Thevenin Equivalents

   Let assume that we have a circuit that contains multiple sources and loads
We are able to use Thevenin's Theorem to reduce a complicated into a simple circuit such as below with a Voltage and a resistor that represent the complicated circuit.














We are given  a circuit and we need to measure the Thevenin equivalent voltage and resistance as shown below




The Thevenin Resistance can be detremined from R_TH=V_oc/I_sc  where I is the current measured as short circuit at the two points

The given element has the theoratical value of

R_c1	R_c2	R_c3	R_L1	Vs1	Vs2
100Ohm	39Ohm	39Ohm	680Ohm	9V	9V

First we used nodal analysis to find the open-circuit voltage

we got a equation:
(Vx-Vs2)/Rc2 + (Vx-Vs1)/Rc1+Vx/Rcl=0

Vx has the value of 8.64V, which is the Thevenin Voltage


Next, we calculate the voltage V_y in the fugure. Then find the short-circuit current use it to compute the Thevenin resistance
(Vx-Vs2)/Rc2 +(Vx-Vs1)/Rc1+Vx/Rc1+Vy/Rc3=0
Vy=5.11V

Isc=5.11/39=0.131A
Rth=66Ohm

After knowing the R_th and V_th. We can know draw another circuit that contains two resistors in series, which are R_th and R_load. We can use a voltage divider since we know the voltage drop across the R_load is 8V to find the theoretical value of R_load
R_L2=825Ohm


After that we just device the experiment and measure all the components and the value of voltage across different components
 

Component	Nominal Value	Measured Value	Power Rating
R_th(Ohm)	66	66.6	0.3W
R_L2,min(Ohm)	825	822	0.3W
V_TH	8.64	8.64	x

Now, we can build the THevenin Equivalent Circuit on the Breadboard and record data

Configuration	Theoretical Value	Measured Value	Percent Error
RL2=RL2,min	VLoad2=8V	7.99	0.12%
R=L2=infinite	VLoad2=8.64	8.64	0%

Measured Value

Component	Nominal Value	Measured Value	Power Rating
R_c1	100Ohm	99.6Ohm	0.25W
R_c2	39Ohm	39.0Ohm	0.25W
R_c3	339Ohm	4.02Ohm	0.25W
R_L1	680Ohm	679 Ohm	0.5W
V_s1	9V	9.07V	18W
V_s2	9V	8.97V	18W

Building the Original Circuit

 

We obtained the following data


 

Configuration	Theoretical Value	Measured Value	Percent Error
R_L2=R_L2min	Vload2=8.64	8.31V	3.90%
R_L2=infinite	Vload2=8.64	8.62V	0.23%

    We can caculate the power that will be supplied to (absorbed by) R_L2 when its equal to R_th P=V^2/4R=0.28W   Next, we set the R_L2 to different value then measured the load voltage to find the power
 

Configuration	V_load2 (V)	P_load2 (W)
0.5R_th	2.87	0.2496
R_th	8.31	1.0463
2R_th	5.74	0.2496