JCECE Engineering JCECE Engineering Solved Paper-2004

  • question_answer
    Reactance of a capacitor of capacitance \[C\,\,\mu F\] for \[AC\] frequency \[\frac{400}{\pi }Hz\] 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                 \[{{X}_{C}}=\frac{1}{\omega \,C}=\frac{1}{2\pi f\,C}\] 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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