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 \[\frac{1}{2}\frac{{{C}_{1}}}{{{n}_{1}}}{{(4V)}^{2}}=\frac{1}{2}({{n}_{2}}{{C}_{2}}){{V}^{2}}\] \[\Rightarrow \] \[{{C}^{2}}=\frac{16{{C}_{1}}}{{{n}_{2}}\,{{n}_{1}}}\]You need to login to perform this action.
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