Circuit Model of DC Machines - Free MCQ Practice Test with solutions, GATE

 The shaft power is sum of mechanical power and rotational losses.

When the electromagnetic torque is in opposite direction, it is of motoring nature.

The conductor emf and current will be in same direction and the developed torque is in opposite direction for a generator.

As the terminal voltage is lesser than armature voltage, the supply is fed to the machine and so it will be acting like a motor.

The power of the machine remains unaltered by the type of connections.

 From the speed and emf relation, E = 250*2950/3000
= 245.8 V
This is less than the terminal voltage. Hence it is a motor.

Correct answer: 342 A (option A).

Use the armature equation: E - I_a R_a = V, hence I_a = (E - V)/R_a.

Substituting E = 127 V, V = 120 V and R_a = 0.02 Ω: I_a = (127 - 120)/0.02 = 7/0.02 = 350 A.

Field current is I_f = V / R_f = 120 / 15 = 8 A.

The armature supplies both field and external load, so I_a = I_load + I_f. Therefore I_load = I_a - I_f = 350 - 8 = 342 A.

Thus the load current is 342 A, so option A is correct.

 This is a motor performance equation.

The voltage drop at brush-commutator contact is variable (1-2V) and independent of armature current.

Concept:
Induced EMF in the generator is given by
E_g = V + l_a x R_a
Induced EMF in the motor is given by
E_m = V - l_a x R_a
Where V = terminal voltage
l_a = armature current
R_a = armature resistance
Calculation:
Difference between induced EMF when machine act as a generator and as a motor = E_g - E_m
= 2 I_aR_a = 2 x 20 x 1 = 40 V