Question

An alternating current source driving a circuit is represented by the voltage equation \( V = 220\sqrt{2}\sin(100\pi t) \). What are the Root Mean Square (RMS) voltage value \( V_{\text{rms}} \) and the alternating frequency \( f \) of this source supply?

• A. \( 220\,\text{V and } 50\,\text{Hz} \)
• B. \( 220\sqrt{2}\,\text{V and } 100\,\text{Hz} \)
• C. \( 110\,\text{V and } 50\,\text{Hz} \)
• D. \( 220\,\text{V and } 100\pi\,\text{Hz} \)

Hint:

When an AC voltage is listed simply as a single value without context (e.g., "220V wall outlets"), it always refers to the RMS value, not the peak value.

Answer 1

The Correct Option is A

Concept: The instantaneous voltage equation for a sinusoidal alternating current source can be written in standard form as: \[ V = V_0\sin(\omega t) \] where \( V_0 \) represents the peak amplitude voltage, and \( \omega \) represents the angular driving frequency (\( \omega = 2\pi f \)).

Step 1: Extract the peak voltage and calculate the RMS value. By comparing our given equation \( V = 220\sqrt{2}\sin(100\pi t) \) to the standard form, we isolate the peak value parameter: \[ V_0 = 220\sqrt{2}\,\text{V} \] The Root Mean Square voltage is defined as the peak voltage divided by the square root of two: \[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{220\sqrt{2}}{\sqrt{2}} = 220\,\text{V} \]

Step 2: Extract the angular frequency and calculate the cyclic frequency. Next, isolate the coefficient of time (\( t \)) to find the angular frequency value: \[ \omega = 100\pi\,\text{rad/s} \] Using the relationship linking angular frequency to linear cyclic frequency (\( \omega = 2\pi f \)): \[ 2\pi f = 100\pi \quad \Rightarrow \quad f = \frac{100\pi}{2\pi} = 50\,\text{Hz} \] This matches the standard values used in commercial grid systems across India.