Equivalent Resistance

[SpartanG345]

http://upload.wikimedia.org/wikipedia/commons/9/9f/Thevenin_equivalent.png
When we are asked to find the equivalent resistance between A and B, i understand we are trying to find the Thevenin resistance between A and B, which is caused by what is in the black box.

However if there is a resistor actually between A and B, do you add this to the Thévenin resistance?

What does the equivalent resistance between 2 points in a circuit mean in a physical sense.
I thought it was the output resistance between 2 points, but if that is the case will a resistor between A and B effect this?

[Averagesupernova]

To Thevenize a circuit you open all current sources, short all voltage sources, remove the external load and compute the resistance between the output terminals. This gives you the Thevenin resistance. You also need to find the Thevenin voltage. To do this you remove the load and compute the voltage that appears at the output terminals with the all voltage and current sources how they are supposed to be. This is the Thevenin voltage. You then take the Thevenin voltage put it in series with the Thevenin resistance and put that in series with the load. Calculate the current/voltage at the load. You will find those calculated values to be the same as if you solved for the problem conventionally. The advantage to Thevenizing is you can substitute any load in and solve quickly.
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This should answer your question about the resistor in parallel with the output.

[karthicknar]

However if there is a resistor actually between A and B, do you add this to the Thévenin resistance?

No it wont be added to Thevenin resistance, because that resistance will be called as load resistance and has to be removed (not considered) while calculating Rth.

What does the equivalent resistance between 2 points in a circuit mean in a physical sense.

In a physical sense it implies the power transfer capability of the source Vth to the load at terminal AB. The more the value of Rth, Vth is a weak source as far as load is concerned because a slight current drawn by load, will create more voltage drop across Rth and resulting in less load voltage.

[Averagesupernova]

No it wont be added to Thevenin resistance, because that resistance will be called as load resistance and has to be removed (not considered) while calculating Rth.

In a physical sense it implies the power transfer capability of the source Vth to the load at terminal AB. The more the value of Rth, Vth is a weak source as far as load is concerned because a slight current drawn by load, will create more voltage drop across Rth and resulting in less load voltage.

If the resistor between A and B is part of the black box then it is to be figured. Not sure if the OP meant it that way.

[n.karthick]

If the resistor between A and B is part of the black box then it is to be figured. Not sure if the OP meant it that way.
Let me put it this way. Even though I believe there could not be a resistor just between A and B and also a part of black box at same time, let us assume if there is a one, it has to be considered for calculation of Vth and Rth. Yet if that resistance is atleast 100 times larger than load resistance, then it can be safely ignored.

[sophiecentaur]

The basic answer to the question is that any network of resistors, and voltage and current sources will have a, Rth. You can measure this by looking at the change (drop) in PD that you measure when you take various currents into your external load - just like you measure the internal resistance (that you can't get at) of a battery in School.
All that's necessary is for the network to be linear.