The plates of a `50(mu)F` capacitor charged to `400(mu)C` are connected through a resistance of `1.0k(ohm)`. Find the charge remaining on the capacitor 1s after the connection is made.

Explanation

The time constant is
`CR=(50(mu)F)(1.0k(Omega))=50ms.`
`At t=1 s,t/CR=(1s)/(50ms)=20.`
`the charge remaining on the capacitor is
`q=Qe^(-t/CR)`
`(400(mu)C)e^(-20)=8.2xx10^(-7)(mu)C.`
We see that in typical charging or discharging corcuit, the time constant is of the order of a millisecond. Also four to five time constants are sufficient for 99% of hte charging or discharging. thus for practical purposes we can assume that charging or discharging is complete in a fraction of a second.