JEE Main On 16 April 2018 Question 18
Question: In the following circuit, the switch S is closed at $ t=0 $ .The charge on the capacitor $ C _1 $ as a function of time will be given by $ ( C _{eq}=\frac{C _1C _2}{C _1+C _2} ). $ [JEE Main 16-4-2018]
Options:
A) $ C _{eq}E[1-\exp (-t/RC _{eq})] $
B) $ C _1E[1-\exp (-tR/C _1)] $
C) $ C _2E[1-\exp (-t/RC _2)] $
D) $ C _{eq}E\exp (-t/RC _{eq}) $
Show Answer
Answer:
Correct Answer: A
Solution:
The formula for charge varying with time in charging capacitor is
$ Q=E\times C _{eq}(1-\exp (\frac{-t}{R\times c _{eq}})) $
as the equivalent capacitor is given in the question.
So the correct option which matches is A.
