**VOLTAGE EQUATION OF DC MOTORS**

**1. DC Shunt Motor:**

In a dc shunt motor, the field winding is connected across the armature and the supply is given to the armature. As such the current for the field winding is taken from the supply. The armature current is therefore equal to the difference of line current and the shunt field current. Therefore,

I

_{a}– I_{L}- I_{sh}
I

_{sh}= V/R_{sh}
E

_{b}= V – I_{a}R_{a}
Where,

I

_{sh}is the shunt field current
R

_{sh}is the shunt field resistance.**2. Back e.m.f. in DC Series Motor:**

In a D.C. series motor, the applied voltage V has to supply the ohmic drop in armature (I

_{a}R_{a}) and also series field voltage drop (I_{se}R_{se}) ie.,
V = E

_{b}+ I_{a}R_{a}+ I_{se}R_{se}
Since the field winding is in series with the armature, the current in the armature and series field are same. Therefore,

I

_{L}= I_{a}= I_{se}
Voltage equation can be rewritten as under,

V = E

_{b}+ I_{a}(R_{a}+ R_{se})
Where,

V is the applied voltage in volts

I

_{a}(R_{a}+ R_{se}) is the voltage drop in the armature and series field winding.
Multiplying both sides by I

_{a}
VI

_{a}= I_{a}E_{b}+ I_{a}^{2}R_{a}+ I_{a}^{2}R_{se}
is the power equation

Where

VI

_{a}is the power input in watts
I

_{a}E_{b}is the electrical equivalent of mechanical power developed
I

_{a}^{2}R_{a}is the copper loss in the armature
I

_{a}^{2}R_{se}is the copper loss in the series field winding

**3. Back emf in DC Compound motor:**

In DC Compound motor there are two possible connections. One when the shunt winding is connected parallel to the armature, it is called short shunt compound motor and when the shunt winding is connected parallel to the supply, it is called long shunt compound motor. The two types of connections are shown in figure.

In the long shunt compound motor, the voltage equation of the armature circuit is given by,

V = E

_{b}+ I_{a}R_{a}+ I_{se}R_{se}
But, I

_{a}= I_{se}
Therefore, V = E

_{b}+ I_{a}(R_{a}+ R_{se})
And the current equation is I

_{L}= I_{sh}+ I_{a}
Multiplying the voltage equation both sides by I

_{a}
I

_{a}V = I_{a}E_{b}+ I_{a}^{2}R_{a}+ I_{a}^{2}R_{se}
Where I

_{a}V is the power input to the armature circuit. The total power taken by the motor is equal to the sum of the power input to armature circuit and shunt field circuit. That is,
I

_{L}V = I_{a}V + I_{sh}V
The electrical power equivalent to mechanical is I

_{a}E_{b}.
In the short shunt compound generator, the voltage equation is given by,

V = E

_{b}+ I_{a}R_{a}+ I_{se}R_{se}
And, I

_{L}= I_{se}= I_{a}+ I_{sh}
Where, I

_{L}= Line current in amps
I

_{a}= Armature current in amps
I

_{sh}= Shunt field current in amps
= V

_{sh}/R_{sh}= E_{b}– I_{a}R_{a}/R_{sh}
The electrical power develop equivalent to mechanical is IaEb. In all the cases the back emf is also equal to,

E

_{b}= ZΦNP/60A**SPEED EQUATION OF DC MOTOR:**

The speed equation of dc motors can be deduced from the voltage equation itself. Thus we can find the voltage and speed relation in dc motor. The back emf equation for a motor is given by,

Eb = V – IaRa

But, Eb = ZΦNP/60A

or Eb = K1ΦN where, K1 = ZP/60A

or Eb ∝ ΦN or N ∝ Eb/Φ

That is, N ∝ (V – IaRa)/Φ

From the above it is seen that for the speed to change either

a. the field flux has to be changed or

b. the armature current has to be changed. Change of armature current means change of load conditions.

For a dc series motor,

N ∝ (V – Ia(Ra + Rse))/ Φ

It should be noted that for the field flux to change, a resistance called the diverter is added in parallel to the series field and for the armature current to change a resistance is connected in series to the series field.

## 0 on: "Voltage and Speed Relation in DC Motor"