Question

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:

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

Answer 1

The correct option is C $\frac{a^{3}R}{3b}$

Given: the charge flowing through a resistance $R$ varies with time $t$ as $Q=at-b{t}^2$, where $a$ and $b$ are positive constants

To find the total heat produced in $R$

Solution

Charge, $Q=at-b{t}^2......\left(i\right)$

Therefore $t\in [0,\frac{a}{b}]$

We know that

$I=\frac{d\theta }{dt}$

So, eqn(i) becomes

$I=\frac{d}{dt}(at-b{t}^2)⟹I=a-2bt.....\left(ii\right)$

Note: $i$ is positive, $t\in [0,\frac{a}{2b}]$

and $i$ is negative, $t\in [\frac{a}{2b},\frac{a}{b}]$

Positive current means current one direction and negative current means current in opposite direction

$\therefore ,dH=i^{2}Rdt⟹dH=(a-2bt{)}^{2}Rdt⟹H={\int }_0^{\frac{a}{b}}(a-2b{)}^{2}Rdt⟹H={\left[\frac{(a-2bt{)}^{3}R}{3(-2b)}\right]}_0^{\frac{a}{b}}⟹H=\frac{1}{-b}[{(a-2b\frac{a}{b})}^3-a^3]R⟹H=-\frac{1}{6b}\left[\right(-a{)}^3-a^3]R⟹H=\frac{a^{3}R}{3b}$

is the required total heat produced