Q117 / 180 · NEET 2016 (Phase 1) · Physics

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:

Correct answer
A
$\frac{a^{3} R}{6 b}$
B
$\frac{a^{3} R}{3 b}$
C
$\frac{a^{3} R}{2 b}$
D
$\frac{a^{3} R}{b}$

Solution

given

$Q = a t - b t^{2}$

$i = \frac{d q}{d t} = a - 2 b t$

$i = 0$

at $t = \frac{a}{2b}$

$∴ d H = i^{2} R d t$

$= {(a - 2 b t)}^{2} R d t$

$H = \int_{0}^{\frac{a}{2b}} {(a - 2 b t)}^{2} R d t$

${= \frac{{(a - 2 b t)}^{3} R}{3 (- 2 b)}|}_{0}^{\frac{a}{2b}}$

$= \frac{1}{-6 b} [{(a - 2 b \frac{a}{2b})}^{3} - {(a)}^{3}] R$

$= - \frac{1}{6 b} [{(0)}^{3} - a^{3}] R$

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

Hence, the correct answer is option: 'A'.

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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:

Solution

Relevant Questions

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: