A `5.0(mu)F` capacitor having a charge of `(20 (mu)C)`is discharged through a wire aof resistance `(5.0 Omega)`. Find the heat dissipated in the wire between 25 to `50 (mu)s after the connections are made.

Explanation

`hat0= RC = 5xx5xx10^(-6) =25xx10^(-6)s=25`
Current in the wire or circuit at any time is given by
`I =(Q)/(RC)e^(-t//tau)`
So heat dissipated is
`U = int_(t_1)^(t_2) i^2 Edt = (Q^2R)/(R^2C^2)int_(25)^(25) e^(-2t//pi)`
`=(Q^2)/(RC^2)(-(hato)/(2))(e^(-2t//pi))_(25)^(50)`
`=-(Q^2)/(2C)[e^(-2xxx50//25) - e^(-2xx50//25)]=(Q^2)/(2C) [e^(-2) - e^(-4)]`
`=(20xx10^(-6))/(2xx5xx10^(-6))[e^(-2)-e^(-4)]=4.7xx1-^(-6)J`