The charge flowing through a resistance R varies with time t as Q = at - bt2 , where a and b are positive constants, The total heat produced in R is :

Question

The charge flowing through a resistance R varies with time t as Q = at - bt2 , where a and b are positive constants, The total heat produced in R is : (A) (a3 R/3b) (B) (a3 R/2b) (C) (a3 R/b) (D) (a3 R/6b)

Tardigrade Admin · Accepted Answer

Q = at - bt2 i = a - 2bt for i = 0 ⇒ t = (a/2b) From joule's law of heating dH = i2Rdt H = ∫ limitsa/2b0 (a - 2bt)2 Rdt H = ((a - 2bt)3 R/ - 3 × 2b) Bigg|(a/2b)0 = (a3R/6b)

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

$\frac{a ^{3} R}{3 b}$

$\frac{a ^{3} R}{2 b}$

$\frac{a ^{3} R}{b}$

$\frac{a ^{3} R}{6 b}$

$Q = a t - b t^{2}$
$i = a - 2 b t$ { for i = 0 $\Rightarrow t = \frac{a}{2 b}$ }
From joule's law of heating
$d H = i^{2} R d t$
$H = 0 \int a /2 b (a - 2 b t)^{2} R d t$
$H = \frac{( a - 2 b t ) ^{3} R}{- 3 \times 2 b} ∣ ∣_{0}^{\frac{a}{2 b}} = \frac{a ^{3} R}{6 b}$

Q. The charge flowing through a resistance R varies with time t as Q=at−bt2 , where a and b are positive constants, The total heat produced in R is :

3ba3R​

2ba3R​

ba3R​

6ba3R​