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 switch disconnected, 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 ?

Explanation

`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))`.