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JEE Mains · Physics · STD 12 -7. Alternating current

एक प्रत्यावर्ती धारा \(\mathrm{I}=\mathrm{I}_{\mathrm{A}} \sin \omega \mathrm{t}+\mathrm{I}_{\mathrm{B}} \cos \omega \mathrm{t}\) द्वारा दी जाती है। वर्ग माध्य मूल (r.m.s.) धारा ______ होगी।

  1. A \(\frac{\left|\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}\right|}{\sqrt{2}}\)
  2. B \(\sqrt{\frac{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}{2}}\)
  3. C \(\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}\)
  4. D \(\frac{\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}}{2}\)
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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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