## 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

E_{b }= 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}