Thevenin Equivalents

For this experiment, the objective is to reduce this network of resistors and power supplies below to its Thevenin equivalent, so we can find the smallest resistance of Load #2 that can be connected to the system.

Prelab:

R_C1	R_C2	R_C3	R_L1	V_S1	V_S2	V_L2,Min
100 Ω	39 Ω	39 Ω	680 Ω	9 V	9 V	8 V

To find V_Th, we remove R_L2 and solve for the V_oc.

Since Voc = V_x, we used nodal analysis which gives us the equation:
(V_x-V_S2)/R_C2 + (V_x - V_S1)/R_C1 + V_x/R_L1 = 0
V_x = 8.6 V

As finding R_Th from reducing resistors is harder than it seems, we rely on the formula R_Th = V_oc/I_sc.

After we create a short circuit from R_L2, we can find I_sc.
Using nodal analysis we get the equation:
(V_y-V_S2)/R_C2 + (V_y - V_S1)/R_C1 + V_y/R_L1 + V_y/R_C3= 0
V_y = 5.1 VI_sc = V_y/R_C3 = 0.1307 mA
R_Th = 65.7 Ω

Given that R_L2,Min = 8V, we use the voltage divider to find R_L2
R_L2 = V_L2*R_Th/(V_Th - V_L2) = 876 Ω 
Short Circuit R_L2 current I_sc = 0.1307 mA
Open Circuit R_L2 voltage V_oc = V_Th = 8.6 V

Procedure:

Build circuit with these components

Component	Nominal Value	Measured Value	Power or Current Rating
R_Th	66 Ω	65.8 Ω	1 W
R_L2,min	876 Ω	873 Ω	1 W
V_Th	9 V	9.21 V	2 A

To perform the experiment, we measure the voltage across Load 2.

Config	Theoretical Value	Measured Value	Percent Error
R_L2 = R_L2,min	V_Load2 = 8 V	8.5V	6.25 %
R_L2 = ∞ Ω	V_Load2 = 8.6 V	9.17 V	6.63 %

Next we disassemble and build the second circuit by replacing the Thevenin Equivalents with the original circuit.

Component	Nominal Value	Measured Value	Power or Current Rating
R_C1	100 Ω	99.1 Ω	1/8 W
R_C2	39 Ω	38.1 Ω	1/8 W
R_C3	39 Ω	38.6 V	1/8 W
R_L1	680 Ω	667 Ω	1/8 W
V_S1	9 V	9.26 V	2 A
V_S2	9 V	9.21 V	2 A

Now remove R_L2 so it becomes infinite resistance.

We perform the experiment again by measuring V_Load2.

Config	Theoretical Value	Measured Value	Percent Error
R_L2 = R_L2,min	V_Load2 = 8 V	8.20 V	2.5 %
R_L2 = ∞ Ω	V_Load2 = 8.6 V	8.84 V	2.79 %

Next put R_L2 back, but change configuration to find power for a set of resistances

Analysis:
The power going into R_L2 = R_Th is V^2/R = 0.280 W
Calculating the first row is 2.97^2/(0.5*R_Th) = 0.267 W

Config	V_Load 2	P_Load 2
R_L2 = 0.5R_Th	2.97 V	0.267 W
R_L2 = R_Th	8.21 V	1.021 W
R_L2 = 2R_Th	5.89 V	0.264 W

Conclusion:
Seeing how we can vary different resistors, we can easily find the voltage in connection with a Thevenin Equivalent. It becomes very useful when just adding a new element to an existing large circuit.