The equation of an alternating current is `I = 50 sqrt(2) sin 400 pi t A`, then the frequency and the root mean square value of current are respectively.

Options

A. `200 Hz, 50 A`
B. `400 Hz, 50 sqrt(2) A`
C. `200 Hz, 50 sqrt(2) A`
D. `500 Hz, 200 A`

Explanation

To solve the problem, we need to find the frequency and the root mean square (RMS) value of the alternating current given by the equation:

\[ I = 50 \sqrt{2} \sin(400 \pi t) \, \text{A} \]

### Step 1: Identify the peak current (\(I_0\))
From the given equation, we can identify the peak current (\(I_0\)):

\[ I_0 = 50 \sqrt{2} \, \text{A} \]

### Step 2: Calculate the RMS value of the current
The RMS value (\(I_{\text{RMS}}\)) of an alternating current is given by the formula:

\[ I_{\text{RMS}} = \frac{I_0}{\sqrt{2}} \]

Substituting the value of \(I_0\):

\[ I_{\text{RMS}} = \frac{50 \sqrt{2}}{\sqrt{2}} = 50 \, \text{A} \]

### Step 3: Identify the angular frequency (\(\omega\))
From the equation, we can see that the angular frequency (\(\omega\)) is given by:

\[ \omega = 400 \pi \, \text{rad/s} \]

### Step 4: Calculate the frequency (\(f\))
The relationship between angular frequency and frequency is given by:

\[ \omega = 2 \pi f \]

We can rearrange this to find \(f\):

\[ f = \frac{\omega}{2 \pi} \]

Substituting the value of \(\omega\):

\[ f = \frac{400 \pi}{2 \pi} = 200 \, \text{Hz} \]

### Final Results
Thus, the frequency and the RMS value of the current are:

- Frequency (\(f\)) = 200 Hz
- RMS Value (\(I_{\text{RMS}}\)) = 50 A

### Summary
The frequency and the root mean square value of the current are respectively:

\[ \text{Frequency} = 200 \, \text{Hz}, \quad \text{RMS Value} = 50 \, \text{A} \]

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