A four-pole, lap-would DC machine has 462 conductors in the armature. The average flux per pole is 0.02 Wb. Determine the induced armature voltage when the armature rotates at 1000 rpm.

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SSC JE EE Previous Paper 12 (Held on: 24 March 2021 Evening)
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  1. 154 V
  2. 196 V
  3. 120 V
  4. 180 V

Answer (Detailed Solution Below)

Option 1 : 154 V
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Detailed Solution

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Concept:

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:

  1. Simplex Lap winding: m = 1

          ∴ A = P

  1. Duplex Lap winding: m = 2 

         ∴ A = 2P

Calculation:

Given that,

Number poles P = 4

Conductors Z = 462

Speed N = 1000 rpm

The flux per pole (ϕ) = 0.02 Wb

As winding is LAP type 

Number of parallel paths A = 4

\(E_g = {(0.02)(462)(1000)(4) \over 60(4)}\) = 154 V

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