Download - ES100 Lecture 2
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Lecture 2Lecture 2
Dr Ahmed AbuDr Ahmed Abu--SiadaSiada
Electrical Systems 100Electrical Systems 100
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ContentsSeries-Parallel Circuits
Wye-Delta Conversion
Ladder Networks
Current Sources
Source Conversion
Current Sources in Parallel (Series?)
Mesh Analysis
Nodal Analysis
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Series-Parallel CircuitsSeries-parallel circuits are networks where there are
both series and parallel elements
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Reduce and Return Approach
Step 1: Take an overall mental look at the problem
Step 2: Examine each section of the network independently beforetying them together
Step 3: Redraw the network as often as you will need to arrive at
reduced branches. Maintain the original unknown quantities to befound for clarity where applicable
Step 4: Take the trip back to the original network to find detailed
solution
(Some time it may help to draw branche/s as blocks and work out
blockwise)
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An Example of Reduced and Return Approach
Finding V4?
1R
243 //)(' RRRR T +=
1' RRR TT+=
SI
SI
T
SR
EI = SI
TR'
2V
TS RIV =2
2V43
424
RR
RVV
+=
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An Example of Reduced and Return Approach
2k //6k12k
( )//6k12k2kIS
+=
E
54V
S2
S1
I6k12k
12kI
I6k12k
6kI
+
=
+
=
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Wye-Delta Conversion
Often we encounter a different kind of network which appears to benot in series or parallel in relation to the rest of the network. Under
these circumstances it is necessary to convert this portion of circuit
from one form to the other to find appropriate branch connectionwhich then appears clearly in series or parallel with rest of the
network.
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Y- ConversionThe purpose is then to be able to convert Y to or to Y.
)(
)(
)//(
31
31
CAB
CB
CAB
AB
CAB
CAB
ca
CABca
RRR
RR
RRR
RR
RRR
RRRRRR
RRRRRR
+++++=
++
+=+=
+=+=
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-Y Conversion
CBA
CB
RRR
RRR
++=1
)(
)(
)(
)(
)(
)(
32
21
31
CBA
CA
CBA
BA
CBA
CBA
cb
CBA
BC
CBA
AC
BAC
BA
ba
CBA
CB
CBA
AB
CAB
CAB
ca
RRR
RR
RRR
RR
RRR
RRRRRR
RRR
RR
RRR
RR
RRR
RRRcRRR
RRR
RR
RRR
RR
RRR
RRRRRR
++
+
++
=
++
+=+=
+++
++=
++
+=+=
+++
++=
++
+=+=
CBA
CA
RRR
RRR
++=2
CBA
BA
RRR
RRR
++
=3
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Y- Conversion
1
313221
RRRRRRRRA
++=
Similarly, for converting Wye quantities to Delta are given as:
2
313221
RRRRRRRRB ++=
3
313221
R
RRRRRRRC
++=
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-Y /Y- Conversion
If all resistors in the or Y are the same (RA = RB = RC):
From -Y eq:
12
2
3
3
3
RRR
RR
RRRRRR
A
A
A
CBA
BA
===
=++
=
This shows that for a Y of three equal resistors the value of each
resistor equivalent is 3 times the Y resistor.
Y
Y
3RR:
YFor
3
RR:YFor
=
=
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Ladder NetworksA Ladder Network is one where a series-parallel section of a network occur
repeatedly within the network. An example of such network is a Low Pass Filter
circuit. Figure below shows a three section Ladder Network.
To solve a ladder network follow the steps:
Calculate the total resistance
Calculate the source current or total current drawn from source
Work back through the ladder until desired current or voltage is obtained
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Ladder Networks
Combining parallel and series elements to reduce the circuit we get :
S1
T
S
T
IIbackwards,Working
A30
8
240V
R
EI
835R
=
===
=+=
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Ladder NetworksUsing current divider I6 can be found.
A15I
A152IIA,30I
2
S
3S
=
===
Finally,
V20210ARIVand
A10I9
6I
3)(6
6I
666
336
===
==+
=
(Current divider)
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Current Sources
A battery supplies fixed voltage and the source current may vary
according to load. Similarly, a current source is one where it
supplies constant current to the branch where it is connected andthe voltage and polarity of voltage across it may vary according to
the network condition.
AII
EVS
437I
KCLApplyingA,3
4
12VI
V12
21
2
===
=
=
==
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Source Conversion
A Voltage source can be converted to a current source and vice
versa. In reality, Voltage sources has an internal resistance Rs and
current sources has a shunt resistance Rsh. In ideal cases, Rs equalto 0 and Rsh equal to .
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Source Conversion
For us to be able to convert sources, the voltage source must have a series
resistance and current source must have some shunt resistance.
Eg.
