Question

An electric current in a conductor varies with time according to the expression i(t) = 110 sin (120πt), where i is in amperes and t is in seconds. what is the total charge passing a given point in the conductor from t = 0 to t = 1/180 s?

Answer 1

As we know that current is defined as rate of flow of charge

$i = \frac{dq}{dt}$

so by rearranging the equation we can say

$q = \int i dt$

here we know that

$i(t) = 110 sin(120\pi t)$

here we will substitute it in the above equation

$q = \int 110 sin(120\pi t) dt$

$q = 110 [- \frac{cos(120\pi t)}{120\pi}]$

now here limits of time is from t = 0 to t = 1/180s

so here it will be given as

$q = \frac{110}{120\pi}( -cos0 + cos(\frac{2\pi}{3}))$

$q = 0.44 C$

so total charge flow will be 0.44 C

Answer 2

Answer:

The total charge passing a given point in the conductor is 0.438 C.

Explanation:

Given that,

The expression of current is

$i(t)=110\sin(120\pi t)$

$\dfrac{dq(t)}{t}=110\sin(120\pi t)$

$dq(t)=110\sin(120\pi t)dt$....(I)

We need to calculate the total charge

On integrating both side of equation (I)

$\int_{0}^{q}dq(t)=\int_{0}^{\dfrac{1}{180}}110\sin(120\pi t)dt$

$q=110(\dfrac{-\cos(120\pi t)}{120\pi})_{0}^{\dfrac{1}{180}}$

$q=-\dfrac{110}{120\pi}(cos(120\pi(\dfrac{1}{180}))-\cos120\pi(0))$

$q=-0.2918(-\dfrac{1}{2}-1)$

$q=0.438\ C$

Hence,  The total charge passing a given point in the conductor is 0.438 C.