Question

What happens if the field winding of a running DC Shunt motor suddenly opens?

• A. The motor stops immediately
• B. The motor speed remains constant
• C. The motor attains dangerously high speed
• D. The motor starts running in reverse

Hint:

Always use a Field Failure Relay (or field release coil in a 3-point/4-point starter) in DC shunt motor starter circuits.
This ensures that if the field current drops or opens, the armature supply is disconnected instantly to prevent a dangerous runaway condition.

Answer 1

The Correct Option is C

Step 1: Understanding the Problem:
This question explores the relationship between the field excitation and speed of a DC shunt motor under running conditions.

Step 2: Key Formula or Approach:
The speed ($N$) of a DC motor is given by the speed equation:
\[ N = \frac{E_b}{k \Phi} \]
where:
$E_b$ = Back EMF
$\Phi$ = Magnetic flux per pole
$k$ = Motor constant

Step 3: Detailed Explanation:

• In a DC shunt motor, the field winding is connected in parallel with the armature across the voltage supply, providing a constant flux.

• If the field winding suddenly opens while the motor is running, the shunt field current drops to zero.

• The magnetic flux ($\Phi$) in the machine does not immediately drop to absolute zero, but falls to a very low value determined only by the residual magnetism ($\Phi_{res}$) of the magnetic pole cores.

• According to the speed relation $N \propto \frac{1}{\Phi}$, when the flux becomes extremely small, the speed must increase dramatically to maintain the back EMF ($E_b = V - I_a R_a$).

• As the speed attempts to rise toward infinity, the motor enters a "runaway" condition.

• This dangerously high speed can lead to catastrophic mechanical failure of the armature due to massive centrifugal forces.

Step 4: Final Answer
Thus, the motor will attain a dangerously high speed, corresponding to option (C).