Question

An alternator is supplying 75% of its rated power to an infinite bus at rated voltage and rated frequency. The excitation is made 1.25 times the rated voltage. The per unit reactance is 0.7. Operating power factor of machine is

• A. 0.9688 (leading)
• B. 0.812 (lagging)
• C. 0.9688 (lagging)
• D. 0.812 (leading)

Answer 1

The correct option is C 0.9688 (lagging)

$\text{Power (P) =}\frac{V{E}_f}{X_s}\sin \delta$

$0.75=\frac{1\times 1.25}{0.7}\sin \delta$

$\delta ={24.83}^{\circ }$

Current is given by,
$\overrightarrow{I}=\frac{\overrightarrow{E_f}-\overrightarrow{V}}{jX}=\frac{1.25\angle {24.53}^{\circ }-1\angle 0^{\circ }}{j0.7}$

$I=0.77\angle -{14.36}^{\circ }$

$\text{Phase angle,}\varphi \text{=}{14.36}^{\circ }$

$\text{Power factor =}cos\varphi =0.9688\text{lagging}$