# 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 **I _{L}** = 60 A

Armature current Ia = 60 A** ( In DC series motor the line current is equal to the armature current i.e I _{L} = I_{a})**

Flux **Φ** = 25 x 10^{-3}

Armature Resistance **R _{a}** = 0.15Ω

Series Resistance **R _{se}** = 0.2Ω

The Back EMF of DC series Motor is given by

**E _{b} = V − I_{a}(R_{a} + R_{se})**

= 250 − 60 (0.15 + 0.2)

= 250 − 21

**E _{b} = 229**

Back EMF in DC Motor **E _{b}. **

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

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.

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