2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Electrostatics & Capacitance

JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    A body of capacity \[4\,\mu \,F\] is charged to \[80\,V\] and another body of capacity \[6\,\mu \,F\] is charged to 30V. When they are connected the energy lost by \[4\,\mu \,F\] capacitor is [EAMCET 2001]

    A)            7.8 mJ                                     

    B)            4.6 mJ

    C)            3.2 mJ                                     

    D)            2.5 mJ

    Correct Answer: A

    Solution :

               Initial energy of body of capacitance 4 mF is \[{{U}_{i}}=\frac{1}{2}\times (4\times {{10}^{-6}})\,{{(80)}^{2}}=0.0128J\] Final potential on this body after connection is \[V=\frac{4\times 80+6\times 30}{4+6}=50V.\] So final energy on it \[{{U}_{f}}=\frac{1}{2}\times 4\times {{10}^{-6}}{{(50)}^{2}}=0.005\,J\] Energy lost by this body \[={{U}_{i}}{{U}_{f}}=\text{ 7}.\text{8}mJ\]


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