Question

In a power network, the supply voltage is $V\angle \delta$ and receiving end voltage is $V\angle 0$. If the impedance of transmission line is $(\sqrt{3}+jX)\Omega$. Then for maximum power transfer, the value of X should be

• A. 3
• B. $\sqrt{3}$
• C. $\sqrt{3}$
• D. $3\sqrt{3}$

Answer 1

The correct option is A 3

$P_R=\frac{V_S{V}_R}{Z}cos(\theta -\delta )-\frac{V^2}{Z}cos\theta$
For maximum power transfer,
$\theta =\delta$
and also it is given that, $V_R=V_s=V$
$P_{Rmax}=\frac{V^2}{Z}-\frac{V^2}{Z}cos\theta$
$Z=\sqrt{R^2+X^2}$
$cos\theta =\frac{R}{Z}$
$P_{R‘max}=\frac{V^2}{\sqrt{R^2+X^2}}-\frac{V^{2}R}{(R^2+X^2)}$

So, keeping $\frac{d{P}_{R‘max}}{dX}=0$
We get $X=\sqrt{3}R$
$X=\sqrt{3}\times \sqrt{3}$
$X=3\Omega$