208 three phase

[domnic]

208 THREE PHASE .120 volt loads . line a 4 amps. line b 3 amps .line c 6 amps. amps on neutral 2.6 amps. is this correct .

 

[adamscb]

Seems reasonable to me. The current flowing on the neutral is showing that there's some load imbalance, but nothing to get worried about.

 

[jumper]

 

[domnic]

208 three pase

208 three pase

208 v . line a 10 amps. line b 10 amps. amps on neutral 5 amps. ?

 

[jumper]

208 v . line a 10 amps. line b 10 amps. amps on neutral 5 amps. ?

Nope. if C phase was 0A, N=10A

 

[Ingenieur]

using 3, 4 and 6
n = 2.65

 

[domnic]

208 three phase

208 three phase. line a 8 amps. line b 3 amps. neutral amps = ?

 

[Ingenieur]

208 three phase. line a 8 amps. line b 3 amps. neutral amps = ?

7

 

[jumper]

7

???

I get ~7.87, max imbalance between any 2 phases,no?

C=0A.

 

[Ingenieur]

???

I get ~7.87, max imbalance between any 2 phases,no?

C=0A.

sqrt(64 + 9 - (3 x 8)) = sqrt(49)
???

 

[jumper]

sqrt(64 + 9 - (3 x 8)) = sqrt(49)
???

:slaphead:

Maybe i shouldn’t have added 8 and 3.:ashamed1:

 

[Ingenieur]

8/0 deg = 8 + 0j
3/120 deg = -1.5 + 2.598j
add
n = 6.5 + 2.598j = 7/22 deg

 

[Ingenieur]

:slaphead:

Maybe i shouldn’t have added 8 and 3.:ashamed1:

I've got plenty of practice being 'wrong' (I'm married)
it doesn't bother me anymore
lol

 

[jumper]

For the OP:

For A=15, B=10, and C=5; what do you get for N?

 

[jumper]

I've got plenty of practice being 'wrong' (I'm married)
it doesn't bother me anymore
lol

Yeah me too.

No calculator on ipad, was using an online one and spaced out doing formula.:dunce:

 

[JFletcher]

208 THREE PHASE .120 volt loads . line a 4 amps. line b 3 amps .line c 6 amps. amps on neutral 2.6 amps. is this correct .

4² + 3² + 6² = 61
4x3 + 3x6 + 4x6 = 54
61-54 = 7
sqrt 7 = 2.6

yes, you're correct

 

[LarryFine]

208 v . line a 10 amps. line b 10 amps. amps on neutral 5 amps. ?

If you had three 10 amp loads, neutral current would obviously be 0 amps. For every amp you deduct from one phase, the neutral current would rise the same one amp. Deduct 10 amps, neutral current rises 10 amps

 

[wwhitney]

The formula posted is only for resistive loads, yes?

Cheers, Wayne

 

[Ingenieur]

The formula posted is only for resistive loads, yes?

Cheers, Wayne

yes
or at least of similar pf, ie, phase shift due to complex Z is ~ the same

 

[jlee405]

In=3Io
Io=(1/3)(Ia+Ib+Ic)
In= Ia<0 + Ib<120 + Ic<240