# The torque developed by a d.c motor is directly proportional to

The torque developed by a d.c motor is directly proportional to

### Right Answer is: Flux per pole × Armature current

#### SOLUTION

By the term, torque is meant the turning or twisting moment of a force about an axis. It is measured by the product of the force and the radius at which this force acts. Consider a pulley of radius r meter acted upon by a circumferential force of F Newton which causes it to rotate at N r.p.m. The angular speed of the pulley is

Work done by this force in one revolution

= Force × distance = F × 2πR Joule

The power developed = Work Done/Time

= (F × 2πR)/60/N

= (F × R) × (2πN)/60

The power developed = T × ω watt or P = T ω Watt

### Armature Torque

Let Ta be the torque developed by the armature of a motor running at N r.p.s. If Ta is in N/M, then
power developed

P =  Ta × 2π N watt

The electrical power converted into mechanical power in the armature = EbIa watt

Power in armature = Armature torque × ω

EbIa = Ta × 2πN/60
or
Ta = 9.55EbIa /N (N-m)

The back EMF (Eb) of the motor is given by

EB = PΦZN/60A

Φ = flux per pole in Weber

P = number of poles

Z = total number of armature conductor

N = rotation speed of the armature in revolution per minute (r.p.m)

A = number of parallel paths

$\begin{array}{l}\dfrac{{P\Phi ZN}}{{60A}} \times {I_a} = {T_a} \times \dfrac{{2\pi N}}{{60}}\\\\{T_a} = \dfrac{1}{{2\pi }}\Phi {I_a} \times \dfrac{{PZ}}{A}\\\\{T_a} = 0.159\Phi {I_a}.\dfrac{{PZ}}{A}N – M\end{array}$

Hence From the above equation it is clear that the torque of the DC series motor is directly proportional to the armature current (Ia) and Flux per pole(Φ).

T ∝ Ia × Φ

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