JEE Physics 2025 Question with Solution

Given: The alternating current is $i = 100\sqrt{2}\sin(100\pi t)$.

Find: The RMS value of current and the frequency.

For an alternating current of the form $i = I_0 \sin(\omega t)$, the peak current is $I_0 = 100\sqrt{2} \, \text{A}$ and the angular frequency is $\omega = 100\pi$.

The RMS current is

$$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100 \, \text{A}$$

The frequency is

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

Therefore, the RMS current is $100 \, \text{A}$ and the frequency is $50 \, \text{Hz}$. The correct option is C.

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

Find: $I_{\text{rms}}$ and $f$.

The general form of alternating current is

$$i = I_0 \sin(\omega t)$$

Comparing with the given equation,

$$I_0 = 100\sqrt{2} \, \text{A}$$

and

$$\omega = 100\pi$$

Now use the RMS relation,

$$I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$$

so,

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

Using $\omega = 2\pi f$,

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

Thus, the RMS current is $100 \, \text{A}$ and the frequency is $50 \, \text{Hz}$. Hence, option C is correct.