JEE Mains · Physics · STD 12 -7. Alternating current
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 ______.
- A \(\frac{\left|\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}\right|}{\sqrt{2}}\)
- B \(\sqrt{\frac{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}{2}}\)
- C \(\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}\)
- D \(\frac{\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}}{2}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{\frac{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}{2}}\)
Step-by-step Solution
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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