ESO 210/ESO203A
Introduction to Electrical Engineering
(Units 3-1-2-13)
(2014-15, First Semester)
Lecture-2
DC Circuits:
Loop Analysis of resistive circuit
in the context of dc voltages and currents
2
Objectives of this lecture:
• Meaning of circuit analysis; distinguish between the
terms mesh and loop.
• To provide more general and powerful circuit analysis
tool based on Kirchhoff’s voltage law (KVL) only.
3
Introduction:
The Series-parallel reduction technique that we learned in lecture-3 for analysing
DC circuits simplifies every step logically from the preceding step and leads on
logically to the next step.
Unfortunately, if the circuit is complicated, this method (the simplify and
reconstruct) becomes mathematically laborious, time consuming and likely to
produce mistake in calculations.
In fact, to elevate these difficulties, some methods are available which do not
require much thought at all and we need only to follow a well-defined faithful
procedure.
One most popular technique will be discussed in this lesson is known as ‘mesh or
loop’ analysis method that based on the fundamental principles of circuits laws,
namely, Ohm’s law and Kirchhoff’s voltage law.
Some simple circuit problems will be analyzed by hand calculation to understand
the procedure that involve in mesh or loop current analysis.
4
Meaning of circuit analysis
The method by which one can determine a variable (either a voltage or a current)
of a circuit is called circuit analysis.
Basic difference between ‘mesh’ and ‘loop’ is discussed in lecture-3 with an
example.
A ‘mesh’ is any closed path in a given circuit that does not have any
element (or branch) inside it.
A mesh has the properties that:
(i) every node in the closed path is exactly formed with two branches
(ii) no other branches are enclosed by the closed path.
(iii) Meshes can be thought of a resembling window partitions. On the other hand,
‘loop’ is also a closed path but inside the closed path there may be one or more
than one branches or elements.
5
Solution of Electric Circuit Based on Mesh (Loop) Current Method
Let us consider a simple dc network as shown in Figure 4.1 to find the currents
through different branches using Mesh (Loop) current method.
Mesh analysis is valid only for circuits that
can be drawn in a two-dimensional plane
in such a way that no element crosses over
another
6
7
8
Note:
To analyse, a resistive network containing voltage and current sources using
‘mesh’ equations method the following steps are essential to note:
• If possible, convert current source to voltage source.
• Otherwise, determine the voltage across the current source and write the
mesh equations as if these source voltages were known. Augment the set of
equations with one equation for each current source expressing a known mesh
current or difference between two mesh currents.
.
9
To Summarize:
Step-I: Draw the circuit on a flat surface with no conductor crossovers.
Step-2: Label the mesh currents (𝐼𝑖 ) carefully in a clockwise direction.
Step-3: Write the mesh equations by inspecting the circuit
(No. of independent mesh (loop) equations=no. of branches (b) - no. of
principle nodes (n) + 1).
Here
b=7
n=5
No. of independent
equations: 7-5+1=3
10
11
12
13
14
Solution: One can easily convert the extreme right current source (6 A)
into a voltage source. Note that the current source magnitude is 6 A and its
internal resistance is 6Ω. The given circuit is redrawn and shown in Figure
4.3 (c)
15
16
17
18
19
20
Practice Problems
21
22
23
24
25
Thevenin’s Theorem
Thevenins Theorem states that
"Any linear circuit containing several voltages and resistances can be replaced by just a
Single Voltage in series with a Single Resistor".
Thevenins Theorem is especially useful in analyzing power or battery systems and other
interconnected circuits where it will have an effect on the adjoining part of the circuit.
In other words, it is possible to simplify any "Linear" circuit, no matter how complex, to an
equivalent circuit with just a single voltage source in series with a resistance connected to a
load as shown below.
RTh
VTh
26
The basic steps for Thevenin’s Equivalent:
1. Remove the load resistor RL or component concerned.
2. Find RTh by shorting all voltage sources or by open circuiting all the current
sources.
3. Find VTh by the usual circuit analysis methods.
4. Find the current flowing through the load resistor RL.
27
• As far as the load resistor RL is concerned, any "one-port" network
consisting of resistive circuit elements and energy sources can be
replaced by one single equivalent resistance RTh and equivalent
voltage VTh, where RTh is the source resistance value looking back
into the circuit and VTh is the open circuit voltage at the terminals
For example, consider the circuit below.
Firstly, we have to remove the centre 40Ω resistor and short out (not physically as
this would be dangerous) all the emf´s connected to the circuit, or open circuit
any current sources. The value of resistor RTh is found by calculating the total
resistance at the terminals A and B with all the emf´s removed, and the value of
the voltage required VTh is the total voltage across terminals A and B with an open
circuit and no load resistor RTh connected. Then, we get the following circuit.
28
Find the Equivalent Resistance (Rs)
1. Remove the load resistor RL or component concerned.
2. Find RTh by shorting all voltage sources or by open circuiting all the
current sources.
RTh
29
Find the Equivalent Voltage (VTh)
3. Find VTh by the usual circuit analysis methods.
so the voltage drop across the 20Ω resistor can be calculated as:
VAB = 20 - (20Ω x 0.33amps) = 13.33 volts.
30
Then the Thevenin’s Equivalent circuit is shown below with
the 40Ω resistor connected.
4. Find the current flowing through the load resistor RL.
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
In some ways Norton's Theorem can be thought of as the opposite to "Thevenins Theorem", in that
Thevenin reduces his circuit down to a single resistance in series with a single voltage.
