Practice 5_AFC - Edgar Bello

Vista previa del material en texto

Index 
Objective 3 
Equipment 3 
Material 3 
Theoretical introduction 4 
Development of the practice 5 
 ​1.- Voltage divider 4 
 2.- Current divider. 9 
Questionnaire 10 
Conclusions 11 
Calculations 12 
 
 
2 
 
Objective: 
The student will identify the circuit known as “Voltage Divider Circuit” in its simplest 
form. Will understand the concept of “Voltage Division” and will make the comparison 
between the measured values in each of the measured values ​​in each of the 
associated elements in the practice circuit. Will understand the utility of these circuits 
both in the analysis of more complex networks and in applications where accuracy 
and high values ​​in current consumption are not required. 
 
Equipment provided by the laboratory 
● 1 digital multimeter 
● 1 variable voltage source 
 
Material 
● 4 alligator-alligator tips 
● 4 banana-alligator tips 
● 1 protoboard 
● 2 1K 𝝮 ½ W resistors 
● 1 470 𝝮 ½ W resistor 
● 1 560 𝝮 ½ W resistor 
● 2 2.2K 𝝮 ½ W resistors 
● 1 3.3K 𝝮 ½ W resistors 
● 1 10K 𝝮 potentiometer 
 
 
 
 
3 
 
“Current and voltage divider” 
Theoretical Introduction. 
A circuit like the one shown in the Figure 1 is called “Voltage Divider Circuit” due to 
in each resistor we can register a voltage drop having an exact submultiple value of 
the value of the source. 
The multiplication factor that defines the submultiple is obtained as a function of the 
resistors that they form the circuit, in the following way. 
The current I in the only mesh of the circuit can be obtained by Ohm’s Law with the 
following expression: 
 
 
 
 
 
Figure 1 
With a known current, a expression for the voltage in each resistor can be found. 
 
In the same way, the voltage drop in the other resistors can be obtained. 
 
With the already obtained expression for the current we can also deduce the nodal 
voltage expressions, in the nodes N1, N2 and N3, with respect to the common node 
or earth, as illustrated in the following figure. 
4 
 
Figure 2 illustrates how to measure the voltage at each of the nodes, with respect to 
the common node, node zero, node reference. 
 
 
Figure 2 
 
A circuit like the shown in figure 3 is called Current divider due to the total current I 
is divided for each resistor. In accordance with Ohm’s law, we have: 
 
Figure 3 
Solving voltage we have 
 
5 
 
Replacing for each current in accordance with Ohm’s law 
 
 
Practice development 
Voltage divider 
1. Adjust the voltage source to 10 V, and adjust the current 
source tho the maximum value. 
2. Without energizing the source, build the circuit shown in 
the figure 4 in a protoboard. 
 
3. Make the voltage measurements V​R1​, V​R2​, V​R3​, V​1​, V​2 and 
V​3​, and record the results in the table 1. 
Fact Theoretical value Simulation value Error ∆V or ∆I 
V​R1 4.9261V 4.926V 0.0001V 
V​R2 2.3152V 2.315V 0.0002V 
V​R3 2.75862V 2.759V 0.00038V 
V​1 10V 10V 0.0V 
V​2 5.07389V 5.074V 0.00011V 
V​3 2.758606V 2.759V 0.0003940V 
I total 4.92611mA 4.926mA 0.00011mA 
 
Table 1. ​In which are record the voltage voltages of the voltage divider circuit, 
measured and calculated. The error it represents is also logged. 
 
 
 
6 
 
 
Simulation. 
 
 
 
4. From figure 2 if R​1 = 1k𝝮 and R​2 = 2.2k𝝮 what value of R​3 is necessary to 
have a voltage V​3 = 5V if Vs = 10 V, build the circuit and if necessary do a 
resistor fix or use a potentiometer and check. 
R3 in theory =3.2K𝝮 
R3 in simulation = 3.3 K𝝮 
 
 
 
 
 
 
 
 
 
7 
 
Simulations. 
 
