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VMMC VMMC Medical Solved Paper-2010

  • question_answer
    0.2 F capacitor is charged to 600 V by a battery, on removing the battery, it is connected with another parallel plate condenser of 1 F. The potential decreases to

    A)  100 V                   

    B)         120 V

    C)  300 V                   

    D)         600 V

    Correct Answer: A

    Solution :

    By using charge conservation \[0.2\times 600=(0.2+1)V\] \[V=\frac{0.2\times 600}{1.2}\] \[=100\,V\]


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