# why would Ix would be 0?

Current coming out of the 5V source would never be 0 so what would make the current coming out of the 3v dependent source 0?

$$\i_x\$$ is not the current coming out of the $$\3v\$$ source - it's actually the current flowing between the negative terminal of the $$\3v\$$ source and the positive terminal of the current source. That is, the current through the conductor below the $$\i_x\$$ arrow.

The reason it is zero is that there is no loop for current to flow. All current flowing through that path must return by another path, but there is no other path available. Thus is it zero.

It ought to be intuitive that Ix must be 0.

• Ix has no loop therefore cannot be anything except 0 in this model.

"Current coming out of the 5V source" only returns in a loop to 5V -ve side.

• 3V is fixed so i must be 1mA

• 20i is a current amplifier.

• ix and i are not related since ix does not exist. (0)

• But in this model since i=1mA , Vth= 500mV
• It's $3 \nu$, not 3 V (volts), where $\nu$ is defined to be the voltage across the 25-ohm resistor. Unfortunate choice of symbols. – Dave Tweed Dec 31 '18 at 22:39
• aha TY Dave and Happy new year – Sunnyskyguy EE75 Dec 31 '18 at 22:56

Current coming out of the 5V source would never be 0

This is an invalid assumption.

Suppose the load on the output is an ideal voltage source with value 5/3 V.