Thevenin and Nortons Theorems

Thread Starter

careless25

Joined Dec 1, 2011
5

Vahe

Joined Mar 3, 2011
75
You should not be removing the 1k resistor.
The open circuit voltage is the voltage across the 1k resistor with
nothing (open-circuit) connected to the terminals that are marked with voltage Vout.

There are a number of ways to do this problem. For example, you can do nodal analysis. Mark the bottom terminal as ground and assign node voltages V1, V2, etc. with respect to ground. One of the node voltage will be Vout. Write the nodal equations and solve for Vout.

You can use superposition by taking one source at a time to find out Vout then you add up all of the results to form the total Vout. Turn off the current sources by open circuiting and the voltage sources by short circuiting.

--Vahe
 

Thread Starter

careless25

Joined Dec 1, 2011
5
Hi
I know I can do nodal analysis and superposition on the question but it specifically asks to use Thevenins theorem, thats why i was trying to break apart the circuit, simplify it(find overall voltage and R_th) and then connect it back together.
Or is my understanding of thevenins theorem completely off?? :S.
 

syed_husain

Joined Aug 24, 2009
61
Hi
I know I can do nodal analysis and superposition on the question but it specifically asks to use Thevenins theorem, thats why i was trying to break apart the circuit, simplify it(find overall voltage and R_th) and then connect it back together.
Or is my understanding of thevenins theorem completely off?? :S.
i am afraid that you did not understand fully how to analyze a circuit using thevein and norton theorem:(. i suggest one reference "Electrical Engineering: Principles and Applications" 4th edition by Alan R. Hambley. if you read the section 2.6 of that book you will get a firm understanding about thevenin and norton theorem (worked for me anyways:D). for your convenient i am giving you the steps to follow to analyze cct using thevenin/norton method from the above mentioned book(page:90):

1. Perform two of these:
a) Determine the open-circuit voltage, Vt = Voc
b) Determine the Short cct current, In = Isc
c) Zero the Independent sources and the find the thevein resistance Rt looking back into the terminals. Do not zero dependent sources.
2. Use the equation Vt = Rt* In to compute the remaining value
3. The Thevenin equivalent consists of voltage source Vt in series with Rt
4. the Norton equivalent consists of a current source In in parallel with Rt
hope this will help.

cheers
 

Thread Starter

careless25

Joined Dec 1, 2011
5
Ok, so this is where I am stuck. I know I have to find Vt = Voc, but I dont know how to go about it.

Heres a question I did from the textbook I am learning from:
https://docs.google.com/leaf?id=0By2PVt4Tdl6VZTNkMzlmN2QtM2MwYS00MDVkLWE0YzItZTNhMzE5NDRiZWI0&hl=en_US

For this question, I break the circuit around 4kΩ resistor, then jut by looking at the voltage sources it is obvious that the Voc = -6V. (6v - 12v).
Rth = 4kΩ.

Now I close the circuit, and perform voltage division to find voltage across the 4kΩ resistor.

I get Vo = -3V.

I am trying to apply the same reasoning to this question but I am clearly missing something.
 

syed_husain

Joined Aug 24, 2009
61
Ok, so this is where I am stuck. I know I have to find Vt = Voc, but I dont know how to go about it.
Simple. just shorted out the port (at V0) and find out the current (either node or mesh). that will give you the short cct current. multiply that current by the thevenin resistance (Rt) will give you the Vt or Voc.

btw, it is better if you show us your working and then we can point out what and where are you doing wrong.

cheers.
 

Vahe

Joined Mar 3, 2011
75
Nodal analysis is only one of the techniques that you can use to find the Thevenin or Norton equivalent circuits. The use of Thevenin or Norton theorems do no necessarily dictate that you *have* to use nodal analysis. Thevenin's theorem states that the network at two terminals of interest may be replaced by a voltage source (thevenin voltage) in series with a resistance (thevenin resistance).

For example, if the circuit only has resistors and independent sources, the Thevenin equivalent resistance may be found by turning off all of the independent sources (voltage sources replaced by shorts, current sources by open circuits). Then you look into the terminals of interest and if the circuit structure allows, you may be able to perform series and parallel combination of resistances to find Rth; however, this is not always possible. In this case, you will have to attach a test source at the terminals of interest. For example, place a test voltage source, Vtest and determine the current that leaves its positive terminal, Itest. Then Rth=Vtest/Itest. Now when you setup this problem, one way of doing it would be to write nodal equations or mesh equations. Does this make sense?

--Vahe
 

syed_husain

Joined Aug 24, 2009
61
For example, if the circuit only has resistors and independent sources, the Thevenin equivalent resistance may be found by turning off all of the independent sources (voltage sources replaced by shorts, current sources by open circuits). Then you look into the terminals of interest and if the circuit structure allows, you may be able to perform series and parallel combination of resistances to find Rth; however, this is not always possible. In this case, you will have to attach a test source at the terminals of interest. For example, place a test voltage source, Vtest and determine the current that leaves its positive terminal, Itest. Then Rth=Vtest/Itest. Now when you setup this problem, one way of doing it would be to write nodal equations or mesh equations. Does this make sense?
--Vahe
thanks a lot. i totally forgot this test current method.
 

Thread Starter

careless25

Joined Dec 1, 2011
5
Hi,

Here's another question I am having trouble with:
https://docs.google.com/open?id=0By2PVt4Tdl6VMjhjY2YwMzUtMzI1Yi00MjhkLWEwOTMtMjU1ZWQ0NmZmNzM0

This is an AC steady state analysis problem.
What I have tried:

I simplify the circuit:

right most loop: Z = 2 - 2j
combine with parallel inductor: 2+2j

left most loop: Z = 4-2j

now current division to get going through Z = 2+2j
I = -0.9428 \(\angle45\)

So now I calculate voltage over the Z(2+2j) and do voltage division to find Vo. I am assuming voltage is equal across parallel Z values, so then voltage across 2Ω resistor and -2jΩ Capacitor is the same as voltage across Z = 2+2j ([capacitor + resistor in series] + inductor in parallel)

so voltage division should give me voltage across capacitor, right?

C25
 
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