nortonthevenin
I am working on finding the equivalent norton and thevenin circuits. To fond the equivalent resistance, I zero out any sources. When this happens, according to my professor. Only the 10 ohm resistor to the left is short circuited as a result of the zeroing. Why is this the case? Why is the 12 ohm resistor not also short circuited by zeroing out any sources?
Since there is no current through the 5 ohm resistance (the one at the right), you can consider that the 10 ohm and 8 ohm resistances are connected in series (two resistors are connected in series if the same current passes through them). Thus, the Thevenin resistance is (10+8) // 5 = 3.9 ohm.
In the example you've posted, we see two Thévenin circuits that share a 40-ohm load resistor. But the question is what differentiates between the "load" resistor and the "Thévenin" resistor.
I've redrawn the circuit below:
simulate this circuit – Schematic created using CircuitLab
The Thévenin equivalent circuits are enclosed in boxes as shown.
Alone both Thévenin and Norton circuits have open-ended terminals. Whatever impedance is found across those terminals is deemed the "load" that the equivalent circuit is driving.
For a Norton equivalent, we would see the following circuit:
Best Answer
When you zero out a current source, you make a 0 A current source. That is, an open circuit, not a short circuit.