Question

What amount of heat will be generated in a coil of resistance `R` due to a charge q passing through it if the current in the coil
a. decreases down to zero uniformly during a time interval `t_0`?
b. decrases down to zero having its value every `t_0` seconds?

Answer 1

a. The corresponding `i.t` graph will be a straight line with i decreasing from a peak value (say `i_0`) to zero in time `t_0`.
i-t equation will be as
`i=i_0(i_0/t_0)t (y=-mx+x)`…..i
Here `i_0` is unknown, which can be obtained by using the fact area under `i-t` graph gives the flow of charge. Hence,

`q=1/2(t_0)(i_0)`
`:. i_0=(2q)/t_0`
Substituting is Eqn. i we get `i=(2q)/t_0(1-t/t_0)`
or `i=((2q)/t_0-(2qt)/t_0^2)`
Now, `t` time `t` heat produced in a short interval `dt` is
`dh=i^2Rdt`
`=((2q)/t_0-(2qt)/t_0^2)^2Rdt`
`=4/3(q^2R)/t_0`
b. Here, current decreases from some peak value `(say i_0)` to zero exponentially with half life `t_0`.

i-t equation in the case will be
`i=i_0e^(-lamdat)`
Here` lambda=(In(2))/t_0`
Now, `q=int_0^ooidt=int-0^ooi_0e^(-lamdat)dt=(i_0/lamda)`
`: i_0=lamdaq`
`:. i=(lamdaq)e^(-lamdat)`
`:. dH=i^2Rdt-lamda^2q^2e^(-2lamdat)Rdt`
`or H=int_0^oo dh=lamda^2 q^2 R int_0^oo e^(-2lamdat)dt=(q^2lamdaR)/2`
substituting `lamda=(In(2))/t_0` we have `H=(q^2RIn (2))/(2t_0)`