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If p (3phase) = Vline x Iline x sqrt(3), how does S = Vline x Iline x sqrt(3)?

FOR 3PHASE: P = v X I X PF X SQRT(3) FOR 1PHASE P = V X I X PF 1) ARE THESE EQUATIONS CALCULATING POWER ABSORBED BY LOAD OR POWER SUPPLIED BY SOURCE? 2) Also, in the following equations: P_phase = E_phase x I_phase x cos(theta).....the book says its the power dissipated in each phase of...

Thanks, I see. One mroe question on this stuff....if its listed as 120/240 single phase...this means V_ll = 240 and V_ln = 120 correct?

What about Delta 3-wire system? It doesnt have a high leg correct? Also, it will be listed something like 120/208-240 for a y-delta correct....thus nothaving the 120/240...or similar ratio?

In the above example, it says the load is 19 KVA.....is there a way to calculate the transformer KVA from the given information? Also, for 3 phase transformers, can you calculate its S value from : S= 3 x phase voltage x phase current and for single phase for both WYE and DELTA systems?

WILDSOLDIER THE 4-WIRE DELTA DELTA ALWAYS HAS THE HIGHLEG ASSOCIATED WITH IT, CORRECT? Or the side that has 4 wires that is? in your diagram you show two of the phases to neutral = 120 while one of the phases to neutral = 208.......phase c in your diagram.....you also show line to line...

Yes, I took the NCEES was all, but I could do the whole exam in under four hours, I remembered it verbatim. I think I got confused on some things like single line diagrams...determinint flow from generators to transformers and vice versa

For 3 phase then: S = V x I* x sqrt(3) P(KW)= S x PF = V x I* x PF x sqrt(3) ?

I asked a question similar to this earlier..but it was w/o a pf....I have a 19KVA load supplie by a 208/120, 4-wire, 3-phase, wye connected system....with a 0.85 pf....what is the current delivered to the load? I know w/o a power factor....I = 19kva/(208*(sqrt(3)))= 52.7 amps..but how is it...

oK, i SEE NOW THAT THE SUM OF THE RMS CURRENTS DELIVERED TO THE LOAD = 0 FOR A 3-PHASE SYSTEM...BASICALLY CANCELING EACH OTHER OUT WITH THEIR NEGATIVE AND POSITIVE HALY CYCLES OF EACH PHASE....BUT WHAT ABOUT SINGLE PHASE SYSTEMS? WHEN THE SINGLE PHASE CURRENT IS ALTERNATING HOW DOES THE LOAD RUN...

Is the line current the rms current or is it the peak current?

Ok, I see what your saying about being out of phase. But the 52.7 amps are on each phase correct? Also, Even if the currents are not in phase, is their a way to calculate how much current is being deliver to the load at any given time? Is there a min current the 3 phases must provide to the load...

Is this 52.7 amps of line current in all three phases of the balanced 3 phase system? Does it mean the total current being delivered to the load is 3*52.7 amps = 158.1 amps?

Im not sure what you mean depending on whether its 208 V or 120 V...to me..V_ll = 208....V_ln = 120 V....and if you have one phase, you model it phase to neutral using the 120 V....If it were on 208 V, that would be either V_ll for a 2 or 3 phase load..bout not a single phase....which is what I...

EX: I have a S= 19 KVA nonlinear load supplies by a 208/120, 4-wire, 3-phase, wye connected system.....what is the current delivered to the load? I did like this I = S3phase/ (V_LL*sqrt(3)) = 19000/(208*1.732) = 52.7 Amps, is this correct? Also, does this mean its the current delivered by all...

tHANKS FOR CLEARING IT UP FOR ME gENTLEMAN.

BASED ON WHATEVERYONE SAYS INT HIS FORUM, YOU HAVE WHAT YOU NEED...I HAVE PPI AND ITS GOOD BUT ITS NO DIFFERENT THAN THE OTHER...ALL PRACTIVE PROBLEMS ARE GOOD.

Thanks for clearing this up for me gentleman

WHAT IS THE PURPOSE OF THE INTERLOCKS? Also, does thie mean, when normally open contact M1 closes that normally closed, contact M1 on the reverse rung opens? Thanks!

Attached is a motor control circuit...with latching. I have a question about how it works. If I push the forward button...the M1 auxiliary contact will close...energizing coil M1...but what happens to normally closed M2 contact when I pust forward button? Does it open? If so would this not keep...