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 \[\left( {{C}_{eq}}=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}} \right).\] [JEE Main 16-4-2018]
A) \[{{C}_{eq}}E[1-\exp (-t/R{{C}_{eq}})]\]
B) \[{{C}_{1}}E[1-\exp (-tR/{{C}_{1}})]\]
C) \[{{C}_{2}}E[1-\exp (-t/R{{C}_{2}})]\]
D) \[{{C}_{eq}}E\exp (-t/R{{C}_{eq}})\]
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.
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