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NEET AIPMT SOLVED PAPER SCREENING 2010

  • question_answer
    A series combination of \[{{n}_{1}}\] capacitors, each of value \[{{C}_{1}},\] is charged by a source of potential difference  4V. When another parallel combination of \[{{n}_{2}}\] capacitors, each of value \[{{C}_{2}},\] is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of \[{{C}_{2}},\] in terms of \[{{C}_{1}},\] is then

    A) \[\frac{2{{C}_{1}}}{{{n}_{1}}\,{{n}_{2}}}\]             

    B)        \[16\frac{{{n}_{2}}}{{{n}_{1}}\,}{{C}_{1}}\]

    C) \[2\frac{{{n}_{2}}}{{{n}_{1}}\,}{{C}_{1}}\]             

    D)        \[\frac{16{{C}_{1}}}{{{n}_{1}}\,{{n}_{2}}\,}\]

    Correct Answer: D

    Solution :

    Case I. When the capacitors are joined in series \[{{U}_{series}}=\frac{1}{2}\frac{{{C}_{1}}}{{{n}_{1}}}{{(4V)}^{2}}\] Case II. When the capacitors are joined in parallel \[{{U}_{parallel}}=\frac{1}{2}({{n}_{2}}{{C}_{2}}){{V}^{2}}\] Given, \[{{U}_{series}}={{U}_{parallel}}\] or                            \[\frac{1}{2}\frac{{{C}_{1}}}{{{n}_{1}}}{{(4V)}^{2}}=\frac{1}{2}({{n}_{2}}{{C}_{2}}){{V}^{2}}\] \[\Rightarrow \]                               \[{{C}^{2}}=\frac{16{{C}_{1}}}{{{n}_{2}}\,{{n}_{1}}}\]


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