# 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 whe

41 views
in General
closed
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.
1. 154 V
2. 196 V
3. 120 V
4. 180 V

by (54.3k points)
selected

Correct Answer - Option 1 : 154 V

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 wave type

Number of parallel paths A = 4

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