# Thevenin-equivalence

Discussion in 'Electronic Basics' started by Rikard Bosnjakovic, Oct 7, 2006.

1. ### Rikard BosnjakovicGuest

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.

2. ### Johnny BoyGuest

*** OK to here
*** Now:
U3 = I1*R3 = 240mA * 50 = 12V
and
U5 = I2*R5 = 80mA * 200 = 16V
therefore A = -4V with respect to B.
Hi Rikard, It's many, many years since I studied this stuff, so I could be
making an error too, but I end up with an Rth of 92.19 ohms, (92.18749896
ohms), and an Ith of 43.389831mA. The error of 0.07 ohms might be due to
rounding during my calculations, or I might be totally wrong here. Does this
correlate with any of your (wrong) answers? (I won't explain how I came to
this answer, since it's wrong.)
I do, however, get the correct Vth of -4v, with A=12V and B=16V. I've
inserted my reasoning in your text above.

.... johnny

3. ### Greg NeillGuest

Something's fishy with your equations. Your diagram shows I1
flowing through R2 and R3 and I2 flowing through R4 and R5,
but you then choose to make a mesh by including Vth at AB
as a circuit element. Then you write your equations with
I2 flowing through R2 and R3 (which it doesn't according to
the diagram, and if you make the meshes as you say, there
should be different mesh currents in R2 and R3.

If you want to write consistent equations you should use a
consistent model.

Here's how I would go about this problem (there are several
ways to arrive at the solution, but everyone has their own
favorite approaches).

First I would note that there's an open circuit at AB, so
we have a couple of voltage dividers formed by the pairs
R2&R3 and R4&R4. So it would be nice to know the voltage
at the top of these dividers. For this purpose we redraw
the circuit as follows (combining R2+R3 and R4+R5):

+----25----+-----+--- V1
+| | |
32V O 100 300
| | |
+----------+-----+

Now, 100||300 is 75 Ohms, so we have a simple divider:

V1 = 32V*75/(25+75) = 24V

Fine. We have 24V at the top of the two voltage dividers.
We can now fine the voltages at A and B quite easily since
they're just trivial voltage dividers also:

Va = 24V*50/(50+50) = 12V

Vb = 24V*200/(200+100) = 16V

So we have:

Vth = Va - Vb = 12V - 16V = -4V

That takes care of the open circuit (thevenin) voltage.

To find the short circuit current, short out AB and
redraw showing the three resulting meshes:

+---25------+---------+ Three meshes, three mesh
| | | currents i1, i2, i3. The
| 50 i2 100 currents are assumed to flow
+| | | clockwise in the meshes.
32V O i1 +---AB----+
| | | The short circuit current
| 50 i3 200 through AB will be given
| | | by Iab = i3 - i2.
+-----------+---------+

The equations:

125i1 - 50i2 - 50i3 = 32V (1)
-50i1 + 150i2 - 0i3 = 0V (2)
-50i1 - 0i2 + 250i3 - 0V (3)

Solve for i2 and i3:

i2 = (32/295)A = 108.475mA

i3 = (96/1475)A = 65.085mA

Then Iab = i3 - i2 = -43.390mA

So the Thevenin resistance is Vth/Iab = 4V/43.29mA = 92.2 Ohms

4. ### Tom BiasiGuest

Do you notice that R1 is in parallel with the other two branches?
You can combine the series resistors and will have three parallel branches.

Solve the parallel combination.

Use this resistance to find the current with the source voltage.

Draw the current flow.
Current times R1 will allow you to determine what voltage you have across
the other two branches.

Find the voltage across R4 and R5. The difference is VAB

Tom

5. ### Rikard BosnjakovicGuest

I would have seen it, if it wasn't for A and B. Since they're between the
R2/R4 and R3/R5 respectively, they can't be in parallel.

Or can they?

Please correct me if I'm wrong. What I mean is, if I replace R2/R4 with a
serial one, then the point A will be lost since it's supposed to be in
between.

6. ### Rikard BosnjakovicGuest

After considering your explaination of finding Vth, I find it perfectly
clear of what I did wrong. The problem is, the lecturer usually includes
the open circuit voltage in his mesh traversions and it seems to work for him.

