I think there's an error here. The equivalent T circuit choice depends only if the dots are on the same side, not if current is flowing into the dots on both sides. If the dots are both on the same side, you use the first T circuit. If the dots are on different sides you use the other T equivalent circuit. This is regardless of whichever way current is flowing. You actually explain it correctly in your video here: th-cam.com/video/QLMMDdnKKB0/w-d-xo.html It's just about whether or not the dots are on the same side, not about whether currents are flowing into the dots on both sides.
(17:40) The current I2 should have the other direction. The system of equations becomes: (1) (700+3700j)*I1- I2*1200j = 300 (2) -I1*1200j + (900-900j)*I2 = 0 Leading to I1 = 0.063 ∠-71.6º and I2 = 0.060 ∠63.4º
Thanks for the explanation!!
I think there's an error here. The equivalent T circuit choice depends only if the dots are on the same side, not if current is flowing into the dots on both sides. If the dots are both on the same side, you use the first T circuit. If the dots are on different sides you use the other T equivalent circuit. This is regardless of whichever way current is flowing.
You actually explain it correctly in your video here:
th-cam.com/video/QLMMDdnKKB0/w-d-xo.html
It's just about whether or not the dots are on the same side, not about whether currents are flowing into the dots on both sides.
Thx very much! Excellent
(17:40) The current I2 should have the other direction. The system of equations becomes:
(1) (700+3700j)*I1- I2*1200j = 300
(2) -I1*1200j + (900-900j)*I2 = 0
Leading to I1 = 0.063 ∠-71.6º and I2 = 0.060 ∠63.4º
Concerning I2 the mistake is that since the current is entering the sign of -1200J you have to write (900+J300)I2+1200J(I2-I1)=0
I think that on 12:36 i2 on equivalent T must go towards left
Sir how did you get L1-M and L2-M?? Please tell
Sir, which book do you refer to?