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Thevenin

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Micheal

New Member
Hi All,

Please help on the attached.

The question need to get the Thevenin voltage and resistance in between point AB.

Kindly advice,

Thanks & Best Regards,
Gilbert
 

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dknguyen

Well-Known Member
Most Helpful Member
For resistance, turn off all sources. THis means voltage sources become 0V (which is a wire) and current sources become 0A (which is an open circuit) ,then calculate the equivelant resistance from the load terminals. The remaining circuit will have only resistors and be VERY simple. You should be able to find the equivelant resistance no problem.

Voltage is more of a pain and I haven't done it with dependent sources in a while, but I would just use circuit analysis (like mesh analysis) to solve for the voltage at the load terminals.
 
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Micheal

New Member
The result I get from your advice is Rth = 1.87 Ohm which is wrong

What I do in my lab the Rth is 6 Ohm and this is the corrent answer.

The only way that I need to manual calculate to get the result, but how?

Someone please help. It is urgently need to pass up.

Thanks
 

EN0

Member
This might help you, notice that there are resistors in a triangular configuration.

I wasn't able to place the resistors in a triangular format, but you can draw it on paper.

Ra = (R1xR2) / (R1+R2+R3)

Rb = (R1xR2) / (R1+R2+R3)

Rc = (R2xR3) / (R1+R2+R3)

Example, in the attached schematic, we want to have the resistance from point Ra to the other point (A & B). Therefore, we solve using the Ra formula. If all the resistors are 1Ω, we can equate the following:

Ra= (1x1) / (1+1+1) = 0.33Ω

Now, to solve for R4 and R5, we can replace our triangular resistive circuit with a simple "Ra." Notice that R4 and R5 are parallel. So 1 / (1/1 + 1/1) = 0.5Ω Thus, the total resistance is 0.33+0.5 = Approximately 0.83Ω.

For your Thevenin equivalent, we have R1 = 2Ω, R2 = 4Ω, and R3 = 6Ω. Using Ra, we get the resistance to equal 0.67Ω Now just add the 2Ω resistor by point A and we get 2.67Ω.

That's my answer.
 

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Micheal

New Member
Hi All,

Just wonder why the lab report answer is 6 Ohm and how about Vth ? Please refer the attached lab report.
 

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dknguyen

Well-Known Member
Most Helpful Member
I actually get 4.4ohms. Am I just missing something about dependent sources when doing THevnin other than turning them off? (it's been a long time)

The current source goes to 0A and becomes open circuit so it just dissapears. But the dependent source goes to 0V which shorts out the left 0 ohm resistor turn into a wire. So the only resistors left are a 4 ohm and 6 ohm resistor in parallel with each other, and the whole combination in series with the right 2 ohm resitor.
 

qdn

New Member
Do a KCL:
5A + 2Vx/2 = Vx/4. Therefore, Vx = -20/3
Do a KVL:
Vx+2Vx+Voc = 0.
Thus, 3Vx + Voc = 0. So, Voc = -3Vx= -3*(-20/3) = 20V.
Voc = Vth.
 

Micheal

New Member
Hi qdn,

Thanks. How about Rth? Is it 4.4 Ohms correct? as the lab result is 6V, how to get it?

Urgently need help.
 

crutschow

Well-Known Member
Most Helpful Member
The dependent source across the 2Ω resistor affects resistance since it mirrors the voltage across the 4Ω resistor. It does effectively remove the 2Ω from the circuit, but it still affects the apparent Thevenin resistance. Since its control is the voltage across the 4Ω resistor and it has a gain of 2X, it will have a voltage drop of twice whatever appears across the 4Ω. It thus acts like an 8Ω resistor in series with the 4Ω. The equivalent Thevenin resistance is then 8Ω + 4Ω = 12Ω in parallel with 6Ω in series with 2Ω = 6Ω.
 
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EN0

Member
Oops, mistake...

Actually R1=2Ω, R2=6Ω, and R3=4Ω. Therefore, Ra=(2 x 6) / (2 + 6 + 4) = 1Ω Thus, the total resistance (Rt) = 3Ω

Michael, do you have an answer? If you do get the answer in the future, please post it. I'm curious as to what it is.
 
In order to find the equivalent Thevenin voltage and Thevenin resistance of a circuit you should first remove all INDEPENDENT active sources (in other words short circuit all voltage sources and open circuit all current sources,but not their impedances)....let the DEPENDANT voltage and current sources stay as they are....Now place a imaginary voltage source (say of value V) across the terminals....calculate the current I flowing through this voltage source....and calculate Rth=V/I...Simple isnt it???Apply this to your problem n you will get the result....
Now just measure the open circuit voltage appearing across the terminals (withour removing any sources) n this will give u your Vth....
 

The Electrician

Active Member
Do a KCL:
5A + 2Vx/2 = Vx/4. Therefore, Vx = -20/3

This is wrong; you left out the current in the dependent source.

Let Id be the current in the dependent source. Then you would have:

5A + 2Vx/2 = Vx/4 + Id

But, interestingly, you got a value for Vx which is exactly the negative of the correct value.

To solve for the open circuit output voltage, Vth, note that the current from the current source must pass to ground through the 4Ω and 6Ω resistors. Let V1 be the same as Vx and let V2 be the voltage at the top of the 6Ω resistor. Thus we can write:

V1/4 + V2/6 = 5

The difference between V2 and V1 is just the voltage provided by the dependent source, so we can write:

V2 - V1 = 2*V1 (since V1 is the same as Vx)

Solving these two equations simultaneously, we get:

V1 = 20/3
V2 = 20

so Vth = 20

Now let's calculate the short circuit output current. This is the current through the 2Ω output resistor when its right end is connected to ground.

This changes our previous first equation to:

V1/4 + V2/6 + V2/2 = 5

We still have for the second equation:

V2 - V1 = 2*V1

Solving these two simultaneously, we get:

V1 = 20/9
V2 = 20/3

The short circuit output current, Isc, is then (20/3)/2 = 10/3

Rth is given by Vth/Isc = 20/(10/3) = 6Ω
 

qdn

New Member
To solve for the open circuit output voltage, Vth, note that the current from the current source must pass to ground through the 4Ω and 6Ω resistors. Let V1 be the same as Vx and let V2 be the voltage at the top of the 6Ω resistor. Thus we can write:

V1/4 + V2/6 = 5

Just want to ask, why did you exclude the dependent source out of the calculation, doesn't the dependent source current also go through the 4Ω and 6Ω?
 

The Electrician

Active Member
I didn't exclude it. The 5 amp source injects current into the circuit and the only way for it to get back to the current source is through the 4Ω and 6Ω resistors. Current which passes through the dependent source can find its way back through those two resistors. We don't know how much through each resistor until we solve the circuit.

The effect of the dependent source is not explicitly dealt with in the first equation, but as I said, current which passes through the dependent source will eventually return to the current source through the two resistors.

It's in the second equation:

V2 - V1 = 2*V1 (since V1 is the same as Vx)

that the dependent source is dealt with explicitly.
 
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