Calculate Currents & Power in 3-Phase Systems w/ Star-Connected Load

[Mathn00b!]

Homework Statement

Hello, I am given the following problem.
A balanced three phase 4 wire supply has r.m.s. line voltage of 415V and supplies a star-connected load made of 3 impedances:
Za = 10 - j10; Zb = 10 + j10; Zc = 0 + j8;

Calculate the current phase and neutral currents, also the total power dissipation. [NOTE: Assume that Va is drawn vertically up, i.e. its phase angle is 90]

Homework Equations

I've used V_line = √3 x V_phase
Also, P = E_a x I _a x cos(Φ_a) + E_b x I _b x cos(Φ_b) + E_c x I _c x cos(Φ_c)

The Attempt at a Solution

Please find the attached file, as my solution to the problem. As it could be seen the answer is quite different. In fact their answer is just the sum of the 1st two bits of the power, ignoring the last one. I assume they used cos(90) to get 0 and hence lower power. If this is the case, then the Z_c should look like 8∠90° .
But isn't it found from the equation : Z_c = √0² + 8² ∠tan^-1 (8/0) ?

http://i1375.photobucket.com/albums/ag441/gl0ck1/StarLoad_zpslnhuotig.jpg

 

[Hesch]

Please find the attached file, as my solution to the problem.

Error: The requested attachment could not be found.

 

[Hesch]

You must find the center voltage, E_n, at the center of the star-load.

Then:

I_a = ( E_a - E_n ) / Z_a , etc.

( I think? I cannot see if the center of the star-load is connected to neutral )

 

[Mathn00b!]

Strange, just sent it to one of my friends, he was able to open it. Will post a photobucket link just in a sec.
I am asked to find the neutral current, which could be found from the sum of all currents, no?

 

[Hesch]

I've been able to open it now (error disappeared).

 

[Hesch]

You may argue: If neutral is not connected to center of the star-load, then I_n = 0.
But I've found an error: Zc = ( 0 + j8 ) = 8 / 90°

 

[Mathn00b!]

Could you please, explain to me how you found it to be 8 ∠90°? What formula did you use? Isnt't it the same I used at the 1st post of the thread Zc = √0² + 8² ∠tan-1 (8/0) ? and tan^-1 (0) = 0.
I presume it is connected since I've been asked to find it.

 

[Hesch]

I'm not using any formula: Draw the vector ( 0 + j8 ) in the complex plane and measure the angle ( or you can see it by intuition ).

( Contrary: ( 8 + j0 ) = 8 / 0° )

 

[gneill]

The angle you've associated with Zc is incorrect. As a result the angle calculated for $I_c$ is incorrect.
The angle calculated for $I_a$ is incorrect.

 

[Mathn00b!]

The angle you've associated with Zc is incorrect. As a result the angle calculated for $I_c$ is incorrect.
The angle calculated for $I_a$ is incorrect.

Yes you are right, it should be 135, but really don't understand the Zc's angle. Should I do it always like that? Drawing a vector, when the real part is zero?

 

[gneill]

Yes you are right, it should be 135, but really don't understand the Zc's angle. Should I do it always like that? Drawing a vector, when the real part is zero?

It certainly can be helpful to sketch a vector for a complex value in order to get an idea of what the angle should be (such as the quadrant it lies in). But you should know that any purely imaginary number must have an angle of either + or - 90 degrees.

 

[Mathn00b!]

Thank you both for the replies!