AIIMS AIIMS Solved Paper-2004

  • question_answer
    A 40\[\mu F\] 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 2 ms. The power delivered to the patient is:

    A) 45 kW    

    B)                       90 kW                   

    C)       180 kW                

    D)       360 kW

    Correct Answer: B

    Solution :

                    A capacitor is a device that stores energy in the electric field created between a pair of conductors on which equal but opposite electric charges have been placed. The energy stored in a capacitor \[=\frac{1}{2}C{{V}^{2}}\] Given, \[C=40\mu F=40\times {{10}^{-6}}F,V=3000V\] \[E=\frac{1}{2}\times 40\times {{10}^{-6}}\times {{(3000)}^{2}}\] \[=180J\] Also \[1\text{ }W=1\text{ }J/s\] \[\therefore \]   \[2\text{ }ms=2\times {{10}^{-3}}\text{ }s\] Hence, power \[=\frac{180J}{2\times {{10}^{-3}}s}=90\times {{10}^{3}}W=90kW\].


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