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 μ C ) 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>&nbsp;0</p>
61 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
5 months ago
Correct Option - 3
Detailed Solution:

i i = 8 4 1 4 0 = 0 . 6 A

i 2 = 8 4 6 0 = 1 . 4 A

p.d across capacitor = 20 v

Q = CV = 5 * 20 = 100    ? C

After S is open

Q = Q . e-t/RC

=100 e-4

 

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Physics Current Electricity 2025

Physics Current Electricity 2025

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