Question

The charge flowing through a resistance $R$ varies with time $t$ as $Q=at-b{t}^2$. The total heat produced in $R$ by the time current ceases is

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

Answer 1

The correct option is A $a^{3}R/6b$

Given,

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

$i=\frac{dQ}{dt}=\frac{d}{dt}\cdot (at-b{t}^2)$

$\Rightarrow i=\frac{d}{dt}\cdot at-\frac{d}{dt}\cdot b{t}^2$

$\Rightarrow i=a\cdot \frac{d}{dt}-b\cdot \frac{d}{dt}{t}^2$

$\Rightarrow i=a-2bt$

we know, $H=\int i^2\cdot Rdt$

Now, for the limits of integration,

when $i=0,a=2bt$

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

$H={\int }_a^{a/2b}{i}^2\cdot Rdt$

$\Rightarrow H={\int }_a^{a/2b}(a^2+4b^2{t}^2-4bt)Rdt=R{[a^{2}t+\frac{4b^2{t}^3}{3}-\frac{4ab{t}^2}{2}]}_0^{a/2b}$

$\Rightarrow H=R[a^2\left(\frac{a}{2b}\right)+\frac{4b^2}{3}\left(\frac{a^3}{8b^3}\right)-\frac{4ab}{2}\left(\frac{a^2}{4b^2}\right)]=R[\frac{a^3}{2b}+\frac{a^3{b}^3}{6b^3}-\frac{a^3}{6b}]=R\left[\frac{3a^3{b}^3+a^3{b}^2-{3a}^3{b}^3}{6b^3}\right]$

$\Rightarrow H=R\left[\frac{a^3}{6b}\right]=\frac{R{a}^3}{6b}$