lesson 3
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Ohms alw • A voltage of 20V is applied across AB shown in
the Figure 1(b).Calculate the total current, the current in each resistor and power dissipated in each resistor and the value of the series resistance to have the total current.( All resistors are measured in ohms.)
Definitions • Electric circuit: A combination of various electric
elements connected in any manner is called an electric circuit.
• Passive network: is one which contains no source of emf. In it.
• Active network: is one which contains one of more source of emf. In it.
• Node: is a junction in a circuit where two or more circuit elements are connected together
• Branch: is a part of a network which lies between two junctions.
Definitions
• Loop : It is close path in a circuit in which no elements or node is encountered more than once.
• Mesh: It is loop that contains no other loop within it.
Network Analysis Direct methods : Determining different voltages and currents in the original circuit.
• KCL and KVL
• Nodal Analysis
• Superposition theorem
Network Reduction Method : the original circuit converted into a much simpler circuit.
• Thevenin’s theorem
• Norton’s theorem
Polarity • Sign of battery e.m.f:
Rise in voltage should be given a +ve sign and fall in voltage a –ve sign
• Sign of IR drop
FOR A GIVEN CIRCUIT LET B NUMBER OF BRANCHES N NUMBER OF NODES
THE MINIMUM REQUIRED NUMBER OF LOOP CURRENTS IS
)1( NBL
MESH CURRENTS ARE ALWAYS INDEPENDENT
AN EXAMPLE
2)16(7
6
7
L
N
BTWO LOOP CURRENTS ARE REQUIRED. THE CURRENTS SHOWN ARE MESH CURRENTS. HENCE THEY ARE INDEPENDENT AND FORM A MINIMAL SET
KVL ON LEFT MESH
REPLACING AND REARRANGING
DETERMINATION OF LOOP CURRENTS
KVL ON RIGHT MESH
2 4 5 30
Sv v v v
USING OHM’S LAW
1 1 1 2 1 2 3 1 2 3
4 2 4 5 2 5
, , ( )
,
v i R v i R v i i R
v i R v i R
SHORTCUT: POLARITIES ARE NOT NEEDED. APPLY OHM’S LAW TO EACH ELEMENT AS KVL IS BEING WRITTEN
1I @KVL
2I @KVL
396
12612
21
21
kIkI
kIkIREARRANGE
add and 2/*
mAIkI 5.0612 22
mAIkIkI4
561212 121
EXPRESS VARIABLE OF INTEREST AS FUNCTION OF LOOP CURRENTS
21 IIIO
1I @KVL
1IIO NOW
THIS SELECTION IS MORE EFFICIENT
996
12612
21
21
kIkI
kIkIREARRANGE
substract and 2/*
3/*
mAIkI4
31824 11
EXAMPLE: FIND Io
AN ALTERNATIVE SELECTION OF LOOP CURRENTS
2I @KVL
Mesh Analysis: KVL
Applying KVL to the three loops. We get
V1-I1R1-(I1-I2)R4=0 or I1(R1+R4)-I2R4=V1 -I2R2—(I2-I3)R5—(I2-I1)R4=0 or I1R4+I2(R2+R4+R5)+I3R5=0 -I3R3—V2-(I3-I2)R5=0 or I2R5-I3(R3+R5)=V2
Take any resistatnce values, write equations and solve the current
Exercise: Try • Determine the loop currents and current
supplied by each battery in the circuit shown in Figure
Answer: I2=542/299A , I3=-1875/598A , II1= 765/299A.
Discharge current of V1=765/299A Discharge current of V2=I1-I2=220/299A. Discharge current of V3= I2+I3=2965/598A Discharge current of V4 =I2=545/299A. Discharge current of V5= 1875/598A.
