SOLUTION

Voltage Equation of DC Motor

The voltage equation of the DC Motor is given as

V = E_b + I_aR_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_aR_a

The above equation can also be written as

E_b = V − I_aR_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_aR_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,

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