JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    An LCR circuit of \[R=100\,\Omega \] is connected to an AC source 100 V, 50 Hz. the magnitude of phase difference between current and voltage is \[{{30}^{o}}\]. The power dissipated in the LCR circuit is :

    A)  50 W                           

    B)  \[86.6\] W

    C)  100 W                         

    D)  200 W

    Correct Answer: B

    Solution :

    Average power dissipated in an AC circuit, \[{{P}_{av}}={{V}_{rms}}\,{{I}_{rms}}\,\cos \phi \]     ?(i) Where, the term \[\cos \phi \] is known as power factor Given \[{{V}_{rms}}=100\,V,\,\,R=100\Omega ,\,\phi =\,{{30}^{0}}\] \[\therefore \,\,{{I}_{rms}}\,=\frac{{{V}_{rms}}}{R}=\frac{100}{100}=1A\] Putting the values in eq. (i) we get \[{{P}_{av}}\,=100\times 1\,\times \,\cos {{30}^{0}}\] \[=100\frac{\sqrt{3}}{2}\] \[=50\,\sqrt{3}\,=86.6\,w\]

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