But by reading your post I understand that since no current flows through
A and B, I cannot use Kirchoff's laws of current to find the voltage and
that I have to use voltage dividing instead.
Reading through this example a couple of times, I'm getting a grip of what
you're doing. I think the big hurdle for me is when I've got a lot of
linear equations like this, and so eliminate all variables except one, and
then going backwards.

Thank you Greg for taking time replying. Atleast you got me halfway back
on track. I think I will have to practice more on the linear equations
kind of math.

-- Rikard.

7. ### Greg NeillGuest

You _can_ include the Vth element in a mesh style of
solution, but it adds one extra unknown (Vth) so one
extra equation is required. In other words, the
set of linear equation to solve is slightly more
complex. This may not be a problem if you've access
to a computer to take care of the grunt work, but
I find that simple, small steps are best when working
by hand.

So, let's take a look at what the mesh equations would
look like if you include the Vth unknown. You still
have three meshes as I detailed previously, only this
time we include the Vth as a circuit element between
points A and B.

125*i1 - 50i2 - 50i3 = 32V
-50i1 + 150i2 = Vth
-50i1 + 250i3 = Vth

[Note: the numerical constants on the left are assumed to
have inits of Ohms]

Since we have four unknowns, we need one more equation.
Since AB is actually an open circuit, we must have

i2 + i3 = 0

You can solve the above four equations to find Vth.

You'd then need to set Vth = 0 in the first three
equations (to "force" a short circuit between A and B)
and solve for the currents i2 and i3 again, since
then i2 will not equal i3, and the short circuit
current from A to B will be given by i3 - i2.

As I said above, personally I like to solve these sorts
of problems in short, simple steps so it's less likely
to make a mistake when under pressure.

Another approach to finding the Thevenin resistance in
this circuit would be to perform a Delta-Y conversion
of R1,R2, and R3 (after replacing the voltage source
with a short circuit) to yield a circuit amenable directly
to reduction to a single resistance:

Ra = R1*R2/(R1 + R2 + R3) = 10 Ohms
Rb = R1*R3/(R1 + R2 + R3) = 10 Ohms
Rc = R2*R3/(R1 + R2 + R3) = 20 Ohms

Giving the following equivalent circuit:

+--------------------------------------+
| |
10 100
| |
+---20------oA Bo-------------+
| |
10 200
| |
+--------------------------------------+

+------------------110-----------------+
| |
+------------------210-----------------+
| |
+-----20----oA Bo-------------+

and

+------------72.188-----------+
| |
+---20----oA Bo------+

so that Rth = 92.2 Ohms as before.

8. ### Tom BiasiGuest

Yes,
Point AB will be lost, until you find the main current. Then you will back
up and reconstruct the original circuit and distribute the current.
The is an infinite resistance between AB. It is not there as far as this
step is concerned.
Do It.

Tom

9. ### Alan BGuest

Referencing "Electric Circuit Analysis," Taber/Silgalis:

"Step 1. Remove the element or elements in the circuit that were
designated as the load."

In your case, the load is between A and B, correct?

"Step 2. Determine the open-circuit voltage across terminals A-B, call it
Voc."

This is what you are attempting to do by your re-write. However, you over
complicate the process by introducing mesh equations and then mixing them
with current node equations. Simplify your thought process by using the
Voltage Divider Law.

----------------------- C
| | |
| | |
| + A + B
| | |
_ 32 > 50 > 200
- E > R3 > R5
| | |
----------------------- D

First, find VC.

Rp = R2+R3 || R4+R5 = 75

VC = E * Rp / (R1 + Rp) = 24

Now find the two voltages, A and B.

VA = VC * R3 / (R2+R3) = 12

VB = VC * R5 / (R4+R5) = 16

Vth = VA - VB = -4
"Step 3. Replace the voltage supply E by its internal resistance... Find
the series resistance...."

In this case, as you found out, the resistance network is difficult to
calculate. But, knowing Vth, and redrawing the circuit, current values can
be set up and Ith found.

