Question

Find the average current for the time interval $t=-1$ to $t=1\text{s}$, if the current is given by, $i=(6t^5+3t^2+1)\text{A}$.

• A. $1\text{A}$
• B. $0.5\text{A}$
• C. $3\text{A}$
• D. $2\text{A}$

Answer 1

The correct option is D $2\text{A}$

Average current is given by,

$i_{avg}=\frac{{\int }_{t_1}^{t_2}idt}{{\int }_{t_1}^{t_2}dt}$

$i_{avg}=\frac{{\int }_{-1}^1(6t^5+3t^2+1)dt}{{\int }_{-1}^{1}dt}$

$i_{avg}=\frac{[t^6+t^3+t{]}_{-1}^1}{[t{]}_{-1}^1}$

$i_{avg}=\frac{[1^6-(-1{)}^6]+[1^3-(-1{)}^3]+[1-(-1\left)\right]}{[1-(-1\left)\right]}$

$\therefore i_{avg}=2\text{A}$

Hence, $\left(\text{D}\right)$ is the correct answer.