Question

Current I versus time t graph through a conductor is shown in the figure. Average current through the conductr in the interval 0 to 15 sec

Answer 1

Answer:

5 A

Explanation:

1) Figure is attached :

We have, current vs time graph through a conductor as shown in figure.

We know,

Average current is given by

$i_{avg}=\frac{\int idt}{\int dt}$

We know ,

Integration of i w.r.t. t gives Area under i-t graph.

And

Integration of i dt = 15-0=15 s

2) So, In triangle

Base = 15-0=15s

Height = 10 A

Area of triangle = 1/2*base * height = 0.5*15*10=75 C

So,

$i_{avg}=\frac{\int idt}{\int dt}\\ \\=\frac{75C}{15s}=5A$

Hence,

Average current is 5A.