Rectifier conduction interval: For a half-wave rectifier delivering a rectified voltage across a load resistor from a sinusoidal AC source, over what portion of each cycle does load current flow?

Introduction / Context:A half-wave rectifier uses a single diode to pass only one polarity of an AC waveform to the load. Knowing the conduction interval helps estimate average DC value, ripple, and transformer utilization in basic power-supply design.

Given Data / Assumptions:

• Sinusoidal input voltage.
• Ideal diode (conducts in one direction, blocks in the other).
• Resistive load.

Concept / Approach:During the positive half-cycle (assuming diode oriented to pass positive), the input drives the diode forward; current flows and the load sees the positive half-sine. During the negative half-cycle, the diode is reverse-biased; current ceases and the load voltage is near zero. Thus, conduction occurs for half the cycle.

Step-by-Step Solution:Positive half-cycle: diode forward-biased → current flows for ~0° to 180°.Negative half-cycle: diode reverse-biased → current ~0 for 180° to 360°.Net: conduction for 180° of each 360° cycle.Average DC output and ripple can be computed from the half-sine waveform.

Verification / Alternative check:Oscilloscope view of the load voltage shows only positive lobes separated by zero-volt intervals, confirming half-cycle conduction.

Why Other Options Are Wrong:0°: no conduction would imply an open circuit.90°: quarter-cycle conduction does not match a simple half-wave rectifier with ideal diode.360°: full-cycle conduction corresponds to full-wave rectification or a resistive AC load without rectification.

Common Pitfalls:Forgetting diode drop and transformer secondary resistance slightly reduce the exact conduction interval around the zero crossings in real circuits; for basic analysis we assume ideal behavior.

Final Answer:180 degrees