Household mains example: if the outlet provides 120 V rms (root-mean-square), what is the corresponding peak voltage of the sinusoidal waveform?

Introduction / Context:
AC specifications often use RMS values because they directly relate to equivalent DC heating effect in resistive loads. However, many design calculations—such as capacitor voltage ratings or rectifier sizing—require the peak voltage. Converting between RMS and peak for a pure sine wave is a basic skill in power electronics and instrumentation.

Given Data / Assumptions:

• Waveform is a sinusoid.
• RMS voltage V_rms = 120 V.
• Standard relationship for sine waves applies: V_peak = √2 * V_rms.

Concept / Approach:
For a sinusoidal signal, V_peak = V_rms * sqrt(2), and V_pp (peak-to-peak) = 2 * V_peak. These arise from the RMS definition V_rms = V_peak / sqrt(2). The conversion is exact for a sine wave and independent of frequency.

Step-by-Step Solution:
Use V_peak = sqrt(2) * V_rms.Compute: V_peak = 1.4142... * 120 ≈ 169.7 V.Round sensibly to 170 V for component selection and reporting.Select 170 V.

Verification / Alternative check:
Reverse the calculation: V_rms = V_peak / sqrt(2) = 170 / 1.414 ≈ 120 V, confirming consistency.

Why Other Options Are Wrong:
100 V and 120 V are below the true peak; 180 V exceeds the precise value and is not the standard conversion. “None” is invalid because 170 V is correct.

Common Pitfalls:
Confusing peak with peak-to-peak (which would be ≈ 340 V); using approximate factors like 1.5 instead of sqrt(2); forgetting that non-sinusoidal waveforms require a different approach.

Final Answer:
170 V