JEE Main 2025 — Alternating Current Question with Solution

To find the root mean square (RMS) value of the given alternating current, follow these steps:

The current is represented as:

$i=5\sqrt{2}+10\cos \left(650\pi t+\frac{\pi }{6}\right)\text{Amp}$

Here, the time-independent DC component is $5\sqrt{2}$ and the AC component is $10\cos \left(650\pi t+\frac{\pi }{6}\right)$.

Calculate the square of the current, $i^2$:

$i^2={\left(5\sqrt{2}\right)}^2+{\left(10\cos \left(650\pi t+\frac{\pi }{6}\right)\right)}^2+2\times 5\sqrt{2}\times 10\cos \left(650\pi t+\frac{\pi }{6}\right)$

Simplifying, we have:

$i^2=50+100{\cos }^2\left(650\pi t+\frac{\pi }{6}\right)+100\sqrt{2}\cos \left(650\pi t+\frac{\pi }{6}\right)$

Find the average value $⟨i^2⟩$:

The average value of $\cos$ terms over a period is zero, simplifying our equation to:

$⟨i^2⟩=50+\frac{100}{2}+0$

This simplifies to:

$⟨i^2⟩=50+50=100$

Calculate the RMS current:

The RMS value is the square root of the mean of the squares of the current:

$⟨i⟩=\sqrt{100}=10\text{Amp}$

Thus, the RMS value of the current is 10 Amps.