Q38. The capacities of two capacitors are C₁ and C₂ and their respective potentials are V₁ and V₂. I

Q: What is the correct answer for CUET UG Physics Q38?

The correct answer is c. When two charged capacitors with capacitances C₁ and C₂ and initial potentials V₁ and V₂ are connected, they reach a common potential (V_common). Initial total energy E_initial = ½C₁V₁² + ½C₂V₂².By conservation of charge, the total charge before connection equals the total charge after connection:C₁V₁ + C₂V₂ = (C₁ + C₂)V_common.So, V_common = (C₁V₁ + C₂V₂) / (C₁ + C₂).Final total energy E_final =

Q: Q38. The capacities of two capacitors are C₁ and C₂ and their respective potentials are V₁ and V₂. If they are connected with a wire, then the loss of energy is given by: (a) C1C2(V1+V2) /…

The correct answer is c. When two charged capacitors with capacitances C₁ and C₂ and initial potentials V₁ and V₂ are connected, they reach a common potential (V_common). Initial total energy E_initial = ½C₁V₁² + ½C₂V₂².By conservation of charge, the total charge before connection equals the total charge after connection:C₁V₁ + C₂V₂ = (C₁ + C₂)V_common.So, V_common = (C₁V₁ + C₂V₂) / (C₁ + C₂).Final total energy E_final =

Q: What are the options for this CUET UG Physics question?

A. C1C2(V1+V2) / (2(C1+C2)) · B. C1C2(V1-V2) / (2(C1+C2)) · C. C1C2(V1-V2)² / (2(C1+C2)) · D. (C1+C2)(V1-V2) / (C1C2)

Question

Q38. The capacities of two capacitors are C₁ and C₂ and their respective potentials are V₁ and V₂. If they are connected with a wire, then the loss of energy is given by:
(a) C1C2(V1+V2) / (2(C1+C2))
(b) C1C2(V1-V2) / (2(C1+C2))
(c) C1C2(V1-V2)² / (2(C1+C2))
(d) (C1+C2)(V1-V2) / (C1C2)

FirstInTest · Accepted Answer

When two charged capacitors with capacitances C₁ and C₂ and initial potentials V₁ and V₂ are connected, they reach a common potential (V_common). Initial total energy E_initial = ½C₁V₁² + ½C₂V₂².By conservation of charge, the total charge before connection equals the total charge after connection:C₁V₁ + C₂V₂ = (C₁ + C₂)V_common.So, V_common = (C₁V₁ + C₂V₂) / (C₁ + C₂).Final total energy E_final = ½(C₁ + C₂)V_common² = ½(C₁ + C₂) * [(C₁V₁ + C₂V₂) / (C₁ + C₂)]².The loss of energy (ΔE) is E_initial - E_final. After algebraic simplification, this energy loss is given by:ΔE = (C₁C₂(V₁ - V₂)²) / (2(C₁ + C₂)).This energy loss is typically dissipated as heat in the connecting wire due to spark or resistance.

Answer

C1C2(V1+V2) / (2(C1+C2))

Answer

C1C2(V1-V2) / (2(C1+C2))

Answer

C1C2(V1-V2)² / (2(C1+C2))

Answer

(C1+C2)(V1-V2) / (C1C2)

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