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BCECE Engineering BCECE Engineering Solved Paper-2006

  • question_answer
    In a series LCR circuit the frequency of a 10 VAC voltage source is adjusted in such a fashion that the reactance of the inductor measures 15 \[\Omega \] and that of capacitor 11\[\Omega \]. If R = 3 \[\Omega \] the potential difference across the series combination of L and C will be:

    A)  8 V                                        

    B)         10 V

    C)         22 V                                     

    D)         52 V

    Correct Answer: A

    Solution :

    Given, \[R=3\Omega ,\,{{X}_{L}}=15\Omega ,{{X}_{C}}=11\Omega ,\] \[{{V}_{R}}=10\,V\] \[\therefore \] Current through the circuit \[i=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}}\] \[=\frac{{{V}_{rms}}}{\sqrt{{{(3)}^{2}}+{{(15-11)}^{2}}}}\] \[=\frac{10}{\sqrt{9+16}}\] \[=\frac{10}{5}=2A\] Since, L, C and R are connected in series combination the potential difference across R is \[{{V}_{R}}=i\times R=2\times 3\] \[=6\,V\]                 Across L,                 \[{{V}_{L}}=i{{X}_{L}}=2\times 15=30\,V\]                 Across C,              \[{{V}_{C}}=i{{X}_{C}}\]                                                 \[=2\times 11=22\,V\] So, potential difference across series combination of L and C is \[={{V}_{L}}-{{V}_{C}}\] \[=30-22=8\,V\]


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