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Method of Circuit Analysis
(Mesh Analysis and Nodal Analysis)
Mesh is a closed loop which does not contain any other loops
within it. In most circumstances, a mesh will contain one or more
voltage sources and one or more type of circuit elements. In dccircuit theory, these elements are limited to resistances only in
steady state analysis. Mesh analysis determines the mesh or loop
currents in the circuit.
i1 i2I1 = i1I2 = i2I3 = i1-i2
By solving i1 and i2
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Mesh AnalysisSteps to determine Mesh Currents:
Identify the n number meshes in the circuit
Assign mesh currents, in clockwise directions
Apply KVL to each of the n meshes. Use Ohms law to
express the voltages in terms of the mesh currents. Take
appropriate voltage drop polarity (+ve clockwise and veanticlockwise) into consideration in writing these equations.
Solve the n simultaneous equations to get the n mesh
currents.
nn iiiii ,1321 ......,,
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Mesh Analysis-An Example
A.1iandA1i
12i36i
12
22
==
=
Find I1, I2 and I3.
i1 i2
(Loop i1)
12i3i
010i15i5
010)i10(i5i15
21
21
211
=
=+
=
12i
1i2i020i10i10
04i6i)i10(i10
21
12
21
2212
=
=
=+
=
i
(Loop i2)
Method 1:Using method of substitution
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Mesh Analysis-An ExampleMethod 2: Using Cramers Rule (Also known as Format Approach):
41311
13
42221
21
42621-
2-3
1
1
i
i
21
23
2
1
2
1
=+=
=
=+=
=
===
=
12ii
12i3i
21
21
=+
=From previous 2 loops :
We obtain the determinant as
A1
i
A1
i
22
11
==
==Thus:
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Mesh Analysis-Super MeshSometime, there may be a current source in one of the mesh or a
current source in common between two meshes.
If the current source involves only one of the meshes, then the
analysis is easier as we have 1 less equation to solve as the current is
defined by the current source in one of the mesh already.
If however, the current source is common between two meshes,
then we need to form a super mesh. This is necessary as we need toapply KVL in solving mesh equations and we do not know the
voltage across the current source in advance.
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Mesh Analysis-Super MeshCase 1:
In this example, A.52 =i
i1 i2
A
ii
2i
0)(64i10-
1
211
=
=++
Applying KVL in Mesh 1
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i1 i2
Mesh Analysis-Super MeshCase 2:
we create a super-mesh by excluding the current source and any
element connected in series with it as shown.
A8.2i2.3i
20146i
0410620
21
21
221
==
=+
=+++
A
i
iii
A6ii 12 =
i1 i2
In super-mesh
super-mesh
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Nodal AnalysisA Node is defined as the junction of two or more branches. In a n
node circuit, 1 node is taken as the reference (usually the ground is
taken as reference) and we need to solve node voltages using KCL.
By solving node v1 and v2
We can solve :
2
212
R
VVI
=
1
11
RVI =
3
23
R
VI =
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Nodal AnalysisSteps to determine Nodal voltages:
Select a node as reference node
Assign voltages to the remaining nodes
Apply KCL to each of the n-1 non-reference nodes. Use Ohms
law to express currents in terms of node voltages.
Solve the n-1 simultaneous equations to get the unknown nodevoltages.
121 ,...., nvvv
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Nodal AnalysisApplying KCL to each nodes
Example in node v1
00
12
2
21
1
1=+
+
II
R
VV
R
V
node v2
00
2
2
12
3
2=
+
I
R
VV
R
V
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Nodal Analysis- An ExampleAt node 1, applying KCL,
6053v-
01054
v
6
0
21
122
=+
=+
+
v
vv
v1v2
20v3v
vv2v20
054
vv
2
0v
21
211
211
=
+=
=
+
At node 2, applying KCL,v2v1
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Nodal Analysis- An ExampleUsing Cramers Rule or Format Approach:
Vv
Vv
v
v
2012
60180603
203
33.1312
60100560
120
1231553
13
60
20
53
13
22
11
2
1
=+
=
=
=
=+
=
=
=
==
=
=
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Nodal Analysis-SupernodeSometime, there may be a voltage source connected between a
reference and nonreference node. If this is the case, the voltage of
the nonreference node is simply set equal to the voltage source and
we have 1 less equation to solve.If however, the voltage source is common between two or more
unknown nodes, then we need to form a super node. This is
necessary as we need to apply KCL in solving node equations and
we do not know the current through the voltage source in advance.
A Supernode is formed by enclosing the voltage source between
two nonreference nodes and any branches in parallel with it.
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Nodal Analysis-SupernodeConsider the following circuit. Nodes 2 and 3 form a supernode. At
Supernode we get,
10Vv
04
vv6
0v8
0v2
vv
1
133212
=
=+++
v1 v3v2 v1
v2 v3
At Supernode, applying KCL,
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Nodal Analysis-SupernodeApplying KVL to supernode we get,
5v,
05
32
32
=
=++
vor
vv
v
2
v
3
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Nodal Analysis-SupernodeWe note the following properties of a supernode,
The voltage source inside the supernode provides a constraint
equation to solve for the node voltages
A supernode do not have a voltage of its own
A supernode requires the application of both the KCL and KVL