Given

Delta Connection

P = 12000W

VL=440V

P.f = 0.7(lead)

Find:

R = ?

X = ?

Q = ?

Formulae

$V_L = V_{ph} I_L= \sqrt{3} I_{ph} \\[2ex] P= \sqrt{3} V_L \ I_L \ cos\phi \\[2ex] p.f = cos \phi $

Solution

$P = \sqrt{3} V_L I _L \ Cos \phi \Rightarrow 12000= 440 \times I_L \times 0.7 \Rightarrow I_L =39A\\[2ex] \therefore I_{ph} = I_L / \sqrt{3} =22.5 A \\[2ex] \therefore Z_{ph} = \dfrac {V_{ph}}{I_{ph}} = \dfrac {440}{22.5}=19.5 \Omega$

from Impendence triangle ,

$\phi = cos ^{-1} \ 0.7 =45 ^\circ\\[2ex] \therefore R = Z \cos \phi = 19.5 \times 0.9 = 13.65 \Omega \\[2ex] \therefore X = Z \sin \phi = 19.5 \times sin45 = 12.66 \Omega$

From Power triangle

$\therefore tan \phi = \dfrac{Q}{P} \Rightarrow = P \ tan \phi\\[2ex] =12 tan \phi =12 tan 45\\[2ex] =12 KVA$