- Thread starter BatmanisWatching1987
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Where is the rest of the work?View attachment 22427View attachment 22428View attachment 22429

With Problem 111

I solved it by using the following formula

S = √3∠-30*V[SUB]L[/SUB]*I[SUB]L[/SUB]* = √3∠-30* (12.5∠0kV) * (70∠20) = 1515.54∠-10.

But in the other problem posted, why isn't there a 30 Degree Phase shift as well.

They are both 3 Phase Balanced Delta Loads.

It matters if you are using I_phase vs. I_line.

V_L in the problem is using V_LL voltage (perhaps the L they use is for "load"?). Maybe the I_L is I_phase.

I am pretty much sure this question was asked in a different thread and answered. When calculating 3 phase power (reactive, real or apparent) forget about the 30 degree shift between phase and line voltage/current. Only angle that matters is your load impedance angle which in this case is negative arc cos(0.85) = -31.8 = θView attachment 22427View attachment 22428View attachment 22429

With Problem 111

I solved it by using the following formula

S = √3∠-30*V[SUB]L[/SUB]*I[SUB]L[/SUB]* = √3∠-30* (12.5∠0kV) * (70∠20) = 1515.54∠-10.

But in the other problem posted, why isn't there a 30 Degree Phase shift as well.

They are both 3 Phase Balanced Delta Loads.

P=3XVφXIφXcos(θ)

Q=3XVφXIφXsin(θ)

S=3XVφXIφ

This isn't at all helpful, but I'm so glad that I don't have to study for the Power PE Exam again! I was doing what you're doing this time last year. Good luck!View attachment 22427View attachment 22428View attachment 22429

With Problem 111

I solved it by using the following formula

S = √3∠-30*V[SUB]L[/SUB]*I[SUB]L[/SUB]* = √3∠-30* (12.5∠0kV) * (70∠20) = 1515.54∠-10.

But in the other problem posted, why isn't there a 30 Degree Phase shift as well.

They are both 3 Phase Balanced Delta Loads.