A particular amplifying device supplies electrons in a manner such that the current passing out of the device is given by I(t) = (Io)e^kt, where I(0) = Io = 2A.The current measured 12 seconds later is 3.4 A.Find
(a)The charge Q that passes out of the device as a function of time..Q(t):
(b)The total number of electrons N that pass out of the device if left on for one minute
First, use the fact that I = 3.4 A at t = 12 to calculate k.
3.2 = 2.0 e^(12k)
12k = ln 1.6 = 0.47
k = 0.03917 sec^-1
(a) Integrate I(t) for Q(t), with Q = 0 at t = 0.
(b) Calculate Q(t=60) for the total charge at 1 minute, and divide by the charge of an electron, 1.6*19^-19 C,
To find the charge Q that passes out of the device as a function of time, we need to integrate the current I(t) with respect to time.
(a) The charge Q(t) can be found by integrating the current I(t) over the time interval [0, t]. Let's calculate it step by step:
Q(t) = ∫[0, t] I(t) dt
Given: I(t) = (Io)e^(kt), where Io = 2A.
Q(t) = ∫[0, t] (Io)e^(kt) dt
Integrating I(t) with respect to t, we get:
Q(t) = (Io/k) * e^(kt) + C
To find the value of C, we can use the initial condition I(0) = Io = 2A:
I(0) = (Io)e^(0) = Io = 2A
So, Q(0) = (Io/k) * e^(k * 0) + C = Io
Substituting Io = 2A and e^(k * 0) = 1, we get:
C = 2A
Therefore, the charge Q(t) passing out of the device as a function of time is given by:
Q(t) = (Io/k) * e^(kt) + 2A
Now, let's move on to part (b).
(b) To find the total number of electrons N that pass out of the device if left on for one minute, we need to calculate the total charge over this time period.
Given: 1 minute = 60 seconds.
Using the equation Q(t) = (Io/k) * e^(kt) + 2A, we can find Q(60):
Q(60) = (Io/k) * e^(k * 60) + 2A
Now, let's substitute the given values: Io = 2A and the current measured 12 seconds later is 3.4A.
Q(60) = (2A/k) * e^(k * 60) + 2A
We do not have the value of k, but we can use the given information to find it. We know that when t = 12 seconds, I(t) = 3.4A:
3.4A = (2A) * e^(k * 12)
Simplifying, we have:
e^(k * 12) = 3.4A / 2A = 1.7
Taking the natural logarithm of both sides, we get:
k * 12 = ln(1.7)
Solving for k, we find:
k = ln(1.7)/12
Now that we have the value of k, we can substitute it into Q(60):
Q(60) = (2A / (ln(1.7)/12)) * e^((ln(1.7)/12) * 60) + 2A
Calculating this expression will give us the total charge passing out of the device over one minute.
To find the total number of electrons N, we need to divide the total charge Q by the charge carried by each electron, denoted by e.
N = Q / e
e is the elementary charge, approximately equal to 1.602 x 10^(-19) Coulombs.
Substituting the value of Q(60) into the equation and dividing by e will give us the total number of electrons N that pass out of the device if left on for one minute.