A 3 - phase, 3000 kVA, 0.95 lagging pf load is supplied by a Vee connection at 400V. The kVA output and operating power factor angle of both transformers respectively are

1000 ∠ {9.1}^{\circ} k V A a n d 100 ∠ {9.1}^{\circ} k V A

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1000 ∠ - {9.1}^{\circ} k V A a n d 1000 ∠ - {9.1}^{\circ} k V A

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1732 ∠ {48.19}^{\circ} k V A a n d 1732 ∠ - {11.89}^{\circ} k V A

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1732 ∠ - {48.19}^{\circ} k V A a n d 1732 ∠ {11.8}^{\circ} k V A

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Solution

The correct option is C $1732 ∠ {48.19}^{\circ} k V A a n d 1732 ∠ - {11.89}^{\circ} k V A$
$kVA supplies by Vee connection transformer,$

$S_{V e e} = \frac{1}{\sqrt{3}} \times S_{3 - ϕ} = \frac{3000}{\sqrt{3}} = 1732 k V A$

$p.f angle of transformer-1$ $= ϕ + 30^{\circ} = c o s^{- 1} (0.95) + 30^{\circ} = {48.19}^{\circ}$

$c o s ϕ_{1} = c o s {48.19}^{\circ} = 0.67 l a g g i n g$

$p.f. angle of transformer-2$ $= ϕ - 30^{\circ} = c o s^{- 1} (0.95) - 30^{\circ} = - {11.8}^{\circ}$

$c o s ϕ_{2} = c o s {11.8}^{\circ} = 0.98 l e a d i n g$

${\to S}_{1} = 1732 ∠ {48.19}^{\circ} k V A$

${\to S}_{2} = 1732 ∠ - {11.8}^{\circ} k V A$

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