a simple electric motor has an armature resistance of 1ohm and runs from a dc source of 12 V . it draws a current of 2A when unloaded . when a certain load is connected to it its speed reduces by 10 % of its initial value . The current drawn by the loaded motor is

Dear Student,

Armature resistance of motor, R=1 ohmvoltage of DC source, E=12 Vcurrent drawn, i=2 AWhen certain load is connected the speed becomes 90100 of its unloaded value.Now, the current in the electric motor is given by,i=E-eRwhere, e=back emfWe know that,e α ωwhere, ω=angular speedIn the first case,2=12-e1e=10 VIn the second case, induced emf=e'as e' α ω'therefore, e'e=ω'ω=910or, e'10=910or, e'=9 VTherefore, current in the second case is given byi'=E-e'Ri'=12-91=3 A

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