BHU PMT BHU PMT (Screening) Solved Paper-2010

  • question_answer
    An\[8\mu F\]capacitor is connected across 220 V, 50 Hz line. What is the peak value of the charge through capacitor?

    A)  \[2.5\times {{10}^{-3}}C\]                          

    B) \[2.5\times {{10}^{-4}}C\]

    C)  \[5\times {{10}^{-5}}C\]                              

    D) \[7.5\times {{10}^{-2}}C\]

    Correct Answer: A

    Solution :

                     Capacitive reactance \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2\pi \times 50\times 8\times {{10}^{-6}}}=398\,\Omega \] \[{{V}_{rms}}=220\,V,,{{V}_{peak}}=\sqrt{2}\times 220\]                 \[=311\,V\]         \[{{q}_{rms}}={{V}_{rms}}V=1.76\times {{10}^{-3}}C\] Peak value of charge                 \[{{q}_{peak}}=\sqrt{2}{{q}_{rms}}=1.76\times {{10}^{-3}}\times \sqrt{2}\]                 \[=2.5\times {{10}^{-3}}C\]


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