# A 4 pole, 250 V, DC series motor has a wave-connected armature with 200 conductors. The flux per pole is 25 mWb when the motor is drawing 60 A from the supply. The armature resistance is 0.15 Ω while the series field winding resistance is 0.2 Ω. Calculate the speed under this condition.

A 4 pole, 250 V, DC series motor has a wave-connected armature with 200 conductors. The flux per pole is 25 mWb when the motor is drawing 60 A from the supply. The armature resistance is 0.15 Ω while the series field winding resistance is 0.2 Ω. Calculate the speed under this condition.

### Right Answer is: 1374 RPM

#### SOLUTION

Given Data

Supply Voltage V = 250 V

Number of Poles P = 4

Total number of armature conductors Z = 200

Number of parallel paths in the armature winding A = 2 (for wave winding number of parallel path is always 2)

Line current IL = 60 A

Armature current Ia = 60 A ( In DC series motor the line current is equal to the armature current i.e IL = Ia)

Flux Φ = 25 x 10-3

Armature Resistance Ra = 0.15Ω

Series Resistance Rse = 0.2Ω

The Back EMF of DC series Motor is given by

Eb = V − Ia(Ra + Rse)

= 250 − 60 (0.15 + 0.2)
= 250 − 21

Eb = 229

Back EMF in DC Motor Eb

${E_b} = \dfrac{{P\Phi ZN}}{{60A}}$

Where

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 (r.p.m).

A – Number of parallel paths in the armature winding.

$\begin{array}{l}N = \dfrac{{{E_b} \times 60A}}{{P\Phi Z}}\\\\ = \dfrac{{229 \times 60 \times 2}}{{4 \times 25 \times {{10}^{ – 3}} \times 200}}\\\\N = 1374{\text{ R}}{\rm{.P}}{\rm{.M}}\end{array}$

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