Practice 1_AFC - Edgar Bello

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Index 
Objective 3 
Equipment 3 
Material 3 
I.-Theoretical introduction 3 
II.- Development of the practice 4 
1.-Ohmmeter use 4 
2.- Voltmeter use 5 
3.- Ammeter use 6 
Questionnaire and conclusions 8 
Calculus and simulations 10 
 First Part 10 
 Second Part 12 
Evidence 15 
 
 
2 
“Use of the Ohmmeter, voltmeter and the ammeter on DC measurements” 
Objective: The student will understand the proper handling of the measuring 
instruments, so at the end of the practice will be able to: use the digital ohmmeter, 
the digital voltmeter and the digital ammeter properly. 
Equipment: 
1 Digital multimeter. 
1 Variable voltage source. 
4 banana-alligator tips. 
2 alligator-caiman tips. 
Material: 
1 Protoboard. 
1 1KΩ resistor at ¼ watt. 
 1 Resistor of 560Ω to ¼ watt. 
 1 Resistor of 680Ω to ¼ watt. 
1 Resistor of 330Ω at ¼ watt. 
 Wires for connections. 
 
I.- Theoretical introduction 
The electric current or voltage can be measured by ammeters or voltmeters, Figure 1 
shows 2 common forms of meters; one of the analog meters has an indicator pointer 
that moves on a calibrated scale whose angular deflection depends on the 
magnitude of the variable it measures. While the other is a digital meter which shows 
a series of digits on the screen, indicating the magnitude of the variable that it 
measures. Figure 2 shows the symbols of the voltmeter and the ammeter that are 
used in the diagrams of electrical circuits. 
 
Figure 1. a) Analog meter Figure 2. Symbol of each meter 
 b) Digital meter 
3 
 
To measure the electric current in the branch of a circuit, that branch must be 
opened and the ammeter must be inserted in such a way that it is connected in 
series with the element from which you want to know its current. It is said that two 
elements are in "series" if one end joins with another’s end, and does not exist some 
driver connected to that union. The electric current that flows through this trajectory 
necessarily passes through the current meter (Ammeter). To measure the voltage 
between two points, the voltmeter is connected in parallel with the electronic device 
that you want to know the voltage drop. Two elements of the two terminals that are 
connected in parallel if the terminals of one are connected to the terminals the other. 
It does not matter if in those unions there is or not another Connection. The essential 
characteristic of a parallel connection, is that through the elements exists the same 
voltage. 
II.- Development of the practice 
1. Ohmmeter use 
Without energizing any circuit element, measure the resistance value presented by 
each resistor, as shown in the Figure 3 and fill the Table 1. 
 
Figure 3. Connection with the Ohmeter 
 
Table 1. Resistive value measurement 
Resistor Digital ohmmeter 
measurement 
Value using color 
code 
Resistance Colors 
R1 994.2 Ω 1000 Ω Brown/Black/Red/
Gold 
R2 329.55 Ω 330 Ω Orange/Orange 
/Brown/Gold 
4 
R3 550.7 Ω 560 Ω Green/Blue/Brown
/Gold 
R4 671.17 Ω 680 Ω Blue/Gray/Brown/
Gold 
 
 
2. Voltmeter use 
Figure 4 shows how the voltage in an element should be measured. With the voltage 
source off, assemble the circuit of figure 5. And once the circuit is armed turn on the 
voltage source and fill the table 2 
 
Figure 4. Example of a connection with the Voltmeter 
 
Figure 5. Serial Circuit 
 
 
 
5 
Table 2 Voltage measurement (Practical) 
Voltage source Digital multimeter 
Voltage in R1 and 
R2 
Voltage in R1 Voltage in R2 
E = 1 V 1.0065 V 0.754 V 0.250 V 
E = 2 V 2.0259 V 1.5198 V 0.5046 V 
E = 3 V 3.0208 V 2.2666 V 0.7526 V 
E = 4 V 4.0148 V 3.0140 V 1.008 V 
E = 5 V 5.0223 V 3.7703 V 1.2528 V 
E = 6 V 6.0480 V 4.5393 V 1.5081 V 
E = 7 V 7.0630 V 5.3010 V 1.7630 V 
E = 8 V 8.0220 V 6.0200 V 2.0028 V 
E = 9 V 9.0260 V 6.7710 V 2.2527 V 
E = 10 V 10.0220 V 7.5160 V 2.5039 V 
E = 11 V 11.0550 V 8.2890 V 2.7635 V 
E = 12 V 12.0300 V 9.0190 V 3.0097 V 
 
