BCECE Engineering BCECE Engineering Solved Paper-2005

  • question_answer
    An LCR circuit of R = 100 0 is connected to an AC source 100 V, 50 Hz. The magnitude of phase difference between current and voltage is \[30{}^\circ \]. 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 \[\cos \text{o }\!\!|\!\!\text{ }\]is known as power factor. Given, \[{{V}_{rms}}=100\,volt,\,R=100\,\Omega ,\phi ={{30}^{o}}\] \[\therefore \]  \[{{I}_{rms}}=\frac{{{V}_{rms}}}{R}=\frac{100}{100}=1\,A\] Putting the values in Eq. (i), we get \[{{P}_{av}}=100\times 1\times \cos {{30}^{o}}\] \[=100\frac{\sqrt{3}}{2}\] \[=50\sqrt{3}\] \[=86.6\,W\] Note: The product of \[{{V}_{rms}}\]and \[{{I}_{rms}}\] the apparent power, while the true power is obtained by multiplying the apparent power by the power factor \[\cos \phi \]


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