Average DC value — repaired stem for a full-wave rectifier: For an ideal full-wave rectifier delivering a resistive load from a sinusoidal source with peak voltage Vp = 8.6 V, what is the average (mean) value of the rectified output voltage?

Introduction / Context:
The mean (average) output of a rectifier is a standard performance metric used for DC conversion. The original question referenced a “given circuit” without sufficient detail; we minimally repair it to a canonical case: an ideal full-wave rectifier with a sinusoidal input of peak Vp = 8.6 V driving a resistive load. This makes the problem solvable and educationally standard.

Given Data / Assumptions:

• Rectifier topology: ideal full-wave (both half-cycles rectified).
• Input waveform: sinusoid with peak Vp = 8.6 V.
• Load: purely resistive; diode drops neglected.
• We seek the average (mean) of the rectified waveform over one full period.

Concept / Approach:
For an ideal full-wave rectified sine, the average value is V_avg = (2/π) * Vp. Numerically, 2/π is approximately 0.6366. Multiply this factor by the given peak voltage to obtain the DC average at the output.

Step-by-Step Solution:

Verification / Alternative check:
As a quick sense check, note that the average of a full-wave rectified sine must be less than Vp and greater than half of Vp. Here, Vp/2 = 4.3 V and the result 5.48 V is between 4.3 V and 8.6 V, which is reasonable.

Why Other Options Are Wrong:

• 2.74 V: This equals about 0.318 * 8.6 V, which is the average for a half-wave rectifier, not full-wave.
• 6.37 V: Approximately 0.707 * 9 V; 0.707 * Vp is the RMS of a sine wave, not the mean of a rectified waveform.
• 0 V: The mean of a rectified waveform is positive, not zero.
• 4.30 V: This equals Vp/2, not the correct average of a full-wave rectified sine.

Common Pitfalls:
Confusing average (mean) with RMS; using the half-wave average factor (1/π) instead of the full-wave factor (2/π); forgetting to use the peak value Vp rather than RMS input when applying rectifier formulas.

Final Answer:
5.48 V