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For a sinusoidal AC voltage with peak $V_m$, the RMS value is
A$V_m$, the peak value itself
B$V_m / \sqrt{2}$
C$2 V_m$, twice the peak
D$V_m^2$, peak value squared
Answer & Solution
Correct answer: B. $V_m / \sqrt{2}$
1. RMS (Root Mean Square): the equivalent DC value that produces the same average power dissipation.
2. For $v(t) = V_m \sin(\omega t)$: $V_{RMS} = \sqrt{\langle v^2 \rangle} = V_m / \sqrt{2}$.
3. Numerically: $V_{RMS} \approx 0.707\,V_m$.
4. Example: 230 V AC mains has RMS = 230, so peak $V_m = 230\sqrt{2} \approx 325\,\text{V}$.
5. Most AC meters report RMS values directly.
_Source: Tony Kuphaldt, "Lessons in Electric Circuits — AC", Vol II, Ch 1 (RMS)._
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