# Process of commutation

## Introduction:

Commutation means the current reversal in a coil when the brush passes over/under the commutator segments connected to the coil under study.

A dc machine can’t be without a commutator. The process of commutation is well understood if we assume it to be purely resistive.

To understand commutation in the following discussion,  pay attention to the coil in the diagrams shown in red colour.

## Assumptions

1. Brush and the commutator segment contact area are of equal size.
2. Commutator segments are separated from each other using very thin layer of mica insulation.
3. Current collected by the brush is proportional to the area of overlap. So the total current collected by the brush at any time is always constant.
4. Each coil except the coil always carries current $i_c$ .
5. Coil is the coil under the commutation process.

## Step 1

• Brush will travel from right to left i.e. from segment 3 to segment 2
• In this step it is completely covering segment 3.
• All the coils carry the current $i_c$. In this step even the coil also carries the current $i_c$.
• Segment 3 will collect current $i_c$ from coil and $i_c$ from the coil on its right. So, the brush will collect the full current $2*i_c$

## Step 2

In this step, brush travels further to cover smaller area of segment 2 and larger area of segment 3.

• 1. All coils except coil carry current $i_c$ .
• Brush coverage tells that brush collects more current from the seg-3 and less from seg-2.
• Current through coil is $i_c-i_2$
• Total current collected by the brush is: $i_2+[(i_c-i_2)+i_c]=2*i_c$
• Current through coil is from Left to Right.

## Step 3

This step is very important. The brush covers exactly half the segment area of seg-3 and seg-2.

• All coils except coil carry current $i_c$ .
• Brush overlap tell us that the brush collects equal currents from each segment.
• So $i_3 = i_2=i_c$.
• So current through coil is $i_c - i_2$ from left to right added to $i_c - i_2$ from right to left. This results in 0 current in coil.

## Step 4

This step is very similar to the step 2 above. But the brush covers more area of seg-2 and lesser area of seg-3.

• All coils except coil carry current $i_c$.
• Brush overlap tell us that the brush collects less current from seg-3 and more from seg-2.
• So Current through coil is $i_c - i_2$. But from right to left.

## Step 5

This step is similar to the step 1 above. But it covers all of seg-2 instead of seg-3 in step 1.

• All coils except coil carry current $i_c$.
• Brush overlap tell us that the brush collects all current from seg-3.
• So Current through coil is $i_c$. But from right to left.

## Conclusion

Thus from step 1, step 3 & step 5 we conclude that the current direction through the coil changes from left to right via zero value, while the brush is passing from segmet-3 to segment-2. This is the action of commutation..

Notice that the current direction in the coil changes but the output current through the brush is always coming out. This is ac do dc conversion.

Because of this action, alternating emf developed across the armature gets converted to dc emf which is available across the output terminals.