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Chapter 9
Chapter 9
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Simultaneous Equations
Circuit analysis methods in Chapter 9 require use of
simultaneous equations.
To simplify solving simultaneous equations, they
are usually set up in standard form. Standard form
for two equations with two unknowns is
a1,1 x1  a1,2 x2  b1
a2,1 x1  a2,2 x2  b2
coefficients
constants
variables
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Simultaneous Equations
A circuit has the following equations. Set up the
equations in standard form.
10  270 I A  1000( I A  I B )  0
1000  I B  I A   680 I B  6  0
Rearrange so that variables and their coefficients
are in order and put constants on the right.
1270 I A  1000 I B  10
1000 I A  1680 I B  6
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Solving Simultaneous Equations
Three methods for solving simultaneous equations
are
 Algebraic substitution
 The determinant method
 Using a calculator
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Solving Simultaneous Equations
Solve for IA using substitution.
1270 I A  1000 I B  10
1000 I A  1680 I B  6
Solve for IB in the first equation:
I B  1.270I A  0.010
Substitute for IB into the second equation:
1000I A  1680(1.270I A  0.010)  6
Rearrange and solve for IA.
IA = 9.53 mA
1134I A  10.8
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Solving Simultaneous Equations
If you wanted to find IB in the previous example,
you can substitute the result of IA back into one of
the original equations and solve for IB. Thus,
1270 I A  1000 I B  10
1270(9.53 mA)  1000 I B  10
I B  2.10 mA
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Solving Simultaneous Equations
The method of determinants is another approach to
finding the unknowns. The characteristic determinant is
formed from the coefficients of the unknowns.
Write the characteristic determinant for the
equations. Calculate its value.
1270 I A  1000 I B  10
1000 I A  1680 I B  6
 1270 1000 

  1.134
 1000 1680 
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
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.
To solve for x2:
Constants
Unknown
 b1 a12 
 a11 b1 
variable




b
a
a
b
22 
x1   2
x2   21 2 
 a11 a12 
 a11 a12 




Characteristic
a
a
a
a
 21
22 
 21 22 
determinant
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Solving Simultaneous Equations
Solve the same equations using determinants:
1270 I A  1000 I B  10
1000 I A  1680 I B  6
 10 1000 



