# Exam Question Help

994 6
Just wondering if someone out there could assist me with this question...

A three phase motor with a FLA of 14A , service factor of 1.15 and PF of .8 lag, is connected to a 460V, 60Hz supply. A capacitor bank of 2386 VAR is connected to the load side of the motor controller at the motor terminals. Calculate the maximum setting of the overload device in this controller.

I want to say 17.5A (14x1.25) but that is not the answer and I can't figure out how to include the capacitor bank in the formula as it should be decreasing the current draw and increasing the PF, any help would be great, thanks.
1 - 7 of 7 Posts

#### sparky=t

· Bill
Joined
·
114 Posts
What is the maximum allowable per article 430?, you have to read the question & understand what they are asking, not what you think they are asking.....

#### macaber

· Registered
Joined
·
22 Posts
Discussion Starter · ·
What is the maximum allowable per article 430?, you have to read the question & understand what they are asking, not what you think they are asking.....
Right, but this doesn't get me any farther ahead.

· Bill
Joined
·
114 Posts
Duty rating?

#### just the cowboy

· Registered
Unemployed
Joined
·
4,946 Posts
Service factor

They don't alway make motors in the size you need, so they use a service factor which is how many % over the rated horsepower it is good for. Try 1.15 times the current first and see if that works

#### macaber

· Registered
Joined
·
22 Posts
Discussion Starter · ·
They don't alway make motors in the size you need, so they use a service factor which is how many % over the rated horsepower it is good for. Try 1.15 times the current first and see if that works
****, forgot to mention that I am working off of the CEC, not sure how much different our books are, I should have posted this in the CEC forum, sorry.

#### sraistrick

· Registered
Joined
·
1 Posts
Hey maybe a bit late but maybe this is it

460v x 14a= 6440VA

make your power triangle with that as the hypotenuse then

6440 x .8pf =5152Watts put this on the bottom of the same triangle then

Pythagoras: 6440(2) -5152(2) = square root the answer = 3864 VARS.. add this to vertical or triangle

those are Vars of inductance so if you subtract the added vars of capacitance:

3864-2386=1478 VARS

if you do pythagoras again with the same VA and the new VARS then:

6440(2) - 1478(2) = sqr rt = 6268 Watts

6268/6440=.97pf

hope it helps a bit, I'm doing 4th year right now, IP in a week

1 - 7 of 7 Posts