Practice 2_AFC - Edgar Bello

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Index 
Objective 3 
Equipment 3 
Material 3 
Practice development 3 
1.- Voltage dependence 3 
2.- Resistance dependency 5 
3.- Calculation of the power in the resistors 8 
Conclusions 9 
Calculus and simulations 10 
Evidence 12 
 
 
2 
“Ohm’s Law” 
Objective: The student will understand and handle the proper interpretation of ohm's 
law, so that at the end of the practice, the student will be able to: 
• Calculate the voltages, currents, powers and resistances that are present in a 
circuit 
• Understand the behavior of the current with respect to the voltage. 
• Understand the behavior of the current with respect to the resistance. 
• Deduce Ohm's law. 
 
Equipment: 
1 digital multimeter. 
1 Variable voltage source. 
 
Material: 
1 Protoboard 
Resistors from 1KΩ to ¼ watt and 1 Ohm at 1 watt. 
Connection wire for the protoboard. 
4 banana-alligator tips. 
2 points alligator-alligator. 
Cutting and tip tweezers. 
1 potentiometer of 2.5KΩ or greater. 
 
Practice development 
1. Voltage dependence. 
Without turning on the voltage source yet, set the potentiometer value to 2.5KΩ. 
Build the circuit illustrated in figure 1 on the protoboard. Once built the circuit、turn 
on the voltage source, and vary its value from zero to 15 V, according to table 1. 
 
3 
Figure 1. 
Table 1 
Voltage source (V) Current value (measured) 
(mA) 
Current value (calculated) 
(mA) 
0 0 0 
1 0.286 0.2857 
2 0.572 0.5714 
3 0.858 0.8571 
4 1.144 1.1428 
5 1.429 1.4285 
6 1.715 1.7142 
7 2.001 2 
8 2.287 2.2857 
9 2.572 2.2514 
10 2.858 2.8571 
11 3.144 3.1428 
12 3.430 3.4285 
13 3.716 3.7142 
14 4.002 4 
15 4.288 4.2857 
4 
Simulation of “​Voltage dependence​“: 
 
 
According to the table 1, draw the next graph 
Figure 1 
2.-Resistance dependency 
With the voltage source turned off, set the potentiometer value to 0Ω. Assemble the 
5 
circuit illustrated in figure 2 on the protoboard. Once built the circuit, turn on the 
voltage source and set it to 15 V; subsequently vary the value of potentiometer 
according to as requested in the following table: 
 
 
Figure 2 
Remember that to measure resistance you have to turn off the voltage source, in 
case of being impossible to do it, disconnect the potentiometer. 
 
 
Potentiometer 
value (​Ω) 
Total resistance 
(Pot.+R) (​Ω) 
Measured current 
value (mA) 
Calculated current 
value (mA) 
0 1000 15.333 15 
250 1250 12.220 12 
500 1500 10.150 10 
750 1750 8.680 8.5714 
1000 2000 7.570 7.55556 
1250 2250 6.725 6.6 
1500 2500 6.050 6 
1750 2750 5.485 5.4545455 
2000 3000 4.872 5 
2250 3250 4.510 4.6153 
2500 3500 4.170 4.2857 
6 
Simulation of “​potentiometer value and R1 measurement​” 
 
Simulation of “​Resistance dependency​” 
 
Draw a graph according to the previous table 
7 
 
 
3. Calculation of the power in the resistors. 
Before connecting the source, set it to 1 volt, then turn it off and without using the 
protoboard, assemble the circuit illustrated in figure 3, for this circuit use the 
1KΩ resistance at ¼ watt, once armed turn on the voltage source. 
Figure 3 
Which is the value of the current? I = 0.9017 mA 
What is the value of the power that dissipates the resistance? P = 0.9017 mW 
 
Again, assemble the previous circuit, but now using the resistance of 1Ω to 1 watt, 
before connecting the voltage source make sure that it is adjusted at 1 volt and that 
the ammeter is at the maximum scale. 
 
8 
What is the value of the current? I = 256.08 mA 
What is the value of the power that dissipates the resistance? P = 256.08 mW 
 
Remember that protoboard is not used in this circuit. 
 
 
Conclusions 
Tell from these experiments how Ohm's law would be determined, in addition 
of the effect of power on resistive elements. 
 
● Bello Muñoz Edgar Alejandro 
As we can see, the importance of Ohm's law is that in a circuit, we can know the how 
the value of resistance, voltage and current will change if we vary any of the other 
parameters, so, the measured value of a component in the circuit can be checked in 
order to avoid connection mistakes. 
Even though it is simple, the correct knowledge about ohm's law will let us 
manipulate correctly different components and measurement devices, to optimize 
the design and structure of electrical circuits. 
 
● Núñez González Ángel Daniel 
 
Ohm's law is one of the most essential laws of electronics, being one of the simplest 
that can be found in this branch of knowledge is very good in practice, because from 
there you can make different analyzes for other more complex formulas . In this 
practice it can be seen that Ohm's law is checked by coinciding with theoretical, 
simulation and practical calculations, varying by a few tenths, but that is because the 
equipment has minimal variations, the resistances are a bit worn out or there are 
parallax errors. 
 
● López Gracia Angel Emmanuel 
In this practice we managed to control the resistance in a more interactive way, a 
way that, in my point of view, is really funny, the total resistance was difficult to make 
precise because the impression of the potentiometer, but at the end we did get 
similar results to the ones that we had already estimated, we learned the potence 
property. 
I really wonder how Ohm’s law was discovered, but I think it could be a result of 
seeing how resistance affected the functionality of circuits and how the current was 
affected for this resistances. 
 
 
9 
Calculus and simulations 
For the both parts we made use of the Ohm’s law 
RV = I 
 V /R∴ = I 
For the first part, we replaced V with the given values and R with (1000+2500), which 
corresponds to the resistance and the potentiometer, respectively, giving us the next 
results: 
 
Voltaje (V) Current (mA) 
0 0 
2 0.2857 
3 0.5714 
4 0.8571 
5 1.1428 
6 1.4285 
7 1.7142 
8 2 
9 2.2857 
10 2.2514 
11 2.8571 
12 3.1428 
13 3.4285 
14 3.7142 
15 4 
 4.2857 
 
 
10 
In the second part we replaced R with (1000+the resistance given) and V with 15v, 
obtaining the next results: 
 
Potentiometer value (​Ω) Total resistance 
(Pot.+R) (​Ω) 
Calculated current value 
(mA) 
0 1000 15 
250 1250 12 
500 1500 10 
750 1750 8.5714 
1000 2000 7.55556 
1250 2250 6.6 
1500 2500 6 
1750 2750 5.4545455 
2000 3000 5 
2250 3250 4.6153 
2500 3500 4.2857 
 
And for the last part we used the next expressions: 
/RI = V 
IP = V 
Replacing V with 1v, and for R we used 1k​Ω​ and 1​Ω​ for each test, that, after making 
numbers, we got that I1=0.001, I2= 1, P1=0.001w and P2=1w 
11 
 
Evidence 
 
12