JCECE Medical JCECE Medical Solved Paper-2011

  • question_answer
    Instantaneous displacement current of 1.0 A in the space between the parallel plates of \[1\mu F\]capacitor can be established by changing potential difference of    

    A) \[{{10}^{-6\text{ }}}V/s\]        

    B)  \[{{10}^{6}}\text{ }V/s\]

    C) \[{{10}^{-8}}\text{ }V/s\]        

    D) \[~{{10}^{8}}\text{ }V/s\]

    Correct Answer: B

    Solution :

     \[\frac{Q}{t}=\frac{CV}{t}\] or \[{{i}_{d}}=C\left( \frac{V}{t} \right)\] \[[\because \,q=it]\] or \[\frac{V}{t}=\frac{{{i}_{d}}}{C}=\frac{1}{{{10}^{-6}}}\] \[={{10}^{6}}\,V/s\]


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