Appendix III
Kirchhoff’s Rules, Matrix Algebra
and your graphing calculator
Use of Kirchhoff’s rules
When using Kirchhoff’s rules to solve for unknowns in a circuit you will need to set up
linearly independent equations using the two rules. The number of linearly independent
equations used will equal the number of unknowns being sought.
Kirchhoff’s 1st rule:
Current Rule
Kirchhoff’s 2nd rule:
Voltage Rule
At any junction in a circuit, the algebraic sum
of the currents entering the junction is equal to
the algebraic sum of the currents leaving the
junction.
The algebraic sum of the change in potential
encountered in a complete transversal of a loop
in a circuit must be zero.
A linearly independent equation in terms of an electric circuit would have at least one unique
component in it (i.e. a component not used in any other equation).
A junction as mentioned in rule 1 is defined in terms of an electric circuit as a point with
three or more components connect together. The number of linearly independent equations
needed using the junction rule is the number of junctions (n) minus 1 or n-1.
When using Kirchhoff’s rules first set a direction for each current and use them consistently
throughout the problem. Current is generally thought to flow from a voltage source from its
positive terminal to its negative terminal and a larger voltage source can cause current to
flow into a smaller voltage source.
b
R1
c
I1
I2
R3
V1
I1
a
I3
e
R2
V2
I2
d
f
Figure 1
In Figure 1 if the resistance values and voltages V 1 and V 2 are given, we can solve for the
three unknown currents. Here the principles of ohms law will not work because the circuit
cannot be simplified down to a single voltage source and a single resistor.
First, directions are given for each current as indicated by the arrows. Then the number of
junctions can be determined. The circuit above has two junctions 1) where R 1 , R 2 and R 3
meet and 2) where R 3 and the negative terminals of V 1 and V 2 meet. Therefore using the
rule n-1 there is one linearly independent equation for the junction rule. Since I 1 and I 2 are
shown entering the junction, they add to become I 3 . Therefore the first linearly independent
equation is
I1 + I2 = I3
[1]
Now that the linearly independent equation using the current rule has found, the remaining
two linearly independent equations can be found using the loop rule.
The essential part of the loop rule is paying attention to signs as you transverse the loop. The
point where the current enters a resistor is thought to be at a higher potential than the point
leaving the resistor. Therefore, you can assign a positive value to the side of the resistor the
current is entering and a negative value to the side it is leaving. (As shown below)
The potential difference across a resistor is the product of its resistance and current or IR.
There are different methods to determine the polarity of the potential difference across a
component, determine the method that suits you best and use it to find each loop.
One method is as you traverse through a resistor in the same direction as the current there is a
potential dropped (pd) across it or V pd = – IR. Whereas if you transverse through a resistor in
the opposite direction as the current V pd = IR. A voltage source, V, would have a potential
difference of V pd = V if you transverse it in the same direction as the current and V pd = –V if
you transverse it in the opposite direction as the current.
b
+
V1
a
R1
c +
–
I1
R3
I3
–
I2 R2
+
e
V2
–
d
f
The other method, which will be used here, is to assign the polarity for the potential
difference the same sign that was assigned to the component as shown in the figure above. So
as you traverse the loop the first sign you come to for any component is the sign assigned to
the polarity of the potential difference.
If we look at the first loop abcd and set our starting point as a, and proceed in the direction
you have set (i.e. from a to b to c to d) you would approach the negative terminal of the
voltage source therefore V pd would be – V 1 . Proceeding through the voltage source towards
b and the next component R 1 which has a + sign at this point therefore V pd would be +I 1 R 1 ,
leaving R 1 and approaching the point c you would proceed through R 3 in order to complete
the loop and end at point d. Since the sign entering R 3 is +, V pd will be +I3 R 3 . The sum of
the loop will equal zero (Kirchhoff’s second rule) and an equation can be written as
-V 1 + I1 R 1 + I3 R 3 = 0
[2]
Using the same principals for the loop fecd will yield an equation
-V 2 + I2 R 2 + I3 R 3 =0
[3]
We now have three linearly independent equations and a solution for the unknowns can be
found by solving simultaneous equations algebraically and using substitution.
If the values in the circuit are V 1 = V 2 , = 5V and R 1 = 600, R 2 = 200 and R 3 = 75 then
solving simultaneously
I 1 + I 2 –I 3 = 0
[1e]
-5+600I 1 + 75I 3 = 0
[2e]
-5+200I 2 + 75I 3 = 0
[3e]
Setting the eq [2e] and [3e] equal to one another.
