Thevenin's equivalent with dependent and independent sources Problem 5.53

example on how to obtain the Thevenin or Norton equivalent of a circuit with dependent and independent sources

For an example of circuit containing only dependent sources, check out my other video

[kurtrowland]

I love how you did just the right amount of math in your head. It's so refreshing to see people who can do that, as it really makes everything easier to follow. Thank you. I just failed a quiz, but now I feel confident that I'll pass the final!

[dawnschafer]

God bless you and your beautiful ECE knowledge! Life saver for ECE 201 exam 2!!!

[kevinsnaterse]

I've been struggling so much in this class & finding your video gives me hope thank you!!!

[AraujoMatt]

Tobias,
The way I learned it, KVL, KCL and Ohm's law are the tripod foundation of linear circuit analysis. Any of them will always work. Developing intuition is an important part of circ analysis, especially once you know how to simplify using thevenin, norton, superposition and source transformation. To decide between KVL and KCL, I usually count the essential nodes and the mesh loops (accounting for "supermeshes"). I will often use whichever will give me the least number of equations

[AraujoMatt]

Darren,
When looking for Voc, keep in mind the terminals of this voltage. The negative terminal is connected to the bottom of the circuit. The key to solving this lies in using KVL or KCL. There a several different approaches. You could also do KVL on the right loop. That would be Voc= -2Va + V(R2k) This solution give Voc = -16+4=-12 which is the same answer I get in the video using KVL around the outside. Hope that helps

[zyt]

you don't know how you saved my grades I mean my life right now ! THANK YOU

[IamGulzow]

How would you find Voc if there was an independent current source in place of the independent voltage source?

[AraujoMatt]

Thanks Nathalia. I use SmoothDraw3 which is a free download (you could use MS Paint.) I also use a Wacom Intuos pen tablet for writing. Camtasia is what I use for the screencapture. My videos are licensed under the creative commons licensed. You'll have this option from YouTube. It means these videos can be used by anyone for educational purposes. No permission is required to use them. Best of Luck!

[Hulnbfrijabl]

my teacher said, and I've tried and worked for me, if you want to find something in a resistance parallel to a current source you use kcl, else you use kvl, because most of the time you lose less time.

Both solve any problem the thing is to find the one that does it faster so you won't lose time in the test.

[johnnolen]

Not to quibble, but if you have the Norton circuit you don't have to back solve to find the voltage drop across the load. V_Load = I_N÷(G_N + G_Load). In this case, - 3mA ÷(1/4k + 1/4k) = - 6 volts.

[quinstermyer]

For V open circuit why could he do (12)(2/2+4) voltage division to find v at ab???

[LarlemMagic]

+ Axceed1 If there is not two sources then you do resistance equivalency till you have the thevinin model.

[AraujoMatt]

Sorry, I don't see the mistake....2k(1m) =2V. Therefore -2k(Isc) = 6V. Then 6/-2k = -3mA. Hope that clarifies things.

[joshman]

Does the method provided in this Video only work for any circuit where you are tasked to find thvenins?
Can you do thevenins if there is only dependent sources?

[jonathang]

Is the 4K ohm next to Vo the load? I did straight mesh analysis on the circuit and got half of the values for your answers for Isc and Voc.

[cagataysunal]

Comments be like "Hey I realize that I don't know SHIT but you're doing it wrong and I don't care if you're doing it right."

[proteo-xu]

Hi ! tanks for this very good course
i have a question wich method can i use to find thevenin when Vos=0 and Isc= 0 ?

[MrSpacebiscuit]

Open terminals will still have a voltage drop/rise, just no current. So you can apply KVL through an open terminal but not KCL from an open terminal.

[vikramanramakrishnan]

Hey any other way to find Rth with both dependent and independent souce...more like a short cut..?

[m]

how could you employ KVL on the open circuit to calculate Voc in the first part ? isn't that wrong ?