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