The current in a wire varies with time according to the relationship 1=55A - (0.65A/s2) t2. (a) How many coulombs of charge pass a cross section of the wire in the time interval between t = 0 and t = 7.5 s? (b) What constant current would transport the same charge in the same time interval?

A) Q = 321 C

B) I = 42.8 A

Explanation:

We are given the relationship between current and time to be;

I = (55 - 0.65A/s²) t²

A) We are to find the charge transferred (Q) from t=0 s to t = 7.5 s.

Q represents the current which is the net charge flowing through the area in this period of time and it's given by the formula;

ΔQ = I•dt

So, ΔQ = (55 - 0.65A/s²) t²•dt

To obtain Q, let's integrate ΔQ;

∫ΔQ = ∫ (55 - 0.65A/s²) t²•dt at boundary of t = 0 and 7.5s

Q = 55At - (0.65/3) (A/s²) t³ at boundary of t = 0 and 7.5s

Thus,

Q = 55A (7.5s) - (0.65/3) (A/s²) (7.5s) ³

Q = 412.5 A. s - 91.4 A. s ≈ 321 A. s

Now, A. s is also known as Columbs (C)

Thus, Q = 321 C

B) Current flow is given by the equation;

I = Q/t

Now, t will be 7.5s because it is the certain time that represents the current.

Thus, plugging in the relevant values;

I = 321/7.5 = 42.8 A

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a) Given I = 55A - (0.65A/s2) t²

I = dQ/dt

dQ = I*dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I*dt = ∫[55 - (0.65) t²]dt

Q = [55t - 0.65/3*t³]

Q between t=0 and t = 7.5s

Q = [55 * (7.5 - 0) - 0.65/3 (7.5³ - 0³) ]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.