Lecture Voltage Divider
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, 8 2014 Ch. 3 Network Analysis- Part I 1
Example 4
For the given circuit, determine
(a) the value of current I,
(b) the power absorbed by the dependent source, and
(c) the resistance seen by the independent voltage source.
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, 8 2014 Ch. 3 Network Analysis- Part I 2
Solution :
(a) Applying Ohms law to the 4- resistor, gives V1= 4I.
Therefore, the value of dependent voltage source is 4.5V1= 4.5(4I) = 18I. By applying KVL,
24 4 2 18 0I I I I 2 A
(b) For the dependent source, the power absorbed is
1
2 2
(4.5 )( ) 4.5(4 )( )
18 18( 2)
P V I I I
I
72 W
What is the meaning of negative sign ?
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, 8 2014 Ch. 3 Network Analysis- Part I 3
(c) The resistance seen by the independentvoltage source,
24 24
2R I 12
What is the meaning of negative sign ?
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Ans. :The remainder circuit supplies power tothe independent voltage source.
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4
1.11 Voltage Divider Rule:-
The voltage across one of the series resistors is the total
voltage times the ratio of its resistance to the total
resistance.
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1.12 Current Divider Rule:-
Current division is dualof voltage division.
21
12
21
22
21
2
121
1
1
RR
Riior
GG
Gii
RR
R
iiorGG
G
ii
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1.13 STAR-DELTA TRASFORMATION:-
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If we were to connect a source between terminals a and b
of the Y,the resistance between the terminals would be
Rab=R1+R2 eq.1
But the resistance between terminals a and b of the is
Rab=RC(RA+RB) eq.2Combining eq. 1 & eq. 2 we get,
Using a similar approach between terminals b and c, we get
Eq. 3
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and between terminals c and a we get
Eq. 4
Eq. 5
If Equation 4 is subtracted from Equation 3 , then
Eq. 6
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Now Adding Equations 5 and 6, we get
Using a similar approach, we obtain
Notice that any resistor connected to a point of the Y is
obtained by finding the product of the resistors connected to
the same point in the and then dividing by the sum of
all the resistances.
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If all the resistors in a circuit have the same value, R,
then the resulting resistors in the equivalent Y network will
also be equal and have a value given as,
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, 8 2014 Ch. 2 Ohm's Law 15
(a) Star (Y) connection (b) Delta () connection
Star-Delta Connections :
Two ways of connecting three resistors across three
points.
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, 8 2014 Ch. 2 Ohm's Law 16
Star-to-Delta TransformationSame way, we can derive,
B C1 B C
A
R RR R R
R
C A2 C A
B
R RR R RR
A B3 A B
C
R RR R R
R
Remember
The sum of the two nearest
resistances plus the
product of the same two
resistances divided by the
third resistance.
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UNIT-2
NETWORK ANALYSIS
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2.1 KIRCHHOFFS CURRENT LAW (KCL):-
It states that the algebraic sum of all currents entering
a node is zero. Mathematically:
Currents are positive if enter ing a nodeCurrents are negative if leaving a node.
Example:
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Applying Kirchhoff's current law:
I1+ I2+ I3+ I4= 0
(the negative sign in I2indicates that I2has a magnitude
of 3A and is flowing in the direction opposite to that
indicated by the arrow)
Substituting:
5 - 3 + I3+ 2 = 0 Therefore, I3= - 4A (ie 4A
leaving node)
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2.2 KIRCHHOFFS VOLTAGE LAW (KVL):-
It states that the algebraic sum of the voltage drops
around any loop, open or closed, is zero. Mathematically
Example:
Going round the loop in the direction of the current, I,
Kirchhoff's Voltage Law gives:
10- 2I - 3I = 0
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- 2Iand - 3Iare negative, since they are voltage drops
i.e. represent a decrease in potential on proceeding
round the loop in the direction of I. For the same reason +
10V is positive as it is a voltage rise or increase in potential.
Concluding:
5 I = 10 Therefore, I = 2A
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, 8 2014 26
Polarity of Voltages
Notethat polarity of the voltage (emf) across abattery does not depend upon the assumed
direction of current.
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, 8 2014 Ch. 3 Network Analysis- Part I 27
Applying KVL
1. Select a closed loop.
2. Mark the voltage polarity (+ and -) across
each element in the closed loop.
3. Go round the selected loop, and add up all
the voltages with + orsigns.
4. Any one of the following two rules can be
followed :
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, 8 2014 Ch. 3 Network Analysis- Part I 28
(i) Rule 1 :While travelling, if you meet a voltage
rise, write the voltage withpositive sign ; if you
meet a voltage drop, write the voltage withnegativesign.
(ii) Rule 2 :While travelling, write the voltage
with positive sign if + is encountered first; writethe voltage with negative sign ifis encountered
first.
We shall be following Rule 1, as it has a stronganalogy with the physical height (altitude)of a
place.
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, 8 2014 Ch. 3 Network Analysis- Part I 29
Example 5 :Use KVL to find vR2and vx.
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, 8 2014 Ch. 3 Network Analysis- Part I 30
For finding vR2, we write KVL eqn. going
around loop abgha clockwise:
If you choose to go around the loopanticlockwise, you get
Giving the same result.
V320436 22 RR vv
V320364 22 RR vv
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, 8 2014 Ch. 3 Network Analysis- Part I 31
There are two ways to determine vx
1)We can consider this voltage as the voltage
across the gap from dtof. Writing KVL
(habcdfgh) :
2) Knowing vR2, apply a short-cut (bcdfgb) :
V6
x
x
v
v 01412364
V6
x
x
vv 0321412
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, 8 2014 Ch. 3 Network Analysis- Part I 33
Solution :
We need not find the currentsI1, I2andI3.
Instead, we reduce the network.
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E l 7 D i h l f
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, 8 2014 Ch. 3 Network Analysis- Part I 34
Example 7 : Determine the value of
current I .
23I4 = 0 or I = -5 A
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, 8 2014 Ch. 3 Network Analysis- Part I 35
Example 8
Using KCL and KVL, determine the currents ixand iyin the network shown.
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, 8 2014 Ch. 3 Network Analysis- Part I 36
Solution : Using KCL, the currents in other branches are
marked as shown. Writing KVL equations for the loops
1, 2 and 3,
503530)(3502
50227
02)(2505
1001005
0510100
1
1
1
1
1
1
IIIIIII
III
IIII
III
II
yx
yxy
yx
yxx
yx
x
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8 2014 Ch 3 Network Analysis Part I 37
;
50
50100
353
2271005
1
I
II
y
x
Writing the above equations in matrix form,Click
Using Calculator, we solve for Ixand Iy,Click
; andx yI I 3.87 A 0.51 A