THE STRATEGY FOR NODE ANALYSIS 1. IDENTIFY ALL NODES AND SELECT A REFERENCE NODE
2. IDENTIFY KNOWN NODE VOLTAGES
3. AT EACH NODE WITH UNKNOWN VOLTAGE WRITE A KCL EQUATION (e.g.,SUM OF CURRENT LEAVING =0)
0:@321 IIIV
a
4. REPLACE CURRENTS IN TERMS OF NODE VOLTAGES
0369
k
VV
k
V
k
VV baasa AND GET ALGEBRAIC EQUATIONS IN THE NODE VOLTAGES ...
REFERENCE
SV
aV
bV
cV
0:@543 IIIV
b
0:@65 IIV
c
0943
k
VV
k
V
k
VVcbbab
039
k
V
k
VVcbc
SHORTCUT: SKIP WRITING THESE EQUATIONS...
AND PRACTICE WRITING
THESE DIRECTLY
EXAMPLE
@ NODE 1 WE VISUALIZE THE CURRENTS LEAVING AND WRITE THE KCL EQUATION
REPEAT THE PROCESS AT NODE 2
03
12
4
122
R
vv
R
vvi
OR VISUALIZE CURRENTS GOING INTO NODE
WRITE THE KCL EQUATIONS
mA6
1I
2I
3I
036
6:@ 222
k
V
k
VmAV 2
12V V
CURRENTS COULD BE COMPUTED DIRECTLY USING KCL AND CURRENT DIVIDER!!
1
2
3
8
3(6 ) 2
3 6
6(6 ) 4
3 6
I mA
kI mA mA
k k
kI mA mA
k k
IN MOST CASES THERE ARE SEVERAL DIFFERENT WAYS OF SOLVING A PROBLEM
NODE EQS. BY INSPECTION
mAVVk
6202
121
mAVkk
V 63
1
6
10 21
116V V
k
VI
k
VI
k
VI
362
23
22
11
Once node voltages are known
1
1@ : 2 6 0
2
VV mA mA
k
Node analysis
VV
VV
4
6
4
1
SOURCES CONNECTED TO THE REFERENCE
SUPERNODE
CONSTRAINT EQUATION VVV 1223
KCL @ SUPERNODE
02
)4(
212
6 3322
k
V
k
V
k
V
k
Vk2/*
VVV 22332
VVV 1232 add and 3/*
VV 385 3
mAk
VI
O8.3
2
3 LAW SOHM'
OIV FOR NEEDED NOT IS
2
LEARNING EXAMPLE
1 1 21@ : 4 0
6 12
V V VV mA
k k
USING KCL
2 2 12@ : 2 0
6 12
V V VV mA
k k
BY “INSPECTION”
1 2
1 1 14
6 12 12V V mA
k k k
1 2
1 1 12
12 6 12V V mA
k k k
LEARNING EXAMPLE
3 nodes plus the reference. In principle one needs 3 equations...
…but two nodes are connected to the reference through voltage sources. Hence those node voltages are known!!!
…Only one KCL is necessary
012126
12322
k
VV
k
VV
k
V
][5.1][64
0)()(2
22
12322
VVVV
VVVVV
EQUATIONS THE SOLVING
Hint: Each voltage source connected to the reference node saves one node equation
][6
][12
3
1
VV
VV
THESE ARE THE REMAINING TWO NODE EQUATIONS
210
1
1
2
4
11
10
13
10
13
1
23
4
32
:3
01
3
2
11
5
1
2
12
1
32
5
2
2
21
:2
2810
3
2
2
10
1
2
1
2
11
10
31
2
21
2
128
:1
VVV
VVVVV
Node
VVV
VVVVV
Node
VVV
VVVVV
Node
Solve equation to find v1,V2 and V3 and find the current.
VOLTAGE SOURCE CONNECTED TO REFERENCE
VV 31
0263
: 212
xI
k
V
k
VV VKCL@
2
CONTROLLING VARIABLE IN TERMS OF NODE VOLTAGES k
VI
x
6
2
REPLACE
06
263
2212
k
V
k
V
k
VVk6/*
VVVV 602212
mAk
VVI
O1
3
21
OI FIND
LEARNING EXAMPLE
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