A single-phase semiconverter supplies a highly inductive R–L load and includes a freewheeling diode across the load. Compare the shapes of output voltage and output current waveforms.

Difficulty: Easy

Correct Answer: are not similar

Explanation:


Introduction / Context:
In semiconverters (half-controlled bridges) feeding highly inductive loads, current tends to be smoothed by inductance while voltage remains pulsed. A freewheeling diode provides a path during source non-conduction, further decoupling current shape from the instantaneous source voltage.



Given Data / Assumptions:

  • Highly inductive load ⇒ large L, current ripple small.
  • Freewheeling diode (FD) across load.
  • Phase-controlled thyristors determine when source applies positive voltage.


Concept / Approach:

Output voltage v_o(t) is a chopped sinusoid (or zero during freewheeling), whereas output current i_o(t) is nearly continuous due to the inductor storing energy and because the FD maintains a current path when the converter is not supplying voltage. Thus v_o(t) and i_o(t) are generally dissimilar in shape.



Step-by-Step Reasoning:

During conduction: v_o(t) ≈ rectified line segment; i_o(t) rises slowly.During freewheeling: v_o(t) ≈ 0 (across load), but i_o(t) continues flowing through FD, decaying slowly.Hence, current is quasi-DC while voltage is pulsed → not similar.


Verification / Alternative check:

Waveform sketches from standard texts show pulsed v_o with near-continuous i_o for large L and an FD.



Why Other Options Are Wrong:

  • “Are similar/identical”: contradict inductive smoothing and freewheeling action.
  • “Only if α = 0”: even at α = 0, v_o is sinusoidal-rectified while i_o is smoothed.
  • “May be similar”: not for highly inductive loads with FD.


Common Pitfalls:

  • Assuming resistive-load intuition carries over to inductive cases.
  • Ignoring FD role in maintaining current when source is off.


Final Answer:

are not similar

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