Electric charge passing through a resistor changes with time t as Q = at - ` bt^(2)` . Then total heat produce in resistor R = ....

Options

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

Answer

D

Explanation

`(a^(3)R)/(6b)`
`Q = at - bt^(2)`
`therefore I =(dQ)/(dt ) = (d)/(dt) " " [at - bt^(2) ]`
`therefore I = a - 2 bt`
`therefore 0 = a - 2 bt`
`therefore t = (a)/(2b)`
Now power P = `int_(0)^(t) I^(2) R dt = int_(0)^((a)/(2b)) ( a - 2bt)^(2) R dt`
`therefore P = int_(0)^((a)/(2b)) (a^(2) - 4abt + 4b^(2) t^(2) ) R dt `
= `[ a^(2) t - 4ab (t^(2))/(2) + 4b^(2) (t^(3))/(3) ]_(0)^((a)/(2b)) R`
= ` [ (a^(2) xx a)/(2b) - (4 ab xx a^(2))/(2 xx 4 b^(2)) + (4b^(2) xx a^(3))/(3 xx 8b^(3)) ]` R
= `[ (a^(3))/(2b) - (a^(3))/(2b) + (a^(3))/(6b) ]R = (a^(3)R)/(6b)`