A) \[\frac{{{C}_{1}}{{C}_{2}}({{V}_{1}}+{{V}_{2}})}{2({{C}_{1}}+{{C}_{2}})}\]
B) \[\frac{{{C}_{1}}{{C}_{2}}({{V}_{1}}-{{V}_{2}})}{2({{C}_{1}}+{{C}_{2}})}\]
C) \[\frac{{{C}_{1}}{{C}_{2}}{{({{V}_{1}}-{{V}_{2}})}^{2}}}{2({{C}_{1}}+{{C}_{2}})}\]
D) \[\frac{({{C}_{1}}+{{C}_{2}})({{V}_{1}}-{{V}_{2}})}{{{C}_{1}}{{C}_{2}}}\]
Correct Answer: C
Solution :
Initial energy \[{{U}_{i}}=\frac{1}{2}{{C}_{1}}V_{1}^{2}+\frac{1}{2}{{C}_{2}}V_{2}^{2}\], Final energy \[{{U}_{f}}=\frac{1}{2}({{C}_{1}}+{{C}_{2}}){{V}^{2}}\] (where \[V=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}{{C}_{2}}})\] Hence energy loss \[\Delta U={{U}_{i}}-{{U}_{f}}=\frac{{{C}_{1}}{{C}_{2}}}{2({{C}_{1}}+{{C}_{2}})}{{({{V}_{1}}-{{V}_{2}})}^{2}}\]You need to login to perform this action.
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