Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • In the circuit shown, the capacitor is initially charged with charge <img alt="\mathrm{q_0}" src=

In the circuit shown, the capacitor \mathrm{C_1} is initially charged with charge \mathrm{q_0}. The switch S is closed at time \mathrm{t=0}. The charge on \mathrm{C_2} after time t is

Option: 1

\mathrm{\frac{q_0 C_1}{C_1+C_2}\left(1-e^{\frac{t\left(C_1+C_2\right)}{C_1 C_2 R}}\right)}


Option: 2

\mathrm{\frac{q_0 C_2}{C_1+C_2}\left(I-e^{\frac{t\left(C_1+C_2\right)}{C_1 C_2 R}}\right)}


Option: 3

\mathrm{\frac{q_0 C_1}{C_1+C_2}\left(1-e^{\frac{t}{C_2 R}}\right)}


Option: 4

\mathrm{\frac{q_0 C_2}{C_1+C_2}\left(1-e^{\frac{t}{C_2 R}}\right)}


Answers (1)

best_answer

\mathrm{I=\frac{d q}{d t} :\text{apply Kirchhoff's law: } \frac{q_0-q}{C_1}=I R+\frac{q}{C_2} }

\begin{aligned} &\mathrm{ \frac{q_0}{C_1}-q\left[\frac{C_1+C_2}{C_1 C_2}\right]=\frac{d q}{d t} R }\\ &\mathrm{ \int_0^q \frac{d q}{\frac{q_0}{C_1}-q\left(\frac{C_1+C_2}{C_1 C_2}\right)}=\int_0^t \frac{d t}{R} }\end{aligned}
Solve to get: \mathrm{q=\frac{q_0 C_2}{C_1+C_0}\left[1-e^{\frac{t\left(C_1+C_2\right)}{C_1 C_2 R}}\right]}

Posted by

Pankaj

View full answer

NEET 2024 Most scoring concepts

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

Similar Questions

Trending Articles/News

anam-khan

NEET UG 2026 leak probe traces guess paper trail from Kerala student to Rajasthan network
anam-khan

'Replace NTA': FAIMA files plea in SC over NEET UG 2026 paper leak
anam-khan

NEET UG 2026 Paper Leak LIVE: Protests erupt across states, paper leak probe expands; re-NEET date soon

Latest Question