Hi,
I have just watched the video: https://www.youtube.com/watch?v=0IIy4vRsvoE
In this exercise, we need to find the Thevenin equivalent for the circuit with both dependent and independent sources bellow.

I know the method to calculate it:
1. Calculate the open circuit voltage Voc = Vth.
2. Calculate the short current Isc.
3. Rth = Voc(open circuit)/Isc(short circuit).
Now what I don't understand is that calculating Isc.
Why we can't cancel the dependent current source 50Ib as two terminals are shorted together?

 

Now what I don't understand is that calculating Isc.
Why we can't cancel the dependent current source 50Ib as two terminals are shorted together?

Because a current source still supplies its current into a short as well as an ordinary load.

It's interesting that the author of that video has called the transistor amplifier pictured a "common emitter" amplifier, when it's plainly a "common collector" amplifier.

 

Because a current source still supplies its current into a short as well as an ordinary load.

Is this also applicable to independent current source?
With all voltage sources both dependent and independent sources, they will have 0V cross them if two terminals are shorted.
I thought this is similar to current source but it seems wrong! With current source if its terminals are left open then I = 0 but it has no problem with terminals shorted.

 

Whether the current source is dependent or independent makes no difference.

One might ask what condition will render a particular kind of source ineffective?

If you short a voltage source (an ideal source and ideal short), its voltage output will then be zero; it will then be supplying an infinite current through the short and the voltage will be zero; or will it? The question is, who wins the contest? The voltage source is supposed to always provide a voltage no matter what. But a short is supposed to have no voltage across it because it's zero ohms. These extreme cases with mathematically perfect things can pose a problem.

If you open a current source (no load connected to it; a short is a load), its current output will then be zero; it will then be supplying an infinite voltage across itself and the current will be zero; or will it? The same problem as with the previous paragraph arises. An open circuited current source shouldn't be able to provide any current through an open circuit because that's what an open is; a circuit element that can carry no current. But the current source is supposed to provide a current no matter what. If the current source is open circuited, its output voltage rises to infinity attempting to force a current through the open circuit.

These extreme cases have no ready solution.

But, if you short circuit a current source or open circuit a voltage source, no problem.

 

Interesting! Thanks!
This is the first time, I know that the current source can't be left open.
To be honest, I only was aware that it isn't a smart thing to short a voltage source but had no idea about current source.