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

Step 1: Understanding the Concept:
The instantaneous voltage of a sinusoidal AC source is represented mathematically as a function of time.
The standard general form of this equation is:
\[ V = V_0 \sin(\omega t) \]
Where \(V_0\) is the peak (maximum) voltage and \(\omega\) is the angular frequency.
Step 2: Key Formula or Approach:
To extract the required values, we compare the given equation to the standard form:
1. Relationship between peak voltage and RMS voltage: \(V_{\text{rms}} = \frac{V_0}{\sqrt{2}}\).
2. Relationship between angular frequency and cyclic frequency: \(\omega = 2\pi f\).
Step 3: Detailed Explanation:
Given equation: \(V = 220\sqrt{2} \sin(100\pi t)\).
Comparing this to \(V = V_0 \sin(\omega t)\), we find:
- \(V_0 = 220\sqrt{2}\text{V}\)
- \(\omega = 100\pi\text{rad/s}\)
Step 3a: Calculate the RMS Voltage:
\[ V_{\text{rms}} = \frac{220\sqrt{2}}{\sqrt{2}} = 220\text{V} \]
Step 3b: Calculate the cyclic Frequency:
\[ \omega = 2\pi f \implies 100\pi = 2\pi f \]
\[ f = \frac{100\pi}{2\pi} = 50\text{Hz} \]
These values (\(220\text{V}\) and \(50\text{Hz}\)) match the standard parameters for domestic power grids in India.
Step 4: Final Answer:
The RMS voltage is \(220\text{V}\) and the frequency is \(50\text{Hz}\).