Question

The charge flowing through a resistance R varies
with time t as Q = a - bt. The total heat produced
in R in t second is ( a and b are constants)

​

Answer 1

Answer:

The solution is:- they differentiate Q to get i=a-2b. then for i=0 t=a/2b. Then from joule's law of heating dH=i^2Rdt.. then H is integrated from 0 to a-2b.. to get H=(a^3)R/6b..which is the answer. but my doubt is I don't understand why do we hav to differentiate first and then integrate..After obtaining i=a-2bt ..why can't we simply use the formula (i^2Rt)=Q?