AC component (ripple) of a rectifier output – RMS value For a rectifier output with measured total RMS value Vrms and DC component Vdc, what is the RMS value of the AC (ripple) component alone?

Difficulty: Easy

Correct Answer: (Vrms^2 - Vdc^2)^0.5

Explanation:


Introduction / Context:
Rectifier outputs contain a DC component plus ripple (AC component). Quantifying ripple is essential for filter design and performance evaluation. RMS values add in a mean-square sense when components are orthogonal (DC vs AC).


Given Data / Assumptions:

  • Total RMS of output voltage = Vrms.
  • DC (average) component = Vdc.
  • AC component is the residual after removing the DC level.


Concept / Approach:

Because the DC component is constant and the AC component has zero average, the mean of the square of the sum equals the sum of the means of the squares: Vrms^2 = Vdc^2 + (Vac_rms)^2. Therefore, Vac_rms = sqrt(Vrms^2 − Vdc^2).


Step-by-Step Solution:

Start with decomposition: v(t) = Vdc + v_ac(t), with average of v_ac(t) = 0.Compute RMS: Vrms^2 = (1/T)∫[Vdc + v_ac(t)]^2 dt = Vdc^2 + (1/T)∫[v_ac(t)]^2 dt.Hence Vac_rms = √(Vrms^2 − Vdc^2).


Verification / Alternative check:

Check limiting cases: If ripple is zero, Vrms = Vdc → Vac_rms = 0. If Vdc = 0 (pure AC), Vac_rms = Vrms, as expected.


Why Other Options Are Wrong:

  • Vrms includes DC; not just ripple.
  • 0.5 Vrms is arbitrary and incorrect.
  • √(Vrms − Vdc) is dimensionally wrong and not based on mean-square addition.


Common Pitfalls:

Subtracting RMS magnitudes directly instead of subtracting power (square) quantities; forgetting that RMS is a quadratic metric.


Final Answer:

(Vrms^2 - Vdc^2)^0.5

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