The charge flowing through a resistance R varies with time t as Q = at – bt^2, where a and b are positive constants. The total heat produced in R is:

Question

Accepted Answer

Correct answer: B. (b) Given, Q= at-bt^2 ∴I= dQ dt=a-2 bt At t=0 ,Q=0 ⇒I=0 Also, I=0 at t=a/2b ∴ Total heat produced in resistance R, H= 0.05em 0.05emI^2 Rdt=R 0.05em (a-2 bt )^2 dt =R 0.05em 0.05em(a^2+4b^2t^2-4 abt) dt =R(a^2t+4b^2 t^33-4 ab t^22)_0^a/2b =R(a^2 a2b+ 4b^23 a^38b^3- 4 ab2 a^24b^2) = a^3Rb( 12+ 16- 12)= a^3R6b

$The charge flowing through a resistance R varies with time t as Q = at - b t^{2}, where a and b are positive constants. The total heat produced in R is:$

Solution

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