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- เผยแพร่เมื่อ 10 ก.ค. 2024
- Zach with UConn HKN presents an example problem solving for the thevenin equivalent. This is a very important concept so we are making another example problem but this time using dependent sources in the circuit example.
questions.hkn.uconn@gmail.com
uconnhkn
uconnhkn - วิทยาศาสตร์และเทคโนโลยี
That was rad! Thanks for posting this helpful tutorial
Very clear example and explanation thank you. Could you short the independent vintage source and place an independent current source into the A, B terminal? From there do a nodal analysis and use the new V/I relationship to determine thev?
Very helpful and straightforward, explained much better than my prof, thanks
Why isn't the equation 5- 10(i1-20i2)?
When you put a short in between point A and B at 9:30, will there still be a voltage in between those two points? Can you then solve the circuit by using the Node Voltage method?
potential difference across short circuit is zero
When do the KCL at beginning, why you put 0.5Vb? It should be current in the KCL equation right?
The triangle looking thing on the left side of the circuit is a dependent current source that is labeled "0.5 Vb", and is what is called a voltage controlled current source. That means that the current produced by that source is dependent on some voltage, in this case Vb. So, 0.5Vb is a current, even though your using the value of the voltage Vb to get it.
Suber how u made thsi
I'm getting some bob ross vibes
here a better video on thevenin with dependent sources rth in this video has been calculated using 3 methods
(so that you can compare which method takes less time and which is easier to understand)
th-cam.com/video/D1QxFA9_U5w/w-d-xo.html
Thanks for explanation.
Awesome explanation 👍 dude
when you get your Vth . y didnt you cut the 20 ohm resistor. ? should be vth =(5- 10(.5Vb)) right?? while Vb is equal to 0.588)
The 20 ohm resistor is only cut out when calculating the current through the short. The Voc does not put a short in parallel with the 20 ohm resistor and therefore the 20 ohm resistor must be included in the calculation for Voc.
5:35, Why I(10 Ohm) + I (20 Ohm) = I (20 Ohm) + I(40 Ohm), why can you ignore the current through 5V source?
i do not believe the 5v current through the source is ignored. He simply used a Super node at the 5v source and only looked at the incoming and outgoing currents into the Super node (that's why we have a Super node is to not worry about the current through it).
what the...the music took my by surprise.
*us
at 14:25 cant you solve for i2 with that one equation?
You can solve it at that point, but only because he already plugged in the equation of i3 = 20i2 that he found in the beginning.
Why don't you calculate R by replacing Vol source with wire (short circuit) and Current source with Open circuit? Does it not work in case dependent source?
Your right
But you please tell me how it can
I think you are asking for a proof why when calculating the Thevenin equivalent (when both independent and dependent sources are present) must we use the rule Rth=Voc/Isc? All I know is that Thevenin equivalents seem to work if you follow the rules!
I was sooo stuck on my homework and it's not the same question as this but as soon as you pointed out that shorting the resistor means you don't count it I realized that's exactly what I missed. Something so simple but changes the entire problem. Thank you!!
4:30 , i don't know why 13Vb =3Va +10 :(( ,help me please
he simplified the equation before
Here's what I did
0.5Vb = (Va-(Vb-5))/10 + (Va - Vb)/2°
0.5Vb = (20Va - 20(Vb-5) + 10Va - 10Vb)/200
= (20Va - 20Vb + 100 + 10Va - 10Vb)/200
= (30Va - 30Vb+100)/200
Vb = (60Va - 60Vb + 200)/200 (We multiply everything by 2 to get from 0.5Vb to Vb)
= (60/200)(Va-Vb) + 1
= (3/10)(Va-Vb) + 1
10Vb = 3(Va-Vb) + 10 (multiply everything by 10 to get rid of the fraction)
13Vb = 3Va + 10
how can i3=0.5*Vo, a current cannot be equal to a voltage
0.5Vo is a current source dependent to the voltage Vo
What happens when a dependent voltage source is parallel with a short circuit? Will it still have current?
It will have an infinite amount of current!
~Andrew
thanks
I just got here, I didn't watch the video yet, but... how do you do that with the "hologram"???
glass
@@galenholland3872 yeah, but like... how? the glass can show things, and can detect touch on it! the heck of glass is that?? new generation things!? the glass can have stuff drawn on it like being "floating" in the air, and he can press on the glass and things will appear like in a phone touch glass. but as a start, how can the glass display things on it without a project?? (if it was a projector, I'd be seeing the things in his t-shirt and stuff) And then something is detecting the key presses. this is weird XD
@@Edw590 there's a wall of glass between him and the camera, he writes backward with a neon dry erase marker
@@galenholland3872 Ah! And then the image must be flipped or something. Makes sense. Thanks! Just curiosity, but still cool to know!
EDIT: 5 months ago! Wow. Very fast.
@@Edw590 5 months ago I didn't need to learn this ;-;
i get i1=-117/340
Why remove the resistor
You always remove everything not being turned into a thevenin circuit. These components are collectively called the "Load" of the thevenized circuit.
-Andrew
I have a small doubt That is why your are finding the resistance in thevininan thereom by using this formula Rth=Voc/Isc
In some cases we are finding that Rth by short circuit the voltage source and open circuit the current...why it is not applicable for all problems
If I am understanding your question, the equation Rth = Voc/ISc will work for all problems. The other methods are shortcuts that work for special situations.
-Andrew
Thank you
thx i dunno why those dislikes :S
Hahaha thanks Guys
bleh
seriously, please decrease that volume in he beginning. Very annoying and unnecessary