Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

(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.