Re: Thevenin-equivalence
- From: mgkelson@xxxxxxxxx
- Date: 8 Oct 2006 15:28:07 -0700
Rikard Bosnjakovic wrote:
Greetings
First, I'm sorry for this long post. I felt that if anyone is going to
help me correcting my mind, I have to show how I'm thinking when solving
the equations.
Now; during the class "electrical analysis" I've stumbled into some heavy
problems that I cannot seem to straighten out.
At the lectures I usually have no problems in following the lecturer when
he's doing things like this (Thevenin and Norton equivalences). I am
perfectly clear of how he's doing this and that, and why, but I cannot
seem to replicate his work on my own.
This thursday, for example, I sat down with an excercise in my book trying
to solve it for no less than 5 hours. Not being able to solve an excercise
is usually common, but the fact that during my ten attempts I got ten
different answers. This gave me a signal that I obviously haven't gripped
the methods correctly, or my mind is confusing things with other things.
I'm therefore hoping that anyone can shed some light for me upon this
particular problem.
Consider this circuit:
R1 = 25
+------/\/\/\-----------------+----------------------------+
| | |
| | |
| | |
| \ \
| / R2 = 50 R4 = 100 /
| \ \
| / /
| | |
| | |
E = ----- | |
32V --- +----------A B-----------+
| | |
| | |
| | |
| \ \
| / R3 = 50 R5 = 200 /
| \ \
| / /
| | |
| | |
| | |
+-----------------------------+----------------------------+
The question is: Find the Thevenin-equivalence between A and B.
My first step is to rewrite the schematic, adding the currents and
polarities of the resistors:
R1 = 25 I
+------/\/\/\--------->-------+----------------------------+
| + U1 - | |
| v I1 I2 v
| + | | +
| \ \
| U2 / R2 = 50 R4 = 100 / U4
| \ M2 \
| / /
| - | | -
| | |
E = ----- | + Vth - |
32V --- M1 +----------A B-----------+
| | |
| | |
| + | | +
| \ \
| U3 / R3 = 50 R5 = 200 / U5
| \ M3 \
| / /
| - | | -
| v I1 I2 v
| | |
+-----------------------------+----------------------------+
Method 1, find the Thevenin-voltage Vth
=======================================
The unknowns are I1, I2 and Vth (which gives a total of three equations).
The circuit has three meshes (I think that's what they're called in
English, although I'm not 100% sure); M1, M2 and M3.
The equations are as follows:
(1) I = I1 + I2
The meshes (clockwise):
(2) M1: E - U1 - U2 - U3 = 0 => U1 = E - I2*R2 - I2*R3 = E - I2(R2 + R3)
(3) M2: Vth + U2 - U4 = 0 => Vth = U4 - U2
U1 (and then I) is calculated using the formula for voltage drop over R1:
U1 = E * (R1 / [R2 + R3]) = 32 * (25 / 100) = 32 * 1/4 = 8V
I = U1/R1 = 8 / 25 = 320mA
=====
From (2) I solve and calculate I2:
U1 = E - I2(R2 + R3) => I2(R2 + R3) = E - U1 => I2 = (E - U1)/(R2+R3)
I2 = (32 - 8)/(50 + 50) = 24 / 100 = 240mA
=====
From (1) I then get I1:
I = I1 + I2 => I1 = I - I2 = 320mA - 240mA = 80mA
====
Now Vth can be calculated using (3):
Vth = U4 - U2 = I2*R4 - I1*R2 = 240m * 100 - 320m * 50 = 8V
==
8V is, therefore, my answer to the voltage of the Thevenin-equivalent
circuit. The book tells me it should be -4V, so I have obviously made some
major error.
Method 2, find Ith (shorted current over A and B)
=================================================
A and B are shorted, and the current Ith is added. This time I have three
unknowns: I1, I2 and Ith.
R1 = 25 I
+------/\/\/\--------->-------+----------------------------+
| + U1 - | |
| v I1 I2 v
| + | | +
| \ \
| U2 / R2 = 50 R4 = 100 / U4
| \ M2 \
| / /
| - | | -
| | |
E = ----- | Ith |
32V --- M1 2+------------->--------------+
| | |
| | |
| + | | +
|1 \ \
| U3 / R3 = 50 R5 = 200 / U5
| \ M3 \
| / /
| - | | -
| v I1-Ik Ik+I2 v
| | |
+-----------------------------+----------------------------+
(1) I = I1 + I2
The meshes (clockwise):
(2) M1: E - U1 - U2 - U3 = 0 => E - R1*I - R2*I1 - R3*(I1-Ith) = 0
(3) M2: R3*(I1-Ith) - R5*(Ith+I2) = 0 => R3*(I1-Ith) = R5*(Ik+Ith)
R2/R4 and R3/R5 are in parallel, which makes the total resistance for the
whole circuit R1 + R2||R4 + R3||R5 = 25 + (50*100)/(50+100) +
(50*200)/(50+200) = 25 + 5000/150 + 10000/250 ~= 98.3 ohms = Rtot.