Norton on the other hand reduces his circuit down to a single resistance in parallel with a constant
current source.
Norton’s Theorem states that "Any linear circuit containing several energy sources and resistances
can be replaced by a single Constant Current generator in parallel with a Single Resistor".
As far as the load resistance, RL is concerned this single resistance, RTh is the value of the resistance
looking back into the network with all the current sources open circuited and IS is the short circuit
current at the output terminals as shown below.
Nortons equivalent circuit.
RTh
46
The value of this "constant current" is one which would flow if the two output
terminals where shorted together while the source resistance would be measured
looking back into the terminals, (the same as Thevenin).
The basic steps for getting Norton Equivalent is as follows:
1. Remove the load resistor RL or component concerned.
2. Find RTh by shorting all voltage sources or by open circuiting all the current
sources.
3. Find IS by placing a shorting link on the output terminals A and B.
4. Find the current flowing through the load resistor RL.
47
Example:
Solution:
1. Remove the load resistor RL or component concerned.
48
2. Find RTh by shorting all voltage sources or by open circuiting all the current
sources.
RTh
3. Find IS by placing a shorting link on the output terminals A and B.
Nortons equivalent circuit
49
Ok, so far so good, but we now have to solve with the original 40Ω load
resistor connected across terminals A and B as shown below.
Again, the two resistors are connected in parallel across the
terminals A and B which gives us a total resistance of:
The voltage across the terminals A and B with the load resistor connected is given as:
Then the current flowing in the 40Ω load resistor can be found as:
which again, is the same value of 0.286 amps,
50
Mesh analysis:
While Kirchoff´s Laws give us the basic method for analysing any complex electrical circuit, there are
different ways of improving upon this method by using Mesh Current Analysis or Nodal Voltage
Analysis that results in a lessening of the math's involved and when large networks are involved this
reduction in maths can be a big advantage.
Mesh Current Analysis Summary.
This "look-see" method of circuit analysis is probably the best of all the circuit analysis methods with the
basic procedure for solving Mesh Current Analysis equations is as follows:
1. Label all the internal loops with circulating currents. (I1, I2, ...IL etc). L is the no. of loops.
2. Write the [ L x 1 ] column matrix [ V ] giving the sum of all voltage sources in each loop.
3. Write the [ L x L ] matrix, [ R ] for all the resistances in the circuit as follows;
R11 = the total resistance in the first loop.
Rnn = the total resistance in the Nth loop.
RJK = the resistance which directly joins loop J to Loop K.
4. Write the matrix or vector equation [V] = [R] x [I] where [I] is the list of currents to be found.
51
EXAMPLE:
SOLUTION:
One simple method of reducing the amount of math's involved is to analyse the
circuit using Kirchoff's Current Law equations to determine the
currents, I1 and I2 flowing in the two resistors. Then there is no need to calculate
the current I3 as its just the sum of I1 and I2. So Kirchoff's second voltage law
simply becomes:
Equation No 1 : 10 = 50I1 + 40I2
Equation No 2 : 20 = 40I1 + 60I2
therefore, one line of math's calculation have been saved.
52
These equations can be solved quite quickly by using a single mesh impedance
matrix Z. Each element ON the principal diagonal will be "positive" and is the
total impedance of each mesh. Where as, each element OFF the principal
diagonal will either be "zero" or "negative" and represents the circuit element
connecting all the appropriate meshes. This then gives us a matrix of:
Where:
[ V ] gives the total battery voltage for loop 1 and then loop 2.
[ I ] states the names of the loop currents which we are trying to find.
[ R ] is called the resistance matrix.
and this gives I1 as -0.143 Amps and I2 as -0.429 Amps
As : I3 = I1 - I2
The current I3 is therefore given as : -0.143 - (-0.429) = 0.286 Amps. which is the same value of 0.286 amps,
53
Nodal Analysis:
The basic procedure for solving Nodal Analysis equations is as follows:
1. Write down the current vectors, assuming currents into a node are positive.
ie, a (N x 1) matrices for "N" independent nodes.
2. Write the admittance matrix [Y] of the network where:
Y = the total admittance of the first node.
Y = the total admittance of the second node.
R = the total admittance joining node J to node K.
11
22
JK
3. For a network with "N" independent nodes, [Y] will be an (N x N) matrix and
that Ynn will be positive and Yjk will be negative or zero value.
4. The voltage vector will be (N x L) and will list the "N" voltages to be found.
54
Nodal Voltage Analysis complements the previous mesh analysis in that it is
equally powerful and based on the same concepts of matrix analysis.
As its name implies, Nodal Voltage Analysis uses the "Nodal" equations of
Kirchoff's first law to find the voltage potentials around the circuit.
By adding together all these nodal voltages the net result will be equal to zero.
Then, if there are "N" nodes in the circuit there will be "N-1" independent nodal
equations and these alone are sufficient to describe and hence solve the circuit.
At each node point write down Kirchoff's first law equation, that is: "the currents
entering a node are exactly equal in value to the currents leaving the node" then
express each current in terms of the voltage across the branch.
For "N" nodes, one node will be used as the reference node and all the other
voltages will be referenced or measured with respect to this common node.
55
Nodal Voltage Analysis Circuit
In the above circuit, node D is chosen as the reference node and the other three
nodes are assumed to have voltages, Va, Vb and Vc with respect to node D.
For example;
As Va = 10v and Vc = 20v , Vb can be easily found by:
again is the same value of 0.286 amps.