# anyone can help me solve this thevenin equivalent circuit. im lost!!

Discussion in 'Homework Help' started by swaggerz95, Dec 9, 2014.

Dec 6, 2014
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Jan 29, 2010
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3. ### swaggerz95 Thread Starter New Member

Dec 6, 2014
17
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hi,
thevenin theorem basic i think i can do that.
but when i want to apply that, im confusing ..
when i apply superposition ( i was short the independent voltage source) then im lost at there.
sorry for my very bad english

4. ### swaggerz95 Thread Starter New Member

Dec 6, 2014
17
0
anyone can show me the calculation n how to solve this question... im really need to understand with this question.

5. ### ericgibbs AAC Fanatic!

Jan 29, 2010
2,581
389
Hi,
As this is a Homework question, please post your attempt at creating the Thevenin circuit.
Then we can explain why you are having a problem.

E

6. ### swaggerz95 Thread Starter New Member

Dec 6, 2014
17
0

i apply superposition n source transformation to find ix .. its correct? n im stuck to find io.. help me bro.

7. ### WBahn Moderator

Mar 31, 2012
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You can use superposition if you want, but the problem statement says that you are expected to use a Thevenin's equivalent circuit in your approach to find the current in a particular element. That means that you want to transform this circuit in a circuit containing that particular element as a load connected to an equivalent circuit for everything else that consists of a voltage source in series with a resistance. You don't appear to be doing that because you are performing an analysis with the load still in the circuit.

So your first step is to remove that particular element from the circuit and identify the terminals at which it connects to the rest of the circuit.

Your second step is to determine the voltage across the terminals to find the open circuit voltage, which will be your Thevenin voltage.

Your third step is to short the terminals together to determine the short-circuit current, which you will use in conjunction with the Thevenin voltage to determine the source resistance.

You can also determine the source resistance by turning off all of the independent (and only the independent) supplies and applying a test voltage to the output and finding the test current.

Looking at your work, I can't follow it very well. Part of the problem is that I can't make out some of the details of your writing as clearly as I would like (I probably could if I downloaded and opened the attachment, so that's not your problem), but there are several things that don't make sense. You do a lot with a current 'In' which is not defined in the original circuit, but it appears that this is basically 'Ix'. But in your diagrams you end up with it being controlled by the current through the 17Ω resistor instead of the current in the 20Ω resistor. Was this a mistake?