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Question: The capacities of two conductors are \(C _ { 1 }\) and \(V _ { 2 }\). If they are connected by a th...

The capacities of two conductors are C1C _ { 1 } and V2V _ { 2 }. If they are connected by a thin wire, then the loss of energy will be given by

A

C1C2(V1+V2)2(C1+C2)\frac { C _ { 1 } C _ { 2 } \left( V _ { 1 } + V _ { 2 } \right) } { 2 \left( C _ { 1 } + C _ { 2 } \right) }

B

C1C2(V1−V2)2(C1+C2)\frac { C _ { 1 } C _ { 2 } \left( V _ { 1 } - V _ { 2 } \right) } { 2 \left( C _ { 1 } + C _ { 2 } \right) }

C

C1C2(V1−V2)22(C1+C2)\frac { C _ { 1 } C _ { 2 } \left( V _ { 1 } - V _ { 2 } \right) ^ { 2 } } { 2 \left( C _ { 1 } + C _ { 2 } \right) }

D

(C1+C2)(V1−V2)C1C2\frac { \left( C _ { 1 } + C _ { 2 } \right) \left( V _ { 1 } - V _ { 2 } \right) } { C _ { 1 } C _ { 2 } }

Answer

C1C2(V1−V2)22(C1+C2)\frac { C _ { 1 } C _ { 2 } \left( V _ { 1 } - V _ { 2 } \right) ^ { 2 } } { 2 \left( C _ { 1 } + C _ { 2 } \right) }

Explanation

Solution

Initial energy Ui=12C1V12+12C2V22U _ { i } = \frac { 1 } { 2 } C _ { 1 } V _ { 1 } ^ { 2 } + \frac { 1 } { 2 } C _ { 2 } V _ { 2 } ^ { 2 },

Final energy Uf=12(C1+C2)V2U _ { f } = \frac { 1 } { 2 } \left( C _ { 1 } + C _ { 2 } \right) V ^ { 2 }

(where V=C1V1+C2V2C1C2)\left. V = \frac { C _ { 1 } V _ { 1 } + C _ { 2 } V _ { 2 } } { C _ { 1 } C _ { 2 } } \right)

Hence energy loss

ΔU=Ui−Uf=C1C22(C1+C2)(V1−V2)2\Delta U = U _ { i } - U _ { f } = \frac { C _ { 1 } C _ { 2 } } { 2 \left( C _ { 1 } + C _ { 2 } \right) } \left( V _ { 1 } - V _ { 2 } \right) ^ { 2 }