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NEET NEET SOLVED PAPER 2016 Phase-I

  • question_answer
    A capacitor of 2μF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is:                                                                                  

    A)  0%                                        

    B)  20%

    C)  75%                                      

    D)   80%

    Correct Answer: D

    Solution :

                     Initial energy stored in capacitor 2 μF \[{{U}_{i}}=\frac{1}{2}2{{(V)}^{2}}={{V}^{2}}\] Final voltage after switch 2 is ON \[{{V}_{f}}=\frac{{{C}_{1}}{{V}_{1}}}{{{C}_{1}}+{{C}_{2}}}=\frac{2V}{10}=0.2\,V\] Final energy in both the capacitors \[{{U}_{f}}=\frac{1}{2}({{C}_{1}}+{{C}_{2}})V_{f}^{2}\,\,\,\,=\frac{1}{2}10{{\left( \frac{2V}{10} \right)}^{2}}=0.2{{V}^{2}}\] So energy dissipated \[=\frac{V{{}^{2}}-0.2{{V}^{2}}}{{{V}^{2}}}\times 100=80%\]


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