Question

Two parallel plate capacitors of capacitance C and 2C are charged up to a potential difference V and 2V respectively. Now, these capacitors are connected in parallel to each other such that the positively charged plate of capacitor C is connected to the negatively charged plate of capacitors 2C. Find the amount of change in the potential energy of the system.

Answer 1

Answer:

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I GUES THAT IS THE ANSWER,,,

A PARLLEL PLATE CAPACITOR HAS EFFECTIVE CAPACITANCE AS WE ADD IT...

ON SUBSTITUTING IT I GOT THE ANSWER I SAID 3CV^2

Explanation:

Answer 2

Answer :

Capacitance of C₁ = C (V)

Capacitance of C₂ = 2C (2V)

After charging, both capacitors are connected in parallel to each other such that the +ve charged plate of C is connected to -ve plate of 2C$.$

We have to find amount of change in potential energy of system.

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◈ Common pd across parallel connection :

➝ V' = (C₂V₂ - C₁V₁)/(C₁ + C₂)

➝ V' = (4CV - CV)/(C+2C)

➝ V' = 3CV/3C

➝ V' = V

◈ Energy stored in C₁ :

➝ U₁ = 1/2 × C₁V₁²

➝ U₁ = 1/2 × CV²

◈ Energy stored in C₂ :

➝ U₂ = 1/2 × C₂V₂²

➝ U₂ = 1/2 × (2C)(2V)²

➝ U₂ = 4CV²

◈ Total initial energy :

➝ Ui = U₁ + U₂

➝ Ui = 1/2×CV² + 4CV²

➝ Ui = 9CV²/2

◈ Total energy of parallel combination :

➝ Uf = 1/2 × (C₁+C₂) V'²

➝ Uf = 1/2 × (C+2C) V²

➝ Uf = 3CV²/2

◈ Change in potential Energy :

⇒ ΔU = Ui - Uf

⇒ ΔU = 9CV²/2 - 3CV²/2

⇒ ΔU = 6CV²/2

⇒ ΔU = 3CV²