Given:

S = 10 kva

P.f = 0.342

$W_1$ = ?

$W_2$ = ?

Formula:

$P=S\cos \phi$

$P.f = \cos \phi$

$\phi =\cos ^{-1}(P.f) = 70^\circ$

Case 1: Lagging P.f

$W_1 +W_2 =P =S.(P.f)\\ =10\times 0.342\\ =3.4kW ...(1)\\ \tan \phi = \left [ \sqrt{3}\dfrac{(W_1-W_2)}{W_1+W_2} \right ]\\ \therefore W_1-W_2 =(W_1+W_2)\dfrac{\tan \phi}{\sqrt{3}}\\ =3.42\times \dfrac{2.75}{\sqrt{3}} = 5.43....(2)$

$W_1 =4.42 kW\\ W_2 = -1 kW$   ..... Solving (1) and (2)

Case 2: Leading p.f

$W_1+W_2 =3.42 kW ....(3)\\ \tan \phi =\left [- \sqrt{3} \frac{(W_1 -W_2)}{W_1+W_2} \right ]\\ \therefore W_1-W_2 = \frac{-(W_1-W_2).\tan \phi}{\sqrt{3}}\\ =\frac{-3.42\times 2.75}{\sqrt{3}}\\ =-5.43 ...(4)$

$W_1 =-1kW\\ W_2 =4.42kW$  Solving (3) and (4)