Practice 3_AFC - Edgar Bello

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Index 
Objective 3 
Equipment 3 
Material 3 
Theoretical Introduction 3 
Practice development 4 
1​.​ Checking Kirchhoff's Law for voltage (with simulations). 4 
2. Checking Kirchhoff’s Law for Current (with simulations) 5 
Questionnaire. 7 
Conclusions 8 
Calculus 9 
Evidence 13 
 
 
2 
“Kirchhoff’s laws” 
Objective: The student will apply Ohm's law and Kirchhoff's circuit laws for voltages 
and currents, to the analysis of electrical circuits, so that in the end of the practice, 
the student will be able to verify and corroborate the calculations obtained through 
techniques and methods already established, such as the following: 
● Kirchhoff's voltage law, in a series of meshes. 
● Kirchhoff's current law, in a series of nodes. 
 
Equipment: 
1 digital multimeter. 
1 C.D. Variable voltage source. 
 
Material: 
1 Protoboard 
(2) 330Ω resistors up to ¼ watt 
(2) 470Ω resistors up to ¼ watt 
(2) 560Ω resistors up to ¼ watt 
Connection wire for the protoboard. 
6 banana-alligator tips. 
4 points alligator-alligator. 
 
Theoretical Introduction 
Our development in Kirchhoff's laws does not include rigorous evidence, only the 
basic context for understanding circuit theory will be studied. Two basic laws for the 
analysis of circuits containing elements of resistive, inductive and capacitive type, 
are those postulated by the German physicist Gustav Kirchhoff (1824 - 1887); the 
well-known "Kirchhoff’s Current Law” (KCL) and the "Kirchhoff’s Voltage Law” (KVL). 
 
Kirchhoff's current law postulates that: 
"The algebraic sum of the currents that affect a node are zero". 
Kirchhoff's law for voltages postulates that: 
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"The algebraic sum of the voltages around any closed path in a circuit is zero at all 
times". 
Understood as closed, the route through a series of nodes that end at the initial node 
without going through any node more than once. A closed path is usually called a 
loop. 
For example, consider the circuit shown in Figure 2. It is a circuit consisting of two 
closed paths. 
 
Practice development 
1. Checking Kirchhoff's Law for voltage. 
Without turning on the voltage source yet, build the circuit illustrated in figure 1 on 
the protoboard. Once armed proceed to set the voltage values ​​in the sources, with 
the values ​​marked in the tables, and connect it to the circuit as shown. 
 
 
Element Value Potency 
Vs1 9v 
Vs2 5v 
R1 470𝛀 ½ Watt 
R2 330𝛀 ½ Watt 
R3 560𝛀 ½ Watt 
Figure 1. ​Single mesh circuit to check Kirchhoff's law for voltage 
 
 
 
4 
Measure
ments 
Theoric 
value 
(Volts) 
Simulatio
n value 
(Volts) 
Measured 
value 
(Volts) 
Theoretic
al potency 
(miliwatt) 
Simulatio
n potency 
(miliwatt) 
Measured 
potency 
(miliwatt) 
Absorbs 
(A) 
supplies 
(S) 
Voltage 
V​0A 
-9V -9V -8.959V -26.47058
82mW 
-26.469m
W 
—- S 
Voltage 
V​AB 
1.382352
94V 
1.382V 1.0397V 4.065743
94mW 
4.064mW —- A 
Voltage 
V​BC 
5V 5V 5.029V 14.70588
24mW 
14.704m
W 
—- A 
Voltage 
V​CD 
0.970588
24V 
970.588m
V 
740.2mV 2.854671
28mW 
2.854mW —- A 
Voltage 
V​D0 
1.647058
82V 
1.647V 1.2494V 4.844290
66mW 
4.843mW —- A 
--.- ΣV=0V ΣV=-4.12
x V01 −4 
ΣV=-0.90
07V 
ΣP=0mW ΣP=-0.04
mW 
—- —- 
Table 1​. Theoretical, simulation and experimental voltage values. 
 
Simulation of “​Checking Kirchhoff's Law for voltage.​” 
 
2. Checking Kirchhoff’s Law for Current 
Without turning on the voltage sources, assemble the circuit in Figure 2 on the 
protoboard. Once armed proceed to set the indicated voltage values ​​for the sources 
and connect them to the circuit. 
 
5 
 
 
Element Value Potency 
Vs1 9v 
Vs2 5v 
R1 470𝛀 ½ Watt 
R2 330𝛀 ½ Watt 
R3 560𝛀 ½ Watt 
Figure 2. ​4-node circuit to check Kirchhoff's law for current. 
 
