JCECE Medical JCECE Medical Solved Paper-2004

  • question_answer
    Reactance of a capacitor of capacitance \[C\,\mu F\]for AC frequency \[\frac{\text{400}}{\text{ }\!\!\pi\!\!\text{ }}\text{Hz}\]is \[25\,\Omega ,\] the value of C is:

    A) \[75\,\mu F\]         

    B) \[100\,\mu F\]

    C)  \[25\,\mu F\]        

    D) \[50\,\mu F\]

    Correct Answer: D

    Solution :

     Capacitive reactance \[{{X}_{C}}\]is given by \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2\pi fC}\] where,\[\omega =2\pi f\] Given, \[{{X}_{C}}=25\Omega ,f=\frac{400}{\pi }Hz\] Hence, \[25=\frac{1}{2\pi \times \frac{400}{\pi }\times C}\] \[\Rightarrow \] \[C=\frac{1}{25\times 800}=50\,\mu F\]


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