An SCR conducts in the positive half-cycle only. If it is fired at α = 40° and the average anode current is 50 A, what will happen to the average current when the firing angle is changed to α = 80° (same conditions)?

Introduction / Context:
In a half-wave controlled rectifier with a resistive load, the average output (and hence average current at fixed load resistance) depends on the firing angle α through the term (1 + cos α). Increasing α delays conduction and lowers the average value. This question examines how much it drops when α is increased from 40° to 80°.

Given Data / Assumptions:

• Half-wave controlled rectifier, conduction in the positive half only.
• Average current at α = 40° is 50 A.
• Load and supply remain the same; proportionality to (1 + cos α) holds.
• Resistive behavior so Idc ∝ Vdc.

Concept / Approach:

For a half-wave controlled rectifier: Vdc ∝ (1 + cos α). So the ratio of average currents equals the ratio of (1 + cos α) values for the two firing angles. Compute this ratio and scale the known current to estimate the new average current at α = 80°.

Step-by-Step Solution:

Verification / Alternative check:

As α increases, conduction window shrinks, so current must drop. A value between 25 A and 50 A fits physical expectations and the computed ratio.

Why Other Options Are Wrong:

• 50 A: implies no change with α, incorrect.
• 25 A or less than 25 A: too low; would require a much larger α or different load.
• More than 50 A: contradicts the monotonic decrease of (1 + cos α) with α in 0 to 180°.

Common Pitfalls:

• Using Vdc ∝ cos α (full-wave bridge intuition) instead of the half-wave form.
• Forgetting to compute the ratio rather than absolute values.

Final Answer:

less than 50 A but more than 25 A