6 a] A dc motor running with a speed of N rpm obtain an expression for EMF induced in the armature winding.
Solution :
When the Armature of D C motor starts rotating with a speed of „N‟ rpm and armature conductor cuts the magnetic flux, hence an EMF is induced in the Conductor called Back EMF.
The induced emf acts in opposite direction to the applied voltage „V‟ (Lenz‟s law)
The induced EMF in the motor is given by
Let,
Φ = Flux produced by each pole in weber (Wb) and
P = number of poles in the DC motor.
N = speed of the armature conductor in rpm.
Consider a one revolution of the conductor
Total flux produced by all the poles =∅×𝐏
Time taken to complete one revolution=60/N
Now, according to Faraday’s law of induction, the induced EMF of the conductor is equal to rate of change of flux.
e=\frac{d\phi }{dt} and e=\frac{total flux}{time take}
therefor,
Induced EMF of one Conductor is
e=\frac{\phi P}{\frac{60}{N}} = \phi P\frac{N}{60}Let us suppose there are Z total numbers of conductor in a motor, and arranged in such a manner that all parallel paths are always in series.
Here, Z = total numbers of conductor A = number of parallel paths
Then, Z/A = number of conductors connected in series
Therefore,
Induced EMF of DC motor
Eb = EMF of one conductor × number of conductor connected in series.
Induced EMF of DC generator is
e= \frac{N\phi P}{60}\times \frac{Z}{A} volts E_{b}=\frac{\phi PNZ}{60A}