RAJASTHAN ­ PET Rajasthan PET Solved Paper-2012

  • question_answer
    An AC source of frequency 50 Hz is connected to 50 mH inductor and a bulb. To get the maximum brightness from the bulb, what should be the capacitance of a capacitor connected in series with the circuit?

    A)  \[50\times {{10}^{-4}}F\]

    B)  \[1\times {{10}^{-4}}F\]

    C)  \[2.03\times {{10}^{-4}}F\]

    D)  \[3\times {{10}^{-4}}F\]

    Correct Answer: C

    Solution :

     For maximum brightness maximum current is required in the circuit. To get maximum current, \[{{X}_{L}}={{X}_{C}}\] (at resonance) \[\Rightarrow \] \[2\pi \,vL=\frac{1}{2\pi vC}\] \[\Rightarrow \] \[C=\frac{1}{4{{\pi }^{2}}{{v}^{2}}L}\] \[=\frac{1}{4{{(3.14)}^{2}}{{(50)}^{2}}(50\times {{10}^{-3}})}\] \[=2.03\times {{10}^{-4}}F\]


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