Ans. In [fig. 304] during commutation, that is, while the brush contacts with any two segments, as 2 and 3, the currents coming up through the winding on either side of the neutral plane are offered two paths to the brush: 1, direct to brush through the connecting segment, or 2, across the short circuited coil and adjacent segment. Thus, on the right side: to brush through segment 3, or across coil C and adjacent segment 2.
The current will take the path of least resistance.
At the beginning of commutation, almost the entire brush area being in contact with segment 3, the contact resistance of this segment will be much less than for segment 2; hence, not only will the current at the right flow through 3, but also the current at the left after first traversing the short circuited coil. As commutation progresses, the area of contact of 3 decreases while that of 2 increases, and the respective resistances vary in inverse proportion. Likewise the tendency of the current in the left half of the winding to take the longer path through coil C and segment 3 to the brush gradually decreases, becoming zero when the two contact areas become equal. During the second half of the period of commutation, the contact area of segment 2 becomes greater and of 3, less; thus the resistance of 2 is lowered, and that of 3 increased. Accordingly, all of the current at the left will flow through segment 2, and the current at the right will flow through C and 2 rather than through 3. In this way the current is reversed in C, and, if the brush be broad enough to allow a sufficient time interval, the current in C is built up to its full value before segment 3 leaves the brush, thus securing sparkless commutation.
This contact resistance factor in sparkless commutation is illustrated in [figs. 314] to [318], it being assumed that during commutation, the brush contact resistance is inversely proportional to the area of contact, and that the winding is free of resistance and inductance. The current is taken as 40 amperes, in which case 20 amperes will flow from each side of the winding to the brush.
In [fig. 314] the instant before commutation begins all the current will flow through segment A. At the end of the first quarter of the period of commutation, [fig. 315], 30 amperes will flow from the right to brush through A, and from the left, 10 amperes through the short circuited coil via A and 10 amperes through B.
At the end of the second quarter or half period, [fig, 316], the current through each half of the winding will flow to the brush through these respective segments.
At the end of the third quarter, [fig. 317], the current from the right will divide, 10 amperes going through A, and 30 amperes traversing the short circuited coil and out through B. The entire current from the left will flow through segment B.
At the end of the fourth quarter, [fig. 318], or completion of the period the current from each half of the winding will flow to the brush through B.
Figs. 314 to 318.—Brush contact resistance theory of commutation, neglecting self-induction and resistance in the coils. The total current is assumed to be 40 amperes made up of 20 amperes flowing toward the brush from the coils on the right and 20 amperes from the coils on the left. During commutation, that is, the interval during which the brush contacts with any two adjacent segments of the commutator, the current is assumed to vary directly as the contact area.
Fig. 314.—Beginning of commutation; segment A is entirely under the brush, and B is at the initial point of contact. For this position the currents from both sides flow to the brush through segment A.
Fig. 315.—One-quarter period of commutation. One quarter of the brush area is in contact with B and three quarters in contact with A; hence, 10 amperes will flow through B and 30 amperes through A.
Fig. 316.—Second quarter of commutation period. The brush now contacts equally with both segments, hence 20 amperes will flow through each segment.
Fig. 317.—Third quarter of commutation period. Three quarters of the brush area is in contact with segment B and one quarter with segment A; accordingly, 30 amperes will flow through B and 10 amperes through A.