Question

A synchronous motor is drawing 50 A from 400 V, 3-phase supply at unity pf with a field current of 0.9 A. The synchronous reactance of motor is $1.32\Omega$ with the mechanical load remaining constant, the value of field current which would result in 0.8 leading pf is______ A. (Assuming linear magnetization)

• A. 1.077

Answer 1

The correct option is A 1.077
Excitation emf, $\overrightarrow{E_{f1}}=\overrightarrow{V_t}-j{I}_a{X}_s=\frac{400\angle 0^{\circ }}{\sqrt{3}}-50\angle 0^{\circ }\times 1.3j$

$\overrightarrow{E_{f1}}=239.913\angle -{15.72}^{\circ }$

Given load remains constant,
$I\cos \varphi$ is constant

$I_1\cos {\varphi }_1=I_2\cos {\varphi }_2$

$50\times 1=I_2\times 0.8$

$\Rightarrow I_2=\frac{50}{0.8}=62.5A$

Now, $E_{f2}=V_t-j{I}_{a2}{X}_s$

$=\frac{400}{\sqrt{3}}-62.5\angle {36.86}^{\circ }\left(1.3j\right)$

$E_{f2}=287.13\angle -{13.085}^{\circ }V$

Due to linear magnetization
$\frac{I_{f1}}{I_{f2}}=\frac{E_{f1}}{E_{f2}}$

$\Rightarrow \frac{0.9}{I_{f2}}=\frac{239.913}{287.13}$

$\Rightarrow I_{f2}=1.077A$