JEE Main & Advanced Sample Paper JEE Main - Mock Test - 17

  • question_answer
    A series \[R-C\]combination is connected to an AC voltage of angular frequency \[\omega =500\] radian/s. If the impedance of the \[R-C\]circuit is \[R\sqrt{1.25}\] the time constant (in millisecond) of the circuit is

    A) 5

    B)        4

    C)  6                    

    D)        8

    Correct Answer: B

    Solution :

    Time constant = RC Impedance \[=\sqrt{{{R}^{2}}+{{\left( \frac{1}{\omega C} \right)}^{2}}}\] Given impedance \[=R\sqrt{1.25}\] \[\therefore \,R\sqrt{1.25}=\sqrt{{{R}^{2}}+{{\left( \frac{1}{\omega C} \right)}^{2}}}\] \[\therefore \,{{R}^{2}}\sqrt{1.25}={{R}^{2}}+{{\left( \frac{1}{\omega C} \right)}^{2}}\] \[\therefore \,\,\frac{{{R}^{2}}}{4}={{\left( \frac{1}{\omega C} \right)}^{2}}\] \[\therefore \,\,\frac{R}{2}=\frac{1}{\omega C}\] \[\therefore \,\,RC=\frac{2}{\omega }=\frac{2}{500}\times 1000\,ms\] \[\therefore \,\,RC=4\,ms\]


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