Practica

Páginas: 8 (1792 palabras) Publicado: 21 de agosto de 2012
INSTITUTO POLITECNICO NACIONAL
ESCUELA SUPERIOR DE CÓMPUTO

“Practice 5”

Practice N°2 “Voltage and Current Divider”

Subject: Fundamental Analysis of Circuits
Teacher: Durán Camarillo Edmundo René
Team: 9
Students: Solís Gonzales Fernanda
Guerrero Gómez Jorge Raúl

Group: 1CM1
Mexico D,F April 7 of 2012

INDEX
N° | LESSONS | Page |
1 | BODY REPORT | 3 |
2 | CALCULATIONS| 8 |
3 | SIMULATIONS | 9 |
4 | CONCLUSIONS | 10 |

BODY REPORT
Objective
The student will identify the circuit known as "voltage divider circuit" in its simplest form. Understand the concept of "voltage divider" and will make the comparison between the calculated values ​​with measured values ​​in each of the associated elements in the circuit in practice. Understand the usefulness ofthese circuits in both the analysis of more complex networks and in applications where accuracy is required or high values ​​of current consumption.
Material team

The laboratory proportionate:

* 1 Digital Multimeter
* 1 Battery of voltage
The students need get:
* 4 points caiman - caiman
* 4 tips banana - alligator
* 1 breadboard
* 2 resistors 1KΩ a ½ de W
* 1resistor 470Ω a ½ W
* 1 resistor 560Ω a ½ W
* 2 resistors 2.2KΩ a ½ W
* 1 resistor 3.3KΩ a ½ W
* 1 potentiometer 10K
* wire
* cutting pliers
* nose pliers

Theoretical framework
A circuit such as that displayed in the figure 1 is called "voltage divider circuit" because in each divider can register a voltage drop which is a submultiple exact value of the value of thesource. The multiplication factor which defines the submultiple is obtained as a function of the resistors that form the circuit, as follows. The current I in the grid circuit can only be obtained by Ohm's law with the following expression:

Figure 1
Figure 1

Current known, one can find an expression for the voltage and each resistor in absolute value.

Of the form may be obtainedthe same voltage drop in the other resistors.

With the expression already obtained for the current can also deduce the nodal voltages expressions in nodes N1, N2, and N3 with respect to the common node and ground, as illustrated in the figure below.

Figure 2 illustrates how to measure voltage in each of the two nodes with respect to common node, node zero or reference node.

Figure 2
Figure2

A circuit such as that displayed in Figure 3 is called the current divider as the total current I is divided across each resistor. According to Ohm's law we have to

Figure 3
Figure 3

Solving for the voltage we

Substituting for each stream according to Ohm's law

Development of practice
1. Voltage Divider.
Figure 4
Figure 4
1) Adjust the voltage source to 10V, and adjust thesupply current to the maximum.
2) Connections on the tablet without energizing the source, construct the circuit shown in Figure 4.

3) Make voltage measurements VR1, VR2, VR3, V1, V2 and V3 recording the results in Table 1.
Data | Theoretical Value | Calculated Value | Error V ó I |
Vr1 | 4.9 V | 4.9 V | 0 |
Vr2 | 2.3 V | 2.32 V | 0.02 |
Vr3 | 2.7 V | 2.8 V | 0.1 |
V1 | 10 V | 10.05 V| 0.05 |
V2 | 5 V | 5.12 V | 0.12 |
V3 | 2.7 V | 2.8 V | 0.1 |
Total | 27.6 V | 27.99 V | 0.39 |
Table 1
Table 1

4) From Figure 2 if R1 = R2 = 1K and 2.2K that value is required to have a voltage V3 = 10V Vs = 5V if, construct the circuit and if necessary to have a voltage make an arrangement of resistors or use a potentiometer and check.

R3 = 3,117 KΩ

5 ) Construct a voltagedivider circuit as shown in Figure 2 with a Vs = 10V so as to obtain the voltage values ​​FOLLOWING requested in the table No. 2.
Data | R1 | R2 | R3 |
V2 = 5V | 1KΩ | 9.94KΩ | 10.70KΩ |
V3 = 3V | 1KΩ | 9.99KΩ | 10.70KΩ |
Table 2
Table 2

2. Current Divider.
1) Adjust the voltage source to 10V, and adjust the supply current to the maximum.
Figure 5
Figure 5

2) Connections on the...
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