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− 4t2 + 6t + 2. [See this example. The unit of current is an ampere 1 A = 1 C/s.](a) Find the current when t- 0.6 s. (b) Find the current when t1 s X A At what time is the current the lowest?

Answer 1

Answer:

(a) 2.28 A

(b) 1 A

(c) 1.333 s

Step-by-step explanation:

As function of charge Q in term of time t, to get the equation of current (charge per second) we can take the derivative in term of t

$I = \frac{dQ}{dt} = 3t^2 - 8t + 6$

(a) At t = 0.6

$I = 3*0.6^2 - 8*0.6 + 6 = 2.28 A$

(b) At t = 1s

$I = 3*1^2 -8*1 + 6 = 1 A$

(c) To find the time where current is lowest we can take the derivative and set it to 0

$\frac{dI}{dt} = 6t - 8 = 0$

t = 8/6 = 1.333s

At this time I = 3*1.33^2 - 8*1.333 + 6 = 0.667 A

Answer 2

a). The current with $t = 0.6 s$ would be as follows:

$2.28 A$

b). The current with $t = 1s$ × $A$

$1 A$

c). The time at which the current would remain the minimum

$1.333 s$

Find the current

a). Given that,

Equation:

$Q(t) = t3 - 4t2 + 6t + 2$

Since Q(charge) in the form of $t$, the derivative can formed for t

$I = dQ/dt$

$= 3t^2 - 8t + 6$

Since $t = 0.6 s$

$I = 3$ × $0.6^2 - 8$ × $0.6 + 6 = 2.28A$

b) Here,

$t = 1s$

$I = 3$ × $1^2$ $- 8$ × $1 + 6$ $= 1A$

c). The time at which the current would remain the minimum

$dI/dt = 6t - 8 = 0$

∵ $t = 1.333s$

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