The charge flowing through a resistance `R` varies with time `t as Q = at - bt^(2)`. The total heat produced in `R` is

Options

A. `(a^(3)R)/(3b)`
B. `(a^(3)R)/(2b)`
C. `(a^(3)R)/(b)`
D. `(a^(3)R)/(6b)`

Answer

A

Explanation

Charge becomes zero when t = a/b after the start. So we need to calculate heat developed in this interval. Current `I = (dQ)/(dt) = a- 2bt`
Total heat loss:
`H = underset(0)overset(a//b)int i^(2) Rdt= R underset(0)overset(a//b)int (a^(2) + 4b^(2) + t^(2) - 4abt) dt`
`rArr H= R [(a^(3))/(b) + (4a^(3))/(3b) - (2a^(3))/(b)] = (Ra^(3))/(b) [(4)/(3) -1] = (Ra^(3))/(3b)` Hence option (a) is correct