Question

current in a conductor varies with time T as I=3t+4t^2. where I in amp and t in sec. The electric charge flows through the section of the conductor between t=1s and t=3s

Answer 1

Hello...

i = 3t + 4t²

=> dq/dt = 3t + 4t²

=> dq = (3t + 4t²) dt -----------(1)

=> integrating eqn (1) from limit 1 to 3,

=> q = 3/2 [9 -1] + 4/3 [27 -1]

=> q = 12 + 4/3 × 26 = 46.67 C

Answer.......

Answer 2

Answer:

46.67 coulombs of electric charge flow through the section of the conductor between t=1s and t=3s.

Explanation:

The current is calculated as,

$I=\frac{Q}{dt}$     (1)

Where,

I=current in a conductor

Q=electric charge that flows

dt=time during which the charge flows

From the question we have,

I=3t+4t²  (2)

t₁=1second

t₂=3 second

By placing equation (2) in equation (1) we get;

$3t+4t^2=\frac{Q}{dt}$

$Q=(3t+4t^2)dt$    (3)

On integrating equation (3) we get;

$Q=3tdt+4t^2dt$

$Q=\frac{3t^2}{2} +\frac{4t^3}{3}$

$Q=\frac{3(t_2^2-t_1^2)}{2} +\frac{4(t_2^3-t_1^3)}{3}$      (4)

By placing the values of "t₁" and "t₂" in equation (4) we get;

$Q=\frac{3(3^2-1^2)}{2} +\frac{4(3^3-1^3)}{3}$

$Q=\frac{3(9-1)}{2} +\frac{4(27-1)}{3}$

$Q=\frac{3\times 8}{2} +\frac{4\times 26}{3}$

$Q= 12+34.67$

$Q=46.67\ Coulomb$

Hence, 46.67 coulombs of electric charge flow through the section of the conductor between t=1s and t=3s.

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