 
5. Build a voltage divider circuit like the one shown in Figure 2 with Vs = 10V so 
that you get the following requested voltage values in table No. 2 (the 
resistance values ​​must be calculated before enter the laboratory to be able to 
buy them, propose a value in R1 and calculate the other two) 
 
Fact R1 in theory R1 in 
simulation 
R2 R3 
V2​= 5V 1K𝝮 1K𝝮 400𝝮 600𝝮 
V3​= 3V 1K𝝮 1K𝝮 400𝝮 600𝝮 
Table 2 
 
 
 
 
 
 
 
 
8 
 
Simulations. 
 
 
current divider. 
1.- Adjust the voltage source to 10V, and adjust the 
source current to maximum. 
2.- On the connection board and without energizing the 
source, build the circuit shown in figure 5. 
 
3.- Make the current measurements IR1, IR2, IR3, 
IR4. Using the total current IR1 for the formula 
from the current divider calculate the values for IR2, 
IR3, IR4 recording the results in Table 3. 
 
 
9 
 
 
 
 
Datum Simulation value Theoretical value Error ∆I 
I​R1 6.37 mA 6.373626 mA 0.003626 mA 
I​R2 3.626 mA 3.62637 mA 0.00037 mA 
I​R3 1.648 mA 1.64835 mA 0.00035 mA 
I​R4 1.099 mA 1.0989 mA 0.0001mA 
I Total 12.743 mA 12.7472475 mA 0.0042475 mA 
Table 3 In which they record the voltages obtained from the current, measured 
and calculated. The error that represents 
 
∆I = | I​Medido​ - I​Calculado​ | 
 
Simulations. 
 
 
Questionnaire 
1.- What is the reason of the existence of an error or deviation of the value measured 
respect the value calculated? 
10 
 
The main error factor in this type of experiments is that the measures are not 100% 
accurate, this makes the theoretical results be a bit different to the ones measured at 
the end. 
 
2.- Which is the utility of the ​voltage divider ​for the electric circuits analysis? 
In some cases, if makes easier to obtain circuit data in a faster way, because, in 
simple words, this method is an abbreviation of Ohm’s Law for serial circuits. 
 
3.- Which is the utility of the ​current divider ​for the electric circuits analysis? 
Like in the other case, it makes easier to obtain data of the circuit (especially of the 
current), and it’s is an adjustment of Ohm’s Law, in this case for parallel circuits. 
 
4.- Can the voltage and current divider be applied to a mayor number of 
resistances? 
Yes, it can, we just need to make the adjustments of Ohm’s Law. 
 
5.- If the voltages in each node were required with specific predetermined values, 
what should be done to obtain the values? 
We need to do the same calculations as exercises 4 and 5 (better explained below) 
 
Conclusions 
 
Bello Muñoz Edgar Alejandro 
The current and voltage dividers are fundamental elements for the analysis of 
electrical circuits, since they greatly facilitate the analysis of more complex circuits. 
They are especially useful because it is not necessary to know at all the values ​​of 
the voltage (current divider) or current (voltage divider) variables to be able to 
calculate the value of specific resistors, which is one of the most common problems 
at the time of analyze and solve circuits. 
 
 
 
11 
 
Núñez González Ángel Daniel 
The voltage and current divider is a very complete and easy topic to apply to both 
direct and continuous current circuits. These two divisors taught, shown and applied 
to this practice give you an idea of how easy it is to obtain these data without having 
to complicate more in the resolution of the circuit. They are very simple formulas to 
learn and if you do not learn them, just guide yourself by the concept of the formula, 
this will help you a lot if you do not want to learn the formulas because they are 
directly related to Ohm's Law and consequently it is easy and simple to make the 
substitutions directly in them. 
 
López Gracia Angel Emmanuel 
In this practicewe learned a new technique to analyze circuits, seems that this tool 
could be really useful i the future, specially because it reduces Ohm’s Law a lot and 
makes calculus easier to specific parts of the circuit, as long as you understand what 
you are doing, you can find this useful and really fast. 
The planning of the math expressions was a bit long, but I think that went pretty 
aceptable. I honestly learned much in this practice. 
 
