A half-wave rectified sine has Vp = 169.7 V and an average (DC) value of 54 V. What is the average value of the corresponding full-wave rectified sine (i.e., the average of two full-wave peaks)?

Introduction / Context:
For a sinusoid, the average (DC) value after rectification depends on whether the rectification is half-wave or full-wave. This question asks you to convert a known half-wave average into its full-wave counterpart using standard rectifier averages.

Given Data / Assumptions:

• Sine-wave peak Vp = 169.7 V.
• Given half-wave average V_avg(half) = 54 V, consistent with Vp/π.
• Ideal rectification assumed.

Concept / Approach:
For a sinusoid: V_avg(half-wave) = Vp/π and V_avg(full-wave) = 2 * Vp/π. Since the given half-wave average equals Vp/π, the full-wave average is exactly twice that value.

Step-by-Step Solution:
Compute full-wave average: V_avg(full) = 2 * (Vp/π).Given Vp/π = 54 V, then V_avg(full) = 2 * 54 = 108 V.Therefore, the average of the full-wave rectified output is 108.0 V.

Verification / Alternative check:
Direct calculation: V_avg(full) = 2 * 169.7 / π ≈ 339.4 / 3.1416 ≈ 108.0 V, matching the above result precisely.

Why Other Options Are Wrong:

• 119.9 V or 115.7 V: Not equal to 2 * Vp/π for the stated Vp.
• 339.4 V: That is 2 * Vp, not an average value.

Common Pitfalls:

• Confusing average with RMS; for a full-wave sine, V_rms = Vp/√2, not 2 * Vp/π.
• Forgetting the “2” multiplier moving from half-wave to full-wave averages.

Final Answer:
108.0 V