Norton and Thevenin equivalents

Thread Starter


Joined Oct 3, 2009
I am having a TON of trouble figuring out how to find norton and thevenin equivalents..

When my teacher goes over it in class it makes sense... but I just have no idea how to start either of these problems

The first one they want us to find the Norton equivalent... so what I did first was short circuit the voltage source and the Vab and I assumed Vab=24v since thats the potential difference from top to bottom

Then afer that I saw that the current goes through the short circuit then the 12 ohm resistor (but omits the 6 ohm), therefore, the current was 24/12=2 amps

Idk if thats right or not.. but thats how I approached it

for problem #2, I didn't know where to start since my teacher only taught us how to deal with independent sources.. I just assumed you can change the current source to an independent voltage source with voltage 100I*50k with a 50k resistor in series with that..

I would REALLY appreciate any help! thank you



Joined Jul 23, 2009
Hey nirvanaguy,
I'll be glad to help you, but I need to see the ORIGINAL PROBLEMS, and from your sketches, I'm not convinced that that's what you drew for us. If it is, please verify that and I'll help you. But you're making it much harder than it really is. Just follow the rules and it's a breeze. For the first one, before shorting the voltage source, compute the output voltage at (a to b), which is normally a voltage divider. In your case, if the problem is really the way you drew it, Vab=24V. Next short the voltage source, which gives a parallel combination of 0 & 6 ohms, which is still zero ohms. In other words, neglect the 6 ohms for this problem, it might as well not be there. With a-b shorted, compute the required current to get 24V with the 12 ohm resistor, which of course is 2A. That's it. A 2A current source with a 12 ohm parallel resistor.
The next problem isn't totally clear to me as to whether the task is to do a thevenin equivalent on both the input and output sides, or just the output side. I'm going to guess that it's both the input and output sides.First open ckt I1 & you get the 1K, I1=V1/1K. On the other side, 100I1*50K = a voltage of (5X10^6)*(I1) or I1*5 million. Combining the input & output sides, Vout=(V1/1K)*(5X10^6) = V1*5X10^3 volts in series with the 50K resistor. One of the greatest things about N & T equivalents is that they are so easy to double check yourself to make sure you got it right. Use a numerical example of I1=1mA, on the original problem, Iout=100mA and Vout=100mA * 50K = 5,000 volts. Now, take V1=1V to get the same primary side current. Vout=5,000*V1 = 5,000 volts. It checks out and those are the correct answers.


Joined Mar 6, 2009
Hi nirvanaguy,

The equivalents would look like those in the attachment.

You were on the right track with the first problem - just needed to follow it through.

kkazem gave good advice - but he didn't explicitly note the output polarity in the second example.