Question
Download Solution PDFA 4-pole wave wound DC shunt generator has 35 slots with 12 conductors/slot. Its armature supplies a current of 35 A. If the brushes are given an actual mechanical lead of 10°, what is the demagnetising ampere turns/pole at full load?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The demagnetizing ampere-turns per pole in a DC machine is calculated using:
\( AT_d = \frac{\theta_e \times Z \times I_a}{360 \times A} \)
Where:
- \( \theta_m = 10^\circ \) (mechanical lead)
- Electrical lead \( \theta_e = 10^\circ \times \frac{4}{2} = 20^\circ \)
- Total conductors \( Z = 35 \times 12 = 420 \)
- Armature current \( I_a = 35 \, A \)
- Wave winding ⇒ \( A = 2 \)
Final Answer: 402 AT
Last updated on Jul 15, 2025
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