Electronic – Finding the Thevenin Equivalent Resistance

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}\$:

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

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

You need to consider the sources separately, one at a time, and then combine (add together) their effects to determine the final result.

For example, when considering the 16V source (reference designators would be helpful here), replace the current source with an open circuit. What is the Thevenin equivalent now?

Next add the current source back in. What kind of offset does this introduce into the Thevenin equivalent resistance?