An alternating current is represented by the equation $$i = 100\sqrt{2} \sin(100\pi t)$$ ampere. The RMS value of current and the frequency of the given alternating current are

The instantaneous current of a sinusoidal a.c. source is generally written as$$i = I_0 \sin(\omega t)$$where

• $$I_0$$ is the peak (maximum) current.
• $$\omega$$ is the angular frequency related to the ordinary frequency $$f$$ by$$\omega = 2\pi f$$.

Comparing the given expression$$i = 100\sqrt{2}\,\sin(100\pi t)$$with$$i = I_0 \sin(\omega t)$$we identify

$$I_0 = 100\sqrt{2}\quad\text{and}\quad\omega = 100\pi\ \text{rad s}^{-1}$$.

Step 1: RMS (root-mean-square) value
For a pure sine wave, the RMS current is related to the peak current by$$I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$$Applying this formula:

$$I_{\text{rms}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100\ \text{A}$$.

Step 2: Frequency
Using the relation$$f = \frac{\omega}{2\pi}$$substitute $$\omega = 100\pi$$:

$$f = \frac{100\pi}{2\pi} = 50\ \text{Hz}$$.

Result
RMS current $$= 100\ \text{A}$$ and frequency $$= 50\ \text{Hz}$$.

These match Option C.