Question

The capacities of two conductors are C1 and C2 and their respectively potentials are V1 and V2. If they are connected by a thin wire, then the loss of energy will be given by

A. C₁C₂V₁+V₂/2(C₁+C₂)
B. C₁C₂V₁−V₂/2(C₁+C₂)
C. (C₁C₂V₁−V₂)²/2(C₁+C₂)
D. (C₁C₂)(V₁−V₂)/C₁+C₂

Answer 1

The loss of energy when two conductors with capacities C1 and C2 are connected and their potentials are V1 and V2 respectively is given by (C1C2(V1−V2)2)/(2(C1+C2)).

Step-by-step explanation:

The loss of energy when two conductors with capacities C1 and C2, and potentials V1 and V2, are connected by a thin wire can be calculated using the formula for energy stored in a capacitor and the principle of conservation of charge. The final common potential (V) after they are connected can be found by using the formula V = (C1V1 + C2V2)/(C1 + C2). The loss of energy (ΔE) is the difference between the initial and final energy stored in the system. Therefore, the loss of energy is calculated as ΔE = ½C1V12 + ½C2V22 - ½(C1 + C2)V2, which simplifies to ΔE = ½(C1C2(V1 - V2)2)/(C1 + C2). So, the correct option that represents the loss of energy is C. (C1C2(V1−V2)2)/2(C1+C2).