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 + 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.

Answer 1

The current flowing through the wire at t = 0.7 seconds is approximately 2.67 C/s.

What is the current when t = 0.7 s?

Recall the relationship between current and charge:

Current (I) is the rate of change of charge (Q) over time. Mathematically, it's represented as:

I(t) = dQ(t)/dt

Differentiate the charge function:

Given the function for Q(t) = t³ - 2t² + 4t + 1, we can find the current I(t) by differentiating it with respect to time t:

I(t) = 3t² - 4t + 4

Calculate the current at t = 0.7 s:

Now, plug t = 0.7 into the expression for I(t):

I(0.7) = 3(0.7)² - 4(0.7) + 4

I(0.7) ≈ 2.67 C/s

Therefore, the current flowing through the wire at t = 0.7 seconds is approximately 2.67 C/s.

Answer 2

current = 2.67 A

Step-by-step explanation:

we have given,

Q(t) = t³− 2t² + 4t + 1

to find I at t = 0.7s

we know

`\frac{\mathrm{d} Q(t)}{\mathrm{d} t}=I`

so,

`\frac{\mathrm{d} Q(t)}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t}{(t^3-2t^2+4t+1)}`

`\frac{\mathrm{d} Q(t)}{\mathrm{d} t}=3t^2-4t+4`

current at t= 0.7s

I = 2.67 A

hence, current comes out to be 2.67 A