JEE Main & Advanced Sample Paper JEE Main - Mock Test - 12

  • question_answer
    For the circuit as shown in figure, if the value of current is 2.2 A, the power factor of the box is

    A) \[\frac{1}{\sqrt{2}}\]                 

    B) 1

    C) \[\frac{\sqrt{3}}{2}\]                 

    D) \[\frac{1}{2}\]

    Correct Answer: A

    Solution :

    [a]: As rms value of current,\[{{I}_{rms}}={{V}_{rms}}/Z\] So, net impedance across LCR circuit, \[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}=\sqrt{{{(100)}^{2}}+{{(\omega L-{{X}_{C}})}^{2}}}\] \[=\sqrt{{{(100)}^{2}}+{{\left[ \left( 100\times \pi \times \frac{1}{\pi } \right)-({{X}_{C}}) \right]}^{2}}}\] \[{{\left( \frac{220}{2.2} \right)}^{2}}={{(100)}^{2}}+{{\left( (100)-({{X}_{C}}) \right)}^{2}}\Rightarrow {{X}_{C}}=100\Omega \] As,\[\tan \theta =\frac{{{X}_{C}}}{R}=\frac{100}{100}=1\] \[\Rightarrow \]\[\theta ={{\tan }^{-1}}(1)\]or\[\theta ={{45}^{o}}\] Power factor, \[\cos \theta =\frac{1}{\sqrt{2}}\]


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