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EAMCET Medical EAMCET Medical Solved Paper-2009

  • question_answer
    A 8 \[\mu \]F capacitor is charged by a 400 V supply through 0.1 MO resistance. The time taken by the capacitor to develop a potential difference of 300 V is (Given\[lo{{g}_{10}}4=0.602\])

    A)  2.2s                                      

    B)  1.1 s                

    C)  0.55 s                                   

    D)  0.48 s

    Correct Answer: B

    Solution :

     Charge on capacitor \[Q=CV\] \[Q=8\times {{10}^{-6}}\times 400\]       ?(i)                 Since,                    \[Q=It=\frac{V}{R}t\]                                                 \[Q=\frac{300}{0.1\times {{10}^{6}}}t\]                  ?(ii)                 Equate Eqs. (i) and (ii), we get                                 \[\frac{300}{{{10}^{5}}}t=8\times {{10}^{-6}}\times 400\]                                 \[t=\frac{8\times {{10}^{-6}}\times 400\times {{10}^{5}}}{300}\]\[=1.1\,s\]


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