Calculate the Thevenin voltage Vt and the Thevenin resistance Rt.

I think that my Rt will be equal to the total resistance minus the 1 ohm resistor, and Vt will be equal to the voltage drop across the open circuit ab; however, I am not sure how to find those. Any advice to get me goin in the right direction would be greatly appreciated.

 

What is the voltage across the 4 ohm resistor?

 

I'm not sure, should I be able to tell just by looking at the diagram?

 

No you can't tell the voltage drop across the 4 ohm resistor by looking at the diagram.
The Thevinen equivalent voltage will be equal to the voltage drop across the 4 ohm resistor. Temporary short the points a,b and then use voltage divider rule to calculate the voltage across the 4 ohm resistor. Thats your Vth.
For the thevinen resistance Rth, kill all the independant power sources, i.e the voltage source by shortening it and then calculate the total resistance as seen from the point a,b. That will be your Rth.
note: the 8 ohm resistor will be bypassed.

 

How can I use the voltage divider rle if the 4 ohm resistor isn't in series with the top 2 ohm one?

 

How can I use the voltage divider rule if the 4 ohm resistor isn't in series with the top 2 ohm one? I can't just ignore the 1 ohm resistor, and if I consider the 4 and 1 ohm resistors in parallel then the voltage divider rule won't work will it?

 

Temporary short the points a,b and then use voltage divider rule to calculate the voltage across the 4 ohm resistor. Thats your Vth.

Sorry waqar - shorting a,b to find Vth is not the correct method. One simply needs to find Vab for the circuit "as is" - then you have Vth. Presumably you really meant leave a and b open as shown and apply the voltage divider rule.

With a little bit of applied logic you can indeed find Vth virtually by inspection.

Rgds,

t_n_k

 

Redraw the diagram removing the unnecessary components. Find the IR drop of 4 ohm resistor. It becomes a simple series circuit at that point.

Put the 1 ohm resistor back in the circuit and label points A and B. You should be able to figure out the voltage measured between A and B at that point.

Yes, some can do it by inspection. Do enough problems and you will have that ability also.

You will have Vth at that point.

Rth is the second issue. Place an ideal ammeter between A and B and calculate what it would read.

 

I'm terrible when it comes to Thevenin and Norton equivalents, but I'm giving this a try. I have to get better sometime, righ? So if anyone see's something wrong with my approach, please tell me! I want to be directed in the right way.

1. Convert the 8Ω resistor and the 12V source into a current equivalent; so \(\frac{12}{8}=1.5A\)
2. Next, the 2Ω, 4Ω, and the other 2Ω resistors can be added in series and then converted to a voltage equivalent. \(\frac{(2+4+2)}{1.5A}=12V\)
3. Now we simply have a 12V source with a 1Ω resistor, correct?

 

Electronerd,

I might be wrong in my judgement of this too,

But just looking at the circuit, we could figure the voltage across A B terminals

First remember the 8 ohm resistor is in parallel with the resistors on the other side of the 12v. source.

So the 12v. is sourcing current to both branches, so to find the voltage at A B term.

Divide the 3 resistors to the right, into the 12v. source, then take that current times the 4 ohm resistor and that should give the open circuit voltage. Shouldn't it?


I think that 8 ohm resistor is throwing you off a bit,

A voltage source in series with a resistor, can be converted to a current source in parrallel with a resistor.

 

To the Ineffable All,

Ouch! You folks are making a big project out of this problem. First, nuke the east 8 ohm resistor. The 12 volt source will be not change no matter what its resistance value is. And it will be shorted out by the voltage source no matter what its value is. So the open source a,b voltage is easily seen as a series voltage divider to be 6 volts. Its Thevenin value is easily seen as a series parallel resistance branch with a total resistance of 3 ohms. Therefore it can be represented as a 6 volt source in series with 3 ohms.

Ratch

 

To the Ineffable All,

Ouch! You folks are making a big project out of this problem. First, nuke the east 8 ohm resistor. The 12 volt source will be not change no matter what its resistance value is. And it will be shorted out by the voltage source no matter what its value is. So the open source a,b voltage is easily seen as a series voltage divider to be 6 volts. Its Thevenin value is easily seen as a series parallel resistance branch with a total resistance of 3 ohms. Therefore it can be represented as a 6 volt source in parallel with 3 ohms.

Ratch

Hi Ratch,

I'm afraid I still don't understand your method (I told you I was terrible, didn't I?). Perhaps an actual step-through process with diagrams would help me.

Do you know of any great books and such on how to learn this stuff?

Thanks!

Austin

 

Electronerd

First do you see how the 8 ohm resistor is in parrallel with the resistors to the right of the volt. source.

Now if you have a 12v. battery across a resistor what will the voltage of the battery be.

If you put anothewr resistor across that same battery, what will the voltage of the battery be then.

If you had 20 resistors across this same battery what would the voltage of the battery be then.

That's the same thing as with resistors on both sides of the battery as the diagram shows.

That battery voltage will supply the amount of current across each branch while remaining at it's 12v.

The currents will change coming from the battery depending on how many parrallel branches it must serve.

Understand a liuttle better now what were saying about this diagram?

 

ELECTRONERD,

I'm afraid I still don't understand your method (I told you I was terrible, didn't I?). Perhaps an actual step-through process with diagrams would help me.

Did I not do that in the previous post?

Do you know of any great books and such on how to learn this stuff?

That knowledge is everywhere, but look at this link. It is as good as it gets.

Ratch

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/Thevenin.html

 

In addition to the material to which Ratch has linked, you can find further info on Thevenin's Method in the AAC E-book.

hgmjr

 

ELECTRONERD...

Rab=[(2+2)||4]+1=(4||4)+1=2+1=3 ohms (the 8-ohm is short circuited by the killed 12-V source)

Vab is simply is the voltage across the 4-ohm resistor 'cause the 1-ohm resistor sustains no voltage (current passsing through it=zero, open circuited).

to find Vab=V(4-ohm)==> use voltage divider rule 'cause it is in series with two 2-ohm resistors with a toltal voltage of 12 V.

Vab=V(4-ohm)=12 (4/(4+2+2))=12*(4/8)=6 V.

Hope this helps to explain where some answers above came from.