JEE Main 2025 (Online) 7th April Morning Shift | Alternating Current Question 2 | Physics

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

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

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

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

$100 \mathrm{~A}, 50 \mathrm{~Hz}$

JEE Main 2025 (Online) 3rd April Evening Shift

MCQ (Single Correct Answer)

-1

An electric bulb rated as $100 \mathrm{~W}-220 \mathrm{~V}$ is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is :

0.32 A

0.64 A

0.45 A

2.2 A

JEE Main 2025 (Online) 24th January Morning Shift

MCQ (Single Correct Answer)

-1

An alternating current is given by $\mathrm{I}=\mathrm{I}_{\mathrm{A}} \sin \omega \mathrm{t}+\mathrm{I}_{\mathrm{B}} \cos \omega \mathrm{t}$. The r.m.s current will be

$\frac{\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}}{2}$

$\sqrt{\frac{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}{2}}$

$\frac{\left|\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}\right|}{\sqrt{2}}$

$\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}$

Questions Asked from Alternating Current (MCQ (Single Correct Answer))

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