--> Ith
A --------------------------
| --> |
Vth > R2 > R3
| R1 |
c +---------^^^---------+ D
| 25 |
-4 > 100 > 200
| <-- |
B --------------------------

This is a resistor bridge, and I think there is a shortcut way to calculate
the equivalent resistance, but if so, I can't recall it right now. Anyway,
now mesh equations can be applied (bear with the graphics, I intend the
arrows to indicate mesh current).

-4 = 150Ith - 50I2 - 100I3
0 = - 50Ith + 125I2 - 25I3
0 = -100Ith - 25I2 + 325I3

And solve as you please. I solved by substitution and got -43.36mA.

Rth = Vth/Ith = -4 / -43.36 = 92.25 Ohms.

Ok, now on to your problem of solving mesh equations. You've posted this:

----
(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)
---

In each case where you posted mesh equations, you only posted two equations
for a three mesh network. Well, that isn't going to work. Do you know how
to proceed solving by substitution the three equations I posted? Give it a
try and post your results. Do you know how I came up with the equations?
Let's work on it.

10. ### Fred AbseGuest

On Sat, 07 Oct 2006 12:00:53 +0000, Rikard Bosnjakovic wrote:

<lots of working snipped>

Why go to all that complication? The solution is obvious and trivial:

Firstly, R2, R3, R4, and R5 are a bridge circuit, so we can treat it as
two potential dividers, R2/R3, and R4/R5.

We need to know the voltage at point R1/R2/R4, relative to the
negative terminal of the battery.

R2/R3, and R4/R5 simplify to 300 ohms in parallel with 100 ohms, or 75
ohms.

So, in effect we have 75 ohms in series with 25 ohms, and the voltage at
R1/R2/R4 is 75/(75+25) = 75/100 = 3/4 times 32 volts, which is 24 volts.

The voltage at point A will be 50/100, ie. half this, which is 12 volts.

The voltage at point B will be 200/300. or 2/3 times 24 volts, which is 16
volts.

Hence the voltage at A with respect to B is 12 - 16, or -4 Volts.

No need for complicated mesh or nodal analysis.

Maybe they don't teach circuit recognition anymore :-(

Now tell me what the voltage between A and B would be if R4=200 ;-(

Now for the Thevenin equivalent resistance:

This is defined as the Thevenin equivalent voltage divided by the current
which will flow if the output is shorted, so:

Short points A and B

The circuit now reduces to a simple potential divider, 25 ohms, 33.33
ohms, and 40 ohms, from which we can calculate the voltages at R1/R2/R4,
and (R2/R3, shorted to R4/R5)

The voltage across R3 will be 13.017V
The voltage across R2 will be (23.864-13.017) = 10.847V

The current in R2 will be 10.847/50 = 216.94mA
The current in R3 will be 13.017/50 = 260.34mA

Hence the current in the short circuit is 260.34 - 217.94 = 43.4mA

The Thevenin output resistance is therefore 4 / 43.4e-3 = 92.166 ohms.

Which is as near the published result as makes no difference, allowing for
rounding errors.

11. ### Rikard BosnjakovicGuest

Agreed. I think I used mesh equation because the book (or lecturer, I
don't remember) told us to. The way you and others showed it, by using
voltage divider instead, resulted in a much simpler circuit.

I do however expect that I lack knowledge for mesh equations and I need to
study them more. I fear that some of the exam questions will be something
like "Do this and that with mesh equations", and if I solve it in other
ways I will most likely fail that question.

[...]
This description and your path of thought made perfect sense to me, and I
finally managed to solve this part on my own. I even tried with different
values and got the answer correct.

Thank you.
I did however have some problems in trying to find out how you got these
three equations. When I drew the same circuit as above, I got no less than
6 unknown currents:

At top-left point; Ith entering, I1 going right, I2 going down.
At point C, I2 entering from above, I3 going right, I4 going down.
At point D, I1 entering from above, I3 entering from right, I5 going down.
At bottom-left point, Ith leaving left, I4 entering from above, I5
entering from right.

