Back EMF and torque explained in DC machine with numerical example

⚡ Back EMF (Electromotive Force) in DC Machine

Definition: When the armature of a DC motor rotates in the magnetic field, it cuts the magnetic flux and an EMF is induced in the armature conductors which opposes the applied voltage. This is called Back EMF (E_b).

Formula:

E_b = (Φ × Z × N × P) / (60 × A)

• Φ = Flux per pole (in Weber)
• Z = Total number of armature conductors
• N = Speed of motor (in RPM)
• P = Number of poles
• A = Number of parallel paths in armature

⚙️ Torque in DC Motor

Definition: Torque is the rotating force developed on the armature of a DC motor which drives the mechanical load.

Formula:

T = (Φ × Z × I_a × P) / (2 × π × A)

• I_a = Armature current
• Other symbols same as above

🔗 Relationship Between Back EMF, Torque & Speed

The speed of a DC motor is given by:

N ∝ (V − I_a R_a) / Φ

Where V = Supply voltage and R_a = Armature resistance.

📐 Numerical Examples on Back EMF & Torque

✅ Example 1: Back EMF Calculation

Problem: A 4-pole, 230 V DC shunt motor has 888 armature conductors. The flux per pole is 20 mWb. The motor runs at 1000 RPM. The armature is wave connected (A = 2). Find the back EMF.

Solution:

Given: Φ = 20 mWb = 20 × 10^-3 Wb, Z = 888, P = 4, N = 1000 RPM, A = 2

Formula: E_b = (Φ × Z × N × P) / (60 × A)

Substituting values:

E_b = (0.02 × 888 × 1000 × 4) / (60 × 2)

E_b = (71,040) / 120 = 592 V

Answer: Back EMF = 592 V

---

✅ Example 2: Torque Calculation

Problem: A 4-pole DC motor has 600 conductors on its armature. The flux per pole is 25 mWb. If the armature current is 50 A and the motor is lap connected (A = 4), calculate the torque developed.

Solution:

Given: Φ = 25 mWb = 25 × 10^-3 Wb, Z = 600, P = 4, I_a = 50 A, A = 4

Formula: T = (Φ × Z × I_a × P) / (2 × π × A)

Substituting values:

T = (0.025 × 600 × 50 × 4) / (2 × π × 4)

T = (3000) / (8 × 3.1416)

T = 119.37 Nm

Answer: Torque developed = 119.4 Nm