Question

Given \( v(t) = 100 \sin(100\pi t) \), the RMS value is:

• A. \( 50\text{ V} \)
• B. \( 100\text{ V} \)
• C. \( 70.7\text{ V} \)
• D. \( 141.4\text{ V} \)

Hint:

Memorize the constant scaling factor multiplier for sinusoidal inputs: \[ \frac{1}{\sqrt{2}} \approx 0.707 \] Thus, you can instantly find the result by calculating: \(V_{\text{rms}} = 0.707 \times V_m = 0.707 \times 100 = 70.7\text{ V}\).

Answer 1

The Correct Option is C

Concept: The Root-Mean-Square (RMS) value of a time-varying periodic voltage function \(v(t)\) with period \(T\) represents its effective DC heating equivalent value, calculated as: \[ V_{\text{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} v^2(t) \, dt} \] For standard symmetrical sinusoidal expressions that conform to the format \(v(t) = V_m \sin(\omega t + \phi)\), where \(V_m\) represents the peak maximum value, evaluating this integral over a full cycle consistently reduces to: \[ V_{\text{rms}} = \frac{V_m}{\sqrt{2}} \]

Step 1: Identifying the peak value \(V_m\).
We map our specific given voltage expression to the canonical sinusoidal format: \[ v(t) = 100 \sin(100\pi t) \quad \longleftrightarrow \quad v(t) = V_m \sin(\omega t) \] By direct inspection, we establish the parameters:
• Peak value, \(V_m = 100\text{ V}\)
• Radian frequency, \(\omega = 100\pi\text{ rad/s}\)

Step 2: Performing the calculation.
We substitute our identified peak parameter \(V_m = 100\text{ V}\) into the RMS formula: \[ V_{\text{rms}} = \frac{100}{\sqrt{2}} \] To remove the radical from the denominator, we rationalize the fraction by multiplying both top and bottom by \(\sqrt{2}\): \[ V_{\text{rms}} = \frac{100\sqrt{2}}{2} = 50\sqrt{2}\text{ V} \] Substituting the decimal numerical approximation for the square root of two (\(\sqrt{2} \approx 1.4142\)): \[ V_{\text{rms}} = 50 \times 1.4142 = 70.71\text{ V} \] Rounding this off to one decimal place yields \(70.7\text{ V}\), corresponding directly to Option (C).