Final practice - Edgar Bello

Vista previa del material en texto

1 
Index 
Objective 3 
Equipment 3 
Material 3 
Theoretical introduction 3 
Practice development 4 
Conclusions 9 
Questionnaire 9 
Calculations and simulations 10 
 
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“Thevenin’s Theorem” 
Objectives​: 
➢ The student will demonstrate in a practical way the validity of the Thevenin theorem. 
➢ The student will experimentally obtain the value of the power delivered by the circuit 
Thevenin's equivalent. 
 
Equipment and Material: 
 
Provided by the laboratory: 
↠ 1 Digital Multimeter. 
↠ 1 Variable voltage source. 
 
For the students: 
↠ 6 banana - alligator tips. 
↠ 4 alligator-alligator tips. 
↠ Prototyping board (breadboard). 
↠ 3 1 KΩ to 1/4 watt resistors. 
↠ 1 0.47 KΩ to 1/4 watt resistor. 
↠ 1 2.2 KΩ to 1/4 watt resistor. 
↠ 1 3.3 KΩ to 1/4 watt resistor. 
↠ Connecting wire for the breadboard. 
↠ Cutting pliers. 
↠ Tweezers. 
↠ 1 Potentiometer 2.5 KΩ or greater. 
 
 
 
Theoretical introduction: 
In very large circuits, with a large number of meshes and nodes it would be very difficult, if 
not impossible, find branch currents or node voltages using the techniques of mesh analysis 
or node analysis. However, when you want to analyze the voltage, the current and power in 
some particular element, a method that turns out to be very powerful in very complex linear 
circuits is the method based on the theorem of Thevenin. 
 
This theorem states that any linear circuit of C.D. two-terminal you can replace with an 
equivalent circuit consisting of a voltage source and a resistor serially. The value of this 
source is the Thevenin voltage (VTh) and that of the resistor is the Thevenin resistance 
(RTh). 
 
To calculate the voltage and resistance of Thevenin, proceed as follows: 
continuation: 
1. Separate the portion of the circuit through which the equivalent of 
Thevenin. 
2. Mark the 2 terminals between which the removed portion of the 
circuit in step 1. 
3. Calculate the RTh by short-circuiting the voltage sources and open circuit the 
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current sources. Its value is that of the equivalent resistance seen from the pair 
of marked terminals. 
4. Calculate the voltage that appears between the marked terminals, open circuit, the 
which is VTh. 
 
Development: 
Before going to the laboratory, the student should theoretically solve the circuit in which 
practice is based, to make a comparison between the values ​​obtained theoretically and the 
experimental results obtained. Include in your report the mathematical development full of 
these. The theoretical analysis that will be used will be seen in class. 
 
I.- Obtain theoretically the Thevenin equivalent circuit starting from the circuit shown in figure 
1, between points A and B, to calculate the voltage and current at across the load resistor 
RL and record the results in Table 1. 
 
Figure 1 
 
II.- In the circuit shown in figure 1 measure the voltage and current through the branch of the 
load resistor RL and record your results in table I. 
 
Measures Theoretical value Measured value 
IRL Current 456/90475A≈5.04006mA 5.0401 mA 
VRL Voltaje 5472/329v≈16.63221884v 16.632V 
Potency in RL 0.083827485W 83.826mW 
Table I 
 
 
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Simulations​: 
 
 
 
III.- Turn off the sources and disconnect the RL load resistor. Then proceed to perform the 
measurements listed below. 
 
IV.- Activate the sources and measure the open circuit voltage and the short circuit current 
between points A and B as shown in Figures 2 and 3, respectively, and write down their 
results in table 2. 
 
Figure 2 
 
5 
 
Figure 3 
 
 
Measures Theoretical value Measured value 
IAB=IN Current 228/10835A≈0.021042916A 21.043mA 
VAB=VTH Voltaje 3040/139v≈21.875036V 21.871V 
RTH 433400/417Ω≈1039.3285Ω 1039.348Ω 
 
Table 2 
 
Simulations: 
 
 
 
 
 
V.- From the previous measurements, obtain the Thevenin resistance. 
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Figure 4 
 
VI.- Build the Thevenin equivalent circuit, as shown in figure 4 and with the results obtained 
experimentally. Then make the measurements that they ask in table 3. 
 
