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CET Karnataka Engineering CET - Karnataka Engineering 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 \[200\text{ }rad\text{ }{{s}^{-1}}\]. 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                 \[i=\frac{{{V}_{R}}}{R}\]                 \[=\frac{100}{1000}=0.1\,A\] At resonance,                 \[{{V}_{L}}={{V}_{C}}=i{{X}_{C}}=\frac{i}{\omega C}\]                 \[=\frac{0.1}{200\times 2\times {{10}^{-6}}}\]                 \[=250\,V\]


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