Answer: the line-to-line voltage at the high-voltage terminals of the transformer and the sending-end of the feeder is 249.71∠1.8° kV
Explanation:
First we find the phase voltage per phase at the primary side connected in Y, so we say
V₂ = 240K/√3 = 138.56 kV
Now we find the primary current
<em>I</em>₁ = ((400 × 10⁶) / 3(138.56 × 10³)) ∠ -cos⁻¹ (0.8)
<em>I</em>₁ = 962.28∠ -36.87° A
To find the voltage V₁, we say
V₁ = ( 1.2 + j6) <em>I</em>₁ + V₂
we substitute
V₁ = ( 1.2 + j6) 962.28∠ -36.87° + 138.56 × 10³
V₁ = 143∠1.57° kV
Now we find the phase voltage at the sending end
Vₓ = ( 0.6 + J1.2 )<em>I</em>₁ + V₁
Vₓ = ( 0.6 + J1.2 ) 962.28∠ -36.87° + 143∠1.57° K
Vₓ = 144.17∠1.8° kV
So to Determine the line to line voltage at the sending end, we say:
Vₓ (line to line) = √3 × 144.17∠1.8° kV
Vₓ (line to line) = 249.71∠1.8° kV