Question

The current through a wire depends on time as $i=i_0+3\alpha t^2$, where $i_0=10A$ and $\alpha =\frac{1}{2}A/se{c}^2$. Average current for that interval $t=0$ to $t=2s$ is:

• A. $16.28A$
• B. $10.28A$
• C. $12A$
• D. $10A$

Answer 1

Answer: (C)

$12A$

The current flowing through the wire depending on time is given by $i=i_0+3\alpha t^2$
Since current $i=\frac{dq}{dt}$, we write $\frac{dq}{dt}=i_0+3\alpha t^2$

Total charge flowing through the wire in the time interval $t=0s$ to $t=2s$,
$q={\int }_o^{2}idt={\int }_0^2(i_0+3\alpha t^2)dt$
Or $q={\int }_0^2(10+\frac{3}{2}{t}^2)dt$
We get $q=10t{|}_o^2+\frac{3}{2}\times \frac{t^3}{3}{|}_o^2$

Or $q=10(2-0)+\frac{3}{2}\times \frac{(2-0{)}^3}{3}=24$

Average current $i_{avg}=$ total charge$/$total time

Average current $i_{avg}=\frac{24}{2}=12A$