# 3 Phase Series Equivalents (Wye & Delta)

Discussion in 'Homework Help' started by jegues, Oct 12, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
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See first figure attached for my solution to the series portion of the problem.

The answer they give to the problem is,

$Z_{series} = (1.84 - j1.38) \Omega$

I would agree if this was a Y connection but it asked for the equivalent for a delta connection.

I first calculated the equivalent for a Y connection and converted it to a delta afterwards.

I based my reasoning on one of the other examples given in my text. (See second figure attached)

Is the solution they give incorrect or am I misunderstanding something?

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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My approach would be to find the parallel delta configuration first. I assumed the 208V is the line-to-line rms voltage.

Given the power per phase is 5kW then the delta (per phase) parallel resistance will be

Rp=208^2/5000=8.6528Ω

Zp=Rp||-jXc [capacitive since it's a leading power factor]

$Z_p=\frac{1}{\frac{1}{R_p}+\frac{1}{-jX_c}}=\frac{R_pX_c}{X_c+jR_p}$

$arccos (pf)=36.87^o=arctan(\frac{R_p}{X_c})$

or

$R_p=tan(36.87^o)X_c$

or

$X_c=\frac{R_p}{tan(36.87^o)}=\frac{8.6528}{0.75}=11.537 \Omega$

So Zp = 8.6528Ω||-j11.537Ω = 5.538-j4.153 Ω

The per phase delta series equivalent of the parallel combination would therefore simply be

Rs=5.538Ω and Xs=4.153 Ω (capacitive reactance)

So our answers agree, although you didn't derive the parallel equivalent for part (b)

Last edited: Oct 13, 2011