A transmission line has following parameters:
$A = D = 0.98 ∠ 3^{\circ} a n d B = 98 ∠ 78^{\circ} Ω$ .
If the maximum power transmitted from the transmission line is 100MW and receiving end voltage is 110kV, then value of sending end voltage will be approximately

112 kv

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125kV

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120kV

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117kV

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Solution

The correct option is D 117kV
$F o r m a x i m u m p o w e r t r a n s f e r p u t t i n g δ = β i n b e l o w e q u a t i o n$
$P_{R} = \frac{| V_{S} | | V_{R} |}{| B |} c o s (β - δ) - ∣ ∣ \frac{A}{B} ∣ ∣ | V_{R} |^{2} c o s (β - α)$
$P_{R m a x} = \frac{| V_{S} | | V_{R} |}{| B |} - \frac{| A | | V_{R}^{2} |}{| B |} c o s (β - α) . . (1)$
$G i v e n P_{R m a x} = 100 M W, V_{R} = 110 k V$

$S o l v i n g f o r s e n d i n g e n d v o l t a g e | V_{S} | u s i n g e q u a t i o n (1)$
$100 = \frac{| V_{s} | \times 110}{98} - \frac{0.98 \times 110^{2}}{98} c o s (78^{\circ} - 3^{\circ})$
$100 = \frac{V_{s} \times 110}{98} - 121 c o s 75^{\circ}$
$100 = \frac{V_{s} \times 110}{98} - 31.317$
$V_{s} = \frac{131.317 \times 98}{110} = 116.991 k V$

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