Thevenin's Theorem-Independent and Dependent Source (Edited)
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- เผยแพร่เมื่อ 7 พ.ค. 2012
- Shows how to find the Thevenin equivalent for a circuit with both dependent and independent sources. More instructional engineering videos can be found at www.engineeringvideos.org.
This video is licensed under the Creative Commons BY-SA license creativecommons.org/licenses/b....
Thank you, super helpful.
I love tutorials. They are very helpful! However, I just want to point out a small mistake you have in this video. The NPN transistor circuit you've shown is not a Common Emitter; it's a Common Collector. :)
Thanks!
This is one of the weirder things that happen when doing Thevenin's theorem computations. When we find the short circuit current, we short the output, which in this circuit, means shorting the resistor. So we have a resistor with resistance 2k Ohms in parallel with a wire with resistance 0 Ohms. This parallel combination has a resistance of 0 Ohms, which is equivalent to a wire. So we can replace the wire in parallel with the resistor with just a wire; the resistor goes away.
Cheers!
Es ist common collector amplifier aber noch sehr gut video!
thankssssssss...........
@Darryl Morrell: Thank you for these great videos, immense utility now with exams coming up for me!
I still do not understand the part at 05:10 though, even after your explanation in the comments. How can the voltage be zero when we have a voltage source in the left end of the circuit? This makes zero sense to me. Does that voltage never reach the right end? :P
An imaginary short circuit path was created in parallel to the 2k resistor. Since current flows in the path of least resistance(in this case, the imaginary short circuit path), the current flowing through the 2k resistor is zero. Since V=IR, and I=0 through the 2k resistor, the voltage across the 2k resistor is zero. Hence, the 2k resistor can be removed from the analysis, as had Dr. Morrell explained.
If I wanted to find just Rth, and I had an independent current and independent voltage source, can I zero both of them? Or can you only zero the ind. voltage source?
you can not set both at the same time, you should do it one at a time
Why is the voltage on the left vin-voc?
there is a case where a current source is injected (1 A) at the place where thevenin's resistance is to be measured. when do you that?
What is going on at 5:10? It seems pretty important - and really counter intuitive - to just skip over it with "These aren't the electrons you're looking for"
where did u get 51 from?
I don't understand why you didn't treat the load as open circuit !!!?
I think youve made a mistake when you wrote the eq Ib+50IbI2 it should be 1000Ib istead of Ib, because the resistance is 1K ohm and not 1 ohm
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+Rafay Shirazi Nah I didnt understand what he was doing, now I do. He is right.
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