Alternating Current | JEE Main 2025 Solution

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JEE MAIN 2025

24-01-2025 SHIFT-1

Question

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

Select the correct option:

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

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

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

$\frac{{\sqrt {I_A^2 + I_B^2} }}{2}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

${i_{{\rm{rms}}}} = \sqrt {\frac{{\int {{{\rm{I}}^2}} {\rm{dt}}}}{{\int {{\rm{dt}}} }}} $
$\sqrt {\frac{{{\rm{I}}_{\rm{A}}^2 + {\rm{I}}_{\rm{B}}^2}}{2}} = {{\rm{i}}_{{\rm{rms}}}}$

Question Tags

JEE Main

Physics

Medium

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