Given:

$V_L$ = 415 V

$Z_{ph}$ = 4.2 -j 5.6 $\Omega$

$V_{ph}$= ?

$I_{ph}$ = ?

P.f = ?

Formulae:

(1) $V_L = \sqrt{3}V_{ph}$

(2) $V_{ph} =I_{ph}.Z_{ph}$

(3) P.f :$= \cos \phi_2$

Solution:

(1) $V_{ph}=\dfrac{V_L}{\sqrt{3}} = \dfrac{415}{\sqrt{3}} = 239.6 V$

(2) $I_{ph}=\dfrac{V_{ph}}{Z_{ph}} = \dfrac{239.6\angle0}{(4.2-j5.6)}=\dfrac{239.6\angle0}{7 \angle -53.13} = 34.24\angle[0-(-53.13)]$

(3) $P.f = \cos \phi_2 = \cot 53.13 =0.6$