Whenever a DC MOTOR starts it requires a current that is equal to 5-6 times its load current to start.Thus it is required to reduce the voltage when the motor starts.This job is done with a starter.In its simplest form, the starter of a dc motor works like a variable resistance in series with the armature circui.Its work is to reduce the starting voltage upto such a value so that the increased current does not burn the armature windings. As the rotating armature of dc motor picks up speed, the starter resistance is gradually reduced to almost zero. At full speed the motor starts running normally, i.e. the job of starter finishes here.
In other words, the starter offers resistance during starting of dc motor only,to provide large current. At full speed,the starter is electrically out of armature circuit of the motor.
In other words, the starter offers resistance during starting of dc motor only,to provide large current. At full speed,the starter is electrically out of armature circuit of the motor.
thank you for giving the idea and mechanism of the starter
ReplyDeletethnks mam
ReplyDelete:)
Deleteţђāŋķ you
ReplyDelete:)
Deletethanks for such a easy language discussued by you
ReplyDeletebecause everyone discussed in heavily the starter
thank you
Thank You. your article helped a lot.
ReplyDelete:)
DeleteThank You.
ReplyDelete<<< the starter offers resistance during starting of dc motor only,to provide large current.>>> To "provide" large current or to "limit" large current.
ReplyDeleteGeneral Emf. Equation, E = Eb + IaRa. At the starting of the DC motor, back-emf, Eb = 0. So, Ia = E/Ra. But the armature Resistance(Ra) is so very small and the current (Ia) would be so high which is dangerous and can pull out the armature winding. Here comes the role of a starter which limit the starting current to a safe level.
Also, you said, the starter of a dc motor works like a variable resistance in series with the armature circuit, so this extra resistance is used to block the unsafe armature current. So, i think the starter offers resistance during starting of dc motor not to provide large current but to limit the large current.
As we know P=I*V
DeleteWhen the motor starts, we need high current and power is constant, right. So, if we want to increase I, we have to reduce V by the same factor, so that their product remains constant.
And initially, when back emf is zero, V=I/(Ra+Rs)
Where, Ra= armature resistance( negligible)
Rs= starter resistance
So, the above equation becomes V=I/Rs
I hope it's clear now, that in order to decrease voltage, we increase Rs initially and then gradually decrease it when the motor gains momentum.
P.S: Starter does not limit the current, it increases the current in the beginning and then brings it back to normal when the motor speeds up.