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thunderhadron
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Homework Statement
The Problem is as follows :-
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In the figure shown initially the switch is open for a long time. Now the switch is closed at t = 0. Find the charge on the rightmost capacitor as a function of time. Given that it was initially uncharged.
Homework Equations
& The attempt at a solution[/B]When the switch is open if we talk only about the first capacitor (Left one), it is connected to the battery for a very long time hence it is fully charged and the potential difference between its plates become the potential difference of the battery. When the switch is closed at t = 0, the resistance of the capacitor (right one) is zero hence the total charge will flow only in the outward circuit. Hence it is surely the charging of the right one capacitor and hence its equation will become as q= cV(1-e^(-t/RC) ) which is quite obvious. But there is big problem the left capacitor was charged for a very long time and we can assume that now it’ll behave like a battery. There’ll be discharging of left capacitor and charging of right capacitor. The net charge on the right capacitor will be due to both – the battery and the discharging current of the left capacitor. Now how to proceed for the charge from the left one capacitor. Or m I doing wrong in this method, Should I go for some another method I am pretty confusion in doing this.
Please friends help me in solving this issue.
The answer of the question is as follows:
q= cV/2 (1-1/2 e^(-t/RC) )
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