A) 3 : 1
B) 5 : 1
C) 3 : 5
D) 5 : 3
Correct Answer: C
Solution :
Initially potential difference across both the capacitor is same hence energy of the system is \[{{U}_{1}}=\frac{1}{2}C{{V}^{2}}+\frac{1}{2}C{{V}^{2}}=C{{V}^{2}}\] ??(i) In the second case when key K is opened and dielectric medium is filled between the plates, capacitance of both the capacitors becomes 3C, while potential difference across A is V and potential difference across B is \[\frac{V}{3}\] hence energy of the system now is \[{{U}_{2}}=\frac{1}{2}\,(3C){{V}^{2}}+\frac{1}{2}\,(3C)\,{{\left( \frac{V}{3} \right)}^{2}}\]\[=\frac{10}{6}\,C{{V}^{2}}\] ??(ii) So, \[\frac{{{U}_{1}}}{{{U}_{2}}}=\frac{3}{5}\]You need to login to perform this action.
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