3. Ammeter use 
Figure 6 shows how the ammeter should be connected for an electric current 
measurement in an element. 
 
Figure 6. Example of a connection on the ammeter 
6 
 
Figure 7. Parallel circuit 
 
Table 3 Current measurement ​(Practical) 
Voltage source Digital multimeter 
Current through 
R1 and R2 
Current through 
R1 
Current through 
R2 
E = 1 V 2.423 mA 1.579 mA 1.257 mA 
E = 2 V 4.994 mA 3.635 mA 2.603 mA 
E = 3 V 9.874 mA 5.366 mA 3.908 mA 
E = 4 V 13.066 mA 7.216 mA 5.915 mA 
E = 5 V 16.567 mA 9.082 mA 7.444 mA 
E = 6 V 19.614 mA 10.820 mA 8.935 mA 
E = 7 V 22.991 mA 12.503 mA 10.476 mA 
E = 8 V 26.167 mA 14.309 mA 11.954 mA 
E = 9 V 29.586 mA 16.241 mA 13.464 mA 
E = 10 V 32.089 mA 17.920 mA 15.017 mA 
E = 11 V 36.450 mA 19.704 mA 16.413 mA 
E = 12 V 39.599 mA 21.426 mA 17.929 mA 
 
 
 
7 
Questionnaire 
 
1.- What is the characteristic of a series circuit? 
 
-The current is the same throughout the circuit and the total voltage is equal to the 
sum of the voltages. 
 
2.- What is the characteristic of a parallel circuit? 
 
-The total current equals the sum of the currents and the voltage is the same 
throughout the circuit. 
 
3.- Why should an ammeter not be connected in parallel? 
 
-Because when the series is connected, the current flow passes through the 
ammeter and is measured without alterations, if the current is connected in parallel, it 
will be diverted causing measurement errors. 
 
4.- Why should the circuit be de-energized when measuring the resistance of an 
electronic circuit? 
Because it can cause measurement errors and even make damage in the 
measurement devices. 
 
Conclusions 
 
● Bello Muñoz Edgar Alejandro 
 
The completed practice was very important because the whole team learnt how to do 
the measurements of voltage, current and resistance in the correct way, avoiding 
mistakes and obtaining the expected results. 
Nevertheless, in a practical way the values ​​of resistance and voltage are not exactly 
the same as those calculated theoretically, certain variations between the theoretical 
and the real originate, by variations in the components and measuring devices 
 
 
● Núñez González Ángel Daniel 
 
It is true that being the first practice is something simple, but important to learn these 
basic concepts and subsequently not have any errors in the conceiving of the 
circuits. Measuring the voltage and current in the exercises done correctly is 
important to know so as not to damage the equipment since this in the labor field is 
something terrible. It is important to say that at the time of practice there will always 
8 
be minimal errors of calculation because the material wears out over time and that is 
why there are small differences between theoretical calculations and practice. 
 
● López Gracia Angel Emmanuel 
 
The development of the practice was not really difficult, our team worked very good 
and with good communication; and well, the teacher provided us with some 
knowledge that helped us out, like how to cut a cable, doing “bridges” and some 
good-engineer habits. 
 
The values gotten in the practical part are not as exact as the theoretical values 
were, but they were really near, since the theoretical calculations do not contemplate 
all the variables that could exist in real life, it might be some errors in the precision of 
the measures, I also learned the importance ofbeware of doing, in other words, 
before trying to make something (like take measures) it is important to first know 
what are you doing, the electrical part of sciences can be a bit delicate sometimes, 
but as we keep practicing, we keep learning, and with all, the objective, learn how to 
use the measurement devices and how resistances work in simple series and 
parallel circuits, were achieved, so the practice, in my opinion, was a total success. 
 
 
9 
Calculus and simulations 
 
First Part 
Before using the voltmeter, we decided to predict the results, and for achieve that we 
made the next list of equations: 
First we used Ohm’s law 
V=IR 
We inferred that the current will be obtained with 
I=V/R 
I=Vi/(R1+R2) 
Using Vi as the initial voltage and (R1+R2) as the total resistance of the circuit. 
With ​I ​defined, we could use it in a new calculation 
V=IR 
Vn=[Vi/(R1+R2)]Rt 
Using Vn as the new voltage and Rt as the resistance related with the new voltage. 
Vn=[Vi/(R1+R2)](R1+R2) 
Vn=[Vi/(R1+R2)]R1 
Vn=[Vi/(R1+R2)]R2 
Also taking R1=1KΩ and R2=330Ω. 
After all the respective calculations we obtained the next table 
 
