Gate 2016 - EE - Set 2 - Question 40

Given, V_{s}=400kV and V_{r}=420kv

At no load \Rightarrow V_{s}=AV_{r}\Rightarrow 400=A(420)

A=1+\frac{YZ}{2}=1+\frac{(r+j\omega L)(g+j\omega C)}{2}

For a lossless line r=g=0

A=1-\frac{(\omega C)(\omega L)}{2}

Using the formula for the product of phase shift constant and length, \Rightarrow (\beta l)=\sqrt{(\omega L)(\omega C)}, where \beta is the phase shift constant and l is the length

\Rightarrow \beta ^{2}l^{2}=(\omega L)(\omega C)

A=1-\frac{\beta ^{2}l^{2}}{2}

0.9524=1-\frac{\beta ^{2}l^{2}}{2}

2(0.9524)=2-\beta ^{2}l^{2}

-0.0952=-\beta ^{2}l^{2}

\beta =\frac{0.3085}{l} _________(1)

We also have the proportion, \frac{v}{f}=\frac{2\pi }{\beta }, from which we isolate the expression for \beta \Rightarrow \frac{2\pi f}{v}

Substituting the above expression of \beta into (1), we get

\frac{2\pi f}{v}=\frac{0.3085}{l}

l=\frac{0.3085v}{2\pi f}

l=\frac{0.3085\times 3\times 10^{8}}{2\times 3.14\times 50}

l=294.745 km