Question

Current is flowing in conductor as i = 6t+4t^2
here t is time second and i is current then find
average current in conductor t = 0 to t = 10 sec.
(1) 50 A
(2) 330 A
(3) 200 A
(4) 420 A​

Answer 1

Answer:

Given :

current through conductor is given as :

$\boxed{i = 6t + 4 {t}^{2} }$

To find:

Average current passing through the conductor in time 0-10 seconds

Concept:

The given Equation actually represents the instantaneous current flowing through the conductor.

In order to calculate the average current, we need to take the ratio of total charge passed to the total time taken.

Calculation:

$\therefore \int \: dq = \int \: i \: dt$

Putting the limits :

$</p><p> \displaystyle \: = > \int_{0}^{q} \: dq = \int_{0}^{10} \: i \: dt$

$</p><p> \displaystyle \: = > \int_{0}^{q} \: dq = \int_{0}^{10} \: (6t + 4 {t}^{2} ) \: dt$

$= > q = \bigg \{3 {t}^{2} \bigg \}_{0}^{10} + \bigg \{ \dfrac{4}{3} {t}^{3} \bigg \}_{0}^{10}$

$= > q = 300 + 1333.33$

$= > q = 1633 .33 \: c$

So average current be I :

$I = \dfrac{total \: charge}{time}$

$= > I = \dfrac{1633.33}{10}$

$= > I = 163.33 \: amp$

So final answer :

$\boxed{ \red{ \huge{ \sf{ \bold{ I = 163.33 \: amp}}}}}$