A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor $\frac{C}{2}$. The energy loss in the process after the charge is distributed between the two capacitors is: Option: 1 Option: 2 Option: 3 Option: 4

\begin{aligned} &\text { Heat loss = }\\ &H=\frac{C_{1} C_{2}}{2\left(C_{1}+C_{2}\right)}\left(V_{1}-V_{2}\right)^{2}\\ &=\frac{C \times \frac{C}{2}}{2\left(C+\frac{C}{2}\right)}\left(V_{0}-0\right)^{2}=\frac{C}{6} V_{0}^{2}\\ &H=\frac{1}{6} C V_{0}^{2} \end{aligned}

Similar Questions

Preparation Products

Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
Test Series JEE Main May 2021

Unlimited Chapter Wise Tests, Unlimited Subject Wise Tests, Unlimited Full Mock Tests, Get Personalized Performance Analysis Report,.

₹ 6999/- ₹ 2999/-
Knockout JEE Main May 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 14999/-