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SSC JE EE Previous Paper 12 (Held on: 24 March 2021 Evening)

Option 4 : 25 A

__Concept:__

Consider a separately excited DC motor as,

Eb is back EMF developed in the motor at speed N rpm

V = Eb + Ia Ra** .....(1)**

V is the supply voltage

Ra is armature resistance

Ia is armature current

Vf is field supply

Rf is field winding resistance

If is field winding current

Let T is the torque developed by the armature of a motor running at N rpm.

Power developed in the armature of motor can be written as,

P = V Ia = Eb Ia+ I2a Ra

Thus the gross mechanical power output provided by the motor is

Pm = Eb Ia = V Ia - I2a Ra(losses)

**The efficiency of the motor is = Output Power(Pm) / Input power(P)**

**Calculation**

P_{m }= 10 HP

= 10 × 746 = 7460 W

V = 400

η = 0.7355

η = 7460 / (400 × I)

⇒ 0.7355 = 7460 / (400 × I)

⇒ I = 25.35 ≈ **25 A**

__Additional Information__

Maximum power condition:

If the value of \(\frac{dP_m}{dI_a}=0,\) the motor will deliver maximum power

\(\frac{dP_m}{dI_a}= V - 2I_aR_a = 0\)

Ia Ra = V / 2 ....(2)

From equation (1) & (2)

Eb = V / 2

The efficiency of the motor is = Output Power(Pm) / Input power(P)

\(η= \frac{E_b I_a }{VI_a }× 100=\frac{E_b }{V}× 100=50\)%

** **