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.
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
