JEE Main 2025 Apr 7 Shift 1, Physics Q9: Introduction to Alternating Currents and Circuits

Compute $⟨i^2⟩$ over one period.

$i^2=(5\sqrt{2}{)}^2+100{\cos }^2\left(650\pi t+\frac{\pi }{6}\right)+2(5\sqrt{2})(10)\cos \left(650\pi t+\frac{\pi }{6}\right)$

Time-averaging term by term:

• Constant: $⟨50⟩=50$
• $⟨{\cos }^2⟩=\frac{1}{2}$, so $⟨100{\cos }^2(\cdot )⟩=50$
• $⟨\cos (\cdot )⟩=0$

$⟨i^2⟩=50+50+0=100$

RMS value:

$i_{\text{rms}}=\sqrt{⟨i^2⟩}=\sqrt{100}=10$ Amp

$\therefore$ Option (3).