SCHOOL OF MATHEMATICS
MATHEMATICS FOR PART I ENGINEERING
Self Study Course
MODULE 27
FURTHER APPLICATIONS TO ELECTRICAL CIRCUITS
Module Topics
1. Inverse of a matrix using elimination
2. Mesh analysis of circuits
3. Nodal analysis of circuits
A:
Work Scheme based on JAMES (FOURTH EDITION)
1. The first topic uses results from module 17 on matrices. Before starting this module you should make
sure that you know how to calculate the inverse of a matrix. This is described in module 17 and in section
5.4 and section 5.5 of James. The most useful method is to use elimination. This method is described in
sections 6 and 7 of module 17.
Example A:
Use the elimination method to solve the matrix equation
µ
Multiply row 1 by 2
µ
1
−2
−2
3
¶µ ¶ µ ¶
x
3
=
y
4
2
−2
−4
3
¶µ ¶ µ ¶
x
6
=
y
4
Form a new row 2 by adding old row 1 and old row 2
µ
2
0
−4
−1
¶µ ¶ µ ¶
x
6
=
y
10
y=
10
= −10
−1
Hence
By backward substitution in row 1
2x − 4y = 6
2x = 6 + 4y
= 6 − 40
= −34
Hence the solution is x = −17 and y = −10.
***Do Exercise 74 on p.370 using the elimination method.***
–1–
2.Mesh Analysis
Mesh analysis is a systematic procedure which uses Kirchhoff’s voltage law (KVL) to find all the currents
in a circuit. The method of analysis is as follows:
1. Clearly label all the known quantities in the circuit.
2. Identify all the “meshes” in the circuit. A mesh is defined as a loop with no sub-loops.
3. Assign mesh currents using the convention that they are measured in a clockwise direction.
4. Apply KVL for each mesh and express the voltages in terms of the mesh currents.
5. Write the resulting system of equations in matrix form. Solve these equations using the elimination
method to obtain the mesh currents.
6. Now the mesh currents are known the voltages may be obtained from Ohm’s law.
The method is best illustrated using a worked example.
Example A:
below.
Use mesh analysis to calculate the currents and voltages in the circuit shown in figure 1
6Ω
2Ω
Ι1
Ι2
4Ω
12V
6V
Fig. 1.
This circuit has two meshes.
Mesh 1: KVL gives
12 = 6I1 + 4(I1 − I2 )
Mesh 2: KVL gives
6 = 4(I2 − I1 ) + 2I2
We can write this as a pair of simultaneous equations
10I1 − 4I2 = 12
−4I1 + 6I2 = 6
or in matrix notation as
µ
¶µ
10
−4
−4
6
20
−20
−8
30
I1
I2
¶
µ
=
12
6
¶
Multiply row 1 by 2 and row 2 by 5
µ
¶µ
I1
I2
¶
µ
=
24
30
¶
Form a new row 2 by adding old row 1 and old row 2
µ
¶µ ¶ µ ¶
20 −8
I1
24
=
0
22
I2
54
–2–
Hence
I2 =
54
= 2.45
22
By backward substitution in row 1
20I1 − 8I2 = 24
20I1 = 24 + 8I2
Hence the solution is I1 = 2.18A and I2 = 2.45A. Note that the current flowing through the 4Ω resistor is
I2 − I1 = 0.27A.
The corresponding voltages are given by Ohm’s law V = IR and are therefore given by v1 = 13.1 V,
v2 = 4.9 V and v3 = 1.1 V.
Note that we can check that we have done the calculation correctly by applying KVL to each mesh and
checking that it gives the correct voltage. In this example v1 − v3 = 13.1 − 1.1 = 12 which is the voltage
source in mesh 1, while v2 + v3 = 4.9 + 1.1 = 6 which is the voltage source in mesh 2.
***Do Exercise A:
2 below.
Use mesh analysis to calculate the currents and voltages in the circuit shown in figure
6Ω
2Ω
Ι1
12V
Ι2
4Ω
6V
2Ω
Fig. 2.
3.Nodal Analysis
An alternative to mesh analysis is nodal analysis. This is a systematic procedure which uses Kirchhoff’s
current law (KCL) to find all the voltages in a circuit. It should be used in preference to mesh analysis for
circuits that have fewer nodes than meshes. It also has the advantage that it can be used to analyse any
circuits not simply planar ones. The method of analysis is as follows:
1. Clearly label all the known quantities in the circuit.
2. Identify all the “nodes” in the circuit.
3. Select a node as the reference node and assign it a potential of 0 volts. All other voltages in the circuit
are measured with respect to the reference node.
4. Label the voltages on all the other nodes.
5. Assign and label the currents in the circuit (including the polarities).
6. Apply KCL at each node to obtain equations for the voltages.
7. Write the resulting system of equations in matrix form. Solve these equations using the elimination
method to obtain the node voltages.
8. Now the voltages are known the currents may be obtained from Ohm’s law.
–3–
We start by looking at a simple example with no voltage sources.
Example B:
Use nodal analysis to calculate the currents voltages in the circuit shown in figure 3 below.
4Ω
A
B
3A
2Ω
4Ω
6A
C
Fig. 3.
This example has 3 meshes and 3 nodes (note that what appears to be a fourth node to the right of node C
is really part of C since there is simply wire with no resistance between them).
We take node C to be the reference node and solve for KCL at node A and node B.
Node A: KCL gives
−3 + (VA − VC )/2 + (VA − VB )/4 = 0
Node B: KCL gives
6 + (VB − VA )/4 + (VB − VC )/4 = 0
Rearranging and recalling that VC = 0 we can write this as a pair of simultaneous equations
3
VA −
4
1
− VA +
4
1
VB = 3,
4
1
VB = −6.
2
Multiplying both equations by 4 we may write this in matrix notation as
µ
¶µ
¶ µ
¶
3 −1
VA
12
=
−1
2
VB
−24
Multiply row 2 by 3
µ
3
−3
−1
6
¶µ
VA
VB
¶
µ
=
12
−72
¶
Form a new row 2 by adding old row 1 and old row 2
µ
¶µ
¶ µ
¶
3 −1
VA
12
=
0
5
VB
−60
Hence
VB = −
60
= −12
5
By backward substitution in row 1
3VA − VB = 12
3VA = 12 + VB
= 12 − 12 = 0
–4–
Hence the solution is VA = 0V and VB = −12V.
***Do Exercise B:
Use nodal analysis to calculate the voltages in the circuit shown in figure 4 below.
2A
2Ω
2Ω
6Ω
4Ω
4A
Fig. 4.
B:
Work Scheme based on STROUD (SIXTH EDITION)
This module is not covered by S., so work through A: Work scheme based on JAMES (FOURTH
EDITION), presented above.
–5–
Specimen Test 27
1.
Use the elimination method to solve the following matrix equations
(i)
µ
(ii)
2.
µ
1 4
2 3
¶µ ¶ µ ¶
x
5
=
y
5
2 4
4 1
¶µ ¶ µ ¶
x
8
=
y
9
Use mesh analysis to calculate the currents and voltages in the circuit shown in figure 5 below. Verify
that the resulting voltages satisfy KVL for each mesh.
3Ω
2Ω
Ι1
Ι2
2Ω
6V
6V
4Ω
Fig. 5.
3.
Use nodal analysis to calculate the voltages in the circuit shown in figure 6 below. Use Ohm’s law to
calculate the currents flowing in each component of the circuit.
6Ω
A
B
4A
2Ω
3Ω
2A
C
Fig. 6.
–6–