I am stuck on a question similar to the delta question I asked:

Three devices are connected in wye to a three-phase four-wire 240 Volt supply.

• A-N Z = 30 ohms pf = .9 leading
• B-N Z = 20 ohms pf = .8 lagging
• C-N Z = 10 ohms pf = 1

Calculate the neutral current.


I know that the neutral current will be the vector difference of currents ABC.

First I need to find the currents:

Ia= 240/_0 deg / 30 /_25.84 = 8_/-25.84A

Ib=240/_120 / 20/_36.87 = 12_/83.13A

Ic=240/_240 / 10/_0 = 24/_0A

Next I find the difference between the currents:

In= 7.2-j3.49 - 1.435+j11.91 - -12-j20.78 = 17.765+j12.36 = 21.64/_34.82A

The answer given for the question is 14.6A????

 

Any help would be appreciated. I realised the Phase to N voltage would be 138.56V instead of 240V but tried the same calculations and still did not get the correct answer. Any ideas anyone?

 

You angles are out of kilter. For instance your A phase current has a lagging phase angle when it is explicitly stated that the A phase has leading power factor.

Referring to the attached solution - While I use different phase rotation than yours, you may be able to see how the solution would evolve using your preferred notation.

 

Thanks t_n_k I was so close but had my +/- on my angles all messed up. I appreciate your time and help.