Discussion
A 3300 V, delta connected synchronous motor has a synchronous reactance per phase (delta) of \(18\ \Omega\). It operates at a leading power factor of 0.707 when drawing 800 kW from the mains. The armature current is:
Answer: B
For a three phase motor the input power is \(P = \sqrt{3}\,V_L I_L \cos\phi\). With a line voltage of 3300 V, 800 kW and a power factor of 0.707:
\[ \begin{aligned} I_L \\ &= \frac{800 \times 10^3}{\sqrt{3} \times 3300 \times 0.707} \\ &= \frac{800000}{4041} = 198\ \text{A} \end{aligned} \]
The synchronous reactance is not needed for this part, since the power and power factor fix the current. The value 330 A would be for a power factor of about 0.42, and the larger values do not follow from the data. The answer is 198 A, option B.
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