The equation of an alternating voltage is `V =100sqrt2 sin 100pit` volt. The RMS value of voltage and frequency will be respectively

Options

A. 100 V, 50 Hz
B. 50 V, 100 Hz
C. 150 V, 50 Hz
D. 200 V, 50 Hz

Answer

To solve the problem, we need to find the RMS (Root Mean Square) value of the voltage and the frequency from the given equation of the alternating voltage:

**Given:**
\[ V = 100\sqrt{2} \sin(100\pi t) \text{ volts} \]

### Step 1: Identify the peak voltage \( V_0 \)
From the equation, we can see that the peak voltage \( V_0 \) is:
\[ V_0 = 100\sqrt{2} \text{ volts} \]

### Step 2: Calculate the RMS voltage \( V_{rms} \)
The formula for the RMS voltage in terms of the peak voltage is:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \]

Substituting the value of \( V_0 \):
\[ V_{rms} = \frac{100\sqrt{2}}{\sqrt{2}} \]

### Step 3: Simplify the expression
The \( \sqrt{2} \) in the numerator and denominator cancels out:
\[ V_{rms} = 100 \text{ volts} \]

### Step 4: Identify the angular frequency \( \omega \)
From the equation, the angular frequency \( \omega \) is:
\[ \omega = 100\pi \text{ rad/s} \]

### Step 5: Calculate the frequency \( f \)
The relationship between angular frequency and frequency is given by:
\[ \omega = 2\pi f \]

Rearranging this to find \( f \):
\[ f = \frac{\omega}{2\pi} \]

Substituting the value of \( \omega \):
\[ f = \frac{100\pi}{2\pi} \]

### Step 6: Simplify to find \( f \)
The \( \pi \) in the numerator and denominator cancels out:
\[ f = \frac{100}{2} = 50 \text{ Hz} \]

### Final Answer:
The RMS value of the voltage and the frequency are respectively:
\[ V_{rms} = 100 \text{ volts} \]
\[ f = 50 \text{ Hz} \]

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