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In the conventional dc machine (with a closed continuous commutator winding on its armature), for example, full-wave rectification of the alternating voltage induced in individual armature coils is achieved by means of a commutator, which makes a unidirectional voltage available to the external circuit through the stationary carbon brushes held against the commutator surface.The armaturewindings of dcmachines are located on the rotor because of this necessity for commutation and are of the closed continuous type, known as lap and wavewindings. The simplex lap winding has as many parallel paths as there are poles, whereas the simplex wave winding always has two parallel paths. The winding connected to the commutator, called the commutator winding, can be viewed as a pseudostationary winding because it produces a stationary flux when carrying a direct current, as a stationarywindingwould. The direction of the flux axis is determined by the position of the brushes. In a conventional dc machine, in fact, the flux axis corresponds to the brush axis (the line joining the two brushes). The brushes are located so that commutation (i.e., reversal of current in the commutated coil) occurs when the coil sides are in the neutral zone, midway between the field poles. The axis of the armature mmf is then in the quadrature axis, whereas the stator mmf acts in the field (or direct) axis. Figure shows schematic representations of a dc machine. The commutator is thus a device for changing the connections between a rotating closed winding and an external circuit at the instants when the individual coil-generated voltages reverse. In a dc machine, then, this arrangement enables a constant and unidirectional output voltage. The armature mmf axis is fixed in space because of the switching action of the commutator (even though the closed armature winding on the rotor is rotating), so the commutator winding becomes pseudostationary.
having looked at the circuit, i first wanted to start my frequency from 0 and plot for 10000 to 20000. so in matlab i used f = 0:10000:20000; but the equation for frequency is f= 1
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a cast steel has minimun diameter of 200 mm and a cross sectional area of 250/1000000 m . calclate the mmf to produce a flux of300/1000000 Wb.
Explain different stage in energy audit?
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