JEE MAIN - Physics (2025 - 4th April Morning Shift - No. 6)

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
$\frac{100}{\sqrt{2}} \mathrm{~A}, 100 \mathrm{~Hz}$
$50 \sqrt{2} \mathrm{~A}, 50 \mathrm{~Hz}$
$100 \sqrt{2} \mathrm{~A}, 100 \mathrm{~Hz}$
$100 \mathrm{~A}, 50 \mathrm{~Hz}$

Explanation

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

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