Two identical capacitors are connected in parallel across a potential difference V. After they are fully charged the battery is removed and the positive of first capacitor is connected to negative of second and negative of first is connected to positive of other. The loss in energy will be -

Question

Two identical capacitors are connected in parallel across a potential difference V. After they are fully charged the battery is removed and the positive of first capacitor is connected to negative of second and negative of first is connected to positive of other. The loss in energy will be - (A) (1/2) C V2 (B) CV 2 (C) ( CV 2/4) (D) Zero

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e2a39ad4c456ff42b59237e083209c57-.png" /> ∴ loss in energy =( C 1 C 2/2( C 1+ C 2))( V 1- V 2)2 =( C 2/2 × 2 C )( V -(- V ))2 =( C /4) × 4 V 2= CV 2

Q. Two identical capacitors are connected in parallel across a potential difference $V$ . After they are fully charged the battery is removed and the positive of first capacitor is connected to negative of second and negative of first is connected to positive of other. The loss in energy will be -

$\frac{1}{2} C V^{2}$

$C V^{2}$

$\frac{C V ^{2}}{4}$

Zero

$∴$ loss in energy $= \frac{C _{1} C _{2}}{2 ( C _{1} + C _{2} )} (V_{1} - V_{2})^{2}$
$= \frac{C ^{2}}{2 \times 2 C} (V - (- V))^{2}$
$= \frac{C}{4} \times 4 V^{2} = C V^{2}$

Q. Two identical capacitors are connected in parallel across a potential difference V. After they are fully charged the battery is removed and the positive of first capacitor is connected to negative of second and negative of first is connected to positive of other. The loss in energy will be -

21​CV2

CV2

4CV2​