---
title: "DC motor working principle and torque equation"
subject: "Electrical Machines I"
chapter: 4
key: "semester_4/electrical_machines_i:dc-motor-principle-torque-equation"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/dc-motor-principle-torque-equation
site: BCT Engineering
---

# DC motor working principle and torque equation

Electrical Machines I · Chapter 4: DC Motor

- Asked in: 2076 Bhadra, 8 marks; 2071 Magh, 5+3 marks; 2070 Magh, 4 marks; 2069 Poush, 8 marks (asked 4 times)
- Video: [Working Principle of DC Motor in Electrical Machines with Diagram and Example |BEE|](https://www.youtube.com/watch?v=yw4jss4w0qU) — Trouble- Free
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/dc-motor-principle-torque-equation
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-4

## Question

Explain working principle of D.C. motor. Derive torque equation of a D.C. motor.

## Short answer

A DC motor uses the force F=BIl on armature conductors in a magnetic field, with Fleming's left-hand rule giving direction; the commutator reverses conductor current to keep torque unidirectional. Torque developed is Ta = P φ Z Ia / (2π A) N·m, or Ta = k φ Ia, with k = PZ/(2πA).

## Answer

A DC motor converts DC electrical energy into mechanical energy. It works on the principle that **a current-carrying conductor placed in a magnetic field experiences a mechanical force**, $F = B I l$ newtons, whose direction is given by **Fleming's left-hand rule**.

#### Working principle
1. The field winding produces a steady flux between N and S poles.
2. DC supply is given to the armature through brushes and commutator, so current flows in the armature conductors.
3. Each conductor carries current in the field and experiences a force $F = BIl$. Conductors under the N pole and under the S pole carry opposite currents (because of the commutator), so their forces act in the same rotational sense and produce a **torque**.
4. When a conductor moves from one pole to the next, the commutator reverses its current, so the torque stays in one direction and the rotation is continuous.
5. As the armature rotates, a back emf $E_b$ is induced which opposes the supply: $V = E_b + I_a R_a$.

```text
        N pole
   ---------------
    (x) (x)  -> F      x = current into page
       armature        . = current out of page
    (.) (.)  <- F
   ---------------
        S pole
  Forces form a couple -> rotation
```

#### Torque equation
**Method 1: from force on conductors**

Let $r$ = armature radius (m), $l$ = conductor length (m), $I_a/A$ = current per conductor. Average flux density:

$$
B = \frac{P\phi}{2\pi r l}
$$

Force per conductor $= B \dfrac{I_a}{A} l$. Torque per conductor $= B \dfrac{I_a}{A} l\, r$.

Total torque for $Z$ conductors:

$$
\begin{aligned}
T_a &= Z \cdot \frac{P\phi}{2\pi r l} \cdot \frac{I_a}{A} \cdot l \cdot r \\
&= \frac{P \phi Z I_a}{2\pi A}\ \text{N·m}
\end{aligned}
$$

**Method 2: from power balance**

Mechanical power developed $= E_b I_a$, and also $= T_a \omega = T_a \dfrac{2\pi N}{60}$. So

$$
T_a \cdot \frac{2\pi N}{60} = E_b I_a = \frac{P\phi Z N}{60 A} I_a
$$

$$
T_a = \frac{P \phi Z I_a}{2\pi A} = 0.159\, \frac{P \phi Z I_a}{A}\ \text{N·m}
$$

#### Conclusions
- For a given machine, $T_a = k \phi I_a$, i.e. $T_a \propto \phi I_a$.
- Shunt motor ($\phi$ nearly constant): $T_a \propto I_a$.
- Series motor ($\phi \propto I_a$ before saturation): $T_a \propto I_a^2$.
- In terms of speed: $T_a = 9.55 \dfrac{E_b I_a}{N}$ N·m. The useful **shaft torque** is $T_{sh} = 9.55 \dfrac{P_{out}}{N}$, which is less than $T_a$ by the torque lost to iron and friction losses.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. On what principle does a DC motor work?
   A. A current-carrying conductor in a magnetic field experiences a force (F = BIl)
   B. A conductor moving in a field has an emf induced in it at standstill
   C. A changing flux induces current in a stationary coil
   D. Opposite charges attract and produce rotation
   Answer: A. Force on a current-carrying conductor in a field gives torque; direction by Fleming's left-hand rule. (Key point: Basic principle)

2. Why does the torque stay in one direction as the armature rotates?
   A. The field polarity reverses every half turn
   B. The commutator reverses the current in a conductor when it moves from one pole to the next
   C. The brushes rotate along with the armature
   D. The back emf reverses along with the current
   Answer: B. Current reversal under each pole keeps the force on conductors in the same rotational sense. (Key point: Commutator role)

3. Which expression gives the armature torque of a DC motor?
   A. Ta = P·φ·Z·N / (60·A)
   B. Ta = 2π·P·φ·Z / (A·Ia)
   C. Ta = P·φ·Z·Ia / (2π·A)
   D. Ta = P·φ·A·Ia / (2π·Z)
   Answer: C. Equating Eb·Ia = Ta·ω gives Ta = PφZIa/(2πA); the other expression is the back emf. (Key point: Torque equation)

4. How does torque vary with armature current in a series motor (before saturation) compared with a shunt motor?
   A. Series: Ta ∝ Ia; shunt: Ta ∝ Ia^2
   B. Both are proportional to Ia^2
   C. Both are proportional to Ia
   D. Series: Ta ∝ Ia^2; shunt: Ta ∝ Ia
   Answer: D. Ta ∝ φIa; in a series motor φ ∝ Ia, in a shunt motor φ is nearly constant. (Key point: Series vs shunt torque)

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