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BHU PMT BHU PMT (Mains) Solved Paper-2010

  • question_answer
    A charged capacitor discharges through a resistance R with time constant\[\tau \]. The two are now placed in series across an AC source of angular frequency\[\omega =\frac{1}{\tau }\]. The impedance of the circuit will be

    A)  \[\frac{R}{\sqrt{2}}\]                                    

    B)  \[R\]

    C)  \[\sqrt{2}R\]

    D)  \[2R\]

    Correct Answer: C

    Solution :

                     RC is time constant, then \[\tau =RC\] According to question,                 \[\omega =\frac{1}{\tau }=\frac{1}{RC}\]                 \[R=\frac{1}{\omega C}\] Impedance of the R-C circuit \[Z=\sqrt{{{R}^{2}}+X_{C}^{2}}\]                 \[Z=\sqrt{{{R}^{2}}+{{\left( \frac{1}{\omega C} \right)}^{2}}}\]                 \[Z=\sqrt{{{R}^{2}}+{{R}^{2}}}=\sqrt{2}R\]


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