Question

The charge $\left(\text{in C}\right)$ flowing through a resistance $R$ varies with time $\left(\text{in s}\right)$ as $Q=2t–8t^2$. The total heat produced in the resistance is (for $0\le t\le \frac{1}{8}s$)

• A. $\frac{R}{6}\text{Joules}$
• B. $\frac{R}{3}\text{Joules}$
• C. $\frac{R}{2}\text{Joules}$
• D. $R\text{Joules}$

Answer 1

The correct option is A $\frac{R}{6}\text{Joules}$

We know that,
$i=\frac{dQ}{dt}=2-16t$
Heat $=R{\int }_0^{1/8}{\left(i\right)}^2.dt$
Heat $=R{\int }_0^{1/8}{(2-16t)}^2.dt$
Heat $=R{\int }_0^{1/8}(4+256t^2-64t).dt$
Heat $=R{[4t+\frac{256t^3}{3}-32t^2]}_0^{1/8}$
Heat $=R\left[4\right(\frac{1}{8})+\frac{256(\frac{1}{8}{)}^3}{3}-32(\frac{1}{8}{)}^2]$
Heat $=R\left[\right(\frac{1}{2})+\frac{1}{6}-(\frac{1}{2}\left)\right]$

Heat=$\frac{R}{6}J$