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Haryana PMT Haryana PMT Solved Paper-2008

  • question_answer
    A coil of inductive reactance 31\[\Omega \] has a             resistance of 8\[\Omega \]. It is placed in series with a condenser of capacitive reactance 25\[\Omega \]. The combination is connected to an AC source of 110 V. The power factor of the circuit is

    A)  0.56                                      

    B)  0.64

    C)  0.80                                      

    D)  0.33

    Correct Answer: C

    Solution :

                    Power factor of (AC) circuit is given by \[\phi =\frac{R}{Z}\]                 ??..(i) where R is resistance employed and Z the impedance of the circuit.                 \[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}\]     ??(ii) Eqs. (i) and (ii) meet to give                          \[\cos \phi =\frac{R}{\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}}\]            ....(iii)      Given,.\[R=8\Omega ,\,{{X}_{L}}=31\Omega ,\,{{X}_{C}}=25\Omega \] \[\therefore \]  \[\cos \phi =\frac{8}{\sqrt{{{(8)}^{2}}+{{(31-25)}^{2}}}}\]                 \[=\frac{8}{\sqrt{64+36}}\] Hence, \[\cos \phi =0.80\]


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