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 sectio

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?

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Thông Đạt 5 months 2021-08-09T08:06:40+00:00 2 Answers 13 views 0

Answers ( )

    0
    2021-08-09T08:07:54+00:00

    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

    0
    2021-08-09T08:08:00+00:00

    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.

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