# Thevenin's Theorem Practice

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#### EN0

##### Member
Hey Everyone,

I'm terrible when it comes to Thevenin's Theorem, so I'd like some practice. Does anyone know of links or PDFs that contain a complete explanation or Thevenin's theorem and also numerous practice problems?

I'd appreciate it!

Austin

#### MrAl

##### Well-Known Member
Hi,

You could use a circuit simulator if you wanted to make up your own problems.
Connect some circuit elements and let it solve for all the node voltages perhaps,
then try to solve them yourself on paper using the theorems you want to understand
better. Just in case you cant find too many on the web, you could make up as
many as you want that way too.

I could probably give you a graphical method which makes Thev and Norton
a bit simpler to remember if you are interested.

Last edited:

#### EN0

##### Member
Hi,

You could use a circuit simulator if you wanted to make up your own problems.
Connect some circuit elements and let it solve for all the node voltages perhaps,
then try to solve them yourself on paper using the theorems you want to understand
better. Just in case you cant find too many on the web, you could make up as
many as you want that way too.

I could probably give you a graphical method which makes Thev and Norton
a bit simpler to remember if you are interested.

Hey MrAl,

Thanks, that's an interesting idea.

Austin

#### mbarazeen

##### Member
my sugestion is to make simple networks and solve it by normal methord(kherchoffs/ mexwell) then try the same by thev or nortons ways and make your self realize how it works. simulation may be a good way of getting the answer to compare, but if you work it out by different methods then you dont want more excercise to keep it in your mind for ever.

#### Grossel

##### Well-Known Member
Here's an easy practice exercise:

Have a voltage source and two resistors in series.

Calculate the voltage over R2 (the lower one in chematic). That is thevenin voltage.

Calculate the parallell value of the two resistors. That will be output impedance.

Calculate current through R1 if R2 is shorted. Use both thevenin eqivalent model, and also just do a V/R1 calculation. Result should be the same

Use an ampere meter and put it over R2. That will show approximately same current as you calculated.

Now you have shown that short circuit current is the same for thevenin model as for the real circuit. Also you can have a resistor in parallell with R2 and calculate new voltage drop. You'll have the same voltage drop for thevenin model and in real life.

#### EN0

##### Member
Ok, I found some practice problems online and I need help with the one in the schematic. I don't see how they get the answer. So the main schematic is shown above and I simplified it below. What I did was: 2.25Ω x 0.8A to find the voltage drop across the 2.25Ω resistor. I get 1.8 and then apparently I add that to the 1.2V source (Don't know why) and I get 3V. Can someone please explain why I do that?

Last edited:

#### EN0

##### Member
Here's an easy practice exercise:

Have a voltage source and two resistors in series.

Calculate the voltage over R2 (the lower one in chematic). That is thevenin voltage.

Calculate the parallell value of the two resistors. That will be output impedance.

Calculate current through R1 if R2 is shorted. Use both thevenin eqivalent model, and also just do a V/R1 calculation. Result should be the same

Use an ampere meter and put it over R2. That will show approximately same current as you calculated.

Now you have shown that short circuit current is the same for thevenin model as for the real circuit. Also you can have a resistor in parallell with R2 and calculate new voltage drop. You'll have the same voltage drop for thevenin model and in real life.

Can I see the schematic?

#### jamesinnewcastle

##### New Member
Ok, I found some practice problems online and I need help with the one in the schematic. I don't see how they get the answer. So the main schematic is shown above and I simplified it below. What I did was: 2.25Ω x 0.8A to find the voltage drop across the 2.25Ω resistor. I get 1.8 and then apparently I add that to the 1.2V source (Don't know why) and I get 3V. Can someone please explain why I do that?

Remember that a current source will attempt to keep its rated flowing through it (by affecting the voltage across itself) and that a Voltage source will attempt to maintain its rated voltage (by affecting the current through itself).

Connecting a voltage source across a current source will result in both devices being happy and achieving their goals. Put a resistor in series and you now have to see what effect that has. The current will stay as the current source wants (as we have a series circuit so it is the same through each component)

The voltage across the resistor is therefore as you calculated (by Ohms law) The voltage across the voltage source is its rated voltage so the only thing missing is the voltage across the current source. As the sum of voltages in a closed circuit equals zero then the voltage across the current source is the (algebraic) sum of the other two voltages.

Any good?
James

#### mbarazeen

##### Member
when you simplify you will get the equvalent a current source 0.8A parellel to the resistor 2.25 ohms. thats what you put as 1.8V. dont change the current source to voltage source, just connect the 2nd current source across, now the total of 1.6 A will flow through the resistor thus the voltage would be 3.6V.

heyy you have made a mistake when calculating the voltage source..it is 1.8V not 1.2.
hope you got where you made msitake

Last edited:

##### New Member
Ok, I found some practice problems online and I need help with the one in the schematic. I don't see how they get the answer. So the main schematic is shown above and I simplified it below. What I did was: 2.25Ω x 0.8A to find the voltage drop across the 2.25Ω resistor. I get 1.8 and then apparently I add that to the 1.2V source (Don't know why) and I get 3V. Can someone please explain why I do that?

1st step : open the terminal under test (Is2)

find Rth
for independent current source - open circuit
for independent voltage source - short circuit

so in this case both current sources Is1 and Is2 are open

by looking from terminal Is2,

6ohm and 3ohm are seen connected in series with equivalent resistance
of 9ohm. The 9ohm is seen parallel with 3ohm resistor(which is also
parallel to the terminal under test

so Rth = (9*3)/(9+3) =27/12 = 2.25ohm

or if its confusing try this method with Is1 and Is2 still open

place a test voltage, Vth, on the terminal under test and Isc going into the
node between both the 3ohm resistor
now you have a mesh circuit and use KVL
draw mesh curent clockwise I1 on mesh 1 and I2 on mesh2
where Isc= -I2
you be able to arrive to these equation
12I1 - 3I2 = 0
-3I1 + 3I2 = -Vth
solve for I2 where I2 = -(12Vth)/27 = - Isc

therefore Vth/Isc = 27/12 =Rth
Rth = 2.25 ohm

to find Vth

Vth is the voltage drop on the 3ohm resistor parallel to the terminal under test
place back Is1 but Is2 remains open and place V on the 6 ohm resistor
use KCL

you should be abale to derive this equation

(V/6) + (V/6) + 0.8 = 0
solve and V = -2.4V

so voltage drop on 3ohm resistor = [ 3/(3+3) ] * (-2.4) = -1.2V

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