# spin up exam -3 Q 17 & 18



## PE blues (Mar 17, 2013)

For calculating neutral currents, In=-(Ia+Ib+Ic). Can anyone explain why B is leading A by 120 and C is leading A by 240? I would assumed B lagging A by 120 and C lagging A by 240 for ABC sequence (assumed as phase sequence not given)


----------



## spinup (Mar 17, 2013)

PE blues said:


> For calculating neutral currents, In=-(Ia+Ib+Ic). Can anyone explain why B is leading A by 120 and C is leading A by 240? I would assumed B lagging A by 120 and C lagging A by 240 for ABC sequence (assumed as phase sequence not given)


PE blues,

Looks like you are using the *1st edition*, we are currently in the 2nd edition. There is an errata on the website www.spinupexams.com under the support tab.

*Solution 3-17*(Page 125) - Reported 3/5/12

Replace with IN =-( IA + IB + Ic)

=-( 54.8&lt;(0 -36.9° )+ 0.25&lt;(120°-5°) + 96&lt;(240°-36.8°))

=-83.4&lt;-122.3°

Change answer from "(A)" to "(E)"

*Solution 3-18*(Page 125) - Reported 3/5/12

Replace with IN =-( IB + Ic,)

=-( 0.25&lt;(120°-5°) + 96&lt;(240°-36.8°) )

=- 96&lt;-15.9°

Change answer from "(D)" to "(E)"

For the *second edition*: there is no change. 3-17 answer is (A), 3-18 answer is (D).

Feedback can be provided to: [email protected]

Good Luck on your studies!

Joan


----------



## PE blues (Mar 20, 2013)

my question is why 120 deg is used and not -120 deg?


----------



## PSUEngineer (Mar 29, 2013)

PE blues said:


> my question is why 120 deg is used and not -120 deg?


Think of it more like, there's a 120 degree separation between the phases;

Va &lt; 0

Vb &lt; 120

Vc &lt; 240

assuming positive sequencing.

When I run the math I get 83.4&lt;57.7...but according to the answers provided that's wrong; so if you divide their answer by that answer I get 1&lt;0 which leads me to say they're both right just a sign difference..

My question is how do you get their answer based off of 83.4&lt;57.7??


----------



## GreenNGold (Mar 16, 2014)

Can someone explain to me how this problem is different from PPI 1-67?

Here is the PPI problem: A distribution system is designed as a 3ph, 4 wire, neutral-grounded system. The phase-to-phase voltage is 12.8kV. The complex load power is expected to be 150+j75 kVA between phase A and neutral and between phase B and neutral. The expected neutral current due to the phase A and B, is most nearly:

A) 0A

B) 13A

C) 23A

D) 33A

Solution: -|In| = |Ia + Ib| = ((150+j75kVA)/(7.39kV&lt;0)) + ((150+j75kVA)/(7.39kV&lt;120)) = 22.70A&lt;-33.....Answer is C

Why is there an additional phase shift in the spin ups problem? Is it from the voltage angle? And if so, shouldn't that have changed the answer for problems 3-11 and 3-16?

Thank you.


----------



## Wheretostart (Apr 1, 2014)

my thought is, the solution is wrong, but the result is right.

for

Va &lt; 0

Vb &lt; 120

Vc &lt; 240

it is negative sequencing.....


----------

