A synchronous motor operating at rated voltage draws 1.0 pu current and 1.0 pu power factor. The machine parameters are: synchronous reactance 1.0 pu, armature resistance negligible. Apart from supplying this rated power, if the motor has to supply an additional leading reactive power of 0.8 pu, then the field current has to be increased by?




Before we proceed with the solution, we need to understand the problem statement. It states that a synchronous motor is operating at rated voltage and draws 1.0 pu current and 1.0 pu power factor. It also gives the machine parameters such as synchronous reactance and armature resistance. The problem statement asks us to find the increase in field current required to supply an additional leading reactive power of 0.8 pu.


We know that the synchronous motor is operating at rated voltage, 1.0 pu current, and 1.0 pu power factor. Therefore, we can calculate the rated power using the following formula:

Rated power = V * I * PF

Where V is the rated voltage, I is the rated current, and PF is the power factor.

Substituting the given values, we get:

Rated power = 1.0 * 1.0 * 1.0 = 1.0 pu

Therefore, the synchronous motor is operating at a rated power of 1.0 pu.


The problem statement asks us to find the increase in field current required to supply an additional leading reactive power of 0.8 pu. Therefore, we need to calculate the total reactive power required by the motor.

We know that the power factor of the motor is 1.0 pu. Therefore, the reactive power is equal to zero. However, the problem statement asks us to supply an additional leading reactive power of 0.8 pu. Therefore, the total reactive power required by the motor is 0.8 pu.


The field current required by the synchronous motor is given by the following formula:

Field current = V / Xs

Where V is the rated voltage and Xs is the synchronous reactance.

Substituting the given values, we get:

Field current = 1.0 / 1.0 = 1.0 pu

Therefore, the field current required by the synchronous motor to operate at a rated power of 1.0 pu is 1.0 pu.


To supply an additional leading reactive power of 0.8 pu, the field current has to be increased.

We know that the reactive power is given by the following formula:

Reactive power = V * I * sin(θ)

Where θ is the angle between the voltage and current phasors.

Since the motor is operating at a power factor of 1.0 pu, θ is equal to zero. Therefore, the reactive power is equal to zero.

To supply an additional leading reactive power of 0.8 pu, the angle θ has to be increased. Since the motor has to supply leading reactive power, the angle θ has to be