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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2025
  • Easy
  • MCQ

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 ______.

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Answer & Solution

Correct Answer

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

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Detailed explanation

\(\begin{aligned} & I=I_{\mathrm{A}} \sin \omega t+I_B \cos \omega t \\ & I_{\mathrm{ma}}=\sqrt{I_A^2+I_B^2} \\ & \text { So, } I_{\mathrm{RMS}}=\sqrt{\frac{I_A^2+I_B^2}{2}}\end{aligned}\)
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