Capítulo 9 - Análisis de Ramas, Lazos y Nodos (1).ppt

Aálisis de Ramas, Lazos , Circuitos Electricos

1. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved

2. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Circuit analysis methods in Chapter 9 require use of
simultaneous equations.
Simultaneous Equations
To simplify solving simultaneous equations, they
are usually set up in standard form. Standard form
for two equations with two unknowns is1,1 1 1,2 2 1
2,1 1 2,2 2 2
a x a x b
a x a x b


coefficients
variables
constants

3. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Simultaneous Equations 
A A B
B A B
10 270 1000( ) 0
1000 680 6 0
I I I
I I I
    
   
A circuit has the following equations. Set up the
equations in standard form.AB
AB
1270 1000 10
1000 1680 6
II
II

   
Rearrange so that variables and their coefficients
are in order and put constants on the right.

4. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Solving Simultaneous Equations
Three methods for solving simultaneous equations
are
Algebraic substitution
The determinant method
Using a calculator

5. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Solving Simultaneous Equations
Solve for I
Ausing substitution.BA
1.270 0.010II
Solve for I
Bin the first equation:
Substitute for I
B into the second equation:AA
1000 1680(1.270 0.010) 6II    
Rearrange and solve for I
A.
I
A= 9.53mAA
1134 10.8I AB
AB
1270 1000 10
1000 1680 6
II
II

   

6. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Solving Simultaneous Equations
If you wanted to find I
Bin the previous example,
you can substitute the result of I
Aback into one of
the original equations and solve for I
B. Thus,AB
B
B
1270 1000 10
1270(9.53 mA) 1000 10
II
I
I



2.10 mA

7. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Solving Simultaneous Equations
The method of determinantsis another approach to
finding the unknowns. The characteristic determinant is
formed from the coefficients of the unknowns.AB
AB
1270 1000 10
1000 1680 6
II
II

   
Write the characteristic determinant for the
equations. Calculate its value.1270 1000
1000 1680




1.134

8. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Constants
Characteristic
determinant
Unknown
variable
SummarySummary
Solving Simultaneous Equations
To solve for an unknown by determinants, form the
determinant for a variable by substituting the constants
for the coefficients of the unknown. Divide by the
characteristic determinant.1 12
2 22
1
11 12
21 22
ba
ba
x
aa
aa






 11 1
21 2
2
11 12
21 22
ab
ab
x
aa
aa







To solve for x
2:

9. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Solving Simultaneous Equations
Solve the same equations using determinants:AB
AB
1270 1000 10
1000 1680 6
II
II

    A
10 1000
6 1680
1270 1000
1000 1680
I








9.53mAB
1270 10
1000 6
1270 1000
1000 1680
I







 2.10mA

10. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Many scientific calculators allow you to enter a set of
equations and solve them “automatically”. The
calculator methodwill depend on your particular
calculator, but you will always write the equations in
standard form first and then input the number of
equations, the coefficients, and the constants. Pressing
the Solvekey will show the values of the unknowns.
Solving Simultaneous Equations1,1 1 1,2 2 1
2,1 1 2,2 2 2
a x a x b
a x a x b



11. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
In the branch current method, you can solve for the
currents in a circuit using simultaneous equations.
Branch current method
Steps:
1.Assign a current in each branch in an arbitrary
direction.
2.Show polarities according to the assigned directions.
3.Apply KVL in each closed loop.
4.Apply KCL at nodes such that all branches are
included.
5.Solve the equations from steps 3 and 4.

12. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Branch current method
1.Assign a current in each branch in an arbitrary
direction.V
S1 V
S2
R
1
R
3
R
2
270 W
1.0 kW
680 W
10 V 6.0 V
A
2.Show polarities according to the assigned
directions.
+
+
+
3.Apply KVL in each closed loop. Resistors are
entered in kWin this example.12
23
0.270 1.0 10 0
1.0 0.68 6.0 0
II
II
  
   
4.Apply KCL at nodes such that all branches are
included.1 3 2
I I I
5.Solve the equations from steps 3 and 4
(see next slide).

13. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Branch current methodV
S1 V
S2
R
1
R
3
R
2
270 W
1.0 kW
680 W
10 V 6.0 V 12
23
1 2 3
0.270 1.0 0 10
0 1.0 0.68 6.0
0
II
II
I I I
  
   
  
In standard form, the equations are
Solving: I
1 = 9.53 mA, I
2= 7.43 mA, I
3= 2.10 mA
The negative result for
I
3indicates the actual
current direction is
opposite to the
assumed direction.
(Continued)

14. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
In the loop current method, you can solve for the
currents in a circuit using simultaneous equations.
Loop current method
Steps:
1.Assign a current in each nonredundant loop in an
arbitrary direction.
2.Show polarities according to the assigned direction
of current in each loop.
3.Apply KVL around each closed loop.
4.Solve the resulting equations for the loop currents.

15. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights ReservedV
S1 V
S2
R
1
R
3
R
2
270 W
1.0 kW
680 W
10 V 6.0 V
SummarySummary
Loop current method
1.Assign a current in each nonredundant loop in
an arbitrary direction.
2.Show polarities according to the assigned
direction of current in each loop.
+
+
+
3.Apply KVL around each closed loop. Resistors
are entered in kWin this example.
4.Solve the resulting equations for the loop
currents (see following slide).
+
Loop A Loop B 
 
10 0.270 1.0 0
1.0 0.68 6.0 0
A A B
B A B
I I I
I I I
    
   
Notice that the
polarity of R
3is
based on loop B
and is not the
same as in the
branch current
method.

16. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights ReservedV
S1 V
S2
R
1
R
3
R
2
270 W
1.0 kW
680 W
10 V 6.0 V
SummarySummary
Loop current method
+
+
+
+
Loop A Loop B1.270 1.0 10
1.0 1.68 6.0
AB
AB
II
II

    9.53 mA
2.10 mA
A
B
I
I


I
1=I
A= 9.53 mA
I
3=I
B= 2.10 mA
I
2= I
AI
B= 7.43 mA
(Continued)
Rearranging the loop equations into standard form:

17. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights ReservedV
S
+15 V
R
4
R
5
R
3R
1
R
2
680 W680 W
680 W 1.0 k W
560 W
SummarySummary
The loop current method can be applied to more
complicated circuits, such as the Wheatstone bridge.
The steps are the same as shown previously.
Loop current method applied to
circuits with more than two loops
Loop A
Loop B
Loop C
The advantage to the
loop method for the
Wheatstone bridge is
that it has only 3
unknowns.

18. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Write the loop current equation for Loop A in the
Wheatstone bridge:V
S
+15 V
R
4
R
5
R
3R
1
R
2
680 W680 W
680 W 1.0 k W
560 W
Loop A
Loop B
Loop C   15 0.68 0.68 0
A B A C
I I I I     

19. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
In the node voltage method, you can solve for the
unknown voltages in a circuit using KCL.
Node voltage method
Steps:
1.Determine the number of nodes.
2.Select one node as a reference. Assign voltage
designations to each unknown node.
3.Assign currents into and out of each node except the
reference node.
4.Apply KCL at each node where currents are assigned.
5.Express the current equations in terms of the voltages
and solve for the unknown voltages using Ohm’s law.

20. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Node voltage methodV
S1 V
S2
R
1
R
3
R
2
270 W
1.0 kW
680 W
10 V 6.0 V
A12
1 2 3
1 2 3

S A S A A
V V V V V
I I I
R R R

  
Solve the same problem as before using the node
voltage method.
1.There are 4 nodes. A is the one unknown node.2.B is selected as the reference node.3.Currents are assigned into and out of node A.4.Apply KCL at node A (for this case).5.Write KCL in terms of the voltages (next slide).
B231 III 

21. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
SummarySummary
Node voltage methodV
S1 V
S2
R
1
R
3
R
2
270 W
1.0 kW
680 W
10 V 6.0 V
A12
1 3 2
+
S A S A A
V V V V V
R R R


B   
10 6.0
+
0.27 0.68 1.0
0.68 10 0.27 6.0 0.183
7.45 V
A A A
A A A
A
V V V
V V V
V


   

(Continued)

22. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Branch
Determinant
Loop
Matrix
Node
Simultaneous
equations
The solution of a matrix consisting of an array
of coefficients and constants for a set of
simultaneous equations.
A closed current path in a circuit.
One current path that connects two nodes.
Key Terms
An array of numbers.
The junction of two or more components.
A set of nequations containing nunknowns,
where nis a number with a value of 2 or more.

23. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Quiz
1.In a set of simultaneous equations, the coefficient
that is written a
1,2appears in
a. the first equation
b. the second equation
c. both of the above
d. none of the above

24. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Quiz
2.In standard form, the constants for a set of
simultaneous equations are written
a. in front of the first variable
b. in front of the second variable
c. on the right side of the equation
d. all of the above

25. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Quiz
3.To solve simultaneous equations, the minimum number of
independent equations must be at least
a. two
b. three
c. four
d. equal to the number of unknowns

26. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Quiz
4.In the equation a
1,1x
1+a
1,2x
2= b
1, the quantity b
1
represents
a. a constant
b. a coefficient
c. a variable
d. none of the above

27. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Quiz
5. The value of the determinant is
a. 4
b. 14
c. 24
d. 3435
28




28. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Quiz
6.The characteristic determinant for a set of simultaneous
equations is formed using
a. only constants from the equations
b. only coefficients from the equations
c. both constants and coefficients from the equations
d. none of the above

29. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Quiz
7.A negative result for a current in the branch method
means
a. there is an open path
b. there is a short circuit
c. the result is incorrect
d. the current is opposite to the assumed direction

30. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Quiz
8.To solve a circuit using the loop method, the equations
are first written for each loop by applying
a. KCL
b. KVL
c. Ohm’s law
d. Thevenin’s theorem

31. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Quiz
9.A Wheatstone bridge can be solved using loop
equations. The minimum number of nonredundant loop
equations required is
a. one
b. two
c. three
d. four

32. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Quiz
10.In the node voltage method, the equations are
developed by first applying
a. KCL
b. KVL
c. Ohm’s law
d. Thevenin’s theorem

33. Chapter 9
Principles of Electric Circuits, Conventional Flow, 9
th
ed.
Floyd
© 2010 Pearson Higher Education,
Upper Saddle River, NJ 07458. • All Rights Reserved
Quiz
Answers:
1. a
2. c
3. d
4. a
5. b
6. b
7. d
8. b
9. c
10. a