creative Commons BY-SA license creativecommons....

[@matsvandamme1]

Thank you! I'm a Belgian Electrical Engineering student and this was very helpful to me

[@md.shajibhossain9999]

Thank you so much! I have a class test tomorrow and you lecture really helped me a lot to understand very easily

[@mihaitensor]

thank you..i'm a very grateful student ..you're the teacher that should explain Electrical Engineering in universities..

[@merseyless]

That 3ohm resistance was a nice resistance. Thankyou for making it that number.

[@alexanderjack5937]

Very good explanation if I'll do problems at home, but if I were at exam I'll struggle with these complex numbers when I put them in a matrix form. In general great work and voice.

[@ltmillerII]

thank you, for that nice number!

[@xoraxera]

Thanks you helped me a lot! Keep up the good work.

[@sirr3d488]

Hi, I have an issue. If i try to change the cos to sin by the formula sin(x)=cos(x-90) and work with sin x, => I1=5

[@HamzasaeedKhan]

hello sir I did all the steps but Walframe didn't show me V1 and V2 values but upto Alternate form, is this any problem? v1(1/(i10)+1/(-i628.3))-v2(1/(-i628.3))=-i2,-v1(1/(-i628.3))+v2(1/(i31.4)+1/(-i628.3))=1.5

[@raghad8475]

same :(

[@DarrylMorrell]

1/(377 633x10^(-6) ) gives the value of 4. 1/j gives the -j.

[@effect2ful]

how we solve it when the sources' frequencies are different????????

[@jo6ex21]

It can not be solved

[@rustman888]

How do you know which omega to use if you have two sources and the impedance isn't given?

[@rustman888]

I know you could use superposition principle, but you can't always do that right? Say if you have dependent sources.

[@EvilMrMahler69]

your tone of voice is perfectly apathetic

[@drastikdynasty]

That's cool, too bad we can't use wolfram alpha on an exam

[@Falcrist]

drastikdynasty Get a calculator with complex matrix capability. The TI 89 and HP 50g are options that most engineering teachers won't bother you about. The HP Prime is what I've been using (VERY highly recommended). The HP 35s is a scientific calculator has a few programs that solve systems of linear equations with complex coefficients. It is currently allowed on the Fundamentals of Engineering exam. The TI-68 and HP-42s also handle complex valued linear systems of equations... but they're expensive/rare/vintage. With such a calculator, the procedure is as follows: 1) Write the equations for each node (or mesh) on paper in standard form: "(a+jb)x + (c+jd)y = z" 2) create a coefficient matrix "A" and a column vector "b" (consists of the z values) 3) Now the system can be written as Ax=b, where x is the variable matrix. You just have to solve for x (so "x=A^(-1)*b" ) Basically, you just enter "A^(-1)*b" into your calculator, and the answer will be given to you in the form of a column vector where the top number is the first variable, and the bottom number is the last variable. If you're using RPN entry, put matrix b on the stack, then put matrix A below it, and hit divide.

[@bishenpatel1676]

+Falcrist Thanks for the info mate... Will try to find those calculators and see if they have it in store..

[@camy1005]

+Falcrist Can you tell me how you solve equations with complex values on your prime? I can not get mine to solve any equations with phasors (either in polar or rectangular form) in them. An example would be: X(50+40+j*497.418+330)+7=0, I can not get any value for X unless I work it by hand and that kinda defeats the point of the calculator.

[@Falcrist]

Cameron Parette The solve function you seem to be looking for is Linear System (linsolve). With that, you simply make a vector, and enter each equation into an element in that vector, then create a vector with each of the variables in it like so. linsolve( [equation1 equation2 equation3], [variable1 variable2 variable3] ) More information can be found by highlighting the Linear Solve function in the menu (toolbox>CAS>Solve>Linear System) and hitting the Help button. However, the easiest way to solve systems of complex equations is to use matrices: First, make an N×1 matrix and an N×N matrix (where N is the number of variables). Each row of the matrices represents one of the equations in the system. The N×N matrix is called the coefficient matrix (denoted as A). It holds the coefficients of each variable in each of the equations. The N×1 matrix (denoted by b) holds the non-homogeneous part of each equation... that is, any number that isn't a coefficient should be manipulated so that it's on the other side of the equals sign from the variables, then those numbers are put into the column matrix. There's actually a third matrix involved that contains the actual variables. It's also N×1, and is denoted by x. The entire system of equations is thus represented as *Ax=b*. If you multiply it out, you'll actually get the original system of equations back. Example (3 variable system with 3 equations): Equation 1: ax+by+cz=0 Equation 2: dx+ey+fz=1 Equation 3: gx+hy+iz=2 NOTE: these equations are in a sort of standard form (each column contains only 1 variable and it's coefficients). You may have to do some minor calculation to get this form. Coefficient matrix (A): a b c d e f g h i Variable matrix (x): x y z Non-homogeneous matrix (b): 0 1 2 Finally we can do the result matrix. By basic algebra, if Ax=b, then it follows that x=b/A... though with matrices, it is proper to write x=(A^-1)b. It doesn't matter, since your calculator will accept fractions with matrices in them anyway. So literally, all you have to do to get the answers is compute b/A or (A^-1)b. The results will pop out in matrix form such that the first (top) element of the result matrix corresponds to the first (top) variable in the variable matrix. In this case if our results are 10 20 30 It means that x=10, y=20, and z=30. Alternatively, you can make A and b into one big matrix (called an augmented matrix) like: a b c 0 d e f 1 g h i 2 and ask for the reduced row echelon form (RREF). The result will be (assuming the same answer as above): 1 0 0 10 0 1 0 20 0 0 1 30 Where that last column holds the answers you're looking for. In this way, you can quickly and easily solve systems of equations with relatively large numbers of variables. All of this is from linear algebra. I've been somewhat careful to use correct terminology, so that it's easy for you to find other versions of this explanation.

[@camy1005]

+Falcrist I am still trying things people have been telling me, but the matrix method you show here gives consistent repeatable results so thank you for giving me a way to pass my tests I appreciate it!

[@pentamk]

Why not the impedance -0.4 ohm? why - 4 ohm ?

[@BLUFFIRL]

what if the frequencies are different

[@daniyal241585]

why is it going to be -j4?

[@kara96raza]

stop using wolform alpha i want to learn how to solve it by my self

[@14macktactoe141]

+Kara Raza just learn how to convert from polar to rectangle and how to solve a matrix

[@2011Thesituation]

lol great vid work the maths wit a scientific calculator an step by step most of my marks is doing that -_-

[@photon2724]

we have to do it by hand in our exam............

[@JohnReiEnriquez]

Hi, instead of wolfram alpha, I would suggest Microsoft Mathematics :D It's Free!

[@cstevenson1234]

What I wanted to learn from this video was how to solve without the use of a computational program: gonna have to dislike.

[@Ghoombagage]

Cramers Rule. It's gonna be a bitch though

[@roteuxwer]

Then go on a different video that will show you how to solve equations. Because this is a Circuit analysis lesson, not a math lesson.

[@garen5594]

"finding the equations are going to be easy". hey buddy, is this or is this not suppose to teach us? im sorry for being cranky, but it must be assumed that we dont know anything. IT would be nice if you could quote kirchoff's laws and explaining more how you get the equations. whelp, back to more research