Respuesta :
As we know that current is defined as rate of flow of charge
[tex]i = \frac{dq}{dt}[/tex]
so by rearranging the equation we can say
[tex]q = \int i dt[/tex]
here we know that
[tex]i(t) = 110 sin(120\pi t)[/tex]
here we will substitute it in the above equation
[tex]q = \int 110 sin(120\pi t) dt[/tex]
[tex]q = 110 [- \frac{cos(120\pi t)}{120\pi}][/tex]
now here limits of time is from t = 0 to t = 1/180s
so here it will be given as
[tex]q = \frac{110}{120\pi}( -cos0 + cos(\frac{2\pi}{3}))[/tex]
[tex]q = 0.44 C[/tex]
so total charge flow will be 0.44 C
Answer:
The total charge passing a given point in the conductor is 0.438 C.
Explanation:
Given that,
The expression of current is
[tex]i(t)=110\sin(120\pi t)[/tex]
[tex]\dfrac{dq(t)}{t}=110\sin(120\pi t)[/tex]
[tex]dq(t)=110\sin(120\pi t)dt[/tex]....(I)
We need to calculate the total charge
On integrating both side of equation (I)
[tex]\int_{0}^{q}dq(t)=\int_{0}^{\dfrac{1}{180}}110\sin(120\pi t)dt[/tex]
[tex]q=110(\dfrac{-\cos(120\pi t)}{120\pi})_{0}^{\dfrac{1}{180}}[/tex]
[tex]q=-\dfrac{110}{120\pi}(cos(120\pi(\dfrac{1}{180}))-\cos120\pi(0))[/tex]
[tex]q=-0.2918(-\dfrac{1}{2}-1)[/tex]
[tex]q=0.438\ C[/tex]
Hence, The total charge passing a given point in the conductor is 0.438 C.
