wye connected loads

[bob]

Not to nit-pick, but that is Nodal-analysis via Kirchhoff's Current Law, not Ohm's Law.

Rick you are right. I was watching the news and typing at the same time. No excuse.

 

[Rick Christopherson]

They will cancel. No need to return on the other phases.

Oh, I didn't notice this last but very major faux pas. As I already said earlier in this thread, currents don't simply cancel and obliterate each other. Kirchhoff states that the sum of the currents entering a node will be zero, but that means that some currents will be entering and some will be leaving. Smart $ was correct in what/how he stated his previous message.

Doing a Nodal analysis will cause many people to draw this erroneous conclusion, so for this situation, you will understand it better if you performed a mesh analysis instead.

I already mentioned this earlier, but I don't know which posting it was from.

 

[LarryFine]

Rick you are right. I was watching the news and typing at the same time. No excuse.

I can't wait to see you type while changing channels!

"According to MSNBS's Law, . . . " :grin:

 

[bob]

This is some more informmation on this subject should you care to look.

http://forums.mikeholt.com/showthread.php?t=93575

 

[david luchini]

Was the example of:

Line A, 5A, pf=1
Line B, 10A, pf=1
Line C, 8.67A, pf=0.867

meant to illustrate an unbalanced condition where the neutral current will be zero, or an unbalanced condition where the neutral will carry the unbalanced load?

By my calculation, if the power factor is LEADING then the neutral will carry 8.67 amps, but if the power factor is LAGGING then the neutral current will be zero amps. The example didn't specify if the power factor was leading or lagging.