When the current-carrying conductor is placed in the magnetic field, the conductor experiences a force that exerts a turning moment or torque F x r. The current and magnetic field produce the torque, and hence, it is called Electromagnetic torque. In a DC motor, the armature current flows in the presence of air gap flux, and due to an interaction of field flux and armature current, the armature conductors experience a force. Hence, the torque is produced at the circumference of the rotor, and the rotor starts rotating.

To understand the torque equation, let us first understand the voltage equation of the DC motor.

Referring to the above diagram, E is the supply voltage fed to the armature, and Eb is the back emf developed across the armature. The back EMF always opposes the armature voltage, and thus, it limits the armature current. Ra is the armature resistance. The IaRa drop takes place in the armature when the armature current Ia flows through the armature. The voltage equation of the DC motor is given below.

First, power needs to be calculated to calculate torque. If we multiply the back EMF(Eb) with armature current(Ia), we get the motor’s power.

**Derivation of Torque Equation of DC motor**

Multiplying both sides of equation(1) by armature current(Ia), We get

I_{a}^{2} R_{a} is the power loss that occurs in the armature and is wasted in the form of heat energy, and thus, the power developed in a DC motor is;

The torque is the rotational force, and it depends on the developed power of the armature at a particular speed. Thus, the torque of the motor depends on the speed of rotation.

Putting the value of Pm from equation(3) in equation(5),

The **back EMF of the DC motor** is mathematically expressed as;

Putting the value of back emf (E_{b}) from equation(8) in equation(7), we get the torque equation of the DC motor.

The mechanical torque developed by DC motor can be calculated by subtracting the mechanical loss from the gross torque.

T_{m} is also called the shaft torque(T_{sh}) of the DC motor.

**Solved Problems on the Torque Equation of DC motor**

**A DC motor takes an armature current of 100 A at 460 V. The armature circuit resistance is 0.2 Ω. The machine has 6 poles, and the armature is lap-connected with 864 conductors. The flux per pole is 0.05 Wb. Calculate-(i) the Speed of the motor and (ii) the gross torque developed by the armature. **

I_{a} = 100 A

V = 460 V

R_{a} = 0.2 Ω

P = 6

A = P= 6 (Lap winding)

Z = 864

Φ = 0.05 Wb

T_{a} =?

E_{b} = ΦZN/60 x (P/A) ——–(1)

Also, E_{b} = V – I_{a}R_{a}

= 460 – 100 x 0.2

= 460 – 20**Eb = 440 Volts**

Putting the value of Eb in equation (1)

E_{b} = ΦZN/60 x (P/A)

440 = 0.05 x 864 x N /60 x (6/6)

440 = 0.72 N**N = 440 /0.72 = 611 r.p.m.**

**Motor Gross Torque**

T = ΦIa/2π x Z (P/A)

T = 0.159 ΦIa x Z (P/A)

T = 0.159 x 0.05 x 100 x 864 x(6/6)

T = 686.55 N-m

## 1 thought on “Torque Equation of DC Motor- Its Derivation”