-5+600I 1 +75I 3 = -5+200I 2 +75I 3
Simplifying using algebra
600I1 = 200I 2
Solving for one current in terms of the other
3I1 = I 2
Substitute the newly defined I 2 into eq. [1e]
I 1 +3I 1 =I 3
4I1 = I 3
Last substitute 4I 1 for I 3 into equation [2e]
600I1 +75(4I 1 )=5
900I1 =5
I 1 =5.55x10-3A
Then from [A]
I 2 =3I 1 =3*5.55x10-3A=16.65x10-3A
and
[B]
I 3 =4I 1 =4*5.55x10-3A=22.2x10-3A
[A]
[B]
Use of matrices to solve for a system of equations
Once all the linearly independent equations are establish they can be set up so that matrix
algebra can be used to find the unknowns. On one side of an equation is all the unknown
variables and they are equated to one of the known values as follows.
I 1 + I 2 –I 3 = 0
I 1 R 1 + I3 R 3 =V 1
I 2 R 2 + I3 R 3 = V 2
The first task is to set up a 3 x 4 matrix to represent the equations above using the values
below.
V 1 = V 2 , = 5V and R 1 = 600, R 2 = 200 and R 3 = 75. Where the current variables that are
being solved for is given an initial value of 1
[4]
1
1
-1
600 0
75
0
200 75
0
5
5
What is shown here is an augmented matrix which consist of 2 matrices the first a 3x3 for the
statements on the left side of the equal sign the second a 3x1 matrix for the statement on the
right side of the equal sign.
A method to solve for the unknowns is called Reduced Row-Echelon Form
A matrix is in Reduced Row-Echelon Form when the following conditions are applied to the
larger portion of the augmented matrix and satisfied.
1. The first nonzero number in any row is the number 1, this is called a leading 1.
2. For a column that contains a leading 1 all other values are zero.
3. The leading 1 of a row is strictly to the right of the leading 1 in the row above it.
4. All rows that only contain zeroes, if there are any are at the bottom of the matrix.
The final result will be in format shown below with the values S 1 , S 2 andS 3 being the
solutions to the unknown variables being solved for.
1
0
0
S1
0
1
0
S2
0
0
1
S3
If you are curious about the method you can Google it using RREF. A site at
www.math.odu.edu offers a good applet that shows a step by step method.
Using the graphing calculators to solve the RREF matrix
Current technology has brought us the graphing calculator and many of them have the ability
to perform this matrix manipulation.
For the TI series of calculators, you have the ability to create, save and solve a matrix.
For the 83/ 84 models
• Start by selecting Matrix.
• Arrow over to edit and select an empty matrix slot, for example [ B ]
• Create a 3 x 4 matrix
• Enter in the coefficients. Use the matrix at the top of the page as your example. Your
complete matrix would resemble it.
• Select quit
• Select Matrix once again.
• Go to the MATH, then arrow down to rref( and select it. It should appear on the screen
• Go back to matrix once again and select the matrix that you had put the coefficients into
• Our example would look like rref([B] , press enter.
Your solution should then resembled the format shown above for the solved matrix.
The TI-89 has a few ways to solve for this as well.
The easiest if the application has be installed is the Simultaneous equation solver found by
selecting the APPS button then choosing Flashapps. If it isn’t on the calculator consider
installing it. It is a free program that can be downloading from the TI site or may be with
material supplied with the calculator.
Using the Simultaneous Equation solver is simple.
• You are first ask the three choices, current, open and new. Select new.
• You are solving for 3 equations (eqns) and 3 unknowns.
• Once these are entered a 3x4 matrix will be displayed. All you have to do is enter in the
information. An example is shown at the top of the previous page.
• Once the information has been enter select F5.
• Voila the solution is displayed.
The main trick here is getting the initial equations set up as shown in equation [4].
Changes can be easily made to the coefficients by selecting F4 within the solutions screen.
Rref( method
If however the simultaneous equation is not install or is unavailable you still can use the rrfe(
function.
• Start by selecting APPS then DATA/Matrix/ Editor
• Select New
• Within the pop up box Type: Matrix, folder : main, variable: create a name for the variable.
Make it short something like xyz.
• Select Alpha to turn off the alpha keys then enter in Row: 3 Col: 4
• A 3x4 matrix is created and the coefficients can be entered into the matrix. The matrix at
the top of the previous page is an example of how it should look.
• Once the information has been entered completely Select QUIT.
• Select the MATH functions 2nd 5.
• Scroll to Matrix and select rref(
• rref( should now appear on the screen. Enter in the matrix variable name our example was
xyz and close the parenthesis, so rref(xyz).
• Press enter to solve. The screen should now display a 3x 4 matrix with the solutions in the
format as discussed in a previous section.