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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2010

  • question_answer
    An L-C-R series circuit with a resistance of 100 ft is connected to an AC source of 200 V and angular frequency 300 rad/s. When only the capacitor is removed, the current lags behind the voltage by \[60{}^\circ \]. When only the inductor is removed the current leads the voltage by \[60{}^\circ \] The average power dissipated is

    A)  50 W                                    

    B)  100 W 

    C)  200 W                  

    D)         400 W

    Correct Answer: D

    Solution :

                    \[\tan \phi =\frac{{{X}_{L}}}{R}=\frac{{{X}_{C}}}{R}\] \[\Rightarrow \]               \[\tan {{60}^{o}}=\frac{{{X}_{L}}}{R}=\frac{{{X}_{C}}}{R}\] \[\Rightarrow \]               \[{{X}_{L}}={{X}_{C}}=\sqrt{3}R\] \[ie,\]    \[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}=R\] So average power                 \[P=\frac{{{V}^{2}}}{R}=\frac{200\times 200}{100}=400\,\,W\]


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