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

JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Capacitance

  • question_answer
    A \[2\mu F\]capacitor is charged to 100\[volt\] and then its plates are connected by a conducting wire. The heat produced is [MP PET 1999; Pb. PET 2003]

    A)                    \[1\ J\]

    B)                                      \[0.1\ J\]

    C)                    \[0.01\ J\]                      

    D)            \[0.001\ J\]

    Correct Answer: C

    Solution :

                       Heat produced = Energy of charged capacitor \[=\frac{1}{2}C{{V}^{2}}\] \[=\frac{1}{2}\times (2\times {{10}^{-6}})\times {{(100)}^{2}}=0.01J\]


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