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At t =0S_w is closed , if initially C_1 is uncharged andC_2is charged to a potential difference2inthen find out following(GivenC_1 =C_2 =C ) (a)Charge onC_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 qcharge 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 ` in -(q)/(c ) -iR ( q-2in C)/( C) =0 ` `in -( q)/(c) -( q)/(c) +2in -iR = 0` `3in =( 2q)/( C) + iR ` ` (##MOT_CON_JEE_PHY_C25_SLV_042_S01.png" width="80%"> ` 3in -iR=(2q)/(C) ` ` (dq)/( dt) RC=3in C - 2q` ` int _0^(q) (dq)/( 2in C -2q) =int _0^(t)(dt)/( RC) ` ` - (1)/(2)ln ((3Cin -2q)/( 3Cin )) =(t)/(RC) ` ` 1n ((3in C- 2q)/( 3in C) ) =-( 2T)/(RC) ` ` 3in C - 2q=3in C e^(=2 tau//RC)` ` 3in C( 1-e^(-2tau//RC)) = 2q` ` q= (3)/(2) in C(1-e^(-2tau//RC))` `i =(dq)/( dt) =(3in )/(R) e^(-2t//RC)` ` i =(dq)/( dt) =(3in)/(R) e ^(-2tau//RC)` on the plate B `q'=2in C -q` ` 2in C -(3)/(2) in C +(3)/(2) in C e^(-2tau//RC) ` ` (##MOT_CON_JEE_PHY_C25_SLV_042_S02.png" width="80%"> ` =(in C)/(2) + (3)/(2) in Ce^(-2t//RC) =(in C)/( 2) [1+ 3e^(-2tau//RC) ]` |
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