If the field circuit of a DC shunt motor is opened
A. DC Shunt Motor
B. The motor will attain dangerously high speed.
C. The speed of motor will be reduced.
D. The armature current will reduce.
Answer: B
Share your understanding of this question with the correct explanation.
If the field circuit of a DC shunt motor is opened, the motor will attain dangerously high speed.
In a DC shunt motor, the field circuit consists of the field winding connected in parallel with the armature winding. The field winding is responsible for creating the magnetic field that interacts with the armature current to produce torque and drive the motor.
When the field circuit is opened, the field winding is disconnected from the power source, resulting in the absence of the magnetic field. Without the magnetic field, the motor loses its ability to generate the necessary torque to counterbalance the mechanical load.
As a result, the armature current continues to flow, but without the opposing electromagnetic force from the field winding. This leads to an unbalanced torque equation, where the torque produced by the armature current is insufficient to overcome the mechanical load.
The motor, in this condition, accelerates rapidly, attaining dangerously high speeds. This can pose significant risks to the motor itself, the driven machinery, and the surrounding environment.
Therefore, the correct answer is B: The motor will attain dangerously high speed. When the field circuit of a DC shunt motor is opened, the motor lacks the necessary torque to counterbalance the mechanical load, leading to a rapid increase in speed that can be hazardous.