Question

The current in a conductor varies with time $t$ as $I=3t+4t^2$,where $I$ in ampere and $t$ in seconds. The elelctric charge flows through the section of the conductor between $t=1\text{s}$ and $t=3\text{s}$ is $\frac{n}{3}\text{C}$. Here value of $n$ is

Answer 1

$I=\frac{dq}{dt}=3t+4t^2$

$\int dq={\int }_1^3(3t+4t^2)dt$

$Q={[\frac{3t^2}{2}+\frac{4t^2}{2}]}_1^3$

$=[\frac{27}{2}+\frac{108}{3}]-[\frac{3}{2}+\frac{4}{3}]$

$=13.5+36-\frac{17}{6}$

$=49.5-\frac{17}{6}=\frac{297-17}{6}$

$=\frac{280}{6}=\frac{140}{3}\text{C}$