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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    A 40 mF capacitor in a defibrillator is charged to 3000 V. The energy stored in the capacitor is sent through the patient during a pulse of duration 2ms. The power delivered to the patient is                                     [AIIMS 2004]

    A)                    45 kW                              

    B)            90 kW

    C)                    180 kW                            

    D)            360 kW

    Correct Answer: B

    Solution :

               Power \[=\frac{\frac{1}{2}C{{V}^{2}}}{t}=\frac{1\times 40\times {{10}^{-6}}\times {{(3000)}^{2}}}{2\times 2\times {{10}^{-3}}}=90\,kW\]


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