Delta Unbalanced Resistive Load

[@oliver_peng]

Thanks for the video! I believe for Line C's current, you meant to write 17.44A @83 (instead of 19.44A)

[@stevebyra7795]

Thank you for this! I was wondering if ai was crazy

[@Patrick-kickass]

Very helpful!

[@franciskailondo920]

This helped alot. Thank you. Definitely subscribing

[@warrenmcdonald7643]

What does this do to the voltage drop across each leg?

[@sohailjanjua123]

Hi, I like your lecture.Thanks

[@alexparadi522]

Thank you for your videos! I'm hoping you (or someone else) can help point me to the answers to some of my questions. What I'm most curious about is how the different lines from the power source (3 phase generator for example) can carry both the 'outbound' current for the loads that are on their line as well as the 'return' current for the other phases without disrupting the outbound current in any way. I'm also not sure I understand why you would get any current between A and C. If electricity always follow the path of least resistance, isn't the current supplied by line A going to go through the A/B resistor and return back along line B? And the similarly the current from Line C is going to go through the C/B resistor and return back along line B? I'm clearly struggling to understand some fundamental concepts that would allow me to mentally understand why a 4th wire is not required with Delta 3 phase loads (balanced or unbalanced).

[@ZackHartle]

Kind of a tough this to explain. Remember most current will take the path of least resistance, but not all the current is going that way. It will still split as there is still a potential difference between those points

[@UnicornAdvisory]

Is there another variation to the law of cosines that's used for finding neutral currents, that can be applied to the Delta systems?

[@jaredpoels]

From the HV Chart, what do you do with those numbers to get the line current?

[@MrTheta..]

I have this exact problem at work. The currents you solved for, are those what I would see with an ammeter on each phase, A, B, C? Looking at the resistance values, the currents are proportional to them. I guess another way of asking would be, can't you just leave off the phase angle? To me, this seems like a snapshot in time, with specific phase angles. Is this the same as using DC equivalent? I can calculate the current if each side, but I know things are not in phase, so the total on one leg will be less than the sum. Is this what that is solving for? The ammeter reading?

[@ZackHartle]

Yes, you're solving for what you would read on an RMS ammeter. The first half is phase currents and the second half is Line currents. Phase angles will play a bigger role in circuits that have inductive or capacitive values