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

the charge on the capacitor at time t after the connections are made is
`Q//Q_(0)e^(-t//RC)`
`or,i=dQ/dt=-(Q_(0)//RC)e^(-t//RC)`.
heat dissipated during the time `t_(1) to t_(2)` is
`U=int_(t_(1))^(t_(2))i^(2)Rdt`
U=int_(t_(1))^(t_(2))(Q_(0)^(2))/(RC^(2))e^(-2t//RC)dt`
`(Q_(0)^(2))/(2C)(e^-(2t_(1))/(RC)-e^-(2t_(2))/(RC)) ...(i)
the time constant RCis `(5 Omega)xx(5.0(mu)F)=25(mu)s.`
Putting `t_(1)=25(mu)s,t_(2)=50(mu)s and other values in (i),
`U=(20(mu)C)^(2)/(2xx5.0(mu)F)(e^(-2)-e^(-4))=4.7(mu)J.`