Question

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

• A. 23.28
• B. 4.82
• C. 46.7
• D. 0

Hint:

In DC series motors, speed reduction affects load torque significantly when torque depends on speed. Use the relation \( T \propto I^2 \) and \( E \propto N I \) when magnetic saturation is not present.

Answer 1

The Correct Option is A

Let:
- Initial speed = \( N \), new speed = \( N/2 \)
- Torque \( T \propto N^3 \Rightarrow T_2 = \left(\frac{N}{2}\right)^3 = \frac{1}{8} T_1 \)

For a DC series motor with negligible internal resistance and assuming a linear magnetic circuit:
- Torque \( T \propto \phi I \propto I^2 \Rightarrow T \propto I^2 \)
- So \( \frac{T_2}{T_1} = \left( \frac{I_2}{I_1} \right)^2 = \frac{1}{8} \Rightarrow I_2 = \frac{I_1}{\sqrt{8}} = \frac{40}{\sqrt{8}} = 14.14 \, {A} \)

Now, for a DC motor:
- \( V = E + I_a R \), and for negligible resistance, initially:
\[ E_1 = V = 400 \, {V} \]
Back EMF is proportional to speed and flux:
\[ E \propto N \phi \Rightarrow E_2 = \frac{1}{2} \cdot \frac{14.14}{40} \cdot E_1 = \frac{1}{2} \cdot \frac{14.14}{40} \cdot 400 = 70.7 \, {V} \]

Now apply KVL with external resistance \( R \):
\[ V = E_2 + I_2 R \Rightarrow 400 = 70.7 + 14.14 R \]