RMS-to-peak-to-peak conversion (sine wave): What is the peak-to-peak voltage of a sinusoidal AC waveform whose RMS value is 56 V?

Difficulty: Medium

Correct Answer: 158 V

Explanation:


Introduction:
Converting between RMS, peak, and peak-to-peak values is essential for power calculations, component voltage ratings, and oscilloscope interpretation. For a pure sine wave, fixed ratios relate these quantities.


Given Data / Assumptions:

  • Sine wave with Vrms = 56 V.
  • Standard sine relationships apply.
  • No distortion or DC offset.


Concept / Approach:

For a sine wave: Vrms = Vp / √2, so Vp = Vrms * √2. Peak-to-peak is Vpp = 2 * Vp. Compute Vp, then double it to obtain Vpp. Keep sufficient precision for rounding to the nearest integer volt if options are whole numbers.


Step-by-Step Solution:

Compute peak: Vp = Vrms * √2 = 56 * 1.41421356 ≈ 79.2 V.Compute peak-to-peak: Vpp = 2 * Vp ≈ 158.4 V.Rounded to the nearest volt: Vpp ≈ 158 V.Therefore, the correct option is 158 V.


Verification / Alternative check:

Reverse check: If Vpp = 158 V ⇒ Vp ≈ 79 V, then Vrms ≈ 79 / √2 ≈ 55.9 V, which rounds to 56 V, confirming consistency.


Why Other Options Are Wrong:

  • 164 V: Would imply Vrms ≈ 58 V; too high.
  • 82 V: That is approximately Vp, not Vpp.
  • 79 V: Exactly the peak, not peak-to-peak.
  • 112 V: Corresponds to 2 * Vrms, not a standard sine relation.


Common Pitfalls:

  • Confusing peak with peak-to-peak (factor of 2).
  • Applying Vrms = Vavg (true only for DC) or mixing rectified-wave formulas.


Final Answer:

158 V

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