A) \[16\frac{{{n}_{2}}}{{{n}_{1}}}{{C}_{1}}\]
B) \[\frac{2{{C}_{1}}}{{{n}_{1}}{{n}_{2}}}\]
C) \[2\frac{{{n}_{2}}}{{{n}_{1}}}{{C}_{1}}\]
D) \[\frac{16{{C}_{1}}}{{{n}_{1}}{{n}_{2}}}\]
Correct Answer: D
Solution :
| [d] Equivalent capacitance of \[{{n}_{2}}\] number of capacitors each of capacitance \[{{C}_{2}}\] in parallel\[={{n}_{2}}{{C}_{2}}\] |
| Equivalent capacitance of\[{{n}_{1}}\] number of capacitors each of capacitances \[{{C}_{1}}\] in series. |
| Capacitance of each is \[{{C}_{1}}=\frac{{{C}_{1}}}{{{n}_{1}}}\] |
| According to question, total energy stored in both the combinations are same |
| i.e.,\[\frac{1}{2}\left( \frac{{{C}_{1}}}{{{n}_{1}}} \right){{\left( 4V \right)}^{2}}=\frac{1}{2}\left( {{n}_{2}}{{C}_{2}} \right){{V}^{2}}\] |
| \[\therefore \]\[{{C}_{2}}=\frac{16{{C}_{1}}}{{{n}_{1}}{{n}_{2}}}\] |
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