DC chopper feeding an R–L–E load at duty cycle a = 1 When a step-down chopper has duty ratio a = 1 (switch always ON), what is the relationship between the minimum and maximum load current over a switching period?

Introduction / Context:In buck choppers feeding R–L–E loads, the inductor current usually ripples because the inductor sees different voltages during ON and OFF intervals. If the duty ratio a equals 1, the switch never turns OFF.

Given Data / Assumptions:

• Ideal buck chopper, a = 1.
• Load model R–L–E (back-emf source E).
• Steady-state operation.

Concept / Approach:With a = 1, the inductor is connected continuously to the input source. There is no alternating ON/OFF in a switching sense within a period, so the inductor voltage is essentially constant after transients. A constant inductor voltage implies no periodic current ripple.

Step-by-Step Solution:For a = 1: v_L = V_in − R*i − E.In steady state, the average di/dt over a period is zero; with no switching intervals, the periodic ripple component disappears.Therefore Imax and Imin over any putative switching period are equal: Imax = Imin.

Verification / Alternative check:Let the converter be replaced by a DC source V_in. The R–L–E circuit reaches a DC current set by (V_in − E)/R with only an exponential transient, not a periodic ripple.

Why Other Options Are Wrong:

• (b) Implies ripple—contradicted by a = 1.
• (c) and (d) discuss sign; the sign depends on V_in and E and is not the point asked.
• (e) Discontinuity cannot occur when the switch is always ON with an inductor in series.

Common Pitfalls:

• Assuming ripple exists for any duty ratio; at a = 0 or a = 1 ideal ripple is zero.

Final Answer:Imax = Imin (no ripple)