"Back EMF" (1 Viewer)

[Poops]

If back emf is created to oppose the supply emf, why does it cause the speed of the motor to increase? :S

If it creates an opposing force on the rotating motor, shouldn't it slow down?? Or does it just decrease the net emf? Help anyone?

 

[umm what]

it causes the speed of motor to decrease because it opposes supplied emf

jeezz

 

[umm what]

if there was no back emf, motor will turn with infinite speed..which is impossible (law of conservation of eneergy wud be violated

 

[someth1ng]

If back emf is created to oppose the supply emf, why does it cause the speed of the motor to increase? :S

If it creates an opposing force on the rotating motor, shouldn't it slow down?? Or does it just decrease the net emf? Help anyone?

- To determine the force on each end of the coil (the force experiencing ends), you need to know the current through the coil (proportional to emf).
- At rest, there is no back emf and so, when you turn the motor on, it will speed up and it will stop speeding up when the net emf is equal to zero.
- When net emf is zero, the rotation speed is at equilibrium - the slower the rotation, the supplied emf>back emf causing speed to increase where the higher speeds cause back emf>supplied emf making the motor slow down.

 

[Poops]

- To determine the force on each end of the coil (the force experiencing ends), you need to know the current through the coil (proportional to emf).
- At rest, there is no back emf and so, when you turn the motor on, it will speed up and it will stop speeding up when the net emf is equal to zero.
- When net emf is zero, the rotation speed is at equilibrium - the slower the rotation, the supplied emf>back emf causing speed to increase where the higher speeds cause back emf>supplied emf making the motor slow down.

Ohh, I understand! Thank you

 

[Poops]

if there was no back emf, motor will turn with infinite speed..which is impossible (law of conservation of eneergy wud be violated

lol yeah stupid question. thanks for helping though

 

[someth1ng]

Ohh, I understand! Thank you

No problem, glad I helped

 

[U MAD BRO]

also when there is load on the motor or when the motor is first starting, there is little relative change between the coil and the magnetic field so the back emf is less.
This means supply emf will be very high causing the motor to heat up, that's why a variable resistor is used.

 

[umm what]

also when there is load on the motor or when the motor is first starting, there is little relative change between the coil and the magnetic field so the back emf is less.
This means supply emf will be very high causing the motor to heat up, that's why a variable resistor is used.

Starting resistance*

 

[someth1ng]

also when there is load on the motor or when the motor is first starting, there is little relative change between the coil and the magnetic field so the back emf is less.
This means supply emf will be very high causing the motor to heat up, that's why a variable resistor is used.

You should say net emf is very high because supply emf is actually constant and you should also refer to Ploss=I^2 R

 

[U MAD BRO]

yes I agree, net EMF increases, which increases the current in the coil and according to Ploss=I^R the coil will heat up etc...
But why do we always call it a starting resistance, what if there is load on the motor, i.e. a drill, don't we need a resistor to protect the coil?

 

[someth1ng]

The thing with a drill is that the slower the rotation, the higher the torque which you need...

 

[U MAD BRO]

The thing with a drill is that the slower the rotation, the higher the torque which you need...

yeah that makes sense.