DC chopper control of a separately excited DC motor A separately excited DC motor is fed from a DC source through a chopper. At 600 rpm and rated torque, the chopper duty cycle is 0.8. At the same 600 rpm but at half the rated torque, what is the most likely duty cycle (assume armature resistance drop is not negligible but smaller than back EMF)?

Introduction / Context:
For a DC motor under chopper control, average armature voltage V_avg ≈ E + I_aR_a at a given speed. At constant speed, back EMF E is approximately constant. Reducing torque reduces armature current I_a, which reduces the resistive drop I_aR_a, hence the required V_avg and duty cycle decrease moderately.

Given Data / Assumptions:

• Speed held at 600 rpm in both cases → back EMF E nearly unchanged.
• Rated torque → I_a,r; half torque → I_a ≈ 0.5I_a,r.
• V_avg = αV_s = E + I_aR_a.
• R_a drop is smaller than E but not negligible.

Concept / Approach:

Duty cycle scales with required V_avg. If current halves, the I_aR_a term halves while E is constant; thus α reduces slightly from 0.8, not drastically. A modest reduction to about 0.75 aligns with typical ratios when E dominates the sum.

Step-by-Step Solution:

Verification / Alternative check:

In the limit R_a → 0, α would be unchanged at 0.8 (since E dominates). For large R_a, α would drop more than a little. Hence 0.75 is a reasonable mid-case.

Why Other Options Are Wrong:

• 0.9: would increase voltage despite reduced torque—illogical.
• 0.8: exact equality would imply zero R_a drop, which is unlikely per the prompt.
• 0.6: indicates excessive drop; inconsistent with “R_a drop smaller than E”.

Common Pitfalls:

Assuming duty cycle scales directly with torque; it actually depends on both E and I_aR_a.

Final Answer:

0.75