JEE Main 2025 — Alternating Current Question with Solution

To solve the problem, we need to determine both the RMS value of the current and the frequency from the given equation:

$$i(t)=100\sqrt{2}\sin (100\pi t).$$

Here's how we do it step by step:

RMS Current Calculation:

For a sinusoidal current of the form $$i(t)=I_0\sin (\omega t),$$ the RMS (root-mean-square) current is given by:

$$I_{\text{RMS}}=\frac{I_0}{\sqrt{2}}.$$

In our case, the amplitude $$I_0=100\sqrt{2}\text{A}.$$ Plugging into the formula:

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

Frequency Calculation:

The argument of the sine function is $$100\pi t.$$ In the general form $$\sin (\omega t),$$ $$\omega$$ is the angular frequency which relates to the frequency $$f$$ by the equation:

$$\omega =2\pi f.$$

Given $$\omega =100\pi ,$$ we can solve for $$f$$:

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

With these calculations, we find that the RMS current is $$100\text{A}$$ and the frequency is $$50\text{Hz}.$$

Thus, the correct option is:

Option D: $$100\text{A},50\text{Hz}.$$