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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2010

  • question_answer
    In a series resonant R-L-C circuit, the voltage across R is 100 V and the value of \[R=1000\,\Omega \]. The capacitance of the capacitor is\[2\times {{10}^{-6}}F\]; angular frequency of AC is 200 rad\[{{s}^{-1}}\]. Then the potential difference across the inductance coil is

    A)  100 V          

    B)  40 V

    C)  250 V          

    D)  400 V

    Correct Answer: C

    Solution :

    The current in the circuit \[i=\frac{{{V}_{R}}}{R}\] \[=\frac{100}{1000}=0.1A\] At resonance, \[{{V}_{L}}={{V}_{C}}=i{{X}_{C}}=\frac{i}{\omega C}\] \[=\frac{0.1}{200\times 2\times {{10}^{-6}}}\]\[=250V\]


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