Question

A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

• A. Zero
• B. \( \dfrac{3}{2}CV^2 \)
• C. \( \dfrac{25}{6}CV^2 \)
• D. (9)/(2)CV²

Hint:

When capacitors are connected with opposite polarity, subtract charges before redistributing.

Answer 1

The Correct Option is B

Step 1: Initial charges: Q₁ = CV, Q₂ = 2C(2V)=4CV
Step 2: Since opposite terminals are connected, net charge: Qₙet = 4CV - CV = 3CV
Step 3: Equivalent capacitance: Cₑq = C + 2C = 3C
Step 4: Final voltage: Vf = fracQₙetCₑq = V
Step 5: Final energy: U = (1)/(2)CₑqVf² = (1)/(2)(3C)V² = (3)/(2)CV²