Lesson 10 – Circuit Analysis I
Important:
This module contains additional material not in the textbook!! These extra
techniques do not contain any “new physics.” Kirchhoff’s laws contain the
physics, but these techniques enable you to analyze more complicated circuits
faster. Furthermore, they provide clarity when one attempts to design or analyze
electrical circuits. You may use any technique that you wish in solving electric
circuits on the first exam.
A.
B.
On the final, all of these individual techniques will be testable and I may specify
that you use a particular technique is solving a circuit!!
I.
Simple Resistor Circuits With One Supply
There are many ways to attack simple resistor circuits with only one power supply
like the circuit shown below. In the problem below, you are asked to solve for
several different currents and voltages in the circuit. In these cases, one of the best
ways to solve the problem is to simplify the circuit by combining resistors.
EXAMPLE: In the circuit below, solve for all currents and voltages.
I1
A
B
5Ω
I3
I2
+
10 V
4Ω
-
D
3Ω
C
Figure 1
4Ω
STEP 1: Simplify the circuit by combining resistors.
I1
5Ω
A
B
+
10 V
R*
-
3Ω
D
C
Figure 2
I1
A
+
10 V
R**
-
D
Figure 3
STEP 2: Solve for I1 using Figure 3.
I1 =
STEP 3: Use Ohm’s Law and Kirchhoff’s to solve for the other currents and voltages.
From Figure 2:
VAB =
VBC =
VCD =
From Figure 1:
I2 =
I3 =
EXAMPLE 2: What power is dissipated by the 5 Ω resistor and supplied by the battery.
PR =
PB =
II.
Solving Circuits Using Kirchhoff’s Law
When a circuit has multiple power supplies, it may not be possible to solve the
circuit by combining resistors. In the following example, we will use the powerful
Kirchhoff’s laws to solve the circuit.
I3
I1
A
B
5Ω
7Ω
D
I2
10 Ω
6V
8V
C
4V
E
EXAMPLE: Solve for all the currents and voltages in the circuit above.
Solution:
At Pont B, we have by KCL that
For the left loop, we have by KVL that
For the right loop, we have by KVL that
For the outside loop, we have by KVL that
We now use algebra to solve for I1, I2, and I3 using any three of these equations:
(1)
I1 = I2 + I3
(2)
5 I1 + 10 I2 = 2 v
(3)
10 I2 – 7 I3 = 4 v
Substituting Eqtn. 1 into Eqtn. 2, we have
15 I2 + 5 I3 = 2 v
10 I2 – 7 I3 = 4 v
Multiplying Eqnt. 2 by 2 and Eqt. 3 by 3, we have
Subtracting Eqnt. 3 from Eqtn. 2, we have
Substituting I3 into Eqnt. 3, we have
Putting our results back into Eqnt. 1, we have
We now use Ohm’s law to find the voltage drops across each resistor.
III.
Analyzing Circuits Using Superposition
In our previous example, we solved a circuit involving multiple power supplies
using combinations of Kirchhoff’s and Ohm’s Law. We now re-do the example
using an alternative approach called “Superposition.”
In Superposition, you will
1) “zero” all but one power supply
2) solve the circuit for all desired voltages and currents due to this power supply
3) Repeat steps 1 & 2 for remaining power supplies
4) Sum the results of each power supply to find the currents and voltages due to
all supplies.
PREVIOUS EXAMPLE:
STEP 1: Zero all sources except the 6 v power supply. (Zero a voltage source by
“shorting” it. Zero a current source by “opening” it.)
I3*
A
I1*
6v
B
5Ω
7Ω
I2*
10 Ω
D
C
E
E
STEP 2: Simplifying the circuit, we have
I1* =
I2* =
I3* =
STEP 3: We repeat our work but Zero all sources except the 4 v power supply.
I3**
B
5Ω
7Ω
I1**
I2**
A
10 Ω
D
E
C
4v
E
Simplifying and solving the circuit, we have
E
We repeat the process again zeroing all sources except the 8 v power supply.
I3***
B
5Ω
7Ω
D
I1***
I2***
A
10 Ω
8v
E
C
E
Simplifying the circuit, we have
STEP 4: We obtain the total currents due to all the sources by adding our previous
current results:
I1 = I1* + I1** + I1*** =
I2 = I2* + I2** + I2*** =
I3 = I3* + I3** + I3*** =
We can now use Ohm’s Law to find the voltage drop across each resistor as we did
in our previous solution of this problem.
IV.
Mesh Analysis
This is just a more efficient way of solving loop (mesh) circuits using Kirchhoff’s voltage
law. It is the same method that we did previously except that we are setting up a
systematic way to do the algebra!!
We will solve our famous example using Mesh Analysis!!
A.
STEPS IN MESH ANALYSIS:
1. Draw and label a current arrow clockwise through each loop.
A
5Ω
B
7Ω
D
10 Ω
6v
8v
C
4v
E
2. Build the resistance matrix, R, whose elements are determined as follows
The diagonal element of the matrix, Rjj, is the sum of all the resistors through
which current Ij passes through.
The non-diagonal matrix element, Rjk is the negative of the sum of the
resistors seen by both the current Ij and the current Ik.
R=
3. Build the source column matrix, V, where the element Vj1 is the algebraic sum
of the voltage sources in the loop of the current Ij.
V=
4. We now find the unknown currents by solving the following matrix equation.
This is solving simultaneous linear (Kirchhoff’s voltage law) equations and is
covered in standard College Algebra textbooks. In addition, many calculator’s
have built in matrix math capabilities. Furthermore, this approach is ideal for
computer solution (using array).
V=RI
Determinant Solution Method: (Solving Simultaneous Equations)
We first find the determinant of the R matrix.
Det ( R) =
To find the jth current, we create a new matrix R* by replacing the jth column
of the R matrix with the source column matrix. The jth current is then found by
dividing the determinant of matrix R* by the determinant of the matrix R.
For Current I1, we have
R* =
I1 =
For current I2, we have
I2 =
Note: Mesh analysis is extremely powerful for solving large electrical circuits.