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# What is Speed Control of DC Series Motors?

The speed of a DC series motor is given by,

$$\mathrm{𝑁 \varpropto\frac{𝐸_{𝑏}}{\varphi}}$$

$$\mathrm{⇒ 𝑁 = 𝐾 (\frac{𝑉 − 𝐼_{𝑎}(𝑅_{𝑎} + 𝑅_{𝑠𝑒})}{\varphi}) … (1)}$$

Hence, it is clear from the eq. (1) that the speed of a DC series motor can be changed by using any one of the following two methods −

- Field Control Method
- Armature Resistance Control Method

## Field Control Method

The *field control method* is based on the fact that by varying the field flux in the series motor, its speed can be changed, as,

$$\mathrm{N \varpropto\frac{1}{\varphi}}$$

The change in the flux can be achieved by in the following ways −

## Field Diverter

In this method, a variable resistance called field diverter is connected in parallel with the series field winding as shown in the figure.

The field diverter shunts some portion of the line current from the series field winding, hence weakening the field and increasing the speed of the motor. The lowest speed that can be obtained using this method is corresponding to the zero current through the diverter which is the normal speed of the motor. Therefore, the field diverter method only provides speed above the normal speed. The series field diverter method is mainly used in traction work.

## Armature Diverter

In this method, a variable resistance called *armature diverter* is connected in parallel with armature, which shunts some portion of the line current from the armature winding and hence reducing the armature current (see the figure).

For a given load, if armature current (Ia) is decreased, the flux must increase to maintain the load torque (τ_{a} ∝ ϕI_{a}) constant. Since the speed (N ∝ 1/ϕ), so the motor speed must decrease. By using armature diverter method, any speed below the normal speed can be obtained.

## Tapped Field Control

In this method, to decrease the flux, the number of turns in the series field winding is reduced and hence the speed is increased.

A switch (S) is used to short any part of the series field winding as shown in the figure. With the full turn of the field winding, the motor runs at normal speed and by reducing the number of turns in the series field winding, speeds above the normal speed can be obtained.

## Armature Resistance Control Method

In the armature resistance control method, a variable resistance is connected in series with the supply to complete the motor circuit as shown in figure.

This series resistance reduces the voltage available across the armature and thus the speed is decreased. By changing the value of variable resistance, any speed below the normal speed can be obtained. This method has poor speed regulation, but this has no significance since the series motors are used in varying speed applications.

The armature resistance control method is the most common method employed to control the speed of DC series motors.

## Numerical Example

A 250 V DC series motor runs at 1000 RPM when drawing a line current of 50 A. The armature and series field resistances are 0.08 Ω and 0.05 Ω respectively. If the current taken by the motor remains the same, determine the value of series resistance required to reduce the speed to 800 RMP.

## Solution

Let R_{series} be the series resistance.

Here,

$$\mathrm{𝑅_{𝑎} + 𝑅_{𝑠𝑒} = 0.08 + 0.05 = 0.13 \Omega}$$

$$\mathrm{𝑁_{1} = 1000 RPM; \:𝑁_{2} = 800\:RPM}$$

$$\mathrm{𝐼_{𝑎1} = 𝐼_{𝑎2} = 50 A}$$

Therefore,

$$\mathrm{𝐸_{𝑏1} = 𝑉 − 𝐼_{𝑎1}(𝑅_{𝑎} + 𝑅_{𝑠𝑒}) = 250 − (50 × 0.13) = 243.5 V}$$

Also,

$$\mathrm{𝐸_{𝑏2} = 𝑉 − 𝐼_{𝑎2}(𝑅) = 250 − 50𝑅 V}$$

Where,

$$\mathrm{𝑅 = 𝑅_{𝑎} + 𝑅_{𝑠𝑒} + 𝑅_{𝑠𝑒𝑟𝑖𝑒𝑠}}$$

Now,

$$\mathrm{\frac{𝑁_{2}}{𝑁_{1}}=\frac{𝐸_{𝑏2}}{𝐸_{𝑏1}}}$$

$$\mathrm{⇒\frac{800}{1000}=\frac{250 − 50𝑅}{243.5}}$$

$$\mathrm{⇒ 𝑅 = 1.104 \Omega}$$

$$\mathrm{\therefore R_{serires} = 1.104 − 0.13 = 0.974 \Omega}$$

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