# The speed of a d.c. motor is

### Right Answer is:

Inversely proportional to flux per pole

#### SOLUTION

**Voltage Equation of DC Motor**

The voltage equation of the DC Motor is given as

**V = E _{b} + I_{a}R_{a}**

The voltage V applied across the motor armature has to

(i) overcome the back e.m.f. E_{b} and

(ii) supply the armature ohmic drop I_{a}R_{a}

The above equation can also be written as

**E _{b} = V − I_{a}R_{a}**

### Back EMF of DC Motor

When the armature of a d.c. the motor rotates under the influence of the driving torque, the armature conductors move through the magnetic field and hence e.m.f. is induced in them as in a generator. The induced e.m.f. acts in opposite direction to the applied voltage V (Lenz’s law) and is known as **back or counter e.m.f. 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.

From the above equation, it is clear that the EMF of DC Motor is Directly proportional to the Number of poles of the machine **(P)**, Flux per pole in Weber **(ϕ)**, Total number of armature conductors**(Z)**, and Speed of armature **(N)**

From the voltage equation and Back EMF equation, we can conclude that

$\begin{array}{l}\dfrac{{P\Phi ZN}}{{60A}} = V – {\rm{ }}{I_a}{R_a}\\\\N = \dfrac{{V – {\rm{ }}{I_a}{R_a}}}{\Phi } \times \left( {\dfrac{{60A}}{{PZ}}} \right)r.p.m\\\\{E_b} = V – {\rm{ }}{I_a}{R_a}\\\\\therefore N = K\dfrac{{{E_b}}}{\Phi }\end{array}$

The above equation shows that speed is directly proportional to back e.m.f. E b and inversely to the flux Φ

**N ∝ E _{b} /Φ.**

$N = \dfrac{{{E_b}}}{\Phi } \propto \dfrac{{V – {\rm{ }}{I_a}{R_a}}}{\Phi }$

But as the value of armature resistance R_{a} and series field resistance R_{se} is very small, the drop I_{a}R_{a} and Ia (R_{a} + R_{se}) is very small compared to applied voltage V. Hence, neglecting these voltage drops the speed equation can be modified as,

$N \propto \dfrac{V}{\Phi }$