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When the load is connected to a DC generator, the load current flows through the armature conductors and set up a magnetic flux called armature flux. This armature flux is always 180° phase apart from the main field flux and distorts or opposes it. This opposition of armature flux on the main field flux is known as armature reaction.

The armature reaction has two undesirable effects. The first is it reduces the main field flux called as demagnetising effect and the second is it distorts the main field flux called as cross magnetising effect.

In order to calculate the demagnetising and cross-magnetising ampere turns consider the figure shown below.

If the brushes are shifted by an angle of β° electrical, then the direction of currents in the conductors between the lines AB and CD in the interpole zones spread over 2β° electrical at the top and bottom of the armature is such as to produce a flux opposing the main field flux. Thus, these are known as demagnetising armature conductors. The rest of the conductors spread over the angle (180-2β°) carrying currents, which produces only cross-magnetising effects.

Let,

$$\mathrm{P\:=\:total\:number\:of\:poles}$$

$$\mathrm{Z\:=\:total\:number\:of\:armature\:conductors}$$

$$\mathrm{A\:=\:Number\:of\:parallel\:paths}$$

$$\mathrm{I_{a}\:=\:Total\:armature\:current}$$

$$\mathrm{\beta°\:=\:Brush\:shift}$$

Therefore,

$$\mathrm{Total\:number\:of\:conductors\:per\:pole\:=\frac{Z}{P}}$$

As the one turn consists of two conductors, thus,

$$\mathrm{Number\:of\:turns\:per\:pole\:=\:\frac{1}{2}\:\frac{Z}{P}}$$

If each armature conductor carries a current of Ic amperes, then

$$\mathrm{Total\:ampere\:turns\:per\:pole\:=\:\frac{1}{2}\:\frac{Z}{P}*I_{c}}$$

Since, these ampere-turns are spread over one pole pitch i.e. 180° electrical, hence

$$\mathrm{Ampere\:turn\:per\:degree\:electrical\:=\:\frac{1}{2}\:\frac{Z}{P}*\frac{I_{c}}{180°}}$$

Therefore, the demagnetising ampere turns per pole is given by,

Since, these ampere-turns are spread over one pole pitch i.e. 180° electrical, hence

$$\mathrm{Ampere\:turn\:per\:degree\:electrical\:=\:\frac{1}{2}\:\frac{Z}{P}*\frac{I_{c}}{180°}}$$

Therefore, the demagnetising ampere turns per pole is given by,

Now, if β = 0, then the brushes lie on the GNA and the demagnetising ampere-turns are zero and the whole ampere-turns are cross-magnetising ampere-turns.

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