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Consider the shown network. Initially both swiches are open and the capacitor are uncharged. Both the switches are then closed simultaneously at t=0 (a) Obtain the current through switch S_(2) as function of time. (b) If switch S_(2) is opened again after a long time interval, find the total heat that would dissipate in the resistor and the charge that would flow through switch S_(1) after S_(2) is opened? |
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Answer» Solution :Consider the charges on capacitors and currents through various branches, as shown in the figure For loop `1`, we have `R_(1)(I_(2)+I_(3)-I_(1))=(q_(1))/(C_(1))` `(1)` For loop`2`, `R_(2)i_(3)=(q_(2))/(C_(1))` `(2)` For the outer loop, `i_(3)R_(2)+(q_(1))/(C_(1))=epsilon` `(3)` Also, `i_(1)=(dq_(1))/(dt)` and `i_(2)=(dq_(2))/(dt)` `(4)` Putting the values of `i_(1)` and `i_(2)` from Eq. `(4)` and of `i_(3)` from Eq. `(2)` in `(1)`, we get `(d)/(dt)(q_(2)-q_(1))=(q_(1))/(R_(1)C_(1))-(q_(2))/(R_(2)-C_(2))` `(5)` From Eqs. `(2)` and `(3)`, we get `(q_(2))/(C_(2))+(q_(1))/(C_(1))=epsilon` `(6)` From Eqs. `(5)` and `(6)`, we get `int_(0)^(q_(2))(dq_(2))/(((epsilonC_(2)R_(2))/(R_(1)+R_(2)))-q_(2))=int_(0)^(t)(dt)/(R_(eq)(C_(1)+C_(2)))`, where `R_(eq)=(R_(1)R_(2))/(R_(1)+R_(2))` `rArrq_(2)=(epsilonR_(2)C_(2))/((R_(1)+R_(2)))[1-e^((1)/(R_(eq)(C_(1)+C_(2))))]` Similarly, `q_(1)=(epsilonR_(1)C_(1))/((R_(1)+R_(2)))[1-e^((1)/(R_(eq)(C_(1)+C_(2))))]` `rArr i_(1)=(dq_(1))/(dt)=(epsilonC_(1))/(R_(2)(C_(1)+C_(2)))e^((1)/(R_(eq)(C_(1)+C_(2))))` Similarly, `i_(2)=(epsilonC_(2))/(R_(1)(C_(1)+C_(2)))e^((-1)/(R_(eq)(C_(1)+C_(2))))` From equation `(1)` Current through `S_(2)=(i_(1)-i_(3))=i_(2)-(q_(1))/(R_(1)C_(1))` Putting the values, we get `q_(1)=(12muC)(1-e^((1)/(12mu)))`, `q_(2)=(48muC)(1-e^((1)/(12mu)))`  `i_(1)=(1A)e^((1)/(12mu))`, `i_(2)=(4A)e^((1)/(12mu))` Current through SWITCH `s_(2)=-[2-e^((1)/(12mu))]A` along the indicated direction as shown in fig `(i)`, With both the switches closed the steady state charges and currents are as shown in Fig `(ii)`.  With switch `S_(2)` open and `s_(1)` closed, the steady state charges are as shown in Fig.`(iii)`. HENCE, the charge flown through switch `S_(1)=[(36+72)-(12+48)]muC=48muC`. Total heat dissipated in the resistors `=` [INITIAL ENERGY `+` work done by battery when `48muC` flows through it after switch `S_(2)` is opened ] `-` [FINAL energy] `={(1)/(2)C_(1)V_(1)^(2)+(1)/(2)C_(2)V_(2)^(2)}+epsilon(DeltaQ)-{(1)/(2)C_(1)V'_(1)^(2)+(1)/(2)C_(2)V'_(2)^(2)}=136muJ`. |
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