Hi,
You could use a circuit simulator if you wanted to make up your own problems.
Connect some circuit elements and let it solve for all the node voltages perhaps,
then try to solve them yourself on paper using the theorems you want to understand
better. Just in case you cant find too many on the web, you could make up as
many as you want that way too.
I could probably give you a graphical method which makes Thev and Norton
a bit simpler to remember if you are interested.
Here's an easy practice exercise:
Have a voltage source and two resistors in series.
Calculate the voltage over R2 (the lower one in chematic). That is thevenin voltage.
Calculate the parallell value of the two resistors. That will be output impedance.
Calculate current through R1 if R2 is shorted. Use both thevenin eqivalent model, and also just do a V/R1 calculation. Result should be the same
Use an ampere meter and put it over R2. That will show approximately same current as you calculated.
Now you have shown that short circuit current is the same for thevenin model as for the real circuit. Also you can have a resistor in parallell with R2 and calculate new voltage drop. You'll have the same voltage drop for thevenin model and in real life.
Ok, I found some practice problems online and I need help with the one in the schematic. I don't see how they get the answer. So the main schematic is shown above and I simplified it below. What I did was: 2.25Ω x 0.8A to find the voltage drop across the 2.25Ω resistor. I get 1.8 and then apparently I add that to the 1.2V source (Don't know why) and I get 3V. Can someone please explain why I do that?
Ok, I found some practice problems online and I need help with the one in the schematic. I don't see how they get the answer. So the main schematic is shown above and I simplified it below. What I did was: 2.25Ω x 0.8A to find the voltage drop across the 2.25Ω resistor. I get 1.8 and then apparently I add that to the 1.2V source (Don't know why) and I get 3V. Can someone please explain why I do that?