please help me. how self excited DC generator initially starts.
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When generator is switched off, a small magnetism is developed in rotor iron which induced electromotive force in the armature due to which current is produced in the field windings. Initially, weak magnetic fieldcreates less current in the coil, but to sustain self-excitation, the additional magnetic flux increases the electromotive force in the rotor, due to which voltage keep on increasing until the machine takes the full load.
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Hello Mate. Thanks for asking this question. your answer is =>
It is the device that converts mechanical energy into electrical energy in the form of Direct Unidirectional current.
Split ring commutator S1 and S2 , armature coil , brushes , magnets, brushes B1 and B2.
Suppose initially the plane of the coil is horizontal. The split ring Si touches the brush Bi and S2 touches B2.
When coil ABCD is rotated clockwise, magnetic field lines passing through it change. The arm AB moves upwards and arm CD moves downwards.
According to Fleming's right hand rule, the induced current in the armature coil flows in the direction ABCD. In the external circuit, current flows from B2 to Bi.
As the coil crosses its vertical position, the split ring Si comes in contact with B2 and S2 with Bi. The arm AB now moves downwards and arm CD moves upwards. The induced current in the armature flows along DCBA. But in the external circuit, the current still flows from B2 to Bi.
Hence, a unidirectional current flows in the external circuit.
It is the device that converts mechanical energy into electrical energy in the form of Direct Unidirectional current.
Split ring commutator S1 and S2 , armature coil , brushes , magnets, brushes B1 and B2.
Suppose initially the plane of the coil is horizontal. The split ring Si touches the brush Bi and S2 touches B2.
When coil ABCD is rotated clockwise, magnetic field lines passing through it change. The arm AB moves upwards and arm CD moves downwards.
According to Fleming's right hand rule, the induced current in the armature coil flows in the direction ABCD. In the external circuit, current flows from B2 to Bi.
As the coil crosses its vertical position, the split ring Si comes in contact with B2 and S2 with Bi. The arm AB now moves downwards and arm CD moves upwards. The induced current in the armature flows along DCBA. But in the external circuit, the current still flows from B2 to Bi.
Hence, a unidirectional current flows in the external circuit.
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