Half-wave rectifier (ideal diode): Given an input of v(t) = 20 sin(ωt) volts, determine the average (DC) value and the root-mean-square (rms) value of the output voltage.

Introduction / Context:
Half-wave rectifiers pass only one half-cycle of a sinusoid. For an ideal diode, the negative half is completely blocked, leading to standard formulae for the average (DC) and rms output values. This problem reinforces those results for a specific amplitude.

Given Data / Assumptions:

• Input: v(t) = V_m sin(ωt) with V_m = 20 V.
• Ideal diode (no drop), pure half-wave rectification.
• Output taken across the load directly.

Concept / Approach:

For an ideal half-wave rectifier: V_avg = V_m/π and V_rms = V_m/2. These come from integrating the rectified waveform over a full period, remembering that the second half of the cycle is zero.

Step-by-Step Solution:

Verification / Alternative check:

Compute rms from definition: V_rms^2 = (1/2π) ∫_0→π (V_m sinθ)^2 dθ = V_m^2/4 → V_rms = V_m/2, confirming 10 V.

Why Other Options Are Wrong:

• Options listing 5 V for average or rms contradict standard half-wave results.
• Average of 10 V implies full-wave or different amplitude.
• Values like 14.14 V correspond to V_m/√2 of an unrectified sine, not a half-wave output.

Common Pitfalls:

• Forgetting the average is over the full period, not just the conducting half.
• Confusing full-wave with half-wave formulae.

Final Answer:

V_avg ≈ 6.37 V and V_rms = 10 V