Electronic – Finding Thevenin Impedance of a circuit

That could only be replaced by an infinite voltage source.

Current sources operate on the basis that they will output their current no matter the load. When you leave it open circuit your open circuit voltage would be infinity. IE, this circuit cannot exist.

The standard replacement will be a current source with a parallel resistance to a voltage source with series resistance. As shown in this wikipedia image.

http://en.wikipedia.org/wiki/Th%C3%A9venin's_theorem">

OK, say we have an unknown circuit and we want to measure its Thevenin equivalent:

We have a resistive load \$R_L\$ connected to our circuit and we don't want to remove it while measuring the necessary quantities. First we measure the short-circuit current by connecting an ammeter between the measurement node and ground. The ammeter will short-circuit the load, which is exactly what we want in order to measure the short-circuit current. Let's call the result of our measurement \$I_{Th}\$. This current defines the ratio of \$V_{Th}\$ and \$R_{Th}\$: $$I_{Th}=\frac{V_{Th}}{R_{Th}} \tag{1}$$

Since we cannot remove the load, we cannot directly measure the open-circuit voltage \$V_{Th}\$. Instead, we measure the voltage \$V_L\$ across the load. The current through the load is then

$$I_L=\frac{V_L}{R_L}=\frac{V_{Th}}{R_{Th}+R_L} \tag{2}$$

Equations (1) and (2) allow us to solve for the unknown quantities \$V_{Th}\$ and \$R_{Th}\$:

$$V_{Th} = \frac{V_LR_LI_{Th}}{I_{Th}R_L-V_L}$$

$$R_{Th} = \frac{V_LR_L}{I_{Th}R_L-V_L}$$