Question

The quantity of charge through a conductor is modeled as Q = (3.00 mC/s4)t4 − (2.00 mC/s)t + 9.00 mC. What is the current (in A) at time t = 4.00 s? A

Answer 1

Answer:

The current at time t = 4.00 s is 0.766 A.

Explanation:

Given that,

The quantity of charge through a conductor is modeled as :

$Q=(3t^4-2t+9)\ mC$

We need to find the current (in A) at time t = 4.00 s. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(3t^4-2t+9)}{dt}\\\\I=12t^3-2$

At t = 4 s

$I=12(4)^3-2=766\ mC/s\\\\I=0.766\ C/s=0.766\ A$

So, the current at time t = 4.00 s is 0.766 A. Hence, this is the required solution.