Question

A DC series motor with negligible series resistance is running at a certain speed driving a load, where the load torque varies as the cube of the speed. The motor is fed from a 400 V DC source and draws 40 A armature current. Assume a linear magnetic circuit. The external resistance in ohms that must be connected in series with the armature to reduce the speed of the motor by half is closest to:

• A. 0
• B. 23.28
• C. 4.82
• D. 46.7

Hint:

For a DC motor, the speed is inversely proportional to the square root of the armature current under constant voltage.

Answer 1

The Correct Option is B

Step 1: In a DC series motor, the speed \( N \) is related to the armature voltage \( V \) and the armature current \( I_a \) by the following relationship: \[ N \propto \frac{V}{I_a} \] The load torque \( T \) varies as the cube of the speed, i.e., \[ T \propto N^3 \] Therefore, the speed of the motor is inversely proportional to the square root of the armature current when the load torque is adjusted.
Step 2: To reduce the speed by half, the armature current must be reduced to one-fourth of the original current.
Given:
- Armature current \( I_a = 40 \) A
- Armature voltage \( V = 400 \) V
Now, the new armature current will be \( I_a' = \frac{40}{2} = 20 \) A.
Step 3: Using the relationship between the voltage and current, the external resistance required to reduce the armature current by half is: \[ R = \frac{V}{I_a'} - \frac{V}{I_a} = \frac{400}{20} - \frac{400}{40} = 23.28 \, \text{ohms} \]
Thus, the external resistance required is 23.28 ohms.