# Neglecting all losses, the developed torque (T) of a DC separately excited motor, operating under constant terminal voltage, is related to its output power (P) as under:

### Right Answer is:

T ∝ P

#### SOLUTION

Relation of output power and torque under constant terminal voltage can be given as

The torque developed by a d.c motor is directly proportional to Flux per pole × Armature Resistance i.e

**T ∝ ΦI _{a}
T ∝ I_{a}———1
**

Output power for the separately excited motor is given as

**P = E _{b}I_{a}**.

If voltage supply is constant then power is

**P** **∝ I _{a}**

**……………. (2)**

From equation (1) and (2),

**T ∝ P.**

or

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

ω = 2πN/60 rad/sec

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**

From the relation of power developed in the armature is equivalent to mechanical torque developed, is

**P = T × ω**

Where

P = Output power of separately excited motor and it is given as **P = E _{b}I_{a}**ω = Angular speed in rad/sec. and it is given as ω = 2πN ⁄ 60

where N = Speed of motor in rpm,

E_{b} = induced back emf,

I_{a} = armature current.

If voltage supply is constant then power is

∴ T∝ power output P will be equal to power developed in the armature.

∴ T ∝ P.