Voltage divider over R1 gives U1:
U1 = E * R1/Rtot = 32 * 25/98.3 = 8.14V
This in turn yields I:
I = U1/Rtot = 8.14/98.3 = 82.8mA
======
Between point 1 and 2 in the schematics, there is a potential:
E - U1 - U2 = 0 => E - U1 = R2*I1 => I1 = (E - U1)/R2
Which gives I1:
I1 = (32 - 8.14)/50 = 477.2mA
=======
This is used in (1) to get I2:
I = I1 + I2 => I2 = 82.8m - 477.2m = -394.4mA
========
Then, I rewrite (2) to get Ith:
R3*(I1-Ith) = E - U1 - U2
R3*I1 - R3*Ith = E - U1 - U2
R3*Ith = R3*I1 - E + U1 + U2
(R3*I1 - E + U1 + U2)
Ith = ------------------- = (U3 - E + U1 + U2)/R3 =
R3
= (23.6 - 32 + 8.14 + 23.86) / 50 = 472mA = Ith
===========
Method 3, calculation of Rth (resistor for the Thevenin-equivalent)
===================================================================
Alternative 1: Rth = Vth / Ith (from method 1 and 2 above)
Rth = 8 / 472m = 16.94 ohms
Alternativ 2:
Short E and calculate the resistance over A and B. Redrawing the
schematic, I get this:
+--------------------+
| |
| \ R1
| /
| \
| /
| |
| +-----+-----+
| | |
| \ R2 \ R4
| / /
| \ \
| / /
| | |
| A -----+ +----- B
| | |
| \ R4 \ R5
| / /
| \ \
| / /
| | |
+--------------+-----------+
And at this point I have honestly no idea how it's supposed to be
calculated. I see no serial or parallel points. R1, R2 and R4 form a
Y-shaped resistor and I can convert it to a D-shape, but if I do that the
top part of the network gets even more confusing and - for me - still
unsolvable.
The result should, according to the book, be Rth = 92.12 ohms.
Conslusions
===========
Method 1:
The resulting answer was yet another resulting answer apart from my ten
other tries. I take this as it seems that I'm only cargo culting the
methods from my lecturer, without actually understanding what's going on.
Method 2
Probably the same in this case. I try to follow the lecturer's thinking,
try to follow the book's "method-solving-model", try to get an opinion on
*what* it is that happens in the solutions, but fail.
Method 3
Alternative 1 failed because both the Vth and the Ith were wrong.
The reason alternative 2 failed was probably due to the fact that I wasn't
able to further shorten the resistor network.
So, where do I go from here?
I have googled for lab reports and all sorts of notes on Thevenin/ Norton-
equivalences, but all I can find are the basic steps; like how to
transform a Norton-circuit to a Thevenin one (and vice versa). So they
really haven't been useful when it comes to more complicatde circuits.
Besides from the fact that I need to learn this for the class, I think
it'll turn out to be useful over all, so I really don't feel like letting
go of the problem yet.
But it's turning really boring just sitting for five hours calculating,
and getting wrong results every time. Different results at that.
-- Rikard.
It's been a long time, but as I recall whenever a teacher or text book
wanted to teach Kirchhoff's laws or mesh circuits, etc, they would
provide a problem with at least 2 power supplies in it. Your problem
only has one power supply, so I doubt if it was meant to be solved
using mesh circuits. This is a simple voltage divider problem. It can
be solved using the basic formulas at:
http://ourworld.compuserve.com/homepages/g_knott/elect61.htm
http://www.aikenamps.com/VoltageDividerRule.htm
The voltage drop behind R1 and over the two parallel branches is:
V(branches) = 32 (75/100) = 24V
The voltage drop over R3 is:
V(R3) = 24 (50/100) = 12V
The voltage drop over R5 is:
V(R5) = 24 (200/300) = 16V
The potential difference is +/- 4V depending on your reference.
.
- References:
- Thevenin-equivalence
- From: Rikard Bosnjakovic
- Thevenin-equivalence
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