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NEET AIPMT SOLVED PAPER SCREENING 2012

  • question_answer
    In an electrical circuit R, L, C and an AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is \[\pi /3.\]If instead, C is removed from the circuit, the phase difference is again \[\pi /3.\]The power factor of the circuit is

    A) 1/2                                        

    B) \[1/\sqrt{2}\]    

    C)        1                             

    D)        \[\sqrt{3}/2\]

    Correct Answer: C

    Solution :

    Here, phase difference \[\tan \phi =\frac{{{X}_{L}}-{{X}_{C}}}{R}\]                 \[\tan \frac{\pi }{3}=\frac{{{X}_{L}}-{{X}_{C}}}{R}\] When, L is removed                 \[\sqrt{3}=\frac{{{X}_{C}}}{R}\]                 \[{{X}_{C}}=\sqrt{3}R\] When C is removed \[\tan \frac{\pi }{3}=\sqrt{3}=\frac{{{X}_{L}}}{R}\] \[{{X}_{L}}=R\sqrt{3}\] Hence, in resonant circuit \[\tan \phi =\frac{\sqrt{3}R=\sqrt{3}R}{R}=0\]                 \[\phi =0\] \[\therefore \]Power factor \[\cos \phi =1\] It is the condition of resonance therefore phase difference between voltage and current is zero and power factor is \[\cos \phi =1.\]


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