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The given RC circuit has two switches S_(1) and S_(2) The switch S_(2) is closed till the capacitor C attains its maximum possible charge 90. Then, Sy is opened and S, is closed simultaneously till the capacitor releases half of its total stored charge go for a time interval ty. Finally S is opened and Sa is closed till the capacitor attains a charge (3//4)qo for a time interval t_(2). Find the ratio (t._(1)//t_(2). |
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Answer» Solution :When the switch `S_(2)` is closed, the capacitor is charged to a potential V. Now the switch `S_(2)` is opened and the switch `S_(1)` is closed The instantaneous charge on the capacitor is `impliesq=q_(0)e^((1)/(RC))` where`q_(0)=CV`Putting`t=t_(1)` for`q=q-(0)//2`,`weobaine^(t_(1)/(R_(1)C))=(1)/(2)` `implies(t_(1)=R_(1)CIn2` Again, the switch `S_(1)` is opened and `S_(2)`is closed. Therefore, the capacitor STARTS charging from charge `q-(0)//2 to (3)/(4)q_(0)` Now, instantaneous charge on the capacitor is `q=q_(0)[1-e^((t)/(R_(1)+(R_(2))C)]]+q_(0)/(2)e^((t)/(R_(1)+R_(2)C))=q_(0)[1-(1)/(2)e^((t)/(R_(1)+R_(2)c)]]` `At,t=t_(2),q^(.)=3q_(0)//4implies3q_(0)/(4)=(q_(0)(1)/(2)e^((t_(2))/((R_(1)+R_(2))C)))` `impliese^(t^(2)/(R_(1)R_(2)C))=(1)/(2)impliest_(2)=(R_(1)+R_(2))CIn2` `therefore` The required ratio of the times =`(t_(1)/(t_(2))(R_(1)CIn2)/((R_(1)+R_(2))CIn2)` or,`(t_(1))/(t_(2))=(R_(1))/((R_(1)+R_(2))` 
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