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 :
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 samei.e.,\[\frac{1}{2}\left( \frac{{{C}_{1}}}{{{n}_{1}}} \right){{\left( 4V \right)}^{2}}=\frac{1}{2}\left( {{n}_{2}}{{C}_{2}} \right){{V}^{2}}\] \[\therefore \]\[{{C}_{2}}=\frac{16{{C}_{1}}}{{{n}_{1}}{{n}_{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
