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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2018
  • Medium
  • MCQ

In an \(A.C.\) circuit, the instantaneous \(e.m.f\) and current are given by. \(e=100\) \(sin\) \(20t\) \(i=20sin\) \(\left( {30t - \frac{\pi }{4}} \right)\) In one cycle of \(A.C.\), the average power consumed by the circuit and the wattless current are, respectively

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Answer & Solution

Correct Answer

(A) \(\frac{{1000}}{{\sqrt 2 }},10\)

Step-by-step Solution

Detailed explanation

As we know, average power \(P_{\text {avg }}=V_{\text {ms }} I_{\text {ms }} \cos \theta\)

\(=\left(\frac{V_{0}}{\sqrt{2}}\right)\left(\frac{I_{0}}{\sqrt{2}}\right) \cos \theta=\left(\frac{100}{\sqrt{2}}\right)\left(\frac{20}{\sqrt{2}}\right) \cos 45^{\circ}\left(\because \theta=45^{\circ}\right)\)

\(P_{\text {avg }}=\frac{1000}{\sqrt{2}} \text { watt }\)

Wattless current \(\mathrm{I}=\mathrm{I}_{\mathrm{rms}} \sin \theta\)

\(=\frac{\mathrm{I}_{0}}{\sqrt{2}} \sin \theta=\frac{20}{\sqrt{2}} \sin 45^{\circ}=10\, \mathrm{A}\)

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