$P=10\ kW\\ V_L=440\ V\\ pf=0.6\ (lead)$

For a delta - connected load,

(i) Values of circuit elements,

$V_L=V_{ph}=440\ V$

$P=\sqrt 3\ V_L\ I_L\ cos\phi \\ 10\times 10^3=\sqrt 3\times 440\times I_L\times 0.6\\ I_L=21.87\ A$

$I_{ph}=\dfrac{I_L}{\sqrt 3}=\dfrac{21.87}{\sqrt 3}=12.63\ A$

$Z_{ph}=\dfrac{V_{ph}}{I_{ph}}=\dfrac{440}{12.63}=34.84\ \Omega$

$\phi =cos^{-1}(0.6)=53.13^{\circ}\\ R_{ph}=Z_{ph}cos\phi =34.84\times 0.6=20.90\ \Omega \\ X_{ph}=Z_{ph}sin\phi =34.84\times 0.8=27.87\ …