Lab 6: Thevenin Equivalents

Introduction:
In the real world, a power system usually contains multiple sources and loads. If we consider a element of a power system, it is wise to model the circuit bu the Thevenin Equivalent. Here we take under the consideration a circuit and only  our attention is the element "load 2". The problem is given a minimum acceptable voltage across "Load 2", we have to determine the smallest smallest "Load 2" resistance can be used, and other questions might be applied are what voltage will exist at "Load 2" when we remove it and what short-circuit current will flow if we replace "Load 2" by a short.

Figure 1.

Procedure:
Step 1:
Given:
Rc1 = 100Ω
Rc2 = Rc3 = 39Ω
RL1 = 680Ω
VS1 = VS2 = 9V


Step 1A) 
V_{load 2, min}  = 8V, determine the R_TH and V_TH :

R_TH = 65.94 Ω , V_TH = 8.6 V

 Step 1B) Calculate the voltage V_y

V_y = 5.07 V.


Step 1C)
1- Find smallest  RL2 :

 

RL2= 879Ω

2- The short-circuit current:

i = V/R = 8.6 / 65.94 = 0.13 A

3- The open-circuit voltage:

V= 8.6 V

Step 2) derive the experiment.

stage 1: build the circuit as showing below.

 Measurement:

Component	Nominal Value	Measure value	Power or current rating
RTH	65.94 Ω	66 +/- 1 Ω	1 W
RL2, min	879 Ω	872 +/- 1 Ω	0.5 W
VTH	8.6 V	8.61 +/- 0.01 V	2 A

Result:

Config	Theoretical Value	Measured Value	% error
RL2  = RL2,min	VL 2 = 8 V	7.75 +/- 0.01 V	3.12 %
RL2 = ∞Ω	VL2  = 8.6 V	8.65 +/- 0.01 V	0.58 %

Stage 2: Build the circuit of figure 1:

 Measure the components:

Component	Nominal Value	Measure Value	Power or Current rating
Rc1	100 Ω	98.5 +/- 0.1 Ω	0.25 W
Rc2	39 Ω	38.3 +/- 0.1 Ω	0.25 W
Rc3	39 Ω	38.3 +/- 0.1 Ω	0.25 W
RL1	680 Ω	667.3 +/- 0.1 Ω	0.25 W
Vs1	9 V	9.05 +/- 0.01 V	2 A
Vs2	9 V	9.16 +/- 0.01 V	2 A

Given that R_L2  = R_TH = 65.95 Ω, The voltage across R_L2 is equal to V_TH/2 = (8.6 / 2) = 4.3 V

so, P_L2  = V²/R_L2 = (4.3)²/ 65.95 = 0.280 W

This turns out to be the maximum power that can be supplied to R_L2 .

Verify this maximum power by setting R_L2 equal to 0.5 R_TH , R_TH , 2R_TH and make measurement.

Configuration	VL2	PL2
RL2 = 0.5 RTH = 33 Ω	2.92 +/- 0.01 V	0.258 W
RL2 = RTH = 66 Ω	4.35 +/- 0.01 V	0.287 W
R­L2  =2 RTH = 132 Ω	5.78 +/- 0.01 V	0.253 W

  

Conclusion:

  Based on the results, we can see that the maximum power supplied to R_L2 when R_L2  = R_TH .The percent error in this calculation was then seen to be 2.5%. The % error could be from equipment error, calibration approximations, and uncertainty. Overall, we see that modeling the original circuit with its Thevenin Equivalent proved to be very useful. Only using a simple schematic, we can then vary Load 2, or whatever load necessary, to obtain desired results.