Question

An electric current in a conductor varies with time according to the expression I(t) = 86 sin (40πt), 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 = 1/60 s?

Answer 1

1.03C

Step-by-step explanation:

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) =
`(dQ)/(dt)` -----------------(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 =
`\int\limits^a_b {I(t)} \, dt` ---------------------------(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 =
`\int\limits^(1)/(60) _0 {86sin(40\pi t)} \, dt`

Q = 86 x
`\int\limits^(1)/(60) _0 {sin(40\pi t)} \, dt` ------------------(iii)

Solve the integral as follows;

let u = 40πt [differentiate both sides]

=>
`(du)/(dt)` = 40π

=> dt =
`(du)/(40\pi )`

=> du = 40π dt

Substitute the values of u = 40πt and dt =
`(du)/(40\pi )` into equation(iii)

Q = 86 x
`\int\limits^(1)/(60) _0 {sin(u)} \, (du)/(40\pi )` -------------------(iv)

Q =
`(86)/(40\pi )` x
`\int\limits^(1)/(60) _0 {sin(u)} \, du`

Now, integrate sin u

Q =
`(86)/(40\pi )` x [-cos (u)] ------------------(v)

Substitute the value of u back into equation (v)

Q =
`(86)/(40\pi )` x [-cos (40πt)]

Q = -
`(86)/(40\pi )` x [cos (40πt)] ------------------(vi)

Now, insert the values of t = 0 and t = 1/60 into equation (vi) as follows;

Q = -
`(86)/(40\pi )` x [ [cos (40π(1/60))] - [cos (40π(0))] ]

Q = -
`(86)/(40\pi )` x [ [cos (
`(4\pi )/(6)`)] - [cos (0)] ]

Q = -
`(86)/(40\pi )` x [ [cos (
`(2\pi )/(3)`)] - [cos (0)] ]

Q = -
`(86)/(40\pi )` x [ [-0.5] - [1] ]

Q = -
`(86)/(40\pi )` x [ -1.5 ]

Q =
`(86)/(40\pi )` x [1.5 ]

Q =
`(86 * 1.5)/(40\pi )`

Take π = 3.142

Q =
`(86 * 1.5)/(40 * 3.142)`

Q =
`(129)/(125.68)`

Q = 1.03C

Therefore, the total charge passing a given point in the conductor from t = 0 to t = 1/60s is 1.03C

Answer 2

1.14 C

Step-by-step 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