1. Field Diverter Method:

A variable resistor, known as a diverter, is connected in parallel with the series field winding. By adjusting the diverter's resistance, a portion of the current is shunted away from the field winding, reducing the field current and, consequently, the magnetic flux. This reduction in flux leads to an increase in motor speed, as speed is inversely proportional to flux.

2. Tapped Field Control:

In this method, the field winding is divided into sections with taps at various points. A selector switch allows the operator to choose different numbers of turns in the field winding, effectively varying the total ampere-turns. Selecting fewer turns reduces the magnetic flux, resulting in higher motor speed. This method provides discrete speed settings based on the available taps.

3. Armature Diverter Method:

Here, a variable resistor is connected in parallel with the armature winding. Adjusting this resistor changes the current distribution between the armature and the diverter. For a constant load torque, reducing the armature current increases the field current, enhancing the magnetic flux and decreasing the motor speed. Conversely, increasing the armature current reduces the field current, decreasing the flux and increasing the speed.

$$R_a \space = \space 0.025Ω$$

$$R_{sh} \space = \space 230Ω$$

$$I_1 \space = \space 5A$$

$$V \space = \space 230V$$

$$N_1 \space = \space 1000rpm$$

$$I_{sh} \space = \space {{230} \over 230} \space = \space 1A$$

$$I_{a1} \space = 5 - 1 \space = \space 4A$$

$$E_{b1} \space = \space V - I_{a1}R_a$$

$$= \space 230 - 4 \times 0.025$$

$$= \space 229.9V$$

On load,

$$I_2 \space = \space 41A$$

$$I_{sh} \space = \space 1A$$

$$I_{a2} \space = \space 41 - 1 \space = \space 40A$$

$$N_2 \space = \space ?$$

$$E_{b2} \space = \space V - I_{a2}R_a$$

$$= \space 230 - 40 \times 0.025$$

$$= \space 229V$$

Since Ф remains the same,

$$E_b \space ∝ \space N$$

$${{N_2} \over N_1} \space = \space {{E_{b2}} \over E_{b1}} \space$$

$$N_2 \space = \space {{229} \over 229.9} \times 1000$$

$$N_2 \space = \space 996.08rpm$$

Drop in speed:

$$= \space 1000 - 996.08 \space = \space 3.92rpm$$