What Are the Effects of Counter-EMF on Electric Motors?

[jim hardy]

right

one of the formulas you can derive from that basic understanding is

e = Blv

e = voltage
l = length of conductor
v = velocity of conductor relative to field
B = magnetic flux density

"""after long hours thinking about it...""" bravo !
cross check your microscopic understanding against the formulas in your book and make them meld. Once you have the basic physics imprinted you can derive the formulas with ease, which beats cramming for exams.

old jim

 

[Momento]

right

one of the formulas you can derive from that basic understanding is

e = Blv

e = voltage
l = length of conductor
v = velocity of conductor relative to field
B = magnetic flux density

"""after long hours thinking about it...""" bravo !
cross check your microscopic understanding against the formulas in your book and make them meld. Once you have the basic physics imprinted you can derive the formulas with ease, which beats cramming for exams.

old jim

Makes sense!

Because a conductor with many turn's will generate a significantly powerful C-EMF, if there is a small portion exposed to the magnetic field it would generate a weak C-EMF compared to INPUT-EMF,

Thanks Jim! I appreciate all you're efforts everyone thanks!

 

[CGOLDING]

When C-EMF is produced - is it true that some of the energy is dissipated as heat - if so is there an equation which can be used to find out how much is lost in heat.

 

[NascentOxygen]

When C-EMF is produced - is it true that some of the energy is dissipated as heat - if so is there an equation which can be used to find out how much is lost in heat.

The energy losses as heat arise from friction and I².R losses in conductors (that includes eddy currents in metal). Primarily, it's the I².R copper losses that are load-dependent; if there is no armature current, these losses are a minimum. The counter emf determines the current, this in turn determines the Ohmic losses.