Thevenin and Norton Question

thevenin voc

Could somebody please help me with this question? I don't understand when they say determine the thevenin equivalent as viewed from 2 different terminals. How do you look into a terminal to determine what the circuit looks like?
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ways to determine effective resistance thevenin

When you look in to two different terminals, you need to find two things for a Thevenin equivalent circuit:

Resistance
Voltage

For resistance, turn OFF all independent sources (so your voltage source is a short, and your current source is an open). Find the resistance that you see (in your circuit, you will have parallel combinations of resistors). This resistance is both the Thevenin resistance and Norton resistance (there is no difference).

For the Thevenin voltage, you can use superposition:
Find the voltage at X-X' with the voltage source on and the current source off (open circuited). Then find the voltage at X-X' with the current source on and the voltage source off (short circuited). Add your two results (keeping in mind polarity), and that is your Thevenin voltage.

calculate thevenin voltage from voc and isc

In the example I attached, the solution gave two different resistance equivalent when looking at terminal xx' and when looking at terminal yy'. I guess my question is I don't understand what to look for when I look into a terminal.
Why would the combined resistance have different values when looking at different terminals?

thevenin resistance

I think ur assumption is wrong to think that the effective resistance looking from two nodes will be same.Forget about Thevenin's equivalent, even if u take a resistor cube u will see diferent effective resistance looking from the diagonally opposite points and the two diagonal points in a plane.You can refer HC verma for that.And thevenin's equivalent is just like an extension of effective resistance.

thevenin resistance for a cube of resistors

Whenever you are required to calculate the impedance (resistance) viewed from a certain circuit, think about what you would measure between the two points.
When required to find the Thevenin or Norton equivalent between two points, you should replace the original circuit with just two elements: a voltage source plus a series impedance (resistance) for the Thevenin, a current source in parallel with an impedance (or admittance) for the Norton circuit.
And yes, the two equivalent circuits for the two different views, XX and YY will be different.

You can use the method described by rtarbell to find the resistance and voltage.
Or, you can use this method:
1. calculate open-circuit the voltage between the required points, say XX first. This will be the voltage of the voltage source in the Thevenin circuit, Voc
2. calculate the short-circuit current that would flow between the two points, if you shorted them together. In the Norton circuit, this would be the value of the current source, Isc
3. calculate the series resistor of the Thevenin circuit
Rthevenin=Voc/Isc. The Thevenin circuit is now fully determined. It consists of a voltage source of magnitude Voc, in series with a resistor Rthevenin.
4. calculate the admittance of the Norton circuit
Ynorton=Isc/Voc. The Norton circuit thus consists of a current source of magnitude Isc in parallel with an admittance Ynorton. Since this admittance is simply the inverse of the above Rthevenin, it follows that once you know Rthevenin, you can just say that the Norton equivalent circuit is a current source of magnitude Isc in parallel with a resistor Rthevenin, calculated above already.

The method I described can be used in any linear circuit, even if it contains controlled current or voltage sources, which cannot be "passivised" (shorted for voltage sources, open for current sources), as described by others, in order to calculate the equivalent resistance.

thevenin resistance in cube

First of all, you can thevenize the 40ohm resistor and the 1A source. Then you have a 40V voltage source in series with a 40ohm resistor.
Now this 40 ohm resistor is in series with the 20 ohm one, so you have just a 60ohm resistor. You have reduced the circuit to a two-node, two-mesh one. You can go further and thevenise the 88V voltage source and the 10 ohm and the 50 ohm resistor to obtain a 73.333V source in series with an 8.3333 ohm resistor. The circuit now becomes just one mesh, with two voltage sources and one resistor, 68.333 ohm. Calculate the current through it: (73.333-40)/68.333=0.487A
The voltage drop across the 8.333 ohm resistor would be 8.333*0.487=4.065V. This gets subtracted from the 73.333V, to get the open-circuit voltage of 73.333-4.065=69.27V This is Voc.

For the short-circuit current, the 50ohm resistor is zero, so the 88V source will supply a current through the 10 ohm resistor: 88/10=8.8A.
The 1A source will supply 1A, which will be divided between the 20ohm and the 40ohm resistors. Through the 20ohm one you will have then two thirds of the 1A, that is 0.666A. Thus, the total SC current is: Isc= 8.8+0.666=9.466A

With that, Rth=69.27V/9.466A=7.317 ohm.

Does this result seem right? Let's see, if there was only the 88V source and the 10ohm resistor we would have just that, an 88V source and 10ohms of resistance. Since we have more resistance in parallel, the total resistance will be lower. Since the 10ohm is the lowest, it seems OK that the total resistance is not much lower than 10 ohms.
The voltage is 69.27V. That is close to, but lower than 88V, which seems OK, given the divider involving a 10ohm resistor and the rest; and the 1A times 40ohm is 40V, so if this was alone in the circuit, the final voltage would have been lower than 69.27V. So the 69.27V looks OK.

example solution thevenin and norton question

vvv,
Thank you for all you help. I have one more question though. In determining the voltage at XX', why did you include the 50 ohm resistor? I thought you said you open the circuit in order to determine the voltage. How would you calculate Rth from YY'? Would you open the circuit to calculate the voltage and then short circuit it to calculate the current?

example solution for thevenin and norton question

Open-circuit does not refer to taking something out of the circuit, but to the fact that the two points of the circuit between which you want the equivalency are left open, that is nothing is connected to them and under those conditions you measure the voltage. Short-circuit means just that: you short the two points together and see what current flows thorugh the short.

For the YY case, you can use a similar approach: thevenise the 40ohm/1A combination and you get 40V/ 40ohm. Careful, the voltage you will calculate at YY' will be 40V plus the voltage drop across this 40 ohm resistor.
We thevenised befoer the 88V with the 10 ohm and 50 ohm resistors and we got 73.33V, with 8.333 ohms of resistance. We will do the same now.
so we now have a 73.33V source, in series with 8.33 ohm, in series with 20ohms and in series with 40ohms and another source, 40V in magnitude. The current through the circuit is again (73.33-40)/68.333=0.487A
With that, the voltage drop across the thevenised 40ohm resistor is 40*0.487=19.48V. So the open-circuit voltage at YY' is 40+19.48=59.48V

The SC current is: 1A from the source plus the current flowing from the 88V through the equivalent resistor, divided by the ratio of the 20 ohm and 50 ohm resistors.
88/(10+50||20)=88/24.28=3.62A

The current through the 20 ohm resistor is then 2.58A
So the total current flowing through the short at YY' is 3.58A.
With that, Rth=59.48/3.58=16.61 ohm.

I hope I got them all correct. You better double-check the results, but that is the idea.

question on thevenin and norton

They are correct, thank you. The way my professor did it was combining the resistors looking at terminal XX then finding Vth. I like your way better though.