Question

A system with two buses are connected through a lossless transmission line with reactance value X. Bus 1 has a voltage magnitude $V_1=1.2pu$ and bus 2 has voltage of 1 pu. When bus 1 is sending reactive power to bus 2 of magnitude $Q_{12}$. The reactive power, $Q_{12}$ magnitude is increased by 20% by varying bus 1 voltage, keeping bus 2 voltage fixed. Assuming there is no real power flow through the transmission line, the new value of voltage magnitude in pu of bus 1 will be _______.

• A. 0.80 pu
• B. 1.02 pu
• C. 1.23 pu
• D. 1.80 pu

Answer 1

The correct option is C 1.23 pu

Given system can be drawn as,

As real power flow is zero, load angle is, $\delta =0$
$Let,Bus1voltage=1.2pu=V_1$
$Bus2voltage=1pu=V_2$

$Q_{12}=\frac{V_1^2}{X}-\frac{V_1{V}_2}{X}cos\delta$

$=\frac{(0.12{)}^2}{X}-\frac{\left(1.2\right)\times \left(1\right)}{X}cos{0}^{\circ }$

$=\frac{1.44}{X}-\frac{1.2}{X}=\frac{0.24}{X}$

When reactive power requirement is increased by 20 %

$Q_{12}^{\prime }=1.2Q_{12}andBus2voltageremainsunchanged$

Using reactive power relation again,

$1.2Q_{12}=\frac{V_1^{\prime 2}}{X}-\frac{V_1^{\prime }\times \left(1\right)}{X}=1.2\times \frac{0.24}{X}$

Resolving,

$V_1^{\prime 2}-V_1^{\prime }-0.288=0$

$So,V_1^{\prime }=1.23,-0.233$
Ignoring second value as it is impractical.

$\therefore V_1^{\prime }=1.23pu$