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

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

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

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

$$l=294.745 km$$