Get Answers to all your Questions

header-bg qa

.A capacitor of capacitance C has charge Q. It is connected to an identical capacitor through a resistance. The heat produced in the resistance is

Option: 1

\mathrm{ \frac{Q^2}{2 C}}


Option: 2

\mathrm{ \frac{Q^2}{4 C}}


Option: 3

\mathrm{ \frac{Q^2}{8 C}}


Option: 4

Dependent on the value of the resistance


Answers (1)

best_answer

As the capacitors are identical, each of them finally have charge \mathrm{\frac{Q}{2}}

Initial energy of the system = \mathrm{E_i=\frac{Q^2}{2 C}}
Final energy of the system \mathrm{=E_f=2\left[\frac{\left(Q / 2^2\right)}{2 C}\right]=\frac{Q^2}{4 C} }

Heat produced = loss in energy = \mathrm{E_i=E_f=\frac{Q^2}{4 C}}

Posted by

Divya Prakash Singh

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks