Question
Download Solution PDFA 400V, 3-phase, star-connected synchronous motor has an armature current of 200A at an effective resistance of 0.04 OHMS. The short circuit load loss at half-full load will be _________.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept
The copper loss is given by:
P1 = I2 × R
At half load:
P2 = (0.5I)2 × R = 0.25(I2 × R)
The short circuit load loss at half-full load will be:
P = 0.25(I2 × R) - (I2 × R) = -0.75(I2 × R)
By considering only magnitude:
P = 0.75(I2 × R)
Calculation
Given, I = 200A
R = 0.04Ω
P = 0.75 × (200)2 × 0.04
P = 1200 W
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