MGIMS WARDHA MGIMS WARDHA Solved Paper-2010

  • question_answer
    Two capacitors, one 4 pF and the other 6 pF, connected in parallel, are charged by a 100 V battery. The energy stored in the capacitors is

    A) \[1.2\times {{10}^{-8}}J\]                            

    B)  \[2.4\times {{10}^{-8}}J\]

    C) \[5.0\times {{10}^{-8}}J\]                            

    D)  \[1.2\times {{10}^{-6}}J\]

    Correct Answer: C

    Solution :

    The energy stored in capacitor is given by \[E=\frac{1}{2}C{{V}^{2}}\] Resultant capacitance \[C'={{C}_{1}}+{{C}_{2}}=4+6=10\,pF\] \[E=\frac{1}{2}\times 10\times {{10}^{-12}}\times {{(100)}^{2}}\] \[(1\text{ }pF={{10}^{-12}}F)\] \[=5\times {{10}^{-8}}J\]


You need to login to perform this action.
You will be redirected in 3 sec spinner