RAJASTHAN PMT Rajasthan - PMT Solved Paper-2004

  • question_answer
    The graphs given below depict the dependence of two reactive impedances\[{{X}_{1}}\]and\[{{X}_{2}}\] on the frequency of the alternating emf applied individually to them. We can then say that:           

    A)   \[{{X}_{1}}\] is an inductor and \[{{X}_{2}}\] is a capacitor

    B)  \[{{X}_{1}}\] is a resistor and \[{{X}_{2}}\] is a capacitor

    C)  \[{{X}_{1}}\] is a capacitor and \[{{X}_{2}}\] is an inductor

    D)  \[{{X}_{1}}\] is an inductor and \[{{X}_{2}}\] is a resistor

    Correct Answer: C

    Solution :

    The inductive reactance in A.C. circuit \[{{X}_{L}}={{X}_{2}}=\omega L\] \[\Rightarrow \]\[{{X}_{2}}\propto \omega \]      \[{{X}_{2}}\propto f\]                               \[(\therefore \omega =2\pi f)\] The capacitive reactance                                 \[{{X}_{C}}={{X}_{1}}=\frac{1}{\omega C}\] \[\Rightarrow \]               \[{{X}_{1}}\propto \frac{1}{\omega }\]                               \[{{X}_{1}}\propto \frac{1}{f}\]    \[(\therefore \omega =2\pi f)\] Hence, it is obvious that \[{{X}_{1}}\] is a capacitor and \[{{X}_{2}}\] is an inductor.


You need to login to perform this action.
You will be redirected in 3 sec spinner