Discontinuous current in a DC chopper with an R–L–E load Under which condition is the inductor current most likely to become discontinuous?

Introduction / Context:
Current continuity in choppers depends on how long the inductor sees a positive or negative voltage each cycle. If during part of the cycle the inductor voltage drives current toward zero long enough, the current can reach zero, leading to discontinuous conduction mode (DCM).

Given Data / Assumptions:

• Buck-type chopper feeding R–L–E load.
• Ideal components for conceptual reasoning.

Concept / Approach:
Low duty ratio means the switch is ON for a short time and OFF for a long time. During OFF time, the inductor current freewheels and decays (especially if E opposes current). If the OFF interval is sufficiently long, the current can fall to zero before the next ON interval, causing DCM.

Step-by-Step Solution:
When a is low, average inductor voltage is small.The prolonged OFF-time lets i(t) decay exponentially via the freewheel path.If decay reaches zero before the next ON, conduction becomes discontinuous.

Verification / Alternative check:
Time-constant analysis: with small a, OFF interval dominates. For modest L and non-negligible R or opposing E, i(t) may hit zero within one period.

Why Other Options Are Wrong:

• (a) High L resists current change and favors continuous conduction.
• (c) High E can contribute to decay but DCM is most predictably caused by insufficient ON-time—hence (b) is the primary condition.
• (d) Incorrect because (b) is valid.
• (e) Low R leads to slower decay, favoring CCM.

Common Pitfalls:

• Confusing the roles of L and R; large L supports continuity, large R speeds decay.

Final Answer:
Duty ratio a is low