Tuesday, June 23, 2009

Circuit Analysis II Lecture 6

Disclaimer: This series of posts is to serve as notes for myself as well as any others interested in the subject of Circuit Analysis II. This is the course ECE 213 at UNM.
The course will follow the book:
Electric Circuits (8th Edition)

The links on the right, will contain all the links to this course's lecture notes.

First we worked on problem 13.8c
it turns out that there is only one node so the equation is:
v1 - 20/s / 2 + v1 / 1.25s + v1 / 20 /s - 5/s = 0

Now we can simply solve for v1

Next we went over Thevinin starting on page 520, figure 13.17 - 19

First we migrate to the s-domain, so we end up with
480/s for the source
Since we are looking at the circuit and the 60ohm resistor is not connected so it has no current running through it, so we can get rid of that particular resistor.

We end up using voltage division because we want to know the voltage at the inductor in this particular example.

vth = (480/s / 20 + .002s * .002s)
so pulling things out and canceling things out we get
vth = 480 / s+104

Next we want to convert it to look at it from the a and b terminals, so what we'll want to do is get rid of the source, remember:

voltage = short
current = open
doing that we can get zth
zth = (20s / s + 10000) + 60

Cleaning it up we get:

zth = (60s + 6*105 + 20s/s + 4)

and finally we can get:

Now we add a 5uF capacitor to the a and b terminals, so looking for the current we get:
Ic = (480/s+104) / (80(s+7500)/(s+104))

Basically what we're looking at is
Ic = vth / zth + zc

after solving it we get 65/(s+5000)2

We would be able to do partial fraction decomposition to continue but that is outside the scope of this class period, so we didn't continue.

Next we worked on problem 13.20 on page 552
1. We want to move to the s-domain
2. Next we find Req
note: we use voltage division in this case.
3) Using Partial fraction decomposition we want to get the k's
4) Convert back to t-domain

Next we worked on Problem 12.42a page 504
F(s) = (10s2 85s +95) / (s2 6s + 5)

First reaction might be to try to reduce it, but it turns out this is not a proper fraction and therefore we will need to do long division to make it a proper fraction.

Once we do that we get
10 + (25s + 45)/(s2 + 6s + 5)
reducing it we get:
10 + (25s + 45)/(s+1)(s+5)

using partial fraction decomposition for the second part we get,
20 and 5 for the k's

since 10 is a constant we use (delta)(t)
so our equation ends up as:
10(delta)(t) + 5e-t + 20e-5t

Finally we tried one last one and that was:
problem 12.21a
which using the laplace transform from the table f(t-a)u(t-a)
we get

Continue on to Lecture 7 6/24/09