JEE Main 2025 (Online) 3rd April Evening Shift | Alternating Current Question 6 | Physics

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

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

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

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

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

JEE Main 2025 (Online) 22nd January Evening Shift

MCQ (Single Correct Answer)

-1

A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is $I_0$. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be

$\frac{\mathrm{I}_0}{2}$

$\frac{\mathrm{I}_0}{\sqrt{2}}$

$2 \mathrm{I}_0$

$\mathrm{I_0}$

JEE Main 2024 (Online) 9th April Morning Shift

MCQ (Single Correct Answer)

-1

A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb :

becomes zero

remains same

increases

decreases

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