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A circuit consists of a capacitor with capacitance C and a coil of inductance L connected in series, as well as a switch and a resistance equal to the critical value for this circuit. With the switchdisconnected, the capacitor was charged to a voltage V_(0), and at the moment t=0, the switch was closed . Find the corrent I in the circuit as a function of time t.What is I_(max) equal to ? |
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Answer» Solution :`o=(q)/(C)+L(dI)/(dt)+RI, I=+(dq)/(dt)` For the critical case `R=2sqrt((L)/(C))` Thus `LCddot(q)+2 sqrt(LC)dot(q)+q=0` Look for a solution with `q prop E^(alphat)` `alpha=-(1)/(sqrt(LC))` An independent solution is `t e ^( alpha t)`. Thus `q=(A+Bt)e^(-t//sqrt(LC))`, At `t=0 q=CV_(0) ` thus `A=CV_(0)` Also at`t=0 dot(q)=I=0` `0=B-A(1)/(sqrt(lC))implies B=B_(0)sqrt((C)/(L))` Thus FINALLY `I=(dq)/(dt)=V_(0)sqrt((C)/(L))e^(-t//sqrt(LC))` `-(1)/(sqrt(LC))(CV_(0)+V_(0)sqrt((C)/( L))t) E^(-t//sqrt(LC))` `=-(V_(0))/( L)te^(-t//sqrt(LC))` The current has been defined to increase the CHARGE. Hence the MINUS sign. ltbr. The current is maximum when `(dI)/( dt)=-(V_(0))/(L) e^(-t//sqrt(LC))(1-(t)/(sqrt(LC)))=0` This GIVES `t= sqrt(LC)` and the magnitude of the maximum current is `|I_(max)|=(V_(0))/(e)sqrt((C)/(L))`. 
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