Through a coil of resistance R charge is flowing in such a manner that in every time interval t 0 the amount of charge flow is reduced by a factor of 1 2 . If a total of q charge flows through the coil in very long time, then the total amount of heat generated in the coil will be

Step-by-step solution

I ( t ) = I 0 e - λ t ,         λ = l o g e 2 T 1 2 = l o g e 2 T 0 q = ∫ d q = ∫ 0 ∞ I ( t ) ⋅ d t = ∫ 0 ∞ I 0 ⋅ e - λ t d t = I 0 ( - λ ) ⋅ e - λ t 0 ∞ ⇒     q = I 0 λ     ⇒ I 0 = q λ H = ∫ d H = ∫ I 2 R d t = R ∫ 0 ∞ I 0 2 e - 2 λ t d t H = R I 0 2 1 2 λ = R 2 λ × q 2 λ 2 = q 2 R λ 2 = q 2 R 2 × l o g e 2 t 0 &#8658