JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा Question Bank Critical Thinking

  • question_answer
    In an LR-circuit, the inductive reactance is equal to the resistance R of the circuit. An e.m.f. \[E={{E}_{0}}\cos (\omega t)\] applied to the circuit. The power consumed in the circuit is [MP PMT 1997]

    A)            \[\frac{E_{0}^{2}}{R}\]     

    B)            \[\frac{E_{0}^{2}}{2R}\]

    C)            \[\frac{E_{0}^{2}}{4R}\]  

    D)            \[\frac{E_{0}^{2}}{8R}\]

    Correct Answer: C

    Solution :

                       \[P={{E}_{rms}}{{i}_{rms}}\cos \varphi =\frac{{{E}_{0}}}{\sqrt{2}}\times \frac{{{i}_{0}}}{\sqrt{2}}\times \frac{R}{Z}\]                    Þ \[\frac{{{E}_{0}}}{\sqrt{2}}\times \frac{{{E}_{0}}}{Z\sqrt{2}}\times \frac{R}{Z}\]\[\Rightarrow \,\,P=\frac{E_{0}^{2}R}{2{{Z}^{2}}}\]            Given \[{{X}_{L}}=R\] so, \[Z=\sqrt{2}R\]\[\Rightarrow \,P=\frac{E_{0}^{2}}{4R}\]


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