## EMF And Torque Equation of DC Machine

Table of Contents

EMF And Torque Equation of DC Machine, EMF Equation of DC Generator, Torque Equation of DC Motor

## EMF Equation of DC Generator

Consider a DC generator with the following parameters,

P = number of field poles

Ø = flux produced per pole in Wb (weber)

Z = total no. of armature conductors

A = no. of parallel paths in armature

N = rotational speed of armature in revolutions per min. (rpm)

Now,

**Average emf generated per conductor is given by dΦ/dt (Volts) … eq. 1****Flux cut by one conductor in one revolution = dΦ = PΦ ….(Weber),****Number of revolutions per second (speed in RPS) = N/60****Therefore, time for one revolution = dt = 60/N (Seconds)****From eq. 1, emf generated per conductor = dΦ/dt = PΦN/60 (Volts) …..(eq. 2)**

Above equation-2 gives the emf generated in one conductor of the generator. The conductors are connected in series per parallel path, and the emf across the generator terminals is equal to the generated emf across any parallel path.

Therefore, **Eg = PΦNZ / 60A**

For simplex lap winding, number of parallel paths is equal to the number of poles (i.e. A=P),

Therefore, for simplex lap wound dc generator, **Eg = PΦNZ / 60P**

For simplex wave winding, number of parallel paths is equal to 2 (i.e P=2),

Therefore, for simplex wave wound dc generator, **Eg = PΦNZ / 120**

## Torque Equation of DC Motor

Between the armature and the stator, a mechanical torque develops when the armature conductors of a DC motor carry current while the stator field flux is present. The torque is calculated by multiplying the force by the radius at which it acts.

**dc motor torque equation T = F × r (N-m) …where, F = force and r = radius of the armature****Work done by this force in once revolution = Force × distance = F × 2πr (where, 2πr = circumference of the armature)****Net power developed in the armature = word done / time**

= (force × circumference × no. of revolutions) / time

= (F × 2πr × N) / 60 (Joules per second) …. eq. 2.1

But, F × r = T and 2πN/60 = angular velocity ω in radians per second. Putting these in the above equation 2.1

Net power developed in the armature = P = T × ω (Joules per second)

### Armature Torque (Ta)

**The power developed in the armature can be given as, Pa = Ta × ω = Ta × 2πN/60**- The mechanical power developed in the armature is converted from the electrical power,

Therefore, mechanical power = electrical power

That means, Ta × 2πN/60 = Eb.Ia - We know, Eb = PΦNZ / 60A
- Therefore, Ta × 2πN/60 = (PΦNZ / 60A) × Ia
- Rearranging the above equation,
**Ta = (PZ / 2πA) × Φ.Ia**(N-m)

The term (PZ / 2πA) is practically constant for a DC machine. Thus, armature torque is directly proportional to the product of the flux and the armature current i.e. Ta ∝ Φ.Ia

### Shaft Torque (Tsh)

Due to **iron and friction losses** in a dc machine, At the machine’s shaft, the entire developed armature torque is unavailable. Shaft torque is never greater than armature torque because some torque is lost.

A DC motor’s shaft torque is specified as,

Tsh = output in watts / (2πN/60) ….(where, N is speed in RPM)