Table 1.1 Voltage measurement (Theoretical) 
Voltage source Theoretical approximation 
Voltage in R1 and 
R2 
Voltage in R1 Voltage in R2 
E = 1 V 1 V 0.751879699 V 0.248120301 V 
E = 2 V 2 V 1.503759398 V 0.496240602 V 
E = 3 V 3 V 2.255639098 V 0.744360902 V 
E = 4 V 4 V 3.007518797 V 0.992481203 V 
E = 5 V 5 V 3.759398496 V 1.240601504 V 
E = 6 V 6 V 4.511278195 V 1.488721805 V 
E = 7 V 7 V 5.263157895 V 1.736842105 V 
E = 8 V 8 V 6.015037594 V 1.984962406 V 
E = 9 V 9 V 6.766917293 V 2.233082707 V 
E = 10 V 10 V 7.518796992 V 2.481203008 V 
10 
E = 11 V 11 V 8.270676692 V 2.729323308 V 
E = 12 V 12 V 9.022556391 V 2.977443609 V 
 
 
And for the 3th part of the practice, we simulated the circuit just as shown down 
 
Simulation of “Voltage measurement” 
 
 
Getting the next table 
 
Table 1.2 Voltage measurement (Simulation) 
Voltage source Simulation results 
Voltage in R1 and 
R2 
Voltage in R1 Voltage in R2 
E = 1 V 1 V 0.752 V 0.248 V 
E = 2 V 2 V 1.504 V 0.496 V 
E = 3 V 3 V 2.256 V 0.744 V 
E = 4 V 4 V 3.007 V 0.993 V 
E = 5 V 5 V 3.759 V 1.241 V 
E = 6 V 6 V 4.511 V 1.489 V 
11 
E = 7 V 7 V 5.263 V 1.737 V 
E = 8 V 8 V 6.015 V 1.985 V 
E = 9 V 9 V 6.767 V 2.233 V 
E = 10 V 10 V 7.519 V 2.481 V 
E = 11 V 11 V 8.271 V 2.729 V 
E = 12 V 12 V 9.022 V 2.978 V 
 
 
Second part 
 
For the correct use of the ammeter we decided to also make a previous 
approximation of the results with the next equations 
Using again Ohm’s law 
V=IR 
And replacing R with 1/[(1/R1)+(1/R2)] (added in that way because they are 
resistances in parallel) we get 
V=I/[(1/R1)+(1/R2)] 
I=V[(1/R1)+(1/R2)] 
And in order to calculate the other currents by separate we used Ohm’s law in each 
case individually 
V=IR 
I=V/R 
With V as the voltage of the source and R for the resistances individually (using 
R1=560Ω and R2=680Ω) getting the next results. 
Table 2.1 Current measurement ​(Theoretical) 
Voltage source Theoretical approximation 
Current through 
R1 and R2 
Current through 
R1 
Current through 
R2 
E = 1 V 3.25630252 mA 1.78571429 mA 1.47058824 mA 
12 
E = 2 V 6.51260504 mA 3.57142857 mA 2.94117647 mA 
E = 3 V 9.76890756 mA 5.35714286 mA 4.41176471 mA 
E = 4 V 13.0252101 mA 7.14285714 mA 5.88235294 mA 
E = 5 V 16.2815126 mA 8.92857143 mA 7.35294118 mA 
E = 6 V 19.5378151 mA 10.7142857 mA 8.82352941 mA 
E = 7 V 22.7941176 mA 12.5 mA 10.2941176 mA 
E = 8 V 26.0504202 mA 14.2857143 mA 11.7647059 mA 
E = 9 V 29.3067227 mA 16.0714286 mA 13.2352941 mA 
E = 10 V 32.5630252 mA 17.8571429 mA 14.7058824 mA 
E = 11 V 35.8193277 mA 19.6428571 mA 16.1764706 mA 
E = 12 V 39.0756303 mA 21.4285714 mA 17.6470588 mA 
 
And, as said before, we did a simulation of the circuit (shown down) 
 
Simulation of “Current measurement” 
 
 
 
 
 
13 
 
Table 2.2 Current measurement ​(Simulation) 
Voltage source Simulation results 
Current through 
R1 and R2 
Current through 
R1 
Current through 
R2 
E = 1 V 3.256 mA 1.786 mA 1.471 mA 
E = 2 V 6.512 mA 3.571 mA 2.941 mA 
E = 3 V 9.768 mA 5.357 mA 4.412 mA 
E = 4 V 0.013 A 7.143 mA 5.882 mA 
E = 5 V 0.016 A 8.928 mA 7.353 mA 
E = 6 V 0.020 A 0.011 A 8.823 mA 
E = 7 V 0.023 A 0.012 A 0.010 A 
E = 8 V 0.026 A 0.014 A 0.012 A 
E = 9 V 0.029 A 0.016 A 0.013 A 
E = 10 V 0.033 A 0.018 A 0.015 A 
E = 11 V 0.036 A 0.020 A 0.016 A 
E = 12 V 0.039 A 0.021 A 0.018 A 
 
 
14 
Evidence
 
15