I.e.:

1 Ith = I1 + I2
2 I2 = I3 + I4
3 I5 = I1 + I3
4 Ith = I4 + I5

Some of them are not linearly dependant, so I can reduce them:

Ith = I1 + I3 + I4 (1+2)
Ith = I1 + (I5 - I1) + I4 (1+2+3)
Ith = I4 + I5

So I take it you reduced the number of unknowns to three, and put into
three equations. I think this makes sense.

However, can you please elaborate how you got the constants?
The first row for example. -4 on the left, I take it you start the mesh
walk at point A. I'm not at all sure about the right side of the equation.
150Ith looks like Ith(R2+R4) (i.e. a walk straight down from A to B
through R2 and R4), but if you walked that way, why substract I2 and I3?

They're not part of that mesh path, are they?

12. ### Greg NeillGuest

Quick 'n dirty guide to writing mesh equations:

1. Identify and label (number) the meshes in the circuit.
2. Identify the mesh currents by sketching in a clockwise
pointing current arrow in each mesh.
3. Write the mesh equations by inspection as follows.
a. Create a 'blank' matrix equation that you will
proceed to fill in. It will have as many rows as
there are meshes. So for example, for a three mesh
circuit you would write a 3x3 equation:

| | |i1| | |
| |*|i2| = | |
| | |23| | |

b. The entries in the matrix position ij are:
i=j : The sum of the resistances around mesh i
i/=j: Negative sum of the resistances shared by
meshes i and j

c. The entries in the column vector on the right are:
Row i is the sum of the voltage rises due to active
voltage sources as you proceed clockwise around mesh i.

Example:

+---100---+--200--+---300---+
| | | |
50 1 30 2 | 3 |
|+ | |+ |
----- 12V | ----- 16V --
-- | -- ----- 8V
| | | |+
+---------+--60---+---40----+
| 4 |
| |+7V |
+------||-----300-----------+
|

There are four meshes which are numbered. I didn't
sketch in the currents because ascii's a bitch.

The diagonal entries will be: The Voltage vector will be:
(1,1): 50+100+30 = 180 Ohms (1): +12V
(2,2): 200+60+30 = 290 Ohms (2): -16V
(3,3): 300+40 = 340 Ohms (3): +16V + 8V = 24V
(4,4): 60+40+300 = 400 Ohms (4): -7V

The off-diagonal entries:

(1,2): -30 Ohms The complete equation:
(1,3): 0 Ohms
(1,4): 0 Ohms |180 -30 0 0 ||i1| | 12V|
(2,1): -30 Ohms |-30 290 0 -60 ||i2| |-16V|
(2,3): 0 Ohms | 0 0 340 -40 ||i3| = | 24V|
(2,4): -60 Ohms | 0 -60 -40 400 ||i4| | -7V|
(3,1): 0 Ohms
(3,2): 0 Ohms Note that there is a symmetry about
(3,4): -40 Ohms the diagonal that simplifies filling
(4,1): 0 Ohms out the matrix.
(4,2): -60 Ohms
(4,3): -40 Ohms

With a bit of practice you should be able to write
down the matrix equations by inspection.

13. ### Alan BGuest

OK, the problem is that my drawing is crude. I meant the three currents to
mean three mesh currents, and I had a comment to that effect. I mean Ith
to be the leftmost mesh current, so the voltage drops around that mesh are:

-4 (the applied voltage) = 150Ith (100 + 50 ohms multiplied by Ith)

- 50I2 (subtract 50 ohms times I2 because the I2 mesh current is in the
opposite direction)

- 100I3 (subtract 100 ohms times I3 for the same reason)

The other two equations take the positive value as the total of the
respective mesh current times the total resistance in the loop, subtracting
the opposite direction currents through the shared resistances.

Does that make sense?

14. ### Alan BGuest

Redrawn:

A --------------------------
| > 50 ^ I2 V > 50
Vth | > R2 |-------<------| > R3
| | R1 |
-4 V +---------^^^---------+ D
| | 25 |
| > 100 |---->--------| > 200
Ith | > R4 ^ I3 V > R5
<---| | |----<--------| |
B --------------------------

15. ### Guest

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.