 
Measurements Theoretical value Simulation value 
 
Measured value 
Current I1 (Left 
Branch) 
10.5455148 mA 11 mA 10.549 mA 
Current I2 (Center 
Branch) 
12.2533576 mA 12 mA 
 
12.333 mA 
Current I3 (Right 
Branch) 
1.70784283 mA 1.708 mA 1.5805 mA 
Table 2​. Theoretical, simulation and experimental current values. 
 
 
Measure
ments 
Theoric 
value 
(Volts) 
Simulatio
n value 
(Volts) 
Measured 
value 
(Volts) 
Theoretic
al potency 
(miliwatt) 
Simulatio
n potency 
(miliwatt) 
Measured 
potency 
(miliwatt) 
Absorbs 
(A) 
supplies 
(S) 
Voltage 
V​0A 
-9V -9V -8.983V 94.90963
32 mW 
-99mW -94.761m
W 
S 
6 
Voltage 
V​AB 
4.956391
956V 
4.956V 4.881V 52.26770
47 mW 
54.516m
W 
51.489m
W 
A 
Voltage 
V​BC 
4.043608
017V 
4.044V 4.0747V 49.54777
51 mW 
48.528m
W 
50.244m
W 
A 
Voltage 
V​CD 
0.956391
983V 
956.392m
V 
996.2mV 1.633367
19 mW 
1.633mW 1.701mW A 
Voltage 
V​D0 
-5V -5V -5.0817V 8.539214
13 mW 
-8.54mW -8.679m
W 
S 
 ΣP=​-0.000
000285795 
mW 
ΣP=-2.86
3mW 
ΣP=251m
W 
--- 
Table 3. ​Theoretical, simulation and experimental voltage values. 
 
Simulation of “​Checking the current Kirchhoff Law.​” 
 
Questionnaire. 
1. Define what is a node in an electric circuit. 
With the use of the concept in real circuits, what we can say it is, is, a node is 
a part of an electric circuit that do the roll of a dot in mathematics, an abstract 
concept that indicates you a localisation of anything, in the case of a node, it 
indicates the intersection of 2 or more lines of a circuit, an abstract localization of a 
convergence. 
7 
 
2. Define what is an electric circuit 
An electric circuit is a system that involves electric parts, parts that involve the 
basics electric laws (Ohm’s and Kirchhoff’s), and can be associated with each other 
in an order, respecting physics laws and properties of electricity. 
 
3. Express mathematically Kirchhoff’s law for current 
ncoming current utgoing current∑
 
 
I = ∑
 
 
I − ∑
 
 
O = 0 
4. Define what is a closed trayect in an electric circuit. 
A closed trayect in an electric circuit means that the electron flow in it can go 
and come freely, all the ways of traveling in the circuit have a respective way to start 
and to end, both from an electric source. 
 
5. Define what is a voltage drop 
A voltage drop is an event that occurs in an electric circuit element, the 
original voltaje travel through the circuit, but every element make it to be lower and 
lower (because the resistance of each element). 
 
Conclusions. 
● Bello Muñoz Edgar Alejandro 
The knowledge acquired about the laws of Kirchhoff allowed the team to make the 
circuit connections correctly, and it was of great importance to perform the correct 
measurement of specific points in the circuit, so that, each component had the 
expected current, resistance and voltage values. 
The use of multiple voltage sources was interesting because in this way the behavior 
of the voltage and current in the circuit components is more noticeable, and we still 
check that the Kirchoff laws are fulfilled. 
 
● Núñez González Ángel Daniel 
The more we advance in the practices, the more and more they become more 
laborious, because the topics that cover them are of equal complexity and basic 
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concepts are still used to solve them, but at the time of doing the practices, you have 
to have more considerations to solve the exercises, such as checking the polarity of 
the circuits well, making connections between other things well. 
Kirchhoff's laws are essential for the resolution of this type of circuits, knowing how 
to implement them correctly is something that in this coursewe must learn. 
● López Gracia Angel Emmanuel 
The practice indeed was a success, we learned what are watts and how to measure 
they, at the same time how to calculate them theoretically, the difficult part in this 
practice was the second one because we needed to use some knowledges that we 
weren't supposed to use at perfection, in this case was the use of nodes as an entity 
of the circuit, but at the end we learned how to, and as always we still are in the 
process of learning. By doing this kind of practices, in the future, we will become 
experts in these basics themes, and about the Kirchhoff’s of law, we learn how to 
use it in some circuits, now the challenge is to use it in more complex circuits, thing 
that makes me feel excited. 
 
Calculus. 
 
