power of motor

子路

2010-04-05 11:45 pm

given:
when the 10kg load exists,
back emf=100V
mechanical power output=500W

圖片參考：http://i256.photobucket.com/albums/hh182/zilu_photo/sshot-2010-04-05-15-38-20.png

圖片參考：http://i256.photobucket.com/albums/hh182/zilu_photo/sshot-2010-04-05-15-38-41.png

i have reservation for (d) (ii)
the solution is

圖片參考：http://i256.photobucket.com/albums/hh182/zilu_photo/sshot-2010-04-05-15-41-52.png

I think:

(current)(back emf)=(torque)(angular frequency)

when the load is removed,
the current changes and the torque provided by the motor also changes

I don't think we can calculate the new speed by this way
I have no way to find the new speed.....

更新1:

torque/current = constant thanks!!! i missed this

回答 (1)

天同

2010-04-06 4:47 am

✔ 最佳答案

As what you have pointed out, the work done by the back emf equals to mechanical output of the motor.
i.e. (current)(back emf)=(torque)(angular velocity)
hence, back emf/angular velocity = torque/current

But the torque produced by the motor is proportional to the current flows through the coil. Thus, torque/current = constant

That is to say, back emf/angular velocity = constant
therefore, by simple proportion, you could calculate the new angular velocity if you have already calculated/given the back emf and angular velocity under an on-load condition.

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