An oscilloscope shows a sinusoidal signal with a peak excursion of 3 major divisions above the zero reference. If the vertical sensitivity is 5 V/cm (5 volts per major division), what is the RMS value of the AC voltage?

Introduction:Relating oscilloscope deflection to RMS voltage requires converting displayed amplitude (in divisions) to volts and then applying the sinusoidal peak-to-RMS relationship. This problem ensures comfort with scope scaling and AC amplitude definitions.

Given Data / Assumptions:

• Peak height above zero: 3 major divisions.
• Vertical scale: 5 V per division.
• Signal is a pure sine wave; use the standard RMS relation.

Concept / Approach:Convert divisions to peak voltage: Vp = divisions * volts_per_div. For a sine wave, Vrms = Vp / sqrt(2). Do not confuse peak (0-to-maximum) with peak-to-peak (maximum-to-minimum). Here the statement explicitly refers to peak above zero, so Vp is used directly.

Step-by-Step Solution:1) Compute peak: Vp = 3 * 5 V = 15 V.2) Use sine relation: Vrms = Vp / sqrt(2).3) Evaluate: Vrms ≈ 15 / 1.414 ≈ 10.606 V.4) Round sensibly to one decimal: ≈ 10.6 V.

Verification / Alternative check:If the display had been 3 divisions peak-to-peak, Vpp = 15 V and Vrms would differ (Vpp / 2√2). But the problem says 'peak' above zero, confirming the calculation used.

Why Other Options Are Wrong:5.3 V: corresponds to using half the correct peak or mis-scaling.15.0 V: that is peak, not RMS.21.2 V: would be Vpp for 3 divisions at 5 V/div, not applicable here.

Common Pitfalls:Confusing peak with peak-to-peak, or forgetting that RMS for a pure sine is Vp / sqrt(2), not Vp / 2.

Final Answer:10.6 V