Electrical Engineering: Ch 12 AC Power (12 of 38) How to Find Power Using Mesh Analysis Part 1

[@derechosapollo6362]

VL is (105.8, -169.11degrees)

[@BentHestad]

Superb stuff, like allways! I bow and thank! This channel really has helped me a lot, I am happy to see the number of subs are increasing very steadily, highly deserved so indeed. Why not every even slightly curious person on earth is a sub is a puzzle for me. For the time beiing I struggle quite a bit with magnetically coupled circles, like two inductors coupled within same circuit, hope for something around this theme as well at some point. Excellent professors here at NTNU in Trondheim also, I am both proud of them and grateful for their great work, but to be on professor van Biezens level you really got to get up early in the morning, every day:-)

[@MichelvanBiezen]

Thank you for the comment. If you are looking for mutual inductance you can find that here: PHYSICS 47 INDUCTANCE & INDUCTORS

[@affection231]

Hey Michel, I believe you cannot factor out the negative from the complex number at the end of the video, because magnitude was + and phase angle was - but after factoring out again the magnitude is + and phase angle is - so it means you have disrupted the values and added a redundant negative

[@MichelvanBiezen]

There is mathematically nothing wrong with factoring out a negative since it still has the exact same meaning. It becomes a matter of personal preference.

[@noahelekhtra9456]

Hello I really admire your work and hope you continue to upload great lessons. I write because I wanted to ask how to convert current into complex numbers I.e. 8:45 on the video. Or if you have a videos about that then can I have a link please. Thank you again for all the work you do for us.

[@MichelvanBiezen]

Noah, This playlist may help you understand complex numbers and phasors: ELECTRICAL ENGINEERING 10 PHASORS, PHASOR DIAGRAMS IN RCL CIRCUITS

[@Hot-Introduction]

Hello Sir, At 13:14 if I don't pull out the -ve sign and convert it into magnitude/_ phasor representation it comes out to be different (+105/_10.89) which changes the voltage drop across the current source (calculated in the next part) to be very different. Please tell me if this is correct.

[@MichelvanBiezen]

We should get the same answer. (With these types of problems, it is easy to make a little mistake somewhere).

[@kannanr9016]

Thank you sir, one doubt sir , V (L) = (i1-i2)X(L) sir because V (L) consider loop2 based i2 is absorbed i.e. voltage drop sir but you wrote i2-i1 ,this case i.e. your wrote case i2 is voltage rise (loop2) sir. Please adjust my English conveying sir ..

[@alinoory5236]

Dr. Michel, I have question about mesh 1, if you were doing kcl why you didn't go through the nodes or if it was kvl why you didn't go through each component? in the second mesh you added up all the components and made them equal to zero which mean you did kvl around the loop. please tell me what is going on in mesh 1. thanks

[@MichelvanBiezen]

In mesh 1 we have a current source which will force that current no matter what components you have in that loop or what happens with the components in the loop

[@alinoory5236]

@@MichelvanBiezen I see now. Thank you again.

[@pidaras_pidarasina]

Isn't capacitor should have 90 degrees shift, while induction -90 degrees?

[@MichelvanBiezen]

The numbers used in the video for the capacitor and inductor are correct.

[@ajj7794]

Hello, *THANK YOU FOR YOUR WORK*, But... I am having some problems, you used M2 yo find I2, i used M1 since we know I1 thus i get M1=-20*I1 - J10*I1 + J10*I2 = 0 thus now using I1 = *[ 4¦_0 * 20 ¦_0 ]* - _( 10¦_90 _*_4¦_0 )_** + [ 10¦_90 * I2 ]* =0 thus I2=(80+J40)/10¦_90 = 89.44¦_26.57 / 10¦_90 = 8.94¦_-63.43, At the end you have Current for L= 2+J10.39 - 4 = -2+J10.39 but you says its - (2-J10.39) which is correct but because of this you get the cartesian form to be -10.58¦_-79.11 thus your voltage is -105.80¦_10.9 but if you keep VL as -2+ J10.39 you get the cartesian form =10.58¦_-79.11 thus your voltage 105.8¦_10.9 which is what i have. Where am i wrong,for both, please help :D

[@rockyjoe3817]

Mr please reply to my previous comment :(