Electrical Engineering: Ch 3: Circuit Analysis (12 of 37) Mesh Analysis w/ Voltage Sources: D = -336

In this video I will find the 3 currents of a circuit with 2 current sources using mesh analysis.

Next video in this series can be seen at:

[HamedShahbazi]

Hi professor I really appreciate you. I'm not electrician or not study subject close to but really like electronics and your teaching . I calculate 3 equations and find this 3 answers :
I(1)=-2.5714
I(2)=-7. 7143
I(3)=-4

[monkeyface]

Hello Michel, I checked the determinant of your calculations using MATLAB. MATLAB says the det should be -336. However when i checked your calculations, you calculated it to be -80, this is actually -160 not 80. With this correction your det would be correct. I absolute love the way you calculate the det, so much easier than the way I was shown.

[ericrodrigues]

the determinant of the first matrix should be -336 according to mine and the online determinant calculator. Is there a mistake in the video?

[samueltomol]

THANK YOU sir it really helps me a lot FROM PHILIPPINES

[JuanMorales-ipso]

the currents i4, i5 and i6 also need to be corrected, right?

[eyadsalah-eddine]

Hi Professor Van Biezen, I am a UC Berkeley student and I notices you have made a slight error with calculating your determinant for the system of linear equations. The determinant should equal -336

[johnjoelbiacan]

Sir Michel (6:59) 4X4X10= 160 not 80. and Thank you Sir for these lecture God Bless

[mareklewanski]

Hi all, what I got is I1=-2, 571A; I2=-7, 7142A; I3=-4, 00A; I4=1, 429A; I5=3, 714A; I6=-5, 1432A.

[BushraNaz]

I did this by calculator and have completely different answers for I1 I2 and I3. I1=-2.57A I2=-7.71 I3=-4.

[mareklewanski]

In this example, why did you assume voltage drop in exact node in MESH 3 ? You started from - not from +. I dont get that.

[jov]

can't believe we'd actually need it for 2021 classes hehe

[medeiroscr]

hi professor, should not the mesh I 1 be = to positive 16?

[veys]

Hi Professor, thank you for great work. I think when we add all three equations together at the beginning I1 and I2 cancel therefore I3 falls easily in our hands as -4.

[EmmanuelGbiaye]

Sir i was getting a different D so i decided to pay closer attention and saw that you made a mistake. You said 4×4×-10 was -80 instead of -160

[Tahycoon]

Hello professor,
Why don't we in general draw the + and - around the resistors? Do we assume them or can't we do it all?

best regards, keep up the great work!

[mohamudsali]

The determinant is not correct in that way. At least one product number is assumed as half-value that is why it's incorrect. 4*4*10= 160, but not 80.

[gamefreak]

Hi professor, my university teaches a different sign convention where for example where your one equation is -6I1+2I2+4I3=-16 mine would just be 6(I1)-2(I2)-4(I3)= +16 and this is how we are taught.

Now when I do Cramer's rule for a 3X3 matrix, instead of going to the right with the two extra lines I go downward and everything works out in the end, except my values differ from yours. this is how my university teaches it. except, in your video circuit analysis 7 I also did cramers rule bringing the cross multiplication downwards instead of to the right like you and I got exactly the same determinants as you did there with my sign convention also being the opposite of yours.when I determine all my currents though they all cancel out according to KCL. Is this acceptable?

so my first determinant would be derived from a matrix that looks like this:
|6 -2 -4 |
|-2 10 -8 |
|-4 -8 18 |
6 -2 4
-2 10 -8

Where my signs are just the opposite of yours and then D= (6x10x18 + -2x-8x-4 + -4x-2xx-8)-(-4x10x-4 + -8x-8x6 + 19x-2x-2) this gives me the answer of 464. is this fine?

Then D1= -5024, D2 = -2592 and D3= -1344 which gives me an I1 of -10.83A an I2 of -5.59A and an I 3 of -2.90A through this I get an I4 of -7.93A an I5 of -5.24 and an I6 of -2.69, they all obey KCL in the end so I don't think it is really a problem?

note that all my nodal and mesh analysis answers are the same as yours with the same exact signs apart from this cramers rule example, even though my equation signs differ from yours.

[solomongebreyesus]

I have used matris method and i get -13.58 , -14, 18 , -9, 33

[aishwaryaakavitha]

Sir, why didn't you assign current to each branch.with that only we could assign the signs in the kvl equation

[philippb]

Hello sir.

On an other of your videos "Electrical Engineering: Basic Laws (11 of 31) Kirchhoff's Laws: A Medium Example 2"
Shows us to solve it like this
if we add currents on thew problem it would be
LOOP1 16 -4*i1 - 2*i2 =0
LOOP2 2*i2 -8*i3 -40v = 0

can we solve it with this way ?
thank you