http://img231.imageshack.us/img231/88/thevenin.png

help me solve this question!!
Fine the value of I using thevenin theorem.
the answer I=7A, but i dont get the answer!!!!
someone please help me!!!

 

This is a homework help forum. To get help, you must first show your work, then we can show you where you went wrong.

 

but im not sure what im doing... i dont understand how to determine the open source Voltage.. !!

 

Let's start with deciding an analysis method. Thevenin's theorem simply states that any complex network of resistors, voltage sources, current sources, etc. can be simplified to a single voltage source in series with a resistor. We may use any analysis method (nodal analysis, mesh analysis, source transformations, etc.) to analyze your circuit. Which analysis method do you prefer?

 

im not really sure because im studying circuit in japanese, i use method which

I=E/Ri+R

calculated
Ri= 1/(1/4+1/4) = 4/3Ω
R= 12

but i dont know how to determine the open circuit voltage..
so...

 

could someone teach me the easiest way to solve this problem

 

Find Vth by any method you prefer to circuit analysis by remove from the circuit 12 Ohm resistor.
To find Rth
1. Replace voltage sources with short circuits and current sources with open circuits.
2. Find equivalent resistance seen at AB point.

 

i still dont get the right answer.. could someone just explain the solution

 

Have a look at this:

http://www.allaboutcircuits.com/vol_1/chpt_10/8.html

and this:

http://en.wikipedia.org/wiki/Thévenin's_theorem

In your circuit if you remove the 12Ω resistor from the a-b terminals, and then replace the 80 volt source with a short circuit, and the 8 amp source with an open circuit, the resistance looking into the a-b terminals will be 4Ω; this is Rth (or Req).

To calculate the open circuit voltage, notice that the 2Ω resistor has no effect. Then calculate the voltage drop across the 4Ω resistor. If the 12Ω resistor is absent, how much current will there be in the 4Ω resistor? If you know the current in the 4Ω resistor, you know the voltage drop across it. That voltage drop is in series with the 80 volt source, so what will be the voltage seen at the a-b terminals? This will be Vth (or Veq).

Finally, if you know Rth and Vth, what is the value of Vth/(Rth + 12)?

 

why the R+h is not 1/(1/4+1/4) = 4/3Ω ?
why it is only 4Ω?
i dont understand that

 

Becaues to find Rth we replace voltage sources with short circuits and current sources with open circuits.
Ans in your circuit 12Ω resistor is in series with current source.

 

why the R+h is not 1/(1/4+1/4) = 4/3Ω ?
why it is only 4Ω?
i dont understand that

Please post a picture of the circuit with the 8A current source replaced with an open circuit, and with the 12Ω resistor removed, and we will discuss it further.

 

is there an easy way of finding Eth with kirchoffs laws? I have found it with mesh/nodal analysis, which would be the best way of finding Eth??

 

To find Eth we disconnect the load resistor (12Ω)
So we end up with simple circuit where all components connect is series.
And in series circuit the current through each of the components is the same.
In in our circuit the current that will be flow is equal 8A.
So A-B terminals voltage is equal 80V plus voltage drop across 4Ω resistor.
So only thin we must do to find Eth is to use Ohms law to fond voltage drop on a 4Ω resistor.

 

To find Eth we disconnect the load resistor (12Ω)
So we end up with simple circuit where all components connect is series.
And in series circuit the current through each of the components is the same.
In in our circuit the current that will be flow is equal 8A.
So A-B terminals voltage is equal 80V plus voltage drop across 4Ω resistor.
So only thin we must do to find Eth is to use Ohms law to fond voltage drop on a 4Ω resistor.

I'm trying to teach myself using the Horowitz and Hill text along with the Student Manual. I've been following this problem and Jony's explanation finally brought me to the answer. Thank You!!!!! The only problem I have is that it seems unintuitive (to me) to ADD the voltage drop with the 80V. I would want to subtract it. Cold someone explain the reasoning behind that? If it upsets the thread you can contact me privately. Thanks Again!!!

Dan Carter

 

I'm trying to teach myself using the Horowitz and Hill text along with the Student Manual. I've been following this problem and Jony's explanation finally brought me to the answer. Thank You!!!!! The only problem I have is that it seems unintuitive (to me) to ADD the voltage drop with the 80V. I would want to subtract it. Cold someone explain the reasoning behind that? If it upsets the thread you can contact me privately. Thanks Again!!!

Dan Carter

If you measured the voltage across the 4Ω resistor what polarity would it have? Would the top be more positive than the bottom or vice-versa? Mark the polarity you expect on the 4Ω and compare that with the polarity of the 80V source. Would the voltages add or subtract? If they have the same sense [top more positive than bottom on 4Ω] then they would add.

What polarity do you expect across the 4Ω?

 

Have a look at this:


To calculate the open circuit voltage, notice that the 2Ω resistor has no effect.

what do you mean by this?? no effect on what?

 

what do you mean by this?? no effect on what?

The Electrician is correct to say this.

In calculating the open circuit voltage across a-b it doesn't matter whether the 2Ω is there or not.

In fact if you think about it, you can ignore it for the whole exercise ....

 

Can you show why? I can see why by using nodal anlysis and mesh analysis, is there another method??

 

If you have a voltage source with a resistor across it, it can be ignored as far as its effect on the rest of the circuit. Its only effect is to draw some extra current from the voltage source, but that current only passes from the voltage source into the resistor in parallel with the voltage source. It has no effect on the rest of the circuit.

Similarly, a resistor in series with a current source causes some extra voltage drop in the branch containing the current source and the resistor in series with it, but it does not change the current coming from the current source. It has no effect on the rest of the circuit (except perhaps for some very unusual circuit topology).