Question

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+4

Find the current when t = 1s. At what time is the current lowest?

Answer 1

Explanation:

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 :

$Q=t^3-2t^2+4t+4$

We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :

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

At t = 1 s,

Current,

$I=3(1)^2-4(1)+4\\\\I=3\ A$

So, the current at t = 1 s is 3 A.

For lowest current,

$\dfrac{dI}{dt}=0\\\\\dfrac{d(3t^2-4t+4)}{dt}=0\\\\6t-4=0\\\\t=0.67\ s$

Hence, this is the required solution.