contestada

The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by Q(t) = t3− 2t2 + 4t + 1. [See this example. The unit of current is an ampere 1 A = 1 C/s.] (a) Find the current when t = 0.7 s.

Respuesta :

Answer:

current = 2.67 A

Explanation:

we have given,

Q(t) = t³− 2t² + 4t + 1

to find I at t = 0.7s

we know

[tex]\frac{\mathrm{d} Q(t)}{\mathrm{d} t}=I[/tex]

so,

[tex]\frac{\mathrm{d} Q(t)}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t}{(t^3-2t^2+4t+1)}[/tex]

[tex]\frac{\mathrm{d} Q(t)}{\mathrm{d} t}=3t^2-4t+4[/tex]

current at t= 0.7s

I = 2.67 A

hence, current comes out to be 2.67 A