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# 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

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

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

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

$l=294.745 km$

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