 
Measures Theoretical value Measured value 
IRL Current 456/90475A≈5.04006mA 
VRL Voltaje 5472/329v≈16.63221884v 
Potency in RL 0.83827485W 
 
 
Table 3 
 
Simulations: 
 
 
 
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VII.- Compare the results of point VI with those obtained in point I of the circuit theoretically 
calculated and comment. 
 
 
 
 
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Questionnaire 
 
1. What does Thevenin's theorem establish? 
Thevenin's theorem states that any combination of batteries and resistors between two 
terminals can be replaced by a single voltage source and a single series resistor r. 
 
2. For what purpose was the open-circuit voltage and short-circuit current measured 
between points A and B at point IV of development? 
To verify that the values of voltage and current are the same in the equivalent Thevenin's 
circuit. 
 
3. With which measured values ​​in Table 2 would you implement the Norton equivalent 
circuit? 
I​N​ and R​th 
 
Conclusions​: 
 
Bello Muñoz Edgar Alejandro 
 
Thevenin's theorem is a useful tool for making an analysis of an electrical circuit because it 
allows to replace a complex circuit with a simpler one. In this way, we use other important 
topics of the subject, such as resistor simplification and Ohm's law to make the circuit easier 
to work with. 
 
Núñez González Angel Daniel 
In the world of electronics and circuits there will always be many circuits which will become 
eternal or very complex to solve for any reason, the issues previously seen such as: source 
exchange, Thevenin's and Norton's theorem, these topics help us a lot in solving these 
complex problems, since they greatly simplify the circuits and make with a few calculations 
and ohm's law they can be concluded in a precise and less stressful way . 
Finally, it is very important to know these topics because they are the basis for all other 
electronic subjects at the university. 
 
López Gracia Angel Emmanuel 
Thevenin’s theorem, such as the manipulation of voltage/current sources have been really 
useful in the analysis of electric circuits, I’m grateful that those tools exist, they make the 
process really easy to carry on, it must be said that I would have preferred see those themes 
before mesh/nodes analysis, but well, they weren’t that difficult, maybe in future assignments 
it will be really useful. 
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Calculations​: 
As an algorithm, we followed the next steps: 
● Cut the circuit in the resistance 
● Find RTH 
○ Turn off the sources (doing the respectives changes) 
● Find VTH 
○ Find the voltaje difference between the node A and B 
● Substitute in the Thevenin’s circuit 
● Find RL, VL, IL and WL 
We cut the circuit 
 
To find RTH we transformed the circuit: 
 
And we reduce the resistances: 
thR = 1 +1
+470+10001
+11000
1
1000
1
2200
 
th Ω 039.328537ΩR = 417
433400 ≈ 1 
Now, we return to the cut circuit: 
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To make thing easier, we transformed the sources a few times, for that, we used 
equivalent structures: 
 
 
 
 
 
 
We had to transform the sources with Ohm’s Law: 
R V = I 
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Now, we used Kircchoff’s law 
1 R2 2 R3 R5 V + V + V + V + V = 0 
.5 70I 0 000I 200I − 2 + 9 − 2 + 1 + 2 = 0 
170I 2.5 4 = 2 
A I = 22.54170 
Therefore, to calculate , we just applied Ohm’s lawR5 V 
R5 R5 (2200) 1.8705036v V = I = 22.54170 = 139
1650 ≈ 1 
But, as we want to calculate the voltage difference between A and B, me need to 
add the source of 10v, and as the sourceis parallel, we can add them in this way: 
ab V = 0 v 1.8705036v th 139
1650 + 1 = 139
3040 ≈ 2 = V 
Now we can substitute these values in the Thevenin’s circuit: 
And we use Ohm’s law again to find the voltage 
in R8: 
 
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R→ 3040/139v (3300 33400/417) (4339.328537) V = I = I + 4 = I 
Therefore: 
56/90475 A (which is IRL) I = 4 
Therefore: 
r8 8 300(456)/90475 472/329v 6.63221884v V = I · R = 3 = 5 ≈ 1 
Other values such as W and I were calculated using Ohm’s law in each requested 
components. 
 
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