The circuit shown in the figure is in steady state for a long time. The connection to battery is suddenly broken (switch S is opened up). What is the charge (in ) on the capacitor after 0.001 sec.?
The circuit shown in the figure is in steady state for a long time. The connection to battery is suddenly broken (switch S is opened up). What is the charge (in ) on the capacitor after 0.001 sec.?
Option 1 - <p>100</p>
Option 2 - <p>100 e<sup>-2</sup></p>
Option 3 - <p>100 e<sup>‑4</sup></p>
Option 4 - <p> 0</p>
61 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
Answered by
5 months ago
Correct Option - 3
Detailed Solution:
p.d across capacitor = 20 v
Q = CV = 5 * 20 = 100
After S is open
Q = Q . e-t/RC
=100 e-4
Similar Questions for you
Ohm's law is valid if I depends on V' linearly.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers
Learn more about...

Physics Current Electricity 2025
View Exam DetailsMost viewed information
SummaryDidn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
or
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
or
See what others like you are asking & answering







