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At t =0 S_w is closed, if initially C_1 is uncharged and C_2 is charged to a potential difference 2epsi then find out following ("Given "C_1= C_2= C) (a) Charge on C_1 and C_2 as a function of time. (b) Find out current in the circuit as a function of time. (c) Also plot the graphs for the relations derived in part (a) |
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Answer» Solution :Let q charge FLOW in time 't' from the BATTERY as shown. The charge on various plates of the capacitor is as shown in the FIGURE. Now applying KVL `epsi-q/C-iR -(q-2epsiC)/(C)=0` `epsi-q/C-q/C+2e-iR=0` `3epsi=(2q)/(C)+iR`  `3epsi-iR=(2q)/(C)` `3epsi-iRC=2q` `(dq)/(dt)RC=3epsiC-2q` `underset(0)OVERSET(q)int (dq)/(2epsiC-2q)=underset(0)overset(t)int (dt)/(RC)` `-1/2ln ((3Cepsi-2q)/(3Cepsi))=t/(RC)` `ln ((3epsiC-2q)/(3epsiC))=-(2t)/(RC)` `3epsiC-2q=3epsiC e^(-2t//RC)` `3epsiC (1-e^(-2t//RC))=2q` `q=3/2epsiC(1-e^(-2t//RC)` `i=(dq)/(dt)=(3epsi)/(R) e^(-2t//RC)` On plate B `q' =2epsiC-q` `=2epsiC-3/2epsiC+3/2epsiCe^(-2t//RC)`  `=(epsiC)/(2)+3/2epsiCe^(-2t//RC) =(epsiC)/(2) [1+3e^(-2t//RC)]` 
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