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 connected in series \[{{U}_{series}}=\frac{1}{2}\frac{{{C}_{1}}}{{{n}_{1}}}{{(4V)}^{2}}\] Case II. When the capacitors are connected in parallel \[{{U}_{parallel}}=\frac{1}{2}({{n}_{2}}{{C}_{2}}){{V}^{2}}\] According to question, \[{{U}_{series}}={{U}_{parallel}}\] or \[=\frac{1}{2}\frac{{{C}_{1}}}{{{n}_{1}}}{{(4V)}^{2}}=\frac{1}{2}({{n}_{2}}{{C}_{2}}){{V}^{2}}\] \[\Rightarrow \] \[{{C}_{2}}=\frac{16{{C}_{1}}}{{{n}_{2}}{{n}_{1}}}\]You need to login to perform this action.
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