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 :
| [d] 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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