Download Muddiest Points Week 3

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Multimeter wikipedia , lookup

Schmitt trigger wikipedia , lookup

Ohm's law wikipedia , lookup

Power MOSFET wikipedia , lookup

Switched-mode power supply wikipedia , lookup

Flexible electronics wikipedia , lookup

Surge protector wikipedia , lookup

Integrated circuit wikipedia , lookup

Opto-isolator wikipedia , lookup

Resistive opto-isolator wikipedia , lookup

Rectiverter wikipedia , lookup

Two-port network wikipedia , lookup

RLC circuit wikipedia , lookup

Network analysis (electrical circuits) wikipedia , lookup

Transcript
Muddiest Points Week 3
I am confused with delta and y circuits.
>> What I want you to know:





Recognize delta and Y circuits
These occur when no resistors are in series or parallel. Converting will often make it
possible to simplify the circuit by combining parts in series and parallel.
Be able to look the equations up in the table and convert them (Xerox the table and
include it in your notes)
They are also commonly used in the Wheatstone Bridge (two deltas) configuration. This
is used to measure small changes in R, often associated with small changes in sensors.
They are also used to hook up three phase power systems (which we will not cover in this
class).
>> This student gave a good way to remember the formulas: So, in order to calculate the
resistance from delta to y circuit from a point, I need to multiply the resistors coming from the
same point in the delta circuit and divide them by the total resistors. In order to calculate
resistance from y to delta circuit from a point, I need to sum up the multiplication of 2 different
resistors of each 3 of them (that would be 3C2, or R1R2+R2R3+R1R3) and divide it with the
resistor on the opposing point.
Resistors in series and parallel and VI dividers: I realize that iterations of problems are the key
to doing these things correct. But the one question I have is on the HW 2.43 this came across
very confusing even with looking at the answer key. It seems very doable but I did stare at it for
about 20 minutes trying different things that just didn’t seem right. I was wondering if you have
any patterns or indicators to solve these kinds of problems. The power source where I start was
the biggest hurdle and then stair stepping up to the unknown we are looking for was kind of
confusing. So I guess the question is, is there a good way to work through these things that are
setup like this?
>> There are several ways of doing this problem (and most problems). I usually start by looking
at (highlighting?) what is KNOWN. Then I see if there is anything I can calculate based on what
is known (which is exactly how they solved this problem). I think highlighting the parameters in
the circuit that you know, and when you find another, highlight it too, really helps. You were
trying to start from the source, because you know that. The ‘trick’ in this one was to start at the
load, because you also know THAT (you know Vout and Rout, so of course you know Iout….
And it proceeds from there).
>>Another perfectly valid solution for this problem would be to combine all of the Rs in series
and parallel, start from the SOURCE, and keep ‘unwrapping’ the Rs as you move to the right
doing V and I dividers.
>> Another valid solution would be to write all of the VI equations, even though you would have
a lot of unknowns, and toss them all into a matrix equation, and solve it with calculator or matlab
The most confusing point for me this week would have to be the difference between current
dividers and voltage dividers.
>> The key is in the DIVIDER part.




Current dividers take your original current and divide it up between parallel branches in a
circuit. Voltage dividers take your original voltage and divide it up between series
resistors in a series branch.
We use them to divide V and I so we can use parts in different parts of the circuits. We
also use them to determine values of resistive sensors (because we can measure just V to
do this).
Current divider is in parallel and voltage divider in series.
The MATH looks the same IF you do V dividers with R and I dividers with G=1/R
Vn = Vs Rn / (sum of R’s)
In = Is Gn / (sum of G’s)
When I was working on Multisim creating my own circuits or an example from the homework.
The resistors would break. I don’t know what I am doing wrong and I was hoping you would tell
me. For example, the picture below is of a circuit I created. It would run perfectly but then the
resistors would break shortly.
>> They break because you are putting in too much power (P=VI) for the resistors you
specified (0.25W by default). Calculate (or measure, in the very early moments of multisim!) the
VI values, then edit the resistor power rating to be above this.