1. Chapter 9

2. Summary 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
constants
coefficients
variables

3. Summary Summary Simultaneous Equations Example: Solution:
A circuit has the following equations. Set up the equations in standard form.
Solution:
Rearrange so that variables and their coefficients are in order and put constants on the right.

4. Summary Summary Solving Simultaneous Equations
Three methods for solving simultaneous equations are
Algebraic substitution
The determinant method
Using a calculator

5. Summary Summary Solving Simultaneous Equations Example: Solution:
Solve for IA using substitution.
Solution:
Solve for IB in the first equation:
Substitute for IB into the second equation:
Rearrange and solve for IA.
IA = 9.53 mA

6. Summary 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,
2.10 mA

7. Summary Summary Solving Simultaneous Equations Example: Solution:
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.
Example:
Solution:
1.134

8. Summary 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 variable
Characteristic determinant

9. Summary Summary Solving Simultaneous Equations Example: Solution:
Solve the same equations using determinants:
Solution:
9.53 mA
2.10 mA

10. Summary 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.

11. Summary Summary Branch current method
In the branch current method, you can solve for the currents in a circuit using simultaneous equations.
Steps:
Assign a current in each branch in an arbitrary direction.
Show polarities according to the assigned directions.
Apply KVL in each closed loop.
Apply KCL at nodes such that all branches are included.
Solve the equations from steps 3 and 4.

16. Summary Summary Branch current method Example: Solution:
Solve the equations from steps 3 and 4
(see next slide).
Apply KVL in each closed loop. Resistors are entered in kW in this example.
Apply KCL at nodes such that all branches are included.
Show polarities according to the assigned directions.
Assign a current in each branch in an arbitrary direction.
Solution: A + + +

17. Summary Summary Branch current method Solution:
In standard form, the equations are
(Continued)
Solving: I1 = 9.53 mA, I2 = 7.43 mA, I3 = mA
The negative result for I3 indicates the actual current direction is opposite to the assumed direction.

18. Solving 3x3 Matrix

19. Summary Summary Loop current method OR Mesh Analysis
In the loop current method, you can solve for the currents in a circuit using simultaneous equations.
Steps:
Assign a current in each nonredundant loop in an arbitrary direction.
Show polarities according to the assigned direction of current in each loop.
Apply KVL around each closed loop.
Solve the resulting equations for the loop currents.

20. Summary Summary Loop current method Example: Solution:
Apply KVL around each closed loop. Resistors are entered in kW in this example.
Solve the resulting equations for the loop currents (see following slide).
Show polarities according to the assigned direction of current in each loop.
Assign a current in each nonredundant loop in an arbitrary direction.
Solution:
Notice that the polarity of R3 is based on loop B and is not the same as in the branch current method. + + +
Loop A
Loop B +

21. Summary Summary Loop current method Solution:
Rearranging the loop equations into standard form:
(Continued)
I1=IA = 9.53 mA
I2= IA-IB = 7.43 mA
I3=IB = 2.10 mA + + +
Loop A
Loop B +

22. Solution has been given in Class

23. Summary
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.
The advantage to the loop method for the bridge is that it has only 3 unknowns.
Loop A
Loop B
Loop C

24. Summary Summary Example:
Write the loop current equation for Loop A in the Wheatstone bridge:
Loop A
Loop B
Loop C

25. Summary Summary Node voltage method
In the node voltage method, you can solve for the unknown voltages in a circuit using KCL.
Steps:
Determine the number of nodes.
Select one node as a reference. Assign voltage designations to each unknown node.
Assign currents into and out of each node except the reference node.
Apply KCL at each node where currents are assigned.
Express the current equations in terms of the voltages and solve for the unknown voltages using Ohm’s law.

26. Summary Summary Node voltage method Example: Solution:
Solve the same problem as before using the node voltage method.
Solution:
Write KCL in terms of the voltages (next slide).
Apply KCL at node A (for this case).
There are 4 nodes. A is the one unknown node.
Currents are assigned into and out of node A.
B is selected as the reference node. A B

27. Summary
Summary
Node voltage method
Solution:
(Continued) A B

28. Key Terms Branch One current path that connects two nodes. Determinant
Loop
Matrix
Node
Simultaneous equations
One current path that connects two nodes.
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.
An array of numbers.
The junction of two or more components.
A set of n equations containing n unknowns, where n is a number with a value of 2 or more.

29. Quiz
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

30. Quiz
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

31. Quiz
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

32. Quiz 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

33. Quiz
5. The value of the determinant is
a. 4
b. 14
c. 24
d. 34

34. Quiz
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

35. Quiz 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

36. Quiz
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

37. Quiz
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

38. Quiz
In the node voltage method, the equations are developed by first applying
a. KCL
b. KVL
c. Ohm’s law
d. Thevenin’s theorem

39. Quiz
Answers:
1. a
2. c
3. d
4. a
5. b
6. b
7. d
8. b
9. c
10. a