Calculations 
For the first part (subsection 3) we need to calculate ​I​, which is going to be 
calculated by Ohm’s Law for serial circuits ΣRV = I 
Then we have that , an expression that will be useful below, ins/(R1 2 3)I = V + R + R 
other subsections, for that reason we’ll call it ​ecuation 1​. 
In this subsection 0/(1000 70 60) .00492611 AI = 1 + 4 + 5 = 0 
And using voltage divider for each resistance an node, we can calculate each one 
voltage. 
(R1 1I 000V = R = 1 .00492611) .92611 V0 = 4 
And, in the same way for each resistance, will give us the results registered before. 
For the nodes have been a bit more difficult, each case was different: 
For ​V1​ was easy, this because was a node with all resistances that we used to 
calculate the general conditions of the circuit, so .1 R1 2 3)I sV = ( + R + R = V 
12 
 
But for the others, we had to recalculate each one: 
2 R2 3)IV = ( + R 
And for ​V3 
3 3IV = R 
Giving us the results before registered. 
 
For subsection 4, we reused the last equations, setting them in this form: 
3 3I vV = R = 5 
0/(1000 200 3)I = 1 + 2 + R 
Joining them… 
v 3(10/(1000 200 3) 0R3/(3200 3)5 = R + 2 + R = 1 + R 
(3200 3) 0R3 6000 R35 + R = 1 = 1 + 5 
6000 R31 = 5 
=3.2K𝝮3 200R = 3 
 
For subsection 5, we first calculated the relations between the resistances, also with 
the help of before equations. 
=5v ...12 R2 3)IV = ( + R 
=3v ...23 3IV = R 
 ...30/(R1 2 3)I = 1 + R + R 
 ...4R1 1IV = R 
 ...5R2 2IV = R 
 ...6R3 3IV = R 
⊦ 
Using the 3rd in the 1st: 
5 = R1+R2+R3
10(R2+R3) 
1 2 3 R2 R3R + R + R = 2 + 2 
...71 2 3R = R + R 
We do the same steps of before but with equations 2 and 3… 
3 = 10R3R1+R2+R3 
R1 R2 R3 0R33 + 3 + 3 = 1 
13 
 
R1 R3 R23 = 7 − 3 
...81 R3 2R = 3
7 − R 
Then we join 7 and 8 
2 3 1 1R + R = R = R = R3 23
7 − R 
R2 R32 = 3
4 
R2 R36 = 4 
.5R2 31 = R 
 
Now we can decide an arbitrary value to R1 in order to set R2 and R3, in this case, 
we used 1K𝝮, and gave us that R2 should be 400𝝮 and R3 600𝝮. 
 
And for the subsection 3 for the Table 3, we used a current divider 
We first calculated ​I ​with Ohm’s Law 
0 (1000 )1 = I + 1+ +11000 12200 13300
 
(this is also ​IR1​).006373626AI = 101000+ 1
+ +11000
1
2200
1
3300
= 294550 ≈ 0 
Then we calculate VR1 
R1 .37vV = 1000V s1000+ 1
+ +11000
1
2200
1
3300
= 100001000+ 1
+ +11000
1
2200
1
3300
≈ 6 
Then VR2,3,4 
VR2,3,4= .626v
V s1
+ +11000
1
2200
1
3300
1000+ 1
+ +11000
1
2200
1
3300
=
10
+ +11000
1
2200
1
3300
1000+ 1
+ +11000
1
2200
1
3300
= 91
330 ≈ 3 
And at the end we can calculate the currents in the rest of the resistances (​I2,I3,I4​): 
2 000 .003626AI = 91
330 ÷ 1 ≈ 0 
3 200 .001648AI = 91
330 ÷ 2 ≈ 0 
4 300 .001098AI = 91
330 ÷ 3 ≈ 0 
 
14