6
1680

IA  
 1270 1000 



1000
1680


 9.53 mA
Principles of Electric Circuits - Floyd
 1270 10 



1000

6

IB  
 1270 1000 



1000
1680


 2.10 mA
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Solving Simultaneous Equations
Many scientific calculators allow you to enter a set of
equations and solve them “automatically”. The
calculator method will 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 Solve key will show the values of the unknowns.
a1,1 x1  a1,2 x2  b1
a2,1 x1  a2,2 x2  b2
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Branch current method
In the branch current method, you can solve for the
currents in a circuit using simultaneous equations.
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.
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Branch current method
5.4.
3.
2.
1. Solve
Apply
Show
Apply
Assign
the
polarities
KCL
KVL
aequations
current
at
innodes
each
according
infrom
each
closed
suchsteps
branch
that
toloop.
the
3alland
in
assigned
Resistors
branches
an4 arbitrary
are
are
(see
included.
directions.
direction.
nextin
slide).
entered
kW in this example.
0.270 I1  1.0 I 2  10  0
1.0 I 2  0.68I 3  6.0  0
I1  I 3  I 2
+ R1
VS1
10 V
Principles of Electric Circuits - Floyd
270 W
A
R3 +
+ R2 680 W VS2
6.0 V
1.0 kW
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Branch current method
(Continued)
In standard form, the equations are
0.270 I1  1.0 I 2
 0  10
0  1.0 I 2  0.68I 3  6.0
I1
 I2
 I3  0
Solving: I1 = 9.53 mA, I2 = 7.43 mA, I3 = 2.10 mA
The negative result for
I3 indicates the actual
VS1
current direction is
10 V
opposite to the
assumed direction.
Principles of Electric Circuits - Floyd
R1
270 W
R3
680 W
R2
1.0 kW
VS2
6.0 V
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Loop current method
In the loop current method, you can solve for the
currents in a circuit using simultaneous equations.
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.
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Loop current method
1. Show
2.
3.
4.
Solve
Assign
Applythe
polarities
KVL
a current
resulting
around
according
in equations
each
eachnonredundant
closed
to the
for loop.
assigned
the loop
Resistors
loop in
an
direction
currents
(see
of in
current
direction.
following
each
slide).
loop.
arearbitrary
entered
kW ininthis
example.
10  0.270 I A  1.0  I A  I B   0
1.0  I B  I A   0.68 I B  6.0  0
+ R3
+ R1
VS1
10 V
Principles of Electric Circuits - Floyd
270 W
Loop A
Notice that the
polarity of R3 is
based on loop B
and is not the
same as in the
branch current
method.
680 W V
S2
6.0 V
1.0 kW Loop B
+ R2
+
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Loop current method
Rearranging the loop equations into standard form:
(Continued)
1.270 I A  1.0 I B  10
I1=IA = 9.53 mA
1.0 I A  1.68I B  6.0
I A  9.53 mA
I2= IAIB = 7.43 mA
I3=IB = 2.10 mA
I B  2.10 mA
+ R1
VS1
10 V
Principles of Electric Circuits - Floyd
270 W
Loop A
+ R3
680 W V
S2
6.0 V
1.0 kW Loop B
+ R2
+
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Loop current method applied to
circuits with more than two loops
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 A
R1
680 W
VS
+15 V
560 W
R2
680 W
Principles of Electric Circuits - Floyd
R3
Loop B 680 W
R5
Loop C
R4
1.0 kW
The advantage to the
loop method for the
bridge is that it has
only 3 unknowns.
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Write the loop current equation for Loop A in the
Wheatstone bridge:
15  0.68  I A  I B   0.68  I A  IC   0
Loop A
R1
680 W
VS
+15 V
560 W
R2
680 W
Principles of Electric Circuits - Floyd
R3
Loop B 680 W
R5
Loop C
R4
1.0 kW
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Node voltage method
In the node voltage method, you can solve for the
unknown voltages in a circuit using KCL.
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.
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Node voltage method
Solve the same problem as before using the node
voltage method.
1.
There
are
4 in
nodes.
isinto
the this
one
unknown
2. Write
3.
4.
5.
B
Apply
Currents
is selected
KCL
KCL
are
atassigned
asterms
node
theAreference
A
of
(for
the
voltages
and
node.
case).
out
of
(next
nodenode.
slide).
A.
VS 1  VA
I1 
R1
VA
I2 
R2
I1  I 3  I 2
VS1
10 V
VS 2  VA
I3 
R3
R1
270 W
A
R3
680 W
R2
1.0 kW
VS2
6.0 V
B
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Summary
Node voltage method
(Continued)
VS 1  VA
VS 2  VA
VA
+

R1
R3
R2
10  VA
6.0  VA
VA
+

0.27
0.68
1.0
0.68 10  VA   0.27  6.0  VA   0.183VA
VA  7.45 V
VS1
10 V
R1
270 W
A
R3
680 W
R2
1.0 kW
VS2
6.0 V
B
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Key Terms
Branch One current path that connects two nodes.
Determinant The solution of a matrix consisting of an array
of coefficients and constants for a set of
simultaneous equations.
Loop A closed current path in a circuit.
Matrix An array of numbers.
Node The junction of two or more components.
Simultaneous A set of n equations containing n unknowns,
equations where n is a number with a value of 2 or more.
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Quiz
1. In a set of simultaneous equations, the coefficient
that is written a1,2 appears in
a. the first equation
b. the second equation
c. both of the above
d. none of the above
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
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
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
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
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Quiz
4. In the equation a1,1x1 +a1,2x2 = b1, the quantity b1
represents
a. a constant
b. a coefficient
c. a variable
d. none of the above
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Quiz
 3 5
5. The value of the determinant 
 is
2
8


a. 4
b. 14
c. 24
d. 34
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
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
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
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
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
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
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
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
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
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
Principles of Electric Circuits - Floyd
© Copyright 2006 Prentice-Hall
Chapter 9
Quiz
Answers:
Principles of Electric Circuits - Floyd
1. a
6. b
2. c
7. d
3. d
8. b
4. a
9. c
5. b
10. a
© Copyright 2006 Prentice-Hall