Answer:
1.14 C
Explanation:
Recall,
i = dq/dt
q = ∫idt......................... Equation 1
Where q = charge, i = current, t = time.
Given: i = 86sin(40πt)
Substitute into equation 1
q = ∫[86sin(40πt)]dt
q = 86(-cos40πt)/40π
q = 86/40π[-cos(40πt)]..................... Equation 2
Given the time limit,
0 and 1/60 s.
Substitute into equation 2
q = -86/40π[cos(40π×1/60)-cos0]
q= -86/40π[cos(2π/3)-1)
q = -86/40π[(-2/3)-1]
q= -86/40π(-5/3)
q= -0.684(-1.667)
q = 1.14 C
1.03C
The current (I) flowing through a conductor is related to the charge (Q) passing through the conductor at a given range of time, t, as follows;
I(t) = [tex]\frac{dQ}{dt}[/tex] -----------------(i)
Now, to express the charge (Q) in terms of I and t, we need to integrate equation (i) as follows;
dQ = I(t)dt
Q = [tex]\int\limits^a_b {I(t)} \, dt[/tex] ---------------------------(ii)
Where, according to the question;
I(t) = 86 sin (40πt)
a = final time t = 1/60s
b = initial time t = 0
Substitute these values into equation(ii) as follows;
Q = [tex]\int\limits^\frac{1}{60} _0 {86sin(40\pi t)} \, dt[/tex]
Q = 86 x [tex]\int\limits^\frac{1}{60} _0 {sin(40\pi t)} \, dt[/tex] ------------------(iii)
Solve the integral as follows;
let u = 40πt [differentiate both sides]
=> [tex]\frac{du}{dt}[/tex] = 40π
=> dt = [tex]\frac{du}{40\pi }[/tex]
=> du = 40π dt
Substitute the values of u = 40πt and dt = [tex]\frac{du}{40\pi }[/tex] into equation(iii)
Q = 86 x [tex]\int\limits^\frac{1}{60} _0 {sin(u)} \, \frac{du}{40\pi }[/tex] -------------------(iv)
Q = [tex]\frac{86}{40\pi }[/tex] x [tex]\int\limits^\frac{1}{60} _0 {sin(u)} \, du[/tex]
Now, integrate sin u
Q = [tex]\frac{86}{40\pi }[/tex] x [-cos (u)] ------------------(v)
Substitute the value of u back into equation (v)
Q = [tex]\frac{86}{40\pi }[/tex] x [-cos (40πt)]
Q = - [tex]\frac{86}{40\pi }[/tex] x [cos (40πt)] ------------------(vi)
Now, insert the values of t = 0 and t = 1/60 into equation (vi) as follows;
Q = - [tex]\frac{86}{40\pi }[/tex] x [ [cos (40π(1/60))] - [cos (40π(0))] ]
Q = - [tex]\frac{86}{40\pi }[/tex] x [ [cos ([tex]\frac{4\pi }{6}[/tex])] - [cos (0)] ]
Q = - [tex]\frac{86}{40\pi }[/tex] x [ [cos ([tex]\frac{2\pi }{3}[/tex])] - [cos (0)] ]
Q = - [tex]\frac{86}{40\pi }[/tex] x [ [-0.5] - [1] ]
Q = - [tex]\frac{86}{40\pi }[/tex] x [ -1.5 ]
Q = [tex]\frac{86}{40\pi }[/tex] x [1.5 ]
Q = [tex]\frac{86 * 1.5}{40\pi }[/tex]
Take π = 3.142
Q = [tex]\frac{86 * 1.5}{40 * 3.142}[/tex]
Q = [tex]\frac{129}{125.68}[/tex]
Q = 1.03C
Therefore, the total charge passing a given point in the conductor from t = 0 to t = 1/60s is 1.03C