For the first circuit we used Kirchhoff’s law to deduce that 
∑
 
 
V = 0 
so we found every single voltage in the circuit 
Naming ​R1 R2 R3 ​with ​ V1 V2 V3 ​(applying Ohm’s law)​, ​and ​Vs1 Vs2 ​maintained 
their names, giving us the next expression: 
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with Vn=IRn1 2 3 s1 s2 ∑
 
 
V = V + V + V + V + V = 0 
we continued replacing each Vn with its Ohm’s law equivalent 
1I 2I 3I s1 s2 R + R + R + V + V = 0 
then, to find ​I​ we used Ohm’s law 
RV = I 
/R V s1 s2)/(R1 2 3) I = V = ( + V + R + R 
After finding ​I​, we replaced it in Ohm’s law for each case 
replacing Rn with the given resistances and ​i​ with the previous result.n Rn;V = I 
Once finished, we calculated the Potency by multiplying each Vn by ​I​. 
 
For the second circuit was a different challenge, 
 
To start we decided to separate the circuit in 3 parts, left part, right part and B (as 
indicated in the illustration); to point the direction of the voltage in each line (B0, BC, 
AB), and to point the direction of the currents, getting I1 for the left current, I2 for the 
middle one, and I3 for the right current. 
 
For the left part we proposed the use of Kirchhoff Law for voltages as shown below 
with Vn=ImRn− s1 1 2 ∑
 
 
V = V + V + V = 0 
Replacing Vn with ImRn… 
…(1)s1 1I1 2I2 − V + R + R = 0 
 
And also we deduced that V1=R1I1 (this will important below) 
 
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For B, we used Kirchhoff’s Law for currents... 
ncoming current utgoing current∑
 
 
I = ∑
 
 
I − ∑
 
 
O = 0 
So then 
b I1 3) I2)∑
 
 
I = ( + I − ( = 0 
1 3 2... (2) I + I = I 
 
For the right part we did the same as the left part with the respective values 
with Vn=ImRns2 3 2 ∑
 
 
V = V − V − V = 0 
Replacing Vn with ImRn... 
…(3)s2 3I3 2I2 V − R − R = 0 
Using (2) in (3)... 
s2 3I3 2(I1 3) V − R − R + I = 0 
s2 3I3 2I1 2I3 V − R − R − R = 0 
s2 2I1 R3 2)I3 V − R − ( + R = 0 
R3 2)I3 − s2 2I1 − ( + R = V + R 
3 − s2 2I1)/ R3 2)...(4) I = ( V + R − ( + R 
(R3 just depends in I1 because the rest of the numbers are constants) 
 
Continuing with equation (1) 
…(1)s1 1I1 2I2 − V + R + R = 0 
Replacing (2) in (1) 
s1 1I1 2(I1 3) − V + R + R + I = 0 
s1 1I1 2I1 2I3 − V + R + R + R = 0 
Replacing I3 with (4) 
➔ s1 1I1 2I1 2((− s2 2I1)/ R3 2)) − V + R + R + R V + R − ( + R = 0 
➔ s1 R1 2)I1 2(− s2 2I1)/ R3 2) − V + ( + R + R V + R − ( + R = 0 
➔ R1 2)I1 2(− s2 2I1)/ R3 2) s1 ( + R + R V + R − ( + R = V 
➔ R1 2)I1 − 2V s2 2R2I1)/ R3 2) s1 ( + R + ( R + R − ( + R = V 
➔ R1 2)I1 − 2V s2)/ R3 2) R2R2I1)/ R3 2) s1 ( + R + ( R − ( + R + ( − ( + R = V 
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➔ R1 2)I1 R2R2I1)/ R3 2) s1 − 2V s2)/ R3 2) ( + R + ( − ( + R = V − ( R − ( + R 
➔ R1 2)I1 1((R2R2)/ R3 2)) s1 − 2V s2)/ R3 2) ( + R + I − ( + R = V − ( R − ( + R 
➔ 1(R1 2 R2R2)/ R3 2)) s1 − 2V s2)/ R3 2) I + R + ( − ( + R = V − ( R − ( + R 
➔ …(5)1 V s1 − 2V s2)/ R3 2) ]/[(R1 2 R2R2)/ R3 2))] I = [ − ( R − ( + R + R + ( − ( + R 
 
Now, with I1 calculated, we could calculate I3 replacing I1 with (5) in (4) 
3 − s2 2{[V s1 − 2V s2)/ R3 2) ]/[(R1 2 R2R2)/ R3 2))]})/ R3 2) I = ( V + R − ( R − ( + R + R + ( − ( + R − ( + R
…(6) 
For I2 was easier, replacing (5) and (6) in (2) 
(not indicated for space reasons)2 3 1 I = I + I 
Then, for the voltages we used the Ohm’s Law 
n nRnV = I 
Getting… 
1 1R1 V = I 
2 2R2V = I 
3 3R3V = I 
And for the last part, calculate the potentials, we just applied the definition 
n nV nP = I 
Giving us the table of above. 
 
 
 
 
 
 
 
 
 
 
 
 
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Evidence 
 
 
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