Question
Download Solution PDFA 6-pole, wave-wound armature has 600 conductors and is driven at 700 rpm. Determine the generated EMF in the armature if the flux per pole is 10 mWb.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
EMF equation of a DC Generator:
As the armature rotates, a voltage is generated in its coils, which is called Generated EMF or Armature EMF and is denoted by Eg.
\({E_g} = \frac{{ϕ ZNP}}{{60A}}\)
Where,
Eg = Generated EMF
P = Number of poles of the machine
ϕ = flux per pole in weber
Z = total number of armature conductors
N = speed of armature in revolution per minute (rpm)
A = number of parallel paths in the armature winding
Also,
A = P × m
Where,
m = multiplexity (simplex/duplex)
In wave winding, multiplexity is always 2 (two)
Therefore, A = 2
While in lap winding, there are two types:
- Simplex Lap winding: m = 1
∴ A = P
- Duplex Lap winding: m = 2
∴ A = 2P
Calculation:
Given that,
Number poles P = 6
Conductors Z = 600
Speed N = 700 rpm
The flux per pole (ϕ) = 0.01 Wb
As winding is wave type
Number of parallel paths A = 2
\(E_g = {(0.01)(600)(700)(6) \over 60(2)}\) = 210 V
Last updated on Jun 16, 2025
-> SSC JE Electrical 2025 Notification will be released on June 30 for the post of Junior Engineer Electrical/ Electrical & Mechanical.
-> Applicants can fill out the SSC JE application form 2025 for Electrical Engineering from June 30 to July 21.
-> SSC JE EE 2025 paper 1 exam will be conducted from October 27 to 31.
-> Candidates with a degree/diploma in engineering are eligible for this post.
-> The selection process includes Paper I and Paper II online exams, followed by document verification.
-> Prepare for the exam using SSC JE EE Previous Year Papers.